analytic geometry
graphing lines
slope
geometry
rectangle
geometry unsolved

Does someone know how to prove this:

An angle of x such that 0 < x < 90 degrees can be constructed with a ruler if and only if tan(x) is a rational number.

I can't find any information on ruler-only constructions online.
I know tan(x) = sin(x)/cos(x) = a/b which can be constructed with ruler and compass for a,b in Z (set of integers). But how does this work for just a ruler?

Thanks in advance.

I don't see how you prove this because you can't draw perpendiculars or anything with just a ruler.

Is the ruler marked or unmarked? Because if it's unmarked, I can't imagine how you could construct any sort of angle. If it's marked then I think it can be solved, though I'm not sure how yet.

In either case, we can see that the slope of any angle constructed must be rational because every step of any construction can be described analytically by linear equations with integer coefficients.

[quote="Euclid's Turtle"]Just for a ruler because one may not physically draw a line segment with an irrational length. It can be close, however close (considering the thickness of the lead) but never exact. So rationals are just an implication I guess.[/quote]
These doesn't matter. We can't use a compass. That matters.

Actually I think a ruler can make perpendicular lines.
Consider a thick ruler. They are made with the edges perpendicular so you can draw a $ \frac{\pi}{2}$ angle using that.
Consider a thin ruler. Since you can't use the edge, you may flip the ruler over and over up or down, whichever and mark a point. Say both initial and marked points are the edge of the ruler. This would make a perpendicular line.

a ruler is a rigid "rectangle" such that one of its edges is a straight line and the 3 other are notched. You know the points $ 0$ and $ 1\in{\#OX}$.
    Then if the ruler is unmarked, what can you construct ? Only a random line which has "always" an irrational slope !
    Now assume the ruler is marked and $ OY$ is not given; if you could construct $ OY$ then  you could construct $ \sqrt{2}$: you could solve an irreducible equation of degree 2. I think that it's impossible.


On an escalator which is not moving, a person descends faster than he ascends. Is it faster for this person to descend and ascend once on an upward-moving escalator, or to descend and ascend once on a downward-moving escalator? (It is assumed that all the speeds mentioned here are constant, that the speed of the escalator is the same no matter if it is moving up or down and that the speed of the person relative to the escalator is always greater than the speed of the escalator.)

let, descending speed be 'x' and ascending be 'y'. elevator's speed be 'n'. and let $ x \minus{} y \equal{} l$ and $ y \minus{} n \equal{} l$
let, effective speed be $ E$. In CASE I,$ E_1 \equal{} x \plus{} n$ and $ E_2 \equal{} y \minus{} n$ and in CASE II,$ E_1 \equal{} x \minus{} n$ and $ E_2 \equal{} y \plus{} n$
adding both speeds in both we get as $ x \plus{} y$                                                 $ ..........(1)$
$ \frac {distance}{speed} \equal{} time$ so, in CASE I, $ \frac {k}{x \plus{} n} \plus{} \frac {k}{y \minus{} n} \equal{} (k)[\frac {x \plus{} y}{(x \plus{} n)(y \minus{} n)}]$
similarly for the other. and then u can compare denominators to get the answer


vector
linear algebra
linear algebra unsolved

If $L$ is the line through $A=(1,2,1)$ and $C=(3,-1,2)$, while $M$ is the line through $B=(1,0,2)$ and $D=(2,1,3)$, what is the distance between lines $L$ and $M$?

Is this question asking for the shortest distance between two lines or just general formula for the distance at any point??

Please help me in this qustion,, i am really stuck. I don't know what to do.....

They want the shortest distance. Here's one method: the vector $\vec n=\vec{AC}\times\vec{BD}$ is perpendicular to both lines. Imagine that this vector is vertical and the lines are horizontal. It should become clear that the distance between the lines is equal to the scalar projection of $\vec{AB}$ onto $\vec n$. Namely, $d=\frac{|\vec{AB}\cdot\vec n|}{|\vec n|}$.


A circle of radius r is inscribed in a square.

(a) Express the area A of the square as a function of the radius r of the circle.

(b) Express the perimeter p of the square as a function of x.

NOTE: NO LENGTH has been given in terms of one side of the square.

[hide]

Therefore the lenght of the square is $2r$ it follows that

$A(r)=4r^{2}$ and $P(r)=4(2r)=8r$[/hide]

[quote="Interval"]I don't follow.  Did you answer parts a and b?[/quote]

If you draw a circle inside a square you'll notice that the square's two opposite sides is the diameter of the circle. So the side length is $2r$.

I drew a square and placed a circle inside.

There is a diameter, which equals 2 times radius.

So, the answer would have to be 2r as both of you stated.

Thanks!


If a,b,c >0 , a+b+c=1 and n integer find the max of
$ a^n+b^n+c^n-\frac{1}{3^n} $

Please siuhochung explain your answer analitically

I don't think that value can be attained since $a$, $b$ and $c$ are positive (not nonnegative). However that value should be the supremum.

Maybe. We wait siuhochung's analitical solution

it's obvious that there is only a supremum. The result is immediate because $(a+b+c)^n \ge a^n + b^n + c^n$. Equality "holds" when a->1, b->0, c->0.

[quote="silouan"]Supremum means that there is a value that the RHS could not take it? Someone explain to me because my English are not good[/quote]
1. not really can't take it. It varies from case to case
2. Supremum is a mathematical term. http://mathworld.wolfram.com/Supremum.html

Silouan, mabye you should post this type of inequalities in the pre-olympiad forum?


trigonometry
calculus
calculus computations

Solve $ \frac {d^2y}{dt^2} \minus{} 2\frac {dy}{dt} \plus{} 5y \equal{} e^{ \minus{} t}\cos 2t$.

I feel like something similar was recently posted.  Use undetermined coefficients and "guess" $ y(t)\equal{}e^{\minus{}t}(A\cos(2t)\plus{}B\sin(2t))$ and try to solve for the coefficients.  If that doesn't work, try $ y(t)\equal{}te^{\minus{}t}(A\cos(2t)\plus{}B\sin(2t))$.

yeah, it was [url=http://www.mathlinks.ro/viewtopic.php?t=242997]Differential Equation 8[/url].

Oh, the problem looks almost just like Today's Differential Equation 8, but the problelm 13 is easy to solve than problem 8. :wink:


http://mathcircle.berkeley.edu/mpgeo.pdf

I'm getting very confused about mass points. In the article above, I don't understand the postulates at all... what is meant by $nP+mQ+kR$, for example? (3)

Also....

-how do you do #6 in the Basics section?
-Angle Bisectors #1: why can you put on the weights suggested by the hint?
-Angle Bisectors #3... how do you prove concurrency with mass points?
-Angle Bisectors #5: how do you do it?

Thanks to anyone who answers these questions, I'm really very confused after reading this article :(

[hide="6"]
If we assign weight $1$ to each vertex of the original quadrilateral, the weights of the midpoints are $2$.
Then, we can observe that the diagonals of the new quadrilateral (whose vertices are the midpoints of the original quadrilateral) bisect each other, so it is a parallelogram.
[/hide]

Isn't mass point a simple variation of Menelaus theorem?

I'm going to learn these right before the Aime. I just might need em.  :)

[quote="Altheman"]mass points are easy if you just accept the postulates as true, i.e. if you place the masses to balance the side ratio, and you add the masses to the point in the middle[/quote]

Well all I know about mass points is that if you have a point $C$ on line segment $\overline{AB}$, then $w_A\cdot AC = w_B \cdot CB$, and $w_C = w_A+w_B$. :( And I don't really understand what is meant by any of the postulates...

Here, I drew some images of basic layman's mass points instead of all the fancy wording and lettering.

[url=http://img.photobucket.com/albums/v734/Elemennop/massp1.jpg]1st Mass Points Info[/url]
[url=http://img.photobucket.com/albums/v734/Elemennop/massp2.jpg]2nd Mass Points Info[/url]

amm I also didn't under stand this--Angle Bisectors #1: why can you put on the weights suggested by the hint?  thanks. I search the forum and it didn't help at all.

As far as I know, "Mass Point" has no use if you are farmiliar with Ceva, Menelaus and Area Ratios.

Yeah, but sometimes Mass Point geometry is still useful even if you know Ceva, Menelaus and area ratios. For instance, there's a case where you have to use Ceva/Menelaus twice, whereas you only need one operation if you use Mass Points.

[quote="frt"]For instance, there's a case where you have to use Ceva/Menelaus twice, whereas you only need one operation if you use Mass Points.[/quote]
In my experiences, the problems that involves more than on Ceva or Menelaus can be easily solved by Area Ratios.


induction
modular arithmetic
geometric series

Prove the following by induction:
$ 10^{n \plus{} 1}  \minus{} 10(n \plus{} 1) \plus{} n|81, n>\equal{}0$

Here's my failed attempt. I wrote the 2nd step of induction like this:
$ 10 \cdot 10^k  \minus{} 10(k \plus{} 1) \plus{} k|81$
Then did the following with the last step:
$ \begin{array}{l}
 10^{k \plus{} 2}  \minus{} 10(k \plus{} 2) \minus{} k \plus{} 1 \equal{}  \\
 10^2  \cdot 10^k  \minus{} 10(k \plus{} 1) \minus{} 10 \plus{} k \plus{} 1 \equal{}  \\
 (81 \plus{} 9) \cdot 10^k  \minus{} 10(k \plus{} 1) \minus{} 10 \plus{} k \plus{} 1 \equal{}  \\
 81 \cdot 10^k  \plus{} 9 \cdot 10^k  \plus{} 10 \cdot 10^k  \minus{} 10(k \plus{} 1) \plus{} k \minus{} 9 \\
 \end{array}$
So, now I have $ 81 \cdot 10^k|81$ and $ 10 \cdot 10^k  \minus{} 10(k \plus{} 1) \plus{} k|81$ but don't know what to do with $ 9 \cdot 10^k  \minus{} 9$

Err, I guess you meant $ 81 \mid 10^{n\plus{}1} \minus{} 10(n\plus{}1) \plus{} n$?

[hide="With induction"]
For $ n \equal{} 0$, we get $ 81 \mid 0$, which is true.

Suppose $ 10^n \minus{} 10n \plus{} n \minus{} 1 \equal{} 10^n \minus{} 9n \minus{} 1 \equal{} 81k$ (case $ n\minus{}1$). Then $ 10^{n\plus{}1} \minus{} 90n \minus{} 10 \equal{} 81(10k)$, and also $ 10^{n\plus{}1} \minus{} 9n \minus{} 10 \equal{} 10^{n\plus{}1} \minus{} 10(n\plus{}1) \plus{} n \equal{} 81(10k\plus{}n)$. This shows that $ 81 \mid 10^{n\plus{}1} \minus{} 10(n\plus{}1) \plus{} n$ (case $ n$), so we're done.
[/hide]

[hide="Without induction"]
Substitute $ m \equal{} n\plus{}1$. So for positive integers $ m$ we want to prove $ 81 \mid 10^m \minus{} 9m \minus{} 1$.

Since $ \frac{10^m\minus{}1}{10\minus{}1} \equal{} 1 \plus{} 10 \plus{} 100 \plus{} \ldots \plus{} 10^{m\minus{}1}$ by geometric series, $ 10^m \minus{} 1 \equal{} 9(1 \plus{} 10 \plus{} \ldots \plus{} 10^{m\minus{}1})$. Rewriting the original problem, we want to show $ 81 \mid 9(1 \plus{} 10 \plus{} \ldots \plus{} 10^{m\minus{}1}) \minus{} 9m$, equivalent to showing that $ 9 \mid 1 \plus{} 10 \plus{} \ldots \plus{} 10^{m\minus{}1} \minus{} m$. But notice that $ 1 \plus{} 10 \plus{} \ldots \plus{} 10^{m\minus{}1} \equiv 1 \plus{} 1 \plus{} \ldots \plus{} 1 \equiv m \pmod{9}$. Since $ a \equiv b \pmod{m} \Leftrightarrow m \mid a \minus{} b$ by definition, we're done.
[/hide]

[hide="To answer your previous post"]
It turns out that $ 9 \cdot 10^k \minus{} 9$ is always divisible by $ 81$. This fact inspired and is explained by the "Without induction" solution in this post. To summarize, realize that showing this is equivalent to showing that $ 10^k \minus{} 1$ is divisible by $ 9$. Then look at $ 10^k \minus{} 1$ modulo $ 9$.
[/hide]


Hello, can someone help me on this problem:

If $ \frac{2y\plus{}2z\minus{}x}{a} \equal{} \frac{2z\plus{}2x\minus{}y}{b} \equal{} \frac{2x\plus{}2y\minus{}z}{c}$ then show that
$ \frac{x}{2b\plus{}2c\minus{}a} \equal{} \frac{y}{2c\plus{}2a\minus{}b} \equal{} \frac{z}{2a\plus{}2b\minus{}c}$

Hints, anyone?

[b]Lemma:[/b]  If $ \frac{a}{b} \equal{} \frac{c}{d}$ then $ \frac{a}{b} \equal{} \frac{c}{d} \equal{} \frac{a \plus{} ck}{b \plus{} dk}$ for any $ k$.

