geometry
calculus
inequalities
analytic geometry
quadratics
function
ratio

Triangle $ABC$ has $AB=9$ and $BC: AC=40: 41$. What's the largest area that this triangle can have?

This is a great problem to show to people, and it is nice to have the problems for the resources section. Here is the topic
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=aime&t=51673[/url]
that I was thinking of. It is also interseting to dig up some of my old posts to see how I have changed mathematically.

i got an equation,
 x^2+5 = (7+y)(y) = 7y+y^2

How do i find the biggest x and y value that can make this equation true?

[hide="Whoops"]I got $667.87...$  :o I've already looked at the correct answer (that and I can't possibly be right if this is an AIME). Someone offer an explanation?

EDIT: You're basically trying to solve for the largest value of this:

$\sqrt{(\frac{81x+9} {2})(\frac{81x+9} {2}-41x)(\frac{81x+9} {2}-40x)(\frac{81x+9} {2}-9)}$

(Horrendously distorted usage of Heron's formula)

I decided to take the geometric average of every term under the square root and after some work, it gets me:

$\frac{(81x+9)(81x-9)(x+9)(-x+9)} {6}$

I guess what you could do then is do:

$\frac{-3(81x^2-1)(x^2-9)} {2}$ where $x<9$

I'm not even sure if this is right or if I know what I'm doing at all   :oops: [/hide]

I don't want to revive a dead thread (the thread altheman showed us). But when you used the AM-GM Inequality, the only time they are equal is when the two terms are equal. In this case, it is when $81x^2-1=81-x^2$. However, this gives us $1$ and $-1$, which is obviously incorrect.

[quote="4everwise"]Yeah, I solved this problem with calculus, so I won't post my solution. :diablo:[/quote]

The calculus solution is really messy. :D

i agree that was an incorrect use of the am-gm, obviously that is not correct way to solve for "x"

Heron's Formula:
$A=\sqrt{(\frac{81x+9}{2})(\frac{81x+9}{2}-41x)(\frac{81x+9} {2}-40x)(\frac{81x+9}{2}-9)}$

You want to maximize that tremendous squareroot.
Now, rewrite each factor inside the squareroot as one fraction, factor out a $\sqrt{\frac{81}{16}}=\frac{9}{4}$, rewrite using differences of squares, and let $z = x^2$ to achieve

$A=\frac{9\sqrt{(81-z)(81z-1)}}{4}$

To maximize this, you must clearly maximize $(81-z)(81z-1)$. The roots of this are $z=81$ and $z=1/81$, and the maximum point (vertex) occurs at the z-value halfway between these two roots: or $z=\frac{81+\frac{1}{81}}{2}$

Plugging back into our formula, $A=\frac{9\sqrt{(81-z)(81z-1)}}{4}=\boxed{820}$

[quote="4everwise"][hide]First, consider the triangle in a coordinate system with vertices at $(0,0)$, $(9,0)$, and $(a,b)$. Applying the distance formula, we see that $\frac{ \sqrt{a^2 + b^2} }{ \sqrt{ (a-9)^2 + b^2 } } = \frac{40}{41}$.

We want to maximize $b$, the height, with 9 being the base. Simplifying gives $-a^2 -\frac{3200}{9}a +1600 = b^2$. Recalling that the max/min value of a quadratic is $\frac{4ac-b^2}{4a}$, we see that $\frac{4(-1)(1600) - \left(\frac{3200}{9}\right)^2 }{(-1)(4)} = b^2$. Solving for $b$, $b = \frac{40\cdot41}{9}$ Thus, the area is $9\cdot\frac{1}{2} \cdot \frac{40*41}{9} = \boxed{820}$.[/hide][/quote]

Isn't the max/min of a quadratic thing derived straight from calculus (the very easy method)?...
$\\y=ax^2+bx+c\\y'=2ax+b\implies 0=2ax+b\implies x=\frac{-b}{2a}\\\\ f\left(\frac{-b}{2a}\right)=a\frac{b^2}{4a^2}-\frac{b^2}{2a}+c\\=\frac{b^2-2b^2}{4a}+c=c-\frac{b^2}{4a}$

Masoud Zargar

It can be, but it doesn't have to be

Consider $y = ax^2 + bx + c = a(x^2+\frac{b}{a}x+\frac{c}{a})$

Completing the square:
$y=a((x+\frac{b}{2a})^2+\frac{c}{a}-\frac{b^2}{4a^2})=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}$

The minimum value of $(x+\frac{b}{2a})^2$ is obviously 0, so the extremum of the function will occur at $y=c-\frac{b^2}{4a}=\frac{4ac-b^2}{4a}$

without calculus?
9-40-41 is a pythagorean triple.... :|
oh.... [b]ratio[/b]
 :blush:

[quote="quantum leap"]Can't you just use the triangle inequality for this one?  Since two sides have a ratio of 40:41, and the other side is nine, the two sides can have max lengths of 320 and 328.[/quote]Well, does that guarantee a maximum area?  ;)

[hide]$40x+9 > 41x \implies 9 > x$
Triangle ABC is therefore a $9,359.999...,268.999....$ triangle, and then I can't solve for the area because I suck at geometry.[/hide]

[quote="cdymdcool"][quote="TZF"]Heron's Formula:
$ A = \sqrt {(\frac {81x + 9}{2})(\frac {81x + 9}{2} - 41x)(\frac {81x + 9} {2} - 40x)(\frac {81x + 9}{2} - 9)}$

You want to maximize that tremendous squareroot.
Now, rewrite each factor inside the squareroot as one fraction, factor out a $ \sqrt {\frac {81}{16}} = \frac {9}{4}$, rewrite using differences of squares, and let $ z = x^2$ to achieve

$ A = \frac {9\sqrt {(81 - z)(81z - 1)}}{4}$

To maximize this, you must clearly maximize $ (81 - z)(81z - 1)$. The roots of this are $ z = 81$ and $ z = 1/81$, and the maximum point (vertex) occurs at the z-value halfway between these two roots: or $ z = \frac {81 + \frac {1}{81}}{2}$

Plugging back into our formula, $ A = \frac {9\sqrt {(81 - z)(81z - 1)}}{4} = \boxed{820}$[/quote]

How could you tell that the maximum point occurs at the z-value halfway between the two roots?  Also how did you solve this equation $ A = \frac {9\sqrt {(81 - z)(81z - 1)}}{4}$?[/quote]
The answer to your first question is conceptually obvious: picture a parabola pointing down. By symmetry, its vertex is at the x-coordinate that is the average of 2 roots, and of course, this is the maximum.
To "solve" that equation, he simply plugged in the value of $ z$ that he just found and evaluated.

[hide="Solution"]
Let the three sides be $ 9,40x,41x$, so the area is $ \frac14\sqrt{(81^2\minus{}81x^2)(81x^2\minus{}1)}$ by Heron's formula. By AM-GM, $ \sqrt{(81^2\minus{}81x^2)(81x^2\minus{}1)}\le\frac{81^2\minus{}1}2$, and the maximum possible area is $ \frac14\cdot\frac{81^2\minus{}1}2\equal{}\frac18(81\minus{}1)(81\plus{}1)\equal{}10\cdot82\equal{}\boxed{820}$. This occurs when $ 81^2\minus{}81x^2\equal{}81x^2\minus{}1\implies x\equal{}\frac{4\sqrt{205}}9$.
[/hide]

[quote="LordoftheMorons"][hide]$ 40x \plus{} 9 > 41x \implies 9 > x$
Triangle ABC is therefore a $ 9,359.999...,368.999....$ triangle, and then I can't solve for the area because I suck at geometry.[/hide][/quote]
I believe such a triangle would be the minimum area? The triangle would be practically a line unless I am imagining it wrong.

Yes, because it is approaching a degenerate triangle. The $ 0$ area is also achieved when $ x\equal{}\frac19$.

[quote="4everwise"][hide]First, consider the triangle in a coordinate system with vertices at $ (0,0)$, $ (9,0)$, and $ (a,b)$. Applying the distance formula, we see that $ \frac {\sqrt {a^2 \plus{} b^2} }{\sqrt { (a \minus{} 9)^2 \plus{} b^2 } } \equal{} \frac {40}{41}$.

We want to maximize $ b$, the height, with 9 being the base. Simplifying gives $ \minus{} a^2 \minus{} \frac {3200}{9}a \plus{} 1600 \equal{} b^2$. Recalling that the max/min value of a quadratic is $ \frac {4ac \minus{} b^2}{4a}$, we see that $ \frac {4( \minus{} 1)(1600) \minus{} \left(\frac {3200}{9}\right)^2 }{( \minus{} 1)(4)} \equal{} b^2$. Solving for $ b$, $ b \equal{} \frac {40\cdot41}{9}$ Thus, the area is $ 9\cdot\frac {1}{2} \cdot \frac {40*41}{9} \equal{} \boxed{820}$.[/hide][/quote]

$ \frac {\sqrt {a^2 \plus{} b^2} }{\sqrt { (a \minus{} 9)^2 \plus{} b^2 } } \equal{} \frac {40}{41}$. should be $ \frac {\sqrt {a^2 \plus{} b^2} }{\sqrt { (a \minus{} 9)^2 \plus{} b^2 } } \equal{} \frac {41}{40}$.

Solution without Heron:

Let $F$ be the foot of the altitude from $C$ to $AB.$ Suppose that $CF = h, BF = y, AF = \pm (9 - y)$ (note that the $+$ is when $F \in \overline{AB}$ and the $-$ accounts for when $F$ lies outside of segment $AB.$ Then, by the Pythagorean Theorem on $\triangle BCF, \triangle ACF$, it follows that \[\frac{h^2 + y^2}{h^2 + \left(\pm\left(9 - y\right)\right)^2} = \frac{\left(40x\right)^2}{\left(41x\right)^2} = \frac{40^2}{41^2}. \quad \text{(1)}\] Note that the $\pm$ condition is conveniently irrelevant due the the fact that $\pm(9 - y)$ is squared.

Now, we wish to maximize $[ABC] = \frac{1}{2} \cdot AB \cdot CF = \frac{9}{2} \cdot h.$ This is equivalent to maximizing $h.$

By $(1)$, it follows that
\begin{align*}
& 41h^2 + 41y^2 = 40^2h^2 + 40^2\left(9 - y\right)^2\\
& \implies h^2\left(41^2 - 40^2\right) = y^2\left(40^2 - 41^2\right) - 18 \cdot 40^2 y + 9^2 \cdot 40^2 \\
& \implies 81h^2 = -81y^2 - 18 \cdot 40^2 y + 9^2 \cdot 40^2 \\
& \implies 9h^2 = -\left(9y^2 + 2 \cdot 40^2 y - 9 \cdot 40^2\right).
\end{align*}

Now, recall that a quadratic $f(x) = ax^2 + bx + c$ is maximized/minimized when $x = -\frac{b}{2a}$, at which point $f(x) = c - \frac{b^2}{4a}.$ These two facts can be easily verified by completing the square or taking a derivative.

By applying our maximization/minimization criterion to the above quadratic, it follows that the maximum value of $9h^2$ is \[-\left(-9 \cdot 40^2 - \frac{4 \cdot 40^4}{4 \cdot 9}\right) = 40^2\left(\frac{40^2}{9} + 9\right) = 40^2\left(\frac{40^2 + 9^2}{9}\right) = \frac{40^2 \cdot 41^2}{3^2}.\] By taking the square root (we may do this since we know that $h$ is positive), it follows that the maximum value of $3h$ is $\frac{40 \cdot 41}{3}.$ Now, recall that $[ABC] = \frac{9}{2} \cdot h.$ Therefore, the maximum value of $[ABC]$ is \[\frac{9}{2} \cdot h = \frac{3}{2}\left(3h\right) = \frac{3}{2} \cdot \frac{40 \cdot 41}{3} = \boxed{820}.\]

[hide="Solution"]
You have a triangle with a base length of $9$ and some height which you want to maximize.  Let $B$ be at $(0,0)$, $A$ be at $(9,0)$, and the coordinates of $C$ to be $(x, y)$.

