geometry
analytic geometry
trigonometry
absolute value

It says in the Geometry Moise-Downs book on page 37 that "If we know the distances AB and BC, then we can calculate the distance AC, because AB + BC = AC." So in the definition of B is between A and C, it says, "B is between A and C if (1) A, B,  and C are different points of the same line, and (2) AB + BC = AC..." But isn't it assuming that AB + BC = AC (shouldn't it be a postulate)?     In the McDougal Littell Geometry by Ray C. Jurgensen, Richard G. Brown, and John W. Jurgensen, it says on page 11 that "In the diagram, point B is between points A and C. Note that B must lie on line AC." as the definition of the term between. And then on page 12, it says, " Postulate 1 Ruler Postulate: (1) The points on a line can be paired with the real numbers in such a way that any two points can have coordinates 0 and 1. (2) Once a coordinate system has been chosen in this way, the distance between any two points equals the absolute value of the difference of their coordinates. Postulate 2 Segment Addition Postulate If B is between A and C, then AB + BC = AC."    I am confused because initially, I thought that in the Moise-Downs book page 33 postulate 2 The Ruler Postulate "The points of a line can be placed in correspondence with the real numbers in such a way that... (3) the distance between any two points is the absolute value of the difference of the corresponding numbers." (or the Postulate 1 part 2 in the McDougal Littell book) assured that the distances AB ? BC = AC. By the Ruler Postulate two, we get l b-a l ? l c-b l = l c-a l. By realizing that the absolute values are not needed because they are positive by the Ruler Placement Postulate of the Moise-Downs book (or the Ruler Postulate, namely part #1, in the McDougal Littell book), we get b-a ? c-b =  c-a. It then becomes apparent that the ? is the operation +, so AB + BC = AC. So did I do something wrong in my "proof"? If not, then why is the AB + BC = AC listed as a postulate in the McDougal Littel book? If yes and indeed AB + BC = AC is really a postulate, then why isn't it a postulate in the Moise-Downs book (and why is it just accepted)?

I'm too confused to tackle all of this at once...could you write each question separately? (I'm using the Moise-Downs book.)

AB+BC=AC only if B is between A and C.

[quote]It says in the Geometry Moise-Downs book on page 37 that "If we know the distances AB and BC, then we can calculate the distance AC, because AB + BC = AC." So in the definition of B is between A and C, it says, "B is between A and C if (1) A, B, and C are different points of the same line, and (2) AB + BC = AC..." But isn't it assuming that AB + BC = AC (shouldn't it be a postulate)?[/quote]

I believe they said that because of the diagram (along with the corresponding text) above what you're asking.
Also, you can think of this in terms of the union of two segments...maybe. :maybe: I think this might be more of an intuitive approach (kinda strange for this book), supported by the previous postulates.

Also, the number of a Postulate or Theorem in one book may not be the same in another. (They're only relevant to the book.)

Thanks for replying, fishythefish  :lol:
[quote]Also, the number of a Postulate or Theorem in one book may not be the same in another. (They're only relevant to the book.)
[/quote]
Well, if you read on in my passage, I do juxtapose the two books and the postulates I used to "prove" (probably wrong) the so-called Segment Addition Postulate are in both books (that's the problem)
I'm sorry, I should have listed out the questions  separately  :blush:

I did read the entire post (maybe that's why I'm confused :P ), and I agree that your argument is valid. Where do they prove that AB+BC=AC, right? And if not, why isn't it a postulate?

I've tried using Google for other places people have asked this. All I can gather is that this is an intuitive argument that falls just short of a postulate (in the book). The most reasonable one (and the most concise) is that this is just another variation of

"The sum of all of the parts equals the whole."

I am perplexed by this too, but I think you can accept this just like a postulate. Really, when you get down to it, there is nothing wrong with the statement.

It is also worth noting that in other sources, they do refer to this as "the Segment Addition [i]Postulate[/i]". Maybe it's just a printing error, but I doubt it. This method of teaching geometry claims that you can prove everything with these 24 postulates. I don't know why it isn't there, but it isn't.

I asked my math teacher today. He agreed with me that the proof with the diagram was supposed to appeal to your intuition. He also agreed that this is circular reasoning:

If you look at the paragraph above the diagram, they use the word "between" to show that AB+BC=AC. But then they defined "between" afterward, and used the segment addition to define it!

No problem. When you asked that question, I started to wonder the same thing. But then again, most books seem to have a few errors, don't they?

I'm just surprised that this wasn't an error in the solutions or something, but an error in the teaching method. Geometry should name some postulates, and leave nothing else to intuition, right? Maybe that's why this is referred to as the Segment Addition Postulate sometimes.

[quote="fishythefish"]No problem. When you asked that question, I started to wonder the same thing. But then again, most books seem to have a few errors, don't they?[/quote]

You're right

In my math class, we learned that it [i]is[/i] a postulate. So is it, or isn't it?

To me, it uses intuition, so I would make it a postulate.

Hmmm, how about the "Segment Addition Postulate"?

That's how I learned it.

That's what I was wondering. I'm using this book, and right at the start of the trimester, my geometry teacher explained that this is how the book works:

The book gives you 24 postulates.
Using those, you can prove all the theorems.
Then, you can use those to prove (or disprove) anything else.

The Segment Addition Postulate was not included.

Again, this would make more sense if the definition was printed before those diagrams. It would still be kinda intuitive (but for some reason, not a postulate), but you could see that AB+BC=AC is more of a condition for B to be between A and C than a postulate in this book.

Can we use  the law of cosines to proof that olds in a degenerate triangle?  :maybe:

Um...What exactly are you asking? Could you clarify what "olds" means?

I am saying that we can concidere a triangle which have an angle equals to 0 or 180, them we use the cosine law to proof what we want.

Maybe I was trying to say holds :P I guess

Good books do not have errors. I have read most of Art of Problem Solving's Intermediate Algebra and I have not found any errors in it except for a minor grammatical error and (maybe) a mathematical error.


calculus
integration
inequalities
real analysis
function
abstract algebra
group theory

Let $A\subseteq[a,b]$ be a measurable subset of the real line, and let $h>0$ be a positive real. Prove that $\frac 1{2h}\int_a^b\mu[A\cap(x-h,x+h)]\mbox dx\le\mu(A)$, where $\mu$ is the usual Lebesgue measure on $\mathbb R$.

Let $f(x)=\chi_A(x),$ the characteristic function of $A.$

Let $g(x)=\frac1{2h}\chi_{(-h,h)}(x).$

Then $\frac1{2h}\mu\left[A\cap(x-h,x+h)\right]=f*g(x),$ the convolution of these two functions.

It is well known that (for $L^1$ functions, which we do have)

$\int_{-\infty}^{\infty}f*g(x)\,dx=\int_{-\infty}^{\infty}f(x)\,dx\cdot \int_{-\infty}^{\infty}g(x)\,dx=\mu(A)\cdot 1=\mu(A).$

Since everything here is nonnegative, the integral from $a$ to $b$ must be no larger than the integral over the whole line.

Yes, the inequality was backwards. I apologize. :oops:

That's OK. I had a typo to fix in my solution anyway.

Why is

$\int^{\infty}_{-\infty} (f*g)(x)dx=\int^{\infty}_{-\infty} f(x)dx\int^{\infty}_{-\infty} g(x)dx$

I haven't come across this before.

For nice enough functions, we have
$\int_{\mathbb{R}}(f*g)(x)\,dx=\int_{\mathbb{R}}\int_{\mathbb{R}}f(t)g(x-t)\,dt\,dx$
$=\int_{\mathbb{R}}\int_{\mathbb{R}}f(t)g(x-t)\,dx\,dt= \int_{\mathbb{R}}\int_{\mathbb{R}}f(t)g(s)\,ds\,dt= \int_{\mathbb{R}}f(t)\,dt\cdot \int_{\mathbb{R}}g(s)\,ds$

There are some details to fill in justifying these manipulations, but that's the basic calculation.

This is an immediate, and compellingly useful, consequence of Fubini's Theorem.

If $f, g\in L^1{\mathbb{R},}$ then $f*g\in L^1(\mathbb{R}).$

Note that it is not immediately obvious why the inetegral defining $f*g(x)$ should converge for any $x$ at all; we certainly can't conclude that it converges for all $x.$

But we can prove (the formalities look very much like what jmerry wrote above - and "nice enough" need only mean $L^1$ - only with absolute values and inequalities as appropriate) that

$\|f*g\|_1\le \|f\|_1\|g\|_1,$

and the fact that $f*g$ is integrable implies that it exists and is finite almost everywhere.

Side comment #1: replace $\mathbb{R}$ by the locally compact abelian group of your choice (with Haar measure). We can even make a version of this argument work in a nonabelian group, although we have to take great care with the order of operations.

Side comment #1: this is the beginning of the presentation of $L^1(\mathbb{R})$ - or $L^1(G)$ for any locally compact abelian group $G$ - as a very nice example of a commutative Banach algebra.


A WOMAN has suffered severe burning to her anus after being struck by lightning which hit her in the mouth and passed right through her body.

Natasha Timarovic, 27, was cleaning her teeth at home when lightning struck the building.

She said: "I had just put my mouth under the tap to rinse away the toothpaste when the lightning must have struck the building.

I don't remember much after that, but I was later told that the lightning had travelled down the water pipe and struck me on the mouth, passing through my body.

It was incredibly painful, I felt it pass through my torso and then I don't remember much at all." Doctors at the city hospital where she was treated for burns to the mouth and rear said: "The accident is bizarre but not impossible.

She was wearing rubber bathroom shoes at the time and so instead of earthing through her feet it appears the electricity shot out of her backside," a medic told local television news channel, 24 Sata.

"It appears to have earthed through the damp shower curtain that she was touching as she bent over to put her mouth under the tap. If she had not been wearing the shoes she would probably have been killed by the blast."

That is amazingly weird/freaky/weird/bad. And really weird too.

Wow. Let's calculate the chance of that happening again.

I feel badly for her.


x, y, z are positive reals with sum 3. Show that 6 < √(2x+3) + √(2y+3) + √(2z+3) < 3√5

Are you sure it's right?  Take $ x\equal{}y\equal{}z\equal{}1$ and you get $ 6>3\sqrt{5}$.

Yeah, I don't think it's right because the problem says 6 < () < 3 :wink:

Cosinator has found the maximum, which can be easily derived via Jensen's Inequality.

hehe, sorry guys, the right side of the inequality is 3√5 instead of 3 :blush:
thanks kunny for pointing out the error :)


Let $a,b,c$ be positive reals such that $abc=2.$ Find the minimum value of the following equality.

\[ \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}.  \]

Let $ A$=$ \sum$ $ \frac{1}{a(1+b)}$ we have $ 3A+3$=$ (1+abc)A+3$=$ \sum$ $ \frac{1+a}{a(1+b)}$+$ \sum$ $ \frac{b(1+c)}{1+b}$
Using AM-GM Inequality,we find that$ 3A+3$ $ \geq$...Thus the minimum is attained for $ a$=$ b$=$ c$.

From [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6044]a well-known inequality[/url] we have that
$ \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\geq\frac{3}{\sqrt[3]{abc}(1+\sqrt[3]{abc})}$.

The equality holds if $ a=b=c$   generally.
So If we put where $ abc=2$  we have that the minimum price is
$ \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\geq\frac{3}{\sqrt[3]{2}(1+\sqrt[3]{2})}$
and the equality attainted when $ a=b=c=\sqrt[3]{2}$


Anyone can post problems right?

Anyway:

Problem 412

What three digit number in quinary (base 5) has an equal value when flipping the digits of the number and turning it into octinary (base 8)?

For example, 123 in base five has a value of 38 in the decimal system. 321 in base eight has a value of 209 in the decimal system, so 123 is not the correct answer.

You mentioned that 331 was the only number that worked. I know this is the Mathcounts forum, but could you explain how you would prove that?

uhh, actually, i dont know  :oops: . my dad showed me this question and took out the part that had to do with proving it.  sorry.

ill ask him when i have time (so much homework to do...)

Proof of this is not difficult. I'll use a simple casework method to 'prove' it.

[hide]From $63a+3b=24c$, we can just test cases of a. We know that a,b,c are all less than 5 because they are digits in quinary.

For a = 1; b = 3 and c = 3 works.

For a = 2; we have 126 + 3b = 24c. We know 3b is nonnegative, so c would have to be greater than 5. However, this would make it not a digit in quinary, so this won't work. Since a = 2 would make c have too high of a value, increasing the value of a would increase the value of c, and c is too large. Thus, a = 1, b = 3, c = 3 is the only answer.[/hide]

I meant proof that this number is the only integer (not just 3-digit) that satifies the condition.

Well, let's see what we get for other-digit numbers.
[hide]1-digit numbers work, but obviously not in the way the problem wants.

For 2-digit numbers, we have $5a+b = 8b+a$, or $4a = 7b$. The lowest solution is $a = 7, b = 4$, but this won't work in quinary, so 2-digit won't work.

We have already seen the one solution for 3-digit numbers.

For 4-digit numbers we get $512a+64b+8c+d = 125d+25c+5b+a$. Rearrange: $511a+59b = 124d+17c$. Let's break this into cases of a.

$a = 1$. $511+59b = 124d+17c$. Obviously, d would have to be 4 or 5 for the RHS to equal the LHS. 5 will make the digit not work in quinary, so let's try 4. $511+59b = 496+17c$. Or, $15+59b = 17c$. $b = 1$ doesn't make c an integer, and any larger value of b will make c greater than 5.

$a = 2$. $1022+59b = 124d+17c$. There is no value of d less than 5 that will bring the RHS even close to the LHS, so this is scrapped. That means 4 digit numbers won't work.

We can try 5-digit numbers, but it should be fairly clear that the coefficient of a will be too large for the other values. We get $4096a+512b+64c+8d+e = 625e+125d+25c+5b+a$. This is $4095a+507b+39c = 624e+113d$. Even if a is 1, e would have to be [u]at least[/u] 5 (If e was 4, d would be greater than 10). So 5-digit won't work.

