function
geometry
integration
calculus
calculus computations

Let $ f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $ f(x) \equal{} x^3 \plus{} x$, and let $ D \equal{} \{(x,y) | (f^{\minus{}1}(x))^2 \plus{} (f^{\minus{}1}(y))^2 \leq 1\}$

Find the area of $ D$.

[hide="Solution"]

Question can be restated with,

$ D \equal{} \{(x^3 \plus{} x,y^3 \plus{} y)\, |\, x^2 \plus{} y^2 \le 1\}$

$ \displaystyle A \equal{} \int \int_{D} d(x^3 \plus{} x)\, d(y^3 \plus{} y) \equal{} \int_{ \minus{} 1}^1 \int_{ \minus{} \sqrt {1 \minus{} y^2}}^{\sqrt {1 \minus{} y^2}} (3x^2 \plus{} 1)(3y^2 \plus{} 1)\,dx\,dy \equal{} \frac {23\pi}{8}$

as required.

[/hide]


determine the smalest M and the bigest m such as for all a,b,c>0 with abc=1 and given that

M>1/(1+a)+1/(1+b)+1/(1+c)>m

Let $ a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{x}{z}$. The expression is equivalent to
\[ \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\]
We will show that
\[ 1<\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}<2 \]
The left side follows from

$ \frac{x}{x+y}> \frac{x}{x+y+z}, \frac{y}{y+z}> \frac{y}{x+y+z},$ and $ \frac{z}{z+x}> \frac{z}{x+y+z}$.

For the RHS, note that $ \frac{x}{x+y}<\frac{z+x}{x+y+z}\iff yz>0$. Deriving similar expressions for $ y$ and $ z$ and adding them, we get our desired inequality.


Let $ a_0 \equal{} 2$, $ a_1 \equal{} 9$, and $ a_n \equal{} 2a_{n \minus{} 1} \minus{} 4a_{n \minus{} 2}.$ Compute $ a_{15}$.

How do you build the recursion?

Generally you go by the characteristic equation, then solve for coefficients etc. (standard procedure).

But since the explicit formula for $ a_n$ isn't required here, you can simplify things thus:

[hide]$ a_{n \plus{} 3} \equal{} 2a_{n \plus{} 2} \minus{} 4a_{n \plus{} 1} \equal{} 2(2a_{n \plus{} 1} \minus{} 4a_n) \minus{} 4a_{n \plus{} 1} \equal{} \minus{} 8a_n$

Hence $ a_{15} \equal{} \minus{} 8a_{12} \equal{} ( \minus{} 8)^2a_9 \equal{} ( \minus{} 8)^3a_6 \equal{} ( \minus{} 8)^4a_3 \equal{} ( \minus{} 8)^5a_0 \equal{} \minus{} 2^{16}$[/hide]


Assume that $ a,b,c\in R^{+}$ and $ abc=1$, try to prove $ (a+b)(b+c)(a+c)\ge4(a+b+c-1)$

[color=indigo]We have: \[ (a+b)(b+c)(c+a)\ge\frac{8}{9}\cdot (a+b+c)\cdot (ab+bc+ca)\] So, we have to prove that: \begin{eqnarray*}2\cdot (a+b+c)\cdot (ab+bc+ca)&\ge& 9\cdot (a+b+c-1)\\ \iff\frac{2}{9}\cdot (ab+bc+ca)+\frac{1}{a+b+c}&\ge& 1\\ \iff 3\cdot\sqrt[3]{\frac{\left(ab+bc+ca\right)^{2}}{81(a+b+c)}}&\ge& 1\\ \iff \left(ab+bc+ca\right)^{2}&\ge& 3\cdot (a+b+c)\end{eqnarray*} What is truly, because: $ \left(ab+bc+ca\right)^{2}\ge 3\cdot abc\cdot (a+b+c)=3\cdot (a+b+c)$[/color]

[quote="April"][color=indigo]We have:
\[ (a+b)(b+c)(c+a)\ge\frac{8}{9}\cdot (a+b+c)\cdot (ab+bc+ca) \]
So, we have to prove that:
\begin{eqnarray*}2\cdot (a+b+c)\cdot (ab+bc+ca)&\ge& 9\cdot (a+b+c-1)\\ \iff\frac{2}{9}\cdot (ab+bc+ca)+\frac{1}{a+b+c}&\ge& 1\\ \iff 3\cdot\sqrt[3]{\frac{\left(ab+bc+ca\right)^{2}}{81(a+b+c)}}&\ge& 1\\ \iff \left(ab+bc+ca\right)^{2}&\ge& 3\cdot (a+b+c) \end{eqnarray*}
What is truly, because: $ \left(ab+bc+ca\right)^{2}\ge 3\cdot abc\cdot (a+b+c)=3\cdot (a+b+c)$[/color][/quote]
yeah , your solution is stronger than the problem , but for this problem we have
$ (a+b+c)(ab+bc+ac)-1=(a+b)(b+c)(a+c)$
so we only have to prove a weaker inequality
$ ab+bc+ac+\frac{3}{a+b+c}\geq 4$


HMMT inequalities quadratics conics parabola

Find the set of solutions for x in the inequality $ \frac{x\plus{}1}{x\plus{}2}>\frac{3x\plus{}4}{2x\plus{}9}$ when $ x\not\equal{}2, x\not\equal{}\minus{}\frac{9}{2}$.

[quote="Smartguy"]Find the set of solutions for x in the inequality $ \frac {x \plus{} 1}{x \plus{} 2} > \frac {3x \plus{} 4}{2x \plus{} 9}$ when $ x\not \equal{} \minus{} 2, x\not \equal{} \minus{} \frac {9}{2}$.[/quote]

Ok, first i go about equating this inequality. ie

$ \frac {x \plus{} 1}{x \plus{} 2} \equal{} \frac {3x \plus{} 4}{2x \plus{} 9}$

Then solve,

$ \equal{}(x\plus{}1)(2x\plus{}9) \equal{} (x\plus{}2)(3x\plus{}4)$

$ \equal{}2x^2 \plus{} 11x \plus{} 9 \equal{} 3x^2 \plus{} 10x \plus{} 8$

$ x^2 \minus{} x \minus{} 1 \equal{} 0$

Solving this quadratic through the quad formula gives

$ x \equal{} \frac{\minus{}b \pm \sqrt(b^2 \minus{} 4ac)}{2a}$

$ x \equal{} \frac{1 \pm \sqrt5}{2}$

Now we plot $ x \equal{} \frac{1 \plus{} \sqrt5}{2}, \frac{1 \minus{} \sqrt5}{2}$ and $ x \not \equal{} \minus{}2, \minus{} \frac{9}{2}$ on the number line and test for numbers inbetween and we find that

$ \minus{} \frac{9}{2} < x < \minus{}2, \qquad \frac{1 \minus{} \sqrt5}{2} \le x \le \frac{1 \plus{} \sqrt5}{2}$

Or you could solve it using casework (the long way). But it relies more on knowing what the inequality means.

[hide="My Solution"]
$ \frac {x \plus{} 1}{x \plus{} 2} > \frac {3x \plus{} 4}{2x \plus{} 9}\\
\frac {x \plus{} 1}{x \plus{} 2} \minus{} \frac {3x \plus{} 4}{2x \plus{} 9} > 0\\
\frac {2x^2 \plus{} 11x \minus{} 9 \minus{} (3x^2 \plus{} 10x \plus{} 8)}{(x \plus{} 2)(2x \plus{} 9)} > 0\\
\frac {x^2 \minus{} x \minus{}1}{(x \plus{} 2)(2x \plus{} 9)} < 0$
Now we start the cases. We know that the expression on the left will be negative when either the numerator is negative and the denominator is positive or vice-versa. We have to make sure [b]both[/b] are true so we need to make use of the intersection operation. ([i]and[/i] instead of [i]or[/i])

Case 1: The numerator is greater than zero and the denominator is less than zero.
$ \left(x \minus{} \frac{1 \plus{} \sqrt{5}}{2}\right)\left(x \minus{} \frac{1 \minus{} \sqrt{5}}{2}\right) > 0$
Thinking about it graphically, we have a parabola opening upwards with $ x$-intercepts at those corresponding roots. The value of the quadratic will be greater than zero outside the roots. This, the solution to that inequality is $ x \in \left(\minus{}\infty,\frac{1 \minus{} \sqrt{5}}{2}\right) \cup \left(\frac{1 \plus{} \sqrt{5}}{2}, \infty\right)$.
We must also have the denominator less than zero, so we have:

$ (x \plus{} 2)(2x \plus{} 9) < 0$

Doing the same method as last time, we have $ x \in \left(\minus{}\frac{9}{2},\minus{}2 \right)$.

Since we must have both solutions coincide we have to perform the intersection operation on both sets of solutions and we have $ x \in \left(\minus{}\frac{9}{2},\minus{}2 \right)$.

Case 2: Opposite of last case. Numerator is less than zero and the denominator is greater than zero. We obtain $ x \in \left(\frac{1 \minus{} \sqrt{5}}{2},\frac{1 \plus{} \sqrt{5}}{2}\right)$ as the solution for the numerator and $ x \in \left(\minus{}\infty,\minus{}\frac{9}{2}\right) \cup \left(\minus{}2, \infty \right)$ for the denominator. Performing the intersection operation once again, we have $ x \in \left(\frac{1 \minus{} \sqrt{5}}{2},\frac{1 \plus{} \sqrt{5}}{2}\right)$.

We take the union of the sets of solutions to both cases since both cases yield valid solutions. Thus the solution of the inequality is $ x \in \left(\minus{}\frac{9}{2},\minus{}2 \right) \cup \left(\frac{1 \minus{} \sqrt{5}}{2},\frac{1 \plus{} \sqrt{5}}{2}\right)$.
[/hide]


let $ x,y,z \in [\minus{}1.1]$ and $ x\plus{}y\plus{}z\equal{}0$ prove that:
$ \sum \sqrt{1\plus{}x\plus{}\frac{7}{9}y^2} \ge 3$

[quote="gauravpatil"]is the sum symetrc or cyclic[/quote]
$ \sqrt {1 \plus{} x \plus{} \frac {7}{9}y^2}\plus{}\sqrt {1 \plus{} y \plus{} \frac {7}{9}z^2}\plus{}\sqrt {1 \plus{} z \plus{} \frac {7}{9}x^2} \ge 3$


trigonometry
limit
calculus
calculus computations

All help appreciated!

Find the interval of convergence for the series:

$\sum_{n=1}^{\infty}\sin{(\frac{\pi}{n})}x^{n}$

As

$\lim_{n\to\infty}\left| \frac{\sin \left(\frac{\pi}{n+1}\right) x^{n+1}}{\sin \left(\frac{\pi}{n}\right) x^{n}}\right|$

converges to $|x|$, we see that this series converges if $|x|<1$ and divergens if $|x|>1$. So the radius of convergence is $1$. For endpoints, we may use other methods to determine whether it converges or not.

1) For $x=1$, $\lim_{n\to\infty}\frac{\sin \left(\frac{\pi}{n}\right)}{\frac{1}{n}}= \pi$, so by limit comparison test with p-series, this series diverges.

2) For $x =-1$, this series becomes an alternating series, so it converges.

Therefore, the interval of convergence is $[-1, 1)$.


Got these questions:

If you have the question:

"The cube of the three-digit natural number A7B is 108,531,333. What is A+B?"

how do u do it without a calculator or without finding the cube root?

and this one 2:

"The number 1 is both a smute and thripe. If the integer s is a smute, then the next smute is s+5. If the integer t is a thripe, then the next thripe is 2t+1. What is the smallest whole number greater than 1, that is both a smute and a thripe?"

How do u do this in algebraic form without listing them out, finding a pattern for the smutes, and then find the matching thripe (or is that the only way to do it?)?

Thx!  :D

1st question:
you know that B is 7 since 7 is the only number that the cube of it ends in 3
then just estimate the cubes of 300, 400, 500...
400^3 = 64,000,000
500^3 = 125,000,000
108,531,333 is between these two.
so, A must be 4
4+7 = 11
yay

a smute is 1+5k.  so just list thripes until you get one ending in 1.(6 wont work because its 2n+1, which is odd)


function
integration
real analysis
real analysis unsolved

Let $f: X\to [0,\infty]$ be measurable and $0<p<\infty$. Prove that
\[\int_{X}f^{p}\,d\mu=p\int_{0}^\infty t^{p-1}\mu\left (\{x: f(x)>t\}\right )\,dt.\]


I'm gonna make a joke game, where people make their own jokes. You [i]may[/i] use Google. I will post a category that will be in the form "fun things to do at/with/to/in -----". For example, if the category was "fun things to do at WalMart", someone could say "Ask people for Grey Poupon." Only I can change the category, and I will change it every few days(according to participation).

First category: Fun things to do to pizza restaurants

My first joke: Call, and say "Oh just surprise me!", give them your address, and hang up.

It may be a bit more specific but

Fun things to do at CiCi's Pizza.

Pay 5 dollars and eat twenty whole pizzas.

Grow a big, protruding butt, then cut in lines.


Do any of you know when the GAME (Garfield Annual Mathematics Extravaganza) is going to be this year?

I haven't heard anything about it.  I'm not sure if it will even be held this year...

were getting lazier and lazier each year...
too lazy for problems lol
i think its on June 9th
itll be horribly sad if im wrong :huh:

Yes, we are currently planning to hold GAME on June 9.   Adam will send out an official announcement once we are sure.

Not really, but I LOST in the GAME. Just thought I would let you know.

I lost!  (GAME is still set for June 9th.)

Only three interested schools contacted us by the deadline--most thought it was to late in the year. Therefore, GAME is postponed until next schoolyear.


Let $ a,b,c$ be positive real numbers such that : $ a\plus{}b\plus{}c\equal{}3$ .

Prove that :

$ \frac{a^2}{b\plus{}3c^3}\plus{}\frac{b^2}{c\plus{}3a^3}\plus{}\frac{c^2}{a\plus{}3b^3}\geq\frac{3}{4}$

[quote="quykhtn-qa1"][quote="Materazzi"]Let $ a,b,c$ be positive real numbers such that : $ a \plus{} b \plus{} c \equal{} 3$ .

Prove that :

$ \frac {a^2}{b \plus{} 3c^3} \plus{} \frac {b^2}{c \plus{} 3a^3} \plus{} \frac {c^2}{a \plus{} 3b^3}\geq\frac {3}{4}$[/quote]
This inequality by can_hang 2007 :wink:[/quote]

Yes ! See here : http://www.maths.vn/forums/showthread.php?t=12749

:)

[quote="2424"][quote="Materazzi"]Let $ a,b,c$ be positive real numbers such that : $ a \plus{} b \plus{} c \equal{} 3$ .