Using the condition, you get the equation $\frac{x^2+y^2}{(x-9)^2+y^2}=\frac{40^2}{41^2}$.  Expanding and simplifying the equation gives an equation in the form of a circle with a center on $y=0$.  You can easily find the two points on the circle where $y=0$, which are $(-360,0)$ and $(\frac{40}{9},0)$ (I got this easily by using the ratios given in the problem rather than using the equation and plugging in $y=0$).  Those points form the diameter of the circle.

The maximum height of the triangle is the $y$-coordinate of the highest point on the circle, which is the same as the circle's radius, so the maximum height of the triangle is $(\frac{40}{9}+360)\div 2=\frac{1640}{9}$.  Therefore, the area of the triangle is $\frac{1640}{9}\times 9\div 2 = \boxed{820}$.
[/hide]

because no one has overkilled it yet

[hide=Apollonian circles]
The locus of points $C$ satisfying the given condition is an Apollonian circle of segment $AB$, so we may compute that circle's diameter and halve that to find the maximum height from $C$ to $AB$.

Let $A = (0,0)$ and $B = (9,0)$. Let $D$ exist on $AB$ such that $DB : DA = 40 : 41$. Then $D = (41/9, 0)$.

The harmonic conjugate of $D$, $E$, is the point lying on the extension of $AB$ such that $BE : AE = 40 : 41$. This yields $\tfrac{x}{9+x} = \tfrac{40}{41} \implies x = 360$.

Therefore the height is $\dfrac{360 + 40/9}{2}$, so the area of the triangle is $9 \cdot 90 + 10 = 820$.

[hide=Sketch of solution with calculus]Let angle $C$ be $\theta$, then express $AB=9$ in terms of $BC=40x$ and $AC=41x$ and $\theta$ by LoC. Then solve for $x$ in terms of $\theta$ and then substitute that value into the equation for area with $\left[ABC\right]=\frac{BC\times AC\times \sin{\theta}}{2}$. After this is done, you get the area of $ABC$ as a function of $\theta$, and to find the maximum, take the derivative and set that to $0$. Then plugging $\theta$ into the area formula yields the maximum area of $\boxed{820}.$[/hide]

Kind of boring.

Let $BC=40x, AC=41x$, so by Heron's the area evaluates to $$\frac 12\sqrt{\left(\frac{9+81x}2\right)\left(\frac{9+x}2\right)\left(\frac{9-x}2\right)\left(\frac{81x-9}2\right)} = \frac 18 \sqrt{(81-x^2)(81^2x^2-81)}.$$ Set $y=x^2$, such that the expression inside the square root equals $$81(81-y)(81y-1) = 81\left[-\left(9y-\frac{81^2+1}{18}\right)^2 - 81 + \frac{(81^2+1)^2}{324}\right].$$ Note that the maximum is then $$\frac 18 \sqrt{(81^2+1)^2 - 81 \cdot 324} = \frac 18\sqrt{(81^2-1)^2} = \boxed{820}$$ via the algebraic substitution $81=x$.


i am having some trouble with these problems. could you please provide me with some hints.

1) Prove that there exist integers $a,b,c$ not all 0 and each of absolute value less than 1,000,000 such that $|a+b\sqrt{2}+c\sqrt{3}|<10^{-11}$.
for some reason i do not think that there is anything special about the number $10^{-11}$

2) Show that in any group of six people there must be three mutual friends or three mutual strangers.

3) Let $a_{1},a_{2},\dots,a_{n^{2}+1}$ ne a sequence consisting of $n^{2}+1$ distinct integers. Prove that this sequence has a monotone (increasing or decreasing) subsequence of length $n+1$.
my problem with this one is that i am not really sure what is being asked.

2

[hide]This is pretty well known. Look at one person, and his relation to the five other people. Since he has five relations, either 3 must be friends or 3 must be strangers. Now look at these 3.... what can you say about them?[/hide]

[hide="3: a clarification"] This is also well-known.

A subsequence of a sequence $a_{1}, a_{2}, ... a_{m}$ is a sequence $a_{ \delta(1)}, a_{ \delta(2)}, ..., a_{ \delta(k) }$ where $k \le m$ and where $\delta$ satisfies

$i < j \Leftrightarrow \delta(i) < \delta(j)$

A concrete example:  Given the sequence

$\{ 1, 2, 19, 200, 40, 30, 20 \}$

One possible subsequence is

$\{ 1, 19, 40, 30, \}$

And another is

$\{ 2, 200, 20 \}$

Essentially, we take a subset of the sequence, and then arrange it in the same order as it appeared in the original sequence.

The problem can be generalized to $a_{mn+1}$, where the statement becomes this:

Either there exists a monotonically increasing subsequence of length $n+1$, or there exists a monotonically decreasing subsequence of length $m+1$. [/hide]

[quote="paladin8"][quote="maokid7"]before i stat i would like to thank [b]t0rajir0u[/b] for his clearification of number 3.
[quote="paladin8"][hide="1"]
Consider all possible values of $a+b\sqrt{2}+c\sqrt{3}$ with $-\frac{1}{2}\cdot 10^{6}< a, b, c \le \frac{1}{2}\cdot 10^{6}$. How can we bound these values? How many of these are there? If this many have to fit in a certain bound, what can we say about how close any two of them must be?[/hide][/quote]

ok so for #1 is somewhat like how it would be done?
since $\frac{-10^{6}}{2}<a,b,c<\frac{10^{6}}{2}$ there are $(10^{6})^{3}$ possible values for $a+b\sqrt{2}+c\sqrt{3}$ but since one of these is where all a,b,c are 0 there is really a total of
$10^{18}-1$ total values.
then we find the maximum and minimum value of $a+b\sqrt{2}+c\sqrt{3}$ and divide the difference by $10^{18}-1$ and this leads us to show that there are 2 numbers that are withing $10^{-11}$ of each other.

is this right or did i totally miss?  :maybe:[/quote]

Yup, that's pretty much it. One thing I found strange is that

$\frac{2(1+\sqrt{2}+\sqrt{3})}{10^{12}}< \frac{1}{10^{11}}$

so I was looking for that to come up, but it seems that we can improve the bound to just

$\frac{1+\sqrt{2}+\sqrt{3}}{10^{12}}$,

or less than half of the desired quantity.[/quote]

i don't think this proof is done, how do you know the difference is still a valid value of a, ex if you had 10^6-1 for the first a, and -10^5 for the second a, the difference would not be an acceptable value of a

i tired to apply this idea to the following problem but am unable to find a way to bound the value of $b$. any thoughts?

Prove that for and positive real $x$ and any positive integer $n$ there exist integers $a,b$ such that $|ax-b|\leq{\frac{1}{n}}$, where $1\leq{a}\leq{n-1}$.

It's only true for irrational $x$, which follows from some principles of continued fractions.

[quote="t0rajir0u"]

The problem can be generalized to $a_{mn+1}$, where the statement becomes this:

Either there exists a monotonically increasing subsequence of length $n+1$, or there exists a monotonically decreasing subsequence of length $m+1$.[/quote]
 [hide="proof"]For each $a_{i}$ in the sequence consider the ordered pair $(k, j)$ where $k$ is the length of the longest ascending sequence starting  at $a_{i}$ and $j$ is the length of the longest descending one ending at $a_{i}$. Two of these pairs are the same if the ascending sequences have length less then $n+1$ and descending less then $m+1$ by the pigeonhole principle. So two numbers have the same ordered pair. Let them be $b$ and $c$. But then $b<c$ WLOG. This means we can put $b$ before the longest ascending sequence starting at $c$ and get a contradiction, as this sequence has length greater then the longest ascending sequence starting at $c$ and hence at $b$.[/hide]

[quote="paladin8"]That's why I bounded it with $-\frac{10^{6}}{2}$ and $\frac{10^{6}}{2}$, so the maximum absolute value of any difference is $10^{6}$.[/quote]

oh alright, i missed that, p.s. good proof!

[quote="WatsonLadd"][quote="t0rajir0u"]

The problem can be generalized to $a_{mn+1}$, where the statement becomes this:

Either there exists a monotonically increasing subsequence of length $n+1$, or there exists a monotonically decreasing subsequence of length $m+1$.[/quote][/quote]

The pigeonhole proof of this statement is beautiful. There is also a more direct approach:

Suppose every monotonically increasing sequence is of length at most $n$. Let $s_i$ denote the longest monotonically increasing sequence originating at $a_{in+1}$ for $0 \le i < m$. Then the last terms of $s_i$ plus $a_{nm+1}$ form a monotonically decreasing sequence of length $m+1$.


geometry
circumcircle
inequalities unsolved
inequalities

If the quadrilateral of sides $a,b,c,d$ and diagonals $p,q$ has both an incircle and a circumcircle then

$ (a+b+c+d)^2 \geq 8pq $

Because ABCD is circumscribable it follows that a+c=b+d  thus:
$(a+b+c+d)^{2}=4(a+c)^{2}=2(a+c)^{2}+2(b+d)^{2} \geq
8(ac+bd).$ (AM $\geq$ GM) Use Ptolemy and we are done : $ac+bd \geq pq$


calculus
derivative
function
integration
limit
Support
real analysis

[quote="cimabue"]Assume $f(x)$ is a continuously differentiable function such that $f, f'\in L^2(\mathbb R)$,
i.e. $\int_{\mathbb R}|f(x)|^2<+\infty$ and $\int_{\mathbb R}|f'(x)|^2dx<+\infty$.
Prove that
\[ \lim_{t\to 0}\int_{\mathbb R}\left|\frac{f(x+t)-f(x)}{t}-f'(x)\right|^2dx=0  \][/quote]
First note that for $g \in L^2(\mathbb{R})$ we have $\lim_{h \to 0} \int_{\mathbb{R}} |g(t+h)-g(t)|^2 dt = 0$. Indeed, this is true if $g$ is smooth (say $C^1$) and has compact support, since then $g$ is uniformly continuous. For general $g$ it follows by a $3 \epsilon$ approximation argument.

Then set $g = f'$, estimate the integral in question with as

$\int_{\mathbb R}\left|\frac{f(x+t)-f(x)}{t}-f'(x)\right|^2dx = \int_{\mathbb R}\left|\frac{1}{t}\int_0^t \left(f'(x+s)-f'(x)\right) ds \right|^2dx$
$... \le \frac{1}{t} \int_0^t \int_{\mathbb R} |f'(x+s) - f'(x)|^2 dx ds$

and use the obvious argument.


How can I draw all the diagonals of:

[asy]draw(polygon(9));[/asy]
without drawing a bunch of lines?

For loops and expi.

[asy]for(int i=0;i<7;++i){
for(int j=i;j<7;++j){
draw(expi(2*pi*i/7)--expi(2*pi*j/7));
}
}[/asy]

Basically what this code does is it takes all the vertices of the polygon and connects them to all the other vertices (except those it has already connected to). The expi(real r) returns the point 1 unit away from the origin with an angle of r (in radians) from the positive x-axis.

[asy]for(int i=0;i<9;++i){
for(int j=i;j<9;++j){
draw(expi(2*pi*i/9)--expi(2*pi*j/9));
}
}[/asy]

Thanks!


inequalities
Euler
geometry proposed
geometry

Given $\triangle ABC$ inscribed $(O,R)$ with centroid $G$ prove that
$\frac{OG}{2}+R \ge \frac{GA+GB+GC}{3}$

yes you are right too, it can be done stronger:
$GA+GB+GC+\frac{3OG^{2}}{R}\ge 3R$
thank you  :)  :)

Dear gemath,if you have a solution to this inequality post it.I managed to get one but it is not so nice. If you didn't have one you should have told. :D :D

Dear mateivld, I don't understand you so much but it has nice proof:
$R(GA+GB+GC)=\sum OAGA\ge \sum \vec{OA}.\vec{GA}=\sum \vec{OA}.(\vec{GO}+\vec{OA})=(\sum \vec{OA}).\vec{GO}+3R^{2}$
we all know $\sum \vec{OA}=\vec{OH}=3\vec{OG}$ this is result of Euler line .Then we have
$R(\sum GA)\ge-3OG^{2}+3R^{2}\leftrightarrow GA+GB+GC+\frac{3OG^{2}}{R}\ge 3R$
OK??  :D  :D

I also have a solution with tons of calculations wich is not so nice.So I won't post it. :D


calculus logarithms trigonometry derivative function integration algebra

The rule that the derivative of a sum is the sum of the derivatives is far easier to use in practice than the product or quotient rules. Do an algebraic step first, and write the function as the difference of two logarithms before you differentiate it.

