geometry
3D geometry
conics
ellipse
perimeter
function
Pythagorean Theorem

A right circular cone of base radius $ 17$ and slant height $ 34$ is given.  $ P$ is a point on the circumference of the base and the shortest path from $ P$ around the cone and back is drawn.  If the length of this path is $ m\sqrt {n}$ in simplest form, find $ m \plus{} n$.

[quote="Yongyi781"]I think your question was a bit ambiguous... what does "around the cone" exactly mean?[/quote]

a path from P across the surface of the cone, and back around to itself.  its like a circle on a slanting plane that intersects the cone.

Oh I see... so the shorest path is [b]not[/b] just around the circumference of the base, it's somewhat slanting up.

I would say the circle is perpendicular to the slant height. Pythagorean Theorem, anyone?

actually it might be an ellipse. im not sure.  (that's why i posted this problem =D)

Well, based on conic sections, the intersection of an oblique plane and a cone is an ellipse, and the intersection of a cone and a plane parallel to the base is a circle...

... so yeah, the "circle" in question is actually an ellipse. However, finding the circumference of an ellipse is a bit problematic... Is this an actual question or one you made up? Since it seems next to impossible to express the circumference of an ellipse with a radical.

[quote="JavaMan"][hide="Hint"]

Note that the distance from the apex to any point on the base is constant.

This means if we made a straight cut (on one side) from the apex to a point on the base and flattened out the object we'd get a sector of a circle.

[/hide][/quote]

That means the perimeter of the path is the distance between the two endpoints of the sector!

[quote="JavaMan"][hide="Hint"]

Note that the distance from the apex to any point on the base is constant.

This means if we made a straight cut (on one side) from the apex to a point on the base and flattened out the object we'd get a sector of a circle.

[/hide][/quote]

sorry, could you explain how this would help with the problem?

It converts the 3D problem into a 2D one, where the shortest distance is simply the length of a straight line between the two endpoints of the sector (as Yongyi781 suggested).

[quote="The Zuton Force"][hide="Graphical"]
Basically, your cone is like a rolled up slice of pizza, where the two edges meet perfectly.

You want to get from one endpoint of the crust to the other endpoint of the crust by going around this cone shape.

To find this distance, flatten out the slice of pizza and draw a straight line (a chord, if you will) connecting the two endpoints of the crust. Since this is the shortest distance when the pizza is flat, it is also the shortest distance when you roll it back into a cone.

The trick now is to find the radius (given) and central angle (easy to calculate) of this "slice of pizza"[/hide][/quote]

oooohhh nice application example!!!
i got it now  :D  thanks!

[quote="JavaMan"][quote="Yongyi781"][hide]

Technically, the correct answer is going from $ B$ to $ A$ and back, for which the total distance is $ 68$. But then, the format is incorrect and we are left with $ m \equal{} 68$ and $ n \equal{} \text{undef}$ :D

Therefore, there is no correct answer, since we could make an infinitesimally thin ellipse without being a line, but I can say that I can make an ellipse that is even thinner. (The ellipse will look very close to a line between $ A$ and $ B$).

[/hide][/quote]

What about $ n \equal{} 1$  :D[/quote]

But that wouldn't be in simplest [b]radical[/b] form :D  :D

[quote="the future"]Is the answer like $ 34\sqrt2$ so $ 36$?[/quote]Isn't it double that? So $ 68\sqrt{2}$, so $ m\plus{}n\equal{}070$?

[quote="123456789"][quote="the future"]Is the answer like $ 34\sqrt2$ so $ 36$?[/quote]Isn't it double that? So $ 68\sqrt {2}$, so $ m \plus{} n \equal{} 070$?[/quote]

Incorrect for both.

Unraveling the cone would create a semicircle. Therefore, the shortest path from $ A$ to $ B$ is along the diameter, which when translated to cone-land is going from a point on the base to the apex and back. Which is $ 68$. Which is less than $ 68\sqrt {2}$.

But wait, that wouldn't be "around the cone," then. To make it around the cone, we could "widen" the line to make a very thin ellipse. But no matter how thin the ellipse is, I can always make an ellipse that is thinner than that one. Think of it as a discontinuity at $ x \equal{} 2$ in the function $ \frac {x^2 \minus{} 4}{x \minus{} 2}$, and trying to find out how close along the line you can get to the point $ (2, 4)$.


I'm a LaTeX noob, just finishing up my first assignment. I'm finding that having to think about formatting and typesetting is getting in the way of thinking about the math I want to express. A lot of people feel this way about non-WYSIWYG formatting languages, which is precisely why WYSIWYG environments were invented and popularized.

So now I'm looking for recommendations about which WYSIWYG environment I should choose. Is LyX good?

You may try to convince me that I should love writing bare LaTeX. You will not succeed.

Well in my basic experience, I'm using TeXnicCenter and it has a good interface but I didn't tried other programs in fact. It has some LaTeX commands just on your click but I believe you must learn how to code as well if you want a good work. Try to follow guides and ask questions.

Experts experiences are required here

LyX is, as far as I know, the only free such editor. All you need to do is add it to your LaTeX distribution and try it to see if you like it. Otherwise you need to pay for software such as [url=http://www.bakoma-tex.com/]BaKoMa TeX[/url] (you can try it for 28 days before you buy it) or [url=http://www.mackichan.com/]Scientific Word[/url] (free 30 day trial).

hasan, WYSIWYG stands for "what you see is what you get."  TeXnicCenter is not a WYSIWYG environment: you write your code, then you process it to get your result.

Microsoft Word is an example of a WYSIWYG editor. With WYSIWYG, you don't type your text and commands into a file and then have it processed. Rather, you manipulate your document right on the screen. You can see immediately what's happening.


How many different ways can you select a block of five adjacent squares on a 8x8 chessboard?

Two squares are adjacents if they have a common side.

(the answer would be 49 for a 3 x 3 board )

I arrived at the answer 2028 by brute force (considering each pentomino), although there is plenty of scope for error in such a method. There seems to be little significance about the answer.


inequalities
algebra
polynomial
parameterization
function
inequalities proposed

1.
Prove that
$\sum_{sym} x_1^4 + 3\sum_{sym} x_1^2x_2^2 \geq 4\sum_{sym} x_1^3x_2$
is true for $n$ positive real numbers $x_1, x_2, x_3, \ldots, x_n$ and determine when equality holds.

2.
Prove that
$\sum_{sym} x_1^4 + 9\sum_{sym} x_1^2x_2^2 + 6\sum_{sym} x_1x_2x_3x_4 \geq 4\sum_{sym} x_1^3x_2 + 12\sum_{sym} x_1^2x_2x_3$
is true for $n$ positive real numbers $x_1, x_2, x_3, \ldots, x_n$ and determine when equality holds.

1.

For $n=2$,
\[x_{1}^4+x_{2}^4 + 6 x_{1}^2x_{2}^2 \ge 4x_{1}^3x_{2}+4x_{1}x_{2}^3 \Leftrightarrow (x_{1}-x_{2})^4 \ge 0.\]
Assume the inequality is true for $n=k$, that is $(k-1)\displaystyle\sum_{i=1}^{k} x_{i}^4 +3\displaystyle\sum_{1 \le i,j \le k, i \neq j}x_{i}^2x_{j}^2 \ge 4\displaystyle\sum_{1 \le i,j \le k, i \neq j}x_{i}^3x_{j}$.

When $n=k+1$,

$k\displaystyle\sum_{i=1}^{k+1} x_{i}^4 +3\displaystyle\sum_{1 \le i,j \le k+1, i \neq j}x_{i}^2x_{j}^2-4\displaystyle\sum_{1 \le i,j \le k+1, i \neq j}x_{i}^3x_{j}$
$=\displaystyle\sum_{i=1}^{k}(x_{k+1}^4+x_{i}^4-4x_{k+1}^3x_{i}-4x_{k+1}x_{i}^3)+(k-1)\displaystyle\sum_{i=1}^{k} x_{i}^4 +3\displaystyle\sum_{1 \le i,j \le k, i \neq j}x_{i}^2x_{j}^2 - 4\displaystyle\sum_{1 \le i,j \le k, i \neq j}x_{i}^3x_{j}$
$\ge 0$,

so the inequality is true for $n=k+1$.

By MI the inequality is true for all natural n.

I guess problem 2 can also be proved by MI, though it will be quite complicated...  :huh:

I got a hint for you for problem 2:
[hide]Try to transform the inequality into the form $a_1^4 + a_2^4 + \ldots$[/hide]

EDIT: yptzoi, determine when equality holds :P

Hint 2:
[hide]Prove that $\sum_{sym} x_1^2x_2^2 + \sum_{sym} x_1x_2x_3x_4 \geq \sum_{sym} x_1^2x_2x_3$

for $n$ variables $x_1, x_2, \ldots, x_n$[/hide]

The first inequality can be generalized to
\[
\sum_{sym} x^4 -  \lambda \sum_{sym} x^3y + (\frac{1}{4}\lambda^2-1) \sum_{sym} x^2y^2 + (\lambda-\frac{1}{4}\lambda^2)\sum_{sym} x^2yz \geq 0,
\]
for any $\lambda \in \mathbb{R}$, and symmetric polynomials in at least 3 variables.
The special case $\lambda=4$ is the one given by Jinbi. The one with $\lambda=2$ is also quite nice.

I think I can proof that all inequalities of symmetric polynomials of degree four in three variables are combinations of this inequality of this one (for a suitable value of $\lambda$ and Muirhead's inequalities. I am trying to generalize this result and will subsequently hopefully find time to write that down.

:roll:  The second inequality is not true, say, when n=4, x1=x2=x3=x4=1.

Guest, I think there may be a confusion of normalization involved. In Jinbi's normalization (which I guess is the one I told him :-)), the value of
the symmetric sum of any monomial is 1 if all values of the parameters are 1. Hence in that case the second inequality gives an equality ($1+9+6=4+12$).

:roll:  I do not think so.   Please strictly clarify the meaning of the symmetric sum notations.

Let $S_n$ denote the symmetrical group on $n$ variables. $S_n$ acts on functions of $n$ variables by
$\sigma f(x_1,\ldots,x_n) = f(x_{\sigma(1)},\ldots,x_{\sigma(n)})$ for $\sigma \in S_n$.
Now we define
\[
\sum_{sym} f = \frac{1}{n!} \sum_{\sigma \in S_n} \sigma f.
\]

Hence for $n=3$ (to avoid too many terms) we say
\[
\sum_{sym} x_1 = \frac{1}{6}(x_1 +x_1+x_2+x_2+x_3+x_3) = \frac{1}{3}(x_1+x_2+x_3).
\]
Note that if $f(1,1,\ldots,1)=1$ then $(\sigma f)(1,1,\ldots,1) = f(1,1,\ldots,1)=1$, and hence
\[
\left(\sum_{sym} f\right) (1,1,\ldots,1) = \frac{1}{n!} \sum_{\sigma \in S_n} (\sigma f)(1,1,\ldots,1) = \frac{1}{n!}
\sum_{\sigma \in S_n} 1 = 1.
\]

Note that sometimes symmetric polynomials are normalized such that the factor before each term equals 1, hence
$\sum_{sym'} x_1 = x_1+x_2+x_3$, however in my humble opinion this normalization is usually not as convenient to write down inequalities.

[quote="Tasar"]\[
\sum_{sym} f = \frac{1}{n!} \sum_{\sigma \in S_n} \sigma f.
\]
[/quote]

Where did you see this?!?!?! Everywhere else it's
\[
\sum_{sym} f = \sum_{\sigma \in S_n} \sigma f.
\]

Let me first say that quite often the factor $n!$ is irrelevant (as it is in the inequalities discussed here). I can't remember where I saw this normalization (except in my own notes of course) (and I must admit I have seen the one with an extra factor $n!$ also before). In practice I don't think there is much difference in working with an extra $n!$ or not, but one advantage of the normalization I wrote down is that the operator $\sum_{sym}$ becomes a projection (i.e. $\sum_{sym} \sum_{sym} f = \sum_{sym} f$).

The most important thing to remember is that each symmetric sum should have as many terms, as long as you do that, the rest will be okay.

:roll:  I agree with billzhao.  Nowhere "symmetric sum" notation is used as Tasar did except himself.


[b]1. The problem statement, all variables and given/known data[/b]
Determine whether the series $ \sum_{n\equal{}2}^{\infty}$ is absolutely convergent,conditionaly convergent or divergent.If

$ a_n\equal{}\frac{1\plus{}n\plus{}n^2}{\sqrt{1\plus{}n^2\plus{}n^6}}$

[b]3. The attempt at a solution[/b]
I get
$ \frac{1\plus{}n\plus{}n^2}{\sqrt{1\plus{}n^2\plus{}n^6}}>\frac{1}{n}$

so I think it's divergent.