Prove that :

$ \frac {a^2}{b \plus{} 3c^3} \plus{} \frac {b^2}{c \plus{} 3a^3} \plus{} \frac {c^2}{a \plus{} 3b^3}\geq\frac {3}{4}$[/quote]
I think we can prove it by CS and pqr! :)[/quote]
What do you mean? The estimation  $ \sum_{cyc}\frac{a^2}{b\plus{}3c^3}\geq\frac{(a^2\plus{}b^2\plus{}c^2)^2}{\sum(a^2b\plus{}3a^3b^2)}$ does not help. :wink:

[quote="arqady"][quote="2424"][quote="Materazzi"]Let $ a,b,c$ be positive real numbers such that : $ a \plus{} b \plus{} c \equal{} 3$ .

Prove that :

$ \frac {a^2}{b \plus{} 3c^3} \plus{} \frac {b^2}{c \plus{} 3a^3} \plus{} \frac {c^2}{a \plus{} 3b^3}\geq\frac {3}{4}$[/quote]
I think we can prove it by CS and pqr! :)[/quote]
What do you mean? The estimation  $ \sum_{cyc}\frac {a^2}{b \plus{} 3c^3}\geq\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum(a^2b \plus{} 3a^3b^2)}$ does not help. :wink:[/quote]
Can_hang 2007 use Cauchy-Schwarz inequality:
$ LHS.(\sum \ ((2a\plus{}b)^2(b\plus{}3c^3)) \ge (2 \sum \ a^2\plus{} \sum \ ab)^2$
Then,use $ pqr$ :)


geometry
rectangle
circumcircle
ratio
perpendicular bisector
geometry solved

Let E be on CB extended of rectangle ABCD, and that the circumcircle of ABE meets diagonal BD at P.  Denote the midpoint of EC and PD be M and M' respectively.  show 1) circumcenter of AEC lies on PE
2) AM' is perpendicular to MM'

first about the part one. APE is rightangled at p and so PE is parallel to prependicular bisector of PA ,so how possibly can the circum center of APC which lies on bisector of AP lie on PE.

part two: notice that using the dot product method it suffices to prove that:
DC.AP =PE.BC obviously since APBE is cyclic BAP=PEB and so angle between PE and Bc equals that of DC and AP
Now notice that we have to prove that DC times AP = PE times BC (in magnitude)
but since APE and BAD are similar we are done.

Well, in that case:

Let O be the circumcenter of AEC. We know that angle AEO = Pi/2-ACE = Pi/2-ACB, but AEP = ABP = ABD = Pi/2-ACB = AEO, so E, O, P are collinear.

A little bit different proof (I guess figure independent: E could stay on segment BC, outside BC nearer C or nearer B).

Let's call F the (second) point where line AC intersect c(ABE) and let K be the midpoint of CF. It results that < EFA = < EAB = 90 then for the similitude of triangles EFC and MKC (this means that the claim is more generally true , also if the point M and M' divide the segments EC and PD in the same rate) it holds < MKC = < MKA = 90. Then ABKM is ciclic. But M' is simmetric of K whith respect to the perpendicular bisector of AB, then also M' is on the circle  c(ABKM) and therefore < AM'M = 90.

Let I be the midpoint of AE, i.e. the center on c(ABE). Let's considere the circle c(IME) this circle is homothetic respect to the circle c(ACE) [center E, ratio=1/2]. The center O of c(ACE) is then on the circle c(IME). We have that OM is parallel to DC and IM is parallel to AC then <OMI = <DCA. But <OEI = <OMI and <PBA = <DCA then <OEI = <PBA and we are done.


Let $ a,b,c$ be positive reals such that $ abc \equal{} 1$. Prove that
$ a\plus{}b\plus{}c \leq a^2\plus{}b^2\plus{}c^2$

[quote="FelixD"]What about Muirhead? $ [2$, $ 0$, $ 0] \succ [4/3$, $ 1/3$, $ 1/3]$.[/quote]

Hello,
Can you post full solution,please ?
PS:[i don't get it at all]

Homogenise the inequality, i.e. multiply the LHS of the inequality with $ (abc)^{\frac{1}{3}}$. Then write it as a symmetric sum and use Muirhead as written above.

hello
we have $ 3(x^{2}\plus{}y^{2}\plus{}z^{2})\geq(x\plus{}y\plus{}z)^{2}$
and $ x\plus{}y\plus{}z\geq3\sqrt[3]{xyz}\equal{}3$
so $ 3(x^{2}\plus{}y^{2}\plus{}z^{2})\geq3(x\plus{}y\plus{}z)$.....
it's too easy.

[quote="Obel1x"]Let $ a,b,c$ be positive reals such that $ abc \equal{} 1$. Prove that
$ a \plus{} b \plus{} c \leq a^2 \plus{} b^2 \plus{} c^2$[/quote]

It is Kiran Kedlaya's old result .


Find all $ a,b,c,d \in R\plus{}$ such that
$ a(1\minus{}b^2)\equal{}b(1\minus{}c^2)\equal{}c(1\minus{}d^2)\equal{}d(1\minus{}a^2)>0$

Is $ R\plus{}$ the set of positive real number?

Posted before
See here : http://www.mathlinks.ro/viewtopic.php?t=177981


Let $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}1$, prove that $ \frac{a^2\plus{}b^2c}{b\plus{}c}\plus{}\frac{b^2\plus{}c^2a}{c\plus{}a}\plus{}\frac{c^2\plus{}b^2a}{b\plus{}a}\geq\frac{2}{3}$

$ \sum \frac{a^2\plus{}b^2c}{b\plus{}c} \equal{} \sum \frac{a^2\plus{}bc}{b\plus{}c} \plus{} \sum \frac{b^2c\minus{}bc}{b\plus{}c} \equal{} \sum \frac{a^2\plus{}bc}{b\plus{}c} \minus{} \sum \frac{bc(a\plus{}c)}{b\plus{}c}$, we want to show that it is greater than $ \frac{2}{3}$.

Claim 1: $ \sum \frac{a^2\plus{}bc}{b\plus{}c} \ge 1$
Claim 2: $ \sum \frac{bc(a\plus{}c)}{b\plus{}c} \leq \frac{1}{3}$

Proof of Claim 2:
Note that $ \frac{bc(a\plus{}c)}{b\plus{}c} \leq \frac{(b\plus{}c)^2(a\plus{}c)}{4(b\plus{}c)} \equal{} \frac{(b\plus{}c)(c\plus{}a)}{4}$
and $ \sum \frac{(b\plus{}c)(c\plus{}a)}{4} \equal{} \frac{a^2\plus{}b^2\plus{}c^2\plus{}3(ab\plus{}bc\plus{}ca)}{4} \leq \frac{1}{3}$

Proof of Claim 1: Rewrite it in this form:
$ \sum \left(\frac{a^2\plus{}bc}{b\plus{}c} \minus{} a \right) \ge 0$
$ \sum \frac{(a\minus{}b)(a\minus{}c)}{b\plus{}c} \ge 0$
This is an old inequality which can be settled by Vornicu-Schur.

[quote="Soarer"]$ \sum \frac {a^2 \plus{} b^2c}{b \plus{} c} \equal{} \sum \frac {a^2 \plus{} bc}{b \plus{} c} \plus{} \sum \frac {b^2c \minus{} bc}{b \plus{} c} \equal{} \sum \frac {a^2 \plus{} bc}{b \plus{} c} \minus{} \sum \frac {bc(a \plus{} c)}{b \plus{} c}$, we want to show that it is greater than $ \frac {2}{3}$.

Claim 1: $ \sum \frac {a^2 \plus{} bc}{b \plus{} c} \ge 1$
Claim 2: $ \sum \frac {bc(a \plus{} c)}{b \plus{} c} \leq \frac {1}{3}$

Proof of Claim 2:
Note that $ \frac {bc(a \plus{} c)}{b \plus{} c} \leq \frac {(b \plus{} c)^2(a \plus{} c)}{4(b \plus{} c)} \equal{} \frac {(b \plus{} c)(c \plus{} a)}{4}$
and $ \sum \frac {(b \plus{} c)(c \plus{} a)}{4} \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} 3(ab \plus{} bc \plus{} ca)}{4} \leq \frac {1}{3}$

Proof of Claim 1: Rewrite it in this form:
$ \sum \left(\frac {a^2 \plus{} bc}{b \plus{} c} \minus{} a \right) \ge 0$
$ \sum \frac {(a \minus{} b)(a \minus{} c)}{b \plus{} c} \ge 0$
This is an old inequality which can be settled by Vornicu-Schur.[/quote]
Prove the inequality than stronger:
$ \frac {a^2 \plus{} 3b^2c}{b \plus{} c} \plus{} \frac {b^2 \plus{} 3c^2a}{c \plus{} a} \plus{} \frac {c^2 \plus{} 3a^2b}{a \plus{} b} \ge 1$
With the same condition
I have an nice solution,goodluck


Mafia
FTW
search
function
\/closed
very old unlocked thread

1. Mods are responsible experienced members, admins choose them. Mods don't spam, they rather provide helpful information. You never know when they choose one. Everyone thinks they aren't choosing mods... but then somehow a new green name pops up. They don't just make you a mod right off the bat, they just ask you as you go. A stage on AoPS where only some people reach.

2. That depends on how many posts you've posted;
0-19 posts: 0 stars(New Member)
20-49 posts-1/2 star(P vs. NP) <- you at the moment
50-99 posts - 1 star(Hodge Conjecture)
100-249 posts-2 stars(Poincare Conjecture)
250-499 posts-2 1/2 stars(Reinman Hypothesis)
500-999 posts-3 stars(Yang Mills Theory)
1,000-2,499 posts-4 stars(Navier Strokes Equations)
2,500+ posts-5 stars (Birch Swinnerton&Dyer)

3. Are your posts helping others? Think if posting the message your posting will help someone. Unless your in the FF, Games or Mafia thread where you have marathons, discussion and strategy.

[quote="$ LaTeX$"][quote="jjfun1"][quote="meewhee009"]Everyone thinks they aren't choosing mods... but then somehow a new green name pops up. [/quote]
They are no longer choosing mods for FTW.[/quote]

Actually, meewhee is correct. AIME15 wasn't a mod until maybe a week ago.[/quote]

Isn't AIME15 just a mod for the forums? I meant that they are not choosing mods for FTW anymore. Levans mentioned that a couple of times.

There's a search function built in to the forum.  Try searching the Questions, Suggestions and Announcements forum for any of the words "Riemann," "moderator," or "rating" to find past instances of your questions.

[quote="MathIsGreat"]I have a few questions about this website...

1.  How do you become a moderator?

2.  Above our avatar picture, there are five stars.  I have half stars, and above that, it says like P versus NP.
What are those?

3.  When I go to my profile, it says my rating is 1.8.  What am I doing wrong?

THANKS!![/quote]

1. Don't try to be one. AoPS doesn't need one now.
2. The # stars are based on posts.

New Member= 0 stars = <20 posts
P vs NP= 1/2 star = 20-49 posts
Hodge Conjecture = 1 star = 50-99 posts
Poincare Conjecture = 2 stars = 100-249 posts
Riemann Hypothesis = 2 1/2 stars = 250-499 posts
Yang-Mills Theory (me) = 3 stars = 500-999 posts
Navier-Stokes Equations = 4 stars = 1000-2499
Birch and Swinnerton Dyer = 5 stars = 2500+ posts

3. Either people are rating your posts spam (1) and below average (2) or you are not making good posts (or both)

I don't mean this post to be unhelpful, but I believe that these...

P vs NP
Hodge Conjecture
Poincare Conjecture
Riemann Hypothesis
Yang-Mills Theory
Navier-Stokes Equations
Birch and Swinnerton Dyer

..are the 7 Millenium Problems introduced in 2000 by the CMI (Clay Mathmatics Institute) which can be viewed on their website.
Plus, I think that the Poincare Conjecture has been proved, so maybe we should get a new one. Any suggestions?

Also, don't carp about your low rating. I mean, don't fret about it. In many instances, the post ratings on AoPS are highly distorted-- they don't give an accurate picture of the quality of posts. Also, remember the fact that a high percentage of people over here (AoPS) are middle-schoolers or teenagers.

Unlike most people with a rating below 2 say, the values are most times rather correct (with an accuracy of $ \pm 0.5$ or $ \pm 1$, being enough). The only problem is that you get a bad rating more easily than a good one, but only if you give chances to get a good amount of bad ratings.

[quote="ZetaX"] The only problem is that you get a bad rating more easily than a good one, but only if you give chances to get a good amount of bad ratings.[/quote]

I acquiesce with your statement (only to some extent). Most times the rating of a post isn't done objectively-- which leads to distortion. Most people on AoPS are teenagers or middle-schoolers, so while rating posts they tend to "rate" only posts which they don't like or posts which they perceive to be "bad".

The more experienced and "mature"* AoPSers don't care to rate posts which further exacerbates the problem.

Also, a high percentage of middle-schoolers and teenagers on AoPS tend to be (at most times) sophomoric. :huh: But I agree with your statement that it is very easy to get a "bad" rating on AoPS. All you need to do is provoke** a bunch of sophomoric middle-schoolers or teenagers-- your rating will sure nosedive. (No offense!)

* People who rate posts objectively. At most times, they turn out to be adults.

** It could also mean making statements which opposes, or is in conflict their your views/opinions.

[b]P.S.:[/b] :D  Enough of carping about post ratings.

[quote="BOGTRO"]By the way, 6 stars above your name indicate an Administrator.[/quote]

Just to be slightly obnoxious by being picky, below.