$ \ln \left|\frac{\cos x}{\cos x \minus{} 1}\right| \equal{} \ln \left|\frac{1}{1 \minus{} \sec x}\right| \equal{} \minus{} \ln|1 \minus{} \sec x|$

[quote="igiul"]I've re written it about fifteen hundred ways including the one above, which I don't see a formula for.[/quote]

The key to differentiating any logarithmic function is to recognize the whatever the argument is, you're going to end-up with a factor of its reciprocal sitting out front. As for example: $ \frac{d\ln\left|u\right|}{dx} \equal{} \frac{1}{u}\frac{du}{dx}$.

The absolute value bars are an artifact of integration. That is, $ \int_{1}^{u} \frac{d\omega}{\omega} \equal{} \ln\left|u\right|$.

Reduced to its simplest form, the one given by Carcul, above this means:

$ \frac{d}{dx} \ln\left|\frac{\cos x}{\cos x \minus{} 1}\right|\equal{} \minus{}\frac{d}{dx} \ln \left|1\minus{}\sec x\right| \equal{} \minus{}\frac{1}{1\minus{}\sec x}\frac{d}{dx}\left(1\minus{}\sec x\right)$

$ \frac{d}{dx}\left(1\minus{}\sec x\right) \equal{} \minus{}\tan x\, \sec x$.

Therefore, $ \frac{d}{dx} \ln\left|\frac{\cos x}{\cos x \minus{} 1}\right|\equal{}\frac{\tan x\, \sec x}{1 \minus{} \sec x}$.

[quote="Dr. No"]The absolute value bars are an artifact of integration. That is, $ \int_{1}^{u} \frac {d\omega}{\omega} \equal{} \ln\left|u\right|$.[/quote]

Ohhh I love you. I feel foolish though: I was trying to get into the form of one of the integration formulas first. Eventually I realized that I just had to use a chain rule combo and there wasn't anything special about this problem.

I got $ \frac{\sec{x}\tan{x}}{|1\minus{}\sec{x}|}$, which is close.

Tell me, are there any instances in which I should retain the absolute value if its inside the natural log?

The way to look at absolute values: don't cross the boundary. The function we're differentiating is $ \ln\left(\frac{\cos x}{\cos x\minus{}1}\right)\equal{}\ln\left(\frac{\minus{}\cos x}{1\minus{}\cos x}\right)$ where $ \cos x$ is negative and not equal to $ \minus{}1$ and $ \ln\left(\minus{}\frac{\cos x}{\cos x\minus{}1}\right)\equal{}\ln\left(\frac{\cos x}{1\minus{}\cos x}\right)$ where $ \cos x$ is positive and not equal to $ \plus{}1$. As it turns out, both formulas give the same formula for the derivative in this case, so we don't need to write the final answer piecewise in this case. The derivative still blows up at every multiple of $ \frac{\pi}{2}$, and no improper integral of the derivative across those points converges.

Let $ f(x) \equal{} \ln \left|\frac {\cos x}{\cos x \minus{} 1}\right|\equal{}\ln |\cos x|\minus{}\ln |\cos x\minus{}1|$,

$ f'(x) \equal{} \minus{} \frac {\sin x}{\cos x} \plus{} \frac {\sin x}{\cos x \minus{} 1}$

$ \equal{} \minus{} \frac {\sin x}{\cos x(1 \minus{} \cos x)}$

$ \equal{} \minus{} \frac {2\sin \frac {x}{2}\cos \frac {x}{2}}{\cos x\cdot 2\sin ^ 2 \frac {x}{2}}$

$ \equal{} \minus{} \frac {1}{\cos x\tan \frac {x}{2}}$

[quote="igiul"][quote="Dr. No"]The absolute value bars are an artifact of integration. That is, $ \int_{1}^{u} \frac {d\omega}{\omega} \equal{} \ln\left|u\right|$.[/quote]

Ohhh I love you. I feel foolish though: I was trying to get into the form of one of the integration formulas first. Eventually I realized that I just had to use a chain rule combo and there wasn't anything special about this problem.

I got $ \frac {\sec{x}\tan{x}}{|1 \minus{} \sec{x}|}$, which is close.

Tell me, are there any instances in which I should retain the absolute value if its inside the natural log?[/quote]

Let’s consider the general case, as in

Find,   $ \frac{d\log\left|u\left(x\right)\right|}{dx}$

$ \frac{d\log\left|u\left(x\right)\right|}{dx} \equal{} \frac{1}{\left|u\right|}\frac{d\left|u\right|}{dx}$

$ \frac{d\left|u\right|}{dx}\equal{}\frac{d\left|u\right|}{du}\frac{du}{dx}$

$ \frac{d\left|u\right|}{du} \equal{}\frac{u}{\left|u\right|}$

$ \Rightarrow \frac{d\log \left|u\left(x\right)\right|}{dx} \equal{} \frac{1}{\left|u\right|}\frac{u}{\left|u\right|}\frac{du}{dx} \equal{} \frac{u}{u^2}\frac{du}{dx}$

$ \therefore \frac{d\log\left|u\left(x\right)\right|}{dx} \equal{}\frac{1}{u}\frac{du}{dx}$

The answer to your question is, no there isn't such a circumstance. Of course the argument of a logarithm can never be negative anyhow, and that's why the absolute value bars show up when you integrate.

I prefer never putting the absolute value bars in in the first place. They give the false impression that we can integrate across the singularity.
The most general antiderivative of $ \frac1x$ on an interval is $ \ln(kx)$ with $ k\neq 0$; $ k$ is positive if $ x$ is positive on our interval and negative if $ x$ is negative. The absolute value mashes two different antiderivatives ($ k \equal{} 1$ and $ k \equal{} \minus{} 1$) together into one function.

[quote="jmerry"]I prefer never putting the absolute value bars in in the first place. They give the false impression that we can integrate across the singularity.[/quote]

It's your choice. Wether you do or whether you don't is a matter of style really. Further there are circumstances when you might want to calculate the logarithm of a negative number; but the result is a family of complex numbers. ( I have no desire to open that can of worms in a discussion of first semester calculus.)

[quote="jmerry"]The most general antiderivative of $ \frac1x$ on an interval is $ \ln(kx)$ with $ c\neq 0$; $ k$ is positive if $ x$ is positive on our interval and negative if $ x$ is negative. The absolute value mashes two different antiderivatives ($ k \equal{} 1$ and $ k \equal{} \minus{} 1$) together into one function.[/quote]

What you've just described are the constraints that define $ \left|kx\right|$.

But consider, $ \frac{d\ln\left|kx\right|}{dx} \equal{}\frac{d\ln\left|x\right|}{dx} \plus{}\frac{d\ln\left|k\right|}{dx} \equal{} \frac{1}{x}$, since $ \frac{d\ln\left|k\right|}{dx} \equal{} 0$.

If you're going to invoke complex numbers, the form with the absolute value isn't even differentiable anywhere. I'm also willing to use a single value of $ k$ in that case- but there is a branch cut somewhere.

As for the second part- you're missing my point entirely. I know that there are other ways to write the same thing, and I don't care. I'd rather treat absolute values as the piecewise defined functions they are, and acknowledge the very real barrier at $ x\equal{}0$.

Just use $ \ln \left|\frac{b}{a}\right|\equal{}\ln |b|\minus{}\ln |a|$

[quote="jmerry"]If you're going to invoke complex numbers, the form with the absolute value isn't even differentiable anywhere. I'm also willing to use a single value of $ k$ in that case- but there is a branch cut somewhere.[/quote]

I explicitly said that I was not going to do that.

[quote="jmerry"]As for the second part- you're missing my point entirely. I know that there are other ways to write the same thing, and I don't care. I'd rather treat absolute values as the piecewise defined functions they are, and acknowledge the very real barrier at $ x \equal{} 0$.[/quote]

No, I'm not missing your point. 1) The conditions you placed on $ k$ define $ \left|kx\right|$. 2) The antiderivative of $ \frac{1}{x}$ is $ \ln\left|x\right| \plus{}\text{const.}$.  One is free to specify that $ \text{const.} \equal{} \ln \left|k\right|$, but in so doing the argument of the logarithm must be positive; thus the absolute value bars.  Recasting the antiderivative in this way, one obtains $ \ln\left|x\right| \plus{} \ln\left|k\right| \equal{} \ln \left|kx\right|$, and we're right back where we started.

As I said, you're free to include the absolute value bars or not as you see fit, but you do have to recognize that logarithms cannot take negative arguments. Furthermore, $ x \equal{} 0$ is an unavoidable problem, because $ \frac{1}{x}$ is not defined there.

[quote="Dr. No."]The conditions you placed on $ k$ define $ \left|kx\right|$[/quote]They don't, because $ x$ is either always positive or always negative on the interval of definition. I don't cross zero, and there's no need for absolute values. That's the point you're missing.

This is my last post of this argument- we've taken up too much space already talking past each other.

Wow thanks guys. The rest of the HW was no problem.

 :D

[quote="JBL"][i]Edit: I started writing this comment before jmerry posted.[/i]

The most general antiderivative of $ \frac {1}{x}$ over its domain is [i]not[/i] $ \ln |x| \plus{} c$ for arbitrary constant $ c$, it's the function given by $ \ln x \plus{} c_1$ for $ x > 0$ and $ \ln \minus{} x \plus{} c_2$ for $ x < 0$ for arbitrary constants $ c_1, c_2$.  This is closely related to the point jmerry's was making.[/quote]

I concede the point that defining the antiderivative the way you all have excludes zero from the domain a priori. But why is that important? You can't integrate $ \frac{1}{x}$ over zero in any case, because it's undefined there.

I suppose there's no reason to start a new topic for this reply, as it is slightly related to the integration of the natural logarithmic function. Try to bear with me, the following is a result of my over-analytical notes from class.

Consider the following indefinite integral: $ \displaystyle\int \frac {x(x \plus{} 2)}{x^3 \plus{} 3x^2 \minus{} 4}\,dx$
You may notice that the sum of the coefficients in the denominator is zero, so $ x \equal{} 1$ is a solution to the denominator, and also a point at which the expression as a whole is undefined. Performing synthetic division we find that $ x^3 \plus{} 3x^2 \minus{} 4$ factors to $ (x \minus{} 1)(x \plus{} 2)^2$. And that means we can simplify the expression. Thus:

$ \displaystyle\int \frac {x(x \plus{} 2)}{x^3 \plus{} 3x^2 \minus{} 4}\,dx \equal{} \int \frac {x}{x^2 \plus{} x \minus{} 2}\,dx$

Here's the rub: the first expression can be integrated, whereas the second cannot. (By methods at the disposal of a calculus I student at my level, although I haven't really considered expressing the numerator in terms of $ u$, which would be the quadratic formula and no fun at all. I don't even know if that would work.) What I find to be peculiar about this is that we effectively did nothing more that pull out a one, yet the result cripples our ability to integrate.

This leads me to question whether or not the antiderivative of these two cases would be the 'same' at all? And if they are the same (which I doubt), can a reversal of this property be implemented to stabilize other integrals? The method by which one would do such a thing seems so excessive for the task that I doubt it would be of any use...

But it would probably be stunningly elegant.

EDIT: Thank you   :lol:

P.S. (Is the simplified form of this integral 'improper'? If not, what is it called when you can't integrate an expression?)

Actually, it can be integrated. All you need is the algebraic technique known as partial fractions: $ \frac{x}{x^2\plus{}x\minus{}2}\equal{}\frac23\frac{1}{x\plus{}2}\plus{}\frac13\frac{1}{x\minus{}1}$.