I think I got it:

Let $ \frac{2y\plus{}2z\minus{}x}{a} \equal{} (A)$
$ \frac{2z\plus{}2x\minus{}y}{b} \equal{} (B)$
$ \frac{2x\plus{}2y\minus{}z}{c} \equal{} (C)$

Add $ (B)$ and $ (C)$ then subtract $ (A)$

$ \frac{2z\plus{}2x\minus{}y\plus{}2x\plus{}2y\minus{}z}{b\plus{}c} \equal{} \frac{4x\plus{}y\plus{}z}{b\plus{}c} \equal{} \frac{2(4x\plus{}y\plus{}z)}{2(b\plus{}c)}$

$ \frac{8x\plus{}2y\plus{}2z}{2b\plus{}2c} \equal{} \frac{8x\plus{}2y\plus{}2z\minus{}(2y\plus{}2z\minus{}x)}{2b\plus{}2c\minus{}a} \equal{} \frac{9x}{2b\plus{}2c\minus{}a}$

Similarly, add $ (A)$ and $ (B)$ and then subtract $ (C)$

$ \frac{2y\plus{}2z\minus{}x\plus{}2z\plus{}2x\minus{}y}{a\plus{}b} \equal{} \frac{4z\plus{}x\plus{}y}{a\plus{}b} \equal{} \frac{8z\plus{}2x\plus{}2y}{2a\plus{}2b}$

$ \frac{8z\plus{}2x\plus{}2y\minus{}(2x\plus{}2y\minus{}z)}{2a\plus{}2b\minus{}c}$ = $ \frac{9z}{2a\plus{}2b\minus{}c}$

Once again, add $ (A)$ and $ (C)$ and subtract $ (B)$

Since each of these fractions is equal to the sum of numerators and denominators:

$ \frac{9x}{2b\plus{}2c\minus{}a} \equal{} \frac{9y}{2a\plus{}2c\minus{}b} \equal{} \frac{9z}{2a\plus{}2b\minus{}c}$

Or $ \frac{x}{2b\plus{}2c\minus{}a} \equal{} \frac{y}{2a\plus{}2c\minus{}b} \equal{} \frac{z}{2a\plus{}2b\minus{}c}$

We can assume x+y+z=1 and a+b+c=1, because any scale of them would also work.
So we have $ \frac {2 \minus{} 3x}{a} \equal{} \frac {2 \minus{} 3y}{b} \equal{} \frac {2 \minus{} 3z}{c} \equal{} n$
We need to show $ \frac{x}{2\minus{}3a}\equal{}\frac{y}{2\minus{}3b}\equal{}\frac{x}{2\minus{}3c}$
Then $ n \equal{} n(a \plus{} b \plus{} c) \equal{} na \plus{} nb \plus{} nc \equal{} 6 \minus{} 3(x \plus{} y \plus{} z) \equal{} 3$
So $ a \equal{} \frac {2 \minus{} 3x}{3}$, etc.
Subbing all of them into the second, we get 1/3=1/3=1/3, clearly true.


[hide]5/8, there are 8 numbers and you want to roll 5 of them[/hide]

[hide]
wanted outcome=5
possible outcome=8
so the answer is obviously 5/8.[/hide]

[quote="colts18"]Some board games come with special dice. One game in particular uses an 8-sided die with the numbers 1 through 8 printed on the sides. What is the probability of rolling a number other than a 1, 2, or 5 with this die?[/quote]

If you want to roll a 1, 2, or 5, then the probability would be $\frac{3}{8}$.

But if you don't want those numbers, then the probablilty would obviously be $\frac{5}{8}$


Sa se arate ca nu exista functii continue neconstante $f:R\to R$ cu proprietatea:\[
f(x + \sqrt 2 ) = f(x + \sqrt 3 ) = f(x),\forall x \in R.
\]

-----------------------------------------------------------------------------------------------------------
[b][color=blue]
Să se arate că nu există funcţii continue $ f: \mathbb{R} \to \mathbb{R}$, neconstante  cu proprietatea:
$ f\left( {x \plus{} \sqrt 2 } \right) \equal{} f\left( {x \plus{} \sqrt 3 } \right) \equal{} f\left( x \right),\forall x \in \mathbb{R}$
[/color][/b]
------------------------------------------------------------------------------------------------------------
[b][u]Soluţie:[/u][/b]
Din ipoteze rezultă că  funcţia are periodă  pe $ \sqrt 2$ dar  şi pe  $ \sqrt 3$.
Se poate demonstra că această funcţie are ca perioadă orice număr de forma
\[ m\sqrt 2 \plus{} n\sqrt 3 ,\forall n,m \in \mathbb{Z}
\]
Deci avem:
\[ f\left( x \right) \equal{} f\left( {x \plus{} m\sqrt 2 \plus{} n\sqrt 3 } \right),\forall m,n \in \mathbb{Z},\forall x \in \mathbb{R}
\]
în particular avem:
\[ f\left( 0 \right) \equal{} f\left( {m\sqrt 2 \plus{} n\sqrt 3 } \right),\forall m,n \in \mathbb{Z}
\]
Dacă se defineşte mulţimea  $ A \equal{} \left\{ {m\sqrt 2 \plus{} n\sqrt 3 \left| {m,n \in \mathbb{Z}} \right.} \right\}$  aceasta este densă in $ \mathbb{R}$, adică:
$ \forall a \in \mathbb{R},\exists \left( {x_n } \right)_{n \in \mathbb{N}}$  astfel încât  $ x_n \in A$  şi $ \mathop {\lim }\limits_{n \to \infty } x_n \equal{} a$

Din  $ f\left( 0 \right) \equal{} f\left( {m\sqrt 2 \plus{} n\sqrt 3 } \right),\forall m,n \in \mathbb{Z}$ avem: $ f\left( {x_n } \right) \equal{} f\left( 0 \right)$ şi deci  $ \mathop {\lim }\limits_{n \to \infty } f\left( {x_n } \right) \equal{} f\left( a \right) \equal{} f\left( 0 \right)$

[u]In concluzie[/u] funcţiile căutate sunt funcţii constante.


trigonometry
limit
calculus
calculus computations

Prove that the series $ \sum_{i \equal{} 1}^{\infty} \frac{\sin i}{i}$ converge.

Set $ a_{k} \equal{} \sin k$ and $ b_{k} \equal{} \frac {1}{k}$

Let $ S_{k} \equal{} \sum_{i \equal{} 1}^{k}\sin i \equal{} \frac {1}{2\sin \frac {1}{2}}\left( \cos \frac {1}{2} \minus{} \cos \left( k \plus{} \frac {1}{2}\right) \right)$

$ \left| S_{k}\right| \leq \frac {1}{2\sin \frac {1}{2}}\Rightarrow \lim_{n\rightarrow \infty }S_{n}b_{n} \equal{} 0$

Now,

$ \left| \sum_{i \equal{} 1}^{\infty }\frac {\sin i}{i}\right|$

$ \equal{} \lim_{n\rightarrow \infty }\left| \sum_{i \equal{} 1}^{n}\frac {\sin i}{i}\right|$

$ \equal{} \lim_{n\rightarrow \infty }\left| \sum_{i \equal{} 1}^{n}a_{i}b_{i}\right|$

$ \equal{} \lim_{n\rightarrow \infty }\left| S_{n}b_{n} \minus{} \sum_{i \equal{} 1}^{n \minus{} 1}S_{i}\left( b_{i \plus{} 1} \minus{} b_{i}\right) \right|$

$ \leq \sum_{i \equal{} 1}^{\infty }\left| S_{i}\left( b_{i} \minus{} b_{i \plus{} 1}\right) \right| \leq \frac {1}{\sin \frac {1}{2}}\sum_{i \equal{} 1}^{\infty }\left( \frac {1}{i} \minus{} \frac {1}{i \plus{} 1}\right) \equal{} \frac {1}{\sin \frac {1}{2}}$

Hence the given series converge.

Converges, and its value is $ \tfrac{{\pi  \minus{} 1}}
{2}$.

Prove ... is a fun challenge.  :rotfl:

the last challenge is easy if you use Cèsaro means.

if I'm not mistaken, using arithmetic means of Fourier series (or Cèsaro sums), we have $ \dfrac{\pi\minus{}x}{2} \equal{} \sum_{n\equal{}1}^{\infty}\dfrac{\sin{nx}}{n}$
So, take $ x\equal{}1$.

This example is due to Fejer, in the begin of early 20.

^^


Study $ \lim_{n\to\infty} a_n$ such that $ na_{n \plus{} 1} \equal{} (n \plus{} 1)a_n \minus{} \max(a_n,\ n^2)\ (n \equal{} 1,\ 2,\ 3,\ \cdots).$

$ na_{n \plus{} 1} \equal{} (n \plus{} 1)a_n \minus{} \max(a_n,\ n^2) \leq (n \plus{} 1)a_n \minus{} a_n \equal{} na_n$, hence $ a_{n \plus{} 1} \leq a_n$, so the sequence is decreasing. Moreover, for $ n$ such that $ a_1 < n^2$ we have $ na_{n \plus{} 1} \equal{} (n \plus{} 1)a_n \minus{} n^2 \leq a_1 \plus{} na_1 \minus{} n^2$, hence $ a_{n \plus{} 1} \leq a_1/n \plus{} a_1 \minus{} n$, therefore the sequence is lower unbounded. Finally $ \lim_{n \to \infty} a_n \equal{} \minus{} \infty$.


Hi all, I am stuck trying to simplify or evaluate the following two sums, N is odd:

\sum\limits_{i=0}^{(N-1)/2}{N\choose i}(N-i)

\sum\limits_{i=0}^{(N-1)/2}{N\choose i}{N-i\choose 2}

(also see the pdf-file attached)

The problem would be much easier were the sums taken over i=0 to N instead of (N-1)/2. I would really appreciate you help!

Best,
serge


My friend thought up a problem and gave it to me, but...

Three people, A,B, and C are running a race.  The probability that A finishs before B is 60%.  The probability that B finishes before C is 90%.  What is the probability that C finishes before A?

I thought that the question as is was impossible, but he thinks that since there is a set number of cases, the probability can be calculated.  Doing it that way, however, I got -.08%  :?   Without at least defining bounds for this race, is it solvable?  And if not, would giving the track bounds make it solvable?

*Finishing before doesn't necessarily mean directly before.  If it went A then C then B, A finishes before B

[hide="solution"]

there are only three ways C can finish before A:

cases: CAB, CBA, BCA (leftmost finish first)

$CAB = 0.1 * 0.6 = 0.06$
$CBA = 0.4 * 0.1 = 0.04$
$BCA = 0.9 * 0.4 = 0.36$

add them all, we get 0.46. the probability that C beats A is $ \frac{0.46}{total}$ =
$ \frac{0.46}{1.42} = \frac{23}{71}$[/hide]

i hope that is somewhat right.

This is a very interesting problem. I don't believe that reddevil's answer is correct right now. And I think the problem needs to specify that the respective placements of the individuals are independent of one another (or if they aren't, how they are related).

[hide="Hint"]Try a Venn-Diagram, or Inclusion-Exclusion.[/hide]


$ x\plus{}\frac{1}{y}\equal{}y\plus{}\frac{1}{z}\equal{}z\plus{}\frac{1}{x}$

Prove that $ x\equal{}y\equal{}z$ or $ (xyz)^2\equal{}1$

 :maybe:

If $ (x, y, z) \equal{} \left(3, 1, \frac{1}{3}\right)$

$ (xyz)^2\equal{}1^2\equal{}1$.

But, $ \left(x\plus{}\frac{1}{y}, y\plus{}\frac{1}{z}, z\plus{}\frac{1}{x}\right) \equal{} \left(4, 4, \frac{2}{3}\right)$ :D

from given condition we have $ x\minus{}y \equal{} \frac{y\minus{}z}{yz} ; y\minus{}z \equal{} \frac{z\minus{}x}{xz} ; x\minus{}z \equal{}\frac{y\minus{}x}{xy}$
multiply all of them together gives $ (x\minus{}y)(y\minus{}z)(z\minus{}x) \equal{} \frac{(x\minus{}y)(y\minus{}z)(z\minus{}x)}{x^2y^2z^2}$
if $ x\minus{}y \equal{}0$ then $ y\minus{}z \equal{} 0 \minus{}> x\equal{}y\equal{}z$
if $ x\minus{}y <> 0$ and $ y\minus{}z <> 0$ and $ z\minus{}x <> 0$ then $ (xyz)^2 \equal{}1$

Hi, HTA, Bachukas :)

Is the problem correct?
Did my post read?

[quote="HTA"]from given condition we have $ x \minus{} y \equal{} \frac {y \minus{} z}{yz} ; y \minus{} z \equal{} \frac {z \minus{} x}{xz} ; x \minus{} z \equal{} \frac {y \minus{} x}{xy}$
multiply all of them together gives $ (x \minus{} y)(y \minus{} z)(z \minus{} x) \equal{} \frac {(x \minus{} y)(y \minus{} z)(z \minus{} x)}{x^2y^2z^2}$
if $ x \minus{} y \equal{} 0$ then $ y \minus{} z \equal{} 0 \minus{} > x \equal{} y \equal{} z$
if $ x \minus{} y < > 0$ and $ y \minus{} z < > 0$ and $ z \minus{} x < > 0$ then $ (xyz)^2 \equal{} 1$[/quote]

Sure, nice & elegant.. :D  :roll: :clap:

[quote="Bugi"]
You are wrong.
It's not if and only if, it's just if.
If x+(1/y)=... then (xyz)^2=1 or...

And have you read mine?
You were going in a wrong direction.[/quote]
Oops :(
I misunderstood a problem.

$ \color{red}\times\color{black}\boxed{x\plus{}\frac{1}{y}\equal{}y\plus{}\frac{1}{z}\equal{}z\plus{}\frac{1}{x} \color{blue}\iff\color{black}  x\equal{}y\equal{}z \ \text{or} \ (xyz)^2\equal{}1}$

$ \color{red}\bigcirc\color{black}\boxed{x\plus{}\frac{1}{y}\equal{}y\plus{}\frac{1}{z}\equal{}z\plus{}\frac{1}{x} \color{blue}\Longrightarrow\color{black}  x\equal{}y\equal{}z \ \text{or} \ (xyz)^2\equal{}1}$

Thank you :)


Asymptote limit algebra polynomial LaTeX

I have this fraction i have to find all the limits for:

$f(x)=\frac{x^3+1}{x^2+1}$

I found out (and checked with my calculator) that there was no vertical limit, but for this fraction i can't find any diagonal limits (as this is the only one left). My calculator also tells me there is a diagonal limit.

Thx in advance for your help.

You mean oblique asymptotes? Try reading what your textbook has to say.

you just divide it as much as you can, and the quotient is the asymptote, and the remainder drops out since the degree of the remainder is higher on the bottom, so it will approach 0

In this case it might work, but generally an oblique asymptote is defined like this:

If there exist both $k=\lim_{x\rightarrow\infty}{f(x)\over x}$ and $n=\lim_{x\rightarrow\infty}({f(x)-kx})$, then $y=kx+n$ is an oblique asymptote. Of course, "$\infty$" can be either $+\infty$ or $-\infty$, and those asymptotes can be different.