Actually, many people above a post rating of 2 also believe that it's a bad indicator. And human nature dictates that many of those with good enough post ratings wouldn't care to mention anyway, not to mention the possible "loss" (of course, not really) of no more post rating if the system is ever canceled due to an uprising. I just simply ignore what my post rating is, like many others have suggested, and attempt to make good posts. Most things in life are much better when let be.


modular arithmetic
analytic geometry
number theory
greatest common divisor
function
national olympiad

2004 USA Team Selection Test

[u]Day 1[/u]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=79775]Problem 1[/url]
Suppose $a_1, a_2, ..., a_n$ and $b_1, b_2, ..., b_n$ are real numbers such that

$(a_1 ^ 2 + a_2 ^ 2 + ... + a_n ^ 2 -1)(b_1 ^ 2 + b_2 ^ 2 + ... + b_n ^ 2 - 1) > (a_1 b_1 + a_2 b_2 + ... + a_n b_n - 1)^2$

Prove that $a_1 ^ 2 + a_2 ^ 2 + ... + a_n ^ 2 > 1$ and $b_1 ^ 2 + b_2 ^ 2 + ... + b_n ^ 2 > 1$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=79772]Problem 2[/url]
Assume $n$ is a positive integer. Considers sequences $a_0, a_1, ..., a_n$ for which $a_i \in \{1, 2, ..., n\}$ for all $i$ and $a_n = a_0$.

(a) Suppose n is odd. Find the number of such sequences if $a_i - a_{i-1} \not \equiv i \pmod{n}$ for all $i = 1, 2, ..., n$.
(b) Suppose n is an odd prime. Find the number of such sequences if $a_i - a_{i-1} \not \equiv i, 2i \pmod{n}$ for all $i = 1, 2, ..., n$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=79777]Problem 3[/url]
Draw a $2004 \times 2004$ array of points. What is the largest integer $n$ for which it is possible to draw a convex $n$-gon whose vertices are chosen from the points in the array?

[u]Day 2[/u]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=79779]Problem 4[/url]
Let $ABC$ be a triangle. Choose a point $D$ in its interior. Let $\omega _ 1$ be a circle passing through $B$ and $D$ and $\omega _ 2$ be a circle passing through $C$ and $D$ so that the other point of intersection of the two circles lies on $AD$. Let $\omega_1$ and $\omega_2$ intersect side $BC$ at $E$ and $F$, respectively. Denote by $X$ the intersection of $DF$, $AB$ and $Y$ the intersection of $DE$, $AC$. Show that $XY || BC$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=79781]Problem 5[/url]
Let $A = (0, 0, 0)$ in 3D space. Define the [i]weight[/i] of a point as the sum of the absolute values of the coordinates. Call a point a [i]primitive lattice point[/i] if all of its coordinates are integers whose gcd is 1. Let square $ABCD$ be an [i]unbalanced primitive integer square[/i] if it has integer side length and also, B and D are primitive lattice points with different weights. Prove that there are infinitely many unbalanced primitive integer squares such that the planes containing the squares are not parallel to each other.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79782]Problem 6[/url]
Define the function $f: \mathbb{R}_0 ^ + \longrightarrow \mathbb{Q}$ as follows: $f(0) = 0$ and

$f(3n+k) = -\frac{3f(n)}{2} + k$, for $k = 0, 1, 2$

Show that $f$ is one-to-one and determine the range of $f$.

Does USA have only one set of TST problems?

[b]cauchyguy[/b] Do you have USA TST 2005 and 2006 problem? :)


trigonometry
geometry
perpendicular bisector
trig identities
Law of Cosines
AMC

Given square ABCD, draw a circle with radius AB and center A. This circle intersects the perpendicular bisector of BC in two points. The point closer to C is labeled E. Find measure of angle AEC.

This is a nice question.  Is there an easy way to do it?  My methods involve either the law of cosines or some pythagorean or trigonometric work around the vertices.  Either way makes it a multi-step problem, I believe.  Was there a quick solution?

Yup...

http://www3.sympatico.ca/yuethomas/images/amc.gif

[hide]AE = 2AN, so <AEN = 30o.  So Triangle ADE is equilateral. Hence DE = DC, <DEC = <DCE = 75o. Then, <AEC = <AED + <DEC = 60o + 75o = 135o.[/hide]

[hide]

You have square ABCD. You know that the distance from A to any point on the circle is 1, so AE = 1. Let the perpendicular bisector be GF, where G and F are on BC and Ad respectively. AF = 1/2, so there's a 30-60-90 triangle. So AEF = 30. DEF = 30 as well. Triangle DEC is iscocoles, and angle D = 30, so angle DEC = 75. 75+30+30 = 135

[/hide]


probability
limit
integration
absolute value
geometric sequence
probability and stats

Hi everybody here is a funny exercise :

Let $ \alpha_1$ and $ \alpha_2$ two reals numbers such that, $ \alpha_1$ at least is irrational.

Let $ X_i$ i.i.d (independent identically distributed) sequences of bernouilli variables :
\[ \mathbb{P}(X_0 \equal{} \alpha_1) \equal{} p_1 \equal{} 1 \minus{} \mathbb{P}(X_0 \equal{} \alpha_2) \in ]0,1[
\]
We have to prove that with probability 1,
\[ S_n \equal{} \sum_{k \equal{} 0}^{N} X_k
\]
is equireparted (modulo 1)

I recall the définition, we say that $ u_n$ is equireparted modulo(1) if for all continuous $ \phi$ of the torus $ \mathbb{R} / \mathbb{Z}$ (that is to say $ \phi$ is 1-periodic) :
\[ \lim_{N \to \infty} \frac {1}{N}\sum_{k \equal{} 0}^{N} \phi(u_k) \equal{} \int_{0}^{1}\phi(t)dt
\]
[/b]

The English term is "equidistributed" (or "uniformly distributed").

Let's just use Weil's criterion, which reduces here to showing that $ \sum_{n\equal{}1}^N exp\Bigl\{2\pi i\sum_{k\equal{}1}^n X_k\Bigr\}\equal{}o(N)$ with probability $ 1$ (the pair $ (m\alpha_1,m\alpha_2)$ is just as good as the pair $ (\alpha_1,\alpha_2)$ when $ m\ne 0$).

Let us denote this sum by $ Y_N$ and compute $ E|Y_N|^2$. Squaring and opening the parentheses, we see that $ E|Y_N|^2\equal{}\sum_{k\equal{}0}^{N\minus{}1} (N\minus{}k) E\left[exp\Bigl\{2\pi i\sum_{k\equal{}1}^n X_k\Bigr\}\right]$. The expectations on the right hand side are easy to compute: they are just $ \left(\frac{e^{2\pi i\alpha_1}\plus{}e^{2\pi i\alpha_2}}2\right)^n$.

If $ \alpha_1\not\equiv\alpha_2\mod 1$, the number in parentheses is less than $ 1$ in absolute value, so we see that $ E|Y_N|^2\equal{}O(N)$ and, thereby, $ P\{|Y_N|\ge N^{2/3}\}\equal{}O(N^{\minus{}1/3})$. Since the condition $ Y_N\equal{}o(N)$ can be checked along any sequence that is denser than any geometric progression, this estimate is more than enough to apply Borel-Cantelli.

If $ \alpha_1\equiv\alpha_2\mod 1$, we have no randomness at all in $ Y_N$, it just equals $ \frac{e^{2\pi i(N\plus{}1)\alpha_1}\minus{}e^{2\pi i\alpha_1}}{e^{2\pi i\alpha_1}\minus{}1}$, which is bounded independently of $ N$.

hello fedja!

Indeed this works, i surmise you did not take more than ten minutes to solve it  :D


Hello everyone. As some of you have heard, our coach Mr. holbrook has passed away.
Here we will be able to give our thanks and honor him.

Hello Everyone. Saluton. It would be a disservice to our great Coach if we were to attempt to express his unbelievable accomplishments and endeavors in words. He is like a valiant crusader, fearless and bold, forever proclaiming the gospel of Mathematics without any signs of fatigue or loss of spirit. May he join the heavenly Father and His choir of angels. He will be forever hailed as a hero in our lives.

Estus kaŭzo de malutilo nia granda Trejnisto se ni ..us prov(ad)i esprimi liajn nekredeblajn plenumojn kaj klopodoj en vortoj. Kiel li estas kuraĝa krucmilitisto, sentima kaj aŭdaca, (por) ĉiam proklamanta la Evangelio de Matematiko tute sen signoj de laciĝo aŭ perdo de spirito. Li ..u aliĝas al la ĉiela Patro kaj Lia ĥoro de anĝeloj. Li estos (por) ĉiam aklamis kiel heroo en nia vivoj.

Farewell, and remember him in your prayers.

Adiaŭ.

Mr. Holbrook was a father figure and a friend.  He motivated us to work harder.  He was a teacher and mentor to many students.  I was astounded when I learned of his death and am still saddened.  He will always be in my prayers and I will never forget him.

As a human, he was unable to move around very efficiently, and that sometimes became a problem for him. Now, he has been released from his earthly burdens, and his soul is free to roam anywhere it wants.

May he live forever in the hearts and minds of any mathematician.

Kiel homa, li ne povis ir de loko al loko tre efike, kaj ke foje fariĝis problemon por li. Hodiaŭ, li estas liberigita de liaj surteraj ŝarĝoj, kaj lia animo estas libera por vagas ie ajn ĝi volas. Li ..u vivas (por) ĉiam en la koroj kaj mensoj de iu matematikisto.

He has left his body behind to roam the heavenly skies.  May he prosper and enjoy what he finds there.

Thank you, Subtraction.

Dankon, Subtraction.

He will never be forgotten. Mr. Holbrook was always so encouraging, and I can say without hesitation that he was the best teacher I have ever had. We were all lucky to meet him. If it wasn't for him...I don't even know.

I was shocked to see the email I received entitled "In Memory of Mr.Holbrook". I am deeply saddened. In the short time that I have known him (about a year), he has helped me tremendously. Whether it be recommending what book to read next, or discussing about a competition, Mr. Holbrook was someone I could always count on. I consider myself extremely lucky to have him as a coach. We must try our best to honor his ways, and hope that the BCA math team will continue to prosper following his core ideologies.

We will miss him. May he rest in peace.

Holbrook's passing was likely more surprising than the bombing at Hiroshima.
I never contemplated what would happen if Holbrook died--the math team would have lost its head, leading to all sorts of questions as to what would happen. Even now, everyone's confused for the most part.
Anyways, may he rest in peace, and may there be donuts and chocolate and candy and sweets in heaven.

There will always be sadness. But happiness and sadness are always parts of our life.

Yes.  Unfortunatly, all good things have too end.

[quote="bluecarneal"]Excuse me, but who is Mr. Holbrook?[/quote]

The best coach who ever lived. See rrusczyk's blog entry, "Sad News", and Zhero's blog entry with the quoted email for more details.

[url=http://www.unl.edu/amc/f-miscellaneous/f1-archive/2005-06-fa/06HSnominations/Holbrook2006.html]Mr. Holbrook[/url] was the coach of the AAST math team.  AAST, a school in New Jersey, is one of the few individual schools in the country that fields its own ARML team (actually, several ARML teams, the most of any single organization, in recent years).

I did not know Mr. Holbrook personally, but I knew who he was because of ARML.  I have also had to lose an inspiring math coach.  My condolences to you.

I am very sorry to hear this news, but I read about it on Mr.Ruscyks blog. I am sorry, and rest in peace.

[quote="HydraPyros"]They have updated the BCA site (go to bergen.org) with the sad news of Mr. Holbrook's passing. The exquisitely written article commemorates his life and talks about his accomplishments.[/quote]

Yes, there is a memorial for Mr. Holbrook on the BCA site.  This is the link: [url]http://bcts.bergen.org/index.php?option=com_content&task=view&id=1908&Itemid=1453[/url]

He will always be remembered, where ever we are.

~Subtraction


The numerator and denominator of the common fraction $ \frac{5}{6}$ are each
increased by 3. What is the positive difference between the resulting
fraction and $ \frac{5}{6}$? Express your answer as a common fraction.

Increase them both by 3 to yield $ 8/9$.
8/9-5/6
Get common denominator-
16/18-15/18= $ 1/18$


function
calculus
derivative
algebra
domain
rotation
complex analysis

my professor said this in class today morning and i cant figure out how- a non constant analytic function cannot take a region R into a line or into a circular arc.

 i've been thinking on these lines- now what happens if it does take a region into a line? if i consider $f'$ and suppose it turns out that is 0, then since R is region, f has to be a constant. But i cannot find any reasons for why that should be the case. or if my approach is wrong, can anyone here lead me to the right one? hints should be fine

WLOG, that line is the real line. What do the Cauchy-Riemann equations say about a real-valued analytic function?

yes from the cauchy riemann equations it follows that f is constant.why may we suppose that the line is the real line? or else make a co-ordinates change? if thats the case, then is analyticity still preserved? it seems like it should but can anyone tell me how. sorry for asking so many questions i am just starting on complex analysis

Looks to me this illustrates the importance of dimension:  since the complex derivative is formulated from the perspective of a "region" about a point, f(z) really has no chance of being analytic if it's image has no neighborhood.  This is the case when f(z) maps to a curve in the image space.  I think this happens only when the real and imaginary components of f(z) are functions of x or y alone:

$f(z)=u(x)+iv(x)$

or:

$f(z)=u(y)+iv(y)$

for $z=x+iy$.

Then it follows from the Cauchy-Riemann equations f(z) can be analytic only if u and v are constant functions.

Is this correct?

You've got the domain and range reversed there, shawtoos.

If $f$ maps a region into a line $L$, we can find complex numbers $a$ and $b$ (with $|a|=1$) so that the rotated and translated line $aL+b$ is the real axis. Then $af+b$ maps the region into the real axis, and has the same analyticity properties.

[quote="jmerry"]You've got the domain and range reversed there, shawtoos.

[/quote]

Dang it!  I see that now.  Maybe need to go back and review some basics and stop messin' around with the zeta function.  Thanks. :)


In triangle ABC, the medians drawn to sides AC and BC are perpendicular. Find AB if AC=b and BC=a


A quadrilateral inscribed in a circle has sides (in order) of 7, 15, 20, 24. Determine the area of the quadrilateral, the length of the two diagonals, and the radius of the circle.

PS: Samuel Greitzer died on February 22, 1988. He was the first coach of the USA IMO team. He was the coach for the years 1974-1980. He was a really passionate geometer which can be seen in many of his problem-solving publications, including Arbelos and many others.  :)

OK, I think I got all the stuff (unless I misread something):

Notice 7 and 24 are legs of a right triangle with hypotenuse 25, and 15 and 20 are also legs of a right triangle with hypotenuse 25. Thus, diameter = 25 --> radius = [b]25/2[/b].

Area = 1/2*(7)(24) + 1/2*(15)(20) = [b]234[/b].

Length of 1st diagonal = diameter of circumscribed circle = [b]25[/b].