If you work through it, the antiderivative you get is the same as the original form.

[quote="igiul"]I suppose there's no reason to start a new topic for this reply, as it is slightly related to the integration of the natural logarithmic function. Try to bear with me, the following is a result of my over-analytical notes from class.

Consider the following indefinite integral: $ \displaystyle\int \frac {x(x \plus{} 2)}{x^3 \plus{} 3x^2 \minus{} 4}\,dx$
You may notice that the sum of the coefficients in the denominator is zero, so $ x \equal{} 1$ is a solution to the denominator, and also a point at which the expression as a whole is undefined. Performing synthetic division we find that $ x^3 \plus{} 3x^2 \minus{} 4$ factors to $ (x \minus{} 1)(x \plus{} 2)^2$. And that means we can simplify the expression. Thus:

$ \displaystyle\int \frac {x(x \plus{} 2)}{x^3 \plus{} 3x^2 \minus{} 4}\,dx \equal{} \int \frac {x}{x^2 \plus{} x \minus{} 2}\,dx$

Here's the rub: the first expression can be integrated, whereas the second cannot. (By methods at the disposal of a calculus I student at my level, although I haven't really considered expressing the numerator in terms of $ u$, which would be the quadratic formula and no fun at all. I don't even know if that would work.) What I find to be peculiar about this is that we effectively did nothing more that pull out a one, yet the result cripples our ability to integrate.

...

But it would probably be stunningly elegant.

EDIT: Thank you   :lol:

P.S. (Is the simplified form of this integral 'improper'? If not, what is it called when you can't integrate an expression?)[/quote]

Consider:

$ \int \frac {x(x \plus{} 2)}{x^3 \plus{} 3x^2 \minus{} 4}\,dx$

This can be easily integrated by a technique called "u substitution."

My old prof would advise, "To do u substitution, 'pick the ugliest part of the integral, and that's u'."  :)

In this case, let $ u \equal{} x^3 \plus{} 3x^2 \minus{} 4$.

 $ du \equal{} \left(3x^2 \plus{} 6x\right) dx \Rightarrow \int \frac {x(x \plus{} 2)}{x^3 \plus{} 3x^2 \minus{} 4}\,dx \equal{} \frac {1}{3}\int \frac {du}{u} \equal{} \frac {1}{3}\left(\ln\left|u\right| \plus{} C\right) \equal{} \frac {1}{3}\left(\ln\left|x^3 \plus{} 3x^2 \minus{} 4\right| \plus{} C\right)$

[quote="igiul"]Consider the following indefinite integral: $ \displaystyle\int \frac {x(x \plus{} 2)}{x^3 \plus{} 3x^2 \minus{} 4}\,dx$[/quote]

Just use $ \boxed{\int \frac{f'(x)}{f(x)}dx\equal{}\ln |f(x)|\plus{}Const.}$

Does anyone actually read the [b]whole[/b] post being responded to? That's the part igiul knew how to do already.

umm wow.  :oops:
I didn't input the numerical value of x when solving for A and B in the partial fraction decomposition. It's in my notes and everything lol.

sad.


limit
integration
trigonometry
calculus
real analysis
real analysis unsolved

${ \lim_{x\to0} \frac { \int_{0}^{x} \frac {t^2}{\sqrt{a+t}}\, dt }{bx-sinx}}=1$

Can i know how to find the constants a and b?

How I see it, your limit doesn't depend on values $ a,b$  (but I haven't checked ) so I see no point in finding them. Or did you want to ask: "for which $ a,b$ does this limit converge"?

Expand in power series- at least the first couple terms. There are two cases for $ a$ ($ a\equal{}0?$) and two for $ b$ ($ b\equal{}1?$).

We have:
$ \int{\frac {t^2}{\sqrt {a \plus{} t}}dt} \equal{} \frac {2}{15}\sqrt {a \plus{} t}\left(8a^2 \minus{} 4at \plus{} 3t^2\right) \plus{} C$
Now:
$ I \equal{} \int_{0}^{x}{\frac {t^2}{\sqrt {a \plus{} t}}dt} \equal{} \frac {2}{15}\sqrt {a \plus{} x}\left(8a^2 \minus{} 4ax \plus{} 3x^2\right) \minus{} \frac {2}{15}\sqrt {a}\cdot 8a^2$
Over $ x\equal{}0$ we have:
$ I \equal{} \frac {1}{3\sqrt {a}}x^3 \plus{} \frac {1}{8a^{\frac {3}{2}}}x^4 \plus{} \frac {3}{40a^{\frac {5}{2}}}x^5 \plus{} O(x^6)$
We know that:
$ \sin(x) \equal{} x \minus{} \frac {1}{6}x^3 \plus{} \frac {1}{120}x^5 \plus{} O(x^6)$
So we have:
$ \lim_{x\to 0}{\frac {\frac {1}{3\sqrt {a}}x^3 \plus{} \frac {1}{8a^{\frac {3}{2}}}x^4 \plus{} \frac {3}{40a^{\frac {5}{2}}}x^5 \plus{} O(x^6)}{(b \minus{} 1)x \plus{} \frac {1}{6}x^3 \minus{} \frac {1}{120}x^5 \plus{} O(x^6)}} \equal{} 1$
We must have here: $ b \equal{} 1$ and $ a \equal{} 4$ :)

[quote="jmerry"]That's not promising- it obscures what's really going on behind a wall of algebra. We're only interested in the limiting behavior at zero, not the exact values away from zero.[/quote]
I`ve been edited my post... In old I tried to show the same one...

Just a stylistic note: When you're using power series in this way to solve a problem, try using fewer terms rather than more. For instance, write $ bx\minus{}\sin x\equal{}(b\minus{}1)x\plus{}\frac16x^3\plus{}O(x^5)$ rather than include as much detail as you did. And I would have simply written $ \frac{t^2}{\sqrt{a\plus{}t}}$ as $ \frac{t^2}{\sqrt{a}}\plus{}O(t^3)$ and hence its integral as $ \frac{x^3}{3\sqrt{a}}\plus{}O(x^4).$

If you try this and discover that you need a term that you hadn't gotten to yet, then it's no big deal to go back and include it - in particular, your previous work should still stand.

More substantively: you have tacitly assumed that $ a\ne0.$ To make the argument complete, you should go back to consider what happens if $ a\equal{}0.$

[quote="fedja"][quote="learn3r"]
And if i want to define the $ O$ inside $ O(x^6)$, can it be any real number?
[/quote]
This question makes me suspect that you are not familiar with $ O$ and $ o$ notation. If it is, indeed, the case, read Kent's sticky in Calculus Computations and Tutorials :).[/quote]

yup, i am not familiar with it. Never learn about it before. Anyway, i am reading the sticky, it is very helpful, thanks alot. :)  :)

[quote="learn3r"][quote="Yuriy Solovyov"]We have:
$ \int{\frac {t^2}{\sqrt {a \plus{} t}}dt} \equal{} \frac {2}{15}\sqrt {a \plus{} t}\left(8a^2 \minus{} 4at \plus{} 3t^2\right) \plus{} C$
Now:
$ I \equal{} \int_{0}^{x}{\frac {t^2}{\sqrt {a \plus{} t}}dt} \equal{} \frac {2}{15}\sqrt {a \plus{} x}\left(8a^2 \minus{} 4ax \plus{} 3x^2\right) \minus{} \frac {2}{15}\sqrt {a}\cdot 8a^2$
Over $ x \equal{} 0$ we have:
$ I \equal{} \frac {1}{3\sqrt {a}}x^3 \plus{} \frac {1}{8a^{\frac {3}{2}}}x^4 \plus{} \frac {3}{40a^{\frac {5}{2}}}x^5 \plus{} O(x^6)$
We know that:
$ \sin(x) \equal{} x \minus{} \frac {1}{6}x^3 \plus{} \frac {1}{120}x^5 \plus{} O(x^6)$
So we have:
$ \lim_{x\to 0}{\frac {\frac {1}{3\sqrt {a}}x^3 \plus{} \frac {1}{8a^{\frac {3}{2}}}x^4 \plus{} \frac {3}{40a^{\frac {5}{2}}}x^5 \plus{} O(x^6)}{(b \minus{} 1)x \plus{} \frac {1}{6}x^3 \minus{} \frac {1}{120}x^5 \plus{} O(x^6)}} \equal{} 1$
We must have here: $ b \equal{} 1$ and $ a \equal{} 4$ :)[/quote]

I am sorry that I am a bit slow :blush: ....can i know how you get this>>>

Over $ x \equal{} 0$ we have:
$ I \equal{} \frac {1}{3\sqrt {a}}x^3 \plus{} \frac {1}{8a^{\frac {3}{2}}}x^4 \plus{} \frac {3}{40a^{\frac {5}{2}}}x^5 \plus{} O(x^6)$

And if i want to define the $ O$ inside $ O(x^6)$, can it be any real number? What should i write if i prefer not to use this notation?

Thanks a lot for your help. Or else I am still puzzle upon the question![/quote]

We know the Taylor`s formula:
$ f(t) \equal{} f(0) \plus{} \frac {f^{(1)}(0)}{1!}\cdot t \plus{} \frac {f^{(2)}(0)}{2!}\cdot t^2 \plus{} \frac {f^{(3)}(0)}{3!}\cdot t^3 \plus{} ... \plus{} \frac {f^{(n)}(0)}{n!}\cdot t^n \plus{} ...$
Let we put here $ f(t) \equal{} \frac {1}{\sqrt {a \plus{} t}}$. We will have:
$ \frac {1}{\sqrt {a \plus{} t}} \equal{} \frac {1}{\sqrt {a}} \plus{} O(t)$

and

$ \frac {t^2}{\sqrt {a \plus{} t}} \equal{} \frac {t^2}{\sqrt {a}} \plus{} O(t^3)$

and

$ \int{\frac {t^2}{\sqrt {a \plus{} t}}dt} \equal{} \frac {t^3}{3\sqrt {a}} \plus{} O(t^4)$

 :)

If $ b\ne 1$ then our limit is $ 0$ because $ \lim_{x\to 0}\frac {x^3}{x} \equal{} 0$

If $ a \equal{} 0$ then $ \int_{0}^{x}{t^{\frac {3}{2}}dt} \equal{} \frac {2}{5}x^{\frac {5}{2}}$ and if $ b\ne 1$ limit is $ 0$, else limit is $ \infty$

[quote="learn3r"][color=darkred]${ \lim_{x\to 0} \frac {\int_{0}^{x} \frac {t^2}{\sqrt [p]{a + t^r}}\ \mathrm {dt} }{bx - \sin x}} = 1$, where $ r>0$ , $ p\in\mathcal N$ , $ p\ge 2$ .

Find the constants $ a > 0$ and $ b\ne 0$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Show easily that $ \lim_{x\to 0}\frac {bx - \sin x}{x^3} = \{\begin{array}{ccc} - \infty & \mathrm {if} & 0\ne b < 1 \\
 \\
\frac 16 & \mathrm {if} & b = 1 \\
 \\
\infty & \mathrm {if} & b > 1\end{array}$ ,

${{ \lim_{x\to 0} \frac {\int_{0}^{x} \frac {t^2}{\sqrt [p]{a + t^r}}\ \mathrm {dt} }{x^3}}\ \stackrel {l'H}{\ = = = \ }\ \lim_{x\to 0} \frac {\frac {x^2}{\sqrt [p] {a + x^r}} }{3x^2}} = \frac {1}{3\sqrt [p]a}$ .

${ \lim_{x\to 0} \frac {\int_{0}^{x} \frac {t^2}{\sqrt [p]{a + t^r}}\ \mathrm {dt} }{bx - \sin x}} =$ ${ \lim_{x\to 0} \frac {\int_{0}^{x} \frac {t^2}{\sqrt [p]{a + t^r}}\ \mathrm {dt} }{x^3}\cdot\frac {x^3}{bx - \sin x}} =$ $ \{\begin{array}{ccc} 0 & \mathrm {if} & b\not\in \{0, 1\} \\
 \\
\frac {2}{\sqrt [p]a} & \mathrm {if} & b = 1\end{array}$ .