The problem is i can't divide it any further than this, unless i'm doing it wrong:

$(x^3+1): (x+1)=x^2-x+1$

And if i then try to find the x's for this 2nd degree equation:

$x^2-x+1=0$

$\frac{1\pm\sqrt{-3}}{2}$

This means the 2nd degree equation doesn't intersect with $x=0$.

And it is the same i end up with if i try with the denominator:

$x^2+1$

So i can't split it up more than this:

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

Am i doing anything wrong in these calculation or did i just miss the point on how to continue from here?

I'm a bit confused as to what you're doing here.  As $x\rightarrow\infty$, only the leading terms in the numerator and denominator really matter.  Here you have

$y=\frac{x^{3}+1}{x^{2}+1}\approx\frac{x^{3}}{x^{2}}=x\quad\text{as}\quad x\rightarrow\infty$

Oh yeah, that makes perfect sense to do it this easy instead of following the book exactly :D

Thx for the help.

Oh, BTW: Found out it wasn't enough to find what $f(x)$ goes towards when it goes towards $x$, i have to find the 'oblique asymptotes' as Chess64 called it.

Then i will have to ask you to show me step by step how to do it. I have tried that with this fraction but it didn't work well for me.

ok, here is the long division

[code]
       x
      ----------
x^2+1| x^3+1
      -(x^3+x)
       ----------
          1-x
[/code]

now we now know by our division that $f(x)=\frac{x^3+1}{x^2+1}=x+\frac{1-x}{x^2+1}$
now look at the last term, as x gets very large, the whole fraction approaches 0 because there is $x^2$ on the bottom, and $x$ on the top (i.e. the bottom has a higher degree than the top)

so our answer is $h(x)=x$

a simple definition of oblique asympote:

if we have $f(x)=h(x)+g(x)$ where for large $x$, $g(x)$ approaches $0$, then our oblique asymptote is $h(x)$, make sure you understand this and can replicate this, since this seems like a standard school problem that you will see in your class again

OK, so long division is just to divide the whole fraction without splitting it into fractions first. I've never learned that, but i can surely believe it is true - why wouldn't it...

Yes, i have learned about the $f(x)=h(x)+g(x)$ though it was described in another way. And $g(x)$ is usally, if not always, the remaining fraction. So when $x$ becomes a great number it is reduced to almost 0 if it is a degree bigger in the denominator (it either has to be bigger in the denominator or the same degree as the nominator). That's the reason why the asymptote is found in $h(x)$.

Great, think i got the point now. Thx a lot everyone, this really helped ;)

Well,
   the problem was not formulated in the best way, but finally is was solved. Please, read and learn what  was written in posts 4 and 11. These are basic results to all similar problems.

  [u] Babis[/u]

Number 4 wasn't too easy, but i think i got the point now. I guess the biggest problem with this forum for me is that english isn't my main language. I know this is how it is for many of you ppl, but i guess you expierence this often everytime a new member joins. You might be good in english, but when i studied english we read texts without math words. Guess that is what makes many the many problems with us foreigners :)

[quote="Sporally"]BTW: How do you make that unlimited sign using LaTeX amcavoy made?[/quote]

\infty


calculus
function
inequalities
inequalities unsolved

Question: Prove that
$ \frac {a}{b \plus{} 2c \plus{} 3d} \plus{} \frac {b}{c \plus{} 2d \plus{} 3a} \plus{} \frac {c}{d \plus{} 2a \plus{} 3b} \plus{} \frac {d}{a \plus{} 2b \plus{} 3c}\ge \frac {2}{3}$, given that $ a,b,c,d > 0$.
Answer: Using calculus, LHS is a convex function of $ a,b,c,d$. So minimum occurs when $ a \equal{} b \equal{} c \equal{} d$, so $ LHS\ge 4\cdot \frac {a}{6a} \equal{} \frac {2}{3}$.

I have another question:

Given that $ x,y,z\ge \minus{} 1$, show that $ \frac {1 \plus{} x^2}{1 \plus{} y \plus{} z^2} \plus{} \frac {1 \plus{} y^2}{1 \plus{} z \plus{} x^2} \plus{} \frac {1 \plus{} z^2}{1 \plus{} x \plus{} y^2}\ge 2$.

Your first one is easy by Cauchy-Schwartz:
\[ \sum_{cyc}\frac{a}{b\plus{}2c\plus{}3d}\geqslant\frac{(a\plus{}b\plus{}c\plus{}d)^2}{4(ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd)}\stackrel{!}{\geqslant}\frac{2}{3}\mbox{,}\]
where the last one is eqivalent to
\[ \frac{a\plus{}b\plus{}c\plus{}d}{4}\geqslant\sqrt{\frac{ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd}{6}}\mbox{,}\]
which is MacLaurin's inequality.  :lol:

[quote="knol"]
I have another question:

Given that $ x,y,z\ge \minus{} 1$, show that $ \frac {1 \plus{} x^2}{1 \plus{} y \plus{} z^2} \plus{} \frac {1 \plus{} y^2}{1 \plus{} z \plus{} x^2} \plus{} \frac {1 \plus{} z^2}{1 \plus{} x \plus{} y^2}\ge 2$.[/quote]
It is JBMO proposed by Panaiotopol.
Use the AM-GM inequality on $ x,y,z$ at the denominators to acquire $ \frac{1\plus{}x^2}{2},\frac{1\plus{}y^2}{2},\frac{1\plus{}z^2}{2}$ and then use CS to kill it.
see also here:
http://gbas2010.wordpress.com/2009/12/07/inequality-8laurentiu-panaitopol-jbmo-2003/


What is the largest integer $ n$ for which $ \binom{8}{3} \plus{} \binom{8}{4} \equal{} \binom{9}{n}$?

By Pascal's Identity, $ \binom{8}{3}\plus{}\binom{8}{4}\equal{}\binom{9}{4}$.  However, we want the largest value of $ n$, and since $ 9\minus{}4\equal{}5>4$, the answer is $ \boxed{5}$.


[quote="vardanch"]$It is very easy$  :D

By $A(m)\geq G(m)$ we have that $(3=a^2+b^2+c^2\geq 3sqrt[3]{{abc}})$
=> $abc\leq 1$
$By A(m)\geq G(m)$
$2(a+b+c) + \frac{1}{abc} =a+b+c+a+b+c+ \frac{1}{abc}\geq 7sqrt[7]{{abc}}\geq 7$ because $abc\leq 1$ .

$Q-E-D$[/quote]

if $abc\leq 1$ then $7\sqrt[7]{abc}\leq 7$ and not $\geq 7$

[quote="ptung"]Given $a,b,c$ positive number, $a^2 + b^2 + c^2 = 3$. Prove that:

$2(a+b+c) + \frac{1}{abc} \geq 7$

Thanks in advance.[/quote]

Unfortunately, this inequality is not true.  For example, take $a = b = \sqrt{9/10}$, $c = \sqrt{12/10}$.  Then $a^2 + b^2 + c^2 = 3$ and
\[ 2(a + b + c) + \frac{1}{abc} \approx 6.999924 < 7. \]

I ask Mr.ptung for help , that problem is wrong but this one is right
Given a,b,c positive numbers,  a^2+b^2+c^2=3.prove that
5(a+b+c)+3/(abc) >= 18

I have some problem like this (certainly they're true )

[color=blue] I'm waiting for a solution for secondary school like me [/color]

just put: Find the infinum value of:

$\frac{\frac{1}{abc} - 1}{3 - a - b - c}$

$PROBLEM$

[quote="ptung"]someone needs help for this question. Personally, I am not interested in this type of inequality. So I came here to ask for help. He told me that he is secondary-school pupil. So please find an elementary proof for the problem.

Given $a,b,c$ positive number, $a^2 + b^2 + c^2 = 3$. Prove that:

$2(a+b+c) + \frac{1}{abc} \geq 7$

Thanks in advance.[/quote]

$SOLUTION$
By $A(m)\geq G(m)$ we have $abc\leq 1$.
$1)$  If $a^2+b^2+c^2\leq a+b+c$ => $2(a+b+c)+1/abc\geq 2(a^2+b^2+c^2)+1/abc\geq 3*2+1=7$
$2)$  If $a^2+b^2+c^2\geq a+b+c$ => $a+b+c\leq 3$
By $A(m)\geq G(m)$ we have that for positive real x,y,z
$sqrt(x)+sqrt(y)+sqrt(z)\leq sqrt(3*(x+y+z))$ => $sqrt(a)+sqrt(b)+sqrt(c)\leq sqrt(3*(a+b+c))\leq a+b+c$ =>

$2(a+b+c)+1/abc\geq 2(sqrt(a)+sqrt(b)+sqrt(c))+1/abc$
 By $A(m)\geq G(m)$

$2(sqrt(a)+sqrt(b)+sqrt(c))+1/abc=sqrt(a)+sqrt(a)+sqrt(b)+sqrt(b)+sqrt(c)+sqrt(c)+1/abc\geq 7$

$Q-E-D$

$Now$ $I$ $will$ $proof$ $that$
$\sqrt x + \sqrt y + \sqrt z \le \sqrt{3(x + y + z)}$
if x,y,z are positive real numbers.
By $A(m)\geq G(m)$ we have tha $x+y\geq \sqrt xy$ $\Rightarrow$ $x+y+z\geq \sqrt xy+\sqrt yz+\sqrt xz$ $\Rightarrow$
$3(x+y+z)\geq  (\sqrt x + \sqrt y + \sqrt z)^2$ => $\sqrt x + \sqrt y + \sqrt z \le \sqrt{3(x + y + z)}$

[quote="nsato"][quote="vardanch"]
$2)$  If $a^2+b^2+c^2\geq a+b+c$ => $a+b+c\leq 3$
By $A(m)\geq G(m)$ we have that for positive real x,y,z
$sqrt(x)+sqrt(y)+sqrt(z)\leq sqrt(3*(x+y+z))$ => $sqrt(a)+sqrt(b)+sqrt(c)\leq sqrt(3*(a+b+c))\leq a+b+c$
[/quote]

The last inequality is false; it should be $\sqrt{3(a+b+c)} \ge a+b+c$.

A proof of this inequality was posted here:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86904]www.artofproblemsolving.com/Forum/viewtopic.php?t=86904[/url][/quote]

What is FALSE????

I proved that $\sqrt x + \sqrt y + \sqrt z \le \sqrt{3(x + y + z)}$.
What is wrong in that proof???

[quote="ptung"]someone needs help for this question. Personally, I am not interested in this type of inequality. So I came here to ask for help. He told me that he is secondary-school pupil. So please find an elementary proof for the problem.

Given $a,b,c$ positive number, $a^2 + b^2 + c^2 = 3$. Prove that:

$2(a+b+c) + \frac{1}{abc} \geq 7$

Thanks in advance.[/quote]
Let b=c and a=1,1, this inequality is wrong.

We can prove $5(a+b+c) + \frac{3}{abc} \geq 18$ by mixing variable.
A great question : Find all positive real number k satisfying the following inequality is true : $k(a+b+c)+\frac{1}{abc} \ge 3k+1$

[quote="yookon88"]just as nsato said... if \[ a+b+c\leq 3  \] then $\sqrt{3(a+b+c)} \geq a+b+c$ and not $\leq a+b+c$ and... He already gave you a counter example...
He didn't say that your proof of \[ \sqrt{a}+\sqrt{b}+\sqrt{c}\leq \sqrt{3(a+b+c)}  \] was wrong...[/quote]

OK. I understand my mistake.  :D


lol hello? dude this is a spam post, and must be removed by the moderator of the board!

OK, if not hello, then ... hi everybody, how do you think about math?

Spam post ... isn't good to say everything here.. we discute about mathematics...


I am wondering if there is a quick way to sum an exponential series of base b from 0 to n.

b^0 + b^1 + b^2 + ... + b^n

A few things:

When b=1, the sum is n (e.g., 1+1+1=3)
When b=2, the sum is 2^n-1 (e.g., 1+2=3=2^2-1)
When b=3, the sum is [3^(n+1)/2], where [x] is the greatest integer not exceeding x (the floor function).

Maybe I'm getting out of my league here, but I'm curious :)

Thanks in advance.

Wow...

I didn't think it would be so simple.

[quote="belenos"]If b != 1 we have :

 :Sigma:(k=0..n) b^k = (b^(n+1)-1)/(b-1)

It can be proved a lot of ways, the simplest being perhaps induction.[/quote]

Note that this is just using the formula for geometric series, with 1st term (t_0) equal to 1 and common ratio b.

A geometric series is a series of the form
t_0, t_0*r, t_0*r^2, t_0*r^3, ...

In this case, t_0 is the first term and r is the ratio between any two consecutive terms.

Now, the sum of this series t_0, t_0*r, t_0*r^2, t_0*r^3, ... is simply t_0*(1-r^(n+1)/(1-r) if r is not 1, or n*t_1 if r is 1.

Plugging in t_0 = 1 and r = b into this yields the formula that belenos gave.

For a finite sum, the nicest way to prove it is probably this:

Look at 1 + b + b^2 + ... + b^n = S (what you are looking for), and compare it to b + b^2 + ... + b^(n + 1) = b*S.  Do you see how to go on from there?


limit
factorial
algebra
binomial theorem
binomial coefficients

Find $1.01^{100}$ to the nearest tenth[i] without using a calculator[/i].

[hide="read story behind it"]This is a great example of the usefulness of problem-solving in  everyday life. My dad asked me to go on the computer and calculate that because he  needed it to figure out something regarding stock. Instead, I thought for a moment and used problem-solving  :yup: [/hide]

[hide]2.7?  It's very close to $e$...[/hide]

[hide]
$e=\displaystyle\lim_{x\to\infty}(1+\frac{1}{x})^x$ which will give you infinite digits of e.