Length of 2nd diagonal:

Applying Ptolemy, length * 25 = 15 * 24 + 20 * 7 --> length * 25 = 500 -->
length = [b]20[/b].


algebra
polynomial
complex numbers
algebra unsolved

For each positive integer n define the polynomial

$ P_n(z) := 1^3 z + 2^3 z^2 + \dots + n^3 z^n $.

Do the zeros of $ P_n $ lie inside, outside or on the unit circle $ |z| = 1 $ ?

The only thing that I have been able to prove is that apart from 0 (which is obviously one of the zeros of $ P_n $) one more zero must lie inside the unit circle. But I don't have an idea, how to get more information about the other roots to be able to decide, if they are inside, on or outside the unit circle.

Can you help me?

P.S.: This problem is in a chapter about geometric interpretation of complex numbers that also covers some material about roots of unity. So perhaps a geometric interpretation of the problem or the consideration of roots of unity might be helpful, but I don't know...

This problem appeared before as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=12915]AoPS Challenge 18[/url].  I believe jmerry posted a couple of solutions there.


If the six solutions of $x^{6}=-64$ are written in the form $a+bi$, where $a$ and $b$ are real, then find the product of those solutions with $a>0$.

Just find the roots of unity of -64, then see which ones have a>0 and multiply them together =)

you can't really say roots of "unity" anymore, since unity stands for 1... :D

you are finding all the solutions. and you are multiplying them. but we know de'moviers law that $cis(\theta_{1}) * cis(\theta_{2}) = cis(\theta_{1}+\theta_{2})$ hence we can find it directly by addind the angles.

[quote="B-tone_Party"][hide] We use DeMoivre's to find out that the roots of the equation are $2 cis (30), 2 cis (90), 2 cis (150), 2 cis (210), 2 cis (270),$ and $2 cis (330)$ (where $cis(x)=cos(x)+isin(x)$.) Since $a$ must be positive, we can rule out all roots where the angle is between 90 and 270, leaving $2 cis (30)$ and $2 cis (330).$  We convert those into $a+bi$ form $(x=\sqrt3+i, x=\sqrt3-i)$ and multiply them to get our answer, 4.[/hide][/quote]
Instead of converting all of them to $a+bi$ form, we can plot them. And we can see that 2 roots have $a>0$, 2 have $a=0$, and 2 have $a<0$. The 2 with positive $a$ are $2 cis 30$ and $2 cis 330$. The product is $4 cis 360=4$.

[hide]If we let $-64=64(\cos180^{\circ}+\sin180^{\circ})$, then by DeMoivre's theorem, the 6th roots of -64 are

$(2, 180^{\circ}+k\frac{360}{6}^{\circ})$.

Note that in order for a>0, the angle must be between -90 (270) and 90 degrees.

There are only two points: $(2, 330^{\circ})$ and $(2, 30^{\circ})$

Their product is $(2\cdot2, 330^{\circ}+30^{\circ})=4$[/hide]

Here comes my geometric interpretation because I can't stand watching so many solutions based on DeMoivre's (boring) :P

[hide="Geometry"]Note that the roots of $x^{6}=-64$ are the vertices of an inscribed regular hexagon on a circle of radius 2. The hexagon has a main diagonal along the y-axis. By symmetry, the roots are conjugate of each other. There is only one root in the first quadrant. So, $(a+bi)(a-bi)=a^{2}+b^{2}$, which is the square of the length of the radius. Thus, the answer is $2^{2}=4$ :D[/hide]

[quote="mysmartmouth"][quote="B-tone_Party"][hide] We use DeMoivre's to find out that the roots of the equation are $2 cis (30), 2 cis (90), 2 cis (150), 2 cis (210), 2 cis (270),$ and $2 cis (330)$ (where $cis(x)=cos(x)+isin(x)$.) Since $a$ must be positive, we can rule out all roots where the angle is between 90 and 270, leaving $2 cis (30)$ and $2 cis (330).$  We convert those into $a+bi$ form $(x=\sqrt3+i, x=\sqrt3-i)$ and multiply them to get our answer, 4.[/hide][/quote]

How do you use DeMoivre's in this case?  I thought DeMoivre's was:

$(\cos x+i \sin x)^{y}= (\cos xy+i \sin xy)$

How can we apply that to the problem?[/quote]You missed out something. The complete DeMoivre's formula is:

$(r(\cos x+i \sin x))^{y}= r^{y}(\cos xy+i \sin xy)$

Let $r=64$, $x=180^\circ$, $y=\frac16$

My work so far:

$x^{6}=-64$
$x=\sqrt[6]{-64}$
$x=\sqrt[6]{-1 \cdot 64}$
$x=2\sqrt[6]{-1}$

So we want the sixth roots of unity of -1.

I THINK that the sixth roots of unity of 1 are:
$cis 60$ $cis 120$ $cis 180$ $cis 240$ $cis 300$ and $cis 360$

So how do we find the sixth roots of unity for -1?

EDIT: I see how to solve it now.  Thanks!

[quote="mysmartmouth"][quote="B-tone_Party"][hide] We use DeMoivre's to find out that the roots of the equation are $2 cis (30), 2 cis (90), 2 cis (150), 2 cis (210), 2 cis (270),$ and $2 cis (330)$ (where $cis(x)=cos(x)+isin(x)$.) Since $a$ must be positive, we can rule out all roots where the angle is between 90 and 270, leaving $2 cis (30)$ and $2 cis (330).$  We convert those into $a+bi$ form $(x=\sqrt3+i, x=\sqrt3-i)$ and multiply them to get our answer, 4.[/hide][/quote]

How do you use DeMoivre's in this case?  I thought DeMoivre's was:

$(\cos x+i \sin x)^{y}= (\cos xy+i \sin xy)$

How can we apply that to the problem?[/quote]
DeMoivre's Theorem also states that
$(\cos{x}+i\sin{x})(\cos{y}+i\sin{y})=\cos{(x+y)}+i\sin{(x+y)}$.

[quote="mysmartmouth"][quote="ehehheehee"]"So how do we find the sixth roots of unity for -1?"

[hide]-1=e^(i180) in degrees

so we just divide 180 by 6 the same way you would divide the exponent by 6 when taking the root to that power.

Since e^(i540) is equal to e^(i180) we can divide 540 by 6 as well
so the roots are
e^(i30)
e^(i90)
e^(i150)
e^(i210)
e^(i270)
e^(i330)

the only ones with a positive a value are the first and the last so it is just:
2*2*e^(i360) or 4][/hide][/quote]

Nice.  Also, note that $e^{ix}= cis x$.

Is there a way to write cis nicely (like $\sin$) in $\LaTeX$?[/quote]
I think the only thing you can do is to use "\mbox{cis} \ x" so you have $\mbox{cis}\ x$.


inequalities
geometry
geometric transformation
reflection
LaTeX

I made this up earlier today.  I'd be interested to see what people think of this problem.

Given $a^2+b^2+c^2=1$ and ${a\geq{b\geq{c}}}>0$,prove $1-ab-ac-bc\geq{2b^2(a-c)^2}$.

Could you give a hint? I don't quite see it yet.

[quote="xxreddevilzxx"][hide="hint"]

try to substitute some stuff. then try to derive it into trivial inequality. Then you are done[/hide][/quote]

That's not how I came up with it.  Could you post your solution?

Ahh! I still don't get it. :(  I don't know how I even qualified for USAMO... How did you solve it, Magnara?

[hide="solution"]

$1-ab-bc-ca \ge 2b^2(a-c)^2$
$a^2+b^2+c^2-ab-bc-ca \ge 2b^2(a-c)^2$
$\frac{1}{2} (a-b)^2+ \frac{1}{2} (b-c)^2+ \frac{1}{2} (c-a)^2 -2b^2(a-c)^2 \ge 0$

Since $a \ge b \ge c$, $a^2 \ge b^2 \ge c^2$, and since $a^2+b^2+c^2 = 1$,
$b^2 \le \frac{1}{3}$, $2b^2 \le \frac{2}{3}$.

So we have

$\frac{1}{2} (a-b)^2+ \frac{1}{2} (b-c)^2 - \frac{1}{6} (a-c)^2 \ge 0$
$3(a-b)^2 + 3(b-c)^2 - (a-c)^2 \ge 0$

Maximizing the last term, we have (1,0,0).(Maximizing makes proof easier because then we can prove that other values can also work.)

$a-b = 1, b-c=0, a-c = 1$
$3(1)^2 + 3(0)^2 - 1^2 \ge 0$
$2 \ge 0$ QED.
[/hide]

i tried my best... hmm didnt look good at all. :?

His solution has other problems

we have, for example $b^2\leq \frac {1}{2}$, not $\frac {1}{3}$

[quote="seungjun"]I noticed that too. Blahblahblah, do you have a solution?[/quote]

Try and show that we can assume $c=0$ (I think; I haven't really looked at it)

[quote="seungjun"]I think your solution is wrong. you said that a2  >= b2  >= c2 , but you assumed that a, b, and c are positive. Magnara, did you mean to say that a, b, and c are positive?[/quote]

Yes, I did mean to make them positive.   :blush:

It's fixed now.

I'll post how I came up with this inequality soon.  It may be too contrived to be solvable, which would reflect poorly on me.

If the condition exists that they are all positive then it is not hard to show that 1 is greater than a, b, and c.

P.S. What is the LaTex symbol for greater than or equal to.

[quote="mna851"]
P.S. What is the LaTex symbol for greater than or equal to.[/quote]

\ge or \geq

Well, here's my solution to this problem.

[hide]We replace 1 with a^2+b^2+c^2.
$a^2+b^2+c^2-ab-ac-bc\geq{2b^2(a-c)^2}$
$(a-b)^2+(b-c)^2+(a-c)^2\geq{4b^2(a-c)^2}$
Seeing a sum of squares there, and in the a^2+b^2+c^2=1, we think of Cauchy.  We multiply by 1 to get
$(a^2+b^2+c^2)((a-b)^2+(b-c)^2+(a-c)^2)\geq{4b^2(a-c)^2}$, which holds by Cauchy (you can do the algebra if you'd like to verify that.[/hide][/hide]


function
geometry
parallelogram
complex analysis
complex analysis unsolved

Let $f\colon\mathbf C\to\mathbf C$ be an entire function which has two periods $a,b\in\mathbf C$ with $\frac{a}{b}\notin \mathbf R$. Prove that $f$ is constant.

Hint:
[hide]The function attains its global maximum in the parallelogram $0\to a \to a+b \to b \to 0$.[/hide]


integration
real analysis
calculus
derivative
inequalities
function
limit

Let $ g\in L^2(\mathbb{R})$ and define $ f(x)\equal{}\int_0^x g(t)dt$.

a) Must $ f$ be differentiable at 0?
b) Must $ f$ have any differentiable points?
c) Let $ \phi(x)\equal{}f^2(x)$, show that $ \phi'(0)$ exists and determine what it is.

Probably not the exact wording, but I don't think I left out any conditions.

a) No. For example, let $ g(x) = \frac23x^{ - 1/3}$ for $ 0 < x < 1$ so that $ f(x) = x^{2/3}$ near zero. This $ g\in L^2$ but $ f'(0)$ does not exist.

b) Yes. In fact, $ f$ must be differentiable almost everywhere, by the Lebesgue theorem on the differentiation of the integral. This theorem requires only that $ g\in L_{\text{loc}}^1,$ which is true.

c) By the Schwarz inequality,

${ f^2(x) = (\int_0^x}g(t)\,dt)^2\le (\int_0^x 1\,dt) (\int_0^xg^2(t)\,dt).$

At least that's it for $ x > 0.$ Being careful about the other side, we determine that

$ 0\le\phi(x) = f^2(x)\le|x||\int_0^xg^2(t)\,dt|.$

But since $ g^2$ is an integrable function we have (by an argument that uses the Dominated Convergence Theorem) that

$ \lim_{x\to 0}\int_0^xg^2(t)\,dt = 0.$

Hence $ \lim_{x\to0}\frac {\phi(x)}{x} = 0,$ so $ \phi'(0) = 0.$


trigonometry
inequalities unsolved
inequalities

Problem 1.
Let a,b,c are positive numbers such that $ abc\equal{}1$.
Prove that $ \frac{{ab}}{{c^2 (a^2  \plus{} b^2 )}} \plus{} \frac{{bc}}{{a^2 (b^2  \plus{} c^2 )}} \plus{} \frac{{ca}}{{b^2 (c^2  \plus{} a^2 )}} \ge \frac{3}{2}$

Problem 2.
Let a,b,c are positive numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}1$.
Prove that $ \frac{a}{{b^2  \plus{} c^2 }} \plus{} \frac{b}{{c^2  \plus{} a^2 }} \plus{} \frac{c}{{a^2  \plus{} b^2 }} \ge \frac{{3\sqrt 3 }}{2}$

Problem 3.
Let $ \Delta ABC$, find the maximum value of $ S \equal{} \frac{{64\sin ^6 B \plus{} 4\sqrt[4]{{2^{1 \plus{} \tan ^2 A} }}}}{{\tan ^2 A \plus{} 12\sin B}}$

Problem 4.
Let a,b,c are positive numbers, prove that:
$ \frac{{\sqrt a }}{{a^2  \plus{} b^2 }} \plus{} \frac{{\sqrt b }}{{b^2  \plus{} c^2 }} \plus{} \frac{{\sqrt c }}{{c^2  \plus{} a^2 }} \le \frac{1}{2}\left( {\frac{1}{{a^2 }} \plus{} \frac{1}{{b^2 }} \plus{} \frac{1}{{c^2 }}} \right)$

Problem 5.
Let a,b,c are positive numbers, prove that:
$ \frac{1}{{b(a \plus{} b)}} \plus{} \frac{1}{{c(c \plus{} b)}} \plus{} \frac{1}{{a(a \plus{} c)}} \ge \frac{9}{{2(ab \plus{} bc \plus{} ca)}}$

Let's do! ;)

[quote="L_Euler"]Problem 1.
Let a,b,c are positive numbers such that $ abc \equal{} 1$.
Prove that $ \frac {{ab}}{{c^2 (a^2 \plus{} b^2 )}} \plus{} \frac {{bc}}{{a^2 (b^2 \plus{} c^2 )}} \plus{} \frac {{ca}}{{b^2 (c^2 \plus{} a^2 )}} \ge \frac {3}{2}$

Problem 2.
Let a,b,c are positive numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.
Prove that $ \frac {a}{{b^2 \plus{} c^2 }} \plus{} \frac {b}{{c^2 \plus{} a^2 }} \plus{} \frac {c}{{a^2 \plus{} b^2 }} \ge \frac {{3\sqrt 3 }}{2}$