That's the right comparison. I wouldn't worry to much about making it $ >;$ this is a routine case for the limit comparison test. That is, if we divide the terms of this series by $ \frac{1}{n}$ and take the limit, then that ratio tends to $ 1.$ If two positive term series have terms that are the same size in this sense (that is, $ \frac{a_n}{b_n}$ tends to a finite nonzero number), then the two series either both converge or both diverge.


function
real analysis
calculus
integration
real analysis unsolved

I have the following question: if $ f$ and $ g$ are measurable functions in $ \mathbb R^N$. How can I prove that the function $ h(x,y) : \equal{} f(x) \plus{} g(y)$ is measurable in $ \mathbb R^N \times \mathbb R^N$.

Thank you.

I'll give you just a hint, if $ X,Y$ are topological spaces with countable base then $ \mathcal{B}_X\times\mathcal{B}_Y\equal{}\mathcal{B}_{X\times Y}$.

lol? Is that even a hint? or just a slightly related statement...Geez.. if only these concepts were that simple.

This problem comes from Lebesgue repeated integration  :mad: .

I'm sorry for my comment, I may be way off if I didn't understand the question correctly. First of all, now I'm confused with the following:
[quote="bookworm_vn"] if $ f$ and $ g$ are measurable functions in $ \mathbb R^N$.
[/quote]
did you mean by that $ f,g$ have image in $ \mathbb{R}^{\mathbb{N}}$? If that is the case, then my hint is more then enough to prove your statement. Simply take some measurable set $ B$ in $ \mathcal{B}_{\mathbb{R}^{\mathbb{N}}}$, adding is a continuous function, so $ (f\plus{}g)^{\minus{}1}(B)\equal{}B'\in\mathcal{B}_{\mathbb{R}^{\mathbb{N}}\times \mathbb{R}^{\mathbb{N}}}$, now because $ \mathbb{R}^{\mathbb{N}}$ has countable base ( or I'm I wrong with that  :huh: ), $ \mathcal{B}_{\mathbb{R}^{\mathbb{N}}}\times\mathcal{B}_{\mathbb{R}^{\mathbb{N}}}\equal{}\mathcal{B}_{\mathbb{R}^{\mathbb{N}}\times \mathbb{R}^{\mathbb{N}}}$. So, $ B'\equal{}B_1\times B_2$ where $ B_1,B_2\in\mathcal{B}_{\mathbb{R}^{\mathbb{N}}}$, so $ (f,g)^{\minus{}1}(B')\equal{}(f,g)^{\minus{}1}(B_1\times B_2)\equal{}f^{\minus{}1}(B_1)\cap g^{\minus{}1}(B_2)\in \mathcal{F}$, where $ \mathcal{F}$ is $ \sigma$-algebra on their domain.

If $ f,g$ are function defined on $ \mathbb{R}^{\mathbb{N}}$ with image in ( say ) $ \mathbb{R}$, I really don't a problem in proving that $ f\plus{}g$ is measurable.

[quote="Mr.Doe"]I'm sorry for my comment, I may be way off if I didn't understand the question correctly. First of all, now I'm confused with the following:
[quote="bookworm_vn"] if $ f$ and $ g$ are measurable functions in $ \mathbb R^N$.
[/quote]
did you mean by that $ f,g$ have image in $ \mathbb{R}^{\mathbb{N}}$? If that is the case, then my hint is more then enough to prove your statement. Simply take some measurable set $ B$ in $ \mathcal{B}_{\mathbb{R}^{\mathbb{N}}}$, adding is a continuous function, so $ (f \plus{} g)^{ \minus{} 1}(B) \equal{} B'\in\mathcal{B}_{\mathbb{R}^{\mathbb{N}}\times \mathbb{R}^{\mathbb{N}}}$, now because $ \mathbb{R}^{\mathbb{N}}$ has countable base ( or I'm I wrong with that  :huh: ), $ \mathcal{B}_{\mathbb{R}^{\mathbb{N}}}\times\mathcal{B}_{\mathbb{R}^{\mathbb{N}}} \equal{} \mathcal{B}_{\mathbb{R}^{\mathbb{N}}\times \mathbb{R}^{\mathbb{N}}}$. So, $ B' \equal{} B_1\times B_2$ where $ B_1,B_2\in\mathcal{B}_{\mathbb{R}^{\mathbb{N}}}$, so $ (f,g)^{ \minus{} 1}(B') \equal{} (f,g)^{ \minus{} 1}(B_1\times B_2) \equal{} f^{ \minus{} 1}(B_1)\cap g^{ \minus{} 1}(B_2)\in \mathcal{F}$, where $ \mathcal{F}$ is $ \sigma$-algebra on their domain.

If $ f,g$ are function defined on $ \mathbb{R}^{\mathbb{N}}$ with image in ( say ) $ \mathbb{R}$, I really don't a problem in proving that $ f \plus{} g$ is measurable.[/quote]

$ f,g : \mathbb R^N \to \mathbb R$. But how to use $ \mathcal{B}$? This is Lebesgue measure..

[quote="Mr.Doe"]If $ f,g$ are function defined on $ \mathbb{R}^{\mathbb{N}}$ with image in ( say ) $ \mathbb{R}$, I really don't a problem in proving that $ f \plus{} g$ is measurable.[/quote]The OP is [i]not[/i] asking to prove that $ f\plus{}g$ is measurable, but something quite different.

[quote="JoeBlow"][quote="Mr.Doe"]If $ f,g$ are function defined on $ \mathbb{R}^{\mathbb{N}}$ with image in ( say ) $ \mathbb{R}$, I really don't a problem in proving that $ f \plus{} g$ is measurable.[/quote]The OP is [i]not[/i] asking to prove that $ f \plus{} g$ is measurable, but something quite different.[/quote]

Finally, I have proved it. The argument lies on the fact that if $ f(x)$ is measurable over $ \mathbb R^N$ then so is over $ \mathbb R^M$ whenever $ M \geq N$. Thanks for some interesting ideas :)


Let A be an nxn matrix and B the matrix that results when thte rows of A are written in reverse order( last row becomes the first, etc.).  How are det(A) and det(B) related?

Each row swap switches the sign of the determinant.  Firgure out how many times the rows have been swapped.  IOt probably depends on weather n( the number of rows in the matrix)  is even or odd.

Would that be a sufficent enough answer for this problem?

TIA~!~!

It's enough, the determinant is alternated (or antisymetric if you prefer).


If you had to marry a movie star(pretend you're old enough..I'm not too!) who would you marry?

[quote="stupidkid"][quote="probability1.01"]Maybe pick the richest one, then divorce![/quote]
I agree.[/quote]
ditto

[quote="furious"]i'd go for the hottest one.  :lol:[/quote]

Which is...? :lol:

[quote="hatsup123"]Which is...? :lol:[/quote]
your mom

[quote="hatsup123"]:([/quote]
i was shocked too

[quote="sheepwarrior"][quote="djimenez"]Well, given that I don't personally know anyone that is really famous internationally (I know a few that do movies in my country, but those usually are not that famous outside the country) I suppose that we would be talking about the one who is for us the most attractive. In that case, my options would be (with a reference, the one that I think can be widerly known):

Monica Bellucci (Persephone in Matrix Reloaded and Revolution, she is 16 years older than me, but still very beautiful)
Charlize Theron (Mary Ann Lomax in The Devil's Advocate, a beautiful african!)
Leticia Brédice (I suppose that the most famous movie from here would be Nueve Reinas)
Paz Vega (Flor in Spanglish, she has my same birth day, and she is just a little older than me)

After that, well, I can think in at least two famous english actresses, but I will stop there  :D[/quote]

charlize theron is african?!?!?!?[/quote]Yes, she is

[quote="Someone in IMDb"][b]Charlize Theron[/b]:

Date of birth (location)
    7 August 1975
    Benoni, South Africa
[/quote]Charlize Theron grew up on a farm outside Benoni, South Africa. She lived right there until the age of 18, after that she spent a few months in Europe, and after that she moved to LA (and she is gorgeous) :D

i'll take anyone of those hot movie stars
:D  :D  :D  :D  :D  :D  :D  :D  :D  :D

and also those Brazil supermodels

[quote="miraculouspostmaster"]Ooh, Ooh, I pick Queen Latifah![/quote]

then i get LaFawnda, from Napolean Dynamite

[quote="tetrahedr0n"]If it was up to me, I'd go for Natalie Portman. After all, I'd like a girl who I can talk to without having to resort to words of four letters or less.[/quote] Good choice.  :D

[quote="juicybooty911"]
and also those Brazil supermodels[/quote]

I agree.  Gisele is pretty hot.
 :D

[quote="ankur87"][quote="djimenez"][quote="juicybooty911"]charlize theron is white(uhhh, i think so) :D[/quote]So what? she was born in South Africa, from South Africans. Remember that most of Africa (as well as basically all of America, the continent from Horn to Bering) was conquered by some European country sometime during the last five centuries, hence, there are white people that has been african for several generations, as well as there are white people that has been american for several generations (say, in some cases, twenty generations or so)[/quote]

If I remember anything from World History (which is probably not the case), I think S. Africa was colonized by the Dutch. Of course, this could be confirmed easily with a google search, but I'm feeling particularly lazy....[/quote]

Originally by the Dutch, I believe, who proceeded to call themselves the Boers, or Afrikaners. They were eventually displaced by the British who wanted to use the colony as a waypoint on the way to India. The Boers moved to the Transvaal, where they found diamonds and thus the British felt compelled to fight them for it, resulting in the Boer War. These diamonds happened to come into the possession of Cecil Rhodes, who I believe led to British to the Transvaal, and thus he started the Rhodes Scholarship. (This is what I remember from AP World History, it could be radically wrong and, as Ankur, I am too lazy to Google it).

Movie stars are just rich snobs who are selfish and for the most part, not very intelligent. They only care about themselves: how much money they have, how they look, Paparazzi, etc. That's why I would never marry a movie star.

But if I had to, I guess Natalie Portman isn't bad. She's beautiful and pretty smart (from what I hear) :blush:


ratio
complex numbers
number theory proposed
number theory

Let $ \mathbb{D}$ be the set of all nonzero complex numbers. Find all quadruples of integers $ (a,b;c,d)$ such that if $ x,y,z,w$ are complex numbers such that $ x^{a}y^{b} \equal{} z^{a}w^{b}$ and $ x^{c}y^{d} \equal{} z^{c}w^{d}$, then $ x\equal{}z$ and $ y\equal{}w$

I'll assume you mean that $ x,y,z,w \in \mathbb{D}$ otherwise there are no solutions because we can always take $ x \equal{} z \equal{} 0$ and $ y \ne w$

The quadruples that satisfy the condition are $ \{ (a,b;c,d) : ad \minus{} bc \equal{} \pm 1 \}$

Firstly, suppose $ ad \minus{} bc \equal{} \pm 1$, then $ x^{ac}y^{bc} \equal{} z^{ac}w^{bc}$ and $ x^{ca}y^{da} \equal{} z^{ca}w^{da}$

Dividing we get $ (\frac{w}{y})^{ad \minus{} bc} \equal{} 1 \implies y \equal{} w$.

Similarly we have $ x^{ad}y^{bd} \equal{} z^{ad}w^{bd}$ and $ x^{cb}y^{db} \equal{} z^{cb}w^{db} \implies  (\frac{x}{z})^{ad \minus{} bc} \equal{} 1 \implies x \equal{} z$.

For the converse I wont give a complete solution but it is simple enough, just long too spell out counterexamples for each case.

If $ {ad \minus{} bc} \equal{} 0$ then the ratios are not restricted and it is easy to construct counterexamples being careful to consider cases where any of the exponents may be zero.

if $ |ad \minus{} bc| \equal{} n \ge 2$ then the ratios $ \frac{w}{y}$ and $ \frac{x}{z}$ are restricted to be $ n^{th}$ roots of unity, but within those constraints counterexamples
can be constructed.


If $ m$ workers can complete a job in $ d$ days, how many days will it take $ n$ workers, working at the same rate, to complete one-third of the job? Express your answer as a common fraction in terms of $ d$, $ m$ and $ n$, in alphabetical order where applicable.

More workers means more of the job is done; the number of workers and the fraction of the job done is a positive correlation.  Let's call the number of days it takes $ n$ workers to complete $ \frac {1}{3}$ of a job $ x$.  Then, $ 3n$ workers would complete the whole job in $ x$ days.  Now that we have the fraction of the job equal for $ m$ workers and $ 3n$ workers, let's write a ratio:

$ \frac {m}{3n} \equal{} \frac {d}{x}$

Now we cross-multiply:

$ mx \equal{} 3nd$
$ x \equal{} \frac {3nd}{m}$

EDIT:  whoops, alphabetical order.

$ x \equal{} \boxed{\frac{3dn}{m}}$

The workers to days in this problem is actually inversely proportional. :)
Let $x$= the number of days it will take $n$ workers to complete a third of the job

Hence $\frac{md}{3}=nx$

$x=\boxed{\frac{md}{3n}}$

Oh darn, I messed up again, didn't I?  Well, it seems as if I had the right idea that the number of workers and the fraction of the job done is a positive correlation, but I assumed that the number of workers and the time it takes was a positive correlation.  Darn, my bad :oops:

BTW, it's alphabetical order so $\boxed{\frac{dm}{3n}}$ :wink:


Find all functions $ g : \mathbb{R}\to \mathbb{R}$ which satisfies
\[g(x\plus{}y)\plus{}g(x) \cdot g(y)\equal{}g(xy)\plus{}g(x)\plus{}g(y)\]
for all reals $x$ and $y$.