In conclusion, $ \boxed {\ b = 1\ ,\ a = 2^p\ }$ .[/color]


analytic geometry
geometry proposed
geometry

I solved this problem but only with coordinates (it really wasn't that bad).  Could someone offer another solution? I think it's a well-known problem.

Say we have triangle ABC, and D the foot of the altitude from A to BC.  Let P be any point on AD.  Say BP meets AC at E and CP meets AB at F.  Then, angle FBP = angle EBP.

Try this post:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6894]www.mathlinks.ro/Forum/viewtopic.php?t=6894[/url]


integration
logarithms
real analysis
real analysis solved

$x^{-x} = e^{-x\ln x}$, expand it, we get
$x^{-x} = \sum_{n\geq 0} \frac{(-1)^n x^n \ln^n x}{n!}$
$\int x^{-x}dx = \int_0^1 \left(\sum_{n\geq 0}\frac{(-1)^n x^n \ln^n x}{n!}\right)dx = \sum_{n\geq 0}\int_0^1 \frac{(-1)^n x^n \ln^n x}{n!}dx$
We can integrate by terms, because $|x\ln x|\leq \frac 1e$ and $\left|\sum \frac{(-1)^n x^n \ln^n x}{n!}\right| \leq \sum \frac{\left(\frac{1}{e}\right)^n}{n!}$ which is convergent.

Thus
$\int x^{-x}dx = \sum_{n\geq 0}\frac{(-1)^n \int_0^1 x^n \ln^n xdx}{n!}$.

We know that $\int_0^1 x^n \ln^n xdx = (-1)^n \frac{n!}{(n+1)^{n+1}}$ thus the conclusion follows easily.


[quote="apollo"]Let x, y, and z be positive real numbers with x + y + z = 1. Prove that
$ \sum_{cyc}\frac{xy}{\sqrt{xy\plus{}yz}}\leq\frac{\sqrt{2}}{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88439

Hey you friend! Of course, you can prove it easily because a very similiar problem was in IMO 2001. Only you need to do is to open and look into the solution of it. Now I'm giving you a link to reach there easy.
[url]http://www.kalva.demon.co.uk/imo/isoln/isoln012.html[/url]


group theory
abstract algebra
superior algebra
superior algebra unsolved

i got this question for my abstract algebra exam, and i have absolutely no clue.

if we define Z(p^inf)={a/b in Q/Z| a,b E Z and b=p^k,  k>=0, p prime}

its easy to show that Z(p^inf) is an additive subgroup of Q/Z, my problem lies in

1) showing that Z(p^inf) is an infinite subgroup
2) if x is in Z(p^inf) then |x| is finite
3) if H is a proper subgroup of Z(p^inf), then H is cyclic

these questions are already past due, so this is just out of curiousity.  i recognize that the notation is somewhat unique, so if theres a name for this subgroup that would be great i could do some research on my own.

thanks

http://en.wikipedia.org/wiki/Pr%C3%BCfer_group
for the proofs, do you know german? ;-)
http://maddin.110mb.com/pdf/artikel_neu.pdf


Tonight Simon Rubinstein-Salzedo will host a Math Jam on the subject of complex numbers.  This Math Jam should be particularly helpful to those of you thinking about enrolling in the Complex Numbers/Trigonometry course.

The Math Jam will take place at 7:30PM ET/4:30PM PT and should last around an hour.

We hope to see you there!

Not able to view your tonight's transcript.

The transcript for some reason cant be seen...

It takes at least a few hours to prepare a Math Jam transcript, so sometimes the transcripts aren't posted and publicly viewable until a few days later. Read the old ones while you are waiting.

We appreciate your enthusiasm.  The transcript is complete but there is no one still in the office tonight who can upload it so you'll need to wait until the morning.  In the meantime, i'll echo tokenadult's suggestion..

-Tim

There was a small problem with the transcript file that I did not notice last night.  The transcript is now posted.


geometry
geometric transformation
rotation
rectangle
MATHCOUNTS
3D geometry

What is the maximum number of rectangular blocks 8" X 4" X 4" that will fit in a rectangular container 12" X 18" that is 12" deep?

[hide]
Ok I am not sure about this, but the volume of blocks is $8\times4\times4=128$.
The volume of container is
$12\times18\times12=2592$.
Dividing gives 20.25 so 20.[/hide]

[hide]I did it the same way and also got 20.[/hide]

[hide="I got not 20"]18. A block of 2 by 12 by 12 is left empty.[/hide]

That's assuming they all lie flat. I see no reason why they can't be arranged differently to fit more.

The answer was [hide]20[/hide] but I still don't see why it isn't [hide]18[/hide].

I didn't think you could fit [hide]20[/hide] in by packing, even the best packing would still only give [hide]18[/hide] because there are always two left over in the "18" dimention if you try to fill it with 4's and 8's.

The question didn't say that you can't chop them up...
For those types of question, just put total volume/volume of block if you are dealing with rectangles, blocks etc...

But how do you know that it will always work?

It usually does work well, but I'm always iffy about assuming this type of stuff.

I think they just assume that you turn them around and stuff.

I tried that, I still got "you know what".

The answer was [hide]20[/hide].
I checked the official mathcounts answer.
It's called rotating the blocks around.

Can someone describe the rotations please?

I doubt anyone here can prove that you can fit 20, you just have to assume..

Can you just assume?  What if I gave you 2x4x4 boxes to fit in a 3x5x5 container?  Dividing gives $2.34375$, but it's clear that you can only fit 1.

That's a different situation. With the larger dimensions always found in these problems, you have much larger areas of unused space, so it's obvious that you can rotate them around to fit more. With your example, the dimensions are just too small, and also the gap between the actual number and the divided volume is smaller. There isn't enough excess space.

There seems to be a lot of debate about this one.  This is a "packing" problem.  In general, packing problems are hard.  Except for some special cases, there usually isn't a formula to tell you how many objects of Shape A will fit into a box of Shape B.

For this problem, the dimensions of the object (8x4x4) are all multiples of 4.  But the dimensions of the box (12x18x12) are not all multiples of 4.  Based on the volume of the box (2592 cu in) and the volume of the object (128), we know we cannot fit more than 20.25 objects in the box.  If all objects are laid squarely in the box, we know we will fill a volume of 12x16x12 with 18 blocks.

These links show the best known packings of [url=http://www.stetson.edu/~efriedma/cubincub/]cubes inside cubes[/url] and [url=http://www.stetson.edu/~efriedma/squinsqu/]squares inside squares[/url].  Some of the best known packings in our size range (10 to 40 objects) waste 20% to 40% of the space!

If we were to fit 20 of our 8x4x4 objects into our 12x18x12 box, we would have about 1.25% wasted volume.  If we were to fit 19 of our objects into our box, we would have about 6% wasted volume.

I don't know the answer to this one, but I would be *very surprised* if you could fit 20 objects in our box.  What is the source of the problem?  Is it possible there is an errata to the answer key?

You CANNOT fit in more than 18 if the faces of the blocks are parallel to faces of the box.

The dimensions of the blocks are mutiples of 4.  However, the box has one edge of length 18.  Thus, you can only "fill up" 16 of that 18, and will have 2 left over at the end (you can't possible fill up that space, because the blocks are quantized into 4's).  We then get 18 blocks.

To make it easier, you could think of it as 1 x 1 x 2 blocks in a 3 x 3 x 4.5 box.  That 0.5 clearly is going nowhere.

However, the faces of the blocks are not required to be parallel to the faces of the box.

PLEASE:

GIVE REASONS!

An answer is near worthless without how you got to that answer, especially when you got an answer already given, thanks!


Does anyone have a good book that can teach me intermediate to olympiad level number thoery and summations? Thanks.

Dude, i think the book " $Introduction$ $to$ $Number$ $Theory$" from W.W.Adams & L.J.Goldstein would be nice. :)

An Introduction to the Theory of Numbers-G.H.hardy,E.M.wright. I found it in emule and it looks good.

The Theory of Numbers involves some calculus though (like the proof that pi is irrational)

i know calculus. i went out and bought those books. thx


function
algebra
polynomial
quadratics
real analysis
real analysis solved

Find all derivable functions $f:R->R$ such that $f(x)-f(y)=(x-y)f'(\frac {x+y}{2})$.

I think $f(x)=Cx^2$, where $C$ is arbitrary real number.

Differentiate wrt $x$ to get $g(x)=g(\frac{x+y}2)+\frac{x-y}2\cdot g'(\frac{x+y}2)$, where $g=f'$. We get, after replacing $\frac{x+y}2$ with $y$ for simplicity, $g(x)-g(y)=(x-y)g'(y)$. We also get $g(y)-g(x)=(y-x)g'(x)$, and this easily yields $g(x)=g(y),\ \forall x,y\in\mathbb R$, so $g'=f''$ is a constant, meaning that $f$ is a second degree polynomial. Conversely, every such polynomial verifies the equation.

The only such functions are quadratic polynomials. In the proof below, I assumed that $f$ is twice differentiable; I suspect that the proof could be modified to not use that assumption, but I don't have that modification at the moment.

For any $a$ and $x, f(a+x)-f(a-x)=2xf'(a).$

Differentiate that with respect to $x$:

$f'(a+x)+f'(a-x)=2f'(a).$

Go back and differentiate with respect to $a$:

$f'(a+x)-f'(a-x)=2xf''(a).$

Add these two equations and divide by 2:

$f'(a+x)=f'(a)+xf''(a).$

Let $a=0$ and integrate this with respect to $x$:

$f(x)=f(0)+xf'(0)+ \frac{x^2}2f''(0).$

That's an arbitrary quadratic polynomial.

I got interrupted while posting and didn't see grobber's post. But both grobber and I did assume twice differentiable.

@Kent. We have $f'(x)=\frac{f(x+1)-f(x-1)}{2}$ and $\frac{f(x+1)-f(x-1)}{2}$ is differentiable, so $f'(x)$ is differentiable too ;)


What do you prefere: Today's system or globalization?
Are you happy living in your country and not caring about the rest or would you rather live on planet and nation Earth where everybody speaks the same language?
Do you think globalization will ever overcome the cultural differences?

nearly everyone in China is learning English for going abroad. In the universities, the most important part except your subject is English.... Everyone seems to be crazy... GRE, TOEFL, IETLS(it seems so) make everyone busy reciting words and doing exercises.... Even some kids begin to learn English at the age of three or four...

I like the idea of everyone to understand a certain language. for that matter, at this point, english seems the best choice, because

1) already a lot of people ( 1/4 of the globe's population ) speak english
2) it is the most teached language, and the most language teachers in non-english countries are english teachers.
3) it's very easy to learn (this is subjective).

I think English is important for all countries. If a million years from now (or MUCH earlier  :D) everybody will speak the same language all over the world, I think it will be derived from English.

However, I think China is somewhat of a special case. I don't know that much about this, but I think that its writing system makes it a not very flexible language, and flexibility is required because of our growing need for communication. If it weren't for this, I think Chinese would make the perfect candidate for a global language (it's spoken by 1/5 of the entire world!!!  :D ).

And finally, to come to the point of this topic, I think there will come a time when will all be part of one BIG country (well, actually, it makes no sense calling it a country). That's mainly because I'm a big SF fan, and all my favourite writers (Asimov, Arthur C. Clarke, Larry Niven etc.) speak of Empire Earth, he universal gallactic language and so on  :D .

yes. the writing of Chinese is a problem. (also its pronunciation to Westerners)
But its grammar is comparatively easy.

Yes, I've heard about that, but English grammar is pretty easy too compared to other languages. Once more I must say that I'm no expert, but I think Latin languages (including Romanian  :D ) have difficult grammar. I know French is tough (I take that in school) and I know students who study to be teachers who have told me that Italian grammar is a nightmare.

Anyway, the problem with Chinese is more than its writing style, because the writing style of a language also influences the kind of ideas you can express in that language. I read in a book (the writer is Romanian, but he studied Japanese and Chinese and he's a very talented haiku writer! :D ) that it's hard to express general notions in Chinese and that when Chinese say 'old man' (for example) they think of a particular old man, and not the general notion of being old and stuff. I can't explain this very well, because, of course, I don't know what the writer is talking about (or if he's right, for that matter).