In the same way, we see that the value of the coefficient of a will increase at a higher rate (8x) than the coefficient of e,f,g, etc... (5x). Since a is already too large for the RHS to equal it, obviously more digits wouldn't work, so 331 is the only answer.[/hide]

This is not a rigorous proof, but you should be able to see where I'm going with it.

This problem was on this year's Mathcamp qualifying quiz, number 2. Is it okay to post problems now or should this thread be locked?

[hide]You could factor

8C=21A+B

If A=0 then C=1 and B=8 but B<5
If A=1 then C=3 and B=3
If A>1 then B>4 therefore 331 is the only answer.
[/hide]

i think mathcamp is pretty much done collecting their qualifying quizzes so i dont think it is necessary to lock this post.

but, somebody, still check up on this info.

This topic is fine, I checked with a person on the Mathcamp staff.

[hide="answer with rigorous proof"]$ABC_{5}=(25A+5B+C)_{10}=(64C+8B+A)_{10}=CBA_{8}$

where A, B, and C are digits.

$63C+3B=24A$

C can most definitely not be 0, so we can say that it is 1. So

$63+3B=24A$

72 is the nearest multiple of 24, and 96 is the next, but only 72 works.

72-63=9, B=3, A=3

Now let's say that C=2 and B=0. That means that A is not an integer.

The closest multiple of 24 to 126 is 144, 144-126=18, B=6, A=6.

Let's check that:

662 and 266:
182 and 182.

331 and 662 work.

But the next one is above 662, and 662 in quinary is a 4 digit number.

So yes, 331 is the only number that works.[/hide]

But that is only a proof for 3-digit numbers, whereas the proof asked for was for any digit number.

I only did what the original question was asking.

The original question did ask for three-digit numbers. But, could somebody still show one for other-digit numbers too.

[hide]
$ 25a \plus{} 5b \plus{} c \equal{} 64c \plus{} 8b \plus{} a$
$ 24a \equal{} 63c \plus{} 3b$
$ 8a \equal{} 21c \plus{} b$
Octal number would have to be less than or equal to $ 444_8 \equal{} 292$.  Palindromes are also eliminated.
Also, $ a > c$, so octal number would have to be of the form 1_3 or 1_4 Then, the middle digit would be less than or equal to 4.  This leaves us with
113
123
133
143
114
124
134
144
Now, we just find the right one in the 8 numbers provided.  We can narrow it down closer using parity, obtaining
113
133
114
134
Then after a little bit of brute force we get 133 (base 8)
[/hide]


algebra
polynomial
calculus
integration
Vieta
modular arithmetic
Rational Root Theorem

A RR theorem solution, if it even exists, would be unnecessarily messy because all you even know about a_0 is that it's odd, which really doesn't tell you much in the way of what integer values might be a root

[quote="max_tm"]Let
$p(x) = a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$
where the coefficients $a_{i}$ are integers. If $p(0)$ and $p(1)$ are both odd, show that $p(x)$ has no integral roots. [i](Canada 1971)[/i]

Can anybody think of a solution that uses a somewhat direct application of the rational root theorem??

Much appreciated![/quote]

Like [b]Elemennop[/b] said, it'd be very messy if it even exists.  [hide="The solution which is most apparent to me involves..."]
Vieta's formulae.[/hide]

I don't see a solution involving Vieta's formula.
[hide="My Solution"]Let's look at the polynomial modulo $2$.  If the desired integral root is even then we get $0+0+\cdots 0+a_{0}\equiv p(0) \equiv 1 \mod 2$.  Now if the desired integral root is odd then we get $a_{n}+a_{n-1}+\cdots a_{0}\equiv p(1) \equiv 1 \mod 2$.  This is a contradiction either way.[/hide]

[b]Solution.[/b] Assume to the contrary that such a polynomial exists with integral roots and both $f(0),f(1)\equiv 1\pmod 2$.  Let $z\in\mathbb{Z}$ be such a root. Hence,
\[f(x)=(x-z)g(x)\implies f(0)=-zg(0)\wedge f(1)=(1-z)g(1) \]
But $1-z\not\equiv-z\pmod$ so at least one of $f(0)$ and $f(1)$ must be even.$\blacksquare$


A "Z-figure" is a figure composed of five squares, with three squares in the vertical line in the middle, one square adjacent (to the left) to the top square, and one square adjacent (to the right) to the bottom square. (So it looks liks a Z!) How many Z-figures do we need, at least, to cover an $8$x$8$ chessboard? (The Z-figures may overlap)

Can part of the Z-figure lie off of the board?

If they can overlap, I'm guessing that means they're also allowed to lie off the board.

Actually, I don't think it matters.  You can cover the same squares  that you would if it was off the board by reorienting the piece.

I thought so at first, Samath, but it's not quite true -- for example, if pieces can go off the board, I can cover 3 consecutive squares along one edge with a single Z-shape, but I can't do that without going off the board.  Now, it may be (probably is true, in fact) that there are optimal solutions which don't require that any piece be so arranged, but it's not clear [i]a priori[/i] that this is so.


superior algebra
superior algebra unsolved

Suppose that $G$ is a group. Prove or disprove: $\frac{Aut(G)}{Inn(G)}$ is isomorphic to $Z(G)$.

Wenn $Z(G)$ shall be the center of $G$, then already $\mathbb{Z}/ 3 \mathbb{Z}$ is a counterexample.


abstract algebra
Ring Theory
superior algebra
superior algebra theorems

I'm reading this proof from D&F and there's something I dont get. It is theorem 1 of chapter 12 on modules over principal ideal domains. The theorem is the following

"Let R be a ring and let M be a left R-module. Then,

(Ever nonempty set of submodules of M contains a maximal element under inclusion) ==> (Every submodule of M is finitely generated)"

The authors prove this assertion by letting N be a submodule of M and denoting S the collection of all finitely generated submodules of N. The hypothesis guarantees the existence of a maximal element N' of S, and we'd be happy if N'=N. So suppose N' is different from N and consider an element x in N\N'. Then the submodule of N generated by {x}uN' is finitely generated, thus violating the maximality of N'.

How is this so? Notice that R is not assumed to have a 1. So there is nothing I can see that guarantees that N' is contained, let alone properly contained, in <{x}uN'>.

What is the problem? If the submodule $ N^{\prime}$ is generated by $ u_1$, $ u_2$, ..., $ u_k$, then the submodule $ \left\langle x\right\rangle \plus{} N^{\prime}$ is generated by $ u_1$, $ u_2$, ..., $ u_k$, $ x$ (hence, finitely generated) and properly contains $ N^{\prime}$ ("contains" is clear; "properly" is because $ x\not\in N^{\prime}$ but $ x\in \left\langle x\right\rangle \plus{} N^{\prime}$).

The authors mean by $ \left\langle\left\{x\right\}\cup N^{\prime}\right\rangle$ the same as I mean by $ \left\langle x\right\rangle \plus{} N^{\prime}$.

  darij

Why would x be in <{x}uN'>? By definition, <{x}uN'> is the set of all finite R-linear combinations of element of {x} and of N'. But if R is without a 1, what guarantees that x is in <{x}uN'>? We'd have to find scalars r_i from R and elements y_i of {x}uN' such that $ \sum_{i \equal{} 1}^n r_iy_i \equal{} x$. What are they?

Ah, I haven't noticed that your ring need not have a $ 1$. But then it all becomes strange: What does it mean for you that an $ R$-module $ N$ is "generated" by some elements $ u_1$, $ u_2$, ..., $ u_k$ ? Do you mean that
[b](a)[/b] every element of $ N$ can be written in the form $ s_1u_1 \plus{} s_2u_2 \plus{} ... \plus{} s_ku_k$ for $ s_1$, $ s_2$, ..., $ s_k$ being elements of the ring $ R$,
or that
[b](b)[/b] every element of $ N$ can be written in the form $ s_1u_1 \plus{} s_2u_2 \plus{} ... \plus{} s_ku_k \plus{} \lambda_1u_1 \plus{} \lambda_2u_2 \plus{} ... \plus{} \lambda_ku_k$ for $ s_1$, $ s_2$, ..., $ s_k$ being elements of the ring $ R$ and $ \lambda_1$, $ \lambda_2$, ..., $ \lambda_k$ being integers?

In the case of [b](a)[/b], your problem is wrong: Let $ M$ be a finite set, and let $ R$ act on $ M$ in the most trivial way: $ rm \equal{} 0$ for all $ r\in R$ and all $ m\in M$; then, every nonempty set of submodules of $ M$ contains a maximal element under inclusion (since $ M$ is finite), but not every submodule of $ M$ is finitely generated (in fact, only the submodule $ \left\{0\right\}$ is finitely generated). Maybe you could require some additional assumptions to make your assertion correct?

In the case of [b](b)[/b], my argument works if you read $ \left\langle u_1,u_2,...,u_k\right\rangle$ as

$ \left\{s_1u_1 \plus{} s_2u_2 \plus{} ... \plus{} s_ku_k \plus{} \lambda_1u_1 \plus{} \lambda_2u_2 \plus{} ... \plus{} \lambda_ku_k\mid s_1,s_2,...,s_k\in R;\ \lambda_1,\lambda_2,...,\lambda_k\in\mathbb{Z}\right\}$.

I hope I didn't make mistakes this time.

[b]A well-meant advice to all algebraists: Whenever you talk about rings, please specify whether they are required to be commutative or not, and with $ 1$ or not.[/b] Otherwise, readers will assume the word "ring" to mean the kind of ring they are most used to - e. g. commutative algebraists will read it as commutative ring with $ 1$, representation theoretists as non-commutative ring with $ 1$, etc.

  darij

It so happens that I was flipping through the section of Dummit & Foote on modules yesterday.  They explicitly say that whenever they talk about $ R$-modules, they are assuming that $ R$ is unital, specifically in order to avoid the pathology in darij's counterexample.

This is a bit painful, of course, to anyone using the book as a reference rather than reading it straight through.  And at 912 pages, who would ever read it cover to cover!  I agree with you, darij, it would be nice if these things were, in general, pointed out explicitly in theorem statements.

darij, I'm using definition (a), sadly.

LydianRain, can you show me exactly where you saw that written?

What D&F usually do is that at the beginning of a chapter or subchapter, they will writte for instance, "In what follows, R is a ring with identity". But at the begining of section 12.1, they just write "Let R be ring and M be a left R-module."

Very strange... D&H is usually very clear and never ambiguous.

I'm using the 2nd edition, so your mileage may vary, but on p.318, the first page of Chapter 10, I read the following:

'Modules satisfying axiom 2(d) are called [i]unital[/i] modules and in this book all our modules will be unital (this is to avoid "pathologies" such as having $ rm\equal{}0$ for all $ r\in R$ and $ m\in M$).'


linear algebra
matrix
geometry
geometric transformation
linear algebra unsolved

Let A and b are nxn diagonalizable matrices such that AB = BA. Prove that there exists matrix P such that
P^(-1)AP and P^(-1)BP are both diagonal matrices. This is an challenging problem. Can anybody help me?

It's by induction. If my translation is good enough - unlikely - it should be called the simultaneous diagonalization theorem (also works for trigonalization if I remember well).

A first step: $A$ has a basis of eigenvectors. Let $\lambda$ be an eigenvalue, and let $v$ belong the the eigenspace of $A$ corresponding to this eigenvalue.

Then $A(Bv)=ABv=BAv=\lambda Bv.$ Thus, $Bv$ belongs to the eigenspace of $A$ with this eigenvalue. So $B$ maps this eigenspace into itself.

See where that leads you.


Phillip flips an unfair coin eight times. This coin is twice as likely to come up heads as tails. How many times as likely is Phillip to get exactly three heads than exactly two heads?

I did not really understand how to do this problem when I viewed the Alcumus solution. I understood the solution, but what I don't understand is [b]why[/b].

Hi,

Since the probability of heads is twice as much as probability of tails, we can use variables. So, let $ P(\text{tails}) \equal{} t.$ Thus, $ P(\text{heads}) \equal{} 2t.$ We are basically looking for $ P(\text{exactly 3 heads}) \equal{} nP(\text{exactly 2 heads}),$ for $ n \in \mathbb{R}.$

Let's first find $ P(\text{exactly 3 heads}).$ Since there are 8 total flips, there are $ \binom{8}{3}$ ways to choose which three we want to be heads. And for each of those choices, there are $ (2t)^3$ probability of getting the three heads (since for each of the three places, we have $ 2t$ probability for each of those places.) The remaining $ 5$ places, we must have tails. Since each tail has probability $ t,$ the probability that all of them will be tails is $ t^5.$ Therefore, the total probability for $ P(\text{exactly 3 heads})$ is $ \binom{8}{3}(2t)^3t^5.$

For $ P(\text{exactly 2 heads}),$ we have $ \binom{8}{2}$ ways to choose where the two heads go. Since, there is a $ 2t$ probability that each of the two is a heads, we have $ (2t)^2.$ The remaining $ 6$ must be tails, and since there is a $ t$ probability for each tail, we have $ t^6.$ Therefore, the total probability for $ P(\text{exactly 2 heads})$ is $ \binom{8}{2}(2t)^2t^6.$

Thus, from $ P(\text{exactly 3 heads}) \equal{} nP(\text{exactly 2 heads}),$ we have
\[ n \equal{} \dfrac{P(\text{exactly 3 heads})}{P(\text{exactly 2 heads})} \\
 \\
 \\
\equal{} \dfrac{\dbinom{8}{3}(2t)^3t^5}{\dbinom{8}{2}(2t)^2t^6} \\
 \\
 \\
\equal{} \dfrac{56\cdot8t^8}{28\cdot4t^8} \\
 \\
 \\
\equal{} \dfrac{56\cdot8}{28\cdot4} \equal{} 2\cdot2 \equal{} 4.\]
Therefore, getting exactly three heads is $ 4$ times as likely as getting two heads.


Still no response? I'm learning for an exam and i haven't solved yet this exercice. So please any response will be appreciated.

Nea, you'd have a better chance of getting a response if you didn't post this in an inappropriate place. This is not matrix or linear algebra. If you don't know what linear algebra is, it's a good guess that your problem doesn't belong in that forum.


does anyone know what this book says when it says: "Every plane contains at least three noncollinear points; space contains at least four noncoplanar points" I mean, there are infinite points in a plane, right?

Postulates always say the minimum. Example: straight line contains 2 points.
Every plane has infinite points- that's what you prove by postulates and theorems already proven.

I'm sorry, but I don't understand what you're saying, could you clarify yourself???
Thanks for replying, though
could you mean that you need three points for a plane? Oh wait, I think that the Plane Postulate (Moise-Downs book) says that "Any three points lie in at least one plane, and any three noncollinear points lie in exactly one plane"  :(

All planes do contain an infinite number of points.  What the Plane Postulate basically says is that if you are given 3 noncollinear (meaning they do not lie on the same line) points, there is exactly one plane that goes through all three points.  (Again, this plane also contains an infinite number of other points, as well.)  If the three points are collinear (lie on the same line), then an infinite number of planes go through them.

That's not what the first version of the Plane Postulate is saying.  It's just positing the existence of at least three points on a plane.  If you had less than three points, you wouldn't need to talk about a plane.

This also applies to what you're asking, theax.

If you are only working with three noncollinear points, it will generally be easiest to think of them in a plane, rather than space.

The plane having three noncollinear points just discuss that the minimum shape of a plane is a triangle. Also, note the plane as two-dimensions. If you make a straight line from 2 points, then take the minimum(perpendicular) distance of the third point from the line, then it assures you that you have 2 lines making it 2 dimensions. Since a line has infinite points, then the plane must also have infinite number of points in its region.

thanks, but I still don't get it...
wait, could you possibly mean that you need three noncollinear points to DRAW a plane (just like you need 2 points to draw a line)?
Any help in this aspect would be greatly appreciated

It's not meaningful to talk about a collection of two or less points as a plane since one can just discuss it as a line.  This is the short answer.

The long answer is that the notion of a "plane" extends to beyond the Euclidean plane.  In particular, in [url=http://en.wikipedia.org/wiki/Finite_geometry]finite geometry[/url] one applies the axioms of ordinary geometry to an underlying set that is finite and one has constructions like the [url=http://en.wikipedia.org/wiki/Fano_plane]Fano plane[/url], which has a mere seven points.  It is really not meaningful to talk about a plane with two points or a space with three points here; one can just as easily be talking about a line or a plane respectively.

So perhaps these postulates would make the most sense if you instead say "every [i]nontrivial[/i] plane" and "every [i]nontrivial[/i] space" respectively.

could you elaborate?
I think I might get it, but right now, it is eluding my grasp
Thanks

What he's saying (in short) is that if you have two points, it is much easier to think about them (and draw them) in a line rather than a plane.

The same applies to having three points and thinking about them in a plane, rather than space.

Also, think about this. Given two points, you can define a single line, but an infinite number of planes. So it is much easier to think of them in a line, rather than in any (or all) of these planes.

Similarly, 3 points define a plane. Why bother your head with thinking about them in space when a plane will do? You don't need to unless you have a 4th point outside of that plane.

okay, I might need some time to digest this, but thanks!   :lol:


Let $ a,b,c,p,q,r$ be nonnegative integers such that $ a\plus{}b\plus{}c\equal{}p\plus{}q\plus{}r\equal{}1$ and $ p,q,r\leq \frac{1}{2}$. Prove that

$ 8abc\leq pa\plus{}qb\plus{}rc$

[quote="discredit"]Let $ a,b,c,p,q,r$ be nonnegative integers such that $ a \plus{} b \plus{} c \equal{} p \plus{} q \plus{} r \equal{} 1$ and $ p,q,r\leq \frac {1}{2}$. Prove that

$ 8abc\leq pa \plus{} qb \plus{} rc$[/quote]

I'm guessing you didn't really mean "integers"? Reals? Rationals?

Yep, reals. Sorry for the huge typo  :blush:

There exists positive real numbers $ u,v,w$ such that $ p\equal{}v\plus{}w$, etc.

Using the AM-GM inequality twice, $ \left( \frac{a}{2}\plus{}\frac{a}{2}\plus{}b\plus{}c\right)^2 (b\plus{}c)\ge \left( 4\sqrt[4]{\frac{a}{2}\frac{a}{2} b c}\right)^2 2\sqrt{bc}\equal{}16abc$. So

$ \sum u[ (a\plus{}b\plus{}c)^2 (b\plus{}c)\minus{}16abc]\ge 0\implies$

$ (a\plus{}b\plus{}c)^2[(v\plus{}w)a\plus{}(w\plus{}u)b\plus{}(u\plus{}v)c]\ge 16(u\plus{}v\plus{}w)abc\implies$

$ (a\plus{}b\plus{}c)^2(pa\plus{}qb\plus{}rc)\ge (p\plus{}q\plus{}r)8abc\implies$

$ pa\plus{}qb\plus{}rc\ge 8abc$ as desired.

[quote="dicredit"]
Let $ a,b,c,p,q,r$ be nonnegative reals such that $ a\plus{}b\plus{}c\equal{}p\plus{}q\plus{}r\equal{}1$ and $ p,q,r\leq\frac12$. Prove that
\[ 8abc\leq pa\plus{}qb\plus{}rc\]
[/quote]
After a quick application of Cauchy-Schwarz, it is enough to prove that\[ \frac18\geq(pbc\plus{}qac\plus{}rab)\]which is both stronger and nicer to prove :D

[hide="Solution by rearrangement"]
WLOG set $ a\geq b\geq c$. By rearrangement if $ \{p',q',r'\}\equal{}\{p,q,r\}$ with $ p'\leq q'\leq r'$ then $ p'bc\plus{}q'ac\plus{}r'ab\geq pbc\plus{}qac\plus{}rab$... so we drop the primes and consider the case $ p\leq q\leq r$.

Now $ pbc\plus{}qac\plus{}rab\leq pac\plus{}qac\plus{}rab\equal{}(a)([p\plus{}q]c\plus{}rb)$. Now, $ r\leq\frac12\implies p\plus{}q\geq r$. Thus $ (p\plus{}q,r)$ and $ (c,b)$ are sorted oppositely, so another form of rearrangement shows that \[ [p\plus{}q]c\plus{}rb\leq\frac{(p\plus{}q\plus{}r)(b\plus{}c)}{2}\equal{}\frac{b\plus{}c}{2}\]
But \[ \frac18\equal{}\frac12\left(\frac{a\plus{}(b\plus{}c)}{2}\right)^2\geq\frac{a(b\plus{}c)}{2}\geq a\cdot\left(\left[p\plus{}q\right]c\plus{}rb\right)\geq pbc\plus{}qac\plus{}rab\]
i.e. the inequality is proven
[/hide]


Let x and y be positive integers with no prime factors greater than 5. Find all x and y which satisfy:

                    $x^{2}-y^{2}= 2^{k}$

For some non negative k.
Might be totally wrong approach, but heres what I've got so far:
$(x+y)(x-y)=2^{k}$
Then
$x+y=2^{t}$
$x-y+2^{s}$  where     $k=s+t$
If $s=t$ then $x+y=x-y$
$\rightarrow y=0$
But $x,y$ are positive integers thus, no solutions for s=t
Because $t\geq s$
 $\rightarrow s=t-a$ for integer $0<a<t$

Thus, $2x=2^{t}+2^{t-a}$
        $2x=2^{t-a}(2^{a}+1)$

Putting $a=1$ gives$x=3*2^{t-2}, y=2^{t-2}$
Both satisyfy the condition of no prime factor greater than 5 and are solutions with $k=2t-1$
Putting $a=2$ gives the solution $x=5*2^{t-3}, y=3*2^{t-3}$
With $k=2t-2$
Putting $a=3$ gives $y=2^{t-4}(2^{4}-9)$ a multiple of 7

This is where I get stuck, it seems for any $a\geq 4$, x has a prime factor greater than 5. If I can show this then the question is finished. But I don't know how!  :(

Can anyone help? :lol:

See http://www.mathlinks.ro/Forum/viewtopic.php?p=420943 for a different approach, which worked. I'm not entirely sure that yours is patchable. Maybe try hitting it with mods 3 and 5, or mods 2,4,8, or some other kind of mod.

Incidentally, BMO2 2006 is available at http://www.mathlinks.ro/Forum/viewtopic.php?highlight=bmo2&t=72694
if you are training on it, so you can compare your solutions with the ones here.

Thanks, hadn't noticed those threads,

Yes I think the solution in the thread is probably a better way, (Annoying cos spent a long time typing it up in latex format!)

Any other ideas how to prove the last part?

I don't know how different this is from the other thread...

$x+y=2^{a}$
$x-y=2^{b}$
Clearly, these two numbers have the same parity.  Because $k$ is positive (were it zero, we'd have no positive integer solutions), we have $a+b$ is positive.  Thus, $a, b$ are natural (if one were zero and the other natural, the two expressions above would be of different parity).
$2x=2^{a}+2^{b}$
$2y=2^{a}-2^{b}$
Now $a, b>0 \iff a-1, b-1 \ge 0$ when $a, b \in \mathbb{Z}$.  Then, any solution is of the form
$x=2^{m}+2^{n}, y=2^{m}-2^{n}$ for nonnegative $m, n$.  However, we must have $2^{m-n}+1, 2^{m-n}-1$ both having no prime factors greater than 5.  Because the two are relatively prime and odd, one is a power of 3 and one is a power of 5...

$3|2^{m-n}+1 \implies m-n \equiv 1\mod{2}$
$5|2^{m-n}-1 \implies m-n \equiv 0\mod{4}$
Impossible.  So we consider the other case.  Let $m-n=r$.
$2^{r}+1=5^{a}$
$2^{r}-1=3^{b}$
If $r$ were odd, we would have $3|2^{r}+1$.  Thus, $r$ must be even.  Let $r=2t$.
$2^{2t}+1=5^{a}$
$(2^{t}-1)(2^{t}+1)=3^{b}$
But $2^{t}-1, 2^{t}+1$ are relatively prime.  Thus, $2^{t}-1=1 \implies r=2$

So any solution to the original equation has the form \[(x, y)=(5*2^{n}, 3*2^{n})\] where $n$ is nonnegative.


[quote="xethissy"]Factor: 12x^2(4x+5)^7(2x+1)+24x(4x+5)^8[/quote] its hard to read the problem ill fix it. next time, you can put dollar signs around the equation because it looks better and easier to understand.

$ 12x^{2}(4x\plus{}5)^{7}(2x\plus{}1)\plus{}24x(4x\plus{}5)^{8}$.
factor out a $ (4x\plus{}5)^{7}$ and get $ 12x(4x\plus{}5)^{7}[x(2x\plus{}1)\plus{}2(4x\plus{}5)]$. that becomes $ 12x(4x\plus{}5)^{7}(2x^{2}\plus{}8x\plus{}10)$. i cant factor this so that my answer. its probably not correct.


integration
trigonometry
real analysis
real analysis unsolved

The antiderivative is not exactly a simple function.  To view it, copy
"1/Sqrt[(Sin[x])](Sqrt[Cos[x]]^5)" into the following website:http://integrals.wolfram.com/index.jsp

[quote="SoNYDvDPro"]Can anyone solve This ? $ I \equal{} \int {\sin } (x)^{ \minus{} \frac {1}{2}} \cos (x)^{\frac {5}{2}} dx$[/quote]$ \int\frac {\cos^{\frac {5}{2}}x}{\sqrt {\sin{x}}}dx$

yes?


Columbia
function
Olimpiada de matematicas

[b]NIVEL BASICO[/b]

[b]Problema 1[/b] [Teoria de Numeros]

Dada la ecuacion:
\[ x^2 + y^2 + z^2 = 2xyz.  \]
Mostrar que esta no tiene soluciones enteras, excepto $x=y=z=0$.

[b] Extension [/b] Dada la ecuacion anterior, tomemos $k$  un numero entero, mostrar que la ecuacion:
\[ x^2 + y^2 + z^2 = kxyz.  \]
tienen soluciones en los enteros solo para valores $k=1$ y $k=3$, encontrando infinitas soluciones en estos casos.

[b]Problema 2[/b] [Geometria]

Tres circunferencias que se intersecan, pasan por el punto $H$. Demostrar que $H$ es el ortocentro del triangulo, cuyos vertices
coinciden con los otros tres puntos de interseccion de dos en dos de las circunferencias.

[b]Problema 3[/b] [Combinatorio]

Tres tipos de objetos, con $2n$  de cada uno, son repartidos a dos personas, tal que cada persona recibe $3n$ objetos en total.
Probar que esto puede ser hecho en "$3n^2 + 3n + 1$" maneras.

[b]NIVEL AVANZADO[/b]

[b]Problema 1[/b]

$p$ es un primo impar congruente con 2 modulo 3. Pruebe que maximo $p-1$ numeros en el conjunto $S=\{m^2-n^3-1: 0<m,n<p\}$ son divisibles por $p$.

[b]Problema 2[/b]

$a,b,c$ son reales positivos tales que su producto no es mayor que su suma. Pruebe que:
\[ a^2+b^2+c^2\geq \sqrt{3}abc  \]

[b]Problema 3[/b]

Un rectangulo $ABCD$ tiene lados $AB=m$ y $AD= n$ con $m,n$ enteros positivos primos relativos e impares. El rectangulo es dividido en cuadritos unitarios y la diagonal $AC$ intersecta los lados de estos cuadritos en: $A_1=A$, $A_2$...,$A_N=C$. Pruebe que:

\[ A_1A_2-A_2A_3+A_3A_4...+(-1)^{N}A_{N-1}A_N=\frac{AC}{mn}  \]

Disfruten de los problemas  ;)
Los Organizadores

Haber ... aki estoy colgando la lista de problemas que conforman:
El RETO DE LA SEMANA ...

Carlos Bravo ;)
Lima -peru

Aqui unas soluciones por Jose Samper de Colombia, les queda de tarea leerlas y revisarlas.

A simple vista felicitaciones Samper, buen trabajo.

Se podran dar cuenta que el 3 no esta terminado... la verdad es que no se como escribirlo, la idea es que la funcion determina los signosde la suma. Estos se ven segun la pariedad de k. Voy a ver si lo escribo bien y lo posteo luego.

Para el problema uno solo hay que usar lo siguiente:

en $Z_p$ ($p=3k+2$) la aplicacion $f(a)=a^3$ es biyectiva. Es decir, todo numero es un cubo, y dos cubos son distintos.

Para el 3 queda una cota meno que $p-1$. Pues es lo mismo que buscar $m,n$ tales que:
$m^2 - 1 \equiv n^3$ (mod $p$). Pero para $m=1$ y $m=p-1$  se tiene que $m^2 - 1 \equiv 0$(mod $p$) por lo que no tienen solucion.
Entonces seria suficiente probar que  $p\nmid m-n \implies p\nmid m^3 - n^3$ que es lo mismo que probar que $p \nmid m^2 + mn +n^2$
Como $(n,p) =1$ se tiene que existe un $\overline{n}$ tq $n\cdot \overline {n} \equiv 1$ (mod $p$)
sup que $m\overline {n} = x$
Ahora sup que:
$m^2 + mn + n^2 \equiv 0$ (mod $p$) $\iff m^2\overline{n}^2 + m\overline {n} +1 \equiv 0$ (mod $p$) que es lo mismo que:
$x^2 + x + 1\equiv 0$ (mod $p$) $\iff 4x^2 + 4x + 4\equiv 0$ (mod$p$) $\iff (2x + 1)^2 \equiv -3$ (mod $p$)
$\iff \binom{-3}{p}_L = 1$
Pero $\binom{-3}{p}_L = \binom {1}{p}_L \cdot \binom {3}{p}_L = (-1)^{\frac{p-1}{2}} \cdot \frac{(-1)^{\frac{p-1}{2}}}{\binom{p}{3}_L} = \frac{(-1)^{\frac{p-1}{2}}}{\binom{2}{3}_L} = \frac{(-1)^{\frac{p-1}{2}}}{(-1)}= (-1)^{\frac{p-1}{2}} \cdot (-1)^{\frac{p-3}{2}}=-1$
que es una contradiccion. Por eso se cumple lo que queriamos, y a cada $m$ solo le corresponde un $n$ por lo que hay max $p-3$ parejas $(m,n)$.

el problema 2 de geometria de básico esta mal... creo que deberia ser circunferencias de igual radio

$Tipe$

Yo creo que las circunferencias deben ser ortogonales

Aki
la version original del problema:

;)

[Ed. MIR]
[b]Tres circunferencias que se intersecan, pasan por el punto $H$. Demostrar que $H$ es el punto de interseccion de las alturas de un triangulo, cuyos vertices coinciden con los otros tres puntos de interseccion de dos en dos de las circunferencias.[/b]
[/Ed. MIR]

Carlos Bravo ;)
Lima PERU

[color=blue][b][size=150]SOLUCIONARIO DE PROBLEMAS: RETO \#1 - 2006[/size][/b][/color]
[b][color=blue]SEMANA: 02-07 FEBRERO[/color][/b]
[b][color=blue]NIVEL BASICO[/color][/b]

$\blacktriangleright$ [color=blue]Pregunta 01[/color] [Teoria de Numeros]
Dada la ecuacion:
$x^2 + y^2 + z^2 = 2xyz.$
No tiene soluciones en los enteros positivos.
{\bfseries Extensi\'on} Dado la ecuaci\'on anterior, tomemos $k>0$
entero positivo, mostrar que la ecuaci\'on:
$x^2 + y^2 + z^2 = kxyz.$
tinen soluciones en los enteros s\'olo para valores $k=1$ y $k=3$,
encontrando infinitas soluciones en estos casos.

[hide="Solucion"]
Sea $(x,y,z) \neq (0,0,0)$ una solucion entera.
Si $2^k, k\geq 0$ es la mayor potencia de 2, el cual divide a $x,y,z$, entonces:
$x= 2^kx_1, y= 2^ky_1, z= 2^kz_1$
$2^{2k}x_1^2 + 2^{2k}y_1^2 +2^{2k}z_1^2 =2^{3k+1}x_1y_1z_1,$
$x_1^2 + y_1^2 + z_1^2 = 2^{k+1}x_1y_1z_1 ... (1)$
Como el lado derecho de (1) es positivo y ademas el lado izquierdo
es par. Concluimos que todos los 3 terminos en el lado izquierdo
no pueden ser pares debido a la eleccion de $k$. A partir de esto,
existe exactamente un termino par. Tomemos que $x_1 = 2x_2$,
mientras $y_1$ y $z_1$ son impares. Luego tendriamos:
$y_1^2 + z_1^2 = 2^{k+2}x_2y_1z_1 - 4x_2^2 \equiv 0\mod 4.$
Lo que genera una contradiccion con:$y_1^2 + z_1^2 \equiv 2\mod 4$
Por tanto no existen soluciones enteras.
[/hide]
$\blacktriangleright$ [color=blue]Pregunta 02[/color] [Geometr\'ia]
Tres circunferencias que se intersecan, pasan por el punto $H$.
Demostrar que $H$ es el ortocentro del triangulo, cuyos v\'ertices
coinciden con los otros tres puntos de interseccion de dos en dos
de las circunferencias.

[hide="Solucion"] PENDIENTE [/hide]

$\blacktriangleright$ [color=blue]Pregunta 03[/color] [Combinatorio]
Tres tipos de objetos, con $2n$  de cada uno, son repartidos a dos
personas, tal que cada persona recibe $3n$ objetos en total.
Probar que esto puede ser hecho en $3n^2 + 3n + 1$ maneras.

[b][color=blue][Enviada por [i]Fabycv - Fabiola Bravo[/i] - Lima / PERU][/color][/b]
[hide="Solucion"]
Previo: Definamos $CR(n;p)$ como el Nro. de combinaciones con repeticion
de n elementos tomados de p en p. Podemos obtener el resultado siguiente:

$CR(n;p) = {n +p-1\choose p}$            ......(1)

En el problema tenemos el conjunto $\{A,B,C \}$ ($2n$ de cada uno), para
repartirlos a las personas $X$ e $Y$, seria lo mismo como repartir $3n$
de los elemetos totales a $X$ (y el resto seria para $Y$).
El problema es encontrar el numero de formas de elegir $3n$ elementos
 del total: $6n$. Tomemos grupos de $3n$ elementos apartir de un
 conjunto $\{A,B,C \}$ ($2n$ de cada uno):

$CR(3;3n) = {3 + 3n -1\choose 3n}$      ......(2)

De aqui debemos restar aquellos grupos de $3n$ q tienes mas de $2n$
elemntos de un mismo tipo, estos grupos seria asi:
$\underbrace{a,a,a,\cdots,a}_{\"2n\" objetos},a,\underbrace{x,\cdots, z.}_{\"n-1\" objetos}$
Luego, adicionalmente de la cadena de $2n$ objetos de tipo $A$, tomamos
un elemento $a$ mas, lo que asegura que es una cadena no permitida.
Calculemos el numero de estas cadenas:

$CR(3;n-1) = {3 + (n -1) -1\choose n-1}$  ......(3)

Notar que esto solo fue para el caso de exceder los $2n$ objetos tipos $A$,
de manera analoga, obtenemos las cadenas no permitidas para los tipos $B$ y $C$.
Asi logramos lo pedido:

${3+3n-1\choose 3n}-3{3+n-2\choose n-1}= {3n+2\choose 2} - 3{n+1\choose 2} = \frac{(3n+1)(3n+2)}{2} - 3\frac{n(n+1)}{2} = 3n^2 + 3n + 1.$
[/hide]

---
[b][color=blue]NIVEL AVANZADO[/color][/b]

[b]Problema 2[/b]
$a,b,c$ son reales positivos tales que su producto no es mayor que su suma. Pruebe que:
$a^2+b^2+c^2\geq \sqrt{3}abc$

[b][color=blue][Enviada por [i]js131090 - jossy alva gutierrez[/i] - Lima/PERU][/color][/b]
[color=blue]Solucion:[/color]
Como $a,b,c$ son reales positivos, entonces $\sqrt{abc} >0$ ;por lo cual podemos definir a  como :
$x=\frac{a}{\sqrt{abc}}, y=\frac{b}{\sqrt{abc}}, z=\frac{c}{\sqrt{abc}}$ ,por lo cual $x,y,z > 0$

Entonces se observara que:
  $a= \frac{1}{yz}$
  $b= \frac{1}{xz}$
  $c= \frac{1}{xy}$

Ahora como el dato es: $a+b+c \ge abc$ , entonces reemplazando: $\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}\ge \frac{1}{yz}.\frac{1}{xz}.\frac{1}{xy}$, resolviendolo seria:
$xyz(x+y+z)\ge 1...............(*)$

Ahora se sabe la siguiente desigualdad:  $(m+n+p)^2 \ge 3(mn+np+pm), \forall m,n,p \in \Re^+$
Entonces poniendo como: $m=xy, n=xz , p = yz$  seria:
$(xy+yz+xz)^2 \ge 3xyz(x+y+z)$
 y de (*) se tendra que:$(xy+yz+xz)^2 \ge 3 ............(**)$; ahora como:  $xy+yz+xz > 0.$

Se tendra de (**): $xy+yz+xz \eg \sqrt{3}$ ;ahora remplasando a $a,b,c$ seria:
$ab+bc+ca \ge \sqrt{3}abc$,pero: $a^2 + b^2 + c^2 \ge ab+bc+ca$
Entonces seria finalmente:
$a^2+b^2+c^2\geq \sqrt{3}abc$

---
[color=blue][b]Los Organizadores [/b][/color]

[size=150][color=blue][b]SOLUCIONARIO DE PROBLEMAS: RETO \#1 - 2006
SEMANA: 02-07 FEBRERO
NIVEL BASICO
Versión PDF[/b][/color][/size]

Segun mi juicio, el problema 2 de basico esta obviamente mal. Tome 2 circunferencias que se cortan en $A$ y $B$, tome una linea l que no sea perpendicular a $AB$ y que corta a los circulos en dos puntos $C$ y $D$, tome la tercera circunferencia por $B,C,D$, aqui H no es el ortocentro de $ACD$!!!!!!!!!!!!!!!!!!!!!!

Es mas el problema es cierto si y solo si las circunferencias tienen igual radio, pista: el angulo que dos circunferencias forman con su cuerda comun es igual al complemento de un angulo fijo!!!

 :(

[color=blue]
 Asi es ...

 he dejado de lado... el prob 02 -NIVEL BASICO....

Carlos Bravo ;)
Lima Peru[/color]

Esto nos enseña  que aun las buenas fuentes pueden tener errores  :( ; que raro que un libro de MIR tenga un error como ese.

$Tipe$


I am in the ninth grade, should I take 12a and then 10b this year or 10a then 12b? Are the tests very similar or does it matter?

I think the test for the A/B date is chosen randomly, so it wouldn't make a difference. You have to choose based on your situation, because the A is two weeks before the B.

10a first. Why? If you do really well on it (by whatever your standards are) then you can take the 12b as a challenge.

If, however, you do poorly on it, you still have a second shot at making AIME (or improving your score, whatever your goal is)

So I would do 10a, then 10b or 12b, depending on how well you do on the first test.

You also have to consider what your school offers and make sure the tests you want are available on the dates you want. For example, I just found out that my school offers 10a and 12b.

[quote="Demosthenes"]you dont get your score for 10A until after you take the (B) test
last year i though i got a 132 on 10A, so i took 10B hoping for a perfect
When the scores came back, i got a 114 on both :(
So i would suggest taking at least one 12 test, preferrably 12B[/quote]

This year the 10a/12a is on February 9th, and the AoPS math jam on the 10th goes over the hard questions on the 10a/12a. So if your a person who doesnt make a bunch of silly mistakes ( :P ) you should more or less know what your score is.

i didnt know about the math jam last year (this accnt wan't on aops yet)

Actually, if you think you have a shot at making Blue MOP (or are <9th), 12a10b might be a better option as to maximizing the chances of making USAMO instead of USAJMO. (so you can switch from 10b to 12b if you mess up 12a)


integration
real analysis
real analysis unsolved

(a) If $f$ is Riemann integrable on $[a, b]$, prove that $F(x) = \int_{a}^{x}f(t)dt$ is continuous.

(b) Prove it satisfies a Lipschitz condition.

We can assume f is bounded and use a mean value theorem. You should get the following, where c>0

$|F(x)-F(y)| = |\int_{y}^{x}f | = c|x-y|$

A Lipschitz property follows from making 'c' the sup of f on [a,b]

(a). Suppose $f$ is Riemann integrable on $[a,b]$. Then $f$ is bounded on $[a,b]$, say $|f(t)|\leq$$K$ for $t\in[a,b]$. For any $x<y$ s.t. $x,y\in[a,b]$; $|F(c)-F(d)|=|\int_{a}^{d}f(t)dt-\int_{a}^{c}f(t)dt|=|\int_{c}^{d}f(t)dt|\leq~K(c-d)$. Given any $\epsilon>0$ we can find a $\delta=\frac{\epsilon}{K}$ s.t. $|F(c)-F(d)|<\epsilon$ whenever $|c-d|<\delta=\frac{\epsilon}{K}$.

(b). By (a), $|F(c)-F(d)|\leq~K|c-d|$ for all $c,d\in[a,b]$.

to: abdul

It looks more pleasant if you were to say $K>0$

sylow_theory: i get sloppy sometimes. thank you for reminding me.


geometry
perimeter
3D geometry
algebra
factorization
sum of cubes

Jamal's rectangular hood has area 1337 and perimeter 7331. Find the sum of the cubes of the dimensions of his hood.

LMAO Jamal's hood.... Dat be TOO gangsta...
[hide]
First lets factor the sum of two cubes.
$x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$
We know the perimeter is $2(x+y)=7331$
We know the area is $xy=1337$
We know that $x^{2}-xy+y^{2}=(x+y)^{2}-3xy$
So we know that $x+y=\frac{7331}{2}$
So we also know that $(x+y)^{2}-3xy=\frac{7331^{2}}{4}-3\cdot{1337}$
Thus the sum of cubes is $(\frac{7331}{2})(\frac{7331^{2}}{4}-3\cdot{1337})$
Which is $49,234,553,390.875$.
But then again my arithmetic is probably wrong - and I used a calculator!
[/hide]

EDIT: Dangit! I mixed up mah area and perimeter. All fixed now.

Hmm, well yeah assuming the computations are correct, that seems, right, but I was thinking of writing $x^{3}+y^{3}=(x+y)^{3}-3xy(x+y)$, which is a tad simpler, but it's essentially the same, because we can factor out $(x+y)$ and get your factorization. My way just makes it more direct so that you can plug in the values for $x+y$ and $xy$.

heh yeah i probably should have that of your way if i thought of the same thing with $(x+y)^{2}$ - but yeah its the same...


Pythagorean Theorem
geometry
similar triangles

Points X and Y trisect segment AC, which is the hypotenuse of right triangle ABC. If, for some angle :theta:, BX = cos  :theta:  and BY = sin  :theta: , find the square of the hypotenuse AC.

It is late at night, but hopefully I can still add and subtract:

[hide] Let a , b, and c be the sides opposite angles A, B, and C respectively.

By Pythagorean Theorem, a^2 + c^2 = b^2.

Let AX = XY = YC = b/3.

By similar triangles, altitude from Y to BC = c/3, altitude from X to BC = 2c/3, altitude from Y to AB = 2a/3, and altitude from X to AB = a/3.

By Pythagorean Theorem,

          (c/3)^2+(2c/3)^2 = sin^2(:theta:)

          (a/3)^2+(2a/3)^2= cos^2(:theta:)

Adding the two equations, 5c^2/9+5a^2/9=sin^2(:theta:)+cos^2(:theta:).

Simplifying, 5/9(c^2+a^2) = 1. b^2 = a^2+c^2 = 9/5 --> b^2 = 9/5.[/hide]

Sayoonara.

fuzzylogic wrote:Good. It can also be solved using [hide]Stewart's Theorem[/hide], right?

Yes.  I've used this question (and generalizations of it to larger numbers of cevians from the right angle) as an excercize in using [hide]Stewart's Theorem.[/hide]


Is this the equation written in an easier to read way?

$ \frac{53506}{(2^{(a\plus{}3)})(5^{2})(10^{3})(43)}$

As for the question itself....someone else will have to help you. I'm not smart enough.

You want a formula for the answer in terms of $ a$?

[quote="Nikki123"]Can someone please help me with this question ?

Find the length l of the non-repeating part and the length m of the repeating part of the decimal representation of 53506/((2^(a+3)).(5^2).(10^3).(43)).[/quote]

If a=8, what would be the answer ? Can someone please fully show me how to do it with working out please ?

Thank you for your input.


Hello everyone! This is my first post and my english isn't very well, so sorry for mistakes.

$ 3 \minus{} 5 \minus{} 8$ is a game for two people who make moves alternately. We have three rows of matches (3 matches in the first row, 5 in second and 8 in third). Player in his move may take arbitrary number of matches from exactly one row. The game ends when there is no matches to take and lose the player who take the last match.

I have some questions:
1) Who has a winning strategy?
2) Who has the winning strategy if there are more players ($ 3, 4, \ldots$)?
3) Who has the winning strategy for a game $ k \minus{} m \minus{} n$ ($ k,m,n$ - natural numbers)?
4) Who has the winning strategy for a game $ k \minus{} m \minus{} n$ with more players?
5) Who has the winning strategy for a game $ k_1 \minus{} k_2 \minus{} \ldots \minus{} k_n$?
6) Who has the winning strategy for a game $ k_1 \minus{} k_2 \minus{} \ldots \minus{} k_n$ with more players?
7) What is the minimal number of moves necesary to win a game in each case (if there is a winning stategy)?

This game is called Nim, see [url=http://en.wikipedia.org/wiki/Nim]here[/url].


A square table n×n is given. What is the smallest number of squares that need to be colored so that every non-colored square is share at least one side with at least one colored square?


1. When 2-chloromethylfuran is treated with aqueous potassium cyanide the only product obtained is 2-cyano-5-methylfuran. Propose a mechanism that might explain this transformation.

2. When 2-bromocyclobutanone is treated with aqueous base, followed by an acidic workup, cyclopropane carboxilic acid is obtained. Propose a reasonable mechanism.

3. Provide a mechanism for the following reaction:

4. When 1-vinylcyclopropanol is treated with peroxibenzoic acid in ethyl ether, 2-(hydroxymethyl)cyclobutanone is obtained with 32% yield. Propose a mechanim that might explain the formation of this product.

5. When oxazole is treated with acrylonitrile in triethylamine and heated to about 80ºC, 4-cyanopyridine is obtained as the main product. What is the mechanism of this reaction?

6. When the compound $ C_6H_5CH(OH)CH(CH_2CH_2CH_3)COCH_3$ is treated with $ BF_3$, the main product obtained is $ E \minus{} C_6H_5CH \equal{} CH \minus{} CH_2CH_2CH_3$. Propose a mechanism for this transformation.

7. When nitroethane is treated with two equivalents of MVK (methyl vinyl ketone) in the presence of potassium carbonate, and the resulting mixture is refluxed in toluene (and in contact with air) for several hours, 2,5-dimethylacetophenone is obtained as the main product. Propose mechanisms for the reactions involved. (note that two [i]open chain[/i] reagents are converted into a [i]cyclic and aromatic[/i] product!)


number theory
greatest common divisor
inequalities
abstract algebra
superior algebra
superior algebra unsolved

a) If $K$ is a finite extension of the field $F$ and $K=F(\alpha,\beta)$ show that $[K: F]\leq [F(\alpha): F][F(\beta): F]$

b) If $gcd([F(\alpha): F],[F(\beta): F])=1$ then does the above inequality always become equality?

c) By giving an example show that if $gcd([F(\alpha): F],[F(\beta): F])\neq 1$ then equality might happen.

[b]a)[/b] A bit more general (by the existence of primitive elements in separable cases not much more general):
Let $x_i$ , $y_j$ be bases of $A,B$ over $F$. Then the $y_j$ generate $K=F[x_i,y_j]$ over $A$ (but are not necessary a basis). Now see http://www.mathlinks.ro/Forum/viewtopic.php?t=71108 for the rest (if modules are not used to you just change $A,B$ to $F[x_i],K=F[x_i,y_j]$).
Thus we have the desired inequality $[K: F]\leq [F(x_i): F][F(y_i): F]$.

[b]b)[/b] Well, the whole extension $K|F$ must contain sub-extensions of both coprime degrees, thus is a multiple of them, thus yes.

[b]c)[/b] $F=\mathbb{Q}, \alpha = \sqrt 2 , \beta = \sqrt 3$

For 1) $ [K : F] \equal{} [F(\beta)(\alpha) : F(\beta)][F(\beta) : F]$

But certainly $ [F(\beta)(\alpha) : F(\beta)] \equal{} deg(\alpha, F(\beta)) \leq deg(\alpha, F) \equal{} [F(\alpha) : F]$, by definition of minimal polynomial.

Sorry if it's the same as what ZetaX said.


for $ a,b,c \ge 0,$

for what conditions does this inequality hold?

$ \sum_{sym} a^4 bc \ge \sum_{sym}a^3 b^3$  :maybe:

[hide="generalize?"]for $ a_{1}, a_{2},..., a_{n} \ge 0,$

for what conditions does this inequality hold?

$ \sum_{sym} a_{1}^{4}a_{2}a_{3} \ge \sum_{sym} a_{1}^{3}a_{2}^{3}$ :maybe:  :| [/hide]

[quote="earldbest"]for $ a,b,c \ge 0,$

for what conditions does this inequality hold?

$ \sum_{sym} a^4 bc \ge \sum_{sym}a^3 b^3$  :maybe:

[/quote]

assume $ a\ge b\ge c\ge 0$

$ c = 0 \Rightarrow b = 0$

$ c\neq0$ assume $ c = 1$

Then

$ \sum_{sym} a^4 bc - \sum_{sym}a^3 b^3 = 2(a - b^2)(a^2 - b)(ab - 1)\ge 0$

2nd and 3rd factor are always not negative (zero only if $ a = b = 1$)

-> It holds if and only if $ \sqrt {ac}\ge b$

For your generalisation i looked at it with n=4 and had put some random values for 3 variables and unfortunatly they didn't have any nice solution for the 4th variable. Maybe this one is better:

$ a_{1}, a_{2},..., a_{n}\ge 0,$

$ %Error. "displaymode" is a bad command.
\left (\prod^n_{i = 1} a^{n - 2}_i \right) \left ( \sum^n_{i = 1} a^n_i \right)\ge \sum^n_{k = 1} \prod_{i \neq k}a^n_i$

EDIT: With this generalisation there's also a nice condition for n=4: if $ a_1\ge a_2\ge a_3 \ge a_4 \ge 0$ then the inequality is true if and only if $ a_1a_4\ge a_2a_3$

I'm still looking for a factorisation for n=5...

[quote="Rafikafi"]

assume $ a\ge b\ge c\ge 0$

$ c = 0 \Rightarrow b = 0$

$ c\neq0$ assume $ c = 1$

Then

$ \sum_{sym} a^4 bc - \sum_{sym}a^3 b^3 = 2(a - b^2)(a^2 - b)(ab - 1)\ge 0$

2nd and 3rd factor are always not negative (zero only if $ a = b = 1$)

-> It holds if and only if $ \sqrt {ac}\ge b$

For your generalisation i looked at it with n=4 and had put some random values for 3 variables and unfortunatly they didn't have any nice solution for the 4th variable. Maybe this one is better:

$ a_{1}, a_{2},..., a_{n}\ge 0,$

$ %Error. "displaymode" is a bad command.
\left (\prod^n_{i = 1} a^{n - 2}_i \right) \left ( \sum^n_{i = 1} a^n_i \right)\ge \sum^n_{k = 1} \prod_{i \neq k}a^n_i$

EDIT: With this generalisation there's also a nice condition for n=4: if $ a_1\ge a_2\ge a_3 \ge a_4 \ge 0$ then the inequality is true if and only if $ a_1a_4\ge a_2a_3$

I'm still looking for a factorisation for n=5...[/quote]

nice!  :lol:


integration
geometry
vector
calculus
advanced fields
advanced fields unsolved

let $\gamma(x)$is in $R^{2}$ and $|k(t)|<1$ where $k(t)=\frac{\gamma'(t)\times\gamma''(t)}{|\gamma'(t)|^{3}}$
and also
$\int_{a}^{b}|\gamma'(t)|dt=\pi$
prove that
$|\gamma(b)-\gamma(a)|>2$

Peaple this is not hard problem. yesterday in my classwork (in diferential geometry) there was this problem and I have solve it. :lol: it's simpl.
we choose point $A$ in curve wich divide curve in to equal part (two equal side) consider one part. and there consider Vector of speed of our curve.  and in point $A$ we construct tangent line.
we consider proection of vector of this curve of one part. after we note that integral ov this proection will be minimal if curve is circle and find minimal value of this one part wich will be equal to $1$ plus the seccond part  also equal to $1$ and you are done.


Show (without calculus :wink: if possible) that

1/3 + 1/4 + 1/5 + ... < 1/12.

[hide]A nice escape from Bertrand's postulate

Very classical telescoping technique :

1/n:^3: < 1/n(n-1)(n+1) = -1/n + 1/2(n+1)+ 1/2(n-1)

So  :Sigma:(1..N) 1/n:^3: < -1/2(N+1)*N + 1/12   [/hide]

Apry's Constant is zeta(3) or the sum from n=1 to infinity of 1/n3. Roger Apry proved in 1979 that this number is irrational, so it has been named after him.


1) Να βρεθεί μη συνεχής συνάρτηση από έναν μετρικό χώρο σε έναν άλλο η οποία να απεικονίζει ανοικτά σύνολα σε ανοιχτά .
2) Υπάρχει μη συνεχής συνάρτηση από το $ R$ στο $ R$ που να απεικονίζει  ανοικτά σύνολα σε ανοιχτά ?

Πολύ βιαστικά (οπότε μπορεί να μην το σκέφτομαι καλά), και επειδή κάθε ανοικτό σύνολο περιέχει ένα ανοικτό διάστημα, δε σου κάνουν [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=226430]αυτές[/url] οι συναρτήσεις?

Cheerio,

Durandal 1707

Αν όμως ζητήσουμε οι συναρτήσεις να είναι 1-1 τότε η απάντηση αλλάζει.


function
complex analysis
complex analysis unsolved

Prove that the set of natural numbers cannot be [i]partitioned[/i] into a [i]finite[/i] number of arithmetic progressions (sets of the form $ \{a\plus{}kd: k\in\mathbb{N}\cup\{0\}\},\ a,d\in\mathbb{N}$) with distinct differeneces $ d$ (excluding the possibility $ a\equal{}d\equal{}1$).

Consider $ \frac{z}{z\minus{}1}\equal{}\sum_{n\in\mathbb{N}}z^n$.

This has been asked several times before, and solved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23227]here[/url]. From what you wrote, you're probably thinking of vess' solution.

vess's solution can also be phrased in terms of partial fraction decomposition, which suggests the following neat interpretation:

A finite union of disjoint arithmetic progressions is periodic with period some positive integer $ N$.  Consider the indicator function of such a finite union and take its discrete Fourier transform.  If all of the differences $ d_k$ are distinct, let $ d_1$ denote the largest difference.  Since $ d_1 > d_k, k \neq 1$, the coefficient of $ e^{\frac {2 \pi i}{d_1} x }$ is nonzero; contradiction.


modular arithmetic
number theory unsolved
number theory

Let given two primes $p, q$ satisfying: $p-1$ divisible by $q$ and $q^{3}-1$ divisible by $p$. Prove that: $p+q$ is a square number.


[hide]Sophie Germain identity may help    :

(x^4 + 4y^4) = (x-2xy+2y)*(x+2xy+2y)

89/5  [/hide]


Given that $ x, y, z$ are positive real numbers satisfying $ xyz\equal{}32$, find the minimum value of

$ x^{2}\plus{}4xy\plus{}4y^{2} \plus{}2z^{2}$.

$ (x\plus{}2y)^2 \plus{} 2z^2 \geq 8xy\plus{}2z^2\equal{}\frac{128}{z} \plus{} \frac{128}{z} \plus{}2z^2 \geq 96$ by A.M-G.M

Inequality holds if and only if $ x\equal{}z\equal{}4 \: , \: y\equal{}2$  :|


analytic geometry
graphing lines
slope
calculus
derivative
calculus computations

Find the equation of tangent of the curve, $ y\equal{}e^x$ which goes through the origin $ (0,0)$

The equation of tangent of the curve $ y\equal{}e^x$ at the point $ (x_0,\,e^{x_0})$ is $ y\equal{}e^{x_0}(x\minus{}x_0)\plus{}e^{x_0}$.
Because the spesific tangent goes through the origin $ (0,\,0)$, holds:
$ 0\equal{}e^{x_0}(0\minus{}x_0)\plus{}e^{x_0} \ \Leftrightarrow \ 0\equal{}e^{x_0}(1\minus{}x_0) \ \Leftrightarrow \ x_0\equal{}1$.
So the equation of the spesific tangent  is $ y\equal{}e(x\minus{}1)\plus{}e \ \Leftrightarrow \ y\equal{}ex$.

Firstly, differentiate y=e^x, to obtain the slope of the tangent to the curve which is: e^x

dy/dx=e^x;

Since the point is (0,0), now by applying the formula--- y-y1=m(x-x1), we obtain the equation of the tangent to the curve.

y=e^x

[b]Gen8[/b], Your solution is not correct/incomplete.

Have a look at the solution of [b]grigkost[/b] :) He has answered correctly.

Okay, I guess so, But my solution is correct to some extent, maybe I have overlooked some aspect  of the question ( I am a human, so I tend to make mistakes), thanks for pointing out a possible flaw.

The equation of the tangent derived doesn't pass through the origin.

The correct equation is: y=(e^x).x

Haha, I overlooked the 'x' , and it has made a whole lot of difference.

Now the equation is correct, it was a silly mistake. :lol:

[quote="Gen8"]The correct equation is: y=(e^x).x[/quote]
That is not the equation of a line. Since the problem asked for the equation of a line, that cannot possibly be correct.

The mistake you just made is an extremely common student mistake - but it's still a mistake. And if I had asked this question on an exam, you would get zero points for it - no partial credit at all.

You're asked for a tangent line to a curve through a point not on the curve. You have to notice that: $ (0,0)$ is not on the curve $ y\equal{}e^x.$ At this point you should realize that if you don't know the point of tangency, you're going to have to solve for that point of tangency. And whatever you do, do not use the same symbol to stand for the point of tangency and for a generic point on the line. It is very common to try to make $ x$ mean both of those things, but if you do that, you're already lost. Note that grigkost did use distinct symbols: $ x$ for the first coordinate of a generic point on the tangent line, $ x_0$ for the first coordinate for the point of tangency.

In order to make the explanation clearer, I'll put in an intermediate step or two that grigkost skipped over. We want the line tangent to the curve at the point $ (x_0,e^{x_0}).$ The derivative - the slope - at that point is $ e^{x_0}.$ That means the equation for the tangent line is

$ y\minus{}e^{x_0}\equal{}e^{x_0}(x\minus{}x_0)$ or $ y\equal{}e^{x_0}(x\minus{}x_0)\plus{}e^{x_0}.$

Now we put $ x\equal{}0,y\equal{}0$ into that equation because of the condition that the line goes through $ (0,0).$ You can now go back to grigkost's post to pick up the rest of this.

(I'll add that grigkost couldn't possibly have written what he did without at least thinking of these intermediate steps.)


http://www.artofproblemsolving.com/Wiki/index.php/2005_AIME_I_Problems/Problem_1

how are the circles congruent?

The question statement is not consistent with the solution as the solution seems to let K include the area of the region inside the six circles while the question statement specifies K to be the area of the region inside circle C and outside of the six circles.

"Outside the six circles" means "not in the 6 circles," not "outside the boundary of the 6 circles".


function
calculus
integration
topology
real analysis
real analysis theorems

hi,

what are the pre-requites for measure theory ?
is there a book with solved problems in measure theory anywhere that I could try out ?

A good knowledge of sequences of functions probably? Rudin's "Principles of Mathematical Analysis" is a classic for learning real analysis up through measure and integration, you can probably find a used paperback reasonably cheap. I keep it on my person at all times...

I agree. That would be a good prereq (although I would say that learning the whole book is by no means necessary). It takes you right up to his book Real and Complex Analysis. The first couple of chapters of that one is essentially an entire course in measure theory.

You need to be completely comfortable with the standard real analysis course, so you should have no trouble with the $ \epsilon\minus{}\delta$ language, basic topology in metric spaces, suprema and infima, (uniform) continuity, (absolute) convergence of series, and such. The knowledge of Riemann integration is helpful but not strictly necessary. Are you going to take a regular course or are you going just to read some book in measure theory by yourself?


Zan has created this rule for generating sequences of whole numbers:
\begin{tabular}{ c }
   If a number is 15 or less, triple the number.\\
   If a number is more than 15, subtract 13 from it.\\
\end{tabular}
Therefore, if Zan starts with 10, she gets the sequence $ 10, \sim 30, \sim 17, \sim 4, \sim 12, \ldots$ If the
first number in Zan's sequence is 34, what is the $ 8^{\text{th}}$ number in the sequence?

We just continue the pattern.

$ 34 \minus{} 13 \equal{} 21$

$ 21 \minus{} 13 \equal{} 8$

$ 8 \times 3 \equal{} 24$

$ 24 \minus{} 13 \equal{} 11$

$ 11 \times 3 \equal{} 33$

$ 33 \minus{} 13 \equal{} 20$

$ 20 \minus{} 13 \equal{} \boxed{7}$


Given 3 real numbers $a,b,c$, show that $a^{a}b^{b}c^{c}\ge a^{b}b^{c}c^{a}$.

i'm assuming they're positive, otherwise we might end up with $(-\pi)^{-\pi}$ which isn't well-defined.
[hide]
The function $\ln x$ is strictly increasing, so the inequality is equivalent to
\[a\ln a+b\ln b+c\ln c\geq b\ln a+c\ln b+a\ln c\]
WLOG let $\max (a,b,c)=a$ since the inequality is cyclic. If $a\geq b\geq c$, then the inequality is proven by Rearrangement. Otherwise, $a\geq c\geq b$, and we can rewrite the inequality as
\[a\ln a+c\ln c+b\ln b\geq b\ln a+a\ln c+c\ln b\]
which is still Rearrangement.
[/hide]


Hehe...would someone who happens to have their packet with them please LaTeX it up?

Please do not distribute files with the problems on this site.

Hmm...would you mind someone putting them in a post?

I think I'll defer to the AMC Director for policy on this issue.

Just ask your math director or someone for a copy of it. The solutions page also has the problems listed.

i took the test but i lost my booklet!!!
Theres gotta be somewhere to find the questions, anyone know where?

I [i]might[/i] have typed up the packet  :D ...

You could always look at the transcript of tonight's math jam.

I can't read my pamphlet b/c of all the writing on it... haha... o well time to bring out the (big) (wicked good) erasers...


The sum of three consecutive integers is 27. What is the product of the integers?

9*3=27.  Thus, the three integers are 8,9, and 10.  Multipliying yields 72*10=720.

$ x\plus{}x\plus{}1\plus{}x\plus{}2\equal{}3x\plus{}3\equal{}27 \implies x\equal{}8$.

Therefore, $ 8 \cdot 9 \cdot 10\equal{}720$.


Hello!

Determine all functions $f:{\Bbb Q}_+^*\to {\Bbb R}$ such that $f(1)=1$ and
\[
f\left(a_1{1\over 1}+\cdots+a_n{1\over n}\right)=a_1f\left({1\over 1}\right)+\cdots+a_nf\left({1\over n}\right)\]
for all $n\in {\Bbb N}^*$ and for all $a_1,\cdots,a_n\in \{0,1\}$.

Just an observation: I've seen a problem asking to prove that any positive rational can be written as the sum of distinct positive rationals with numerator $1$, so $1,\frac 12,\frac 13,\ldots$ form a system of generators of $Q_{+}^*$ over $\{0,1\}$ (see [url=http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fractions/egyptian.html]Egyptian fractions[/url]). This means that It's enough to know $f(1),f(\frac 12),f(\frac 13),\ldots$ etc.. I don't know how we can find all of these though (still thinking):).

I'll use that result without a proof (the fact that all positive rationals can be written as the sum of finitely many positive rationals with numerator $1$). Call the decomposition an Egyptian decomposition.

Assume we have shown that $f\left(\frac xk\right)=\frac{f(x)}k,\ \forall k\in\overline{2,n-1}$. We can write $f\left(\frac x{n-1}\right)=f\left(\frac xn\right)+f\left(\frac x{(n-1)n}\right)$ (I'll explain below why this is possible, given the conditions of the problem). Since $f\left(\frac x{(n-1)n}\right)=\frac{f\left(\frac {f(x)}n\right)}{n-1}$, we get the desired result: $f\left(\frac xn\right)=\frac{f(x)}n$.

Now to explain why it's possible to write $f\left(\frac x{n-1}\right)=f\left(\frac xn\right)+f\left(\frac x{(n-1)n}\right)$. Assume we have an Egyptian decomposition of $x$: $x=\sum_{i=1}^t \frac 1{a_i}$, with distinct $a_i$. Then $\frac x{n-1}=\sum_{i=1}^t \frac 1{(n-1)a_i}$, with distinct $(n-1)a_i$. The problem is that when we write $\frac 1{(n-1)a_i}=\frac 1{na_i}+\frac 1{(n-1)na_i}$, some of the fractions might turn out to be identical.

However, from the result stated in the first paragraph, it's not hard to show that every positive rational actually has an Egyptian decomposition in which the largest summand is as small as we want it to be: for fixed $x\in\mathbb Q_{+}^*,\ u\in \mathbb N^*$ (where $u$ is considered to be as large as we need it to be in order for $\frac 1u<x$ to take place), there is a unique $k$ s.t. $\frac 1u+\frac 1{u+1}+\ldots+\frac 1k<x\le \frac 1u+\frac 1{u+1}+\ldots+\frac 1k+\frac 1{k+1}$, and an Egyptian decomposition of $x-\left(\frac 1u+\frac 1{u+1}+\ldots+\frac 1k\right)$ has all its summands $<\frac 1k$.

We can apply this last result to our fractions $\frac 1{a_i}$ to decompose them in summands small enough so that they don't interfere with each other after we write every summand $s$ divided by $n-1$ as $\frac sn+\frac s{(n-1)n}$.

Thanks! I only wanted to say that you amaze me: you understood what I wanted to say? I thought it was really unclear. :)


group theory
abstract algebra
IMC
college contests

Does there exist a finite group $ G$ with a normal subgroup $ H$ such that $ |\text{Aut } H| > |\text{Aut } G|$? Disprove or provide an example. Here the notation $ |\text{Aut } X|$ for some group $ X$ denotes the number of isomorphisms from $ X$ to itself.

Let $ G \equal{} (Z_p \times Z_p) \rtimes Z_p$ be the Heisenberg group over $ \mathbb{Z} / p\mathbb{Z}$ and let $ H \unlhd G$ be a normal subgroup such that $ H \cong Z_p \times Z_p$. Note that $ \text{Aut}(G) \cong \text{SL}_2(\mathbb{F}_p)$ and that $ |\text{SL}_2(\mathbb{F}_p)| \equal{} (p\minus{}1)p(p\plus{}1)$. Note also that $ \text{Aut}(H) \cong \text{GL}_2(\mathbb{F}_p)$ and that $ |\text{GL}_2(\mathbb{F}_p)| \equal{} (p\minus{}1)^2p(p\plus{}1)$, and we are done.

What do you mean under $ G \equal{} (Z_p \times Z_p) \rtimes Z_p$???
If $ G$ is a non-abelian group of order $ p^3$, then its center contains $ p$ elements. Hence $ \text{Aut}(G)$ contains the group $ \text{Inn}(G)$ of order $ p^2$, and so $ |\text{Aut}(G)| \ne (p \minus{} 1)p(p \plus{} 1)$ as it is divisible by $ p^2$. (Here $ \text{Inn}(G)$ is a group of inner automorphisms, provided by formula $ \varphi(g) \equal{} h^{ \minus{} 1}gh$, where $ h \in G$.)

I found an example.
Let $ G \equal{} \{a^kb^lc^md^n: \ 0 \le k,l,m,n < 3\}$, where $ a^3 \equal{} b^3 \equal{} c^3 \equal{} d^3 \equal{} e$ and $ ba \equal{} abc,\ ca \equal{} acd,\ da \equal{} ad,\ cb \equal{} bc,\ db \equal{} bd,\ dc \equal{} cd$. It is easy to check, that $ G$ is really a group.
$ G$ can be generated by $ a,b$, so each automorphism of this group is defined by its result on $ a,b$. Hence $ |\text{Aut}(G)| \le |G|^2 \equal{} 81^2$. The subgroup $ H \equal{} \{b^lc^md^n: \ 0 \le l,m,n < 3\}$ is isomorphic to $ \mathbb Z_3^3$, and so $ |\text{Aut}(H)| \equal{} 26 \cdot 24 \cdot 18 > 81^2 \ge |\text{Aut}(G)|$. Obviously, $ H$ is normal in $ G$.

My solution was different. But I made a mistake in the exam and got 11 from this question.
Let $ G$ be the group generated by $ x_1,..., x_{2n}, y$ under the following relations:
$ x_ix_j\equal{}x_j x_i$
$ x_i^2\equal{}y^2\equal{}1$
$ y x_i\equal{}x_{i\plus{}n}y$ (mod 2n)
You can easily verify that the elements of $ G$ are $ x_{i_1}...x_{i_k}$ and $ y x_{i_1}...x_{i_k}$ and the product is associative. Let $ H$ be the subgroup generated by $ x_1,...,x_{2n}$, which is $ \mathbb{Z}_2^n$ and $ |Aut(H)|\equal{}(2^{2n}\minus{}1)(2^{2n}\minus{}2)(2^{2n}\minus{}4)...(2^{2n}\minus{}2^{2n\minus{}1})$. You can easily check that the only subgroup of index $ 2$ is $ H$. So if $ f \in Aut(G)$ then $ f(H)\equal{}H$. $ y$ and $ x_1, ..., x_n$ generate $ G$, so by choosing their image under $ f$ we see that $ |Aut(G)|\leq 2^{2n}(2^{2n}\minus{}1)...(2^{2n}\minus{}2^n)$ and for large $ n$ this example satisfies what we want.


[quote="can_hang2007"]Hi all,
We have just published a new mathematics magazine. This will be a good place for us to share our love with maths. :) We have published the first volume in Vietnamese, but from the second volume till then, we will publish the magazine in boths languages English and Vietnamese. We hope you will enjoy it! :) This is Problems of volume 01/2009.

You can send the solution to us via mathvn2008@gmail.com or journal@mathvn.org. Enjoy! :)[/quote]
Are there any solution of the problem?Thanks!


How would I go about solving an equation such as:

x^x = e

Taking the natural log of both sides gives:

x * ln(x) = 1

Where do I go from there?

We can use the lambert W function.  This function is the inverse of $f(W) = We^W$ and is implemented as $W(x)$.  It is also the solution of $W(x)* e^{W(x)} = x$.

$x^x = e \Rightarrow x = e^{W(ln(e))} = e^{W(1)}$

( see http://mathworld.wolfram.com/LambertW-Function.html for more )

It is also well known that the value of $W(1)$ is the omega constant, roughly 0.56714.  (see http://mathworld.wolfram.com/OmegaConstant.html ).

So $x = e^{\Omega}$ and we are done.


I was wondering if there was any way to purchase the AoPS books online as a pdf format?

sorry if this is the wrong forum...

No, the AoPS books are not available in electronic format.


Does anyone know where the MAML Level II competition will be held? My coach never told me where I had to go.

That's unfortunate because it was today.


Hello everyone, just need some help here...

"The area of the top of a rectangular box is 324 in.2, the area of the front of the box is 135 in.2, and the area of the end is 60 in.2. What is the volume of the box?"

P.S.- I don't care what method you use but please show your work cuz I'm really missing something when I do this on my own. Thanks in advance. Peace.

[hide="Solution"]
\[xyz = \sqrt{ (xy)(yz)(xz) }= \sqrt{ 324 \cdot 135 \cdot 60 }= \boxed{ 1620 \text{ in}^{3}}\] [/hide]

[hide="Solution"]

If we let the dimensions of the box be $x$, $y$, and $z$, we have:
$x\times y=324$
$y\times z=135$
$z\times x=60$

Then the volume of the box is
$xyz=\sqrt{x^{2}y^{2}z^{2}}=\sqrt{(xy)(yz)(zx)}=\sqrt{324\times 135\times 60}=1620$.[/hide]


Find $ a$ and $ b$ such that the following system of equations has infinite real solutions.

$ x\plus{}y\plus{}z\equal{}2$
$ 2x\plus{}4y\plus{}3z\equal{}5$
$ ax\plus{}by\plus{}2z\equal{}9$

for this system to have an infinite number of solutions, the last equation must tell us nothing new.  That is,

$ (2x\plus{}4y\plus{}3z)j \plus{} (x\plus{}y\plus{}z)k \equal{} 5j\plus{}2k\equal{}9$ where $ 3j\plus{}k\equal{}2$ (the coefficient of $ z$).  Solving this system:

$ k\equal{}2\minus{}3j$
$ 5j \plus{} 2(2\minus{}3j)\equal{}9$
$ \minus{}j\plus{}4\equal{}9$
$ j\equal{}\minus{}5$
$ k \equal{} 2\minus{}3*\minus{}5 \equal{} 17$

Hence $ a\equal{}2j\plus{}k \equal{} \minus{}10\plus{}17 \equal{} 7$ and $ b\equal{}4j\plus{}k \equal{} \minus{}20\plus{}17 \equal{} \minus{}3$.


The height $ AA'$ of the acute angled triangle $ ABC$ inscribed in a circle $ C(0,r)$ intersects the circle in $ D$. The perpendiculars $ DM$ and $ DN$ on $ AB$ and $ AC$ intersect the circle in $ E$ and $ F$. Prove that:
$ 1)$ $ \triangle AEF \equiv \triangle ABC.$
$ 2)$ If $ P$ is the point where $ OA$ intersects the second time the circle, then prove that $ MN \perp AP.$
$ 3)$ $ EC \parallel BF.$

all the questions are obvious because MA'N is the simson line of triangle ABC


floor function
logarithms
Irrational numbers

You are given three lists A, B, and C.  List A contains the numbers of the form $10^{k}$ in base 10, with $k$ any integer greater than or equal to 1.  Lists B and C contain the same numbers translated into base 2 and 5 respectively:  \[\begin{array}{lll}A & B & C \\ 10 & 1010 & 20 \\ 100 & 1100100 & 400 \\ 1000 & 1111101000 & 13000 \\ \vdots & \vdots & \vdots \end{array}.\] Prove that for every integer $n > 1$, there is exactly one number in exactly one of the lists B or C that has exactly $n$ digits.

The number of digits of $n$ in base $b$ is given by $\left\lfloor \log_{b}(n) \right\rfloor +1 = \left\lfloor \frac{\log_{10}(n)}{\log_{10}(b)} \right\rfloor+1$.
Especially our numbers $10^k$ have $\left\lfloor \frac{k}{\log_{10}(2)} \right\rfloor$ and $\left\lfloor \frac{k}{\log_{10}(5)} \right\rfloor$ digits.

Now simply use Beatty's theorem ( http://www.problem-solving.be/pen/viewtopic.php?t=235 ) for $\alpha = \frac{1}{\log_{10}(2)}$ and $\beta = \frac{1}{\log_{10}(5)}$ (the condition is obviously fulfilled by $\log_{10}(2)+\log_{10}(5)=\log_{10}(10)=1$).


Find the number of ordered pairs $(x,y,z,t)$ of integers $\geq 0$ and $\leq 36$ such that $x^{2}+y^{2}=z^{3}+t^{3}(\bmod 37)$.
I know the answer is $53317=11\times37\times131$, but how?

Let us find the number of ordered pairs $ (x,y)$ satisfying  $ x^{2}\plus{}y^{2}\equiv a(\bmod 37)$, where $ a$ is an integer.
If $ a\equiv 0(\bmod37)$, for the congruence $ x^{2}\plus{}y^{2}\equiv 0(\bmod 37)$, we get $ y^{2} \equiv (6x)^{2}(\bmod 37)$ and then $ y \equiv \pm 6x (\bmod 37)$. There are $ 2\cdot36\plus{}1\equal{}73$ pairs of $ (x,y)$.
Now suppose $ a \not\equiv 0(\bmod 37)$. $ x^{2}\plus{}y^{2}\equiv x^{2} \minus{} 36y^{2} \equiv (x\minus{}6y)(x\plus{}6y) \equiv a(\bmod 37)$. For each $ a$ and any $(x-6y)$ there is only one $(x+6y)$. $(x-6y)$ can take $36$ different values (except $0$). So there are $ 36$ pairs of $ (x,y)$ for each $a$.

On the other hand, for $ z^{3}\plus{}t^{3}\equiv 0(\bmod 37)$, $ t^{3} \equiv \minus{}z^{3} (\bmod 37)$, thus $ t \equiv \minus{}z$ or $ t \equiv11z$ or $ t \equiv\minus{}10z (\bmod 37)$. For $ z^{3}\plus{}t^{3}\equiv 0(\bmod 37)$, there are $ 36 \cdot 3 \plus{} 1 \equal{} 109$ pairs of $ (z,t)$.

For $ x^{2}\plus{}y^{2}\equiv z^{3}\plus{}t^{3}\equiv 0 (\bmod 37)$, there are $ 109 \cdot 73$ pairs of $ (x,y,z,t)$.
For $ x^{2}\plus{}y^{2}\equiv z^{3}\plus{}t^{3}\not\equiv 0 (\bmod 37)$, there are $ (37^{2}\minus{}109) \cdot 36$ pairs of $ (x,y,z,t)$.
Sum of the two numbers is $ 53317$.


[quote="vipCD_A1"]Let $ a,b,c > 0; a + b + c = 3; k > 0$. Prove that:
$ \sum{\frac {|a^2b - kc|}{\sqrt {(k + 1)c + ab}}\leq{\frac {3|1 - ka^2b^2c^2|}{\sqrt {kabc + 2}}}}$
goodluck!
open inequalities
You will make it, it's true![/quote]
everyone solve it's
very hard, ok
I post also you not solve, i sorrow

[quote="vipCD_A1"]Let $ a,b,c > 0; a + b + c = 3; k > 0$. Prove that:
$ \sum{\frac {|a^2b - kc|}{\sqrt {(k + 1)c + ab}}\leq{\frac {3|1 - ka^2b^2c^2|}{\sqrt {kabc + 2}}}}$
goodluck!
open inequalities
You will make it, it's true![/quote]
Are you sure it was proved? If you solved it, you would post it in Purpose Box.
I am currious how to solve it... because I don't think it is true :D

[quote="vipCD_A1"]Let $ a,b,c > 0; a + b + c = 3; k > 0$. Prove that:
$ \sum{\frac {|a^2b - kc|}{\sqrt {(k + 1)c + ab}}\leq{\frac {3|1 - ka^2b^2c^2|}{\sqrt {kabc + 2}}}}$
goodluck!
open inequalities
You will make it, it's true![/quote]
Put $ a=1/2, b=1/2, c=2, k=4$. Then $ {\frac {3|1 - ka^2b^2c^2|}{\sqrt {kabc + 2}}}=0$.

[quote="Zuko"][quote="vipCD_A1"]Let $ a,b,c > 0; a + b + c = 3; k > 0$. Prove that:
$ \sum{\frac {|a^2b - kc|}{\sqrt {(k + 1)c + ab}}\leq{\frac {3|1 - ka^2b^2c^2|}{\sqrt {kabc + 2}}}}$
goodluck!
open inequalities
You will make it, it's true![/quote]
Put $ a = 1/2, b = 1/2, c = 2, k = 4$. Then $ {\frac {3|1 - ka^2b^2c^2|}{\sqrt {kabc + 2}}} = 0$.[/quote]

Probably he means choose 'k' such that the inequality holds   :huh:


The $ 2007$ AMC $ 10$ will be scored by awarding $ 6$ points for each correct response, $ 0$ points for each incorrect response, and $ 1.5$ points for each problem left unanswered. After looking over the $ 25$ problems, Sarah has decided to attempt the first $ 22$ and leave only the last $ 3$ unanswered. How many of the first $ 22$ problems must she solve correctly in order to score at least $ 100$ points?

$ \textbf{(A)}\ 13\qquad
\textbf{(B)}\ 14\qquad
\textbf{(C)}\ 15\qquad
\textbf{(D)}\ 16\qquad
\textbf{(E)}\ 17$

[hide="Click for solution"]
We have:

$ 6x\plus{}4.5>100 \implies 6x>95.5 \implies x>15.91\overline{6} \implies x\equal{}16$, or $ \boxed{\textbf{(D)}}$.
[/hide]

[hide="Without introducing decimals"]$ \frac{3}{2}\cdot 3\plus{}6x\equal{}100\implies x\equal{}100\minus{}\frac{9}{2}\equal{}\frac{191}{12}$.
If $ a<\frac{191}{12}<b$, $ 12a<191<12b$. It's clear that $ a\equal{}15$ and $ b\equal{}16$, so we get that $ \lceil x\rceil\equal{}\boxed{\textbf{(D)}\ 16}$.[/hide]

Finding $ \frac{95.5}{6}$ wouldn't be much fun on a test that disallows calculators :lol:

But you know that $ \frac{96}{6}\equal{}16$...


function
induction
algebra unsolved
algebra

[b]Problem 2.[/b] Let $R^+$ be the set of the positive real numbers and $f: R^+\to R^+$ is functions, such that $f(x+y)-f(x-y)=4\sqrt{f(x)f(y)}$ for all $x>y>0.$
 a) Proove that $f(2x)=4f(x)$ for all $x\in R^+$.
 b) Find all such functions.

[i]Nikolai Nikolov, Oleg Mushkarov[/i]

[quote="silouan"]If I remember well it was posted again . And the main idea was :

First use limits and then
By induction $f(kx)=k^2f(x)$ where $k$ is natural.Then
$f(\frac mk)=\frac{m^2}{k^2}f(1)$ for all $\frac mk \in Q^+$(positive rational numbers)
Finally $f(x)=x^2f(1)$, because of $f$ is continous[/quote]
Were is your induction?
We have
(1) $f(z)>f(t)$ if $z>t$ (for proved $x=\frac{z+t}{2},y=\frac{z-t}{2}$).
Let $a_n=a_n(x)=\frac{f(nx)}{f(x)}$. We have
$a_{n+k}-a_{n-k}=4\sqrt{a_na_k},a_1=1.$ From
$a_3=1+4\sqrt{a_2},a_4=a_2+4\sqrt{a_2a_3},a_5=a_3+4\sqrt{a_4}=1+4\sqrt{a_2a_3}$ we calculate $a_2$
$1+4\sqrt{a_2}+4\sqrt{a_2+4\sqrt{1+4\sqrt{a_2}}}=1+4\sqrt{a_2+4a_2\sqrt{a_2}} \Longrightarrow a_2=4$.
Therefore by induction $a_k=k^2$. From (1) $f(x)=kx^2.$

Solution by Diogene

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=88573[/url]

Thank you Ramanajan , I had in my mind only the scetch of the solution and I couldn't find the link


In a 12mx12mx30m room, a vertical line runs up the middle of the two square walls. Andrew the ant is on one of these lines, 1 meter from the floor, and Bob the ant is on another of these lines, 1 meter from the ceiling. Ants cannot fly, they can only walk on the walls, floor, and ceiling of the room. What is the shortest distance Andrew must walk to get to Bob?

[hide]I looked at three possible ways the first ant could get to the second.

One was a vertical line along the surfaces. The distance traveled for that path is 42m.
The next was a diagonal line that went across one rectangular wall. By "expanding out" the box to make it one right triangle w/ sides 6+30+6=42 and 11-1=10, using pythagorean theorem the answer came out to be approx 43.17m.
The final one was a path that took the square walls, a rectangular wall, and the rectangular ceiling. Again, expanding out the walls, we obtain a right triangle with sides 6+30+1=37 and 11+6=17, using pythagorean theorem we get approx 40.72m.

So the answer I got was 40.72m. I'm not sure if there's another shorter path that I missed.[/hide]

[hide]If you draw it as a net, and you unfold it you would find that the distance was 40 (24-32-40 right triangle) Of course, we just did this in our geometry class and that's the only reason I know the answer.[/hide]

[hide]1 meter down, 30 meters across, 11 meters up, [u]Total=42 meters[/u][/hide]

The most efficient solution almost never involves perpendicular lines. Draw a net, and look at it again.

Whoops!  I was thinking through the folded box! (Me = :wacko: )
[hide] :approx: 40.72 m2[/hide]

All answers are close but have any two people come up with the same answer yet?


ok. so here's the problem:

A, B, C, D, and E are consecutive points on a line. If AB/BC=1/3, BC/CD=1/4, and CD/DE=1/2, what is AC/BE?

Since AC=AB+BC, and AB/BC =1/3, i thought you could assume AC=1+3=4 (i know the proportion doesn't say the exact value of the line segment (im not that stupid) but usually in these typse of problems, you don't have to find the value exactly. )
i did the same to find the length of BE and then i found the fraction. of course my answer was wrong. can someone help me point out my mistake???

[hide]
Let $ AB=1$
Then $ BC$ is three times larger so $ BC=3$,
$ CD$ is four times larger than $ BC$ so $ CD=12$,
$ DE$ is two times larger than $ CD$ so $ DE=24$.
$ AC=AB+BC=4$
$ BE=BC+CD+DE=39$
$ \frac{AC}{BE}= \frac{4}{39}$
[/hide]

that is indeed the answer....
oh... now i see what i did wrong.. i forgot to multiply my proportions by the scale factor... duh...stupid mistake! :mad:


Can someone please help me with this problem? Thanks!

[hide]The identity with respect to * is $ 0$ because $ a*0\equal{}a\plus{}0a\plus{}0\equal{}a$ for all $ a$. Thus, we need $ 4*x\equal{}0$ so that $ 4x\plus{}x\plus{}4\equal{}0$ or $ x\equal{}\minus{}4/5$.[/hide]

[quote="Temperal"]Thus, we need $ 4*x \equal{} 0$ so that $ 4x \plus{} x \plus{} 4 \equal{} 0$ or $ x \equal{} \minus{} 4/5$.[/quote]

Could you please explain how you got this?  Thanks!


Given a right regular prism with hexagonal bases. The lower base has vertices labeled $A$, $B$, $C$, $D$, $E$, and $F$. The upper base has vertices labeled $A'$, $B'$, $C'$, $D'$, $E'$, and $F'$. Vertex $A'$ is directly above vertex $A$, etc. The regular hexagonal bases have sides of length $6$ and the height of the prism is $12$. Find the length of the segment from $A$ to $C'$.

[hide="Answer"] First find the length of $AC$. The angles of the regular hexagon are $120$[b]º[/b] each (obtained by the formula for the angles of regular polygons where $n$ is the number of sides: $\frac{180(n-2)}{n}$). Draw a perpendicular from point $B$ to meet $AC$ in $G$. $\triangle{CBG}$ has angles of $90$, $60$ and $30$ degrees, and the ratio of the sides of this triangle is always $1: 2: \sqrt{3}$. So if $BC=6$, $CG=3\sqrt{3}$. Multiply by $2$ to give $AC=6\sqrt{3}$. $CC'=12$ and using Pythagoras, $AC'=\sqrt{252}=6\sqrt{7}$.[/hide]


function
superior algebra
superior algebra unsolved

Prove that the field $ \mathbb{Q}(x)$ of rational functions contains two subfields $ F$ and $ K$ such that $ [\mathbb{Q}(x) : F]<\infty$ and $ [\mathbb{Q}(x) : K]<\infty$ but $ [\mathbb{Q}(x) : (F\cap K)]\equal{}\infty$.

Perhaps $ F \equal{} \mathbb{Q}(X^2), K \equal{} \mathbb{Q}((X\plus{}1)^2)$.

I can see that $ [\mathbb{Q}(x): \mathbb{Q}(x^2)]\equal{}2$ and $ [\mathbb{Q}(x): \mathbb{Q}((x\plus{}1)^2)]\equal{}2$ but who can tell me what is $ \mathbb{Q}(x^2)\cap\mathbb{Q}((x\plus{}1)^2)$ ?

Their intersection is just $ \mathbb{Q}$. For if $ f$ were in their intersection, then $ f(\minus{}x)\equal{}f(x)$ and $ f(\minus{}2\minus{}x)\equal{}f(x)$ from which it follows that $ f(2\plus{}x)\equal{}f(x)$. So $ f(x)\minus{}f(0)$ has every even integer as a root, from which it follows that $ f(x)\minus{}f(0)$ is identically zero, making $ f$ a constant.


i dont know if this is true for everyone but i think they should offer partial credit on the aime... that way, it'd be a test of who really is good & maybe just makes a few careless mistakes vs. helping ppl who r not that good but just careful

How would you recommend doing this?

If they made the AIME proof style then it would be very time consuming to grade, and they would have to drastically reduce the number of participants to like USAMO numbers (500 or whatev), which is already proof based, so then you would still have to have an easily gradable qualifier, like the AIME, so...

No.

Besides, how would that be done? It's graded by a machine. Should they do it by how close your answer is? One mistake could land you 500 away (and on some problems I've seen, it will). And what about people who guess close to the correct answer?

[quote="starlove"]maybe just makes a few careless mistakes vs. helping people who are not that good but just careful[/quote]

I take offense at this statement. Bob is not that good but just careful. Bob will not make the USAMO. Bill is amazingly good and rather careless. Bill will still make the USAMO. The people in question are those who are mildly good and careless.

And this is a bad idea. And please don't use chattalk (ppl, r)

Yeah, I agree with jb05.  The only situation I can even see where you could possibly give partial credit is a situation like problem #5 on this year's AIME, when a bunch of people put 538.  But I think that just kills the test, when you think about it.  It's not supposed to be a forgiving test.

I put 439 on that since I computed $4+\frac{5}{9}\cdot963=4+5\cdot107=4+435=439$.

Wouldn't I also deserve partial credit? If so, how would I possibly get it? The machine can't possibly tell careless errors from guesses. But if it could, that would be cool.

Yeah, and then wouldn't 593 (a misbubbling) deserve partial credit?  And the list goes on, which makes it really impractical.

I think what he's saying is to completely revamp the AIME in some way so that people actually grade it, which would probably require it to include solutions as well.

I'm not sure how they would do this (at the current moment), especially considering financial restraints, although I do think that some other kind of intermediary to the USAMO would be a good ideal, whether or not it is feasible.

[quote="Twiz"]Not that I think it is necessary, but I guess it would be possible to have a careless mistake commitee that found common careless mistakes and submitted them to the partial credit committee (or something like that). Another way would be to give partial credit if a significant portion of test-takers had the same wrong answer.[/quote]
No, because that may mean that the large group of people all had a wrong method that may seem to work, but not work.
It also may mean that they all misread the problem, something that doesn't deserve partial credit.


Consider the sequence $ x_{1}\equal{}1$, and $ x_{n\plus{}1}\equal{} x_{n}\plus{}\frac{1}{2x_{n}}$ .Prove that:
             $ [25x_{6}25]\equal{}625$


I think this is a correct proof but it is rushed so could be wrong

---

Consider vectors $A = (x_1, x_2, \dots, x_{n-1}), B = (x_2, x_3, \dots, x_n)$.

Then we have to show $\frac{n-1}{n} (A \circ B) \le \frac{|A|^2 + |B|^2 + x_1^2 + x_n^2}{2}$

Let $\theta$ be the angle between A and B.  We have:

$\frac{n-1}{n}|A||B|\cos \theta \le \frac{n-1}{n}|A||B| \le \frac{n-1}{2n}(|A|^2 + |B|^2) \le \frac{|A|^2 + |B|^2 + x_1^2 + x_n^2}{2}$

It follows that the inequality holds for all n.

Well... it looks difficult. Is there any easier proofs?


[hide="1"][b]The owner of a bakery knows that the daily demand for a highly perishable cheesecake is as follows:
Number/day:           0      1        2       3       4        5
Relative/Frequency:  .05   .15     .25   .25    .20      .10

(A) Use simulation to find the demand for the cheesecake on 30 consecutive business days.

(B) Suppose that it costs the baker 5 dollars to produce a cheesecake, and that the unused cheesecakes must be discarded at the end of the business day. Suppose also that the selling price of a cheesecake is 13 dollars. Use simulation to estimate the number of cheesecakes that he should produce each day in order to maximize his profit.
[/b][/hide]

[hide="2"][b]Two wine tasters rate each wine they taste on a scale of 1-5. From data on their ratings of a large number of wines, we obtain the following probabilities for both tasters' ratings of a randomly chosen wine:

                                               Taster 2
Taster 1                   1           2          3          4        5
1                          .03          .02        .01       .00     .00
2                          .02          .08        .05       .02     .01
3                          .01          .05        .25       .05     .01
4                          .00          .02        .05       .20     .02
5                          .00          .01        .01       .02     .06

(a) How do I read this table????
(b) What is the probability that the tasters agree when rating a wine?
(c) What is the probability that taster 1 rates a wine higher than 3?
(d) What is the probability that taster 2 rates a wine higher than 3?[/b]
[/hide]

[quote="t0rajir0u"]Please attempt to do your own homework before asking questions about it.  Conceptual questions are perfectly okay; trying to skip out on your responsibilities isn't.[/quote]

I have tried, and this is why I'm asking on this forum. I have no idea how to do any of those problems. I did the first part to these problems, but then was stumped. Ihaveno idea how to read the 2nd problem's table and do not understand thefirst problem

Simulations are always the same: you use a random number table/generator to simulate the outcome of a random event by assigning certain numbers to certain events and repeating the experiment by drawing new numbers. The hard part is figuring out what the random variable is and how to assign numbers.

All of problem 2 is reading the table; telling you how to read the table defeats the purpose of the question.

Why did you bold everything? That only makes it a strain to read.

[quote="Mathalete"]I have tried, and this is why I'm asking on this forum. I have no idea how to do any of those problems. I did the first part to these problems, but then was stumped. Ihaveno idea how to read the 2nd problem's table and do not understand thefirst problem[/quote]
This forum can help with homework, but note the following.
1. We will not [b]do[/b] the homework. We will [b]help[/b].
2. You will get said help only if you give us all of your progress and note what is most confusing to you.

With respect to 2, if you find yourself saying "I don't get any of this" or "I can't even start", the topic you should start is one asking for explanations of the concepts that the problem is about, without even saying what the problem stumping you is. Once you have a good enough explanation, you should try yourself before getting more specific assistance.

Note that this forum is not about how to pass math class; it's about understanding math to the fullest possible extent. That is what you will gain from posting here, not just an easy A.


linear algebra
matrix
ratio
absolute value
linear algebra unsolved

Dear friends,

How does the definition of condition number of a matrix $ A$ which is defined as $ \mathcal{K}_A\equal{}\|A\|\|A^{\minus{}1}\|$ fit if $ A$ is rectangular instead of square. I mean what should we do for $ A^{\minus{}1}$? Learned members please clarify.

Bhupala

The definition of the condition number given above is for square non-singular matrices. Normally we use an alternate definition to extend this definition to singular and non-square matrices: condition number is the ratio of the largest singular value (in absolute value) to the smallest singular value (in absolute value).


The four midpoints of adjacent sides of a square are joined to form a smaller square. What is the ratio of the area of the smaller square to the area of the original square? Express your answer as a common fraction.

Call the length of the larger square x. The joined midpoints form four 45-45-90 triangle, so the ratio of the three sides is a-a-a(sqrt2). Therefore, a side of the smaller square is (x/2)(sqrt2) = x(sqrt2)/2. Thus, the ratio of the areas is:
[x(sqrt2)/2]^2/x^2 = (x^2/2)/x^2 = 1/2