[quote="cnyd"]$ g : \mathbb{R}\mapsto\mathbb{R}$
$ g(x \plus{} y) \plus{} g(x).g(y) \equal{} g(xy) \plus{} g(x) \plus{} g(y)$[/quote]

Let $ P(x,y)$ be the assertion $ g(x \plus{} y) \plus{} g(x)g(y) \equal{} g(xy) \plus{} g(x) \plus{} g(y)$

$ P(x,0)$ $ \implies$ $ g(0)(g(x) \minus{} 2) \equal{} 0$ and so a first solution $ g(x) \equal{} 2$ $ \forall x$. From now, consider that $ \exists c$ such that $ g(c)\neq 2$, then $ g(0) \equal{} 0$

Let $ g(1) \equal{} a$
$ P(x,1)$ $ \implies$ $ g(x \plus{} 1) \equal{} (2 \minus{} a)g(x) \plus{} a$
So $ g(2) \equal{} 3a \minus{} a^2$ But $ P(2,2)$ $ \implies$ $ g(2)^2 \equal{} 2g(2)$ and so $ 3a \minus{} a^2 \equal{} 0$ or $ 3a \minus{} a^2 \equal{} 2$ and so $ a\in\{0,1,2,3\}$

1) If $ a \equal{} 0$ : $ g(x \plus{} 1) \equal{} 2g(x)$, so $ g(x \plus{} n) \equal{} 2^ng(x)$ and $ g(n) \equal{} 0$. Then $ P(x,2)$ $ \implies$ $ g(2x) \equal{} 3g(x)$ and so $ g(4x) \equal{} 9g(x)$  but $ P(x,4)$ $ \implies$ $ g(4x) \equal{} 15g(x)$ and so we get $ g(x) \equal{} 0$ $ \forall x$

2) If $ a \equal{} 1$ : $ g(x \plus{} 1) \equal{} g(x) \plus{} 1$. Subtracting then $ P(x,\frac yx)$ from $ P(x,\frac yx \plus{} 1)$, we get $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$ and so $ g(xy) \equal{} g(x)g(y)$. So we have a Cauchy equation and a function such that $ g(x^2) \equal{} (g(x))^2$, so $ g(x) > 0$ $ \forall x > 0$ and so $ g(x)$ is monotonous, so is $ x$.

3) If $ a \equal{} 2$ : $ g(x \plus{} 1) \equal{} 2$ ****[u][b]edited [/b][/u]: "and we got the same constant solution $ g(x) \equal{} 2$ we already found" $ \to$ "and we got a contradiction since we have $ g(0)\equal{}0$ in this part of the demo. (thanks Rafikafi for the remark) ***.

4) If $ a \equal{} 3$, we got $ g(x \plus{} 1) \equal{} 3 \minus{} g(x)$ and so $ g(x \plus{} 2) \equal{} g(x)$. Then $ P(x,2)$ $ \implies$ $ g(x \plus{} 2) \equal{} g(2x) \plus{} g(x)$ and so $ g(2x) \equal{} 0$  $ \forall x$, and so contradiction (since $ g(1) \equal{} 3$)

5) Synthesis
Hence the solutions :
$ g(x) \equal{} 0$
$ g(x) \equal{} 2$
$ g(x) \equal{} x$

Nice solution, there's just a tiny thing:

At point 3) you actually have a contradiction since $ g(0)\equal{}0$, otherwise you would have forgotten to check whether $ g(x)\equal{}2$ if  $ x \neq 0$ and $ g(0)\equal{}0$ is a solution, which is not.

[quote="Rafikafi"]Nice solution, there's just a tiny thing:

At point 3) you actually have a contradiction since $ g(0) \equal{} 0$, otherwise you would have forgotten to check whether $ g(x) \equal{} 2$ if  $ x \neq 0$ and $ g(0) \equal{} 0$ is a solution, which is not.[/quote]

Yes, you're perfectly right. I edited my solution.
Thanks for the careful reading :)


\cancel, \sout, and \xout don't work, as AoPS doesn't have the necessary Include files.
So... I was wondering how to cancel in here?
I asked this last year and promptly forgot, but I know it has to do with using a negative space (forgot how) and then the cross-out symbol (again, forgot how). Can anyone help?
TIA
 :help:

EDIT:
Scrolling through the archives, I have found other queries and answers, but I want this to be up in front because I'm guessing a lot of people want to know but don't, and, like me (the first time around), didn't see the "Next Page" button.

Many people are also aware that this site has a search utility: http://www.artofproblemsolving.com/Forum/search.php

Alright, sorry.
Whoever has the power can delete this, if they wish.


i think it should be A

do u have a problem with marking A or B, or r u getting C or D as the answer??

coz if u r hard on deciding whether R is a correct reason of A, just remember as a thumb rule that:
1. in JEE, the chances of ambiguities in A and R is very low
2. in such a question, i think u should mark A, because R does explain major portion of A, the rest of equating each term to 0 and all are just minor calculations.

Hey Shrids already gave the answer. Its D. Thats what my book said. :( Simply because x,y,z may not be reals. :(

That swat i was able to point out.  :D
This concept of no square being negative is quite powerful, for eg. as in the following problem:
Determine all one-to one functions on $ R \rightarrow R$ with the property for all $ x$
           $ f(x^{2}) \minus{} f^{2}(x) > \equal{} \frac{1}{4}$.


He chose Edwards???  I didn't think he could make that good of a choice   :D .

Vote for [url=http://constitutionparty.com/]these guys[/url] instead (no, that's not a link to the GOP website. It's a link to a party that sticks to its premises, unlike so many others).

[quote="Rep123max"]George Bush came into office, thats what happened to it. Its gone.[/quote]
WTF, mate? Lincoln suspended the writ of habeus corpus without the prior consent of Congress; FDR... well, what didn't he do?; Clinton broke rules with his various military actions. That is by no means an exhaustive list or an attempt at one. The fact remains that presidents have always tried to "work around" the Constitution with this or that, and some, like FDR, have been responsible for very large, permanent deviations from it. It has always been the responsibility of the other branches of government, and most importantly, of the people, to see that the Constitution remains the law of the land. Washington said "Government is not reason and it is not eloquence. It is force! Like fire it is a dangerous servant and a fearful master."

When you say that George Bush ruined the Constitution, I assume that you're referring to the USA PATRIOT Act, which was passed by ... oh yea, Congress. If you're talking about the war in Iraq, congratulations, here's a link to the [url=http://www.house.gov/Constitution/Constitution.html]Constitution[/url]. The purpose of the government is to protect the people from domestic and foreign danger. Nothing more, nothing less.

[quote="confuted"][quote="failure"]From the Constitution Party website:
[quote]
Peroutka: God Family Republic[/quote]
Whatever happened to separation of church and state?  So much for the Constitution.  The irony is stunning.[/quote]
What happened to it is that it never existed. I'll say it to you, too: here's a link to the [url=http://www.house.gov/Constitution/Constitution.html]Constitution[/url]. Nowhere in that document, the Bill of Rights, or the Declaration of Independence will you find mention of a "seperation of Church and state." People commonly interpret the first amendment as this seperation:
"Congress shall make no law respecting an establishment of religion, or prohibiting the free exercise thereof..."
However, that certainly does not say "Government oficials shall be athiests" or anything of the sort. It states that the government shall not establish an official Church like the Church of England.

It seems to me like the irony was introduced when someone who doesn't understand the Constitution tried to criticize a group that does.[/quote]

I was fully aware of the limited capacity in which separation of church and state exists upon posting my previous message. Legislating "God" is very clearly making a law that "respects establishments of religion."

Kerry is a Catholic. Edwards and Bush are both Methodist. I loved Kerry and Edwards. "If you were here long time ago. for the democratic primary thread. I was the only one who had enough ESP to predict to say Kerry would win." Edwards is awesome. He is 51. People's maginize or Time (forgot which one). called him the sexiest politician. His wife is 55. He has a 3 and 1 year old. I love his quotes.

I actually defeated an incumbent Republican senator who was part of the Jesse Helm's political machine in North Carolina, the result of which is I'm now the senior senator from North Carolina instead of Jesse Helms, which is a very good thing for this country.
John Edwards

I didn't get to the Senate by accident.
John Edwards

I forgot the exact quote. But it goes something like this that his voluteers were armed with.

John Edwards will beat President Bush in the North, East, West, South. Some spetical guys (maybe confuted and some republicans) would say but Bush is strong in the south. The South is my backyard, and I will beat George Bush in my backyard," the North Carolinian said. "You give me a shot at George Bush. I'm going to give you the White House!

BTW all of the Democratic Presidential hopefulls in the last 4o years have been from the south, except kerry. GO SOUTH!

Erm, not quite.  Dukakis was from Massachusetts.  Mondale was from Minnesota.  McGovern was from South Dakota.

Sigh.  This is a rather disturbing trend, IMO.

[quote="confuted"]What happened to it is that it never existed. I'll say it to you, too: here's a link to the [url=http://www.house.gov/Constitution/Constitution.html]Constitution[/url]. Nowhere in that document, the Bill of Rights, or the Declaration of Independence will you find mention of a "seperation of Church and state." People commonly interpret the first amendment as this seperation:
"Congress shall make no law respecting an establishment of religion, or prohibiting the free exercise thereof..."
However, that certainly does not say "Government oficials shall be athiests" or anything of the sort. It states that the government shall not establish an official Church like the Church of England.

It seems to me like the irony was introduced when someone who doesn't understand the Constitution tried to criticize a group that does.[/quote]

True.  However, the premise of Congress making "no law respecting an establishment of religion" is that the government's stance on religion is not related to any specific religion or group of religions.  The government should not even be neutral; it is to be totally severed from religion, and should ignore religion in its legislation.  Hence, the interpretation leading to the phrase, "separation of Church and State."  The inclusion of the word "God" is a definite reference to religion, and therefore addresses the concept of religion, clearly unconstitutional.

[quote="confuted"][quote="Rep123max"]FDR... well, what didn't he do?;[/quote]

He didn't get elected via at best shady (Jews for Buchanan) or at worst illegal (purge of non-felons from the voter list, racist-in-application demands for identification, etc.) means.

Oh, and don't forget Jackson.

[quote="confuted"]I assume that you're referring to the USA PATRIOT Act, which was passed by ... oh yea, Congress.[/quote]

Not everything passed by Congress is constitutional, man.

And, of course:

[quote="confuted"]If you're talking about the war in Iraq, congratulations, here's a link to the [url=http://www.house.gov/Constitution/Constitution.html]Constitution[/url]. The purpose of the government is to protect the people from domestic and foreign danger. Nothing more, nothing less.[/quote][/quote]

Foreign danger ... right.  That's why we're nation-building there, right?  That's why we attacked a country without weapons that could damage us or even a method of landing them here.  That's why hundreds of U.S. soldiers are dead, to protect us from a regieme with no power?  Gimme a break.

[quote="dts"]Gee, hindsight is always 20/20, isn't it?  Correct me if I'm wrong, but before we invaded, didn't everyone agree (even some countries like France and Germany) that Iraq had WMDs?  Back then, the question was: should we allow them to keep the WMDs, or should we follow through with a UN resolution requiring them to disarm or face the consequences?[/quote]

Thats exactly it.  Everyone agreed about it.  We all thought there were WMDs.    Faulty intelligence it can happen but you take some chances thats part of military.

Why not?  Even if Saddam posed more of a threat to his own people than to us, a large portion of the American people were practically up in arms for the President to take SOME sort of action in the Middle East.

[quote="Rep123max"]Ya, diplomatic. Over 50% of Americans are now against the war in Iraq. And if it is only because that Saddam was a brutal dictator, why not North Korea, why not Iran why dont we start a nuclear war (thats why we dont attack them, because they have weapons, Iraq however didnt so we decided to attack).[/quote]

I added the emphasis on the word "now" to make my point.  Hindsight is better than 20/20.  It's 20/1.

And since when is not attacking a nuclear power a bad idea?  Attacking when the opponent is armed to decimate a large portion of the continent isn't patriotic; it's potentially foolhardy.  If you had the option of attacking a)Iraq, b)Iran, c)North Korea, or d)President Bush, which of the first three would you choose?  You've already accomplished the latter.

Just some more reading material, Rep:

http://anncoulter.com/

Read the column for July 7th.

And before you start saying what a bigot Coulter is, let me point out that she at least has an education (unlike your beloved Michael Moore, the college droupout).

Rep, have you read John Kerry's book?

I'm reading it now. It's boring.

[quote="dts"]Correct me if I'm wrong, but before we invaded, didn't everyone agree (even some countries like France and Germany) that Iraq had WMDs?[/quote]

Okay, I correct you.  Remember the hundreds of thousands of people protesting in New York, San Francisco, and in a variety of smaller protests around the country?

We didn't think so.

Magnus -- anyone can come up with clever put-downs.  Now, if she actually had something productive to add to the political arena, rather than the poison she spits, she might be worth reading.

The protestors you refer to were not protesting on the grounds of whether there were WMD's or not.  They were protesting on the grounds that they feel that war-any war-is morally wrong.  They did not have access to the same "intelligence" that heads of state did.  If you want to debate about public opinion, remember that over half a century ago, quite a few Americans were opposed to entering WWII.

As a matter of clarification: many had heard of the Holocaust; few believed it as truth.

[quote="dts"]They were protesting on the grounds that they feel that war-any war-is morally wrong.[/quote]

That would be interesting if it were true.  People protested for many reasons, but the reason that united everyone, and the reason I was protesting, was that [i]this[/i] war was wrong.  Why was this war wrong?  Because it hadn't been justified.
Now that we're months passed "Mission Accomplished," the war still hasn't been justified.

BTW, I think JBL should team up with Michael Moore.  That'd be very interesting. (I loved Fahrenheit 9/11, so that was meant as a compliment.)

And I thought this was the Kerry-Edwards thread, not the Bush-Cheney thread. Yeah, I could go on about Halliburton, "F___ yourself," what Bush did on August 6, 2001 (i.e. nothing), etc. But I'm not. If there's a Bush topic, I'll do it there.

And before anyone accuses me of being a liberal who takes things way out of proportion: I'll admit, I'm a proud liberal. But who's out of proportion: you talking about President Bush's greatness on a Kerry topic, or me talking about Kerry on a Kerry topic? I don't know. I'm not a conservative.

I can't see Farenhiet 911 because its rated r. Why would a documentary be rated 4? RRR!

It's rated R because of a lot of war footage.

Anyone involved in politics needs to see [url=http://www.ava.nu/thisland.htm]this[/url] before they make their decisions. Agreed?

[quote="Fermatprime"]BTW, I think JBL should team up with Michael Moore.  That'd be very interesting. (I loved Fahrenheit 9/11, so that was meant as a compliment.)[/quote]

Well, I'm flattered.  I don't think I'm the man for the job, though -- if I had to have a job related to digging through 3 1/2 years of a presidency (or 3 years of war, whichever) to find ... well, to find anything, I think it would just depress me.  (This actually happens to me from time to time just reading the newspaper in the morning -- there was a long while when I couldn't bring myself to read anything about Israel/Palestine.)  Besides, there are hundreds of people more effective than I at expressing their views in a way that would make anyone care to listen.

[quote=samrocksnature]16 year bump...dang politics back then[/quote]

Can you remind me of the purpose of bumping this lol

We see that presidents never fail to be criticized! :P

this goes so far back to the point where idek what they're talking about.

It was a thread on the 2004 presidential election, specifically John Kerry's pick for vice president.

[quote=failure]
Whatever happened to separation of church and state?  [/quote]

we don't know.


binomial coefficients
combinatorics proposed
combinatorics

It's just wrong: for $n = 2$, for example, we have $\sum_{i = 1}^{1}\binom{2}{i}\binom{1}{i}2^{n-2i}= 2 \neq \binom{4}{2}$.  If we begin the sum at $i = 0$ it becomes correct for $n = 0, 1, 2, 3$ but for $n = 4$ we have $\sum_{i = 0}^{2}\binom{4}{i}\binom{3}{i}2^{4-2i}= 16+48+18 = 82 \neq \binom{8}{4}$.

But as I said, it's wrong:  your sequence goes 1, 2, 6, 20, 82, 312, ...  while the central binomial coefficients are 1, 2, 6, 20, 70, 252, ....

Yes It is Wrong . Sorry everyone . That is true :
          $\sum_{i=0}^{[\frac{n}{2}]}C_{n}^{i}.C_{n-i}^{i}.2^{n-2i}=C_{2n}^{n}$
P/S Who rate it is spam ! I don't think that :wink:

How many sequences of length $n$ can be made up of the letters $A, B, C,$ and $D,$ subject to the constraint that the number of occurrences of $A$ and $C$ are equal?

Method one: We construct our sequence as follows: Choose some $i$ which will be the number of occurrences of $A$.  Choose $i$ letters out of the $n$ in the sequence to be $A$.  Choose $i$ letters of the remaining $n-i$ to be $C.$  Pick either $B$ or $D$ for each of the remaining $n-2i$ letters.  For each $i$ this corresponds to the summand on the left hand side, so the total number of ways is the sum on the left hand side.

Method two: Using the correspondence $A=00, B=01, C=11, D=10$, we associate each such sequence with a sequence of 1's and 0's of length $2n$.  This is clearly a bijective correspondence, and the condition that the number of $A$'s and $C$'s are equal corresponds to the condition that the number of 1's and 0's are equal.  This implies the colorings are in bijection with the sequences of exactly $n$ 1's and $n$ 0's, of which there are $C_{2n}^{n}$.

Since both sides count the same thing, they are equal.


geometry
geometric transformation
reflection
projective geometry
geometry theorems

We restate the problem as follows:

PROPOSITION. – A triangle ABC is given and let AD be it’s altitude through vertex A. We define an arbitrary point P, on AD and we denote as E, F, the intersection points of the sidelines AC, AB respectively, from the lines BP, CP. Prove that <ADE = <ADF.

PROOF. – Let S be, the intersection point of the segment lines BC, EF. It is well known (easy to prove by Menelaus’s and Ceva’s theorems), that the points S, S', are harmonic conjugates with respect to E, F, where S' the intersection point of the lines AD, EF. So, the pencil of lines D.ES'FS, is in harmonic conjugation.

We draw a line through vertex A and parallel to the line DF, which intersects the segment lines DE, BC, at points K, L, respectively.

It is well known that the point K, is the midpoint of the segment AL (If a pencil of lines is in harmonic conjugation, then a line parallel to one of it's rays, is intersected from the other three rays, at three points which form two congruent segments).

So, from the right triangle ADL, we have that <ADE = <LAD, (1).

From AL // DF, => <LAD = <ADF, (2).

From (1), (2), => <ADE = <ADF and the proof is completed.

Kostas Vittas.

$\begin{array}{c}\underline{\mathrm{L\ E\ M\ M\ A\ .}}\\\\ \boxed{\ \begin{array}{c}\underline{\mathrm{Hypothesis}\ : \ \triangle ABC}\\\\ A\in d\ ,\ d\parallel BC\\\\ M\in BC\ ,\ BM=MC\\\\ \{\begin{array}{ccc}D\in BC & & X\in d\cap DE\\\\ E\in CA & ; & Y\in d\cap DF\\\\ F\in AB & & Z\in AM\cap EF\end{array}\end{array}\ }\Longrightarrow\boxed{\ \begin{array}{c}\underline{\mathrm{Conclusion.}}\\\\ AX=YA\\\\ \Updownarrow\\\\ AD\cap BE\cap CF\ne\emptyset\\\\ \Updownarrow\\\\ Z\in BX\cap CY\end{array}\ }\end{array}$

[color=darkblue][b]Remark.[/b] If $AD\perp BC$ obtain the proposed problem and ... still something moreover.[/color]

Draw a paralell by $B$ to $AC$, denote by $X$ to the intersection of $DF$ and the paralell and $Y$ the intersection of $DE$ and the paralell $XBF$ and $DAF$ are similar, also $BYE$ and $CDE$ are similar. Then we get: $\frac{XB}{FB}=\frac{AD}{AF}$; $\frac{BY}{BE}=\frac{DC}{EC}$

By Ceva: $(\frac{AF}{FB})(\frac{BE}{EC})(\frac{CD}{DA})=1$

From these we get $XB=BY$, so $\boxed{\measuredangle XDB=\measuredangle BDY}$  $\mathbb{QED}$

[color=darkblue][b]Proof of the previous lemma.[/b] Denote the intersections $U\in BX\cap CY$ and $M'\in AU\cap BC\ .$ From the [u]Pappus' theorem[/u] obtain $U\in EF\ .$ Therefore,

$\{\begin{array}{ccc}|\begin{array}{c}\frac{EC}{EA}=\frac{DC}{XA}\\\\ \frac{FA}{FB}=\frac{YA}{DB}\end{array}\Longrightarrow\frac{YA}{XA}=\frac{DB}{DC}\cdot\frac{EC}{EA}\cdot \frac{FA}{FB}& \blacktriangleleft\maltese\blacktriangleright& XA=AY\Longleftrightarrow \frac{DB}{DC}\cdot\frac{EC}{EA}\cdot\frac{FA}{FB}=-1\Longleftrightarrow AD\cap BE\cap CF\ne\emptyset\ .\\\\ |\begin{array}{c}\frac{UA}{UM'}=\frac{XA}{BM'}\\\\ \frac{UA}{UM'}=\frac{AY}{M'C}\end{array}\Longrightarrow\frac{XA}{BM'}=\frac{AY}{M'C}& \blacktriangleleft\maltese\blacktriangleright& XA=AY\Longleftrightarrow BM'=M'C\Longleftrightarrow M'\equiv M\Longleftrightarrow U\in AM\Longleftrightarrow Z\equiv U\ .\end{array}\end{array}$[/color]

[quote="ashrafmod"]Let $ ABC$ be a triangle, and let $ D$ be the foot of the altitude of this triangle issuing from the vertex $ B$ (in other words, let $ D$ be the foot of the perpendicular from the point $ B$ to the line $ CA$).
Let $ P$ be any point on the line $ BD$. The line $ AP$ meets the line $ BC$ at some point $ E$. The line $ CP$ meets the line $ AB$ at some point $ F$.

Prove that the line $ BD$ bisects the angle $ EDF$.[/quote]

This is quite a well-known fact; I proved it as http://www.mathlinks.ro/Forum/viewtopic.php?t=20809 post #5 problem [b](a)[/b]. I have seen it being called the Blanchet theorem.

  darij

[b][size=100]A SIMPLE ELEMENTARY PROOF.[/size][/b]

We draw a line through $P$ and parallel to $BC,$ which intersects the segment lines $AB,$ $DF,$ $DE,$ $AC,$ at points $K,$ $L,$ $M,$ $N,$ respectively.

So, we have $\frac{PL}{PK}= \frac{DC}{BC}$ $,(1)$ and  $\frac{PN}{PM}= \frac{BC}{DB}$ $,(2)$ and  $\frac{PK}{PN}= \frac{DB}{DC}$ $,(3)$

From $(1),$ $(2),$ $(3)$ $\Longrightarrow$ $\frac{PL}{PM}= 1$ $\Longrightarrow$ $PL = PM$ $,(4)$

From $(4)$ we conclude that the segment line $AD,$ bisects the angle $\angle EDF$ $($ because $AD\perp LM$ $)$ and the proof is completed.

Kostas Vittas.

PS. This proof is based on the same idea as in the solution of a similar problem, by [b][size=100]April[/size][/b],
at http://www.mathlinks.ro/Forum/viewtopic.php?t=120958

Problem: $\triangle ABC$ with $D\in BC$ so $AD\perp BC$. $P\in AD$, denote $E=AC\cap BP$, $F=AB\cap CP$.

Solution: Reflect the diagram over $BC$ such that the image of any point, $X$, is $X'$. By menelaus' theorem $\triangle PCA$, transversal $FED$,

$\frac{PF}{FC}*\frac{CE}{EA}*\frac{AD}{DP}=1$

By virtue of reflection, $CE=CE'$, $E'A'=EA$, $A'D=AD$, hence we have:

$\frac{PF}{FC}*\frac{CE'}{E'A'}*\frac{A'D}{DP}=1$

This is menelaus' theorem for $\triangle PCA'$, transversal $FDE'$. Therefore, $F$, $D$, $E'$ are collinear. Hence $\angle FDP=\angle CDE'$ by vertical angles. By virtue of reflection, $\angle E'DC=\angle EDC$. It follows that $\angle FDP=\angle EDP$, as desired.

[color=darkblue]Denote $L\in EF\cap AC$ and $S\in EF\cap BD$. Is known that the division $(A,C,D,L)$ is harmonically $\implies$
the pencil $B(A,C,D,L)$ is harmonically $\implies$ the division $(F,E,S,L)$ is harmonically. But $DS\perp DL\implies \widehat{FDS}\equiv\widehat{EDS}$, i.e. $\widehat{FDB}\equiv\widehat{EDB}$.[/color]

I proved Blanchett's using the same method that I proved another problem, and then I realised that Blanchett's is actually a special case of that problem. Here's the link [url]http://www.mathlinks.ro/viewtopic.php?t=209078[/url]. Think of the triangle as a quadrilateral with one vertex as the foot of the altitude. I will post the proof again anyways.
We compute $ \frac {[BDF]}{[DFA]} \cdot \frac {[DEA]}{[DCE]}$ in two ways.
The first way, it is $ \frac {BF}{FA} \cdot \frac {AE}{EC}$.
The second way, we use the area formula 1/2 ab sinC on all four triangles. We have
$ \frac {1/2 \cdot BD \cdot DF \cdot cos FDA \cdot 1/2 \cdot DA \cdot DE \cdot sin ADE}{1/2 \cdot DF \cdot DA \cdot sin FDA  \cdot 1/2 \cdot DE \cdot DC \cdot cos ADE}$
Simplifying,
$ \frac {BD}{DC} \cdot \frac {tanADE}{tanFDA}$.
Setting them equal, and using Ceva,
$ tan ADE \equal{} tanFDA$
so $ \angle ADE \equal{} \angle FDA$

this theorem is a particolar case of this:

Let a triangle ABC and a point P on it; denote $ E: BP \cap CA$, $ F: CP \cap AB$ and $ Q: AP \cap EF$. Call K the projection of Q on BC, then $ \angle QKF \equal{} \angle QKE$.

Proof: Call $ D: AP \cap BC$, $ Q': EF \cap BC$, as well-know the pencil A.BDCQ' is harmonic so Q' is the harmonic conjugate of Q wrt EF with $ \angle QHQ' \equal{} \frac{\pi}{2}$ and then $ \angle QKF \equal{} \angle QKE$.

Dear Mathlinkers,
can we prove without calculations (reports) and harmonic division this problem?
A new challenge...
Sincerely
Jean-Louis


During practice every player takes two shots in a row on the basket. This is called a double. Alessandra misses 2/3 of her first shots in the doubles, but makes 2/3 of her second shots. If she makes the first shot in a double, then she makes the second shot, also. In the next double which outcome has the greatest probability:
a) she misses both shots;
b) she makes only one of the shots
c) she makes both of the shots?

[hide="Correct"]  But try solving some of the more difficult problems I posted [/hide]

Revenge of the Sith was awesome  :starwars:

[hide="solution"]Because she misses her first shot 2/3 of the time, she makes it 1/3 of the time. When she makes her first shot (1/3 of the time), she makes her second shot (1/3 of her second shots)

Lets look at it with 2 that she misses first and 1 that she makes first:

miss, make
miss, miss
make, make

This shows that for her to make 2/3 of her second shots, she makes 1/2 of her second shots when she misses her first shot. This means that the probabilities are:

A: 2/3*1/2 = 1/3
B: 2/3*1/2 = 1/3
C: 1/3*1 = 1/3

which means that they are all equally likely.

However, you could interpret the question as when she misses her first shot, she has a 2/3 chance of making her second. This makes them:

A: 2/3*1/3 = 2/9
B: 2/3*2/3 = 4/9
C: 1/3*1 = 1/3 = 3/9

making B have the greatest chance.[/hide]

I agree with the answers, and I agree that Revenge of the Sith was awesome

 :starwars:

the answers are correct but azness has the right numbers

I disagre. I believe that they all have a 1/3 chance of happening. Read my first solution again. If she makes her first shot, that already accounts for half of the second shots that she will make.

[quote]Posted: Sun May 22, 2005 1:36 pm    Post subject: Shoooting probability  #1

---------------------------------------------------------------------------

During practice every player takes two shots in a row on the basket. This is called a double. Alessandra misses 2/3 of her first shots in the doubles, but makes 2/3 of her second shots. [b]If she makes the first shot in a double, then she makes the second shot, also.[/b] In the next double which outcome has the greatest probability:
a) she misses both shots;
b) she makes only one of the shots
c) she makes both of the shots?
[/quote]

My first answer is correct unless you take out the bold sentence in which case aznness does have the correct numbers

G-UNIT asked me to comment on this problem.  As the question is written, I believe Hamster1800 is correct -- the events a, b, and c each have probability 1/3.  Here is my analysis.

Let $A$ be the event "she makes the first shot".  Let $B$ be the event "she makes the second shot".  The third sentence of the question says that $\Pr(A) = 1/3$ and $\Pr(B) = 2/3$.  The fourth sentence says that $A \implies B$; in other words, $A$ and $\overline{B}$ (the complement of $B$) never take place simultaneously.  We can create a two-by-two table of probabilities:
\[
\begin{array}{c|c|c|c}
                     & B    & \overline{B} \\ \cline{2-3}
A         &  & 0  & 1/3 \\ \cline{2-3}
\overline{A} &   &  & 2/3 \\ \cline{2-3}
                     & 2/3 & 1/3
\end{array}
\]
Each row has to add up to the probability on its right edge.  Similarly, each column has to add up to the probability on its bottom.  There is only one consistent way to fill in the table:
\[
\begin{array}{c|c|c|c}
                     & B    & \overline{B} \\ \cline{2-3}
A         & 1/3 & 0  & 1/3 \\ \cline{2-3}
\overline{A} & 1/3  & 1/3 & 2/3 \\ \cline{2-3}
                     & 2/3 & 1/3
\end{array}
\]
Looking at the bottom right cell, we see that event a has probability 1/3.  Looking at the top left cell, we see that event c has probability 1/3.  Looking at the other two cells, we see that event b also has probability 1/3.

I would also agree with this, as I said so in PM.

thanks alot guys, i understand the correct solution now

[quote="4everwise"]Ravi B's comment was interesting.[/quote]
it was, and Revenge of the sith was awesome :starwars:

Hm, for some reason I'm still not getting those answers. Let me explain my thinking:
1/3 chance of making first shot.
2/3 chance of making second.
If she makes the first shot, she has to make the second.

Probability of missing both: 2/3 * 1/3 = 2/9
Probability of making only one shot: has to miss first, which is 2/3, and has to make second, which is 2/3, so 2/3 * 2/3 = 4/9
Probability of making both shots: 1/3 = 3/9

What am I doing wrong?

I don't know, it seems right.  Your logic and wording makes it seem like you have the right answer so I agree with you.

[quote="Phelpedo"]Hm, for some reason I'm still not getting those answers. Let me explain my thinking:
1/3 chance of making first shot.
2/3 chance of making second.
If she makes the first shot, she has to make the second.

Probability of missing both: 2/3 * 1/3 = 2/9[/quote]
The two shots are not independent.  You are trying to use the formula $\Pr(X \textrm{ and } Y) = \Pr(X) \cdot \Pr(Y)$, but that formula is true only when the events are independent.

Given that she misses the first shot, her probability of missing the second shot is actually $\frac{1}{2}$, not $\frac{1}{3}$.  (You can use my previous post to prove it.)

Here is a rough way to think of it.  Whenever she makes the first shot, she makes the second shot, right?  In other words, making the first shot improves her chance of making the second shot.  So, to make the probabilities balance out, missing the first shot has to reduce her chance of making the second shot, below her [i]a priori[/i] probability of $\frac{2}{3}$.

Ah, I see what you are saying, but I interpreted that they would be independent. Like, whenever she shoots the second shot, no matter what she shot before, she has a 2/3 chance, and some miracle happens when she makes the first shot.


Hi,

Let $x,y,z>0$ be three real numbers. Prove that
\[ {{1}\over{x+y}}+{{1}\over{y+z}}+{{1}\over{z+x}}\leq {{(x+y+z)^2}\over{6xyz}}. \]

A little transformation would make it

${{2}\over{x+y}}+{{2}\over{y+z}}+{{2}\over{z+x}}\leq {{(x+y+z)^2}\over{3xyz}}$.

By AM-GM we obtain $\sum{{2}\over{x+y}} \leq \sum{{x+y}\over{2xy}}\leq{{(x+y+z)^2}\over{3xyz}}.$
So we obtain $\frac{3}{2} \sum{(xy+xz)} \leq {(x+y+z)^2}$

Which is equivalent to ${3} \sum{xy} \leq {(x+y+z)^2}$ which is obviously true.


Hello!

I have a problem I've been working with for a few days. Now I know that you can solve these questions with just sheer determination and expanding to match the RHS to LHS, but I am thinking there must be an easier way to do these types of questions?
The question is:

(m+n)C(3) = (m)C(3) + (m)C(2)*(n)C(1) + (m)C(1)*(n)C(2) + (n)C3

Now, I know this means that all the combination's of m+n objects taken 3 at a time is equal to the total combination's of m objects taken 3 at a time, plus m objects taken two at a time times n objects taken 1 one at a time, so on, which makes all the possibilities for the different combinations of n+m, thus the expression is true.

What I am concerned about is if there is any other method of doing this? What about induction?

Thank you for taking the time to read this, and any insight will be much appreciated.

Typically the combinatorial proofs are the most elegant.

Alternatively, the algebraic proof is simply expanding $ \binom{n}{k}\equal{}\frac{n!}{(n\minus{}k)!k!}$ and simplifying one side.

Usually you'll only see the above two. Induction is less common (at least from what I've seen) on these kinds of problems. (One example is AoPSWiki's proof of the Hockey Stick Identity, which uses induction).

I wouldn't do this with induction because it's in two variables.

Is there any other way of solving it bar from expanding? I know it can be easily(strenuously) done like that, but couldn't there be an additional way? Like maybe prove each of the terms and then you can easily say they add to LHS

I can't think of any other way to prove this.
What's wrong with a combinatorial proof?

You could do this:

$ \binom {m\plus{}n} 3 \equal{} \binom m 3 \plus{}\binom m 2 \cdot \binom n 1 \plus{} \binom m 1 \cdot \binom n 2 \plus{} \binom n 3$

$ \frac{(m\plus{}n)(m\plus{}n\minus{}1)(m\plus{}n\minus{}2)}{6}\equal{}\frac{m(m\minus{}1)(m\minus{}2)}{6}\plus{}\frac{mn(m\minus{}1)}{2}$ $ \plus{}\frac{mn(n\minus{}1)}{2}\plus{}\frac{n(n\minus{}1)(n\minus{}2)}{6}$

Just multiply everything and you should get the result (I don't want to, it's too annoying).

The LHS is the coefficient of $ x^3$ in $ (x \plus{} 1)^{m \plus{} n}$ and the RHS is the coefficient of $ x^3$ in $ (x \plus{} 1)^m (x \plus{} 1)^n$.  This observation generalizes to [url=http://en.wikipedia.org/wiki/Vandermonde%27s_identity]Vandermonde's identity[/url], but this is not essentially different from the combinatorial proof.

What's "bulky" about a combinatorial proof?  You won't find a better interpretation of this result.  You could prove this particular result by, for example, letting $ m \plus{} n \equal{} 1, 2$ in the LHS and showing that the RHS is zero and then comparing leading terms, but...


The solutions for Round 2 will be posted Wednesday.  Do not discuss Round 2 until then.

Any idea on when results will come out? Also, when will the next round be posted?

Just wondering...its Wednesday but the solutions are not up yet.


How do you go about solving an inequality question? I mean, when I look at the solutions, people seem to know exactly what to do. Mutiply by this, use this, expand this, factorize this! Plus these new techniques SMV, MV, LCF, ABC....... I just can't get my head around them.

I think this basically comes from experience. If you're experienced enough you can tell by looking at an inequality which inequality can be useful in solving it. Then you try that approach and if it somehow fails you try another approach and thus. And when you're stuck you usually try expanding if the situation seems appropriate for it. Also, one has to be very familiar with a number of algebraic identities that can be useful during calculations. This, as well, follows from experience, or should I say, practice. Practice makes perfect, as we all know! :wink:

[hide="For example, when I see"] an inequality involving square-roots and/or fractions I usually try Cauchy-Schwarz or Holder's inequality, depending on the inequality sign. Not all the time it is fruitful but it is my first choice for such inequalities. An example: http://www.mathlinks.ro/viewtopic.php?t=177947

And for symmetric inequalities assuming $ a\ge b\ge c$ or such is a general first choice.

Take a look at this: http://www.mathlinks.ro/viewtopic.php?t=179027. When I first looked at the inequality it reminded me at once of the weighted AM-GM. I tried with it and luckily got a solution within a minute, and the rest of the total nine minutes I spent on typing out the solution.  :roll:

Also, being familiar with well known inequalities come in handy. Take a look at the topic: http://www.mathlinks.ro/viewtopic.php?t=180322
[/hide]
These, however, are not always useful depending on the difficulty of the problem.

That is just my opinion on this topic which I don't say you need always do, due to my lack of experience and knowledge. :oops:  Different people may have different views on this.

Check out the sources at http://www.cip.ifi.lmu.de/~grinberg/faq.html#ineqsbooks . As for the three letter methods, they tend to get pretty ugly when applied (UMV being an exception), but the ideas are mostly simple (maybe except for ABC), and you can discover many of them on your own.

Here is a small list (I have searched only for posts with myself as author) for the mixing variables variants:

MV = mixing variables [b]in the old meaning of the word[/b] (let's call it finitely mixing variables):
http://www.mathlinks.ro/Forum/viewtopic.php?t=23892
http://www.mathlinks.ro/Forum/viewtopic.php?t=5087
http://www.mathlinks.ro/Forum/viewtopic.php?t=6347
http://www.mathlinks.ro/Forum/viewtopic.php?t=23108
http://www.mathlinks.ro/Forum/viewtopic.php?t=6397

MV = mixing variables [b]in the new meaning of the word[/b] (let's call it infinitely mixing variables):
http://www.mathlinks.ro/Forum/viewtopic.php?t=113248 Theorem 1

UMV, also called EMV = ultimately (or entirely) mixing variables:
http://reflections.awesomemath.org/2006_5/2006_5_entirelymixing.pdf
http://www.mathlinks.ro/Forum/viewtopic.php?p=251869#251869 see this for a non-standard application of the method

SMV = symmetrically mixing variables
http://www.mathlinks.ro/Forum/viewtopic.php?t=113248

I'd ask others to post links for the non-MV methods...

  Darij


Chào tất cả,

Gần đây mình thấy rằng vài thành viên trên trang này là người Việt , rất hay gõ tiếng Việt trong box khác, mình nghĩ các bạn chỉ nên gõ tiếng Việt ở box này thôi.

Đây là một ví dụ

http://www.mathlinks.ro/Forum/viewtopic.php?t=138771

Thân.

[color=purple]Minh` la` evarist minh` rat' hay go~ TV trong ca'c bai` post tren ML  bay jo` minh` xin hua'  se~ ko post TV tren ML nua~ .Min`h rat' hi vong cac' ban cung~ se~ lam` nhu the' vi` no' se~ lam anh? huong? toi' uy tin' cua? thanh nien VN[/color]

Lại thế nữa http://www.mathlinks.ro/Forum/viewtopic.php?p=821466#821466 tôi không lock topic đó bởi vì bạn nguyenphihung là 1 thành viên mới. Bạn hãy chú ý đến các bài viết sau của mình nhé!  :wink:

[quote="N.T.TUAN"]Hôm nay là cậu VANCHANH123 http://www.mathlinks.ro/Forum/viewtopic.php?t=147525 , trong topic đó còn có 2 post dạng spam, nhưng tôi chưa delete, để đấy các mod khác làm, dù sao hôm nay tôi cũng locked 2 topic của cậu này rồi :( Nhớ khi xưa làm Mod cho diendantoanhoc.net anh em cũng rất vất vả với cậu này thời gian đầu. Thắm thoắt đã 1 năm rồi...[/quote]
Quả thật .. khi tham gia diễn đàn toán học thì .. lúc đó em mới biết đến IE
Còn nay em cũng khá tệ tiếng Anh tuy nhiên em  muốn học hỏi nhiều .. vàem sẽ cố học Anh nhiều hơn

[quote="N.T.TUAN"]Vừa rồi anh cũng lock 3 topic của chú, lý do là tên topic đặt lung tung. Cố gắng em nhé!  :wink:[/quote]
Sao lại lung tung anh ??
Thì anh có  thể nhắc em sửa tên lại mà
Mà sao lại lung tung
quả nó là những bái toán mới ( có phần sáng tạo ) không lẽ nó không xứng` đáng với  từ [size=150][b]New[/b][/size] sao anh?

Chú cần đọc ngay topic mà anh đã dẫn trong các topic đã bị lock. Sau đó anh em ta nói chuyện tiếp!  :wink:

[quote="DuongChinh_K41SPHN"]Tieng anh cua em cuc ki la tôi tê .Hoc tieng Anh qua la kho .Khi nao em cân su giup đo ve tieng anh thi cac anh nho giup em nhe[/quote]
Ủa,tiếng Anh của ông kém,sao hôm nọ thằng Thế Anh bảo cô giáo hỏi ai học kém tiếng Anh cô tổ chức cho học miễn phí mà ông không giơ tay,oai nhỉ?
(thằng này nổi tiếng nhiều sách "gay" bên 10 toán 2,lan sang tận cả lớp mình!)

[quote="N.T.TUAN"]Cứ post bừa đi, đừng ngại. Thằng Quang nó biết gì tiếng Anh đâu mà post ầm ầm đấy thây.  :lol:[/quote]
Anh nói độc thật!
Thì cứ 'we have' với 'It follows that' như em cũng được hơn 100 bài rồi anh nhỉ!!

[quote="N.T.TUAN"]Cứ post bừa đi, đừng ngại. Thằng Quang nó biết gì tiếng Anh đâu mà post ầm ầm đấy thây.  :lol:[/quote]

Anh Tuân nói cái gì đây hả  :mad:

Các bạn không nên tranh luận gay gắt trên forum này có gì thì nói bên ngoài. Chúng tôi không muốn như thế.

[quote="KealEzlo"]nhưng trìh độ tiếng anh của em yếu kém thì làm sao ạ :(( đọc thì hiểu nhưng post thì hok bít :(([/quote]
Cứ viết tràn em à, biết gì viết nấy, sai thì sửa rồi cũng quen thôi! :)  :)

E mới học Toán bằng tiếng Anh có ... 1 tuần  :rotfl:  mà post bài ầm ầm đấy thôi.
Công nhận vừa xem vừa tra từ điển, vốn từ của mình lên nhanh phết :D

[quote="Kyoshiro"]Công nhận vừa xem vừa tra từ điển, vốn từ của mình lên nhanh phết :D[/quote]

Ha ha, anh cũng thế, hoặc có thể bật sẵn chương trình dịch Việt - Anh trong máy lên cũng được!!  :D


While standing in line to buy concert tickets, Kit moved 60 feet closer to the ticket window over a period of 30 minutes. At this rate, how many minutes will it take her to move the remaining 70 yards to the ticket window?

We have \[ \frac{70}{20}\equal{}\frac{x}{30}\Rightarrow x\equal{}\boxed{105}\]

I'm confused... wouldn't you do $\frac{70}{x}=\frac{60}{30}$ so $x=\boxed{35}$?

60 FEET vs. 70 YARDS.

That's either 20 yards or 210 feet, and in either case, $105$ is the correct answer.

$\frac{60 ft}{30 min} = \frac{70 yd}{x}$

Multiply both sides by $30$ min and $x$, leaving

$x \cdot 60ft = 70 yd \cdot 30 min$

$x = \frac{70 yd}{60 ft} \cdot 30 min$
From here, you can either:
1)  Convert the yards to feet
2)  Convert the feet to yards (yay, more fractions...sorta.)
3)  Recognise that the fraction $\frac{yd}{ft}$ has a numerical value of $3$.
4)  Multiply the fraction by $\frac{3 ft}{1 yd}$, which is equal to one, cancelling out both the feet and the yards, and leaving minutes as your only units.

Either way, you'll wind up with:
$x = \frac{70 \cdot 30\cdot 3}{60} min = 105 min$


How did you do on MMPC part 1?

I guess this only applies to about five people, such as myself, Ragging and Confuted, but i am going to ask anyway.

Also, the problems and solutions are up. The cutoff for part II is 23 this year. Check out

http://www.delta.edu/math/mmpc

I got 36  :(

Looks like I got a 25, going through the key really quickly. I may have miscounted, of course.

... I got some wrong that I thought I should have gotten right  :(  :(  :(

edit: maybe 24. Gah, this means I'll have to go through the key AGAIN
edit2: yeah, 24. Pretty horrible, but good enough to advance I guess.

Just the number of questions correct out of 40, no guessing penalty. The cutoff for part 2 is based on the score of the top 1,000 kids (of 14,000 taking part 1)... Approx. 1000 kids advance to part 2, and the top 100 or so composites get scholarships.

actually, top 100 get invited to a banquet, but only the top 50 actually get a scholarship. the combined score usually needed for top 100 is 46-48 (last year 46.8). for scholarhips, 50-52. the gold, silver are usually scores of 68 +

whoa a banquet damn now i wish i hadn't moved to NC


Given a quadrilateral $ ABCD$ such that $ AD\equal{}BC$. Let the line $ AC$ meet the line $ BD$ at $ O$, and let the angle bisectors of the angles $ DAB$ and $ CBA$ meet each other at $ I$. Prove that the midpoints of the segments $ AB$, $ CD$, and $ OI$ are collinear.

To Mods: Please lock this topic, this is a problem from M&Y, issue 9, and we can't discuss it now

Let $ X\equal{}AD\cap BC$ and $ M$ be midpoint of $ OX$. By Steiner theorem, line passing midpoint of $ AB,CD$ pass $ M$ and it is easy to show that this line is parallel to bisector of $ \angle AXB$ and since $ I$ is $ X$-excenter of $ \triangle AXB$, $ IX$ is bisector of $ \angle AXB$. So line passing midpoint of $ AB,CD$ coincide with line passing $ M$ and parallel to $ IX$ and this line  obvioulsy pass midpoint of $ OI$. Hence we are done

[quote="Jumbler"][quote="Leonhard Euler"] By Steiner theorem, line passing midpoint of $ AB,CD$ pass $ M$ [/quote]

How does this follow from Steiner?[/quote]

What he calls Steiner theorem is actually the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=79751]Gauss line theorem[/url]. Applying this theorem to the quadrilateral ADBC (not ABCD), we see that the midpoints of the segments AB, DC and XO are collinear. Since the midpoint of XO is M, you obtain what you want.

  darij

[quote="Jumbler"][quote="Leonhard Euler"]it is easy to show that this line is parallel to bisector of $ \angle AXB$ [/quote]

Another thing, why is this?[/quote]

This is pretty much the statement of http://www.mathlinks.ro/viewtopic.php?t=17082 .

  darij


topology
vector
function
advanced fields
advanced fields unsolved

is it true that any smoothe contractible manifolds are parallelizable?

Any vector bundle on a contractible paracompact space is trivial.  See GTM 33 Differential topology by Hirsch.

I don't think, we really need such generalizations.

I hope it works!

Let $M$ be our contractible manifold of dimension $m.$ Since it is contractible, there exists $H: M \times [0,1] \rightarrow M$ s.t. $H(x,0)=C(x)=x_0$ and $H(x,1)=x$ for all $x\in M$ where $C$ was assumed to be the constant function on $M.$ For $\pi: TM \rightarrow M$ as a vector bundle ([url=http://en.wikipedia.org/wiki/Fibration]fibration[/url]) there is a map $\hat{H} :M \times [0,1] \rightarrow TM$ s.t. $\pi \circ \hat {H}=H.$ So we have $\hat{H}(x,0) \in \pi^{-1}(x_0)\simeq \mathbb{R}^m$ and $\hat{H}(x,1) \in T_x M$ for all $x \in M.$

By local triviality of $TM$, there is a neighborhood $U$ at $x_0$ s.t.  for $z \in U$, there are $m$ sections (vector fields) $X_1(z),...,X_m(z)$ that are linearly independent in $T_{x_0}M.$ Now, for a given point $x \in M$, we should find $m$ sections which are linearly independent in $T_xM.$ (See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=100729]this[/url], for example). Look at the path $\hat{H}_x(t): [0,1] \rightarrow TM$ starting at $\hat{H}_x(0) \in T_{x_0}M$ and ending in $\hat{H}_x(1) \in T_xM.$

Claim: It is possible to extend our sections via $\hat{H}_x(t)$ throughout our manifold, in a way that they remain linearly independent in $T_xM,  \; \forall x \in M.$

For any $t \in [0,1], \; \hat{H}_x(t)=(x(t), v(t))$ where $v(t) \in T_{x(t)}M.$ Again, by local trivially of $TM,$ there is a neighborhood $U_t$ at $x(t)$ where $TM$ is trivialized over it. Since $[0,1]$ is compact, there exist a finite cover of $[0,1]$ by $U_t$ namely, $\{U=U_0, U_1,...,U_n \}$ where $U_i \cap U_{i+1} \not= \emptyset.$ Therefore, since $TM$ was trivialized over each $U_i$ in particular over $U=U_0$, then we can extend our sections in a way that they remain linearly independent in $T_x M.$ Since $x$ was arbitrary, thus, that there are $m$ sections, $X_1,...,X_m$ s.t. for any $x \in M,  \; X_1(x),...,X_m(x)$ are linearly independent in $T_xM$ and we're done.

[quote="ma_go"]if i'm not mistaken, that's exactly the proof of the aforementioned theorem. you're not using anything about $TM$ being the tangent space of $M$, as opposed to any vector bundle over $M$.[/quote]

I haven't seen the mentioned theorem. I thought about it, yesterday. And yes, I admit that did not really use $TM$, so it should work for any vector bundle.

P.S. Manifolds are paracompact but not the reverse.

Is there a way to attack this problem by the cohomology of vector bundles? or Characteristic classes?

I'm asking this, since I know very little about them!


find x and y

$ \frac{\tan 8}{1\minus{}3\tan^2 8}\plus{}\frac{3\tan 24}{1\minus{}3\tan^2 24}\plus{}\frac{9\tan 72}{1\minus{}3\tan^2 72}\plus{}\frac{27\tan 216}{1\minus{}3\tan^2 216}\equal{}x\tan 8\plus{}y\tan 72$

[quote="mrmohamed"][quote="matrex"]find x and y

$ \frac {\tan 8}{1 \minus{} 3\tan^2 8} \plus{} \frac {3\tan 24}{1 \minus{} 3\tan^2 24} \plus{} \frac {9\tan 72}{1 \minus{} 3\tan^2 72} \plus{} \frac {27\tan 216}{1 \minus{} 3\tan^2 216} \equal{} x\tan 8 \plus{} y\tan 72$[/quote]
[color=darkred][size=150]
[b]hi all
x= (-1/8) ,  y = ( -81/8)
mrmohamed[/b][/size][/color][/quote]
How do you get that? Can you be more specific?

[quote="matrex"]find x and y

$ \frac {\tan 8}{1 \minus{} 3\tan^2 8} \plus{} \frac {3\tan 24}{1 \minus{} 3\tan^2 24} \plus{} \frac {9\tan 72}{1 \minus{} 3\tan^2 72} \plus{} \frac {27\tan 216}{1 \minus{} 3\tan^2 216} \equal{} x\tan 8 \plus{} y\tan 72$[/quote]

[size=150][color=darkred][b]Hi all

the solution[/b][/color][/size]
[url=http://www.servimg.com/image_preview.php?i=451&u=12899977][img]http://i99.servimg.com/u/f99/12/89/99/77/almonk25.gif[/img][/url]


inequalities
induction
inequalities proposed
109

Let $ a,b,c\ge 0$ and $ a\plus{}b\plus{}c\equal{}3$. Find the best $ k$ constant such that:
\[ \frac{a^{5}\plus{}b^{5}\plus{}c^{5}\minus{}3}{a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3}\ge k\]

I hope the last post of you will be the proof for this problem. I will say good bye to mathlinks forum. Good luck, everybody  :wink:

Let $ a\equal{}b\equal{}c\equal{}1$ , we get $ k\leq\frac{4}{9}$,and when$ k\equal{}\frac{9}{4}$ the inequality is eqivalent to

       $ 864F[5][0,1]\plus{}648F[5][0,2]\plus{}540F[5][0,3]\plus{}1296F[5][1,1]\geq 0$,

in which

       $ F[5][0,1]\equal{}\sum{a^{3}(a\minus{}b)(a\minus{}c)}\geq 0$,
       $ F[5][0,2]\equal{}\sum{a^{2}(b\plus{}c)(a\minus{}b)(a\minus{}c)}\geq 0$,
       $ F[5][0,3]\equal{}\sum{bc(b\plus{}c)(a\minus{}b)(a\minus{}c)}\geq 0$,
       $ F[5][1,1]\equal{}\sum{abc(a\minus{}b)(a\minus{}c)}\geq 0$,
the  inequality holds.Therefore  the best $ k\equal{}\frac{4}{9}$.

I don't think $ k\equal{}\frac{4}{9}$ is the best constant.  With $ k\equal{}\frac{4}{9}$ when does equality hold? $ a\equal{}b\equal{}c\equal{}1$ !? I don't think it is correct.

Actually, by General Induction Method (see my Vietnamese book or oncoming second volume of Secrets in Inequalities), we can prove the following results

For $ a_{1},a_{2},...,a_{n}\ge 0, a_{1}+a_{2}+...+a_{n}=n, k,r>0$ the following expression \[ \frac{a_{1}^{k}+a_{2}^{k}+...+a_{n}^{k}-n}{a_{1}^{r}+a_{2}^{r}+...+a_{n}^{r}-n}\]
attains the maximal and minimal value if and only if $ n-1$ variables equal.

For a particular result, I posted in VI forum that (it appears as a problem of day)

[i](page 150, SII book).[/i] Let $ a_{1},a_{2},...,a_{n}$ be non-negative real numbers such that $ a_{1}+a_{2}+...+a_{n}=n$. Prove that
\[ a_{1}^{4}+a_{2}^{4}+...+a_{n}^{4}-n\ge k(a_{1}^{3}+a_{2}^{3}+...+a_{n}^{3}-n)\]
in which the number $ k$ is determined for each $ n$ from
\[ k_{2}=2\ ;\ \ k_{3}=2\lt(\sqrt{63}-7\rt)\ ;\ \ k_{n}=1+\frac{n^{2}}{(n-1)(2n-1)}\ \mbox{\ for\ }n\ge 4\ ;\]

We have the equality hold on:
\[ a\equal{}b\equal{}3\minus{}\sqrt{3},c\equal{}2\sqrt{3}\minus{}3,k\equal{}\frac{5(5\sqrt{3}\minus{}7)}{\sqrt{3}\plus{}1}\]

The general problem's hungkhtn is really  hard for me :D


Subract $\displaystyle\frac{1}{3a}$ from both sides and you will get:

$\displaystyle\frac{1}{3a-1} + \displaystyle\frac{1}{3a+1} > \displaystyle\frac{2}{3a}$

Make a common denominator on the LHS and multiply by $\displaystyle\frac{3a}{3a}$ on the RHS:

$\displaystyle\frac{6a}{9a^2-1} > \displaystyle\frac{6a}{9a^2}$

Seeing as $a$ is positive, this holds true.

Here's a problem with a similar foundation:

Assume Town B is some distance directly south of Town A.  Now consider two scenarios:

1) A plane travels from Town B to Town A with some velocity and travels back to Town B with that same velocity, with no wind helping or hindering it.

2)  A plane travels from Town B to Town A with that same velocity, but the wind, traveling at another velocity against the plane, slows the plane down.  However, when the plane goes back to Town B at the same velocity, the wind (also at the same velocity) speeds the plane up.

Which scenario would take the least time?

What's the difference between $\mathbb{N}$ and $\mathbb{N}^*$? :?

I think. that the second N, (I mean the one with a star)is the set of natural numbers without zero

I think it is more accurate if we write $\forall a\in\mathbb{Z}^+$

There are many conventions.  When things are unclear, I prefer to write $\mathbb{Z}_{\geq 0}$ or $\mathbb{Z}_{>0}$.

[quote="fourierseries"]I think you mean subtract $\frac{1}{3a}$ from both sides rather than $\frac{1}{a}$[/quote]

whoops, yes I did. my fault.  anybody up for that second problem I posted?

For the second problem:

d=distance
r=plane's usual speed
x=wind's speed

Without wind, the time is
\[ \frac{d}{r}+\frac{d}{r}=\frac{2d}{r}  \]

With wind, the time is
\[ \frac{d}{r+x}+\frac{d}{r-x}=\frac{(r+x)d+(r-x)d}{(r+x)(r-x)}=\frac{2dr}{r^2-x^2}  \]

So
\[ \frac{2d}{r}=\frac{2dr}{r^2}<\frac{2dr}{r^2-x^2}  \]

and the scenario without wind takes less time.

nice :surf:  if any of you haven't taken physics yet, be aware of this problem, hehe. it owned a bunch of my classmates on a recent test.


Sa se gaseasca multimea de convergenta pentru seria de puteri $\sum_{n=1}^\infty{\frac{n+2}{n^2+1}*(x-2)^n}$.


inequalities
function
inequalities unsolved

For $a,b,c,d $ positive real numbers with $d<a^3,b^3,c^3$ prove that

$(a^3-d)(b^3-d)(c^3-d) \geq (abc-d)^3$

Generalize to n variables

The inequality as written is obviously wrong. With the right minus in the right hand side, it is a simple consequence of the AM-GM inequality.

[quote="manlio"]For $a,b,c,d $ positive real numbers with $d<a^3,b^3,c^3$ prove that
$(a^3-d)(b^3-d)(c^3-d) \geq (d+abc)^3$
Generalize to n variables[/quote]
I don't know what harazi meant, but right inequality is
\[(a^3-d)(b^3-d)(c^3-d) \leq (abc-d)^3,\]
and general one is
\[(a_1^n-d)(a_2^n-d)...(a_n^n-d) \leq (a_1...a_n-d)^n.\]

Good judge, Myth. Upps, indeed, the inequality is reversed. At least the simple application of the AM-GM inequality (or Huygens or Jensen or) remains true. Sorry.

Sorry very much for the wrong sign  :blush:  :blush:  :blush:

Actually, I meant that it is banal Jensen for concave function $f(x)=\ln(e^x-d)$.

[quote="blahblahblah"]Or it's generalized Holder with coefficients 1/3,1/3,1/3[/quote]

Not directly. Holder doesn't work for negatives.

Yeah, I didn't think of that.  I guess that shows that Harazi's comment that Huygen's works is incorrect as well?

Thanks a lot for the kind words, but with an obvious substitution Huygens workss and very well.  :D


$ f(x)$ is a polynomial with natural factors.
there is  4 positive integers $ a_1$ , $ a_2$ , $ a_3$ and $ a_4$ such that $ f(a_1)\equal{}1$, $ i\equal{}1,2,3,4$
is there any integer $ b$ such that $ f(b) \equal{}4$

[quote="greatestmaths"]$ f(x)$ is a polynomial with natural factors.
there is  4 positive integers $ a_1$ , $ a_2$ , $ a_3$ and $ a_4$ such that $ f(a_1) \equal{} 1$, $ i \equal{} 1,2,3,4$
is there any integer $ b$ such that $ f(b) \equal{} 4$[/quote]

I hope that is $ f(a_i)$ instead of $ f(a_1)$ and $ a_i$ are distinct and $ f(x)$ has integer coefficients.

Consider the polynomial$ g(x) \equal{} f(x) \minus{} 1$.
$ g(x) \equal{} (x \minus{} a_1)(x \minus{} a_2)(x \minus{} a_3)(x \minus{} a_4)h(x)$ where $ h(x)$ has integer coefficients.
$ g(b)$ must be $ 3$. But $ (b \minus{} a_1)(b \minus{} a_2)(b \minus{} a_3)(b \minus{} a_4)$ can't be prime. Thus $ b$ doesn't exist.

P.S. What are natural factors?

i mean by  natural factors "integer coefficient"
and  $ a_i$ are distinct   :wink:


The number of points equidistant from a circle and two parallel tangents to the circle is:
\[ \text{(A)}\ 0\ \ \ \ \text{(B)}\ 2\ \ \ \ \text{(C)}\ 3\ \ \ \ \text{(D)}\ 4\ \ \ \ \text{(E)}\ \text{infinite} \]

[quote="Jimmy"]Do you mean equidistant from the center of the circle?[/quote]

If you were four feet from the center of a circular room, would you say that you wre four feet from the room?  Or would you be four feet from the room when you are outside -- a few feet from the wall?

[hide]I can only think of 2 points that work.[/hide]

[hide]i think there are three points, two located 2r from the center of the circle, and then the center itself[/hide]

maybe i dont understand the question, but wouldnt it be..[hide]
infinity?[/hide]

[hide] C. Tres

Ok, so we know if the point(s) have to be equidistant between two parallel lines that these points will have to lie on the line that is equidistant from the two parallel tangents on the circle. This line will be parallel to the two tangents and run through the center of the circle. The distance from each of the tangents to the center of the circle is the length of the radius (r) of the mentioned circle.

Let's call the line that is parallel and equidistant from the two tangents, which passes through the center, line 'M'

So we now know that we are looking for points that are length (r) from the circle and that lie on line M.  We can easily see that the center of the circle satisfies this. Since 1 is not an answer, we must assume that in the question they do not mean the distance from every point on the circle, but just one.

We can tell that no more interior points in the circle will satisfy our needs. If we now can tell that the point only needs to be equidistant from one point on the circle we can pick the two points that are length (r) distance away from the two points of circumference that lie on line M.

This gives us three points (the center of the circle, and the two points that are length (r) from the two points of the circumference that lie on line M and are on the exterior of the circle)

Not the best explanation, but had to try[/hide]


Ring Theory
superior algebra
superior algebra theorems

$A$ a commutative ring. $Nil(A)$ the ideal formed by the nilpotent elements of $A$.
$P$ is the set of prime ideals of $A$. Show that:
$Nil(A) = \cap_{I \in P} I$

oops .. wrong forum! Sorry again.  :blush:

Oh, yes, this is definitely well-known. :)

One inclusion is trivial: it's clear that all nilpotent elements lie in every prime ideal.

Let $x$ be an element which is not nilpotent. Consider now the collection of all the ideals which do not contain any of the powers $x^n$. This collection is not void, since $(0)$ belongs to it (this is the only place where we use the fact that $x$ is not nilpotent). It's easy to see that this collection, when ordered wrt inclusion, is inductive, so it has a maximal element $I$.

Suppose $a\not \in I,b\not \in I$. This means that there are powers $x^m,x^n$ s.t. $x^m\in (a)+I,x^n\in(b)+I$ (this is what the maximality of $I$ means). This means that $x^{m+n}\in(ab)+I$, so $ab\not\in I$. We have just shown that $a,b\not\in I\Rightarrow ab\not\in I$, which is another way of saying that $I$ is prime, so we have found a prime ideal which does not contain $x$, as desired.


I wanted to ask how satellites are able to revolve around earth?

Our teacher told us that when it is released,then it tries to fall back at the same point from where it was released...is this right?

Where is the point from which it was released? Is that the point on Earth from which the rocket took off or the point on Earth's orbit or some other point? Right now I can't think of any good reason why would the satellite want to get back to this point or why would it want to fall at all :).

What keeps satellites from falling back to Earth is... uhm, let's say it's the curvature of Earth. I don't feel like explaining this - you probably know how it works. Or is that what you're asking about?

Maybe this is what your teacher meant: if you give a sattelite some original velocity and then do not act upon it, it will follow an orbit, which is either open (it flies away) or closed - in that case it is an ellipse or a circle. Therefore, if the surface of Earth wasn't there, after some time it would come to the same point. Since the surface of earth IS there, we must do something to prevent the satellite coming back. Usually, you don't give a satellite its velocity at once, but accelerate it throughut it ascent into its orbit. Therefore, you change its orbital trajectory into one that does not cross the surface of Earth (but it does come back to the point where it turned off its engines).

Newton's thought experiment shows this quite well.

Imagine firing a cannonball from a very tall tower.  We fire it slowly, so it falls to the ground quickly.  We fire it slightly faster, so it travels a little further before falling to the ground.  In fact, we can fire it fast enough so that it travels so far, it goes around the Earth.

At this speed, it won't ever touch the surface of the Earth again (neglecting atmospheric friction), and so it orbits around the Earth.

As to what allows the satellite to revolve around Earth, this clearly shows that gravity is providing the centripetal force that keeps the satellite in orbit.  (yes, centripetal force.  There is no such thing as centrifugal force.  I'll leave that to you to figure out why.)

might be good to learn the basics of circular motion. the satellite is in a radial gravitational field, so it experiences an acceleration - a change in velocity. however, since velocity is a vector quantity, an acceleration need not result in a change of speed. Because the direction of the velocity of the satellite at any given point in the orbit is perpendicular to the acceleration is experiences, the speed does not change, just the direction.

because the field is radial, the satellite can travel in a circle with a constant height from the Earth's surface. another way of thinking of this is by the equation for work. because the component of the weight of the satellite is zero in the direction of displacement, no work is done and the gravitational potential energy (which is directly proportional to the height) remains constant.


for xyz=1 then prove that:

$\sum\frac{xy+yz}{y^2+3}\geq\frac{3}{2}$ :P  :P

[quote="cuong"]for xyz=1 then prove that:

$\sum\frac{xy+yz}{y^2+3}\geq\frac{3}{2}$[/quote]

This is wrong. A counterexample is $x=2^{-\frac13}$, $y=2^{-\frac13}$, $z=2^{\frac23}$.

  darij

Anyway, how did you find this inequality? It's yours?


inequalities
number theory
prime factorization

How many positive integers less than 2007 are divisible by exactly 7 positive integers (including 1 and the number itself)?

[quote="cognos599"]i am almost sure this is wrong but

[hide="?"]well for the number to have 7 factors, it must a perfect power of 6. so just count.

2^6, 3^6, and that is it so is it 2?. [/hide][/quote]I believe you are correct.
[hide="Solution"]The prime factorization of $ 7$ is $ 1\cdot 7$. From the formula to count the amount of divisors of a number, we determine that the number must be a perfect power of $ 6$.
\[ n^6<2007\]The only positive integers $ n$ that satisfy this inequality are $ 2$ and $ 3$. While $ 1$ satisfies the inequality, it has only one divisor. The answer is therefore $ \boxed{2}$.[/hide]


interesting! please a hint or answer... :lol:

[quote="Rushil"]Positive integers are written on all the faces of a cube, one on each. At each corner of the cube, the product of the numbers on the faces that meet at the vertex is written. The sum of the numbers written on the corners is 2004. If T denotes the sum of the numbers on all the faces, find the possible values of T.[/quote]
This isn't very hard actually.

Let the numbers on the "walls" of the cube be $a,b,c$ and $d$ such that $a$ is "opposite" $c$ and $b$ is "opposite" $d$. Also, let the number on the "top" face be $e$ and the number on the "bottom" face be $f$ ($e$ and $f$ are "opposite" each other.)

Then the products at the eight corners  are $abe, ade, bce, cde, abf, adf, bcf$ and $cdf.$
We now have
$abe+ade+bce+cde+abf+adf+bcf+cdf = 2004$
$\Rightarrow ab(e+f)+ad(e+f)+bc(e+f)+cd(e+f)=2004$
$\Rightarrow (e+f)(ab+ad+bc+cd)=2004$
$\Rightarrow (a+c)(b+d)(e+f)=2^{2}\cdot 3 \cdot 167$

Now, note that each of the three terms on the $LHS$ is greater than or equal to $2.$

The rest is just a matter of evaluating individual cases, such as $a+c=2^{2}, b+d=3$ and $e+f=167,$ in which case $T=4+3+167=174,$ and so on.

[quote=MathleteMA]I did it just like FieryHydra. Here are all the possible answers:
$167+6+2=175$
$167+4+3=174$
$334+3+2=339$
$501+2+2=505$[/quote]

Yes it's a funny easy problem and you aren't going to get any other solution.


Let $M=(0,1)\cap \mathbb Q$. Determine, with proof, whether there exists a subset $A\subset M$ with the property that every number in $M$ can be uniquely written as the sum of finitely many distinct elements of $A$.

[quote="erdos"] ... can be uniquely written as the sum of [b]finitely many[/b] distinct elements of $A$.[/quote]

I wanted to erase my post like two seconds ago :). I forgot about periods, that's all :).

One more try :).

I believe the answer to be negative. Assume $\{a_n\}_n=A$, and that $a_j>a_j$. We then have $a_i-a_j=\sum a_k$, and in order for $a_j+\sum a_k$ not to be a different representation of $a_i$ as a sum of finitely many distinct elements of $A$, $a_j$ must be one of the $a_k$'s which means that $a_i\ge 2a_j$. This happens whenever we have $a_i>a_j$, so we can order our elements $a_1\ge 2a_2\ge 4a_3\ge 8a_4\ge\ldots$ .

Now, assume there is a $k$ s.t. $a_k>2a_{k+1}$. Take a rational $t$ strictly less that $a_k$ but still larger than $2a_{k+1}$. Then $t>a_{k+1}+a_{k+2}+\ldots$, so $t$ ahs no representation as the sum of finitely many distinct elements of $A$. This shows that for all $k\ge 1$, we must, in fact, have $a_k=2a_{k+1}$. This is impossible, because, for instance, $\frac{a_1}3$ cannot be represented as the sum of finitely many $a_k$'s.
I hope it's correct this time.


probability
ratio
geometry
expected value

Alice has 20 dollars and Bob has 10 dollars. They decide to play a game where they flip a coin and if it is heads, Alice gives Bob 1 dollar, and if it is tails, Bob gives Alice one dollar. The game ends when someone has all of the money. What is the probability that when the game ends, Alice will have all of the money? (Note: I just made this problem up so I have no idea if there is actually an answer. :lol: )

I don't have a authentic solution for this, but is it $ \frac{1}{1024}$?

I made the problem up so I don't know the answer but I don't see how the answer could be less than 1/2. Can you explain?

The [i]expected[/i] number of flips until the game ends is 200. Don't know if that helps or anything.

My idea was that you need 10 more flips for Alice or whatever the name was to lose, and 10 more flips 1/2^10

[quote="azn6021023"]Um...Alice has more money, therefore the probability she wins should be much greater than $ 1/1024$.

If each player has $ 15$ dollars, then the probability to win is $ 1/2$.  There is a five flip difference, so Bob has a $ 1/32$ chance to reach it, then a $ 1/2$ chance to win.  This implies that Alice has a $ 63/64$ chance to win.[/quote]
Uh, I'm not that stupid. When I said "win" I didn't state which person I meant.

I posted a problem almost exactly like this not too long ago, so I think I can do it correctly now...

[hide="solution?"]

The expected value of a flip = 0.

Therefore, $ 0\equal{} \minus{}20 * (1\minus{}$probability of Alice gaining all of the money$ ) \plus{} 10 *$ probability of Alice gaining all of the money$ ).$

Solving for p gives $ \boxed{\frac{2}{3}}.$[/hide]

Thank you. I think erdogankerem's solution makes the most sense.

azn6021023, I don't think the probability that Bob reaches 15 dollars is $ 1/32$.

Oops.

If we look at it in terms of mass points, we see that the win ratio is clearly $ 2: 1$.  I just wasn't sure if this was correct.

How did you apply mass points to this? I've only seen it used in geometry problems...  :maybe:

It sort of works.  Alice has twice the money Bob has...in a long run, she has twice the probability to win.

I think the idea is that Bob has to win twice as much money as Alice does, so it should be twice as hard for Alice to win.

It turns out that this conjecture is true for a general random walk along a line -- if you start at 0 and move +1 or -1 with equal probability, then the probability of getting to q before getting to -p, where p,q are positive integers, is p/(p+q), and the probability of getting to -p first is q/(p+q). This can be proved very easily with induction.

[b]Follow up:[/b] A particle moves along the x-axis, starting from 0 at t=0, such that, at any point in time, its speed is either 10 or -10. If it reaches 20 before it reaches -10, then Bob wins. If it reaches -10 before it reaches 20, then Alice wins. Find, with justification, the probability that Alice wins.

Let $ P$ be the probability that Alice wins when the particle is at the origin. Then $ P\equal{}\frac{1}{2}\plus{}\frac{1}{4}P$ because there is a 1/2 chance she will win right away and a 1/4 chance that after 2 moves it will move back to the origin. Solving that, gives $ \boxed{P\equal{}\frac{2}{3}}$


trigonometry
geometry
geometric transformation
rotation
geometry proposed

Let $ABC$ be a triangle with $\angle C=60^\circ$ and $AC<BC$. The point $D$ lies on the side $BC$ and satisfies $BD=AC$. The side $AC$ is extended to the point $E$ where $AC=CE$. Prove that $AB=DE$.

Trivial by cosine theorem.

Let $ AC \equal{} CE \equal{} BD \equal{} y$ and $ CD \equal{} x$.

In triangle $ ECD$:
$ ED^2 \equal{} x^2 \plus{} y^2 \minus{} 2xy \cos 120 \equal{} x^2 \plus{} y^2 \plus{} xy$.

In triangle $ ACB$:
$ AB^2 \equal{} x^2 \plus{} (x \plus{} y)^2 \minus{} 2x(x \plus{} y) \cos 60 \equal{} x^2 \plus{} x^2 \plus{} 2xy \plus{} y^2 \minus{} x(x \plus{} y) \equal{} x^2 \plus{} x^2 \plus{} 2xy \plus{} y^2 \minus{} x^2 \minus{} xy \equal{} x^2 \plus{} y^2 \plus{} xy$.

Hence $ AB \equal{} DE$ as wanted.

[hide="Another solution"][quote="anonymous111"]Let $ ABC$ be a triangle with $ \angle C \equal{} 60^\circ$ and $ AC < BC$.The point $ D$ lies on the side $ BC$ and satisfies $ BD \equal{} AC$.The side $ AC$ is extended to the point $ E$ where $ AC \equal{} CE$.Prove that $ AB \equal{} DE$[/quote]
Let K on CB such that CK=CA, so triangle ACK is equilateral. Since CK=CA=BD, BK=CD, $ \angle AKB\equal{}\angle ECD\equal{}120^o$, we have that $ \triangle AKB\equal{}\triangle ECD$, hence DE=AB.[/hide]

Draw the circle (ABC), name P the middle of the arc AB containing C. Since a 60 degs clockwise rotation about P maps the triangles PCA and PDB, hence PC = PD = CD and <PDB = 120 degs, so the triangles PDB and DCE are equal, hence DE = PB = AB.

Best regards,
sunken rock


calculus
integration
function
real analysis
real analysis theorems

This question does not have much to do with mathematics. The author of a book I am reading uses this notation for iterated integrals:
$\int d\mu(x) \, \int h(x,y) \, d\nu(y)$
I don't understand at all how the idea of this notation occurs. The normal one for me would be
$\iint h(x,y) \, d\nu(y) \, d\mu(x)$
This notation conveys the idea that you just add a paranthesis to show what it really means, i.e. it means
$\int \left(\int h(x,y) \, d\nu(y)\right) \, d\mu(x)$

Is the other notation "explainable"? Sorry if it's a vague question, I can't think of how to pose it better

This notation seems to be more common in physics and probability. It's not so much about ideas as it is about keeping track of the variables of integration:
\[\int_{0}^{\infty}\,dx\int_{-1}^{1}\,dy\int_{-\infty}^{\infty}\,dz\int_{0}^{1}f(x,y,z,w)\,dw\] may be easier to read (with some practice) than
\[\int_{0}^{\infty}\int_{-1}^{1}\int_{-\infty}^{\infty}\int_{0}^{1}f(x,y,z,w)\,dw\,dz\,dy\,dx\] or
\[\int_{0}^{\infty}\left(\int_{-1}^{1}\left(\int_{-\infty}^{\infty}\left(\int_{0}^{1}f(x,y,z,w)dw\right)dz\right)dy\right)dx\]  Also, we write something like $Tf$ to indicate that a linear functional $T$ is applied to function $f$. A definite integral is a linear functional, denoted as $\int_{a}^{b}dx$ rather than $T$.