European languages have `cases'. You need to change the suffix of a word. There's no such problem in Chinese. We always write the same characters... We use some words in sentences to express the tense.

Yup, and when I mentioned languages with tough grammar, I completely forgot about Romanian! I've met foreigners who say that it's awful :D Frankly, sometimes I think that's true, even if it's my native language ... :D

Chinese is a typical language of analytic/isolating languages: There are no endings - words are not inflected; all grammatical relationships are shown through the use of word order and/or independent grammatical units (particles, words).

But most European languages appear to be inflectional languages: grammatical relationships are expressed by changing the internal structure of the words through inflections; inflectional endings typically express several grammatical meanings at once.

But English is generally considered to be an isolating language(so its grammar is easy and it's somewhat like Chinese) with some inflectional features.

I don't know much about chinese grammar, but I do know (hope i'm right) that every word has a symbol of its own. Thus it is very very hard to learn how to write. From this point of view Romanian (which is the only phonetical language in the world!) and somewhat English are much better.

Wow! Romanian is the ONLY phonetical language in the world? I'm sooo proud! :D That's pretty cool, isn't it? If an universal language ever exists, I'm sure it'll have some roots in today's Latin languages, even though I maintain my opinion that English wll definitely be the universal language's main ancestor.

I agree that English should become the universal language. It is very important that as many people as possible know one language. Particularly in research topics some results were rediscovered independently as scientists have difficulties reading the research topics from other countries involving languages he is not capable of.
Learning various languages provides you with the opportunity to learn more about the culture, feelings and attitudes of a certain countries. Especially their history gives us the chance to learn why certain matters works this way today. If all only know English as a foreign languages interest in other cultures might be reduced. But on the other hand
I can talk to others via electronic devices and get information from people and not only books. There are many people who speak Russian. I feel it should be taken into consideration to establish it as a universal language. What do the Russian guys Fedor, myth and Alexander (guas) think ? Or maybe German ?

Russian is used in UN as a working language.
but French is said to be the most accurate language in the world, that's why many treaty articles are written in French.

I was impressed when I first heard that efforts have been made to create a universal language. It is called Esperanto, or at least that's what I've heard. I don't really know more about the subject, but the fact that people did attempt to do this is quite impressive in itself. You can learn more here:[url]

http://www.esperanto-usa.org/

[/url]

This site is from the USA, but many countries have websites dedicated to this language.

well Romanian isn't the only phonetic language in the world anylonger :D:D:D

[quote]In Esperanto, every letter has just one sound; every sound has just one letter representing it. The emphasis is always on the next-to-last syllable. So once you learn the alphabet, you can pronounce anything you read, and you can spell anything you hear.[/quote]

But considering that Esperanto, much as Klingonian, is an artificial language, I'm still going to use the "ONLY ONE" adjective :D

there was a joke:
[quote]
Having chosen English as the preferred language in the EEC, the European Parliament has commissioned a feasibility study in ways of improving efficiency in communications between Government departments.

European officials have often pointed out the English spelling is unnecessarily difficult; for example: cough, plough, rough, through and thorough. What is clearly needed is a phased programme of changes to iron out these anomolies. The programme would, of course, be administered by a committee staff at top level by participating nations.

In the first year, for example, the committee would suggest using "s" instead of the soft "c." Sertainly, sivil servants in all sities would resieve this news with joy. Then the hard "c" could be replaced by "k" sinse both letters are pronounsed alike. Not only would this klear up konfusion in the minds of klerikal workers, but typewriters kould be made with one less letter.

There would be growing enthusiasm when in the sekond year, it would be announsed that the troublesome "ph" would henseforth be written "f." This would make words like "fotograf" twenty persent shorter in print.

In the third year, publik akseptanse of the new spelling kan be expekted to reash the stage where more komplikated shanges are possible.Governments would enkourage the removal of double leters whish have always been a deterent to akurate speling.

We would al agre that the horible mes of silent "e"s in the languag is disgrasful. Therefor we kould drop them and kontinu to read and writ as though nothing had hapend. By this tim it would be four years sins the skem began and peopl would be reseptive to steps sutsh as replasing "th" by "z." Perhaps zen ze funktion of "w" kould be taken on by "v," vitsh is, after al, half a "w." Shortly after zis, ze unesesary "o" kould be dropd from vords kontaining "ou." Similar arguments vud of kors be aplid to ozer kombinations of leters. Kontinuing zis proses yer after yer, ve vud eventuli hav a reli sensibl riten styl. After tventi yers zer vud be no mor trubls, difikultis and evrivun vud find it ezi tu understand ech ozer. Ze drems of ze Guvermnt vud finali have kum tru.
[/quote]

It's hard to introdius sach a language, becaz pipal wud hav to stop lafing first!  :D  :D  :D

Albanian is very phonetic. Each letter has a specific sound and if you know how to pronounce each letter, then you can pronounce anything.
As well, I think English is much harder to learn than some European languages. I mean, I found Italian and French extremely easy to learn compared to English. They have a lot of rules which you can follow, whereas English has a lot of irregularities in pronouciation and stuff. Of course, there are exceptions in every language.

[quote="grobber"]Italian grammar is a nightmare.[/quote]
Oh, really? That's why when I have nightmares, all the people in my dream talk Italian.   :?

I don't really no much about Italian grammar, so what I said could be way off, but that's what several friends of mine who study Italian at the university told me, so I just assumed they knew what they were talking about :).

well, in fact italian grammar it's a real nightmare!
i can just tell you that there's not a fixed way to make past participles :)
they are all irregular! and then there are plenties of different tenses...
but i guess it's another phonetic language, because all signs have a different pronouciation... except for "gl", "gn" and some other...

[quote="ma_go"]but i guess it's another phonetic language, because all signs have a different pronouciation... except for "gl", "gn" and some other...[/quote]
He meant that any symbol has a fixed pronunciation, no matter what other symbols are around it.
So sorry about him...   :blush:

 ;)     :P

indeed, italian is the sister language of romanian :) I never learnt italian, but I can understand what an italian is saying many times :D

in my opinion English is the easiest to learn language, because you can practice it a lot ...

[quote="grobber"]Well, an artificial language might be exactly what we need. If we don't develop fast enough to meet our growing need for communication, we might have to 'speed up' evolution and nature and develop such an
artificial language. What puzzles me is this: if Esperanto is such an easy to learn language and it's so efficient and all that stuff written on that website, why haven't I heard of it being used in international political debates and other things of this sort? I had heard of Esperanto before and I did a Google search to find that website; it's not a famous website which everybody knows about or anything like that.

I think human kind isn't really ready for global communication and only a few realize the huge advantages which come with using a universal language. This means that Esperanto isn't a universal language (yet), but just a project. We'll have to wait much longer to see this dream come true: universal comnmunication.[/quote]

The reason that Esperanto never caught on is because languages are VERY cultural. Languages tie people to their history and to their culture. Esperanto has no cultural history. It is very very hard, if not impossible, to get entire bodies of people to voluntarily leave their language for another. Languages do die however. I forget how many, but there are quite a few languages that go extinct each year. Mostly this happens to languages that are spoken by people who have been defeated in war or oppressed, and the winners or oppressors actively try to extinguish the language.

But let us not forget that there once was a "universal language". At first it was Greek, and then later it was Latin. Latin was used for a long time as the universal language of communication for scientific and intellectual ideas throughout Europe. The problem was, that it was not spoken by people from day to day; so only the well educated were able to learn it. This left the rest of the people virtually illiterate to intellectual ideas.

Fortunately, people now have much more access to education than they did back then. But it would still be impossible to get everyone to accept any one language as their own "home" language. Because even if you could - languages evolve and spread like viruses. A language may begin with one strain, but eventually it will branch. First it will be as different dialects of the same language and then it will evolve or fuse with other languages into an entirely different language just as Latin did.

You cannot stop this from happening. People have a habit of making languages their own. This is because languages are intrinsicly cultural and belong to people who share a common culture.

A better idea, may be to develop very efficient universal translators. That way, any person can read any document in their own language. Of course, there are many obvious problems with this. But I believe it has some practicality.

In the end, if America remains such a strong cultural and economic influence, it will probably be that English is used by even more scientific journals. But it will never become the universal "home language" of all people in the world. (Not unless we conquer the world and force it!).  But I think the availability and practicality of universal translators will also increase.

[quote="liyi"][quote="Valentin Vornicu"]well Romanian isn't the only phonetic language in the world anylonger :D:D:D
[/quote]
I don't know if some east asian langauges are phonetic, such as Chinese, Japanese and Korean. Each letter in these languages have a fixed sound and you can read the whole passage if you know how to read the letters although you don't understand the passage at all.[/quote]
Japanese is phonetic, while Chinese, not being phonetic, could be considered partly phonetic in some sense (the pronounication of each Chinese character is sometimes similar to the symbols consisting that character)

I've learnt both English and Chinese in school, and speak both naturally in my daily life. So it may come as no surprise that I'm on the side of culture; I don't support a worldwide language, neither do I think it likely that one will come into being anytime soon.

As gauss202 noted, "Languages tie people to their history and to their culture." I agree. It's impossible for any language to capture all the nuances built into the many languages around the world. In fact, as cultures fragment, what "common tongues" peoples do have tend to become dialects, absorbing peculiarities of the surrounding society or subgroup. Science fiction is still fiction: any language simple enough for most people to use fluently can never be wide enough to represent even a majority of cultures.

Having said that, I can understand easily how it would be more convenient for certain groups of people to communicate with a common language. The only way round this, I suppose, would be to isolate commonalities and build a jargon bridging languages. A new notation, so to speak. Since this would be for narrow and specific areas like organic chemical engineering, say, there wouldn't be as much cultural baggage to discard.

Comments would be appreciated =) or if you have a problem with what I suggested, please say so.


Let H,K,N be nontrivial normal subgroups of a group G and suppose $ G \equal{} H \times K$. Prove that N is in the center of G or N intersects one of H,K nontrivially. Give examples to show that both possibilities can actually occur when G is nonabelian. (Hungerford 1.8.7)

Attempted Solution:

So, I was going to assume that the $ H \cap  N \equal{} K \cap N \equal{} \{e\}$ and try to prove that N is in the center. If I could show G is the internal weak direct product of some subset of those subgroups, then there is a theorem in that section that might help... BTW, do we know that $ H \cap K \equal{} \{e\}$? It seems like that should be true...

Yes, we know that $ H\cap K\equal{}{e}$.  The condition $ G\equal{}H\times K$ can be defined as: $ H$ and $ K$ commute, their intersection is trivial, and $ HK\equal{}G$.

It is enough to show that if $ N$ trivially intersects $ H$, then $ N$ commutes with $ H$.

Choose $ n\in N$, $ h\in H$, and let $ m\equal{}h^{\minus{}1}n h$.  We wish to show that $ m\equal{}n$.  Write $ n\equal{}ab$ with $ a\in H$, $ b\in K$.  Then $ m\equal{}h^{\minus{}1} a b h\equal{}h^{\minus{}1} a b h \equal{} h^{\minus{}1} a h (h^{\minus{}1} b h) \equal{} h^{\minus{}1} a h b$.  But then $ mn^{\minus{}1} \equal{} h^{\minus{}1} a h a^{\minus{}1} \in H \cap N$, so $ mn^{\minus{}1}\equal{}e$.

Haven't found such a group with such a subgroup in the center, but the other example is easy: $ A_3 \times \{e\} \lhd S_3 \times Z_2$.

[quote="LydianRain"]
It is enough to show that if $ N$ trivially intersects $ H$, then $ N$ commutes with $ H$.

[/quote]

Sorry, why is that enough? We need to show ng=gn for all g in G and n in N. We can write g = hk according to your observation that HK = G but I still don't see how that is enough.

Repeating an identical argument as above, $ N$ commutes with $ K$ also.  And if $ N$ commutes with $ H$ and $ K$, then it commutes with $ HK$.