Problem 3.
Let $ \Delta ABC$, find the maximum value of $ S \equal{} \frac {{64\sin ^6 B \plus{} 4\sqrt [4]{{2^{1 \plus{} \tan ^2 A} }}}}{{\tan ^2 A \plus{} 12\sin B}}$

Problem 4.
Let a,b,c are positive numbers, prove that:
$ \frac {{\sqrt a }}{{a^2 \plus{} b^2 }} \plus{} \frac {{\sqrt b }}{{b^2 \plus{} c^2 }} \plus{} \frac {{\sqrt c }}{{c^2 \plus{} a^2 }} \le \frac {1}{2}\left( {\frac {1}{{a^2 }} \plus{} \frac {1}{{b^2 }} \plus{} \frac {1}{{c^2 }}} \right)$

Problem 5.
Let a,b,c are positive numbers, prove that:
$ \frac {1}{{b(a \plus{} b)}} \plus{} \frac {1}{{c(c \plus{} b)}} \plus{} \frac {1}{{a(a \plus{} c)}} \ge \frac {9}{{2(ab \plus{} bc \plus{} ca)}}$

Let's do! ;)[/quote]
First, problem 4 is wrong :wink:

[quote="L_Euler"]Problem 1.
Let a,b,c are positive numbers such that $ abc \equal{} 1$.
Prove that $ \frac {{ab}}{{c^2 (a^2 \plus{} b^2 )}} \plus{} \frac {{bc}}{{a^2 (b^2 \plus{} c^2 )}} \plus{} \frac {{ca}}{{b^2 (c^2 \plus{} a^2 )}} \ge \frac {3}{2}$

Problem 2.
Let a,b,c are positive numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.
Prove that $ \frac {a}{{b^2 \plus{} c^2 }} \plus{} \frac {b}{{c^2 \plus{} a^2 }} \plus{} \frac {c}{{a^2 \plus{} b^2 }} \ge \frac {{3\sqrt 3 }}{2}$

Problem 3.
Let $ \Delta ABC$, find the maximum value of $ S \equal{} \frac {{64\sin ^6 B \plus{} 4\sqrt [4]{{2^{1 \plus{} \tan ^2 A} }}}}{{\tan ^2 A \plus{} 12\sin B}}$

Problem 4.
Let a,b,c are positive numbers, prove that:
$ \frac {{\sqrt a }}{{a^2 \plus{} b^2 }} \plus{} \frac {{\sqrt b }}{{b^2 \plus{} c^2 }} \plus{} \frac {{\sqrt c }}{{c^2 \plus{} a^2 }} \le \frac {1}{2}\left( {\frac {1}{{a^2 }} \plus{} \frac {1}{{b^2 }} \plus{} \frac {1}{{c^2 }}} \right)$

Problem 5.
Let a,b,c are positive numbers, prove that:
$ \frac {1}{{b(a \plus{} b)}} \plus{} \frac {1}{{c(c \plus{} b)}} \plus{} \frac {1}{{a(a \plus{} c)}} \ge \frac {9}{{2(ab \plus{} bc \plus{} ca)}}$

Let's do! ;)[/quote]
Problem 2; we have : $ \frac {a}{b^2 \plus{} c^2} \equal{} \frac {a}{1 \minus{} a^2} \ge \frac {3\sqrt {3}a^2}{2}$
Problem 5; it is equivalent to: $ \sum _{cyc} {\frac {c \plus{} b}{b}} \plus{} \sum_{cyc}{\frac {a}{a \plus{} b}} \ge \frac {15}{2}$
By AM-GM,we have: $ \frac {a \plus{} b}{4a} \plus{} \frac {a}{a \plus{} b} \ge 1$
Then $ LHS \ge 3 \plus{} \frac {3}{4}(\frac {a \plus{} b}{b} \plus{} \frac {b \plus{} c}{b} \plus{} \frac {c \plus{} a}{a}) \ge \frac {9}{2}$
We have Q.E.D

Problem 1 and problem 2, seems to be related :)
In problem 1, set $ a \to \frac {1}{a}$, etc., then we have:
$ \frac {\frac {1}{a}\frac {1}{b}}{\frac {1}{c^2}(\frac {1}{a^2} \plus{} \frac {1}{b^2})} \equal{} \frac {abc^2}{a^2 \plus{} b^2} \equal{} \frac {c}{a^2 \plus{} b^2}$.
Since $ \sqrt {a^2 \plus{} b^2 \plus{} c^2} \ge \sqrt {3}\sqrt [3]{abc}$
So problem 2, follows from problem 1 :)

[quote="L_Euler"]
Problem 4.
Let a,b,c are positive numbers, prove that:
$ \frac {{\sqrt a }}{{a^2 \plus{} b^2 }} \plus{} \frac {{\sqrt b }}{{b^2 \plus{} c^2 }} \plus{} \frac {{\sqrt c }}{{c^2 \plus{} a^2 }} \le \frac {1}{2}\left( {\frac {1}{{a^2 }} \plus{} \frac {1}{{b^2 }} \plus{} \frac {1}{{c^2 }}} \right)$
[/quote]

Problem 4.
Let a,b,c are positive numbers, prove that:
$ \frac{{\sqrt a }}{{a^3  \plus{} b^2 }} \plus{} \frac{{\sqrt b }}{{b^3  \plus{} c^2 }} \plus{} \frac{{\sqrt c }}{{c^3  \plus{} a^2 }} \le \frac{1}{2}\left( {\frac{1}{{a^2 }} \plus{} \frac{1}{{b^2 }} \plus{} \frac{1}{{c^2 }}} \right)$

[quote="L_Euler"][quote="L_Euler"]
Problem 4.
Let a,b,c are positive numbers, prove that:
$ \frac {{\sqrt a }}{{a^2 \plus{} b^2 }} \plus{} \frac {{\sqrt b }}{{b^2 \plus{} c^2 }} \plus{} \frac {{\sqrt c }}{{c^2 \plus{} a^2 }} \le \frac {1}{2}\left( {\frac {1}{{a^2 }} \plus{} \frac {1}{{b^2 }} \plus{} \frac {1}{{c^2 }}} \right)$
[/quote]

Problem 4.
Let a,b,c are positive numbers, prove that:
$ \frac {{\sqrt a }}{{a^3 \plus{} b^2 }} \plus{} \frac {{\sqrt b }}{{b^3 \plus{} c^2 }} \plus{} \frac {{\sqrt c }}{{c^3 \plus{} a^2 }} \le \frac {1}{2}\left( {\frac {1}{{a^2 }} \plus{} \frac {1}{{b^2 }} \plus{} \frac {1}{{c^2 }}} \right)$[/quote]

By AM-GM $ \sum\frac {{\sqrt a }}{{a^3 \plus{} b^2 }}\le \sum\frac {{\sqrt a }}{{2\sqrt{a^3b^2} }}\equal{}\sum\frac {1}{2ab}\le \sum\frac{1}{2a^2}$  Last by CS


number theory
relatively prime
number theory solved

Determine the natural numbers A such that A^2-A is divisible by 10^n. Let n be the number of digits of A in decimal expansion.

[i]German IMO Selection Test 1983, problem 2[/i]

A^2-A = A(A-1), and A is relatively prime to A-1, so if 10^n|A(A-1), 2^n must divide A or A-1, and 5^n must divide the other factor.  Therefore we get two systems of congruences, either A = 0 mod 2^n and A = 1 mod 5^n or A = 1 mod 2^n and A = 0 mod 5^n, and by the chinese remainder theorem, we will be able to find two unique congruence classes mod 10^n that satisfy these systems, call these congruence classes r1_n and r2_n respectively.  Now, if n is the number of digits of A, clearly A <10^n, since 10^n is the smallest positive number with n+1 digits.  Therefore A = r1_n or A=r2_n would work if r1_n or r2_n had n digits.  Use r1_n for example, and suppose that r1_n had k digits, k < n.  We note that r1_n would satisfies the congruence A = 1 mod 2^k and A = 0 mod 5^k, and r1_k = r1_n because of the uniqueness of the chinese remainder theorem, and we do not have to worry if ri_n has n digits, for if not then it is just a repeat..  Therefore the set S := {A | 10^n|A^2-A with n the number of digits of A} is just the set {r1_k, r2_k with k a natural number}

in other words, it is just the solutions of A = 0 mod 2^m and A = 1 mod 5^m for all positive integers m and the solutions of A = 1 mod 2^l and A = 0 mod 5^l for all positive integers l.


limit
real analysis
real analysis unsolved

Let $ f$ be defined on $ [a,b]$ and let $ g$ be continuous on $ [\alpha,\beta]$ and differentiable at $ \gamma$ where $ g(\gamma)\equal{}c \in (a,b)$.

Show that:
(a) If $ g'(\gamma)>0$, then $ D^{\plus{}}(f \circ g)(\gamma)\equal{}D^{\plus{}}f(c)g'(\gamma)$.
(b) If $ g'(\gamma)<0$, then $ D^{\plus{}}(f \circ g)(\gamma)\equal{}D_{\minus{}}f(c)g'(\gamma)$.
(c) If $ g'(\gamma)\equal{}0$ and all derivates of $ f$ are finite at $ c$, then $ D^{\plus{}}(f \circ g)(\gamma)\equal{}0$.

[Defn: $ D^{\plus{}}f(c)\equal{}\overline{\lim}_{h \rightarrow 0^{\plus{}}}\frac{f(c\plus{}h)\minus{}f(c)}{h}, D_{\minus{}}f(c)\equal{}\underline{\lim}_{h \rightarrow 0^{\plus{}}}\frac{f(c)\minus{}f(c\minus{}h)}{h}$.]

I'm pretty sure these are word-for-word the standard proofs

I was thinking of starting with $ \frac{f(g(\gamma\plus{}h))\minus{}f(g(\gamma))}{h}\equal{}\frac{f(g(\gamma\plus{}h))\minus{}f(g(\gamma))}{g(\gamma\plus{}h)\minus{}g(\gamma)} \frac{g(\gamma\plus{}h)\minus{}g(\gamma)}{h}$ and then applying $ \overline{\lim}_{h \rightarrow 0^{\plus{}}}$ to both sides. My rationale (correct me if I'm wrong) is that if $ g'(\gamma)>0$, then the first term on the right will tend to $ D^{\plus{}}f(c)$ (probably continuity of $ g$ is needed in the argument) and the second term will tend to $ g'(\gamma)$. However I am concerned about whether $ \overline{\lim}$ of the product equals the product of the $ \overline{\lim}$ of each term.

Sorry I forgot about this thread! Maybe I am too late.

Anyway, in general the limsup of a product of sequences is not the product of the lim soups, but when one of the sequences has a non-zero limit it is true, which is the case here.

However, this proof is not correct, because $ g(\gamma \plus{} h) \minus{} g(\gamma)$ may be zero very often. Look in your first analysis textbook for a correct proof  :wink:

Thank you Kalle! Actually I realised the mistake earlier and then did an $ \varepsilon$-$ \delta$ proof, modelled after a proof of the usual chain rule.


abstract algebra
Ring Theory
superior algebra
superior algebra solved

Let $R$ be a local ring. If $M \oplus R^n \simeq R^{n+m}$, $M \simeq R^m$.

\textbf{Definition:} Let $A$ be a commutative ring with 1. A projective $A$-module is a direct summand of a free $A$-module.

The problem admits a mild generalization:

Let $R$ be a local ring with maximal ideal $m$ and let $M$ be a projective $R$-module of finite type. Then $M$ is a free
$R$-module of rank $\dim_{R/m}(M/mM)$.
\par\textbf{Proof:} Let $x_1,\ldots,x_n$ be elements of $M$ that yield a basis of $M/mM$ over the field $R/m$. Let $\varphi: R^n\to M$ be an
$R$-homomorphism that sends the standard basis element $e_i$ of $R^n$ to $x_i$ for all $i=\overline{1,n}$. $\varphi$ is obviously surjective. Let
$N=\ker\varphi$. Tensoring by $R/m$ we get the exact sequence $0\to N\otimes_RR/m\to(R/m)^n\stackrel{\overline{\varphi}}{\to}(M/mM)\to 0$, where
$\overline{\varphi}$ is the natural application extending $\varphi$. By hypothesis, $\overline{\varphi}$ is an isomorphism, hence $N\otimes_RR/m=0$.
Because $M$ is projective, the exact sequence $0\to N\to R^n\to M\to 0$ splits, so $R^n=N\oplus M$, therefore $N$ is of finite type. Nakayama's lemma
proves $N=0$, hence $M$ is free of rank $n=\dim_{R/m}(M/mM)$.


Determine all positive integers n such that
$2^{n}+105$
is a square of integer

Just checking - I hope this problem is NOT in the current round of the Polis MO?

Unfortunately it is...

[b]Problem 1 due October 10.[/b]

Overlooked it before.

I deleted the posts which contained a solution (or parts of a solution) and I locked the topic, but I'm afraid it's way too late...


The lock on a safe has three wheels, each of which has 8 possible positions. It is known that the lock is defective and will open if any two wheels are in the correct position. How many combinations must be tried to guarantee opening the safe?

Is the answer by any chance 64? It works for 64 but maybe there's a smaller one? However, it looks to me like the minimum one is 64. Could you tell me if I'm right? (somehow I think I'm wrong.. :D ).

I guess it is [n / 3] + 1  for an alphabet with n letters (n != 0 mod 3)
and n/3 for an alphabet with n letters (n = 0 mod 3), (here n = 8 and the
'letters' are 1 .. 8)

Each combination of 3 digits gives 3 combinations (but not always different) of 2 digits.
If it's possible to have three NEW combinations of three digits each time (except for the 1 or 2 last combinations if n not divisible by 3), only [8 / 3] + 1 combinations are necessary. And this number can't be less.

For n = 3 or 4 it is easy to construct by hand such list :

n=3 : 121, 232, 313
n=4 : 121, 232, 343, 414, 313, 242

For n = 5, it becomes quite cumbersome (but still possible)

I'm sure it's possible to construct such list for every n.

Maybe it has something to do with Eulerian cycles  :?

Well, it's obviously clear that I was wrong.. :D

Wait a second! What if, for n=3, the right combination is 112? None of your 3 combinations 121, 232, 313 contain 2 correct digits..

Well, it's obviously clear that I didn't read carefully the terms of the problem !  I didn't realize that the place of the digits was important ....