$(1+\frac{1}{100})^{100}=2.7...$ (two digits of $e$)

I double-checked with a calculator and for each 10 you multiply to the denominator and the exponent, you get another digit of $e$.[/hide]

edit:
eryaman beat me.  ;)

[quote="calc rulz"][quote="lotrgreengrapes7926"][hide="Hint"] Find a pattern. Do $1.01^2$, $1.01^3$, etc. [/hide]
That might not help, actually. You might as well use the binomial theorem.[/quote]

Nope, see if you can find a much slicker way  ;)[/quote]
Actually, using Binomial Theorem is not that bad.
If you didn't know $\lim_{x\to \infty} \left(1+\frac{1}{x}\right)^x =e$, then this would be one of the best approaches.

[quote="frt"]Actually, using Binomial Theorem is not that bad.
If you didn't know $\lim_{x\to \infty} \left(1+\frac{1}{x}\right)^x =e$, then this would be one of the best approaches.[/quote]
Oh.  :rotfl:

[quote="surge"]Wouldn't binomial theorem depend on $100!$ (without a calculator)?[/quote]Well, for the first couple of values most of it cancels out with the denominator...makes it easier but it is still kind of tedious.

[quote="surge"]Wouldn't binomial theorem depend on a large factorial (without a calculator)?[/quote]

I agree... you'd still have to calculate around 50 binomial coefficients, and it would be a lot of multiplication to do without a calculator.

Wouldn't you need only a few binomial terms to get an accurate enough answer?

Yeah, you only need to calculate the first 4 or 5 terms.
(Remember, we only need to approximate it to the nearest tenth.)

After the first 5 terms, each term becomes so small that they have no effect on the tenth (or hundredth) term.

oh wow, I didn't realize it was to the nearest 10th... that makes it way easier...  :blush:  :blush:


$10$ friends rented $3$ apartments, each with maximum occupation of $4$ people.

[b]1[/b]. In how many ways the friends could organize themselves, such that:

[b]a)[/b] $2$ apartments have $4$people each and the other only $2$ people.

[b]b)[/b] $2$ apartments have $3$people each and the other $4$ people.

[b]
2.[/b] Assume that exist $2$ antagonist people who cannot stay in the same group. How many ways could we organize the group of friends?

Thanks in advance.  :lol:

[quote="nutz_for2.718281828"]uhh... there can't be 4 people in a group of three.[/quote]

Sorry, you are right. I just corrected the number of people.


A fun little puzzle I found in a book of mine.

My house number is a positive integer.
These three statements concerning my house number are true:

If it is a multiple of 3, then it is a number from 50 to 59 inclusive.
If it is not a multiple of 4, then it is a number from 60 to 69 inclusive.
If it is not a multiple of 6, then it is a number from 70 to 79 inclusive.

What is my house number?

[hide][size=200]76[/size]

Size 29! :D[/hide]

[hide]It would be 76. If the number was divisible by 3, it would have to be either 51, 54, or 57. It can't be 51 or 57, because if it isn't a multiple of 6, it has to be between 70 and 79. It isn't 54 because if it isn't a multiple of 4, it must be between 60 and 69. If it isn't a multiple of 4, then it must be between 60 and 69. If it is divisible by 3, it must be between 50 and 59, so 60, 63, 66, and 69 are out. If it isn't divisible by 3, it isn't divisible by 6 either, and it must be between 70 and 79. That means that the number has to be between 70 and 79, but not a multiple of 6 or 3, and it has to be a multiple of 4. That means that the only number possible is 76[/hide]


[quote="GameBot"]In triangle $ ABC$, $ AB$ is congruent to $ AC$, the measure of angle $ ABC$ is $ 72^{\circ}$ and segment $ BD$ bisects angle $ ABC$ with point $ D$ on side $ AC$. If point $ E$ is on side $ BC$ such that segment $ DE$ is parallel to side $ AB$, and point $ F$ is on side $ AC$ such that segment $ EF$ is parallel to segment $ BD$, how many isosceles triangles are in the figure shown?

[asy]draw((0,0)--(5,15)--(10,0)--cycle,linewidth(1));
draw((0,0)--(8,6)--(6.5,0)--(9.25,2.25),linewidth(1));
label("B",(0,0),W);
label("A",(5,15),N);
label("D",(8,6),E);
label("E",(7,0),S);
label("F",(9,3),E);
label("C",(10,0),E);[/asy][/quote]

ABC, BED, DEF, FEC, DEC, BDC, ABD.

7 triangles.


Prove that there is a power of 2 who's last 1000 digits are composed entirely of 1's and 2's.

help me on this one  :blush:

please?

Anyone? Any help would be appreciated...


Actually I worked backwards and counted the pairs that did have a common prime factor but I only got 6 pairs
2,4
2,6
2,8
3,6
3,9
4,8

but I forgot 3 you said

4,6
6,8
6,9

thanks

Since those numbers are small, its easier.  Try numbers from 30 to 40.

[hide="hint to solution"] How many have two in common, three, five?  [/hide]

[quote="mathgeniuse^ln(x)"][quote="Ignite168"][hide]
9C2 or 36 ways to choose 2 different integers.
1 works with anything so 8 possibilities.
2 works with the odd numbers (don't count 1 as we already did it) so 4 more possibilities.
3 works with 4, 5, 7, and 8 so 4 more possibilities
4 works with 5, 7, 9 so 3 more possibilities
5 works with 6, 7, 8, and 9 so 4 more possibilities
6 works with 7 so 1 more possibility
7 works with 8 and 9 so 2 more possibilities
8 works with 9 so 1 more possibility
8+4+4+3+4+1+2+1 = 27
Answer is $\frac{27}{36}$ or $\boxed{\frac{3}{4}}$.[/hide]
[hide="What I think you did wrong"]
You might've counted 4 and 6, 6 and 8, and 6 and 9.[/hide][/quote]

You mean $\frac{3}{4}$ right[/quote]

That's what I wrote?  :?

Oh, right, on my computer I only saw a box, not a number itself, sorry.


Fie A o matrice patratica cu elemente intregi si pentru care exista k natural nenul cu $A^k=I_n$. Ce valori poate lua cel mai mare divizor comun al elementelor matricii $A-I$?

Obviously, I can't fully understand the whole statement, but I think I got idea.
I don't see where you wrote $A\neq I$, otherwise the answer is clear...

ahh... ce mirat am fost cand am cautat un post si am vazut "O problema frumoasa-pentru Kuba, de la Vali Vitejie" I was like :o:o
:))
din pacate doar astazi [acuma] am dat peste ea :) mersi harazi :)

am exclus $A=I_n$.. dar.. nu am reusit mai mult decat 2  :? (care oricum era destul de evident pt $A=-I_n$...


linear algebra
matrix
group theory
abstract algebra
linear algebra unsolved

$A \in GL_n(\frac{\mathbb{Z}}{p \mathbb{Z}})$, and $n \leq p$.
Prove that $ord(A) \neq p^2$, where $ord(A)$ is the least integer $k \geq 2$ such that $A^k=Id$.

The exact same problem was posted and solved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53879]here[/url].

I found yesterday a weird proof, but I prefer grobber's simple one.
Anyway, here's how it goes:

lemma:
every upper triangular matrix $T$ with $1$ on the main diagonal
satisfies: $T^p=1$.
Proof:
One can work in an algebraic closure of $\mathbb{F}_p$. In this extension,
one can suppose that $T$ is in its Jordan form. Each jordan cell is of the
form $M=I+J$. Conclusion follows because $(I+J)^p=I$ if $n \leq p$.

Now, consider a matrix $A$, such that $A^{p^2}=I$. Then $A$ generates a subgroup $H$
of $GL_n(F_p)$ of order $p^2$. A classical theorem says that $H$ is a subgoup of
a $p$-sylow $S$ of $GL_n(F_p)$. But every Sylow subgoups are conjugated and the subgroup
$U$ of upper triangular matrices with $1$ on the main diagonal is a Sylow subgroup. Hence
$A$ is similar to a matrix $T \in U$. But $T^p=Id$, so $A^p=I$. Hence $ord(A) \neq p^2$.


search
complex numbers
superior algebra
superior algebra solved

Hi,

is the additive group of the complex numbers isomorphic with the additive group of the reals?

I have no idea, I find this a weird problem?

Thx

I think they are. Just regard them as vectorial spaces over $Q$. They have the same dimension thus they are isomorphic but this easily implies they are isomorphic as groups.

harazi: They are both of infinte dimensions over Q?  I don't believe that property implies isomorphism.

They are indeed isomorphic, but concrete relation cannot be constructed, since the proof of isomorphism relies on the axiom of choice.

I did not say that the fact they have infinite dimension over $Q$ implies isomorphism. Read more carefully the post and you'll se I said the same dimension, which is the cardinality of R in fact. I know it uses axiom of choice, but if you want to find another solution without using it, so be it. I think mathematics is more interesting if we assume the axiom of choice.  ;)

I think I didn't communicate clearly, but I meant that since the proof of isomorphism relies on the axiom of choice, if we were able to construct a direct relation, then it would show the axiom is redundant.  So even though they are isomorphic, it maybe a waste of time to try to search for a mapping.  I too like that axiom.
Sorry T_______________T


what's national MAO.
I know it's a math contest but specifically what is it?

It's Mu Alpha Theta

their website is http://www.mualphatheta.org/

[quote="msmswatson"]Our school is considering getting up a group to go to 2004 National MAO Convention.  How does National MAO rate with respect to other renown competitions such as AMC or ARML,in terms of organization, quality of problems, etc.?[/quote]

I have attended 8 national MA :theta:  conventions as a student and organizer.  The quality of the problems varies because there are different test writers, though the test-writing has become more organized and more centralized for most recent conventions.  The problems are not as difficult as AMC12 problems.  Basically, if you took the first 15 problems on the AMC12 and extended that level of difficulty to x-problem tests, you would have something close to MA:theta:.  Of course, the quality even varies from test to test since there are different test writers (who have to come up with thousands of problems overall).

I personally recommend the convention for students who are anywhere below USAMO qualification level.  The convention is also a very social environment and I think it's good for students to try to meet as many of their peers nationally as possible.

I enjoy going to the conventions and will probably be in Huntsville this summer for the convention.

[quote="MCrawford"]
I enjoy going to the conventions and will probably be in Huntsville this summer for the convention.[/quote]

Is that Huntsville, Texas? or another huntsville?


function
real analysis
real analysis unsolved

[quote="mli"]Prove that the class of functions mapping $[0, 2\pi]$ to $\mathbb{C}$ with finite Holder norm is complete under the norm: Holder norm + sup norm.[/quote]For a Cauchy sequence in $C^\alpha$, find a subsequence that converges under the sup norm. Then show that the limit is Hoelder and the convergence occurs in the Hoelder norm.


Make sure your sound is turned up and the sound on the video is turned up.

Can you hear sound on videos from YouTube and other sources?

Anyone else having sound problems?

I doubt this has something to do with the vids, but I could only hear sounds when I watch vids. Opposite...


Consider the tiling of the plane by hexagonal cells. For $n\in\mathbb{N}$, denote by $T_n$ the triangular-shaped board of side $n$, i.e. containing $\frac{n(n+1)}2$ cells.
We are interested in non-overlapping tilings of $T_n$ by tiles consisting of three adjacent cells. As those tiles can have three different shapes, let's call $I$-tiles thoses consisting of three hexagons 'in a row', $\Delta$-tiles those of shape $T_2$ and $U$-tiles those of u-shape (like a hydrogen molecule if you want). Prove:

(1) There is no tiling of $T_n$ consisting only of $I$-tiles.
(2) There is no tiling of $T_n$ consisting only of $\Delta$-tiles.
(3) There is no tiling of $T_n$ consisting only of $U$-tiles.

I remember Thurston wrote an article called "Conway's Tiling Groups", which showed a technique that solved these kinds of problems.

In searching for this article, I came across these reference:
[url=http://faculty.plattsburgh.edu/don.west/tiling/]faculty.plattsburgh.edu/don.west/tiling/[/url]
[url=http://www.claymath.org/library/senior_scholars/stanley_ardila_tilings.pdf]www.claymath.org/library/senior_scholars/stanley_ardila_tilings.pdf[/url]

Edit: Two more references:
[url=http://www.math.ubc.ca/~elissa/tilinggrps/Tilinggrps.pdf]www.math.ubc.ca/~elissa/tilinggrps/Tilinggrps.pdf[/url]
[url=http://www.mat.puc-rio.br/~nicolau/publ/papers/resenhas.pdf]www.mat.puc-rio.br/~nicolau/publ/papers/resenhas.pdf[/url]


trigonometry
quadratics
algebra
polynomial
analytic geometry
graphing lines
slope

1)Find the sum of all values of x that satisfy the following equation $2 \tan^{-1}x-\tan^{-1}(\frac{1}{2})=\frac{\pi}{4}$

2)A regular n-gon has 350 diagonals. Find n.

3)Find $\sum_{k=1}^{\infty} \frac{k}{5^k}$

4) ABCD is a quadrilateral with right angles at A and C. Points E and F are on AC. Also DE and BF are perpendicular to AC. AE=3, DE=5, and CE=7, then find BF.

5) Find all roots to $(x+5)(x+6)(x+7)(x+8)=-1$

2:

[hide]

the number of diagonals in an n-gon is C[n,2] - n.

C[n,2]=n(n-1)/2-n=700

simplifying into a quadratic, and solving for the positive value for n gives n=28.

[/hide]

3:

[hide]

S=1/5 + 2/25 + 3/125...

5S = 1 + 2/5 + 3/25 ...

 5S - S = 1 + 1/5 + 1/25 .. = 5/4

 4S = 5/4

   S = 5/16

[/hide]

Need help with 4 (I previously said 5 by accident)

[hide]CD is sqrt74, AD is sqrt34, and those are the only sides I've found.  ABCD is cyclic, so I drew a circle around it.  But I'm not too sure where to go from here.  I thought maybe Ptolemy would help, but I get the equation asqrt74 + bsqrt34 = 10e, where a is side AB, b is side BC, and e is DB.  I used some trig on the right triangles to find all of the angles.  Also, minor arc BC is 62, minor arc CD is 118, minor arc DA is 72, and minor arc BA is 108.  Is there some trig identity I don't know about that would help me here?  If only I knew what FC and BC were...

Hmm since all the angles in triangles BFC and DEC are the same, they are similar.  Also, ABF and DEA are similar.[hide][/hide][/hide]

5: Set x+6.5=a. Then you get a biquadratic equation in a :D  :D  :D .