Oh, and here's an example for the other case.  Let $ G\equal{}D_8 \times Z_2$ where $ D_8\equal{}\langle r,s \rangle$ is the (nonabelian) dihedral group of order $ 8$ and $ Z_2\equal{}\langle x\rangle$.  Then $ \langle r^2 \rangle \lhd D_8$, so $ \langle (r^2, x) \rangle \lhd G$ is a normal subgroup of order $ 2$ in the center of $ G$ that intersects both $ H$ and $ K$ nontrivially.

I see.

When you say "N commutes with K" do you mean that every element in N commutes with every element in K or only that $ n \in N$ and $ k \in K \Rightarrow \exists n' \in N$ and $ k' \in K$ so that nk = k'n' ?

The former--sorry for the lack of clarity.


Hi hope its ok to post here I have a math problem, please can someone help?
I have spent several days on it but just cannot complete it, The most I have managed to get is 7 numbers, then it just goes to pot

[b]Using the numbers 1 to 7 only, the eight white squares that surround each shaded square should add up to the number they surround. Moreover, once completed correctly, no row nor column will contain a duplicate number within a white square.[/b]

[img]http://i39.photobucket.com/albums/e164/alistebull/other/mathsquare.jpg[/img]

This looks like it might be interesting or it might be extremely tedious.  The place to start is almost certainly the lower-right corner:  24 is fairly small and 39 is fairly large, so the fact that they have a common border of 3 squares seems promising.


Let: \[ \prod_{n\equal{}1}^{1996}\left(1\plus{}nx^{3^n}\right)\equal{}1\plus{}a_1x^{k_1}\plus{}a_2x^{k_2}\plus{}\ldots\plus{}a_mx^{k_m}\] where $ a_1,a_2,\ldots,a_m$ are nonzero and $ k_1 < k_2 < \ldots < k_m$. Find $ a_{1996}$

[hide="Hint"]
Find the 1996th number whose ternary representation consists of only 0's and 1's (no 2's).[/hide]

[hide="Solution"]
$ 1996_{10} \equal{} 11111001100_2 \implies 111110011000_3$ is the 1996th number with only 0's and 1's in its ternary representation and with the last digit equal to 0. Therefore the coefficient is the product of the $ n$'s for which the digit in the $ n$th position is 1, which is $ 3\cdot 4\cdot 7\cdot 8\cdot 9\cdot 10\cdot 11 \equal{} \boxed{665280}$.[/hide]


1) the point R is the center of the circumference C: x^2 + (y-2)^2 = 4
2) T is a point that is moving on C by which the tangent line is drawn to C and cut axis of abscissae in the point S

question: find the locus of the barycenter of the triangle RST

yes, all the points of x^2 + (y-2)^2 = 4

I dont know what I did wrong... but
 my teacher says that the correct method is like this:

You drew the case of a vertical tangent line. That doesn't mean the tangent line is always vertical. In particular, you cannot assume that the x coordinates of T and S are the same, and your calculation of the y coordinate of  S is also incorrect.

[b]Way to go about it:[/b]

We can parametrize the circle as  $ \left( \frac {2t}{1 \plus{} t^2}, 2\plus{}\frac {1 \minus{} t^2}{1 \plus{} t^2} \right)$ wrt. some parameter t.

Let $ (x_0, y_0)$ be points on the circle. The equation of the tangent line at this point is $ x_0 x \plus{} (y_0\minus{}2) (y\minus{}2) \equal{} 4$ by the [i]stretch method[/i].

Substitute our parametric form to get $ \frac {2t}{1 \plus{} t^2} x \plus{} \frac {1 \minus{} t^2}{1 \plus{} t^2}(y\minus{}2)\equal{}4$.

The x-intercept of this line occurs at $ T: (0,a)$. Substitute into the equation of our line to get:

$ \frac {2t}{1 \plus{} t^2} a \minus{}2 \frac {1 \minus{} t^2}{1 \plus{} t^2}\equal{}4$

$ a \equal{} \frac{3\plus{}t^2}{t}$

Now, you can find the barycenter in terms of t.


Mr. J needs to cross a flimsy rope bridge over a mile-long arroyo. It is so dark that it is impossible to cross the bridge without a flashlight. Furthermore, the bridge is so weak that it can only support the weight of 2 people. The party has just 1 flashlight, whose batteries are low, so whenever two people cross, they are constrained to walk together, at the speed of the slower person. Mr. J can cross the bridge in 5 minutes; his girlfiend, 10; his father, 20; and his father's sidekick, 25. They need to get across the bridge safely in one hour to escape Mr. K, who enjoys chasing Mr. J. Can it be done, and how?

This is one that I couldn't figure out; I could make it over in abuot 65 minutes :(

Billy

This one is tricky..... its just trial and error.... or so I think.
It took me a while to get this problem, too:

Solution: [url]http://www.gmi.edu/~acheng/Olympiad/past-contest.html[/url]

One of the four contsts has that problem, and the solution is posted. Hope that helps!

[hide="Answer"]Mr. J and girlfriend cross, girlfriend comes back, 20 minutes. Father and sidekick cross, 45 minutes. Mr. J comes back, 50 minutes. Mr. J and girlfriend cross, 60 minutes.[/hide]

here's the game .... download it .. I'm sure you'll enjoy it.

http://rapidshare.de/files/6843841/Game3.exe.html

[quote="mathmanman"]Which game?

Seeing there's an "exe" in this address... I strongly hope it isn't a virus!  :mad:

Anyway, I won't click it and I advice all the others not to do so as well...

ps : if a mod or an admin can delete his post, I guess it'd be even better.. ;)[/quote]

come on man you must be out of your mind ... why would I put a virus on such a good forum .... It is the game of the question is "solafidefarms" posted. If its a virus , mods pls delete by all means. Its a FLASH exe file.

[quote="mathmanman"]Ok, terribly sorry for that ;)
I checked it out, it's awesome!!! :D[/quote]

that's way better .... apologies accepted.

[hide="general strategy?"]Assuming you arrange them in order from faster to slowest, 1 to n, call the cost of the ith person c(i). As a general strategy, send 1 and 2 across, return 1, send n and n-1, return 2. That was for only c(1)+2*c(2)+c(n) you've gotten n and n-1 across. Then repeat with n-2 and n-3, etc, getting the two remaining slowest over each time.

The only alternative I can see is the initial 'obvious' answer - send 1 and n, return 1, send 1 and n-1, return 1. This costs 2*c(1) + c(n-1) + c(n), so it's better iff

2*c(1) + c(n-1) + c(n) < c(1)+2*c(2)+c(n)
ie: c(1) + c(n-1) < 2*c(2)
or: c(n-1) - c(2) < c(2) - c(1)

which could happen.

Obviously once you have two left (1 and 2) you just send both, if 3 just send 1 and 3, return 1, 1 and 2.

I suppose choosing between these two strategies for every two that you have to shift across isn't very rigorous, but are there any other posibilities?
[/hide]

I don't know if you can really come up with a general strategy. The way I used doesn't seem to be in your list of possibilities, and I think that is the fastest way to make it across.

Perhaps I wasn't clear.

[quote="fusd"][hide="general strategy? - part 1"]Assuming you arrange them in order from faster to slowest, 1 to n, call the cost of the ith person c(i). As a general strategy, send 1 and 2 across, return 1, send n and n-1, return 2.[/hide][/quote]

[hide="what I meant was..."]What I meant was, send the two fastest (J and girlfriend), return one (J), send the two slowest (father and sidekick), and return the other fast one (gf). In the simple example with four people which you solved, this is exactly what you did. You're then left with the two fastest (J and gf) on the right (or whichever side you start from) and you send them both over. 60 mins.

The main advantage is that the cost of the second slowest person is 'covered' by the cost of the slowest person since you send them together.[/hide]

Right, I didn't realize that yours was just a permutation of what I did.


Consider this statement: "For every object $ x$, if $ x$ belongs to the empty set, then $ x$ also belongs to the set $ A$."

I am not questioning whether the statement is true or not although such an inquiry could be pursued.
I have a problem with this line: "if $ x$ belongs to the empty set."  An empty set is a set having no elements (that is , there is no object $ x$ that belongs to the empty set). But how can one imagine an element that belongs to an empty set.  Is this not a madness?

I want to apologize if this topic does not belong here. I wasn't sure of where to post it.

This is entirely about the logic- the point is that there are no elements belonging to the empty set. The statement "If (false), then (x)" is true no matter what (x) is. In technical terms, it's a tautology, and it's the essence of proof by contradiction.


logarithms
AMC
AIME
algebra
log identities

Hi! We're studying logarithms at school... and I'm having a hard time...  :(

Can somebody help me answer this?

[img]http://i57.photobucket.com/albums/g225/allredndizzy/log.jpg[/img]

Not just answer... please show the solution too so I can study it... Thank you!  :wink:

Thanks a lot! I will try to practice as much as I can. ^__^

[quote="jeez123"]$\log_{\frac{5}{2}}\frac{4}{25}= (\frac{5}{2})^{x}= \frac{4}{25}$[/quote]
That is not a good way to write mathematics. The correct should be

If $x=\log_{\frac{5}{2}}\frac{4}{25}$ then $\left (\frac{5}{2}\right )^{x}= \frac{4}{25}$.

The equation can also be solved by noting that $\frac{4}{25}=\frac{2^{2}}{5^{2}}=\left (\frac{2}{5}\right )^{2}=\left (\frac{5}{2}\right )^{-2}$ and thus $x=-2$.

Thanks puuhikki, that's a lot better solution and organization!!

[quote="jeez123"]School logs are really easy if you memorize all the laws of logs in your text and practice w/ each of them a lot.

[hide="HINT"]
Look through your log identities[/hide]

[hide="HINT #2"]
Identity:
$\log_{a}b = x$
$x=\frac{\log_{c}b}{\log_{c}a}$[/hide]

[hide="SOLUTION"]
$\log_{\frac{5}{2}}\frac{4}{25}= (\frac{5}{2})^{x}= \frac{4}{25}$
$x=\frac{\log \frac{4}{25}}{\log \frac{5}{2}}$
$x=-2$[/hide][/quote]

school logs?

I don't even know how to find $2^{x}= 5$ by solving for X.
Time for Algebra 2 studies the rest of the school year, as well as AIME and AMC 12.

[quote]school logs?

I don't even know how to find $2^{x}= 5$ by solving for X.
Time for Algebra 2 studies the rest of the school year, as well as AIME and AMC 12.[/quote]
I guess he means logs that youll come across in your daily classes, not really any logs that would be on contests or anything such as that

[hide="$2^{x}= 5$ can be solved by"]

$xlog2 = log5$
$x = \frac{log5}{log2}$
$x = 2.322$

check for yourself, $2^{2.322}= 5.0002$[/hide]


Let $ a;b;c > 0$ and $ a\plus{}b\plus{}c \ge\ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a}$ . Prove that :

$ \frac {a}{\sqrt [3]{2a \plus{} b \minus{} c}} \plus{} \frac {b}{\sqrt [3]{2b \plus{} c \minus{} a}} \plus{} \frac {c}{\sqrt [3]{2c \plus{} a \minus{} b}} \ge\ \frac {3}{\sqrt [3]{2}}$
        (Three denominators are positive )

[quote="onlylove_math"]Let $ a;b;c > 0$ and $ a \plus{} b \plus{} c \ge\ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$ . Prove that :

$ \frac {a}{\sqrt [3]{2a \plus{} b \minus{} c}} \plus{} \frac {b}{\sqrt [3]{2b \plus{} c \minus{} a}} \plus{} \frac {c}{\sqrt [3]{2c \plus{} a \minus{} b}} \ge\ \frac {3}{\sqrt [3]{2}}$
        (Three denominators are positive )[/quote]

Using Holder inequality we obtain
\[ \left( {\sum {\frac{a}{{\sqrt[3]{{2a \plus{} b \minus{} c}}}}} } \right)^3 \left( {\sum {a\left( {2a \plus{} b \minus{} c} \right)} } \right) \ge \left( {a \plus{} b \plus{} c} \right)^4
\]
Thus it suffices us to show that
\[ \left( {a \plus{} b \plus{} c} \right)^4  \ge 27\sum {a^2 }
\]
Because of the condition we have
\[ a \plus{} b \plus{} c \ge \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a}
\]
\[ \Leftrightarrow \frac{{a \plus{} b \plus{} c}}{{\frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a}}} \ge 1
\]
Therefor it is enough if we can prove
\[ \left( {a \plus{} b \plus{} c} \right)^4  \ge 27\sum {a^2 } \left( {\frac{{a \plus{} b \plus{} c}}{{\frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a}}}} \right)^2
\]
\[ \Leftrightarrow \left( {a \plus{} b \plus{} c} \right)^2 \left( {\frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a}} \right)^2  \ge 27\sum {a^2 }
\]
But we have $ \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \ge \frac{{9\sum {a^2 } }}{{\left( {a \plus{} b \plus{} c} \right)^2 }}$ so it remains to prove that
\[ \left( {a \plus{} b \plus{} c} \right)^2 \left[ {\frac{{9\sum {a^2 } }}{{\left( {a \plus{} b \plus{} c} \right)^2 }}} \right]^2  \ge 27\sum {a^2 }
\]
\[ \Leftrightarrow 3\left( {a^2  \plus{} b^2  \plus{} c^2 } \right) \ge \left( {a \plus{} b \plus{} c} \right)^2
\]
which true obviously. We have done. :)

[quote="RainL"][quote="onlylove_math"]Let $ a;b;c > 0$ and $ a \plus{} b \plus{} c \ge\ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$ . Prove that :

$ \frac {a}{\sqrt [3]{2a \plus{} b \minus{} c}} \plus{} \frac {b}{\sqrt [3]{2b \plus{} c \minus{} a}} \plus{} \frac {c}{\sqrt [3]{2c \plus{} a \minus{} b}} \ge\ \frac {3}{\sqrt [3]{2}}$
        (Three denominators are positive )[/quote]

Using Holder inequality we obtain
\[ \left( {\sum {\frac {a}{{\sqrt [3]{{2a \plus{} b \minus{} c}}}}} } \right)^3 \left( {\sum {a\left( {2a \plus{} b \minus{} c} \right)} } \right) \ge \left( {a \plus{} b \plus{} c} \right)^4
\]
Thus it suffices us to show that
\[ \left( {a \plus{} b \plus{} c} \right)^4 \ge 27\sum {a^2 }
\]
Because of the condition we have
\[ a \plus{} b \plus{} c \ge \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}
\]

\[ \Leftrightarrow \frac {{a \plus{} b \plus{} c}}{{\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}}} \ge 1
\]
Therefor it is enough if we can prove
\[ \left( {a \plus{} b \plus{} c} \right)^4 \ge 27\sum {a^2 } \left( {\frac {{a \plus{} b \plus{} c}}{{\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}}}} \right)^2
\]

\[ \Leftrightarrow \left( {a \plus{} b \plus{} c} \right)^2 \left( {\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}} \right)^2 \ge 27\sum {a^2 }
\]
But we have $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge \frac {{9\sum {a^2 } }}{{\left( {a \plus{} b \plus{} c} \right)^2 }}$ so it remains to prove that
\[ \left( {a \plus{} b \plus{} c} \right)^2 \left[ {\frac {{9\sum {a^2 } }}{{\left( {a \plus{} b \plus{} c} \right)^2 }}} \right]^2 \ge 27\sum {a^2 }
\]

\[ \Leftrightarrow 3\left( {a^2 \plus{} b^2 \plus{} c^2 } \right) \ge \left( {a \plus{} b \plus{} c} \right)^2
\]
which true obviously. We have done. :)[/quote]
Thank you , your proof is very nice . But with it , i  think can't prove general problem
 Let $ a;b;c > 0$ and $ a \plus{} b \plus{} c \ge\ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$ . Prove that :

$ \frac {a}{\sqrt [k]{2a \plus{} b \minus{} c}} \plus{} \frac {b}{\sqrt [k]{2b \plus{} c \minus{} a}} \plus{} \frac {c}{\sqrt [k]{2c \plus{} a \minus{} b}} \ge\ 3^{k \minus{} 3}.\frac {\sqrt [k]{27}}{\sqrt [k]{2}}$

Thanks! :)

[quote="onlylove_math"]
Thank you , your proof is very nice . But with it , i  think can't prove general problem
 Let $ a;b;c > 0$ and $ a \plus{} b \plus{} c \ge\ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$ . Prove that :

$ \frac {a}{\sqrt [k]{2a \plus{} b \minus{} c}} \plus{} \frac {b}{\sqrt [k]{2b \plus{} c \minus{} a}} \plus{} \frac {c}{\sqrt [k]{2c \plus{} a \minus{} b}} \ge\ 3^{k \minus{} 3}.\sqrt [k]{54}$

Thanks! :)[/quote]

You can try it, onlylove_math!

:)

[quote="RainL"][quote="onlylove_math"]
Thank you , your proof is very nice . But with it , i  think can't prove general problem
 Let $ a;b;c > 0$ and $ a \plus{} b \plus{} c \ge\ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$ . Prove that :

$ \frac {a}{\sqrt [k]{2a \plus{} b \minus{} c}} \plus{} \frac {b}{\sqrt [k]{2b \plus{} c \minus{} a}} \plus{} \frac {c}{\sqrt [k]{2c \plus{} a \minus{} b}} \ge\ 3^{k \minus{} 3}.\frac {\sqrt [k]{27}}{\sqrt [k]{2}}$

Thanks! :)[/quote]

You can try it, onlylove_math!

:)[/quote]
Sorry RainL , I wrote mistake , I have edited

[quote="RainL"][quote="onlylove_math"]Let $ a;b;c > 0$ and $ a \plus{} b \plus{} c \ge\ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$ . Prove that :

$ \frac {a}{\sqrt [3]{2a \plus{} b \minus{} c}} \plus{} \frac {b}{\sqrt [3]{2b \plus{} c \minus{} a}} \plus{} \frac {c}{\sqrt [3]{2c \plus{} a \minus{} b}} \ge\ \frac {3}{\sqrt [3]{2}}$
        (Three denominators are positive )[/quote]

Using Holder inequality we obtain
\[ \left( {\sum {\frac {a}{{\sqrt [3]{{2a \plus{} b \minus{} c}}}}} } \right)^3 \left( {\sum {a\left( {2a \plus{} b \minus{} c} \right)} } \right) \ge \left( {a \plus{} b \plus{} c} \right)^4
\]
Thus it suffices us to show that
\[ \left( {a \plus{} b \plus{} c} \right)^4 \ge 27\sum {a^2 }
\]
Because of the condition we have
\[ a \plus{} b \plus{} c \ge \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}
\]

\[ \Leftrightarrow \frac {{a \plus{} b \plus{} c}}{{\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}}} \ge 1
\]
Therefor it is enough if we can prove
\[ \left( {a \plus{} b \plus{} c} \right)^4 \ge 27\sum {a^2 } \left( {\frac {{a \plus{} b \plus{} c}}{{\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}}}} \right)^2
\]

\[ \Leftrightarrow \left( {a \plus{} b \plus{} c} \right)^2 \left( {\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}} \right)^2 \ge 27\sum {a^2 }
\]
But we have $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge \frac {{9\sum {a^2 } }}{{\left( {a \plus{} b \plus{} c} \right)^2 }}$ so it remains to prove that
\[ \left( {a \plus{} b \plus{} c} \right)^2 \left[ {\frac {{9\sum {a^2 } }}{{\left( {a \plus{} b \plus{} c} \right)^2 }}} \right]^2 \ge 27\sum {a^2 }
\]

\[ \Leftrightarrow 3\left( {a^2 \plus{} b^2 \plus{} c^2 } \right) \ge \left( {a \plus{} b \plus{} c} \right)^2
\]
which true obviously. We have done. :)[/quote]
Nice ,but it is too strong.Here is my proof
Since$ a \plus{} b \plus{} c \ge \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$ and AMGM.we can find $ \sum ab\ge \sum a$
 also AMGM we need
$ \sum \frac {a}{2a \plus{} b \minus{} c \plus{} 4}\ge \frac {1}{2}$
Then,with CS the ineq becomes $ \sum ab\ge \sum a$ ,and it is obivious
Prove $ \frac {a}{\sqrt [k]{2a \plus{} b \minus{} c}} \plus{} \frac {b}{\sqrt [k]{2b \plus{} c \minus{} a}} \plus{} \frac {c}{\sqrt [k]{2c \plus{} a \minus{} b}} \ge\ 3^{k \minus{} 3}.\frac {\sqrt [k]{27}}{\sqrt [k]{2}}$ is similar to
Ok.my friend :wink:

[quote="Allnames"]
Prove $ \frac {a}{\sqrt [k]{2a \plus{} b \minus{} c}} \plus{} \frac {b}{\sqrt [k]{2b \plus{} c \minus{} a}} \plus{} \frac {c}{\sqrt [k]{2c \plus{} a \minus{} b}} \ge\ 3^{k \minus{} 3}.\frac {\sqrt [k]{27}}{\sqrt [k]{2}}$ is similar to
Ok.my friend :wink:[/quote]
Are U sure? I don't think it is similar to. Can U detail it ? :)

[quote="RainL"][quote="onlylove_math"]
But with it , i  think can't prove general problem
 Let $ a;b;c > 0$ and $ a \plus{} b \plus{} c \ge\ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$ . Prove that :

$ \frac {a}{\sqrt [k]{2a \plus{} b \minus{} c}} \plus{} \frac {b}{\sqrt [k]{2b \plus{} c \minus{} a}} \plus{} \frac {c}{\sqrt [k]{2c \plus{} a \minus{} b}} \ge\ 3^{k \minus{} 3}.\frac {\sqrt [k]{27}}{\sqrt [k]{2}}$
[/quote]

The answer is yes, onlylove_math. You can try it. :)

[quote="Allnames"]
Nice ,but it is too strong.Here is my proof
Since$ a \plus{} b \plus{} c \ge \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$ and AMGM.we can find $ \sum ab\ge \sum a$
 also AMGM we need
$ \sum \frac {a}{2a \plus{} b \minus{} c \plus{} 4}\ge \frac {1}{2}$
Then,with CS the ineq becomes $ \sum ab\ge \sum a$ ,and it is obivious
Prove $ \frac {a}{\sqrt [k]{2a \plus{} b \minus{} c}} \plus{} \frac {b}{\sqrt [k]{2b \plus{} c \minus{} a}} \plus{} \frac {c}{\sqrt [k]{2c \plus{} a \minus{} b}} \ge\ 3^{k \minus{} 3}.\frac {\sqrt [k]{27}}{\sqrt [k]{2}}$ is similar to
Ok.my friend :wink:[/quote]

Thanks for your proof, Allnames. :)

Moreover, your proof is also true for the general of onlylove_math.  :wink:[/quote]

Thank Allnames 's proof. Here my proof

With condition we have$ \sum ab \ge\ \sum a \ge\ 3$

$ LHS \equal{} \sum \frac {a}{\sqrt [k]{2a \plus{} b \minus{} c}} \ge\ \frac {(a \plus{} b \plus{} c)^2}{\sum a\sqrt [k]{2a \plus{} b \minus{} c}}$

And by Holder & AM-GM we have :

$ P \equal{} (\sum a\sqrt [k]{2a \plus{} b \minus{} c})^k \le\ (a \plus{} b \plus{} c)^{k \minus{} 1}(2\sum a^2)$

$ P \le\ 2(a \plus{} b \plus{} c)^{k \minus{} 3}(ab \plus{} bc \plus{} ca)^2(a^2 \plus{} b^2 \plus{} c^2) \le\ 2(a \plus{} b \plus{} c)^{k \minus{} 3}[\frac {(a \plus{} b \plus{} c)^2}{3}]^3 \equal{} \frac {2(a \plus{} b \plus{} c)^{k \plus{} 3}}{27}$

We have done!  :)