\downto  \varepsilon

I'm not so sure either, but it should be important (it's a safe for crying out loud!  :D  :D  :D )

We discussed this problem some time ago. Yes, Grobber?
I don't know whether we should leave it in Unsolved Section. As far as I remember official solution takes four pages. So I have supposition that we will never see solution here.
Maybe we should give appropriate link, Orlando?

Well, Myth I see that you are cleaning the section. That's cool.
I'm agree with you concerning the link.

Pierre.

Here's what I remember about the problem: it was posted somewhere else, and someone guessed (maybe me? :)) the answer was $32$, but there was no solution. Then orl gave a link to a file in the ISL project containing the proof, or something like that, a proof which impressed you, Myth, because you said you had only seen a very long solution (probably this $4$ page solution you mentioned here). Is this accurate? :)

@Pierre! Of course, 150 unsolved problems - that beats everything!


Să se arate că există şi este unic un punct $X$ situat în interiorul unui triunghi $ABC$, astfel încât sumele pătratelor laturilor triunghiurilor $XAB$, $XBC$ şi $XCA$ să fie egale.

Fie $M,N,P$ mijloacele segmentelor $BC$, $CA$, $AB$.
Cu teorema medianei avem:
$XM^{2}=\frac{2(XB^{2}+XC^{2})-BC^{2}}{4}$
$XN^{2}=\frac{2(XC^{2}+XA^{2})-CA^{2}}{4}$
$XP^{2}=\frac{2(XA^{2}+XB^{2})-AB^{2}}{4}$
Atunci avem de demonstrat ca
$4XM^{2}+3BC^{2}=4XN^{2}+3CA^{2}=4XP^{2}+3AB^{2}$
De aici se ajunge la
$XM^{2}-XN^{2}$ constant si $XM^{2}-XP^{2}$ constant.
$M,N,P$ puncte fixe
Se stie ca locul geometric al punctelor $T$ cu proprietatea ca $TA^{2}-TB^{2}$ constant, unde $A$ si $B$ sunt puncte fixe este o dreapta perpendiculara pe $AB$ (daca nu se stie, se poate demonstra  :D )
Inseamna ca $X$ se afla la intersectia dintre respectivele perpendiculare pe $MN$ si $MP$, deci exista si este unic.


For the past years, what scores did people need in order to qualify?

I got 666 in 7th grade and made it. Nah, 06 was 38, 07 was 38?

It was definitely 37 last year, but it could be 42 this year or it could be 31.  It depends both on who's out there and how hard the test is.  You may have some idea off the first, but NO idea of the difficulty, so forget about the "what do I need" and just get as many points as you can.

And get back to work peijin!!

Ok....
I'll go mope about my TEST score and do the State tests.

It's just a guess, but I assume states will be even harder than last year this year.

I THINK YOU'RE KEVIN!
Though I may be wrong.

You should have said :

I do not like green eggs and ham, I do not like them Sam I am!

But then I would have been lying, as I am indeed carl. Perhaps that CatalystOfNostalgia can say that. (Methinks his name is sam...)

Btw, kevin(if I'm thinking of the correct kevin)=MoldyTape.

Oh, maybe not. I wasn't so sure about Sterling and Peter.

I don't think perfect628 is Carl. Carl only has two user names and perfect628 isn't one of them. Sorry to say Peijin, but I hope you know that you are an alternate and your score doesn't actually count, but still try your hardest! Hopefully Alan and Jason will XXXXXXXXXXXXXXXX Go Clarke!!!!!!!!!!!!!!!!!!!!!!

To bad I can't go to States.........

*cough* Peijin made it as an individual says carl *cough*

oh can we stop this nonsense!!!!

perfect is in 9th grade and went to nats with carl in 7th.  If he wants to remain nameless, let him.

shentang, Peijin's score counts for him and if he does well he can make it to nationals.  now get back to work so you have the test done for tomorrow.....unlike this morning!!

perfect, please stop messing with my kids :D in case you hadn't figured it out, allan, peijin, and shentang are all from my school.

frost

Aw, fine...

My name is on the forum countless times, I'm sure you can find it. (No, I'm not cat nose<->carl)[size=9]Sam Trabucco. See, there it is again![/size]

Wait....
We're allowed to discuss scores now...?

You spelled individual wrong the first time.  :P

NO YOU ARE NOT!!!!!!!!!!!!!!!!!

That should be edited ASAP

shentang, please edit your post concerning scores.

good job to Acton and Alan......but you CAN NOT POST SCORES OF ANY TYPE!!!!!!!!!!

Messages will be deleted if they contain scores!!

Best of luck to everyone at states!

Frost they should make you a mod here. Or perhaps me...Or perhaps a responsible person...Yeah you.

I was just made a mod today after those scores were posted :D

So look out :lol:

Btw, @OP, i can ask Boping on Tuesday what the winning scores were in '04.

Sorry, thanks for telling me, but I guess someone did that for me.
I wonder how everyone knows each other.......I only know like 6 people.

In the world? That's unfortunate...

I only know Peter and Boping b/c I go to school with them(Peter's still trying to convince me Diamond>Clarke)...And Cat Nose from nats.

[quote="CatalystOfNostalgia"]clarke >> diamond[/quote]

Why thank you sir.

Diamond beat us at chapter, so they deserve the credit!  We shall have to wait until states to see if Diamond, Clarke, Acton(is that RJ Grey), or some other school comes out on top.  Until then, best of luck to all.

Frost

Diamond is in big trouble next year though... There's only one great 7th grader this year who does well in Mathcounts and IMLEM. Also, there are like no 6th graders in the Regular or Alternates for IMLEM. I expect Clarke to beat us by a larger margin next year and by then, I think Clarke>Diamond, *shivers at thought*.

From perfect628,[quote]In the world? That's unfortunate...
[/quote]

Haha, I meant in Massachusetts.
Hopefully Clarke will always be better than Diamond, after like a 12 year winning streak in IMLEM.

How do you do that box thing, and put what other people said?

[code][quote="name"]stuff[/quote][/code]

Btw, I was saying that I was arguing Clarke's case, not Diamond's.

[quote="CatalystOfNostalgia"]clarke >> diamond[/quote]

Hmmm....
Last time I checked the >> sign means right shift the number after it. At least in C. Not very familiar with other languages.
If you want to use it I think you should first define your variables...

[quote="shentang"]Hopefully Alan and Jason will XXXXXXXXXXXXXXXX Go Clarke!!!!!!!!!!!!!!!!!!!!!!
[/quote]

I don't wanna XXXXXXXXXXXXXXXX :P

Peijin, why would you get info on the past cutoffs?  If you want to make nationals, then don't just go "oh, I'll get this score so I can get in," cause that won't happen if you do that. just do the best you can.  Also, <intentionally omitted cause I don't feel like discussing chapter scores/difficulty>

Anyone have the qualifying score for 2004?

I see no reason why you of all people would want to know, but I'll find out tomorrow.

I got a 20/10=30 on the 2007 Nationals Mathcounts. I should've gotten one more problem for the target but oh well.


number theory
relatively prime
combinatorics unsolved
combinatorics

Find the smallest positive integer $ K$ such that every $ K$-element subset of $ \{1,2,...,50 \}$ contains two distinct elements $ a,b$ such that $ a\plus{}b$ divides $ ab$.

Yes, but so is every problem in $ \bf P$ ;)

can any body post solution of this problem

[quote="Sage!"]This problem is polynomially reducible to $ \mbox{NP \minus{} hard}$ problem of finding a monochromatic clique in graph.[/quote]
But that does not mean this problem is in NPC.

From $a+b|ab$, we can rewrite $b$ as $ka$. We now have $(k+1)a|ab\implies(k+1)|b$. We can now generate pairs $(a,b)$ be taking $q$, a factor of $a$, and letting $b=a(q-1)$. We can make the following observations. Odds beyond $25$ do not generate any useable pairs. Similarly, $17$, $19$ and $23$ do not generate anything. Evens above $24$ only generate useable pairs of the form $(a,a)$, as do $2$, $3$, $14$, $20$ and $22$. $4$ generates only $(4,4)$ and $(4,12)$, so we can throw this in too. Finally, we also throw in $11$ and $13$. Out total is now $13+3+13+5+1+2=37$. Adding one for $1$, we get $38$. Whatever our next number is, we will have
a pair that satisfy the conditions, so the answer is $\boxed{39}$.

[quote=fclvbfm934]Let $c = gcd(a, b)$, so $a = ca_1$ and $b = cb_1$. Therefore, $ca_1b_1$ is divisible by $a_1 + b_1$. Furthermore, since $gcd(a_1, b_1) = 1$, we see that $a_1 + b_1$ is relatively prime to $a_1$ and $b_1$, so $(a_1 + b_1) | c$.

Since $a + b \le 99    \Rightarrow  a_1 + b_1 \le 9$. Clearly, we must have $a_1 + b_1 \ge 3$, so we can go through the cases to find possible pairs (a, b):

(6, 3); (12, 6); (18, 9); (24, 12); (30, 15); (36, 18); (42, 21); (48, 24);
(12, 4); (24, 8); (36, 12); (48, 16)
(20, 5), (40, 10), (15, 10), (30, 20), (45, 30)
(30, 6)
(42, 7), (35, 14), (28, 21)
(40, 24)
(45, 36)

yielding 23 pairs and 24 numbers. Therefore, there are 26 numbers not listed above. Thus, $K > 26$. Now we need to find the smallest possible value of $K - 26$ such that any group of $K - 26$ we choose from the 24 will result in at least one ordered pair showing up. We see that this minimum value is 13, so we obtain the minimum value of $K$ being $\boxed{39}$.[/quote]

How could a+b<=99 implies a+b<=9?

Holy this one took me so much

Notice $a+b | a^2$ and thus you may check that $(a,ka)$ is a solution for $7>k>1$ and counting the number of pairs you'd get 16 pairs. But some numbers are repeated and thus by pingeonhole and some more couting you get that in the worst case a 39 number subset solves the problem. Then to show that $(a,ka)$ are the only pairs is what took me a lot of time: fix $a$ and vary $b$. When $a$ is a power of a prime then $b | a$ obviously. Thus we are only left with the following numbers from 1 to 50: $28,44,45,50$ (excluding all the pairs $(a,ka)$ that worked as above shown, but case-check work gives that none of $28,44,45,50$ work, and we're done.


Find $ x;y \in Z$ such that $ \frac {x^2 \plus{} y^2}{x \minus{} y} \in Z$ and $ \frac {x^2 \plus{} y^2}{x \minus{} y}|1995$

Sorry for little details, but I'm sure that you wan't need any more...
$ x^2 \plus{}y^2\equal{}a(x\minus{}y)$ where $ a|1995\equal{}3\cdot 5\cdot 7\cdot 19$. If $ a\equiv_4 3$ then $ a|x$ and $ a|y$, so if $ x\equal{}ax_1$ and $ y\equal{}ay_1$, so $ x_1^2 \plus{} y_1^2\equal{}x_1 \minus{} y_1$, then $ x_1\equal{}1$ and $ y_1\equal{}0$. If $ a\equiv_4 1$ then $ a\equal{}5$ so $ x^2 \plus{} y^2 \equal{} 5x \minus{} 5y$ from here $ x\leq 5$ and so on.
Bye


Find two sequence
\[ \{x_n\},\{y_n\}\]
of real number satisfy $ \{x_n\},\{y_n\}$
is not convergence but the sequence $ \{x_ny_n\}$ is convergence?

$ (x_n)\equal{}\left(1,2,1,2,1,2,\dots\right)$
$ (y_n)\equal{}\left(2,1,2,1,2,1,\dots\right)$

... and many variants on the same theme.


trigonometry inequalities quadratics function

Find the minimum value of $k$ such that $\sqrt{x}+\sqrt{y}\le k\sqrt{2x+y}$ for all positive numbers $x$ and $y$.

[hide]
$k = \sqrt{\frac{3}{2}}$.

To prove that this works, we start with AM-GM to get

$\frac{4x+y}{2} \ge 2\sqrt{xy}$

$3x+\frac{3}{2}y \ge x+2\sqrt{xy}+y = (\sqrt{x}+\sqrt{y})^2$

$\sqrt{\frac{3}{2}}\sqrt{2x+y} \ge \sqrt{x}+\sqrt{y}$.

We have equality at $y = 4x$.[/hide]

Your answer is correct, paladin8. This solution may be the ninth solution.
This is the easiest solution, I think.

[quote="nat mc"]Can you prove it is the minimum value of $k$?[/quote]

A smaller value of $k$ would not work at the equality case.

I have posted this problem before.  I remember [b]Myth[/b]'s solution by power mean. Please take a look.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=20245]www.mathlinks.ro/Forum/viewtopic.php?t=20245[/url]

kunny

I hope this is solution number 10 :D
[hide]as x and y  are positive we can substitute $x=r \cos^4\alpha$ and $y=r \sin^4\alpha$
the inequality becomes $1 \leq k^2 (2\cos^4\alpha +\sin^4\alpha)$. We have to find the minimum of $2\cos^4\alpha +\sin^4\alpha$ or of $3\cos^4\alpha -2\cos^2\alpha+1$ this is a quadratic function of $\cos^2$ and minimum is attained when $\cos^2=\frac 13$ and is $\frac 23$[/hide]


Right now, whenever I log in, with the "Log me on automatically each visit" checked, I still have to log in everytime. The problem even goes to the extent that if I log in from the main portal, then I would no longer be logged in when I go to the forums.

Usually clearing the cookies solves the problem, but not this time. No matter how many times I've tried to delete all the cookies, the problem still persists. (the other settings like privacy were never changed)

The second problem may or may not be related. For example, when I click on this link: [url]http://www.mathlinks.ro/viewtopic.php?t=6317[/url], it shows a page with "Please enter your username and password to log in", and after I entered them, it tells me that "The topic or post you requested does not exist". This had happened to me numerous times.

I am currently experiencing the same kind of problems with logging in, many other reports were filed in. Please continue to report strange things.

Since last night I have problem to login on mathlinks from the
main portal, when I put my username and password, on the main
portal it says

"Welcome guest"

my name appears on "who is online"
but I cannot post anything.

On the other way, when I go the main page of Art of Solving Problem
I can login.

But I prefer to get in mathlinks from its main portal.
Can somebody repair this technical problem ?

When I  choose the hexagonal logo Mathlinks Everyone
it says "impossible de trouver la page", this is another technical problem.

When I chose the logo "forum" just to the right of the hexagonal logo
"Mathlinks Every one", I arrive on
http://www.artofproblemsolving.com/index.php?c=1

thx
.o.