Misha

For 5: Hint: Producut of 4 consecutive integers +1 is always a square..

[hide]That is  the expression is (x^2+13x+41)^2  = 0

How do you see that right away:

think of  (x+5)(x+8)  as   (x^2+13x+40)

and        (x+6)(x+7)  as   (x^2+13x +42)

Think of 40 as (41-1) and 42 as (41+1) and  the expression of x^2+13x+41  as y

so the expression now is (y-1)(y+1) +1 = 0  or y^2=0 or y=0

You can solve x^2+13x+41=0  easily.  (x= (-13+- sqrt(5))/2 etc..)

 [/hide]

mathkid wrote:1.
[hide]
Let arctan x be :theta: and let arctan 1/2 be :beta:.

tan :theta: = x, so tan 2:theta: = (2x)/(1-x^2)

Taking the tangent of both sides of the given equation... we get

tan(2:theta - :beta:) = 1
(tan 2:theta: - tan :beta:)/(1 + tan :theta: tan :beta:) = 1

tan 2:theta: - tan :beta: = 1 + tan :theta: tan :beta:
(2x)/(1-x^2) - 1/2 = 1 + 1/(1-x^2)

Simplifying results in a polynomial whose roots sum to 3.

i bet i made a mistake.

[/hide]

Be careful....

[quote="Gyan"][quote]Need help with 4 (I previously said 5 by accident) [/quote]

NF -  You are almost there.
Say assume  AC is X asis  and ED is y-axis You know cooridnates of A (3,0), D and C now.. you know slope (and thus equation) of line AB, and    and also of line BC.. just find out where they meet (actually you need the y value only).

Hope this helps.[/quote]

Good, but can you do this without coordinates?

True... in fact first 3 problems are from AHSMIC, for the #2, #3 I changed the numbers

[quote]Good, but can you do this without coordinates?[/quote]
Sorry did not see it but answer is obviously yes. Once you know that the lines are 'known' rest is easy. You can see similar traingles ( the two right  triangles  on one side of AC are similar to the ones on the other side of AC..) multiply two numbers and diivid by third - some thing like that ((7*3)/5) -  Hope that helps.

beta wrote:1)Find the sum of all values of x that satisfy the following equation

2)A regular n-gon has 350 diagonals. Find n.

3)Find

4) ABCD is a quadrilateral with right angles at A and C. Points E and F are on AC. Also DE and BF are perpendicular to AC. AE=3, DE=5, and CE=7, then find BF.

5) Find all roots to

[hide]2. From any of the n vertices, we can choose n-3 other points to form a diagonal (we exclude the chosen vertex and the two adjacent ones, which when connected form the sides).  Since we counted each diagonal twice - once from each vertex, the total number of vertices is n(n-3)/2 = 350

n^2 - 3n = 700

n^2 - 3n - 700 = 0

(n-28)(n+25) = 0

n = 28, -25

But n > 2, so 28.

3. 1/5 + 2/25 + 3/125 + ... = (1/5 + 1/25 + 1/125 + ...) + (1/25 + 1/125 + ...) + (1/125 + 1/625 + ...) + ... = 1/4 + 1/20 + 1/100 + ... = 5/16.

5. (x+5)(x+6)(x+7)(x+8) = -1

[(x+5)(x+8)][(x+6)(x+7)] = -1

[x^2 + 13x + 40][x^2 + 13x + 42] = -1

Let y = x^2 + 13x + 40

y(y+2) = -1

y^2 + 2y + 1 = 0

(y+1)^2 = 0

y = -1

x^2 + 13x + 40 = -1

x^2 + 13x + 41 = 0

x = [-13 + sqrt(5)]/2 or [-13 - sqrt(5)]/2

(It makes sense that there are two roots, since in order for (x+5)...(x+8) to be negative, one or three terms must be negative, so either (x+5) < 0, (x+6), ... > 0 or (x+5), ... (x+7) < 0, (x+8) > 0.

Bah, I can't do 4.  Uh, yet.[/hide]

A big hint for people who are having trouble doing #4:

[hide]Obviously ABCD is cyclic, so draw the circumcircle of ABCD with center at midpt. BD. And extend DE such that it intersect the circle at another point G.........[/hide]


In countdown round, what is the fastest amount of time to answer one question. I believe it was 45 seconds by Bobby Shen.

This is what I have heard going around on the Today Show with the non-stop talking ladies. My teacher let me watch it during class and we had to mute it because it was getting way too annoying.

 :idea:  thinking is a good thing....

I'm assuming you're implying what the least amount of time in which a question in countdown round was answered correctly? If so, then it is WAY less than 45 seconds (this is actually the max :wink: ).
It even could be less than a second (David Yang) if you don't count the seconds after you press the buzzer.

Also, what in the world was your last portion about?

(AOPS23 is not a scammer -- I traded with him and I got my requested tests: I'm beginning to doubt if he's a multi or not, too -- he's been on a long time [other multis die out very quickly]. My point is: stop being mean to him just because he copied someone a bit and is a bit suspicious -- we'd all be guilty if he actually wasn't a scammer or a multi [sort of like how whites discriminated against black when blacks are awesome])

There! I'm done with my speech! lol...

Multiple times, people have buzzed in insanely quickly. Of course, this is because many problems are quite easy.

[quote="tiger21"][quote="myruggy89"]In countdown round, what is the fastest amount of time to answer one question. I believe it was 45 seconds by Bobby Shen. [/quote]

45 seconds is not the fastest time. It is the slowest time. 45 seconds is too much time to do a count down problem.
Fast time in national count down is about 2 - 3 seconds. Most people barely finishing reading the problems.[/quote]

Max Schindler vs. Bobby Shen (this year) had some pretty quick problems (first 3 especially).

Yes, as you've heard before, people at the National Mathcounts competition are super fast.  However, at my chapter(where I actually won the Countdown Round), we pretty much all, me included, stunk.  75% of the time no one even answered, and I ended up getting through some of the rounds by default, or by getting just 1 right  So obviously, the chapter competition was NOTHING like nationals, and my goal is to not only go to nationals by 8th grade, but also to make it to the countdown round.  Also, how can I improve my reaction time for the problems?  I'm always losing the "For the Win' game.

For maybach:

A.) don't revive an old thread like that...
B.) it kinda depends on your reading speed, so read more books, i guess...

[hide="A long quote"]
[quote="AIME15"]If you've ever had to go against Bobby Shen in a countdown, then you know what it feels like to be in a CD, when the problem is displayed and the buzzer goes off before you can read more than 3 words.

How does he do it?

General rule: Read the last sentence first.

Reading the last sentence tells you exactly what you need to know before you trudge through the problem. Many countdown problems are loaded with useless information that you don't need to know. If you know what you need to know, you can sift through that much faster.

General rule: Learn number tricks.

Obviously I can't explain all of them here, but you should be comfortable with multiplying a large number by 11 in under 1.5 seconds, etc.

General rule: Buzz in early.

You need some practice to get this done correctly; however, when used optimally, allows you to buzz in about 4 seconds before you actually have the answer, and just finish off the arithmetic during this time. This can save you lots of time and can make the difference in a countdown match. Or, if the countdown reader doesn't know the rules, such as in the North California State Countdown Round, you can just buzz in if the question is long. This rule is not a general rule.

General rule: Say your answer correctly.

A countdown moderator will not (or at least, should not) accept an answer of "One thousand...I meant ten thousand!". If your answer is "367887654", make sure you say it as "three hundred sixty-seven million, eight hundred eighty-seven thousand, six hundred fifty-four" and not "three billion, six hundred seventy-eight thousand, oh shoot, I meant..." Take the extra 0.25 seconds to make sure you know how to say your answer.

General rule: "Answer the stinking question."

This is copied from my MATHCOUNTS coach (Mr. Lomas). Many times a problem will ask for "inches" when everything is in "feet", or things like that. Not answering the question is giving away free points to your opponent. Again, take the extra 0.25 seconds to make sure you're answering the question.[/quote]
[/hide]


topology
real analysis
real analysis theorems

I am unable to follow the proof of the following theorem, could sm1 explain:

Let $ \{Q_{1} , Q_{2} ... \}$ be a countable collection of nonempty sets in $ \mathbb{R}^{n}$ such that
1) $ Q_{k \plus{} 1} \subseteq Q_{k}  \forall k$
2) Each set $ Q_{k}$ is closed and $ Q_{1}$ is bounded.

then
$ \cap_{k \equal{} 1}^{\infty} Q_{k}$ is closed and nonempty.

Since $ Q_i$ are closed $ A : \equal{} \cap_{i \equal{} 1}^\infty Q_i$ is closed because any intersection of closed sets is closed in any topology. It remains to show that $ A \neq \emptyset$. Assume the contrary. In the sequel we work in the topology induced on $ Q_1$ by the natural topology on $ \mathbb R^n$. $ Q_1$ is compact since it is a closed bounded subset of $ \mathbb R^n$. Taking complements in $ A \equal{} \emptyset$ we get $ Q_1 \equal{} \cup_{i \equal{} 1}^\infty Q_i^c$. The sets $ Q_i^c$ form an open cover of $ Q_1$ so by compactness there must exist a finite subcover. Thus $ Q_1 \equal{} \cup_{i \equal{} 1}^n Q_i^c$ and taking complements once again we see that $ \emptyset \equal{} Q_1^c \equal{} \cap_{i \equal{} 1}^n Q_i \equal{} Q_n$ which is a contradiction.

I have not looked at the above solution, so here's my solution for some practice with my definitions.  Here's my first formal solution to a topology problem.  Please give some constructive criticism. :)

[i]Proof.[/i] Since $ Q_k$ is closed, $ Q_k^c$ is open, so $ \bigcap_{k \equal{} 1}^{\infty}Q_k \equal{} \left(\bigcup_{k \equal{} 1}^{\infty}Q_k^c\right)^c$ is closed.  Since $ Q_1$ is bounded and closed in $ \mathbb{R}^n$, there exists an n-cell $ S$ such that $ Q_1\subseteq S$.  n-cells are compact, and since $ Q_1$ is closed, $ Q_1$ is compact.  Similarly, each $ Q_k$ is compact.  Now assume to the contrary that $ Q_1\cap\bigcap_{k \equal{} 2}^{\infty}Q_k \equal{} \emptyset$ even though each finite intersection of the $ Q_i$ is nonempty.  Then is follows that $ Q_1\subseteq \left(\bigcap_{k \equal{} 2}^{\infty}Q_k\right)^c \equal{} \bigcup_{k \equal{} 2}^{\infty}Q_k^c$.  Since $ \{Q_k^c\}_{k \equal{} 2}^{\infty}$ is an open cover of $ Q_1$ and $ Q_1$ is compact, $ Q_1\subseteq\bigcup_{k \equal{} 1}^nQ_{a_k}^c$ for some finite $ \{Q_{a_k}^c\}_{k \equal{} 1}^n$.  But this implies that $ Q_1\cap\bigcap_{k \equal{} 1}^nQ_{a_k} \equal{} \emptyset$.  This is clearly a contradiction.  And the conclusion follows. $ \Box$


Hi, I need help finding the x and y intercepts for problems like these..

1.Passes through points (-1,5) and (2,-1)

2.Slope -4 and x-intercept 7.

3. slope 2/3 and passes through point (5,7)

common and easy.

put them in the highschool section! :)

Please don't post the same topic under two sections. Decide which section the topic would be best under and post it there.

Ok, here goes.

1. First, find the slope using $\frac{y_2-y_1}{x_2-x_1}=\frac{-6}{3}=-2$
Then use the point-slope form of a linear equation.
$y-y_1=m(x-x_1)$ where m is the slope, and $y_1, x_1$ are a point that the line passes through.

$y-5=-2(x+1)$
$y=-2x+3$

Then, find the intercepts by pluggin in 0 for x and y and finding the value.
When $y=0$, we find $x=\frac{3}{2}$ When $x=0$, $y=3$


What is the arithmetic mean of all prime numbers between 20 and 30?

The prime numbers between $ 20$ and $ 30$ are $ 23$ and $ 29$.  Thus, the arithmetic mean of all prime numbers between $ 20$ and $ 30$ is $ \frac{23\plus{}29}{2}\equal{}\frac{52}{2}\equal{}\boxed{26}$


dollar/>find a,b,c wich are the length sides of an right-angled triangle and are verifing the next relation: $ \frac{1}{a}\plus{}\frac{2}{b\plus{}c}\equal{}\frac{6}{a\plus{}b\plus{}c}$)

[quote="DonkeyKong"][quote="akech"]$ a \equal{} 4$, $ b \equal{} 3$, $ c \equal{} 5$[/quote]
How did you get that?
Did you start off with multiplying $ abc$ on both side?[/quote]
[hide="Solution"]
Multiply $ a\plus{}b\plus{}c$ on both sides to obtain that \[ \frac{a\plus{}b\plus{}c}{a}\plus{}2(\frac{a\plus{}b\plus{}c}{b\plus{}c})\equal{}6\] \[ 1\plus{}\frac{b\plus{}c}{a}\plus{}2\plus{}\frac{2a}{b\plus{}c}\equal{}6\] \[ \frac{b\plus{}c}{a}\plus{}\frac{2a}{b\plus{}c}\equal{}3\] \[ \frac{b\plus{}c}{a}\equal{}x\] \[ x\plus{}\frac{2}{x}\equal{}3\] \[ x^2\plus{}2\minus{}3x\equal{}0\] \[ (x\minus{}1)(x\minus{}2)\equal{}0\] \[ x\equal{}1, x\equal{}2\] \[ x\equal{}1\rightarrow a\equal{}b\plus{}c\] which is impossible since $ b\plus{}c>a$ in a right triangle.  Next, \[ x\equal{}2\rightarrow 2a\equal{}b\plus{}c\]  If a is the hypotenuse, then we have that \[ 4a^2\equal{}b^2\plus{}c^2\plus{}2bc\rightarrow 4(b^2\plus{}c^2)\equal{}(b^2\plus{}c^2)\plus{}2bc\rightarrow b^2\plus{}c^2\equal{}\frac{2bc}{3}<bc\] which is false, because by the AM-GM, we have that $ b^2\plus{}c^2\ge bc$.  Thus, a cannot be the hypotenuse.  Now, it doesn't matter whether b or c is the hypotenuse (because of the symmetry of the equation.)  So, without loss of generality, call $ c$ the hypotenuse.  Thus, we have that \[ 4a^2\equal{}b^2\plus{}c^2\plus{}2bc\] \[ 4(c^2\minus{}b^2)\equal{}b^2\plus{}c^2\plus{}2bc\] \[ 5b^2\plus{}2bc\minus{}3c^2\equal{}0\] \[ (5b\minus{}3c)(b\plus{}c)\equal{}0\] \[ b\equal{}\frac{3c}{5}, b\equal{}\minus{}c\]  The latter implies that either b or c is negative or 0, which is false.  So, we have that $ b\equal{}\frac{3c}{5}$, giving that $ a\equal{}\frac{4c}{5}$ by the Pythagorean Theorem.  So, $ (a, b, c)\equal{}(4x, 3x, 5x)$ for a positive x (x is not necessarily an integer.) [/hide]


Find all the real numbers x such that the sequence xn=sin((2^n)*x) is convergent.

The solutions are x=k \pi /2^p with k,p from Z.

hmm.. How did you come to that result?! :?

cheers!

If it exists, the limit has to be 0. Indeed, use sin2x=2sinx.cox to get that cos(2nx) -> 1/2 and then use cos2x=2cos2x-1 to get a contradiction.

In order for the limit to be 0 we must have the value of 2nx get close to k.pi with k in Z, that is 2nx/pi must be 'almost' integer. Not sure how to finish the proof, but I guess it works... :D

This is another good one coming from old times. I remember that it appeared a couple more times on the forum though I couldn't find exactly where. Also Myth posted a very similar (though a bit harder) problem at
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33765[/url]. Anyway, I don't remember seeing a complete solution and this thread comes quite close. So can someone finish it?  :)

The problem is to find all $x$ such that the sequence $x_n=\sin(2^nx)$ converges.

Again, if nobody finds a solution in 2 weeks, I'll post mine. If this problem seems too easy, try to solve the version posted by Myth.

This one I can solve, but the one posted by Myth is really awful. I will wait 2 weeks to see your proof for Myth's problem, fedja...  :(


Which is more likely: rolling a pair of dice $ 100$ times and getting a double six at least once or rolling four dice $ 3600$ times and getting a quadruple six at least once?

The probability of getting a double six is $ \frac16\times{\frac16} = \frac {1}{36}$. Thus, the probability of not getting at least one of these in $ 100$ tries is $ (\frac {35}{36})^{100}$, which means that the probability of one showing up is $ 1 - ({\frac {35}{36})^{100}}$. In a similar way the probability of rolling at least one quadruple six is $ 1 - (\frac {1295}{1296})^{3600}$. After computing this with my TI-89, I found that $ 1 - (\frac {35}{36})^{100}\approx.94022$ and $ 1 - (\frac {1295}{1296})^{3600}\approx.93789$. Thus, the double six showing up once is more likely.


For $a,b,c$ are positive numbers. Find minimum of expression:
$A=\frac{x^{2}}{(2y+3z)(2z+3y)}+\frac{y^{2}}{(2x+3z)(2z+3x)}+\frac{z^{2}}{(2x+3y)(2y+3x)}$

It's not hard. By the [b]AM-GM Inequality[/b] we have:
$\sum\frac{x^{2}}{(2y+3z)(2z+3y)}\ge\sum\frac{x^{2}}{\left[\frac{5(y+z)}{2}\right]^{2}}=\frac{4}{25}\sum\left(\frac{x}{y+z}\right)^{2}\ge\frac{4}{25}\cdot\frac{\left(\sum\frac{x}{y+z}\right)^{2}}{3}\ge\frac{4}{25}\cdot\frac{\left(\frac{3}{2}\right)^{2}}{3}=\frac{3}{25}$


Now that the Math Jam is over, it's fair game.  Here are mine: I'm proud of my solution to #5, but I think I was way too wordy with #1.  Feel free to critique as much as you like!  I found it interesting that they downplayed geometry in this round - usually there is at least one hard geometry problem (#3 was fairly straightforward), and sometimes two.

Posted in solafidefarms' thread.
Wow, he beat you to it by 3 minutes...

[quote="1234567890"]Nice, looks like you'll get a 25. :)

I never thought of assuming that m>n>p>q, though. How did you think of that?[/quote]
It's a nice trick with symmetric equations - once you assume some sort of ordering, you can play with inequalities that set bounds on what some of the variables can be.  I find that a symmetric equation is always a prompt to at least try ordering the variables, because more often than not some good can come of it.

[quote="Philip_Leszczynski"][quote="1234567890"]Nice, looks like you'll get a 25. :)

I never thought of assuming that m>n>p>q, though. How did you think of that?[/quote]

I don't think you can assume that - although you can assume $m \ge n \ge p \ge q$. In fact the solution (2,2,2,6) does have some equal terms.[/quote]
I think he probably meant that, and merely left out the equalities for speed.  It is important to note that distinction, though.


In $ \triangle ABC$, $ D$ is a point on $ \overline{AB}$ such that $ CD \equal{} 6$ and $ DB \equal{} 8$. If $ \angle CAD \equal{} \angle BCD$, how many units are in the perimeter of $ \triangle ACD$.

[geogebra]ef28ceeb4636592176dacbc2fdafb2ddee765746[/geogebra]

One possible valid representation:

CAD ~ BCD
AD = 4.5
AC = 7.5
Perimeter is 18.

In mathcounts, since proof is unnecessary I will leave it to here.


inequalities algorithm algebra polynomial

Is there some relatively fast algorithm to solve [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=118813]this problem[/url] or at least how can I find an algorithm that finds bigger and bigger values of the expression.

Or in general, if we have been given $n$ integers $A=\{a_{1},a_{2},\cdots,a_{n}\}=\{1,2,\cdots,n\}$, is there an algorithm to maximize the value of $P(b_{1},\cdots,b_{n})$ where $P$ is a polynomial with respect $b_{1},\cdots,b_{n}$ with coefficients chosen in the set $A$?

If you want progressively larger values, you could try this: http://en.wikipedia.org/wiki/Simulated_annealing


calculus
integration
algebra
polynomial
function
calculus computations

The problem is this:

Integrate $\int{\frac{{x}}{\sqrt{x-1}}dx}$

To evaluate, do we need to use u-subsitution or no? If so, can someone show me how to get started or if not, how would we evaluate it?

The second problem is

Integrate $2\pi\int{x^{2}\sqrt{x+1}}dx$

This I have no idea how to evaulate...

any help is appreciated!

You could do both of these by u-substitutions, to make the thing under the integral a little bit nicer, but the substitution is by no means necessary in either case.  For the first one, try either $u = x-1$ or $u = \sqrt{x-1}$.  You'll probably want to express $\frac 1{\sqrt{(\cdot)}}$ as $(\cdot)^{-\frac12}$ if you use the first substitution.

A similar example:
Suppose we wanted to integrate
$\int (2x-1) \sqrt{x+2}dx$
Method 1: I could do a u-substitution, $u = x+2$ to transform this integral to $\int (2(u-2)-1) \sqrt u du = \int 2u\sqrt u-5 \sqrt u du = \int 2u^{\frac32}-5u^{\frac{1}{2}}du = \frac45u^{\frac52}-\frac{10}3 u^{\frac32}+c$ (which I must then convert back to $x$)

or
Method 2: I could substitute $u = \sqrt{x+2}$, $u^{2}-2 = x$, $2udu = dx$ so the integral becomes $\int (2(u^{2}-2)-1)\cdot u \cdot 2udu$ which is then a big polynomial thing which I can easily integrate.

Integrate $\int{\frac{{x}}{\sqrt{x-1}}dx}$

So if we let $u = x-1$, $du = dx$

Therefore, would we just integrate the following, and substitute $x-1$ for $u$ when we're done? I don't think that'd work because we still need to get rid of the $x dx$ in the integral..

$\int{\frac{{1}}{\sqrt{u}}du}$

Yeah, you killed off the $x$ in the numerator -- what happened to it?  It should have been substituted by $u+1$, right?

Also, something I was going to add at the end of my last post, but you responded to quickly:

I could also do this by parts, although it is much more work than necessary.  The substitution in Method 1 is so simple that you might think of the idea (breaking the $2x-1$ into a sum of powers of $x+2$, which is exactly what the substitution did for us) with without actually making the substitution.

True - but even with this substitution, how does it help us? We still have a square root in the denominator.

$\int{\frac{{u+1}}{\sqrt{u}}du}$

Try writing $\frac 1{\sqrt{u}}$ as a power of $u$ and then multiplying.

I'll do the first one out for you

$\int{\frac{x}{\sqrt{x-1}}}\,dx$

Let $u=x-1$

Then the integral is transformed to
$\int{\frac{u+1}{u^{\frac{1}{2}}}\,du}$

We can separate the numerator, so the whole integral
$=\int{u^{\frac12}}\,du+\int{u^{-\frac{1}{2}}}\,du$

$=\frac{2u^{\frac32}}{3}+2u^{\frac12}+constant$

$=u^{\frac12}\left(\frac{2u}{3}+2\right)+constant$

Fill in the $u$s again:

$=\sqrt{x-1}\left(\frac{2(x-1)}{3}+2\right)+constant$

$=\frac23\sqrt{x-1}(x+2)+constant$

and thats it...I hope  :oops:

[quote="shyong"]you can also try to use

$\frac{x}{\sqrt{x-1}}=\sqrt{x-1}+\frac{1}{\sqrt{x-1}}$
 :)[/quote]

nice i never would have thought about that, I would have done it w/ u-sub.

In teaching a calculus class, I would always show them the substitution, and do what coffeym did.

As for what shyong did: note that he needed to write the numerator of the fraction as $x=(x-1)+1,$ then pull that fraction apart into the sum of two terms. coffeym did the exact same step - only he did so after he had already made the substitution, and for him was much easier to see that $u+1$ should be pulled apart into the sum of two different things.

Don't think that a substitution will necessarily be the last word, or that it will solve the problem all by itself. In many cases, you perform substitutions - espicially "little" substitution where $u$ is a linear function of $x$ and $du$ is a constant times $dx$ - just to streamline the presentation of the problem and to make it easier to see what to do next. And that's exactly what coffeym did.

There's nothing wrong with what shyong said. The only problem with it is what maokid7 said: "i would never have thought of that." Doing the substitution first makes that piece of algebra easier to find.


Here's the story.

About a month ago, there was some discussion on the Etc. forum about an "AoPS Movie" or an "AoPS Story."  I'm pretty sure that no one was being serious about it, but about two weeks ago I decided to write a little dialogue involving some members of AoPS, inspired by Fierytycoon's "Interesting Problem: Christmas Holidays" post.

Today I posted the second part of my work-in-progress, which was met with mixed reviews.  Somebody called it "lovely."  somebody called it "interesting".  But it apparently offended some people, so I had to change it around a little.

I have now come to the realization that:
1) I need an editor before I post anything to ensure that this doesn't happen again.
2) I need to be more careful and be aware that people may take something in the story too personally
and
3) I need to only include people who specifically say "I would like my pseudonym to be a part of this."  Note that I say pseudonym, because you could be [i]anybody[/i] on the internet, and that's fine by me.

If you "would like your pseudonym to be a part of this", please contact me via PM, replying here, or posting a comment in my supposedly slanderous LiveJournal.

From now on, I will post any announcements regarding The Saga under this thread.  I will repost links to Parts I and II, and it is up to The Moderator to decide what to do with the old thread, called "'The AoPS Saga' book".  However, if it were up to me, I would leave it as it not only provides some insightful comments that will help me, but will also show us what we did wrong in making comments to each other.

Feel free to comment on the story whenever you like.  However, please keep your comments free of any expletives, or anything that hints at expletives (including asterisks, filler words, etc.).  In this way, we will create a community of constructive comments and criticism.  Hooray for alliteration!

Your loyal servant and author,
mathfanatic

-edited-

It's a sad, sad world when things come to this.  But as of today, September 1st, 2003, I have chosen to make all previous references to "The AoPS Saga" no longer publically viewable.  This comes as a result of certain complaints from AoPS Community members about my writings.

First off I apologize for using asterisks and segregating gay people.
I think this story is humourous and entertaining. Still, I think the stripper scene is unnecessary because
1.) As mentioned by several people, there are parents reading the material on AoPS as well as things linked to it, and I think some parents would be reluctant to allow their child/ren logging in to AoPS if they see what we wrote, even if it was a joke
2.) Even though you say "asterisk swearing" is a bad thing, you used it in your story too, which is a contradiction. Furthermore, you used it in my line, which is another contradiction since you're telling me not to use asterisk swearing.
3.) I have been recommended to keep away from the following topics: religion, sexuality, politics and I suggest you do the same because anyone could be reading your story.
4.) I don't think materials which needs a disclaimer shouldn't be posted.
5.) Keep in mind that this is a "kids" site, meaning there are people under 18, maybe even 13 for whom this material is completely inappropriate
6.) This part of the story along with the rest is currently in violation of the Livejournal Term of Service Clause XVI: Member Conduct subordinate 1:"Upload, post or otherwise transmit any content that is in LiveJournal.com's opinion to be unlawful, harmful, threatening, abusive, harassing, tortious, defamatory, vulgar, obscene, libelous, invasive to another's privacy (up to, but not excluding any address, email, phone number, or any other contact information with out the written consent of the owner of such information), hateful, or racially, ethnically or otherwise objectionable; " which can result in the enforcement of Clause XV: Journal Content subordinate 2:"Should any Content be reported to LiveJournal.com as being offensive or inappropriate, LiveJournal.com might call upon the author to retract, modify, or protect (by means of private and friends only settings) the Content in question within a reasonable amount of time, as set forth by the LiveJournal.com staff. Should the author fail to meet such a request from LiveJournal.com staff, LiveJournal.com has the full authority to terminate any such reported and verified offensive account holding such inappropriate content. LiveJournal.com, however, is under no obligation to restrict or monitor journal Content in any way; "

P.S. If it's not too much trouble, I candidate for the post of editor for this story.