Yeah, it seems to work fine under the http://www.artofproblemsolving.com domain, but not under the http://www.mathlinks.ro domain.

What happens (try deleting your cookies before trying the mathlinks.ro)

Hmmm... it seems to work again  :)

Thanks!

Pierre.


Does anyone know good site I can see some previous year's SAT Math Questions? I looked into some of the questions on the newspaper some weeks ago and they were pretty good. I doubt they're in maximum level of easy AMC-12..

I didn't have any trouble on doing it but it's still good math! :)

[quote="Silverfalcon"]I doubt they're in maximum level of easy AMC-12..[/quote]

They're probably even easier than that. I'd say they're mainly AMC 8 level, with a couple that might be on the AMC 10B.

As for problems, I personally don't know any sites, but your local library almost surely has some books with practice SATs in them.

Now the verbal... That's some tough stuff...

The definitive source of SAT I practice problems is the College Board Web site's [url=http://www.collegeboard.com/student/testing/sat/prep_one/prep_one.html]SAT Preparation Center[/url]. The only book worthy of buying for SAT I prep (but check your local library first) is [url=http://www.amazon.com/exec/obidos/tg/detail/-/0874477050/learninfreed]10 Real SATs[/url] by the College Board.

The math on the SAT I math section ("old" version) is quite easy, easier than AMC 10 problems for sure. I know an eleven-year-old who got an perfect score on it. The best way to prepare for the verbal section is to READ, READ, READ, and READ. Turn off the TV and read. Read some interesting books and magazines about subjects you are interested in. Turn off the video games and read. Read a variety of different kinds of English writing, so that you will be more familiar with English grammar and more aware of correct usage of English vocabulary. THINK while you read and ask yourself why the author used the words he used. Don't forget to read in your free time, and be sure to find challenging things to read that are harder than what you are assigned to read in school. Learn some grammar, and then learn some more grammar. Learn etymology (look it up) and learn how it can be applied to understand English vocabulary better. Most of all, READ, READ, READ, and READ.

Good luck on the test. By the way, did I tell you how important it is to read to prepare for the verbal section of the SAT I?

[quote="demji"]does anyone know where i can find examples for the new sat coming out in 2005?[/quote]

Once again, the definitive source of practice questions for the "new" SAT I is the College Board Web site's [url=http://www.collegeboard.com/student/testing/newsat/prep/prep.html]New SAT Preparation Center[/url] section. It's really a good idea to become familiar with the College Board Web site early in your college search process.


Hello:
We are interested in participating in any related Math contest in the States. This surely help us in our Mathcounts competition taking place in January 2009.  Meanwhile we will like to warm-up in some other competition.
My name is Jesus Narro from Colegio Ingles in Monterrey, Mexico, that's 2 hours from the border.
Hope someone can invite us to participate.

thanks

Jesus Narro

Yes, we participate in Mathcounts.

The American School Foundation schools from Mexico have been organizing Mathcounts competitions.  The chapter test takes place simultaneously in El Salvador,  Puebla, Mexico, and Monterrey Mexico at the end of January.  The State test competition takes place in Puebla Mexico.  We wish to participate in the international Mathcounts competition, but some of the participant schools are not overseas schools.

We are willing to participate in any other competition taking place in the States to motivate students to continue being in Mathcounts.

there is no international mathcounts only national

Well, yes, the national Mathcounts competition.

For us it's sound like international, since they are selecting the top 4 of all the overseas schools from all around the world and the top 4 of each state.

It would be nice if they can accept 4 delegates from Mexico and central America.   I know that the state department is the one to make that call. It would be nice if they consider Central America and Mexico as one more state to be included in the National competition. But if that is not possible, we are more than happy to participate in any other similar competition in the States, since we are so close to the border.  That will motivate my students very much.

We have won the last three chapter test competitions and the last two state competitions in Puebla.  Best team A and best individual.

[quote="jnarro"]I know that the state department is the one to make that call. .[/quote]

I believe, National MATHCOUNTS Organization is the one to make that call ! If you are serious about it, you should contact http://www.mathcounts.org.

Congratulations on the success that you have had in Mathcounts in Puebla.

Some other competitions that you could compete in are:

[b]The Math League:[/b] It is a paper contest that is taken in late January that is 40 problems.  They have separate contest for 6th, 7th and 8th graders.  These problems are good beginning Mathcounts type problems.

[b]The AMC-8:[/b]  This is a test that is given in November.  It is 25 questions that are very similar to Mathcounts problems.

[b]ASMA (American Scholastic Math Association:[/b]  This is a series of 6 tests that are given about once a month starting in October.  There are 7 questions, which can be sovled using a calulator, you have to turn in the top 8 scores from your school.

A couple of other tests that are similar to this test are The Continental Math League and SIGMA.

If you want more information on these tests I suggest that you GOOGLE the names to get to the respective web sites to find the registration forms.

All of these tests are taken in your classroom but you will be able to compare your scores to other kids from all over the world that take the same test.  They will also help you to prepare for Mathcounts.

Best of luck at this year's Puebla contest.  I gave Mathcounts workshops there on three separate years.  There are a lot of very dedicated coachs from the area.

Alicia Ellingher is our National Mathcounts coordinator.  She is in the American School Foundation of Puebla.  She has been talking with different members of the Mathcounts committe.  Also with people from the department of state. We keep hoping to be accepted as one more state for the Mathcounts organization.  Let's hope this dream can come true.

I coordinate the chapter test in the North part of Mexico.  There are more than 20 schools participating.

Hello Jeff, I met you in Puebla several times.  You are a great couch and I keep learning from the different CDs you gave us.  Can it be possible for one school to participate  in a Texas Math competition, like number sense or any other? I'm Jesus Narro from Monterrey Mexico.


[quote="yokyok"]Is there anybody who can show me that if xyz=1 (x,y,z>0) then

x^3/((y+1)*(z+1)) + y^3/((z+1)*(x+1)) + z^3/((x+1)*(y+1)) >= 3/4       ?

Thanks[/quote]
I think, your inequality is very not strong.
Even Holder helps here:
$ \sum_{cyc}\frac{x^3}{(y\plus{}1)(z\plus{}1)}\geq\frac{(x\plus{}y\plus{}z)^3}{3(xy\plus{}xz\plus{}yz\plus{}2(x\plus{}y\plus{}z)\plus{}3)}.$
Thus, it remains to prove that
$ 4(x\plus{}y\plus{}z)^3\geq9\sum_{cyc}(xy\plus{}2x\plus{}1),$ which true by Muirhead.

apply ineq Am-Gm, we have

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

$ .....$

$ \rightarrow \sum \frac{x^3}{(y\plus{}1)(z\plus{}1)} \ge \frac{x\plus{}y\plus{}z}{2}\minus{}\frac{3}{4}$

$ x\plus{}y\plus{}x\ge 3\sqrt[3]{xyz}\equal{}3 \rightarrow \sum \frac{x^3}{(y\plus{}1)(z\plus{}1)}\ge \frac{3}{4}$


[quote="Interval"]Use the given zero to find the remaining zeros of each function.

[1] g(x) = x^3 + 3x^2 + 25x + 75; zero: -5i

[2] g(x) = 2x^5 - 3x^4 - 5x^3 - 15x^2 -207x + 108; zero: 3i[/quote]
[hide="(1)"]Set it equal to zero:
$x^{3}+3x^{2}+25x+75=x^{2}(x+3)+25(x+3)=(x^{2}+25)(x+3)=0$
so $x^{2}=-25\rightarrow x=-5i, 5i$ or $x=-3$
Therefore, the remaining zeros are $5i$ and $-3$[/hide]

interval, please learn $\LaTeX$. you've been here long enough to still type things like this w/o it.

[quote="perfect628"]interval, please learn $LaTeX$. you've been here long enough to still type things like this w/o it.[/quote]

Yes. Please.

Also, a handy fact. For any polynomial, if $a+bi$ is a root, then $a-bi$ is a root as well.

[quote="Interval"]I will post with LATEX from now on.[/quote]
Really? This is a truly happy moment. I can now read your posts better from now on. THANK YOU.


linear algebra
matrix
algebra
polynomial
abstract algebra
vector
college contests

Let A is square real matrix of order n such that  A^2 = E. Show that
              rank(A+E) + rank(A-E) = n

<i>Source VUMC 1994 </i>

I like this one A^2=E this mean that A is a symetry

A vanihes the polynomial x^2-1=(x-1)(x+1)

x+1, x-1 are prime => kernel decomposition theorem =>

R^n = ker(A-E)@ker(A+E) where @ is direct sum of sub vector space

take dimension

(*) n = dimker(A-E)+dimker(A+E)

rank theorem => dimker(A-E)=n-rg(A-E) , dimker(A+E)=n-rg(A+E) we plug in (*)

rg(A-E)+rg(A+E)=n

I don't know the name of decomposition theorem in english,
 in Paris people say
"thorme de dcomposition des noyaux" or " lemme des noyaux"

Theorem: E an R-vector space finite dimension, A,B polynomial with
real coefficients gcd(A,B) = 1, P=A.B product of A ,B
u:E->E linear map
then kerP(u)=kerA(u)@kerB(u) where @ is direct sum

How do you say this theorem in vietnam langage ?
In romanian ?


Yes people greece just won USA , THE BEST team and with NBA . we are at the big final for the first time in the worldleague...it was a very difficult game :D

[quote="kostas"]...it was a very difficult game  :D [/quote]

Όχι και τόσο .. :D Κατοστάρα φάγανε... :D

σιγουρα το λεω ετσι για να το παιξουμε ανωτεροι.αφου δεν ξερουν οι μαθλινκερς ελληνικα: χαλια επαιξε η αμερικα.it was a good game. 101-95

Η πικρή αλήθεια είναι όμως ότι γενικά η Αμερική δε στέλνει την πρώτη της ομάδα συνήθως (και πιστεύω ότι κατά βάθος το ξέρουμε όλοι αυτό) -- για να είναι τα παιχνίδια δίκαια. Ε ήμασταν και εμείς [i]λίγο[/i] κωλόφαρδοι με κάτι άσχετα τρίποντα που πετούσαμε και μπαίναν.

Αλλάααα... Η κατοστάρα κατοστάρα. :lol:

ναι αλλα ενας ενας οι αμερικανοι παιχτες ειναι ΝΒΑερς και ισως καλυτεροι σαν ταλεντο.αλλα η ομαδα τελικα μετραει οπως και (σαφως) το αποτελεσμα :D

Εννοείται ότι η ομάδα μετράει. Μας το έχουν αποδείξει 2 φορές σε 2 χρόνια σε 2 αθλήματα. :!:

Ρε παιδιά ο τύπος που μας έβαλε 27 πόντους σήμερα ήταν πρώτος σκόρερ του ΝΒΑ  την περασμένη χρονιά ....
Πιστεύω πως κατέβασαν ότι καλύτερο είχαν γιατί ποτέ δε θέλουν να χάνουν (έτσι και αλλιώς την είχαν πάθει στην Ινδιανάπολη που κατέβασαν τους κολλεγιακούς ) ........
Τώρα μακάρι να το σηκώσουμε με την Ισπανία ...... :D

ΠΑΙΔΙΑ ΤΟ ΜΟΝΟ ΠΡΑΓΜΑ ΠΟΥ ΜΕΤΡΑΕΙ ΕΙΝΑΙ ΟΤΙ ΗΜΑΣΤΑΝ ΑΝΩΤΕΡΟΙ ΑΠΟ ΤΙΣ ΗΠΑ ΣΕ ΟΛΟ ΤΟ ΠΑΙΧΝΙΔΙ ΠΛΗΝ ΤΩΝ 10 ΠΡΩΤΩΝ ΛΕΠΤΩΝ ΑΡΑ ΤΙΠΟΤΑ ΜΑ ΤΙΠΟΤΑ ΔΕΝ ΜΕΙΩΝΕΙ ΤΗΝ ΝΙΚΗ ΧΑΙΡΕΤΩ ΚΙ ΤΗΝ ΚΥΡΙΑΚΗ ΧΡΥΣΟ

[quote="silouan"]Πιστεύω πως κατέβασαν ότι καλύτερο είχαν γιατί ποτέ δε θέλουν να χάνουν [/quote]

Που ήταν όμως οι Allen Iverson και Kobe Bryant ?  :o

Λένε ότι μπορεί να μην παίξει ο Gasol την Κυριακή  :D

[quote="silouan"]Ρε παιδιά ο τύπος που μας έβαλε 27 πόντους σήμερα ήταν πρώτος σκόρερ του ΝΒΑ  την περασμένη χρονιά ....[/quote]

Πού το είδες αυτό; Τα stats είναι: [url]http://sports.espn.go.com/nba/statistics?stat=nbascoring&sort=pts&avg=pg&seasontype=2[/url]

Ο Ramanujan αποδεικνύει αυτό που είπα σε άλλο post. (βασικά ο Bryant είναι τραυματίας νομίζω) Φάση θα είχε να κατέβαινε ο Carter και ο Iverson. :D

[quote="kostas"]ποσο χρονων ειναι αυτοι οι 2? εξαλλου και παλι δεν παιζει ρολο.τα ανερχομενα εστερια οπως carmelo Anthony kai LeBron james?[/quote]

Bryant - 28

Iverson - 31

Κοιτάξτε και στο nba.com τους 81 πόντους του Bryant σε ένα παιχνίδι - έχει βιντεάκι  :D

τωρα αν σου πω οτι και παλι θα κερδιζαμε?δηλ και με τους δυο αυτους τους παιχταραδες οι ΗΠΑ αμα βρεθει σε πιεση...τα χανει. πηγα και εγω στο σαιτ του ΝΒΑ. ρε παιδια το CNN εκανε 7,5 ωρες να πει την ειδηση. μεχρι και η γειτονες ειπαν καλα λογια για την Ελλαδα  :P

Συγκλονιστικό παιχνίδι!!! :clap:

Πάμε τώρα για το χρυσό!!!!!!  :first:

Λέτε να το σηκώσουμε και πάλι; :P

Ρε παιδιά μην τρελαθούμε τελείως οι Αμρικάνοι φέρανε τους καλύτερους παικτες τους κατά τον πολύ μεγάλο προπονητή Mike siziewski κι εμεις το μόνο που μπορούμε να πούμε οτι αυτη ο ομάδα της Αμρικής ήταν η καλύτερη δυνατή κι συμφωνώ με την άποψη του stedes νομίζω ΄΄οτι ακόμα κι΄έτσι οι Αμερικάνοι θα χάνανε διότι δεν μπορούσαν απλά να συνέλθουν από το σόκ που τους προκάλεσε η εθνική μας αλλά ι ο Παναγιώτης Γιαννάκης αυτός ο προπονηταράς με τις τεχνικές του Τέλος δυστυχώς για το Μπάσκετ αλλα ευτυχώς για μάς ο Pau Gasol (ένας παίκτης διαφήμιση για το άθλημα) λόγω κατάγματος είναι οριστικά εκτός τελικού
Χαιρετώ

για να πουμε και κατι φοβερο,αυτη η ομαδα οπως ειπε και ενας δημοσιογραφος ειναι η νεα γενεια του αμερικανικου μπασκετ.το μελλον δηλ του ΝΒΑ. :D

Ο τελικός άφησε μια γλυκόπικρη γεύση σε όλους μας.
Παρ'όλα αυτά η Ελλάδα φέτος πέτυχε τους αρχικούς της στόχους και αναμφισβήτητα η δεύτερη θέση σε όλο τον κόσμο είναι τεράστια επιτυχία.