Regards,
Tare

Feel free to include me if you find a way.  I know that I haven't been around here for very long though

WOW Tare... you were [size=200][b][i][u]REALLY BORED!!![/u][/i][/b][/size]

[quote="Syntax Error"]you must be bored replying like that[/quote]

Hey... I only did it like 4 times or so with all the bold and italics and stuff...

And to Tare: sorry.

In that case, I have a new, new policy.

No more.

I completely agree with tare, I would be banned from this site if my parents saw some of the stuff that was going on, and I don't want that to happen. Also, I would be ok with being included in mathfanatics new and improved (and appropriate) story. I'm done now.

o bleh...im not in mathfanatic's story...(or am i...i havent read the second chapter yet)...

However, I agree with the cussing discussion...Moderators, lets get rid of it or at least enforce it.

I can't enforce it unless you agree to stop the swearing. I can edit it out later, but I would much prefer it if all of you just stopped that.

Ftr (for the record), I only sweared (well, wrote asterisks) three times (counting the one forced by mathfanatic) and I've already deleted them. The only swearing involving the AoPS book is the swearing by teratomato, me, and mathfanatic (well, it's me who's saying it, but I didn't want to; he wrote it as [b]my [/b]line...*sigh*) and the only one left is the one by teratomato.
and I stopped swearing when my [color=red] burning anger [/color]cooled off.

Yeah guys, is it really that hard to control your swearing on an online forum? Seriously, I mean, its not as if you instintively say it out of anger, as could happen in real life. And yeah, I like talking about girls and dancing and fashion and stuff like that, but when it gets down to stripping, well...that's just not cool.

Well, even if no one pays any note of my words, at least they'll pay attention to ripply's, simply cause she is ripply.

now we've got three peeps out there with an almost identical pear :)

That can be arranged... :twisted: (but it'd take a while, so be patient; it won't be ready today because of reasons mentioned earlier, but I might get it in tomorrow.)

I agree with Syntax Error.  Please stick to the topic, and if the topic no longer exists, then you can't stick to anything, therefore you shouldn't post here.  Especially 2-person conversations.  Thank you.

well, its not exactly a 2-way since we're inviting everyone to join. Personally, I think it's their fault that they havent joined in. Of course it's not completely right for us to just keep posting by ourselves, but nobody's keeping u guys from sayin stuff too...

:P Nostalgia.

How many 1st/2nd generation AoPSers are still around?

Wow, this is one old thread. From before I even joined :o. The funny thing is I can't imagine little freshmen mathfanatic and MithsApprentice...

I think a lot of the earliest AoPSers are still "active" but just not posting with the same frequency that they once did (I fall into this category).

EDIT: On another note, I think this thread should be locked.

I'm totally inactive, but Jeffrey pointed this out to me.  scary stuff...the ripply-FyTy joke, not so much, but Tare and I were out of control.  (Tare more though!)

[quote=ns0631]Math fanatic you're a legend.[/quote]

yay 15 year bump

[quote=MathNinja234]/adding to feed[/quote]

You can bookmark it next time instead of posting.


For the simultaneous equations
\[ 2x\minus{}3y\equal{}8\]
\[ 6y\minus{}4x\equal{}9\]

$\textbf{(A)}\ x=4,y=0 \qquad
\textbf{(B)}\ x=0,y=\dfrac{3}{2}\qquad
\textbf{(C)}\ x=0,y=0 \qquad\\
\textbf{(D)}\ \text{There is no solution} \qquad
\textbf{(E)}\ \text{There are an infinite number of solutions}$

[quote="Smartguy"]For the simultaneous equations
\[ 2x \minus{} 3y \equal{} 8
\]

\[ 6y \minus{} 4x \equal{} 9
\]
(A) $ x \equal{} 4,y \equal{} 0$
(B) $ x \equal{} 0,y \equal{} \frac {3}{2}$
(C) $ x \equal{} 0,y \equal{} 0$
(D) there is no solution
(E) there are an infinite number of solutions[/quote]

The graphs of the lines have the same slope so there is no solution.

Rather, we have 6y - 4x = 9 implies 4x - 6y = -9 implies 2x - 3y = -9/2 = 8, which is FALSE- no solution. D.

$ 2x \minus{} 3y \equal{} 8$
$ y \equal{} \frac {2}{3}x \minus{} \frac {8}{3}$

$ 6y \minus{} 4x \equal{} 9$
$ y \equal{} \frac {2}{3}x \plus{} \frac {3}{2}$

Thus, they are parallel lines and there is no solution or empty set because they have the same slope. D is the correct answer.

[hide="Solution"]
We rearrange the second equation to get
$ 2x\minus{}3y \equal{} \frac{\minus{}9}{2}$, which is not the same as the first equation. Therefore, the answer is $ \boxed{(D)}$.
[/hide]

To make a rigorously complete solution, there is no solution because the slopes are equal [i]and[/i] the y-intercepts are different.


calculus
integration
derivative
modular arithmetic
real analysis
real analysis unsolved

Let $m$ and $n$ be two natural numbers. Calculate:

$\int_{0}^{\infty}\frac{d^{m}}{dx^{m}}(e^{-x^{2}})\frac{d^{n}}{dx^{n}}(e^{-x^{2}})dx.$

Hmm.  I can solve the case where $m-n \equiv 1 \pmod 2$, but I'm still trying to figure out the case where $m \equiv n \pmod 2$.


LaTeX
trigonometry
real analysis
real analysis unsolved

I need help in the following problem :
Study the convergence of the serie associated to the sequence sin(sqrt(n)*(-1)^n/(n^a+ (-1)^n)) where "a" is a real (there are some trivial cases about it)
Sorry for not using Latex(it's not my computer) , and just to remind you * stands for product and ^ for power ; thank you :)

Here it is in LaTeX: \[(-1)^{n}\sin\frac{\sqrt{n}}{n^{a}+(-1)^{n}}\] Converges when $a>1/2$ by the alternating series test (may requires some work). I'd say it diverges for $a\le 1/2$, but do not have a proof.

(-1)^n is after the bracket of sin , and with leibniz ciriterion we can only say that if the serie converges a must be > 1/2 , don't mess up things !!:)


let ABC be a triangle(angle A=90).draw the circumcircleof ABC and name it (C).
AH is perpendicular to BC.circle(C')is tangent to BC in point D.and is tangent to AH and circle(C).
prove that:angle BAD=angle DAH.

It isn't clearly ! I think that there are two circles C'.

I hope I correctly understood your problem:

[color=blue][b]Problem.[/b] Let ABC be a triangle with < BAC = 90°, and let k be the circumcircle of triangle ABC. Let H be the foot of the perpendicular from the point A to the line BC. Let a circle k' touch the segments BH and AH and the circle k, and let D be the point where this circle k' touches the segment BH. Prove that < BAD = < DAH.[/color]

Since < BAC = 90°, the segment BC is a diameter of the circle k. On the other hand, the assertion < BAD = < DAH is equivalent to the assertion that the line AD bisects the angle BAH. And thus your problem is equivalent to the one posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=23606 .

[b]Ashegh, please try to write your problems in a more readable way.[/b] It is a real pain to read your posts; even if your English is not too good, you can certainly care about putting spaces between words!

  darij

I suppose WLOG $H\in [OB]$ and  note $k'=C(I,x)$ and the midpoint $O$ of the side $[BC]$. I apply the generalizationed Pitagora's theorem to the side $[IO]$ in the triangle $OIH$:
${IH=x\sqrt 2 ,\ OH=\frac {b^2-c^2}{2a},\ OI=\frac a2 -x,\ m(\widehat {IHO})=135^{\circ}}\Longrightarrow $
$(a^2-2ax)^2=8a^2x^2+(b^2-c^2)^2+4ax(b^2-c^2)\Longrightarrow$

$(b^2+c^2)^2-4a^3x=4a^2x^2+(b^2-c^2)^2+4ax(b^2-c^2)\Longrightarrow $

$a^2x^2+2ab^2x-b^2c^2=0\Longrightarrow x=HD=\frac {b(a-b)}{a}$.

$DB=HB-HD=\frac {c^2}{a} -\frac {b(a-b)}{a}=a-b\Longrightarrow \frac {DB}{DH}=\frac ab ;$

$\frac {AB}{AH}=c\cdot \frac {a}{bc}=\frac ab\Longrightarrow \frac {DB}{DH}=\frac {AB}{AH}\Longrightarrow \widehat {BAD}\equiv \widehat {HAD}.$

[b]Generalization (Neculai Roman). [/b]

Let $k=C(O,R)$ be the circumcircle of the triangle $ABC$ with ${A\ge 90^{\circ}}$.
I note the point $H\in OC,\ AH\perp OC$. Let $k'=C(I,r)$ be the circle which
is interior tangent to the circle $k$ and which is tangent to the lines $BC,\ AH$.
I note $D\in BC\cap k'$. Then $\widehat {BAD}\equiv \widehat {HAD}.$

[b]Indication. [/b]I note the point $E\in AH\cap k$. Apply the Cassey's theorem to the degenerate circles $A,\ C,\ E$ and the circle $k$.

now i can prove it with inversion :D  :D


geometry
circumcircle
trigonometry
geometric transformation
homothety
geometry proposed

In triangle $ABC$, let $R$ be the circumradius, and $r_{a}$ the exradius opposite $A$. Show that

$\frac{r_{a}}{R}=(-\cos A+\cos B+\cos C+1)$

Let X, Y, Z be the tangency points of the excircle $(I_{a})$ with BC, CA, AB. Inversion in the excircle $(I_{a})$ carries the triangle vertices A, B, C into the midpoints A', B', C' of the excircle chords YZ, ZX, XY and the circumcircle (O) into the 9-point circle (O') of $\triangle XYZ.$ Thus the radius of (O') is $R' = 2r_{a}.$ The inversion center $I_{a}$ is the homothety center of (O') and (O). The homothety coefficient of (O'), (O) is equal to the power of the inversion center $I_{a}$ to the inverted circle (O') divided by the power of inversion $r_{a}^{2},$ hence

$\frac{r_{a}}{2R}= \frac{R'}{R}= \frac{O'I_{a}^{2}-R'^{2}}{r_{a}^{2}}= \frac{H'I_{a}^{2}-r_{a}^{2}}{4r_{a}^{2}}$

where H' is the orthocenter of the $\triangle XYZ$ and $2O'I_{a}= H'I_{a}.$ According to a well-known formula, the orthocenter-circumcenter distance $H'I_{a}$ of the $\triangle XYZ$ is given by

$H'I_{a}^{2}= 9r_{a}^{2}-(x^{2}+y^{2}+z^{2}) = r_{a}^{2}[9-4 (\sin^{2}X+\sin^{2}Y+\sin^{2}Z)]$

where x = YZ, y = ZX, z = XY and $\angle Y \equiv \angle XYZ =\frac{ \angle B}{2},$ $\angle Z \equiv \angle YZX = \frac{\angle C}{2}$ (because $XY \perp CI_{a},\ YZ \perp AI_{a},\ ZX \perp CI_{a}$) and $\angle X \equiv \angle ZXY = 180^\circ-(\angle Y+\angle Z) = 90^\circ+\frac{\angle A}{2}.$ Therefore,

${\frac{r_{a}}{R}= 4-2 \cos^{2}\frac{A}{2}-2 \sin^{2}\frac{B}{2}}-2 \sin^{2}\frac{C}{2}=$

$= 4-(1+\cos A)-(1-\cos B)-(1-\cos C) = 1-\cos A+\cos B+\cos C$

[hide="Comment"]The formula $HO^{2}= 9R^{2}-(a^{2}+b^{2}+c^{2})$ is usually proved using medians and centroid (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=101687]Orthic triangle[/url], post #6, or Coexter & Greitzer, Geometry Revisited, problem 1.7.2), but I prefer the straightforward use of the cosine theorem:

$HO^{2}= AO^{2}+AH^{2}-2AO \cdot AH\ \cos \widehat{OAH}=$

$= R^{2}+4R^{2}\cos^{2}A-4R^{2}\cos A \cos (B-C) =$

$= 5R^{2}-4R^{2}\sin^{2}A+4R^{2}\cos(B+C) cos (B-C) =$

$= 5R^{2}-4R^{2}\sin^{2}A+2R^{2}(\cos 2B+\cos 2C) =$

$= 9R^{2}-(4R^{2}\sin^{2}A+4R^{2}\sin^{2}B+4R^{2}\sin^{2}C) =$

$= 9R^{2}-(a^{2}+b^{2}+c^{2})$

There are other ways, for example, if N is the 9-pont circle center, AN is the A-median of $\triangle AHO,$

$AN^{2}= \frac{2AO^{2}+2AH^{2}-HO^{2}}{4}$

$HO^{2}= 2AO^{2}+2AH^{2}-4AN^{2}$

If C' the midpoint of AB and F the foot of the C-altitude, power of A to the 9-point circle (N) is

$AN^{2}-\frac{R^{2}}{4}= AF \cdot AC' = \frac{h_{c}c}{2 \tan A}= \frac{|\triangle ABC|}{\tan A}$

$4AN^{2}= R^{2}+\frac{4|\triangle ABC|}{\tan A}= 2bc \cos A = R^{2}+b^{2}+c^{2}-a^{2}$

$HO^{2}= 2R^{2}+8R^{2}\cos^{2}A-(R^{2}+b^{2}+c^{2}-a^{2}) =$

$= 10R^{2}-2a^{2}-(R^{2}+b^{2}+c^{2}-a^{2}) = 9R^{2}-(a^{2}+b^{2}+c^{2})$[/hide]


geometry
3D geometry
vector
rotation
analytic geometry

What is the side length of the biggest square that can fit inside a cube of side length n?

Yuck.  (More to follow later, hopefully.)

Okay, so I've gotten one step so far:  any square that can fit in can fit in with its center at the center of the cube.  Let's just define axes from the center of the square, parallel to the edges of the cube.  The square's outermost extents are the same in the +x and -x directions, the +y and -y directions, and the +z and -z directions.  Any point in the cube is at least as close to three sides of the cube which make a corner as the center is.  These distances from the center of the square determine the maximal extent of the square.  If we move it to the center of the cube, these distances all increase.