Συγχαρητήρια ,λοιπόν στην εθνική μας ομάδα μπάσκετ!!!  :clap2:  :winner_second:

Το σημαντικότερο απ'όλα είναι ότι αυτή η ομάδα έκανε μια πολύ καλή πορεία σ'αυτό το παγκόσμιο πρωτάθλημα και μας επιφυλάσει σίγουρα πολλές ακόμη χαρές στο μέλλον, αφού έχει πολλά ακόμη να δώσει.  :wink:  :P

 :winner_second:  :winner_second:  :winner_second:  :winner_second:


Let V subspace of R3 with inner product V={(x,y,z) in R3 : x+z=0, y=2x}

1. how can we find the projection P onto V
2.  how can we find the projection matrix P as of the normal basis of R3
3.   how can we find a basis of R3 for which the matrix P will be $ \left(\begin{array}{ccc}
1 & 0 & 0\\
0 & 0 & 0 \\
0 & 0 & 0\end{array}\right)$

Thanks in advance!!

V is the space spanned by the vector (1,2,-1) (therefore, a line through the origin). The norm of this vector is $ \sqrt{6}$, and so an orthonormal basis for V is $ \frac{1}{\sqrt{6}}  \left[\begin{array}{c} 1 \\ 2 \\ \minus{}1 \end{array}\right]$.


Should I get both 1 and 2? I am going to nationals.

To have a good book (and use it!) is never a bad idea. I also recomend you "Mathematical Circles".

Best,

[quote="djimenez"]To have a good book (and use it!) is never a bad idea. I also recomend you "Mathematical Circles".
[/quote]

I second this recommendation.  We use this along with AoPS in our Independent Study.

For MATHCOUNTS, Volume 1 covers nearly everything you might see at National MATHCOUNTS.

I'm getting the books.  :) I feel that if I apply myself now, I might be a success someday. Even though I'm in grade 9 and most of you guys have more time than me.  :lol:

thanks. I'll get it. Should I get 1 and 2?? or just 1? or just 2?

[quote="mysmartmouth"]I'm in 7th grade but my math teacher strongly reccomended I get volumes 1& 2, even though vol 2 is well beyond me.  It was the right move![/quote]

Even if you didn't buy Vol 2 now, I think that you would get it later anyways.

Yes.  Get both (and solutions) along w/ Mathematical Circles.

What's this mathematical circles thing?

And i'd love to do indep study, but time for me right now is kinda scarce...especially on that 10hr/wk recommendation

I asked Mr. Rusczykthat question myself and it is under the Resources on the right side of the page.

Billy


If I have the original problems from the 2003 AIME's, how do I add them to [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45]this[/url]?

[quote="Valentin Vornicu"]You contact 4everwise on PM :)[/quote]

I see...


[b]For all you guys who spent hours playing the wondeful Starcraft I and II, with its great story and wonderful (for that period) graphics; how much did you enjoy!!![/b]

Don't you think the story needs to be completed with with [b]Starcraft III[/b]. If you know of plans for its release , please post!!! Post on!!!

This thread screams "fanboy".

...... :?

First of all there is no starcraft 2. There is only Starcraft and Starcraft brood war.
second, we don't know when its coming out. Blizzard says they are thinking about it but currently it is not on their "to do" list. Oh yeah, if you want a continuation of the story get Starcraft Ghost. Its supposed to be released on all consoles lJanuary 2006. (I think)

To the above poster. Wasn't your favorite game ff7? If so I wouldn't be talking much about fanboys...

aaahh!!!  :thumbup: i can't believe someone actually brought up the topic of the world's best game!!

Ok, now that i've let that out.... Starcraft 2 is rumoured to be the next big project on Blizzard's agenda, although those rumours have been existing for years.  With starcraft ghost coming out on console (and i'm not expecting it to match the standard of brood wars - why isn't it coming out on computer?!), at least there's tangible evidence starcraft is back on Blizzard's mind.

The storyline of Starcraft 2 (if it ever comes out) will definitely have references to the so called "Duran" (see the secret mission for more details), and its hinted that there will be a come-back of the ancient race of Xel-Naga (its debatable however, whether blizzard will actually introduce this race as a new playable race... Starcraft already has the perfect balance of protoss, terran, zerg).  The zerg-protoss hybrids will probably be mentioned as an ultimate race, and the final product of the Xel-Naga experiments combining an essence of purity and an essence of form.

The fall of Kerrigan will probably play a part somewhere - my guess however, is that this will be explored in Starcraft Ghost.

I hope $Starcraft$ $2$ is going to be the next Blizzard project.  Meanwhile, I'll be on battle.net playing bgh 3v3 as protoss.  gl hf!

p.s. I know, I am a Starcraft fanatic.  Don't hold it against me.  :P

[quote="sirkuku"]
To the above poster. Wasn't your favorite game ff7? If so I wouldn't be talking much about fanboys...[/quote]
No, it's not.

.........Where did you get this false information?

Thanks zeus_three!!!  It is my hunch too ( as would be of anyone who has played Starcraft) that the Xel-Naga will play a VERY Important role in the Next Starcraft.

ALso, by STarcraft 2 , I was refering to brood wars!!! Sorry!!!


Follow-up στο metric spaces post του cmad :)

Έστω ένας νορμικός (έτσι το λέμε το normed?) χώρος $ X$ (δηλ. ένας διανυσματικός χώρος με μία νόρμα $ \|\cdot\|$, π.χ. Ευκλείδεια, sup, whatever...). Θα λέμε ότι ο χώρος αυτός έχει την ιδιότητα Heine-Borel όταν κάθε κλειστό και φραγμένο υποσύνολο του $ X$ είναι και συμπαγές.

Το θεώρημα Heine-Borel μας λέει ότι κάθε normed space πεπερασμένης διάστασης έχει την ιδιότητα Heine-Borel :P

Να δειχθεί το αντίστροφο: αν ένας normed space έχει την ιδιότητα Heine-Borel, τότε είναι πεπερασμένης διάστασης.

Cheerio,

Durandal 1707

PS: Προφανώς πάει σε όσους δεν το έχουν ξαναδεί! Αν δεν υπάρξει ενδιαφέρον... kill it :)

Είναι αρκετά ενδιαφέρον πρόβλημα για να τεστάρετε την κατανόηση της συμπαγότητας, προσπαθήστε το.

[hide="hint"]sequential compactness[/hide]

Κανείς?:(

[hide="hint"]Αν είναι infinite dimensional, μπορούμε να πάρουμε μια ακολουθία απο μοναδιαία διανύσματα που δεν είναι Cauchy (γιατι?)[/hide]


geometry
3D geometry
tetrahedron
sphere
calculus
calculus computations

Suppose that a tetrahedron $ ABCD$ inscribed in a sphere with radius 1, varies satisfying $ \angle{BAC}\equal{}90^\circ$, find the maximum volume $ V$ of the tetrahedron.

let the face BAC be horizontal and below the centre of the sphere.
The volume is a third of the area of this face times the distance from the plane of the face to the point D.
For fixed BAC this is maximum when the point D is at the top of the sphere.
Let G be the circle formed from the intersection of the plane of BAC with the sphere.
G has radius r.
BAC is a right angled triangle inside G so its hypotenuse lies on a diameter and has length 2r.
The area of BAC is maximum when it is iscocelese with area $ r^2$
If x is the distance from the centre of G to D then $ r^2 \plus{} (x \minus{} 1)^2 \equal{} 1$
$ r^2 \equal{} 1 \minus{} (x \minus{} 1)^2 \equal{} x(2 \minus{} x)$
The volume is $ V \equal{} \frac {xr^2}{3} \equal{} \frac {1}{3} x^2(2 \minus{} x)$
differentiating we find the maximum at x = 4/3 and V = 32/81

edit: corrected a typo

Now the problem has been solved completely. :lol:


Suppose $x,y$ are two nonzero numbers such that $(1-x^2)(1-y^2)=1$.
Find the minimum value of $A=(1+\mid x \mid)(1+\mid y \mid)$

 :)

My solution.
From the condition we have that
$x^2y^2=x^2+y^2$
But we know that$ x^2+y^2\geq2xy$ so $x^2y^2\geq2xy$ so $xy\geq2$
Additionally $A=1+\left|x\right|+\left|y\right|+\left|xy\right|\geq1+x+y+xy\geq3+2\sqrt{2}$

Yes, I solved this by the same way!

Just a point :

Since $x^2y^2=x^2+y^2$ and $x^2+y^2\geq2 \mid xy \mid$
we get $\mid xy \mid \geq2$
so $A=1+\mid x \mid+ \mid y \mid+\mid xy \mid\geq 1+2\sqrt{\mid xy \mid}+\mid xy \mid\geq 3+2\sqrt{2}$

The equality holds $if$ $\mid x \mid=\mid y \mid=\sqrt{2}$

 :)  :)

I think that the two numbers $x,y$
do not have necessarily the same sign
so $\mid xy \mid\geq2 $.....

ok?? :)  :)

$\mid x \mid \geq x$ is true $\forall x \in R$

however see my previous post.......

I want to tell this and I want to write this since 4 minutes but my LaTex has a problem


trigonometry
inequalities
inequalities proposed

Prove that for any quadrilateral ABCD,\[ \cos{A} \plus{} \cos{B} \plus{} \cos{C} \plus{} \cos{D}\leq{2}\]

Apparently, you meant [b]convex[/b] quadrilateral (otherwise it is certainly false: take 3 angles close to $ 0$ and one close to $ 2\pi$). In the convex quadrilateral, the inequality as you stated it is fairly obvious because either we have no mor than 2 acute angles, in which case we have at most 2 positive numbers, both not exceeding $ 1$, or we have 3 acute and one obtuse angle, say, $ A$, in which case it is easy to see that $ A\minus{}\frac\pi 2>\frac\pi 2\minus{}B$, so $ \cos A\plus{}\cos B<0$. It may be interesting to ask what is the best constant you can put in the right hand side of this inequality for convex quadrilaterals and what is the worst quadrilateral itself. I'm not sure it hasn't been already discussed though. I guess that the answer to that is $ \frac 12$ and "the equilateral triangle with a dummy vertex on one of the sides", but I didn't really try to prove it.


Find the function $f: N \rightarrow N$ satisfy:
$\forall n \in N \Rightarrow f(f(n))+f(n)=2n+3$

Look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107265]this one[/url], just a special case of this problem.


logarithms
inequalities
function
calculus
derivative
real analysis
real analysis unsolved

Find the minimum of \[ f(x)=2\frac{1}{(1+2x)^k}+x^k,0 \le x \le \frac 12  \]
where $k=\frac{\ln 3-\ln 2}{\ln2}$
This is a rest problem when I solve an inequality [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=73225]here[/url].
It is obvious by drawing a graph that $f(x)$ get its minimum at the endpoints,but I can't prove it by hand,Can somebody help me?

[quote="kunny"] $0\leq x\leq \frac{1}{2}, x^{k-1}(1+2x)^{k+1}\leq 2^{k+1}x^{k-1}\leq 2^{k+1}\left(\frac{1}{2}\right)^{k-1}=4,$
kunny[/quote]
Sorry,you are wrong here.Because $k-1<0$ :)

How about this. Note that $0$ is a local minimum and that $f'\left(\frac1 2\right)=0$. Thus, if the global minimum is less than $2$, the derivative $f'$ should vanish at at least 3 different points on $\left(0,\frac 1 2\right]$: the point of global maximum, the point of global minimum and the point $\frac1 2$. Therefore, the equation $x^{k-1}(1+2x)^{k+1}=4$ must have at least 3 roots on $\left(0,\frac 1 2\right]$. But then the logarithmic derivative $\frac{k-1}x+\frac{2(k+1)}{1+2x}$ of the left hand side must vanish at  least twice on that interval. But it is a fraction with non-vanishing denominator whose numerator is a non-constant linear function, so this is impossible.

Thanks very much fedja.Very nice proof.I just want to clear your proof.(Although it is clear enough) :)
[quote="fedja"]How about this. Note that $0$ is a local minimum and that $f'\left(\frac1 2\right)=0$. Thus, if the global minimum is less than $2$, the derivative $f'$ should vanish at at least 3 different points on $\left(0,\frac 1 2\right]$: the point of global maximum, the point of global minimum and the point $\frac1 2$. [/quote]
Here I think we should calculate one value,for example $f(\frac 1 4)>2$. :)
[quote="fedja"]Therefore, the equation $x^{k-1}(1+2x)^{k+1}=4$ must have at least 3 roots on $\left(0,\frac 1 2\right]$. But then the logarithmic derivative $\frac{k-1}x+\frac{2(k+1)}{1+2x}$ of the left hand side must vanish at  least twice on that interval. But it is a fraction with non-vanishing denominator whose numerator is a non-constant linear function, so this is impossible.[/quote]
Here I think you used Rolle's Theorem to prove that $\frac{k-1}x+\frac{2(k+1)}{1+2x}$ of the left hand side must vanish at  least twice on that interval?Right.Or you use others?

[quote="zhaobin"]
Here I think we should calculate one value,for example $f(\frac 1 4)>2$. :)
..........
Here I think you used Rolle's Theorem to prove that $\frac{k-1}x+\frac{2(k+1)}{1+2x}$ of the left hand side must vanish at  least twice on that interval?Right.Or you use others?[/quote]
1) It is not necessary to calculate any values: just note that $f'(0)=+\infty$.
2) Yes, it's Rolle's theorem that I used. :)