I hope that was clear -- it definitely is true, even if I haven't explained it well.  Now I "just" need to find the longest length for two equal perpendicular vectors emanating from the center of a cube such that their endpoints are still within the cube.

n!!!!!!!

If the cube side length is n, and a square length could be as long as the cube side length, it would be n.

I agree with JBL, i have a feeling that it is longer than n, but I have no idea as to what it is.

It is definitely longer than n.

The original problem asked for the smallest positive integer n so a square of side length n+1 could fit in a cube side length n. Both problems are exactly the same - the hardest thing seems to be able to find what the optimal arrangement of the square is.

Does anyone have a resolution to this problem?

[hide]
i'm fairly certain we could put the diagonal of the square along a space diagonal of the cube, and tilt the square so it fits
that would give $ \frac{n\sqrt{6}}{2}$ as the sidelength
[/hide]

No, definitely not: if you arrange a cube so that its space diagonal is vertical, no matter how you rotate it it will always be substantially taller than it is wide.  (Your suggestion could only work if two of the space diagonals of a cube were perpendicular.)

It's effectively a lower bound, because it doesn't show whether there are bigger squares:

[hide]
side length of the square = $ s$

As the square is maximum, it must contain some sort of symmetry. Assume it touches 4 edges of the cube. The minimal distance between the corner of the cube and the corner of the square is $ x$. For a cube with length 1, by Pythagoras, we have:

$ (x\sqrt2)^2 \plus{} 1 \equal{} s^2$
$ 2(1\minus{}x)^2 \equal{} s^2$

Compare to get:

$ 2x^2 \plus{} 1 \equal{} 2(1\minus{}x)^2$
$ x \equal{} \frac14$

Therefore, substitute in the first equation to get:

$ (x\sqrt2)^2 \plus{} 1 \equal{} s^2$
$ 2x^2 \plus{} 1 \equal{} s^2$
$ 1/8 \plus{} 1 \equal{} s^2$
$ s \equal{} \frac{3\sqrt2}4$

This can be further generalized to $ s \equal{} \frac{3n\sqrt2}4$.
[/hide]

Let's use JBL's idea and try to find two longest orthogonal vectors whose coordinates are less than $ 1$ in absolute value. Suppose that our vectors are $ (a,b,c)$ and $ (x,y,z)$. Then $ ax\plus{}by\plus{}cz\equal{}0$ so two of the three products $ ax,by,cz$ must have the same sign. Thus, without loss of generality, we may assume that $ a,x,b,y$ are positive. Since $ |cz|\le 1$, we can say that $ ax\plus{}by\le 1$. Now we essentially reduced our problem to finding two vectors in the unit square whose scalar product is $ \le 1$ and maximizing the minimum of their lengths. Note that $ (1.0.5)$ and $ (0.5,1)$ are two such vectors and the following picture shows that if we try increase the length, we must also decrease the angle, thus increasing the scalar product
[asy]size(100);
D((0,0)--(0,1)--(1,1)--(1,0)--cycle);
D((0,0));
D(CR((0,0),sqrt(5)/2,0,90),red);
D((1,0.5)--(0,0)--(0.5,1),black);[/asy]
So, $ (0.5,1,1)$ and $ (1,0.5,\minus{}1)$ is the optimal pair.


[b]1[/b].$\ ABC$ is a triangle. Assume that exist $\ X, Y, Z$ be points on the sides $\ BC, CA, AB$ respectively such that $\ AX=BY=CZ$ and $\ BX=CY=AZ$. Prove that: $\ ABC$ is an equilateral triangle.
[b]2[/b].$\ ABC$ is a triangle. $\ M$ be the point on the incircle of $\ ABC$, $\ H, K, L$ respectively the projections of $\ M$ on $\ BC, CA, AB$. Determine the positions of $\ M$ so that $\ (MH+MK+ML)$ attains its greates value and the least value.

[quote="April"][b]1[/b].$ \ ABC$ is a triangle. Assume that exist $ \ X, Y, Z$ be points on the sides $ \ BC, CA, AB$ respectively such that $ \ AX \equal{} BY \equal{} CZ$ and $ \ BX \equal{} CY \equal{} AZ$. Prove that: $ \ ABC$ is an equilateral triangle.
[/quote]

Edit: go to http://www.mathlinks.ro/Forum/viewtopic.php?p=558585#558585

problem 1. - http://www.mathlinks.ro/Forum/viewtopic.php?p=558585#558585


Prove that there are not existing an infinite real sequence {x_n} that lx_nl \leq  0,6666666 for every n in N
and lx_n - x_m l \geq  1/n*(n+1) + 1/m*(m+1) for every m,n in N.

Let k = 0,6666666
consider seqence :x_1,x_2,..,x_n and each term belongs to [-k,k]
By pigeon-hole principle we have:
2k/(n-1)  \geq |x_i-x_j| \geq 1/i*(i+1) + 1/j*(j+1) \geq 2/n*(n+1)
when n gets to infinity we have contradiction!


calculus
probability
probability and stats

I'm currently a high school student (10th grade) who's not really taking any really difficult math courses right now. I have a tutor in pre-calculus/calculus purely because I'm interested. We're still on pre-calculus but we're moving at a pretty quick rate and after single-variable calculus I plan on going on to multi-variable,etc. This notes aren't a problem here b/c the tutor basically provides notes via lecture/textbook.

However, I want to teach myself Statistics too and not just the basics, I mean the whole way with the calculus stuff too. The problem is that the only books at my local library are readinsg like "Statistics De-mystified" and "Idiot's Guide to Statistics". I checked some of them out and these types of readings are really difficult for me to follow because they don't really explain why in-depth but they just list facts.

Strangely enough, it's much easier for me to follow college-texts and such. I really don't have the money to be spending a fortune even on a used copy. Any ideas on how to go about teaching myself something like this/what to do about texts?

I just think it's going to be really difficult but it's something I'm truly interested in. I can't afford a tutor right now either.

Edit: Actually, if there are any accomplished statisticians here who have lectures or outlines and such already prepared and would like to send them to me, that would be amazing.

You can't really understand much of statistics if you don't know probability. A good book on probability is "A First Course on Probability" by Sheldon Ross.

Here's an online text on probability, but I can't vouch for it because I've never used it:
http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/amsbook.mac.pdf


$ a,b,c \geq 0$
$ a \leq b\plus{}c$

Prove that :

$ (b\plus{}c)\plus{}1\plus{}|a\minus{}1|\geq |b\minus{}1|\plus{}|c\minus{}1|\plus{}a$

[quote="Bugi"]You could just divide into cases of values for expressions under absolute braces. There are 8 cases, some impossible because of the condition, and other easily solved.[/quote]

i know this solution ... i want a nice solution for it ...... thanks you :)


[quote="pvthuan"]Let $a,b,c$ be nonegative real numbers, prove that \[2\leq\frac a{\sqrt{b^{2}-bc+c^{2}}}+\frac b{\sqrt{c^{2}-ca+a^{2}}}+\frac c{\sqrt{a^{2}-ab+b^{2}}}.\] [/quote]
Jensen kills it.

[quote="pvthuan"]Let $a,b,c$ be nonegative real numbers, prove that \[2\leq\frac a{\sqrt{b^{2}-bc+c^{2}}}+\frac b{\sqrt{c^{2}-ca+a^{2}}}+\frac c{\sqrt{a^{2}-ab+b^{2}}}\leq3.\] [/quote]
The right part is certainly wrong,because
$\frac a{\sqrt{b^{2}-bc+c^{2}}}+\frac b{\sqrt{c^{2}-ca+a^{2}}}+\frac c{\sqrt{a^{2}-ab+b^{2}}}$ can tends to $\infty$...
Does $a,b,c$ should to be three sides of an triangle?

[quote="ehku"]Can you post your solution, argady? Thank you![/quote]
[quote="pvthuan"]Let $a,b,c$ be nonegative real numbers, prove that \[2\leq\frac a{\sqrt{b^{2}-bc+c^{2}}}+\frac b{\sqrt{c^{2}-ca+a^{2}}}+\frac c{\sqrt{a^{2}-ab+b^{2}}}.\] [/quote]
Since, it's a homogeneous inequality let's assume that $a+b+c=1.$
$f(x)=\frac{1}{\sqrt{x}}$ is convex function. Hence,
$\frac a{\sqrt{b^{2}-bc+c^{2}}}+\frac b{\sqrt{c^{2}-ca+a^{2}}}+\frac c{\sqrt{a^{2}-ab+b^{2}}}\geq\frac{1}{\sqrt{a(b^{2}-bc+c^{2})+b(a^{2}-ac+c^{2})+c(a^{2}-ab+b^{2})}}.$
Id est, remain to prove that
$\frac{1}{\sqrt{a(b^{2}-bc+c^{2})+b(a^{2}-ac+c^{2})+c(a^{2}-ab+b^{2})}}\geq2.$
But    $\frac{1}{\sqrt{a(b^{2}-bc+c^{2})+b(a^{2}-ac+c^{2})+c(a^{2}-ab+b^{2})}}\geq2\Leftrightarrow$
$\Leftrightarrow1\geq4\cdot\sum_{cyc}(a^{2}b+a^{2}c-abc)\Leftrightarrow(a+b+c)^{3}\geq4\cdot\sum_{cyc}(a^{2}b+a^{2}c-abc)\Leftrightarrow$
$\Leftrightarrow15abc+\sum_{cyc}(a^{3}-a^{2}b-a^{2}c+abc)\geq0,$ which true by Schur.

[quote="pvthuan"]Let $a,b,c$ be nonegative real numbers, prove that \[2\leq\frac a{\sqrt{b^{2}-bc+c^{2}}}+\frac b{\sqrt{c^{2}-ca+a^{2}}}+\frac c{\sqrt{a^{2}-ab+b^{2}}}\] [/quote]
My solution
[hide] \[{2a\sqrt{b^{2}-bc+c^{2}}}\le a^{2}+b^{2}+c^{2}-bc \le a^{2}+b^{2}+c^{2}\] [/hide]

The left part is quite simple. I am wondering about \[\frac a{\sqrt{a^{2}-ab+b^{2}}}+\frac b{\sqrt{b^{2}-bc+c^{2}}}+\frac c{\sqrt{c^{2}-ca+a^{2}}}\leq3.\] What a pity! It is wrong. Try $(0,1,\tfrac14)$

[quote="pvthuan"]The left part is quite simple. I am wondering about \[\frac a{\sqrt{a^{2}-ab+b^{2}}}+\frac b{\sqrt{b^{2}-bc+c^{2}}}+\frac c{\sqrt{c^{2}-ca+a^{2}}}\leq3.\] [/quote]
Mr. Thuan, it is true and here is my solution for it.
To prove it, we will prove the stronger inequality \[\sum\frac{a^{2}}{a^{2}-ab+b^{2}}\leq 3\] Or, \[\sum\frac{1}{x^{2}-x+1}\leq 3\] with $x,y,z>0$ and $xyz=1$
We will peove the following Lemma:
Lemma. \[\sum\frac{1}{x^{2}+x+1}\geq 1\] with $x,y,z>0$and$xyz=1$
Proof.
Since $x,y,z>0$and$xyz=1$, then there exists $m,n,p>0$ such that $x=\frac{np}{m^{2}}, y=\frac{pm}{n^{2}}, z= \frac{mn}{p^{2}}$. And our Lemma becomes \[\sum\frac{m^{4}}{m^{4}+m^{2}np+n^{2}p^{2}}\geq 1\] Applying Cauchy-swharz Inequality, we have \[\sum\frac{m^{4}}{m^{4}+m^{2}np+n^{2}p^{2}}\geq \frac{(m^{2}+n^{2}+p^{2})^{2}}{%Error. "summ" is a bad command.
^{2}n^{2}+%Error. "summ" is a bad command.
^{2}np}\geq 1\] Our Lemma is proven. Now, applying this Lemma by replacing $x,y,z$ with $\frac{1}{x^{2}}, \frac{1}{y^{2}},\frac{1}{z^{2}}$, we have

$\sum\frac{x^{4}}{x^{4}+x^{2}+1}\geq 1$

Or,

$\sum\frac{2(x^{2}+1)}{x^{4}+x^{2}+1}\leq 4 (*)$

Notes that $2(x^{2}+1) =(x^{2}+x+1)+(x^{2}-x+1)$ and $x^{4}+x^{2}+1 =(x^{2}+x+1)(x^{2}-x+1)$, we have

$(*) \Leftrightarrow \sum\frac{1}{x^{2}-x+1}+\sum\frac{1}{x^{2}+x+1}\leq 4$

Now, applying the Lemma again, we will have

$\sum\frac{1}{x^{2}-x+1}\leq 3$

QED. :)


inequalities
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inequalities solved

[quote="fermat3"]my freind faisl gave me a    nice  congecture and i find a solution aexp[n+1]\aexp[n]+bexp(n)  +bexp(n+1)\cexp(n)+bexp(n)    +cexp(n+1)\cexp(n) +bexp(n)   >= a+b+c\2    (E)
    assume that we have a+b+c=1         then     (E)   become    a\(1+(b\a)exp(n)   )   +b\(1+(b\c)exp(n) )       +c\(1+(c\a)exp(n)) >=0.5                            so applying jensen inegualitie  to function 1\(1+(x)exp(n)       then we have

(E)>=1\(1+(ab\a+bc\b+ca\a)exp(n)=0.5

[/hide][/list][/code][/quote][/u][/i][/quote]

use latex please. nobody will untrstand a word...
 :(

:stretcher:  :what?:
Use Latex please.Thank you.

I think he wanted to say that \[ \sum \frac{a^{n+1}}{a^n+b^n} \geq \frac{a+b+c}{2} , \] but I'm not too sure.

The function $\frac1{1+x^n}$ is nor concave nor convex. Jensen doesn't work here  :(

Well I guess it can't be solved in general (as it's false) as already mentioned [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83149]here.[/url]  :P

fermat3, why do you dont use latex in all your posts!!!! :mad: