Prove that for all positive reals $ a,b,c$ :
\[ \frac{a^2b\plus{}b^2c\plus{}c^2a}{a^3\plus{}b^3\plus{}c^3}\plus{}\frac{3(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca} \geq 4\]

We have the flowing identities:

$ \left(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca\right)\equal{}\frac{1}{2}\sum_{cyc}(a\minus{}b)^2$
$ \left(a^3\plus{}b^3\plus{}c^3\minus{}a^2b\minus{}b^2c\minus{}c^2a\right)\equal{}\frac{1}{3}\sum_{cyc}(2a\plus{}b)(a\minus{}b)^2$

Hence:

$ \text{LHS} \minus{} 4 \equal{} \frac {3(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)}{ab \plus{} bc \plus{} ca} \plus{} \frac {a^2b \plus{} b^2c \plus{} c^2a \minus{} a^3 \minus{} b^3 \minus{} c^3}{a^3 \plus{} b^3 \plus{} c^3}$
$ \equal{} \sum_{cyc}\frac {3(a \minus{} b)^2}{2(ab \plus{} bc \plus{} ca)} \minus{} \sum_{cyc}\frac {(a \minus{} b)^2(2a \plus{} b)}{3(a^3 \plus{} b^3 \plus{} c^3)}$
$ \equal{} \sum_{cyc}\frac {(a \minus{} b)^2(9a^3 \plus{} 9b^3 \plus{} 9c^3 \minus{} 4a^2c \minus{} 4a^2b \minus{} 4abc \minus{} a^2c \minus{} a^2b \minus{} abc)}{6(ab \plus{} bc \plus{} ca)(a^3 \plus{} b^3 \plus{} c^3)} \ge 0$

which is true !  :oops:

[quote="Cekaa"]Prove that for all positive reals $ a,b,c$ :
\[ \frac {a^2b \plus{} b^2c \plus{} c^2a}{a^3 \plus{} b^3 \plus{} c^3} \plus{} \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca} \geq 4\]
[/quote]
Even $ a^2b\plus{}b^2c\plus{}c^2a\geq\frac{abc(a\plus{}b\plus{}c)^2}{ab\plus{}ac\plus{}bc}$ helps. :wink:

in my solution i used $ (a^2b\plus{}b^2c\plus{}c^2a)(b\plus{}c\plus{}a) \ge (ab\plus{}bc\plus{}ca)^2$ and then p,q,r works


Let $ a,b,c\ge 0$ such that $ a\plus{}b\plus{}c\equal{}3$. Prove that:
\[ (a^2b\plus{}b^2c\plus{}c^2a)\plus{}2(ab^2\plus{}bc^2\plus{}ca^2)\plus{}3abc\le 12\]

I think it is very nice. 1-2-3 are coefficient  :wink:

:(. ( I'm newbie ) .
I use this ineq :
$ a,b,c\geq 0$ ;$ a \plus{} b \plus{} c \equal{} 3$
$ a^2b \plus{} b^2c \plus{} c^2a \plus{} abc \leq 4$ .
Make the similar ineq  and we have done .
Maybe this ineq can be stronger .
______________
Sorry for English :(


logarithms
inequalities proposed
inequalities

why not try estimating the no. of digits..?
it could simplify things..

Here's my solution in my recent video: [url]https://www.youtube.com/watch?v=tPYQdtey6SY[/url]

the 4 one is bigger.

definitely didn't just guess that and then look at the other solutions to check.


Let $ V$ be a vector space of dimension equal to $ n$. Prove that if $ \phi : V \rightarrow V$ is a linear transformation satisfying:
$ \phi^{n}\equal{}0\not\equal{}\phi^{n\minus{}1}$ then there exist such a base of $ V$ - $ (\alpha_{1},\cdots \alpha_{n})$ that:$ \phi(\alpha_1)\equal{}0$ and $ \phi(\alpha_i)\equal{}\alpha_{i\minus{}1}$ for $ i\equal{}2,3,\cdots n$.

Choose $ \alpha_1$ so that $ \phi^{n \minus{} 1}(\alpha_1)\neq 0$. Define the rest of the $ \alpha$ in the obvious way. Show that they're linearly independent.

This particular result is essential to all "canonical forms", and it has appeared before more than once.


let $ A(m,n)\equal{}\prod_{k\equal{}m}^{m\plus{}n\minus{}1}(1\minus{}x^k), B(n)\equal{}\prod_{k\equal{}1}^{n}(1\minus{}x^k)$,where $ m,n$ are positive integers.
 prove that :$ A(m,n)$ is divisible by $ B(n)$ for all positive integers $ m,n$.

$ B(n) \equal{} (\minus{}1)^n\prod_{k\equal{}1}^n (x^k \minus{} 1) \equal{} (\minus{}1)^n \prod_{k\equal{}1}^n \prod_{d | k} \Phi_d(x)$, while $ A(m,n) \equal{} (\minus{}1)^n \prod_{k\equal{}1}^{n} (x^{k\plus{}m\minus{}1} \minus{} 1) \equal{} (\minus{}1)^n \prod_{k\equal{}1}^n \prod_{d | k\plus{}m\minus{}1} \Phi_d(x)$. But $ n! \mid m(m\plus{}1)\cdots (m\plus{}n\minus{}1)$.


algebra
polynomial
greatest common divisor
Rational Root Theorem
number theory
prime numbers

Even my math tutor couldn't figure THIS out!

 Factor each trinomial~~

4. $ 3y^{2}\plus{}7y\plus{}2$

6. $ 2x^{2}\plus{}x\minus{}3$

8. $ 2n^{2}\minus{}11n\plus{}7$

...

 ...Help?

[quote="Verridith"]Even my math tutor couldn't figure THIS out!

 Factor each trinomial~~

4. $ 3y^{2} \plus{} 7y \plus{} 2$

6. $ 2x^{2} \plus{} x \minus{} 3$

8. $ 2n^{2} \minus{} 11n \plus{} 7$

 ...Help?[/quote]

The "Rational Root Theorem" helps here

4. $ 3y^{2} \plus{} 7y \plus{} 2 \equal{} (y\plus{}2)(3y\plus{}1)$

6. $ 2x^{2} \plus{} x \minus{} 3 \equal{} (x\minus{}1)(2x\plus{}3)$ (Note the sum of the coefficients)

8. $ 2n^{2} \minus{} 11n \plus{} 7$ Doesn't factorise as far as I can tell

[quote="Verridith"]Thank you! Is the last one prime?[/quote]

What do you mean "prime". You could say it has no factors to a lower degree in $ \mathbb{Q}$, but if you mean always outputs prime numbers, then no (try $ n\equal{}7$)

Use this problem as an example:

$ 2x^2\plus{}3x\plus{}1$

First multiply the coefficient of $ 2x^2$ by the last term, $ 1$, which equals $ 2$.

Then find two numbers whose product equals $ 2$ and sum equals the coefficient of the middle term,$ 3x$.

These two numbers are $ 1$ and $ 2$ because $ 2*1\equal{}2$ and $ 2\plus{}1\equal{}3$.

Now put an $ x$ next to both of these numbers: $ 2x$ and $ 1x$ or just $ x$.

Replace $ 3x$ with these two terms: $ 2x^2\plus{}2x\plus{}x\plus{}1$

Now factor it like a regular polynomial: $ 2x(x\plus{}1)\plus{}1(x\plus{}1)$ which equals $ (2x\plus{}1)(x\plus{}1)$

Well, in my book, it says that some problems are 'prime'. I think that it is just saying that the problem cannot be factored.

The correct term is [url=http://mathworld.wolfram.com/IrreduciblePolynomial.html]irreducible[/url] (over the integers).  There is a difference between prime and irreducible, but it's too subtle to go into here :)

A useful lemma:  the trinomial $ ax^2 \plus{} bx \plus{} c$ is factorable over the integers [i]if and only if[/i] the trinomial $ x^2 \plus{} bx \plus{} ac$ is.

t0rajir0u, I know you mentioned that the difference between prime polynomials and irreducible polynomials is very subtle, but I would like understand this topic better. Could you give some more insight.

Prime polynomials and irreducible polynomials are, in fact, the same thing (although we don't use the term "prime" because we also have a notion of [url=http://mathworld.wolfram.com/PrimitivePolynomial.html]primitive polynomial[/url], which has a different focus); the terms [url=http://mathworld.wolfram.com/PrimeElement.html]prime[/url] and [url=http://mathworld.wolfram.com/IrreducibleElement.html]irreducible[/url], on the other hand, don't mean the same thing in general.

[quote="Verridith"]Even my math tutor couldn't figure THIS out!

 Factor each trinomial~~

4. $ 3y^{2} \plus{} 7y \plus{} 2$

6. $ 2x^{2} \plus{} x \minus{} 3$

8. $ 2n^{2} \minus{} 11n \plus{} 7$

...

 ...Help?[/quote]

In the trinomial in the form of $ ax^2 \plus{} bx \plus{} c$ is only factorable (with integers, that is) if $ b^2 \minus{} 4ac$ is a perfect square. You can see that in the first 2, you plug it in and you get 25, or $ 5^2$. On the other hand, for the last problem, plug it in and you get 65. Here's how I learned factoring:

Find a pair of numbers n and m such that nm = ac and n + m = b. Rewrite the trinomial as $ (ax^2 \plus{}nx) \plus{} (mx \plus{} c)$. Factor out the GFC out of the 2 parantheses, so you get 2 terms each with a coefficient. Besides the coefficient , the 2 should be the same. Then just factor using "backwords distributive property."

For example, for the first problem, you would notice that 6 + 1 = 7 and 6 * 1 = 3 * 2. Rewrite it as $ (3y^2 \plus{} 6y) \plus{} (1y \plus{} 2)$. Then pull out the GCFs and you get $ 3y(y \plus{} 2) \plus{} 1(y \plus{} 2)$. Now, obviously you have $ \boxed{(3y \plus{} 1)(y \plus{} 2)}$.

Nerd_of_the_Ages, can you explain that a little more in detail? How did you get $ 3y(y\plus{}2)\plus{}1(y\plus{}2)$? I tried to do this in my head, and I figured that it would start out like $ 3y(y\plus{}3)$... How do you get two???

 Sorry for asking so many questions... XP

 You wrote~~

"For example, for the first problem, you would notice that 6 + 1 = 7 and 6 * 1 = 3 * 2. Rewrite it as (3y^2 + 6y) + (1y + 2). Then pull out the GCFs and you get 3y(y + 2) + 1(y + 2). Now, obviously you have \boxed{(3y + 1)(y + 2)}."

[quote="Verridith"]How did you get $ 3y(y \plus{} 2) \plus{} 1(y \plus{} 2)$?[/quote]

This is just reverse distributive property. If you note that $ (3y\plus{}1)(y\plus{}2)$ can be distributed $ 3y(y) \plus{} 3y(2)$ and similarily, $ 1(y)\plus{}1(2y)$.

Thank you for the information. I'm pretty sure I can do most of this now...

 I'll post back on this board topic if I have more questions!

Well, my tutor was showing me a few ways he found to do this, and now it is really easy. Thank you all for the help! Much appreciated!

[quote="Verridith"]Even my math tutor couldn't figure THIS out!

 Factor each trinomial~~

4. $ 3y^{2} \plus{} 7y \plus{} 2$

6. $ 2x^{2} \plus{} x \minus{} 3$

8. $ 2n^{2} \minus{} 11n \plus{} 7$

...

 ...Help?[/quote]

#8
$ (n \minus{} \frac {11 \plus{} \sqrt {44}}{4})(n \minus{} \frac {11 \minus{} \sqrt {44}}{4})$


I was wondering if anyone could finish where grobber left off for P1:  :)

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=7%5Ex+-+3%5Ey+%3D+4&t=1220

I finally got it, seems like 13 is the magic number....hehe... :D


geometry
trapezoid
rectangle
probability
topology
Ramsey Theory
combinatorics unsolved

[quote="chessfreak"]If all points of a circle are colored by one of n given colors (n >= 2),
then there exists a trapezoid (possibly a rectangle) with its vertices
colored by the same color.[/quote]

Assume that the statement doesn't hold.
We use the fact that N is finite.
That means: there exist a color, say black color, which covers infinitely many points
(because circle has infinitely many points and N is finite).
Now we take every three black points and "mark" all the points that make a trapezoid with some of the black points. Since none of the marked points is black (we assumed that there's no such trapezoid), there is infinitely many marked points. So we can do the same thing with marked points, and we can ignore all black points because there are infinitely many points left. Now the number of colours is decreased by 1 (N := N+1) and we can repeat the proces until we get the contradiction.

I don't think there's any sort of contradiction there.

We're not proving "if we have infinitely many points and we color them with $ N$ colors, there will be a monochromatic trapezoid", as such a statement wouldn't be true.  We're proving "if we have all the points on a circle and we color them with $ N$ colors, there will be a monochromatic trapezoid".

It's vital that the points involved take up an entire circle, and the marked points you color with $ N\minus{}1$ colors don't have this property.  In particular, when you take a set of 3 marked points, there's no reason to believe that the points which make a trapezoid with those points are also marked points and are being colored by your $ N\minus{}1$ colors.

It's equivalent to show there are four points A, B, C, D of the same color such that the distance from A to B equals the distance from C to D.  However, we can't simply use the fact that one color appears infinitely many times: I believe it's possible to have an infinite set of reals such that no two sums of numbers from the set are equal.  So you have to somehow use the fact that every point is colored.

Er ... chessfreak, have you already sent the solutions to CRP? Because if you haven't, you are not supposed to ask for help from an outside source.

If you haven't, however, and need a hint, it has an elegant solution very similar to Prashant's probability one at IMOTC 2007. However, I believe these approaches might work equally well.

rik_sengupta, I am not an IMOTC member. So I am just curious about the solutions to these problems.

Anyways, can we approach this problem in terms of limits?

This is what a friend suggested:

"This seems like a problem in analysis or topology, dealing with infinite
sets. The problem hypothesis would be satisfied if 2 parallel lines can be
constructed with all endpoints of the same color.

One observation: If a color exists as a continuous segment then the
hypothesis is always true.(In that case, you can always draw parallel
lines, with endpoints on the segment).

Therefore, for the problem hypothesis to fail, we must assume that all
colors are 'interleaved' with no continuous segment (or interval) of any
single color. But what do we mean by interleaved? With infinite sets, there
is no 'next point'. Between any 2 distinct points, there are infinitely
many points. So, in that case, we may find a particular color arbitrarily
close to any position.

We can argue the problem hypothesis can be satisfied in the sense of
limits:

Suppose we have chosen any 3 points (of the same color). There exists a
fourth point (of the same color)arbitrarily close to any desired position.
This allows construction of the fourth side, parallel to another side,
within an arbitrarily small error. In other words, smaller than any pre
assigned value, which is the definition of a limit."

Is this argument true/clear?

And rik_sengupta I will be grateful if you post the probability solution you are talking about.

THanK YoU.

Or anyone else, really, it does not matter. please someone, try to find a really nice solution to the problem?

THanK YoU

[quote="chessfreak"]rik_sengupta, I am not an IMOTC member. So I am just curious about the solutions to these problems.

[/quote]
it [b]doesn't[/b] matter whather u r an IMOTC member or not but because of u thos members who visit AoPS/ML get undue advantage over those who don't.

[color=red][size=150]SO PLEASE STOP POSTING THE POSTAL PROBLEMS[/size][/color]

oh ok, ok. i didn't realize that this forum could be an advantage for some members. of course i wouldn't want that. sorry, didn't mean to offend anyone -

and i heard this problem MONTHS ago. isn't the last date over yet?

So can anybody post a complete solution?

This problem is from Belarus Mathematical Olympiad 2007.

Let $ n,r$ be positive integers.
Van der Waerden's theorem states that there exists a number $ M(n,r)$ such that for any $ n$-colouring of the integers $ 1,2,\cdots, M(n,r)$ then there exist a arithmetical monochromatic sequence of length $ r$.
The smallest $ M(n,r)$ with this property is the Van der Waerden number $ W(n,r)$.

Now, inscribe a regular $ W(n,4)$-gon in the circle. Let $ A_1,A_2,\cdots,A_{W(n,4)}$ be its vertices. The colouring of the $ A_i$'s induces a colouring of the integers $ 1,2,\cdots, W(n,4)$. Thus, there are four of these vertices, say $ A_i,A_{i\plus{}d},A_{i\plus{}2d},A_{i\plus{}3d}$ which have the same colour, and they are the vertices of a trapezoid.

Pierre.


Happytown in due west of Sadtown, and Sam is due west of Happytown. At 9 am, Fred leaves Sadtown for Sam. At 1 pm, Leah leaves Happytown for Sam. If Leah travels at an average speed 20 mph faster than Fred and they reach Sam at 4 pm, how fast is each traveling.

[hide] Let $ r_{f}$ be the rate of Fred, $ t_{f}$ be the time for Fred, $ r_{l}$ be the rate for Leah, and $ t_{l}$ be the time for Leah. Since they arrive at the same place at the same time, we have

$ r_{f}(t_{f}) \equal{} r_{l}(t_{l})$.

$ \implies 7(r_{f}) \equal{} 3(r_{f} \plus{} 20) \implies r_{f} \equal{} \boxed{15}$.

This means that $ r_{l} \equal{} 15 \plus{} 20 \equal{} \boxed{35}$.[/hide]


Let $ k$ and $ m$ be positive integers such that $ \displaystyle\frac12\left(\sqrt{k\plus{}4\sqrt{m}}\minus{}\sqrt{k}\right)$ is an integer.

(a) Prove that $ \sqrt{k}$ is rational.

(b) Prove that $ \sqrt{k}$ is a positive integer.

$ \displaystyle\frac12\left(\sqrt{k\plus{}4\sqrt{m}}\minus{}\sqrt{k}\right)$
so $ \sqrt{k \plus{} 4\sqrt{m}} \equal{} 2p \plus{} \sqrt{k}$
squaring and subtracting k , you get
$ \sqrt{m} \equal{} p^2 \plus{} p\sqrt{k}$
and then squaring
$ m \equal{} p^4 \plus{} p^2k \plus{} 2p^3\sqrt{k}$
hence $ \sqrt{k}$ is a rational , and since $ k$ is an integer , $ \sqrt{k}$ is also an integer

Here is a more general problem: show that if $ a,b$ are distinct positive integers such that $ \sqrt{a}\pm\sqrt{b}$ is an integer, then $ a$ and $ b$ are both perfect squares. :wink:

[quote="nayel"]Here is a more general problem: show that if $ a,b$ are distinct positive integers such that $ \sqrt {a}\pm\sqrt {b}$ is an integer, then $ a$ and $ b$ are both perfect squares. :wink:[/quote]

$ \sqrt {a}\plus{}\sqrt {b}\equal{}m$
$ \sqrt {a}\minus{}\sqrt {b}\equal{}n$ then
$ 2 \sqrt {a}\equal{}m\minus{}n$ then $ a$ is a perfect square.

similarly for $ b$

Actually by $ \pm$ I meant two different problems, i.e.

1. If $ \sqrt a \plus{} \sqrt b$ is an integer then $ a,b$ are perfect squares.

2. If $ \sqrt a \minus{} \sqrt b$ is an integer then $ a,b$ are perfect squares.

It think this works

[b]Problem 1[/b]

Proceed in a similar way for problem two.

$ (\sqrt{a}\plus{}\sqrt{b}) (\sqrt{a}\minus{}\sqrt{b} ) \equal{} a\minus{}b \in \mathbb{Z} \Rightarrow \sqrt{a}\minus{}\sqrt{b} \in \mathbb{Q}$

Summing as before

$ \sqrt{a}\plus{}\sqrt{b} \plus{}\sqrt{a}\minus{}\sqrt{b} \equal{} 2\sqrt{a} \in \mathbb{Q} \Rightarrow a \in \mathbb{Z}$ and similarly for $ b$[/b]

I will prove that $ \sqrt {a} \plus{} \sqrt {b} \Rightarrow \sqrt {b} \in \mathbb{Z}$.

First, note a lemma: if $ \sqrt {r} \in \mathbb{Q}$, then $ \sqrt {r} \in \mathbb{Z}$ (i.e. $ r$ is a perfect square.) [u]Pf.[/u] Assume otherwise and let $ \sqrt {r} \equal{} \frac {p}{q}$, then $ q^2 r \equal{} p^2$. But $ (p,q) \equal{} 1$ so $ (p^2,q^2) \equal{} 1$. It follows that $ q \equal{} 1 \Rightarrow \sqrt {r} \in \mathbb{Z}$, contradiction.

Next, $ (\sqrt {a} \plus{} \sqrt {b})(\sqrt {a} \minus{} \sqrt {b}) \equal{} a \minus{} b \in \mathbb{Z}$. So $ \sqrt {a} \minus{} \sqrt {b} \in \mathbb{Q} \Rightarrow (\sqrt {a} \plus{} \sqrt {b}) \plus{} (\sqrt {a} \minus{} \sqrt {b}) \equal{} 2\sqrt {b} \in \mathbb{Q}$. So $ \sqrt {b} \in \mathbb{Q}$. It then follows from the lemma that $ \sqrt {b} \in \mathbb{Z}$. We are done.


Number Three: The circumference of a particular circle is 18cm. In sq. centimeters, what is the area of the circle? Express your answer as a common fraction in terms of $\pi$

Number Four: The hypotenuse EDIT: *and leg* of a particular right triangle are $\sqrt{97}$ inches and $4$ inches respectively. The area of this triangle is what common fraction of a square foot?

Number Seven: What is the sum of all the integers from $80$ to $90$, inclusive?

Number Eight: Simplify (express answer as a common fraction):

$\frac{3^4+3^2}{3^3-3}$

[quote="236factorial"][hide="3"]The circumference is 18, so its radius is $\frac{9}{\pi}$. Its area is $\frac{81}{\pi}$[/hide]

In 4, you can't have two hypotenuses (do you mean legs?)

[hide="7"]There are 11 integers. ((80+90)*11)/2=935 (I used a formula). [/hide]

[hide="8"]$\displaystyle \frac{3^4+3^2}{3^3-3}=\frac{90}{24}=\frac{15}{4}$[/hide][/quote]

Yes. I do. Sorry. :blush:

[hide="number 4"]The hypotenuse is $\sqrt{97}$ and the leg is 4, so the other leg must be 9. the area is 18 sq.in. That's 1/8 of a square foot (1 sq. ft = 144 sq. in.). [/hide]

[quote="hwenterprise"]Number Three: The circumference of a particular circle is 18cm. in sq. centimeters, what is the area of the circle? Express your answer as a common fraction in terms of $\pi$

Number Four: The hypotenuse EDIT: *and leg* of a particular right triangle are $\sqrt{97}$ inches and $4$ inches respectively. The area of this triangle is what common fraction of a square foot?

Number Seven: What is the sum of all the integers from $80$ to $90$, inclusive?

Number Eight: Simplify (express answer as a common fraction):

$\frac{3^4+3^2}{3^3-3}$[/quote]

Number 3 is impossible, because a circle's circumference cant be in sq centimeters

[quote="anirudh"]
Number 3 is impossible, because a circle's circumference cant be in sq centimeters[/quote]

Good point  :P, but you misread the problem  ;)

[quote="anirudh"]
Number 3 is impossible, because a circle's circumference cant be in sq centimeters[/quote]
No its not. Look, its two different sentences.
First sentence:
[quote="hwenterprise"]Number Three: The circumference of a particular circle is 18cm.[/quote]
Second sentence:
[quote="hwenterprise"]in sq. centimeters, what is the area of the circle?[/quote]

:what?:  He never written the circumference was in sq cm.. He asked us to give the [b]area[/b]in sq. cm!

Okay anirudh, you can solve the problems?  :roll:

[quote="anirudh"]Okay, but next time, the author of the questions should then capitalize the first letter of their sentences. In this case, it confused me.

Here are my answers:

[quote="hwenterprise"]Number Three: The circumference of a particular circle is 18cm. in sq. centimeters, what is the area of the circle? Express your answer as a common fraction in terms of $\pi$

[/quote]

[quote="Anirudh"]The circ. of a circle is 2pi r. $2\pi r=18$
$r \pi=9$
$\displaystyle r=\frac{9}{\pi}$
[/quote]

[quote="hwenterprise"]Number Four: The hypotenuse EDIT: *and leg* of a particular right triangle are \sqrt{97} inches and 4 inches respectively. The area of this triangle is what common fraction of a square foot? [/quote]

[quote="Anirudh"][img]http://www.artofproblemsolving.com/Classroom/cbe6/images/lx-98255790.gif[/img][/quote][/quote]

No wonder why you got so many posts. Half are spam.

not to be spamming or anything, but what's with the topic name?  :?

Yeah, this title only suits the 7th problem.

That's what it's meant to suit.

I didn't pay attention. :blush:

[quote="anirudh"]I didn't make that many posts: i just wrote like this:

[code][quote="hwenterprise"]the problem[/quote]

[quote="anirudh"]the answer[/quote]

and on and on[/code]

and i actually answer questions to get that many posts. I just go online and answer 40 at a time  :([/quote]

Oh.

Sorry for the critism. :(


This one is very easy : Just expand to get
x 3  + y 3  + z 3 + ( x 3  + y 3  + z 3 + 3x 2 y + 3x 2 z + 3y 2 x + 3y 2 z + 3z 2 x + 3z 2 y)
and use AM/GM on the 21 terms in the parenthesis to get the result.

Pierre.


I repeat again, if you are faint hearted or if you are about to eat your dinner then I suggest you better not visit it.. Else, well

http://www.trifter.com/Practical-Travel/The-8-Most-Cruel--Controversial-World-Delicacies.91674/1

Wow. Certainly cruel. You're jobless arent you

no he aint, he just portrays himself to be jobless  :D   :rotfl:   :P

Either way, doesnt make any difference.

Edited to make sure that the person who wrote this is alive till Saturday

[quote="shruthibaskaran"]Either way, doesnt make any difference.
[/quote]
Well it does!  :huh:  He doesnt want to portray himself as the nerd he is  :P
[quote="shruthibaskaran"]
 Hey Shreyas, how's your new girlfriend btw? ;)[/quote]
Now who's that id wonder  :wink:

Switch this convo to the metathread. I don't want people's culinary desires to be affected by my girlfriend. But just a note, mind you Shru, If I'd want to kalaichify you, it wouldn't take me more than 5 seconds to do so considering your extreme popularity and the fact that your boyfriends are dime-a-dozen. However I am keeping quiet, so don't get yourself smelled on a public forum.

Ummm let me be the referee to this duel of words  :D  n ofc silent spectator  :rotfl:

Ah Shre, dont we all know that? I did decide to remove that particular comment didnt I? And I dont have dime a dozen boyfriends mistah. I am appalled at your dirty thoughts.

you may have thought you were smart but i quoted your text so it will remain until i edit my post but why should i spoil my free entertainment esp when it comes for [u][b]FREE[/b][/u]  :P   :rotfl:   :D

Ugh.. Shreyas, where do you find such ghastly stuff? Man, you should see a shrink...

Rituraj, this is awesome......???? Umm..I don't think so.

Ashwath: Just found it, thought I should share such wonderful links with my fellowmen (and women, incase shruthi starts blabbering about male chauvinism)

Forgive me if this is gross, but the monkey brain really looks like a jelly. I have a sneaky feeling that it MIGHT be a photo of a jelly after all cause the photographer didn't get a monkey's brain to snap.

[quote="rituraj007"]really awesome!!
but think about the animals--its like an ad for someone(certainly not me) who really liked them[/quote]
  :o i think you shud see a shrink, not shre.

awe (uncountable):-
A feeling of amazement

that's what i meant by awesome

Eager to see the reaction of some of the sanyasis of our group to see it

Especially one boy who starting crying when a small bit of the egg shell fell on his hand during sol prpn

[quote="madness"]Rangava?  :rotfl:   :D[/quote]
 valerium?

yes  :D

did you also have that expt da varun? preparation of egg sol?  :maybe:

no. but did ranga really start cring coz egg fell on his hand

Just guessed  :rotfl:  Funny that this discussion shd be under the title "The 8 Most Cruel & Controversial World Delicacies "

wth varun and madness...

@madnesss- u are going to get it from me....
   varun- all that is not true.... :mad:  :mad:
 :mad:

[quote="madness"]noooo ranga plz dont show me ur infamed mirror of death  :roll:   :rotfl:[/quote]

 :rotfl:  :rotfl:  :rotfl:

actually credit to ya  :D  i copied that from ur comment in the c++ grps da :D
i was just reading thro those fun times again... vetti arent i  :D


A student wants to make up a schedule for a 7-day period during which she will study one subject each day. She is taking four subjects: mathematics, physics, chemistry and economics. The total number of different schedules is 4^7. The number of schedules that devote at least one day to each subject is?

I'd say:

Imagine there are just four days available, and she has to study every subject at least once. Then there are 4.3.2.1=4! ways in which to make the schedule (there are 4 subjects to choose from to study the first day, the second day only three, the third day two en the fourth day only one).
Now all of these schedules can be extended to seven days: the extra three days sche can study what she wants because she has already met the condition to study every subject at least once.
So in total there are 4!.4.4.4 = 1536 possible schedules.

Hint

[hide="I ll give out the answer, look if stuck and then work backwards"]

 8400, Think of the permutations within the 7 day week.

[/hide]


Hi, I was wondering, for the most part, what scores at nats would get you to finish what place. Correct me please if I'm wrong but the average is mid-20s, I think. Last year, I got 140-something with a 20 (pretty terrible, I know). Thanks

Unless this years Nationals is really really hard, the cutoff for countdown will be like 40.

this is pretty much

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=266091

Here are the cutoffs for top 12 the last few years.

2008 12th place was 37
2007 12th place was 36
2006 12th place was 36
2005 12th place was 30 (12th through 17th was 30)  1st was 39 and 2nd was 36
2004 They didn't give out that data that year.
2003 12th place was a 40 (11th through 15 was a 40)
2002 12th place was 38 (10th through 15 was a 38)
2001 12th place was 35

2005 was hard in State and Nats. It's kind of weird. I think the mc prob makers didn't notice it!  :maybe:

I think the states for that year was realtively easy.

Exactly, the states that year was very easy, and then nationals very very hard, an odd year

Ok, thanks. Actually, I'm not the type who has a chance at making nats cd. What about scores for making top 50, or 100, or mean/median? Is that info released?

Statistics summaries are given to coachs, so, they are, I'm not sure how many people on this forum have them though.  I think I only have one from 2007 hidden somewhere within folders and boxes...

dang it almost all of last years cders were in 6th or 7th grade....almost no chance for anybody ...but who knows luck fares well at times..

[quote="gauss1181"]Based on all that info and our knowledge of returning Nats competitors, I have a feeling that the cutoff for countdown would be 40...unless this year's Nats test is really really hard like 2005...which I hope would be. (Easy tests kill me because I always make stupid mistakes :( )[/quote]
I would guess that there's gonna be a moderately hard test this year. So maybe around a 35?

I'd estimate the CD cutoff to be in the very high 30s or maybe perhaps 40 sharp.

[quote="xpmath"]I don't know how stacked previous years have been, but I think this year should be pretty epic at the top. Bobby Shen has beasted [b]every[/b] competition this year (or so it seems), and more middle schoolers seem to be doing extremely well on the AMCs and AIMEs (which of course doesn't necessarily relate to Mathcounts). Also considering all the 6th and 7th graders from last year, the countdown cutoff should be pretty high.

I can't wait for nats.[/quote]

some how, orlando is looking more depressing day by day.....

@jonathanchou711   we have our own club of 7th place failures....

[quote="dynamo729"][quote="xpmath"]Bobby Shen has beasted [b]every[/b] competition this year (or so it seems)[/quote]

Bwahaha! I beat him on [url=http://www.mathleague.com/reports/report0809_g8.htm]this thing[/url][/quote]
Eh, so did I...and the Math League is EXTREMELY trivial...even more trivial than MOEMS or the easiest math contest ever, considering that i easily got a perfect.

and my school wouldn't let me take it because i go to the high school for math. i got an unofficial perfect in about ten minutes.

As beast as you are, you cant finish in 5 minutes. Unless you can do a problem in 7.5 seconds. Do they have the same problems for every state?

The California scores for that are amusing.

7 perfects, 5 from MMS.

[quote="frost13"]2001 12th place was 35[/quote]
Just in case anybody's curious, because I have it in front of me, the top 12 (though only 10 were in countdown at this point) was 42,41,41,40,40,39,38,38,37,37,36,35(,35,35)

[quote="Ihatepie"][quote="echen"]You must also realize that as the test get harder, people get better. It will probably stay the same if not rise a little.[/quote] I wouldn't say people get better. More like people learn more tricks to do the problems.

And there's generally been a harder test every 3 or 4 years. So I'm guessing this'll be it. (considering they might not want Bobby Shen to get a perfect score this year)[/quote]

Bobby's in 9th grade... so is David

you do know these people, most of their posts are in 2009 last year


Hey there! I'm teaching myself statistics right now and there's something that I'm not understanding about sets. I'm about to type out an excerpt from the statistics textbook but the concept itself is something learned in High School Algebra/Pre-Calculus so I figured I'd put it here. Sorry if it's not in the right section though :(

First off, complements. Sorry if some of these questions are really stupid. I'm a high school student still and it's difficult for me to udnerstand some of these things...

If there is a set $ A$, then a complement is a set that does not belong to $ A$? It's notation is $ A^{c}$. I'm confused what this actually means though. I'm going to just try and make up a real number example and can someone tell me if I'm correct?

If $ A \equal{} \left\{1,2,3,4,5 : x\in{Z}\right\}$ and $ B \equal{} \left\{6,7,8,9 : x\in{Z}\right\}$, then it can said that $ A \equal{} \left\{x|0 < x < 6 : x\in{Z}\right\}$ and $ B \equal{} \left\{x|5 < x < 10 : x\in{Z}\right\}$?

Then can we let $ A^{c} \equal{} B$ since the set $ B$ does not belong to set $ A$? Also, isn't this the same thing as saying that any element x of the set B is not an element of the set A? And is there a way to show this using the $ \notin$ symbol?

Okay, now just a little quick thing about unions and intersects. $ A\cup{B} \equal{} \left\{x\in{S} : x\in{A} or x\in{B}\right\}$ according to the textbook. I understand the union is the set that consists of all elements that are either in A or B. But the first part, $ x\in{S}$, what is that? Is that just noting that $ S$ is considered the set of $ A\cup{B}$?

Okay, this is a direct excerpt from the text.

It is often useful to extend the concepts of union and intersection to more than two sets. Let $ \left\{A_{a}\right\}$ denote an arbitary collection of sets. Then $ x\in{S}$ is an element of the union of $ \left\{A_{a}\right\}$, denoted
\[ \bigcup_{a}{A_{a}}
\]
I was able to follow along until here, and now I'm conpletely lost. What does $ x\in{S}$ mean here? And can someone explain what the "big union" symbol means in simple terms. I'll wait to see if I might be able to understand DeMorgan's Laws after someone explains this. Thanks for any help!

[b]Complements:[/b] [i]Two sets are complements [u]if and only if[/u] their union is the universal set and their intersection is the empty set.[/i]

For example, if we are in the set of integers, then:
> non-negative integers and negative integers [b]are [/b]complementary sets, because together, they form the set of integers.
> positive integers and negative integers [b]are not[/b] complementary sets, because their union does not contain the zero element.
> non-negative and non-positive integers [b]are not[/b] complementary sets, because they both contain zero.

[b]Notation:[/b]
$ x \in S$ means that the item $ x$ is an element of $ S$. For example, $ 2 \in \{1,2,3 \}$ and $ 4 \not \in \{1,2,3 \}$.
If you are talking about a subset instead of an element, you write $ \subset$. So, $ \{1,2 \} \subset \{1,2,3 \}$

Thanks for that! I understand that part now. But what about the larger union symbol? I have absolutely no clue where to begin with that. What does $ A_{a}$ denote? Does the large union symbol function like the sumation symbol in a way? I've seen it using where there is an initial value and at the top is another value like:

\[ \bigcup^{\infty}_{i=1}{E_{i}}\]

Do you think you can provide another example with numbers again? Thanks so much

You know how $ \sum$ is basically a bunch of $ \plus{}$ signs? $ \bigcup$ is a bunch of $ \cup$ signs. So

\[ \bigcup_{i\equal{}1}^{\infty} E_i \equal{} E_1 \cup E_2 \cup E_3 \cup \ldots\]


Let x,y,z be positve real of numbers satisfy that add up to 1.Prove that:
             6[a^3(b+c)+b^3(c+a)+c^3(a+b)]+3(a^2b^2+b^2c^2+c^2a^2)+12abc <=1

First, I'll write your problem more clearly
$a,b,c$ are possitive real that satisfy $a+b+c=1$ Prove:
$6 \sum a^3(b+c)+3 \sum(a^2b^2) +12abc \leq 1$
Well, I dehomogenizes it and brutal force:
      $6\sum a^3(b+c)+3\sum(a^2b^2)+12abc(a+b+c) \leq (a+b+c)^4$
<-->$-6(\sum a^4)-18(\sum (a^2b^2+a^3(b+c)) \leq 0$  with is obviously negative since $a,b,c \geq 0$


conics
parabola
quadratics
geometry
probability
power of a point
perpendicular bisector

Someone asked me this simple problem:

In a plane are given two different points $A$ and $B$ and a line $r$ (points are in the same half-plane with border $r$). We must construct a circumference $k$ such that $A, B\in k$ and $r$ tangent to $k$.

The problem is that he doesn't know what Power of a Point is.. So, how to find a solution without using the concept of Power of a point ?

You can derive Power of a Point using similar triangles...so I would either show him POP or I would look for a solution with similar triangles.

If the lines $r \parallel AB$ are parallel, the solution is trivial. Assume that these 2 lines are not parallel and let $P$ be their intersection. Let $(C)$ be a circle with center $C$, passing through the points $A, B$ and touching the line $r$ at a point $T$. The angle $\angle ABT$ spanning the arc $AT$ of the circle $(C)$ is equal to half the central angle $\angle ACT$ of this arc. The triangle $\triangle ACT$ is isosceles with $CA = CT$ being radii of the circle $(C)$. Hence the angle $\angle ATC = \frac{180^o - \angle ACT}{2} = 90^2 - \angle ABT$. The radius $CT$ is perpendicular to the line $r$. Thus the angle $\angle ATP = 90^o - \angle ATC  = 90^o - (90^o - \angle ABT) = \angle ABT$. It follows that the triangles $\triangle PAT \sim \triangle PTB$ are similar, because the angle $\angle P$ is common and the angles $\angle ATP = \angle ABT$ are equal. Consequently,

$\frac{PT}{PA} = \frac{PB}{PT},\ \ PA \cdot PB = PT^2$

The product $PA \cdot PB$, equal to the square of the tangent length $PT$ from the point $P$ to the circle $(C)$ is called the power of the point $P$ to the circle $(C)$, because it is the same for any secant of the circle $(C)$ from the point $P$. To construct the tangent length $PT$, it is necessary to find $\sqrt{PA \cdot PB}$. Draw a circle $(P)$ with center $P$ and radius $PA$, intersecting the line $AB$ at a point $A'$ different from $A$, such that $PA' = PA$. Find the midpoint $M$ of the segment $A'B$. Draw a circle $(M)$ with center $M$ and radius $MA' = MB$. Erect a normal to the line $AB$ at the point $P$ intersecting the circle $(M)$ at points $Q, Q'$, such that $PQ = PQ'$. The angle $\angle A'QB = 90^o$ is right, because it spans the diameter $A'B$ of the circle $(M)$. The line $PQ \perp A'B$ is the altitude of this triangle from the vertex $Q$. The triangles $\triangle AQP \sim \triangle A'BC \sim \triangle QBP$ are similar, because each has a right angle, the angle $\angle QA'B$ is common to the 1st pair and the angle $\angle QBA'$ to the 2nd pair of triangles. Hence,

$\frac{PA'}{PQ} = \frac{PQ}{PB},\ \ PQ^2 = PA' \cdot PB = PA \cdot PB = PT^2,\ \ PQ = PT$

Draw another circle centered at the points $P$ with radius $PQ$ intersecting the line $r$ at 2 points $T, T'$, the tangency points of 2 different circles $(C), (C')$ passing through the given points $A, B$. Erect normals to the line $r$ at points $T, T'$ intersecting the perpendicular bisector of the segment $AB$ at points $C, C'$, the centers of the circles $(C), (C')$.

It is certainly possible to solve the problem without using the power of a point to a circle. The locus of points with equal distance from 2 given points $A, B$ is the perpendicular bisector of the segment $AB$. The locus of points with equal distance from the given point $A$ and the given line $r$ is a parabola with the focal point $A$ and directrix $r$. The circle center $C$ is the intersection of the perpendicular bisector with the parabola. This intersection can be found by solving the set of 2 equations of the perpendicular bisector and of the parabola, which leads to a quadratic equation for the distance of the tangency point $T$ from the foot of a normal dropped from the point $A$ to the line $r$. This quadratic equation can be solved by compass and straightedge, but the construction is tedious.

Fun ;) ... I never have liked geometry, and you just strengthened that opinion. Anyway, yeah, teach him POP. It's not a very hard concept and will help a lot in whatever he does.

Probability 101  wrote:

..............................
        Yetti's solutions sometimes scare me ...
..........................................

    In this problem one should  use simple similarity of circles by  constructing
    a perpendicular thru the middle of segment AB till intersection at X with line r and
   reflecting the line r  in this  perpendicular onto r'.
    Inscribe any circle C centered at Z in the angle made by r and r'
    and draw a line thru X and point A.
    This line will intersect circle C in 2 points  M and N.  Now parallels to MZ and NZ
     thru point A will intersect the perpendicular bisector of AB (which is the angle
     bisector of X) in two points - the centers of two solution circles.

    This is it.

    Pestich.

[quote="probability1.01"]Yetti's solutions sometimes scare me, but they're still fun to look through :D[/quote]

[quote="tetrahedr0n"]They all look so professional ... especially the diagrams. He could just put togehter a bunch of geometry problems he solved and make a book out of them.[/quote]

[quote="probability1.01"]Yetti's 100 Proofs... :D[/quote]

Thank you for the kind words.

Yetti.

[quote="pestich"]Probability 101  wrote:

..............................
        Yetti's solutions sometimes scare me ...
..........................................

   You are right to be scared, because his solutions bring more harm than good.

    It is hard to imagine how his mind works, undoubtedly he tries to impress
    middle school students with his knowledge,  and thus his proofs are significantly
    more complicated then required.

    In this problem one should  use simple similarity of circles by  constructing
    a perpendicular thru the middle of segment AB till intersection at X with line r and
   reflecting the line r  in this  perpendicular onto r'.
    Inscribe any circle C centered at Z in the angle made by r and r'
    and draw a line thru X and point A.
    This line will intersect circle C in 2 points  M and N.  Now parallels to MZ and NZ
     thru point A will intersect the perpendicular bisector of AB (which is the angle
     bisector of X) in two points - the centers of two solution circles.

    This is it.

 ...............................
   but they're still fun to look through ....
..................................

    His solutions are fun only for the first time around, after that they are reusable
    on or around October 31 only.   At other times they may have negative effect on
    mental health.

    Pestich.[/quote]

Don't you think that's kind of harsh? I personally think they might be a bit on the lengthy side, but that doesn't mean they "bring more harm than good." We should be grateful that there are people dedicated enough to spend the time writing these solutions, not be condemning them for it. Props to you, yetti! And nice solution, too, pestich! Everyone wins! :D

yea... that was sort of an unnecessary, inappropriate comments... even if it was so hurtful to you to read, you should not comment like that on something yetti probably spent a lot of time thinking, typing, drawing, etc. It would be pointless for him to do all that work just to impress middle schoolers... Would you feel good if we criticize on your comment?

and BTW, not many of us at intermediate topic are Middle SCholers... at least neither paladin8 nor me...

And as both a user and a moderator for Intermediate Topics, I can assure you that in fact, high school students frequent this forum just as much as middle schoolers do. Yetti's solutions are very high quality. A forum user who doesn't like another's solutions is free to post different ones, not to criticize them for non-mathematical reasons.


$ \begin{equation}
U=\frac{ \sum_{F=1}^{N_F}N_S^{(F)} }{N_S \times N_F}
\label{AvgU}
\end{equation}$

How can I fix the sum such that appears like $ \sum_{F=1}^{N_F}\,?$

Use \sum\limits as in \sum\limits_{F = 1}^{N_F}N_S^{(F)}
$ \begin{equation}
U = \frac {\sum\limits_{F = 1}^{N_F}N_S^{(F)} }{N_S \times N_F}
\label{AvgU}
\end{equation}$

Thanks a lot Steve, always with fast replies!

I have a related question, suppose I have $ N^S,$ how can I reduce the size of the power $ S\,?$

How about $ N^{\scriptscriptstyle{S}}$ instead of $ N^S$?


I only have their past years's papers. Are there anymore similar papers available online including organic chemistry?

check IIT-JEE papers. There somewhere online.

Also, you can check out old IChO exams.


If n is the numbner of positive integers which will leave a remainder of 13 when divided into 3247, then find n.

Source: Math Team team round question

3247-13=3234=2*3*7*7*11

n is the number of divisors of 3234 that are bigger than 13...

n=18...

I got 18 as well...

3234 = 2*3*7^2*11 so there are 2*2*3*2 = 24 factors.

The factors smaller than 13 are 1, 2, 3, 6, 7, and 11. Thus there are 24-6 = 18 factors larger than 13.


For   $ a,b,c > 0$
 Prove that
        $ \dfrac{a^3}{b^2 \plus{} c^2 \minus{} bc} \plus{} \dfrac{b^3}{c^2 \plus{} a^2 \minus{} ac} \plus{} \dfrac{c^3}{a^2 \plus{} b^2 \minus{} bc} \geq \dfrac{3.(ab \plus{} bc \plus{} ca)}{a \plus{} b \plus{} c}$

by chebysev :
$ 2LHS \ge 2(a\plus{}b\plus{}c) \ge \frac{6(ab\plus{}bc\plus{}ca)}{a\plus{}b\plus{}c}$

[quote="mitdac123"]by chebysev :
$ 2LHS \ge 2(a \plus{} b \plus{} c) \ge \frac {6(ab \plus{} bc \plus{} ca)}{a \plus{} b \plus{} c}$[/quote]
If I have correctly understood you, your reasoning the foloowing:
If $ a\geq b\geq c$ then $ a^2\minus{}ab\plus{}b^2\geq a^2\minus{}ac\plus{}c^2\geq b^2\minus{}bc\plus{}c^2.$
But it's wrong. :wink:

Shock @@ .I'm really silly :P
$ LHS \ge \frac {(\sum a^2)^2}{ \sum ab(a \plus{} b) \minus{} 3abc}$
Assume :$ p \equal{} 1$;$ r \ge \frac {4q \minus{} 1}{9}$
<=>$ (3q \minus{} 1)^2 \ge 0$


Prove that any convex pentagon whose vertices (no three of which are collinear) have integer coordinates must have area greater than or equal to 5/2.

Errrr, hello.

I was trying to prove there was always one interior point, but it seems quite hard to do rigirously. Then Picks Theorem sorts it out nicely.


Find a polynomialwith integer co-efficients one of whose roots is $x=(\frac{3}{5})^\frac{1}{6} + (\frac{5}{3})^\frac{1}{6}$


function
invariant
logarithms
advanced fields
advanced fields unsolved

Is there an Elementary function coinside with $ \Gamma(n)\equal{}n!$ at every positive int.

This is a representative of a class of ques. They concerns the "invariant" of a class of funcs where such class is closed under iteration, contrafunction and some other usual transforms.

What is your defintion of elementary¿ Field operations, logarithm and exponentiation¿

Elementary functions here refers to the class of funtions containing y=x ,y=exp(x), y=const and closed under multi, addi, iterating, contrafunctioning.


limit
integration
calculus
real analysis
real analysis solved

proove that $\lim_{x\to \infty} x\int_1^x \frac{e^t}{t}dt -e^x=+\infty$
i tried with mean value theorem and it got me nowhere
also i thought of using Taylor expansion series but then again this exercice
i think is not so hard so there must be a better and elegant solution than this

Integration by parts the integral
$x\int_1^x \frac{e^t}{t}dt - e^{x}=   x(- e + \int_{1}^{x}e^{t}/t^2dt)$

Integrate by parts:

$\int_1^x\frac{e^t}t\,dt=\frac{e^t}t|_1^x-\int_1^x\frac{-e^t}{t^2}\,dt = \frac{e^x}x-e+\int_1^x\frac{e^t}{t^2}\,dt$

So $x\int_1^x\frac{e^t}t\,dt-e^x=x\int_1^x\frac{e^t}{t^2}\,dt-ex.$

It should be clear that this goes to infinity.

With one more integration by parts, I could show the following:

Let $f(x)=x\int_1^x\frac{e^t}t\,dt-e^x.$ Then $f(x)$ behaves like $g(x)=\frac{e^x}x$ in the sense that $\lim_{x\to\infty}\frac{f(x)}{g(x)}=1.$

Because the problem was about a limit at infinity, power series were always a poor choice for illuminating the behavior of this function. What I have actually done is to begin the development of an asymptotic series for it.

Here's the full asymptotic development. If $h(x)=\int_1^x\frac{e^t}t\,dt$, we say that

$h(x)$ has the asymptotic series $e^x\left(\frac1x+\frac1{x^2}+\frac2{x^3}+\cdots \frac{n!}{x^{n+1}}+\cdots\right).$

We have to be careful with the meaning of that, since that series diverges for all $x.$ Here's what it does mean: for each $n,$

$h(x)=e^x\sum_{k=0}^n\frac{k!}{x^{k+1}} + o\left(\frac{e^x}{x^{n+1}}\right)$ as $x\to\infty.$

(See my "big-O, little-o" class handout that's now a sticky on top of the "Calculus Computations and Tutorial" forum for the meaning of that.)

The proof? Just keep going with that integation by parts. Do it over and over.


[quote="CatalystOfNostalgia"]$ a,b,c$ are sides of a triangle, $ R$ is the circumradius. Show that:
$ \displaystyle\sum Ra\sqrt {4R^{2} \minus{} a^{2}} \equal{} abc$.[/quote]
[hide="Solution"]
Okay, so say that we have $ \triangle ABC$ with $ BC \equal{} a$, $ AC \equal{} b$, and $ AB \equal{} c$ with circumcenter $ O$.  So we have that the distance from the $ O$ to $ BC$ is $ \sqrt {R^2 \minus{} (\frac {a}{2})^2} \equal{} \frac {\sqrt {4R^2 \minus{} a^2}}{2}$.  Thus, $ a\sqrt {4R^2 \minus{} a^2}$ is the distance from $ O$ to $ BC$ times $ 2$ times $ a$, which is $ 4$ times the area of $ \triangle BOC$.  $ \displaystyle\sum Ra\sqrt {4R^{2} \minus{} a^{2}} \equal{} R\displaystyle\sum 4[BOC]\equal{}4R[\triangle ABC] \equal{} abc$.  [/hide]


function
induction
calculus
integration
number theory proposed
number theory

Let f:N -> N be a function satisfying f(1) = 1, f(2n) = f(n), f(2n+1) = f(2n) + 1, for all positive integers n. Find the maximum of f(n) when 1 <= n <= 1994.

[hide]
A simpler method which is less general is to note that for all integral $n$, $f(2^{n})=1$, and that for all integers that are not powers of two, it is the number of 1s in its base two representation. The number with the most 1s in it will be the max. That will be 1023 [/hide]

[quote="aviateurpilot"][quote="Rust"]By induction $f(n)$ is number of 1 in 2-adic digits n, therefore $f(n)\le f(1023)=10.$[/quote]
$f(1023)=f(2^{10}-1)=10+1=11$
[/quote]
Let $s(n)$ is number 1 in 2-adic digits of n. We have f(1)=s(1).
Let f(k)=s(k) for all k<n. If n=2k we have f(2k)=f(k)=s(k)=s(2k)=s(n).
If n=2k+1, then f(2k+1)=f(2k)+1=s(2k)+1=s(2k)+1. Therefore f(n)=s(n) for all n.
We have $s(2^{n}-1)=n$. It is obviosly for n=1 $2^{1}-1=1$. Therefore $1023=1111111111_{(2)}\not =11$ and $f(1023)=f(2^{11}-1-2^{k})=10$, were $k=6,7,...,11$.


search
geometry
geometric transformation
reflection

What is the length of the shortest path from (5,6) to (-8,10) that touches the x-axis.

The coordingates of A,B, and C are (5,5) (2,1) and (0,k) respectively.  The value of k that makes AC+BC as small as possible is?

$|AB|=\sqrt{(x_{A}-x_{B})^{2}+(y_{A}-y_{B})^{2}}$.

$B'=(2,-1)$ and search for the intersection between $AB'$ and the $y$ axis.

[hide="hint"]
Find the $k$ that makes the point $(0,k)$ equidistant from the two given points.
[/hide]

[hide="solution"]
First find the equation of the line through $(5,6)$ and $(-8,10)$. This line is, in standard form, $4x+13y = 98$ (I think) Then find the line perpendicular to this that goes through the midpoint of segment $AB$. The midpoint is $(-3/2,8)$. Then the perpendicular is $26x-8y=-103$ To find the x-int of this line, we set $y=0$ and get $x=-\frac{103}{26}$, which is about -4.
[/hide]

Of course, I might have messed up, so feel free to shoot holes through my solution.

[hide="my solution to the first prob"]the length of the shortest path from $(5,6)$ to $(-8,10)$ touching the x-axis is $\sqrt{(5+8)^{2}+(6+10)^{2}}$. to get this, reflect $(-8,10)$ across the x-axis to get $(-8,-10)$. verify that the length of the shortest path from $(5,6)$ to $(-8,10)$ that touches the x-axis equals the length of the shortest path from $(5,6)$ to $(-8,-10)$. then the answer easily follows.[/hide]


discuss  about  roots  of equation :
 $ x \plus{}lnx \equal{}a$ ( where  $ a$  is  real  number)

[quote="Hong Quy"]discuss  about  roots  of equation :
 $ x \plus{} lnx \equal{} a$ ( where  $ a$  is  real  number)[/quote]

It's easy!
$ \bullet$ Let $ y\equal{}x\plus{}lnx$, $ x>0$.

$ \bullet$ $ y'\equal{}1\plus{}\frac{1}{x}>0, \forall x>0$.

So we have variation table of this funtion: (See picture).

From variation table of this funtion $ \Longrightarrow$ equation have only one root $ \forall a \in R$.

who  can find  root  exactly!!!  i think it's hard

[quote="hsiljak"]Is it possible at all, without use of approximations and/or iterations? It seems quite transcendental, doesn't it? Hence it is not hard to find it exactly, it is impossible.[/quote]
yes,  can u help me ?

I don't like numerical methods... But I'm sure [url=http://mathworld.wolfram.com/NewtonsMethod.html]Newton[/url] will help you. If you don't fancy it, you've got a bunch of other methods listed in the end of the article.


group theory
abstract algebra
superior algebra
superior algebra solved

Let G be a finite group and n be its order.  If m is a positive integer such that $gcd(m, n) = 1$, then show that for each x in G there exists a unique y in G such that $y^m = x$.

I have a solution, but there are a couple points that I don't understand about it...

1) If a is an element of G, then why is  $a^n = 1$?

2) If T: G -> G and T(a) = a^m, why is T onto G?

Thanks for any help you can provide.

[quote="Fierytycoon"]
I have a solution, but there are a couple points that I don't understand about it...

1) If a is an element of G, then why is  $a^n = 1$?

2) If T: G -> G and T(a) = a^m, why is T onto G?

[/quote]

2) For any map $T:G\rightarrow G$ (we just need that $G$ is a set) onto is equivalent to one-to-one. Let me just show the converse. If $T$ is one-to-one, since $G$ has $n$ different elements, then the image also should be $n$ different elements in $G$, so the image is all of $G$, i.e $T$ is onto. You can check that the map $T(a)=a^m$ is one-to-one

1) COnsider the subgroup generated by $a$, i.e <a>. Since $G$ is finite then $a^i=a^j$ for some $i>j$. It follows that $a^{i-j}=e$. Now consider the smallest integer $k$ satisfies $a^k=e$. Show that $<a>$ is of order $k$. By Lagrange then we have $k\mid n$. It follows that $a^n=a^{\alpha k}=(a^k)^{\alpha}=e$ for some $\alpha$.


The product of two positive whole numbers is 2005. If neither
number is 1, what is the sum of the two numbers?

2005 has only 4 factors - 1 and 2005, 5 and 401.

Since both numbers can't be 1, our answer is $ 401 \plus{} 5 \equal{} \boxed{406}$.


geometry
incenter
circumcircle
geometry solved

In a triangle $ABC$ let $H$ be the orthocentre, $I$ the incentre, $O$ the cicumcentre, $K$ the point where the incircle touches $BC$. Given, that $IO$ is parallel to $BC$, prove that $AO$ is parallel to $HK$.

Let $M$ be the midpoint of $BC$. I'll show that $IM\|AO\ (*)$ first, and then the result will follow: let $T$ be the midpoint of $AH$. $TH=IK,TH\|IK$, so $HK\|IT=TM\|OA$ ($IT=TM$, which is an equality of lines, not segments, follows from $(*)$ and from the fact that $OM=AT,OM\|AT$, so $AO\|TM$).

To prove $(*)$:

We know $OI^2=R^2-2Rr$, so $IM=\sqrt{OI^2+IK^2}=\sqrt{R^2-2Rr+r^2}=R-r$. On the other hand, if we put $U=OM\cap$ the circumcircle of $ABC$, then $MU=R-OM=R-r$, so $IM=MU=R-r$. Since $AO=OU=R$, we find the triangles $IMU,AOU$ to be similar, thus getting $IM\|AO$, Q.E.D.

This is the same as:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=tournament+2003&t=19370


I know there are two distance formulas, right?

Formula A is D = rt.

Formula B is the long formula used to find the distance between two points on the xy-plane, right?

MY QUESTION:

Why do we call BOTH the distance formula if they are used for TWO different reasons?

Because they are both formulas... for distances...

It's also worthwhile to mention that the first one is primarily physics-focused and is more a definition of $r$ than a formula for $d$, while the second one is more geometrical.

If you're going to nitpick about formula names, look up everything that goes by the name "Euler's formula" for a real headache.

I am not complaining about formula names.

I just wanted to know why use the SAME name to indicate two distinctive purposes.

However, I understood your answer.

I think formula B is more accepted to be the distance formula.

Okay, but when I think of the distance formula, I AUTOMATICALLY think about D = rt.

I've heard other names for A, but B is always called the distance formula.

when you say distance formula, i think most people think of formula B.

D=RT is just called... ummm d=rt. :ninja:

heres a website with all kinds of formulas to help prepare for your test:
http://library.thinkquest.org/20991/gather/formula/index.html

D=rt is usually for sciences* (physics, etc.), while the coordinate plane one is used in math.

*Sometimes you use it in math


geometry
analytic geometry
geometric transformation
trigonometry
homothety
incenter
Ross Mathematics Program

Let ABC be a triangle, and let the incircle, with centre I, meet BC at X. Let M be the midpoint of BC. Prove MI bisects the segment AX.
the solution to this problem, by Daniel Sokolowski, in the book is quite difficult and has lots of constructions ... surely there is a simpler solution ...

I'd love to have given a beautiful solution, but there's no point in searching for one since this is a trivial application of barycentric coordinates:

Let $N$ be the midpoiint of $AX$. $I$ has coordinates $(a:b:c)$, $M$ has coordinates $(0:1:1)$ and it's really easy to show that $N$ has coordinates $(a:p-c:p-b)$. Using these, we construct the determinant $\left |\begin{array}{ccc}0&1&1\\ a&b&c\\ a&p-c&p-b\end{array}\right |$.

By subtracting the third column from the second one it becomes obvious that this determinant is $0$, which is the necessary and sufficient condition for three points to be collinear.

[color=blue][b]Problem.[/b] Let $ ABC$ be a triangle. Let the incircle of triangle $ ABC$ have centre $ I$ and touch the line $ BC$ at a point $ X$. Let $ M$ be the midpoint of the segment $ BC$. Prove that the line $ MI$ bisects the segment $ AX$.[/color]

This is problem 3 in my edition (Russian translation, 1978). Coxeter/Greitzer prove it using some cosine law calculation. Actually, it can be shown synthetically:

Let the A-excircle of triangle ABC touch BC at X'. Then, with the notation s = (a+b+c)/2, we have BX = s-b and CX' = s-b. Thus, BX = CX', and the points X and X' are symmetric with respect to the midpoint M of BC. In other words, M is the midpoint of the segment XX'.

Let S be the point diametrically opposite to X on the incircle of triangle ABC. The tangent to the incircle at S is parallel to BC (since tangents to a circle at two diametrically opposite points are always parallel) and meets the sides AB and AC at two new points B' and C'. Since B'C' || BC, the triangle AB'C' is the image of the triangle ABC in a homothety with center A. But the points S and X' are corresponding points in these two triangles: The point S is the point of tangency of the A-excircle of triangle AB'C' with the side B'C' (remember that the A-excircle of triangle AB'C' is the incircle of triangle ABC !), while the point X' is the point of tangency of the A-excircle of triangle ABC with the side BC. Hence, the homothety with center A mapping triangle ABC to triangle AB'C' must also map X' to S. Thus, S lies on the line AX'. Now, since S is the point diametrically opposite to X on the incircle, the center I of the incircle is the midpoint of the segment XS. On the other hand, M is the midpoint of the segment XX'. Hence, the line MI is a midparallel in triangle SXX' and must therefore be parallel to its side SX', i. e. to the line AX'.

Now, if the line MI intersects the line AX at R, then you immediately get AR : RX = X'M : MX = 1, so that R is the midpoint of AX. In other words, the line MI bisects the segment AX. Proof complete.

Only the very last argument of the above proof was invented by me. The crux was to show MI || AX'; once you know this, you immediately get a lot of results, among them not only the fact that MI bisects AX, but also the more important theorem that the incenter I of triangle ABC is the Nagel point of the medial triangle, or the fact that the line IX' bisects the A-altitude of triangle ABC (try proving this for yourself - it is very easy, and it is, BTW, not so easy using barycentric coordinates!). The above proof MI || AX' is from Ross Honsberger's "Episodes in the Nineteenth and Twentieth Century Euclidean Geometry", Washington 1995, pp. 30-31.

[b]Edit:[/b] See also http://www.mathlinks.ro/Forum/viewtopic.php?t=112509 .

  Darij

If $ Y \equal{} AX \cap MI$ , $ a, b$ and $ c$ being the sides of the triangle and supposing, w.l.o.g. that $ b>c$, we have:
$ \frac {AI}{DI} \equal{} \frac {b\plus{}c}{a}$, $ BX \equal{} s\minus{}a$, where $ s \equal{} \frac {a\plus{}b\plus{}c}{2}$, $ BD\equal{} \frac {ac}{b\plus{}c}$, consequently $ MX \equal{} BM \minus{} BX \equal{} \frac {b\minus{}c}{2}$ and $ MD \equal{} BM \minus{} BD \equal{} \frac {a(b\minus{}c)}{2(b\plus{}c)}$.
Apply Menelaus for $ \triangle ADX$ and transversal $ MIY$ , keep into account the above results and get $ AY \equal{} XY$, q.e.d.

Best regards,
sunken rock

Hello!

$ \left(I\right)$ is the incircle of $ \Delta ABC$ and $ \left(I_{a}\right)$ is the $ A$-excircle of $ \Delta ABC$
$ \left\{Y\right\}\equal{}MI\cap AX$, $ \left\{P\right\}\equal{}AB\cap\left(I\right)$, $ \left\{Q\right\}\equal{}AB\cap\left(I_{a}\right)$,
$ \left\{S,X\right\}\equal{}XI\cap\left(I\right)$ and  $ \left\{T\right\}\equal{}BC\cap\left(I_{a}\right)$

$ sin\left(\frac{\angle A}{2}\right)\equal{}\frac{PI}{AI}\equal{}\frac{QI_{a}}{AI_{a}}\Rightarrow\frac{AI}{AI_{a}}\equal{}\frac{PI}{QI_{a}}\equal{}\frac{IS}{I_{a}T}$   (1)
$ IS\bot BC$ and $ I_{a}T\bot BC\Rightarrow IS\parallel{}I_{a}T$   (2)
(1), (2) and $ I\in AI_{a}\Rightarrow S\in AT$
It's known that $ XB\equal{}TC$ and $ MB\equal{}MC\Rightarrow MX\equal{}MT$
$ IX\equal{}IS$ and $ MX\equal{}MT\Rightarrow IM\parallel{}ST$
$ YM\parallel{}AT$ and $ MX\equal{}MT\Rightarrow YX\equal{}YA$
So, $ MI$ bisects the segment $ AI$.

Best regards, Petrisor Neagoe :)

Denote the intersection $ Z$ between $ MI$ with the $ A$-altitude. Then the proposed problem is equivalently

with proving $ AZ \equal{} r$ . The [u][b]Sunkenrock[/b]'s proof[/u] is nicely and immediately. Prove analogously $ AZ \equal{} r$ .

These two properties, i.e. $ MI$ bisects $ AX$  and $ AZ\equal{}r$  are remarkably and well-known. I say to remember they !

Using the notations in the figure by [b]Petrisor[/b], we can say also $ \frac {AI}{AI_{a}} \equal{} \frac {IP}{I_{a}Q} \equal{} \frac {IS}{I_{a}T}$ $ ,(1)$

From $ (1)$ and because of $ IS\parallel I_{a}T$, based on the [b]Thales theorem[/b], we conclude the collinearity of the points $ A,\ S,\ T.$

Because of now, $ MX \equal{} MT$ and $ IX \equal{} IS$, we have that $ MI\parallel AC$.

Hence, we conclude that the line segment $ MI$ bisects the segment $ AX$ and the proof is completed.

Kostas Vittas.


function
algebra
domain
algebra proposed

Find all functions $ f:\mathbb{R}\minus{}[0,1]\to\mathbb{R}$ such that
\[f(x)+f\left(\frac{1}{1-x}\right) = \frac{2(1-2x)}{x(1-x)},\]
for all $x \in \mathbb R$.

[quote="mathemagics"]Find all functions $ f:\mathbb{R}\minus{}[0,1]\to\mathbb{R}$ such that

$ f(x)\plus{}f\left(\frac{1}{1\minus{}x}\right) \equal{}\frac{2(1\minus{}2x)}{x(1\minus{}x)}$

India Competition[/quote]
1. $ f(x)\plus{}f(\frac{1}{1\minus{}x}) \equal{}\frac{2}{x}\plus{}\frac{2}{x\minus{}1}$.
$ x \equal{}\frac{1}{1\minus{}y}, y\to x$ give
2. $ f(\frac{1}{1\minus{}x})\plus{}f(\frac{x\minus{}1}{x}) \equal{}\frac{2}{x}\minus{}2x$
$ x \equal{}\frac{y\minus{}1}{y}, y\to x$ give
3. $ f(\frac{x\minus{}1}{x})\plus{}f(x) \equal{}\frac{2x}{x\minus{}1}\minus{}2x$
(1.+3.-2.)/2 give
$ f(x) \equal{}\frac{x\plus{}1}{x\minus{}1}$.

[quote="Rust"][quote="mathemagics"]Find all functions $ f:\mathbb{R}\minus{}[0,1]\to\mathbb{R}$ such that

$ f(x)\plus{}f\left(\frac{1}{1\minus{}x}\right) \equal{}\frac{2(1\minus{}2x)}{x(1\minus{}x)}$ (*)

India Competition[/quote]
1. $ f(x)\plus{}f(\frac{1}{1\minus{}x}\equal{}\frac{2}{x}\plus{}\frac{2}{x\minus{}1}$.
$ x \equal{}\frac{1}{1\minus{}y}, y\to x$ (**) give
2. $ f(\frac{1}{1\minus{}x})\plus{}f(\frac{x\minus{}1}{x}) \equal{}\frac{2}{x}\minus{}2x$ (***)
$ x \equal{}\frac{y\minus{}1}{y}, y\to x$ give .....
[/quote]

Rust see that, in (*) $ x$ must be greater than 1, so by (**) $ 0<y<1$ but this implies that $ \frac{1}{2}<\frac{1}{1\minus{}x}<1$ in (***) but it's out of the domaind of $ f$

[quote="mathemagics"]Find all functions $ f:\mathbb{R}\minus{}[0,1]\to\mathbb{R}$ such that

$ f(x)\plus{}f\left(\frac{1}{1\minus{}x}\right) \equal{}\frac{2(1\minus{}2x)}{x(1\minus{}x)}$ (*)

India Competition[/quote]

The domain of $ f$ is $ \mathbb{R}\minus{}[0,1]$ so in the condition (*) we must have $ x\in (\minus{}\infty,0)\cup(1,\infty)$ and $ \frac{1}{1\minus{}x}\in (\minus{}\infty,0)\cup(1,\infty)$ that implies $ x>1$ and $ \frac{1}{1\minus{}x}<0$.
(*) is equivalent to
$ f(x)\plus{}f\left(\frac{1}{1\minus{}x}\right) \equal{}\frac{2}{x\minus{}1}\plus{}\frac{2}{x}$
$ \equal{}\left(1\plus{}\frac{2}{x\minus{}1}\right)\plus{}\left(\frac{2}{x}\minus{}1\right)$
$ \equal{}\left(1\plus{}\frac{2}{x\minus{}1}\right)\plus{}\left(1\plus{}\frac{2}{\frac{1}{1\minus{}x}\minus{}1}\right)$
then
$ f(x)\minus{}\left(1\plus{}\frac{2}{x\minus{}1}\right)\plus{}f\left(\frac{1}{1\minus{}x}\right)\minus{}\left(1\plus{}\frac{2}{\frac{1}{1\minus{}x}\minus{}1}\right) \equal{}0$
Let $ g(x)\equal{}f(x)\minus{}\left(1\plus{}\frac{2}{x\minus{}1}\right)$
then
$ g(x)\plus{}g\left(\frac{1}{1\minus{}x}\right)\equal{}0$
$ g(x)\equal{}\minus{}g\left(\frac{1}{1\minus{}x}\right)$
Now if $ x$ goes from 1 to $ \infty$, $ \frac{1}{1\minus{}x}$ goes from -$ \infty$ to 0, so we can set $ g(x)$ an arbitrary function for $ x>1$, and construct $ g$ for negatives values as $ g(x)\equal{}\minus{}g\left(\frac{1}{1\minus{}x}\right)$, so $ f(x)\equal{}g(x)\plus{}1\plus{}\frac{2}{x\minus{}1}$

[quote="mathemagics"][color=darkred]Find all functions[/color] $ f: A\to\mathbb{R}$ such that $ f(x)\plus{}f\left(\frac{1}{1\minus{}x}\right) \equal{}\frac{2(1\minus{}2x)}{x(1\minus{}x)}$, where $ A \equal{}\mathbb R\setminus\{0,1\}$.[/quote]
[color=darkblue][b]Proof.[/b] Denote the function $ g: A\to\mathbb R$, where $ g(x) \equal{}\frac{2(1\minus{}2x)}{x(1\minus{}x)}$ and the function $ \phi : A\to\mathbb R$, where $ \phi (x) \equal{}\frac{1}{1\minus{}x}$.
Observe that the function $ \phi$ is injective and $ \mathrm{Im}\phi\equiv\phi (A) \equal{} A$, i.e. the function $ \phi : A\to A$ is bijective (it is a [b]dynamic function[/b]).
Prove easily that $ \phi\circ\phi\circ\phi \equal{} 1_{A}$ (the identical function) and $ \left\{\begin{array}{c}f\plus{}f\circ\phi \equal{} g\\ \ f\circ\phi\plus{}f\circ\phi\circ\phi \equal{} g\circ\phi\\ \ f\circ\phi\circ\phi\plus{}f \equal{} g\circ\phi\circ\phi\end{array}\right\|$ $ \implies$ $ f \equal{}\frac{1}{2}\cdot (g\minus{}g\circ\phi\plus{}g\circ\phi\circ\phi )$ a.s.o.[/color]

[quote="Virgil Nicula"][quote="mathemagics"][color=darkred]Find all functions[/color] $ f: A\to\mathbb{R}$ such that $ f(x)\plus{}f\left(\frac{1}{1\minus{}x}\right) \equal{}\frac{2(1\minus{}2x)}{x(1\minus{}x)}$, where $ A \equal{} (\minus{}\infty , 0)\cup (1,\infty)$.[/quote]
[color=darkblue][b]Proof.[/b] Denote the function $ g:\mathbb R\setminus\{0,1\}\to\mathbb R$, where $ g(x) \equal{}\frac{2(1\minus{}2x)}{x(1\minus{}x)}$ and the function $ \phi : A\to\mathbb R$, where $ \phi (x) \equal{}\frac{1}{1\minus{}x}$.
Observe that the function $ \phi$ is injective and $ \mathrm{Im}\phi\equiv\phi (A) \equal{} A$, i.e. the function $ \phi : A\to A$ is bijective (it is a [b]dynamic function[/b]).
Prove easily that $ \phi\circ\phi\circ\phi \equal{} 1_{A}$ (the identical function) and $ \left\{\begin{array}{c}f\plus{}f\circ\phi \equal{} g\\ \ f\circ\phi\plus{}f\circ\phi\circ\phi \equal{} g\circ\phi\\ \ f\circ\phi\circ\phi\plus{}f \equal{} g\circ\phi\circ\phi\end{array}\right\|$ $ \implies$ $ f \equal{}\frac{1}{2}\cdot (g\minus{}g\circ\phi\plus{}g\circ\phi\circ\phi )$ a.s.o.[/color][/quote]

Virgil $ \phi$ is not bijective, supose $ x \equal{}\minus{}1$ then $ x\in A$ but $ \phi(x) \equal{}\frac{1}{2}\notin A$

[quote="mathemagics"]Find all functions $ f:\mathbb{R}\minus{}[0,1]\to\mathbb{R}$ such that

$ f(x)\plus{}f\left(\frac{1}{1\minus{}x}\right) \equal{}\frac{2(1\minus{}2x)}{x(1\minus{}x)}$

India Competition[/quote]

The problem is not well defined, as the scope of values for which the equation is verified is not given.

By the way, I think there is a typo in the problem and the domain is $ f:\mathbb{R}\minus{}\{0,1\}\to\mathbb{R}$ instead of $ f:\mathbb{R}\minus{}[0,1]\to\mathbb{R}$.

Then Rust's solution would be the good one.

I found this problem in the resources section here http://www.mathlinks.ro/Forum/resources.php?c=78&cid=46&year=1992
and the problem was proposed with the domain $ \mathbb{R}\minus{}[0,1]$. Do you have any book that contains Indian Olympiads to check it ?


integration
function
real analysis
real analysis unsolved

If we use functions of the form $ f(t)\equal{}t^{\alpha}$ for $ 0<\alpha<1,$ the quantity to be minimized turns out to be $ \frac{2\alpha^{\alpha}\plus{}1}{\alpha\plus{}1}\minus{}1.$ This has a minimum at $ \alpha\approx0.8571$ and that minimum value is about $ 0.48209.$

I have no reason to think this is the actual infinum, but it gives us a target; the next claim would need to be better.

I wanted to say that the inf is $ 1/2$ but I realized I made an error.

Let $ g(x)\equal{}e^{\minus{}x}f(x)$.

Then we have that

$ \int_{0}^{1}|f'(x)\minus{}f(x)|dx\equal{}\int_{0}^{1}|g'(x)|e^{x}dx\geq \int\limits_{0}^{1}|g'(x)|dx$

$ \geq \biggm|\int_{0}^{1}g'(x)dx|\equal{}|g(1)\minus{}g(0)|\equal{}\frac{1}{e}$.

Thus we get that our infimum is between $ \frac{1}{e}$ and $ 1/2$.

Let $ f(x)\equal{}ax^{2}\plus{}bx$ with $ a\plus{}b\equal{}1$.

$ f'(x)\minus{}f(x)\equal{}h(x)\equal{}\minus{}ax^{2}\plus{}(2a\minus{}b)x\plus{}b$.

Perhaps we can use $ h$ and determine the coefficients $ a$ and $ b$ from the condition

$ \int_{0}^{1}|h(x)|dx\equal{}\frac{1}{e}$ and show that the inf is attained at $ 1/e$.

Excellent! Note that to attain $ 1/e$, we'd have to let $ g'$ be a multiple of Dirac delta at $ 0$. We can't really do that, but we can let $ f(x) \equal{} e^{x \minus{} 1}$ for $ x\in [\varepsilon,1]$ and have $ f$ increasing from $ 0$ to $ e^{\varepsilon \minus{} 1}$ on the interval $ [0,\varepsilon]$. This brings $ \int_0^1 |f' \minus{} f|$ arbitrarily close to $ 1/e$. The infimum is not attained.

[quote="mlok"] The infimum is not attained.[/quote]   ... in $ C^1$, but it is attained in $ BV$. :)


[quote="ln(dx/dy)"]How many points in 3-D space (x,y,z) with all positive corrdinates exist such that the sum of its corrdinates equals 100?[/quote][hide]How many points? I guess that would be $c$, or the continuum.[/hide]

Yeah, lol. I think ln(dx/dy) means $(x,y,z)$ where $x,y,z$ are integers lol. I remember seeing this question somewhere... (was it Mathcounts)? Anyhow, this is the way I would do it:

[hide="Solution"]Let's do some casework. There's probably some amazingly powerful uber way to do this, but I'm too stupid to know it, so we'll do it the old-fashioned way :D (if someone knows the uber way, please post).

Suppose $x=1$. Then, we need to find two numbers for $y$ and $z$ that add to 99. These pairs are (1,1,98); (1,2,97); (1,3,96); ...; (1,98,1) for a total of 98 triplets.

Now, suppose $x=2$. Then, we have (2,1,97); (2,2,96); (2,3,95); ...; (2,97,1) for a total of 97 pairs. This looks like a pattern :D. If we continue, we would expect 96 pairs for when $x=3$ and 95 for when $x=4$ all the way down to 1 pair when $x=98$ [(98,1,1)]. This means the total number of ordered pairs is $98+97+96+95+...+3+2+1$. We can use $\frac{n(n+1)}{2}$ to get the total number of pairs as $4851$.

Not sure if this is the right answer, but I think I covered everything :P
[/hide]

Yes, I have seen this exact problem on a NEML (New England Math League) competition in the past year. I do believe [hide]4851[/hide] is the correct answer.

It was on the first (?) math league this year.  There are a lot of leagues with different names but they are the same contest really.  This confuses people for some reason.  If you don't believe me look here:  http://mathleague.com/reports/report0405_4.htm .

One way to do this problem is look for the pattern.  It's not hard and a good example of how playing around with a problem can break it down.

There is another way to do this problem, however.  This way is more elegant in my opinion.  First step is to translate the problem into an equation:

\[x+y+z=100.\]

Consider puttin 100 dots in a row.  Now if we put in 2 'dividers' between the dots we will have 3 sections.  Let the first section represent $x$'s value, second $y$'s value, and 3rd $z$'s value.

As an example suppose we wanted $x+y+z=5$.  Our setup would look like this:

\[\bullet | \bullet \bullet | \bullet \bullet\]

So consider doing that with 100 dots.  How many ways can we do this?  Well, this is equvialent to [i]choosing[/i] two dividers from 99 possible places.  Thus we have $\binom{99}{2}=4851.$


Let $ a ,b,c$ be positive real numbers . Prove that :

$ \sum\frac{a(b^2\plus{}c^2)}{b^2c^2} \geq \frac{6(a^2\plus{}b^2\plus{}c^2)}{a^3\plus{}b^3\plus{}c^3}$

[quote="nguoivn"]It's obviously trues because $ a^3 \plus{} b^3 \plus{} c^3 \ge\ 3abc$  :maybe:[/quote]

[hide] có lẽ em ko hợp với việc sáng tạo anh ạ :D [/hide]


[quote="mdk"]Determine all real $ x$ that satisfy $ x^{5}-11x^{4}+36x^{3}-36x^{2}+11x-1 = 0$[/quote]

Hmm... Easier than I thought.

[hide]Note that $ x=1$ makes the polynomial equal to $ 0$. So $ x-1$ is a factor.

$ (x-1)(x^{4}-10x^{3}+26x^{2}-10x+1)=0$

Guess the last quartic as the product of two quadratic factors. That gets us
$ (x-1)(x^{2}-4x+1)(x^{2}-6x+1) = 0$

$ \boxed{x=1, 2\pm \sqrt{3}, 3\pm 2\sqrt{2}}$[/hide]

[quote="vishalarul"][hide]Note that $ x=1$ makes the polynomial equal to $ 0$. So $ x-1$ is a factor.

$ (x-1)(x^{4}-10x^{3}+26x^{2}-10x+1)=0$
(...)[/hide][/quote]
[hide]
 $ x^{4}-10x^{3}+26x^{2}-10x+1=0$ we can solve in other way. Ofc $ x=0$ isn't a solution, then we devide by $ x^{2}$:
$ x^{2}+\frac{1}{x^{2}}-10\left(x+\frac{1}{x}\right)+26=0$
$ \left(x+\frac{1}{x}\right)^{2}-2-10\left(x+\frac{1}{x}\right)+26=0$
And let $ x+\frac{1}{x}=t$...[/hide]


The cost at Modern Attire for a shirt and a tie is 60 dollars. The cost for a tie and a
pair of pants is 66 dollars. The cost for a pair of pants and a sweatshirt is 72 dollars. The
cost for a sweatshirt and a jacket is 100 dollars. The cost for a shirt and a jacket is
82 dollars. What is the cost of one sweatshirt?

Add them all up to get $ 2(\text{one of each item})\equal{}\$380$.  Dividing by two yields $ \text{one of each item }\equal{}\$190$, and subtracting (jacket+shirt) and (tie+pants) yields $ 190\minus{}82\minus{}66\equal{}\boxed{42}$.


We know that $ \sqrt{(x\plus{}y)^2}\plus{}\sqrt{(x\minus{}y)^2}\equal{}1$. Find the least and the greatest value of expression $ x^2 \minus{} 6x \plus{} y^2 \minus{} 6y$.


Find the number of all natural numbers n between 4 and 1023 inclusive, whose binary representations don't contain three consecutive equal digits.

[hide]I get 372.

I found that if you have a number 3 digits long, you can find 3 that work.  For a number:

4 digits long you can find 5.

5 digits long you can find 8.

6 digits long you can find 13.

The pattern is the Fibonacci pattern, so you sum 3, 5, 8, 13, 21, 34, 55, 89, and 144, a total of 372.[/hide]

A couple of things:

[hide]

it was only up to 1023 so take off the 144. Otherwise your final answer is correct .

However, as with all olympiad style problems, can you prove that it is a fibonacci problem? The proof is worth most of the marks, usually the final answer may get you a maximum of 2 out of 7.

[/hide]


How many different 5-digit numbers can be made using the numbers 1, 2, 3, 4, and 5
once each?


trigonometry
function
integration
calculus
calculus computations

\[ \begin{gathered}
  {\text{prove that }}f\left( x \right) \equal{} \tan \left( {\sqrt[{}]{{1 \minus{} x^2 }}} \right) \plus{} a{\text{ its a solution of }} \hfill \\
  x\left( {1 \plus{} y^2 } \right) \plus{} \sqrt[{}]{{1 \minus{} x^2 }}\frac{{dy}}
{{dx}} \equal{} 0 \hfill \\
\end{gathered}\]
i think its wrong...
thanks

I believe it's meant to be $ f(x) \equal{} \tan\left(\sqrt {1 \minus{} x^2} \plus{} a\right)$, not $ f(x) \equal{} \tan\left(\sqrt {1 \minus{} x^2}\right) \plus{} a$. Can you make this one work!? (Also, your notation is wrong.. You have $ y$ instead of $ f$)

[hide="Why the new function"]

$ x\,(1\plus{}f^2)\plus{}\sqrt{1\minus{}x^2} \frac{df}{dx} \equal{} 0$

$ \frac{x}{\sqrt{1\minus{}x^2}}\,dx \plus{} \frac{df}{1\plus{}f^2} \equal{} 0$

$ \int \frac{df}{1\plus{}f^2} \equal{} \minus{}\int \frac{x}{\sqrt{1\minus{}x^2}}\,dx$

$ \tan^{\minus{}1} f \equal{} \sqrt{1\minus{}x^2} \plus{} C$

$ f(x) \equal{} \tan\left(\sqrt{1\minus{}x^2} \plus{} C\right)$

[/hide]

[quote="milin"]I believe it's meant to be $ f(x) \equal{} \tan\left(\sqrt {1 \minus{} x^2} \plus{} a\right)$, not $ f(x) \equal{} \tan\left(\sqrt {1 \minus{} x^2}\right) \plus{} a$. Can you make this one work!? (Also, your notation is wrong.. You have $ y$ instead of $ f$)

[hide="Why the new function"]

$ x\,(1 \plus{} f^2) \plus{} \sqrt {1 \minus{} x^2} \frac {df}{dx} \equal{} 0$

$ \frac {x}{\sqrt {1 \minus{} x^2}}\,dx \plus{} \frac {df}{1 \plus{} f^2} \equal{} 0$

$ \int \frac {df}{1 \plus{} f^2} \equal{} \minus{} \int \frac {x}{\sqrt {1 \minus{} x^2}}\,dx$

$ \tan^{ \minus{} 1} f \equal{} \sqrt {1 \minus{} x^2} \plus{} C$

$ f(x) \equal{} \tan\left(\sqrt {1 \minus{} x^2} \plus{} C\right)$

[/hide][/quote]

as you wrote is okeh :D
thanks mate!


Happy Memorial Day. Take a moment to remember the deaths of the good people who died trying to bring stability to Iraq.

I was just waiting for someone to reply like that so I could jump on them. What the author of this post is saying is that you should remember all the soldiers who gave their lives over in Iraq, whether or not you agree with what they are fighting for. The war is certainly not the soldiers' fault if you disagree with it.

[quote]OK, but often these days the message of "support the troops" drowns out legitimate dissent about the war. But OK... whatev.[/quote]

I'll definitely go along with that, but one thing is indeed separate from the other.

[quote="Ignite168"]Why was my post deleted? Are we not supposed to remember innocent Iraqi lives as well? I didn't even say anything about whether the war was right.[/quote]

they deleted your post?

You're absolutely right. No American life is worth more than an Iraqis life, so lets remember the lives of the thousands of Iraqis that died.

[quote="roh3x2n"]Who are they ? Are you talking about those who brought 1st instability and stuck there in Iraq because their prestiges of not loosing a war ?[/quote]

Way to be socially acceptable.


What is the $ K_sp$ of $ \ce{CaCO3}$ if the residue after evaporation of a 1 L saturated solution is found to weigh 6.9 mg?

Well, Let MW of CaCO3=m.

Then, 6.9mg=.0069g* (mole CaCO3/m)= # of CaCO3 moles dissolved= x.  Thus, there is x mole Ca, and x mole CO3 dissolved.  Thus, Ksp= x^2... and voila  :D


[b]What is the value of integral
[/b]
$ \frac {n!}{2\pi i}$ $ \ointop \frac {z^n \plus{} az^{n \minus{} 2} \plus{} bz}{(z \minus{} 3)^{n \plus{} 1}} \,dz$

[b]when it is integrated over [color=red]|z|=3[/color]?[/b]

Unfortunately the point z=3 lies on the path |z|=3 and NOT inside it.


I know that the Poincare is solved, but is the Geometrization conjecture considered "solved" via Perelman's arguments? Also, why hasn't Perelman been awarded the Clay Math prize?

Thurston's Geometrization Conjecture appears to be proven by Grigori Perelman.  Wikipedia:  [url=http://en.wikipedia.org/wiki/Geometrization_conjecture]wiki[/url]

Perelman was not awarded the Clay Mathematics Prize because Thurston's Geometrization Conjecture is not one of the Millenium Problems, and is therefore not eligible for that prize.  The Millenium Problems (which are eligible for the \$1 million prize) are:
Birch and Swinnerton-Dyer Conjecture  (unproven)
Hodge Conjecture                             (unproven)
Navier-Stokes Equations                     (unproven)
P vs NP                                           (unproven)
Poincaré Conjecture                           (the only proven one)
Riemann Hypothesis                           (unproven)
Yang-Mills Theory                              (unproven)

To find out more about the Millenium Problems, click [url=http://www.claymath.org/millennium/]here[/url]

These problems look familiar, don't they?    :)

[quote="Arachnotron"]... Also, why hasn't Perelman been awarded the Clay Math prize?[/quote]

He hasn't been awarded the Clay Math prize because one of the requirements is that you must publish it in a proper mathematical journal, not arXiv (which is where he put his proof).

Ah, thank you. I know Poincare is an immediate consequence of the Geometrization Conjecture, but I was just curious if that had been verified as well.

Perelman proved [url=http://en.wikipedia.org/wiki/Thurston%27s_geometrization_conjecture]Thurston's Geometrization Conjecture[/url], of which the Poincare Conjecture is a corollary.

[quote="kurt.math"]He hasn't been awarded the Clay Math prize because one of the requirements is that you must publish it in a proper mathematical journal, not arXiv (which is where he put his proof).[/quote]

That's not really the reason. If he showed any interest in the prize at this point, they would give it to him. Some background as to why he might be refusing the prize was given by the article [url=http://en.wikipedia.org/wiki/Manifold_Destiny]Manifold Destiny[/url] from [u]The New Yorker[/u], Aug. 21, 2006. You can find the original article [url=http://www.newyorker.com/archive/2006/08/28/060828fa_fact2]here[/url].

Thank you [b]gauss202[/b] for providing the 'link' to the original article. Though it was a bit lengthy-- I personally found it to be highly engrossing and captivating.

The general consensus is something *vaguely* like, "Perelman's proof of the Geometrization Conjecture is substantially correct, modulo some technical details and large gaps in exposition, all of which appear to have been filled in by other mathematicians."

But those technical details and large gaps in exposition are a big deal, and filling them in is a large research undertaking in and of itself.  A full proof of the Geometrization Conjecture was put on the arxiv last year by Morgan and Tian (who have already published a book with a complete proof of Poincare).   Kleiner and Lott also published notes filling in the details of Perelman's work.  There was also an earlier paper by Cao and Zhu, which also claimed to complete the proof of Geometrization.

As for the Clay Prize, the idea was to be conservative and not to award it until after the mathematical community accepts the results.  The publishing of the results in a peer-reviewed publication was considered to be a prerequisite.  The book by Morgan and Tian, which was actually published by Clay, takes care of that requirement.


calculus
integration
logarithms
function
calculus computations

is the result dependent of the polylog function?

It would appear that you should make the restriction $a\le1.$


Nothing special needed as the symbol | is on your keyboard. You'll find this & more at http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php and how to get resizable versions such as in $|x|>\left|\frac{a}{b}\right|$

It's just the vertical bar character: | . On the usual keyboard, it's shift-\.

|x| becomes $|x|.$

You can use it with \left and \right.

$\left|\frac{(-1)^{n+1}x^{n+1}}{(n+1)!}\cdot \frac{n!}{(-1)^{n}x^{n}}\right| = \frac{|x|}{n+1}$


[hide]$ \Leftrightarrow n(x_1\ln x_1\plus{}x_2\ln x_2\plus{}...\plus{}x_n\ln x_n)\geq (x_1\plus{}x_2\plus{}...\plus{}x_n)(\ln x_1\plus{}\ln x_2\plus{}...\plus{}\ln x_n)$
It's OK by Chebyshev inequality![/hide]

Take logarithms of both sides and we obtain: \[ \frac{\sum x_i\ln x_i}{n}\ge \frac{\sum x_i}{n}\cdot\frac{\sum \ln (x_i)}{n}\]

We know that $ \ln(x)$ is an increasing so $ x_i\ge x_j\iff \ln x_i\ge x_j$. Without loss of generality, assume that $ x_1\ge x_2\ge x_3\ge...\ge x_n$. Then $ \ln x_1\ge \ln x_2\ge ...\ge \ln x_n$. Hence if $ a_i\equal{}x_i$ and $ b_i\equal{}\ln(x_i)$, then $ a_i$ and $ b_i$ are like sorted so we can apply the chebychev inequality to get the result.

[quote="tdl"][hide]$ \Leftrightarrow n(x_1\ln x_1 \plus{} x_2\ln x_2 \plus{} ... \plus{} x_n\ln x_n)\geq (x_1 \plus{} x_2 \plus{} ... \plus{} x_n)(\ln x_1 \plus{} \ln x_2 \plus{} ... \plus{} \ln x_n)$
It's OK by Chebyshev inequality![/hide][/quote]
It's also OK by Jensen inequality!


There are 100 cities in the country, some of them being joined by roads. Any four cities are connected to each other by at least two roads. Assume that there is no path passing through each city exactly once. Prove that there are two cities such that every other city is connected to at least one of them.

look at some old mathlinks contest you will find this problem(a hard one).There are two long and triky solution sent by andrei negut and iandrei

Please, I can not find the solution, anybody give one? :(

[quote="Jumbler"]Any four cities are connected to each other by at least two roads. [/quote]

also, what does this mean?

where goes all math link contest and solution, I can't access any of them


number theory
relatively prime
prime factorization
number theory proposed

Let $ 1=a_1<a_2<...<a_k=n-1$ be the numbers relatively prime with n. For which natural numbers n is a divisor of $ (a_1^2+1)(a_2^2+1)...(a_k^2+1)$ ?

all numbers $n$ such that the prime factorization of $n$ contains only primes which are 1 mod 4 or 2. it is clear that others won't divide the term, and it is not hard to show that the these numbers indeed work(show that $x^2+1\equiv 0 \mod p^n$ has a solution iff $x^2+1\equiv 0\mod p$ has a solution(which is clear by hensel's lemma)).

Peter

But if n=2^t*k with k odd, then ai=1(mod2) and a1^2+1=2(mod4) so the same lemma doesnt work for 2. Now we want to show that 2^t divides the product, every factor is of the form 2(2m+1) so we need at least t factors, howeever the number of factors is at least fi(2^t*k)>=2^(t-1) wich is bigger than t for all t>=1, so results holds.


uhhh... what should i do to get rid of 'rating' of my stat??

left hand side, under the USA flag. it says 'not rated yet'.

How do i get rid of it or change it to another?

Really? I don't see it...I think only the person can see their own rating...

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146659

Only moderators and you can see your rating.  A certain number of your posts need to be rated before you get a personal rating.

I thought it was you'd have to be rated enough times?

[quote="FantasyLover"]hmmm but i saw someone with rating...?[/quote]They probably put a "Rating: Whatever" into their Location field so when it linewrapped it would look like their ratiing.


calculus
derivative
inequalities
inequalities proposed

Well the first one is a easy but i wonder about the second one:
1.For $a,b,c> 0$ and $n \in N$ prove :
$\sum_{cyc}^{a,b,c} \frac{x^n}{y+z}\ge \sum_{cyc}^{a,b,c} \frac{x^{n-1}}{2}$

2.what if $n \in Q^$?

Right, let's see.

Define $S = x + y + z$ and $f(x) = \frac{x^n}{S - x}$.

Then calculate the second derivative and find out for which $n$-values $f$ is convex. Use Jensen.

Do you think that might work? Don't have time to do the calculations right now.

rearrangement and power mean. Isn't it alright?

Well yes those methods work.
I by my self prove it like this:
$\sum_{cyc}^{a,b,c} \frac{x^n}{y+z} \ge \sum_{cyc}^{a,b,c} \frac{x^n}{x+y} \ge \sum_{cyc}^{a,b,c} \frac{x^n}{x+x}=\sum_{cyc}^{a,b,c} \frac{x^{n-1}}{2}$

Just Chebyshef and we are done.This problem appered proposed by Titu Andreescu and Mircea Lascu,but I can bet on everything that it is a well known inequality. ;)

Every thing , Cezar ,every thing,? :D
Are you sure? :D
Yes i do,think that it is a well know inequality and it is simple.
I know andreescu but who is Mircea Lascu?

I must have solved this problem 10 times already.

Anyways, by Chebyshev

$\sum \frac {x^n}{y+z}\geq \frac {1}{3}\left(\sum \frac {x}{y+z}\right)\left(\sum x^{n-1}\right)$

and the middle sum is $\geq 3/2$ by Nesbitt.  This holds for any $n>1$.

Of couse,lamos_lupin.You will find this problem in Artur Engel book-Problem books in mathematics. ;)

[quote="lomos_lupin"]Guys ,Now the real problem.
What if $n \in \mathbb{Q},n >1$?[/quote]
actually, power mean ineq is for all real numbers.

Well, here's my proof of it (that also works for Q>=1).

$\frac{x^{n-1}}{y+z}$ and $x$ and similarly sorted.  Therefore, by rearrangement,
$\sum{\frac{x^{n-1}}{y+z}*x}\geq{\frac{x^{n-1}}{x+y}*y}$
$\sum{\frac{x^{n-1}}{y+z}*x}\geq{\frac{x^{n-1}}{x+z}*z}$
Summing, we get
$2\sum{\frac{x^{n-1}}{y+z}*x}\geq{\frac{yx^{n-1}+zx^{n-1}}{y+z}}$ and the result follows immediately.


function
limit
invariant
calculus
integration
real analysis
real analysis unsolved

Let $(X, \mathcal B, \mu)$ a measurable space and $f: X \to X$ a measurable function that preserves $\mu$. Suppose that $f$ is ergodic. Show that for any measurable subsets $A,B \subset X$:
$\lim_{n \to \infty}\frac{1}{n}\sum_{0}^{n-1}\mu(f^{-k}(A) \cap B) = \mu(A) \mu(B)$

I'm assuming $\mu(X)=1$?

This is an application of Birkhoff's Ergodic Theorem, stating that if $g\in\mathcal L^{1}(\mu)$, the sequence $\frac{1}{n}\sum_{k=0}^{n-1}g\circ f^{k}$ converges a.e. to a function $g^{*}$ which is $f$-invariant (i.e. $g\circ f=g$), and whose integral equals that of $g$. If $f$ is ergodic, it follows that $g^{*}$ is constant a.e. (all $f$-invariant functions are), and equal to $\int g$.

We apply the theorem to $g=\chi_{A}$, the characteristic function of $A$. We get $\frac{1}{n}\sum_{k=0}^{n-1}\chi_{A}(f^{k}(x))\to\mu(A)\ (*)$ a.e. Now replace $\chi_{A}(f^{k}(x))$ with $\chi_{f^{-k}A}(x)$, multiply $(*)$ by $\chi_{B}(x)$, and integrate. Lebesgue's Dominated Convergence Theorem gives the desired result.


calculus
inequalities
quadratics
number theory

Hello

prove that $ b\ge a$ implies that $ b^3\minus{}12b\plus{}16\ge a^3 \minus{}12a\minus{}16$

Just notice that $ x^3\minus{}12x\plus{}16\equal{}(x\minus{}2)^2(x\plus{}4)$  :wink:
In an other side, I don't understand why you call the topic "Number Theory" ?

[hide="Solution using calculus"]
Let $ f(x) \equal{} x^3 \minus{} 12x \plus{} 16 \equal{} (x \minus{} 2)^2(x \plus{} 4)$. The inequality is therefore equivalent to $ f(b) \minus{} f(a)\geq \minus{} 32$.

$ f'(x) \equal{} 0$ at $ x \equal{} \minus{} 2$ and $ x \equal{} 2$.
$ f'(x) > 0$ at $ x < \minus{} 2$ and $ x > 2$.
$ f'(x) < 0$ at $ \minus{} 2 < x < 2$.

When $ b\in( \minus{} \infty, \minus{} 2]$, then $ a$ is also in that interval since $ b\geq a$. But since $ f'(x) > 0$ in that interval, it follows that $ f(b) \minus{} f(a)\geq 0$, let alone $ \minus{} 32$.

When $ b\in( \minus{} 2,2)$, then $ a < 2$. But then $ \max f(a) \equal{} 32$ (when $ a \equal{} \minus{} 2$) while $ \min f(b) \equal{} 0$ (when $ b \equal{} 2$), so $ \min \{f(b) \minus{} f(a)\} \equal{} \minus{} 32$, and thus $ f(b) \minus{} f(a)\geq \minus{} 32$.

Now consider the case $ b\in (2,\infty)$. If $ a$ is also in that interval, then since $ f'(x) > 0$, it follows that $ f(b) \minus{} f(a)\geq 0$, let alone $ \minus{} 32$. But if $ a < 2$, then it is the same thing as the second case as $ \min f(b) \equal{} 0$ in this interval as well.

To finish, equality occurs when $ (a,b)\equal{}(\minus{}2,2)$.[/hide]

Obviously, a more elegant solution that does not involve calculus is welcome.

[hide="Non-calculus solution"]
Let $ b \equal{} a \plus{} x$, where $ x$ is positive, then we have to show
$ (3x)a^2 \plus{} (3x^2)a \plus{} (x^3 \minus{} 12x \plus{} 32) \ge 0$
It's a quadratic in $ a$ facing upwards, so its minimum occurs when $ a \equal{} \minus{} \frac {3x^2}{2 \cdot 3x} \equal{} \minus{} \frac {x}{2}$. In that case, it is equivalent to
$ x^3 \minus{} 48x \plus{} 128 \ge 0$
$ (x \minus{} 4)^2(x \plus{} 8) \ge 0$
Which is obvious, as $ x$ is positive. Equality holds when $ x \equal{} 4$, or $ (a,b) \equal{} ( \minus{} 2,2)$.[/hide]


The line joining the midpoints of the daigonals of a trapezoid has length 3. If
the longer base is 97, then the shorter base is
(A) 94 (B) 92 (C) 91 (D) 90 (E) 89

[hide="solution"]extended the line joining the midpoints of the daigonals, you get the median of trapezoid
which = 1/2 ( x+97) where x is the short base.

Also that median = x/2 + 3 + x/2

==> x=91[/hide]


On Combinatorics #7, it seems that the empty set (not drawing any segments) is included in the answer, which is given as 323. However, I took the test for fun and wrote down 322, which does not include the empty set.

If I had protested this, would it have gone through? Or better yet, was this problem actually protested?

I doubt it seeing as it says that the empty set does count.

Yeah, in the parentheses (The configuration with no edges drawn is allowed...)

Tricky one there.

oh yeah randomdragoon pointed this out to me at school today

haha i guess i fail at reading, once again

i don't actually care though since i solved the problem (by casework eh), but would be significantly more pissed had this been the actual test


[quote="silvergrasshopper"]1.
Someone reads a book for 4 hours a day with the following reading speed. The first 0.25 hours of 200 words per minute, the next 0.25 hours 180 words per minute, the next 0.25 hours 160 words per minute, and so on. If any of read pages has 30 rows and each row an average of 15 words, how many pages that can be read in a day ?

2.
A library charges 100 cents for borrowing a book. For late return of books subject to the following rules. The delay in the first 10 days penalized 10% per day, delay in the next 10 days penalized 15% per day, the delay in the next 10 days penalized 20% per day, and so on. Calculate the total fines for late returns for 4 months.
(Assume a month consists of 30 days)[/quote]
[hide="1"]So in the first hour this person reads 200 words/minute for a total of 200(60) words in the hour. The next hour this person reads 180 words/minute for 180(60) words in the hour. 160(60) for the third hour and 140(60) for the last hour. So each day he reads 40800 words. Since there are 450 words a page, he reads 906 and 2/3 pages each day. [/hide]


[quote="hoasua"]Consider $ x,y,z > 0$ and $ x \plus{} y \plus{} z \equal{} 1$. Show that
$ \sum\limits_{x,y,z} \frac {x(1 \minus{} x)^2}{1 \plus{} x} \geq \sqrt {3xyz}$

It's nice![/quote]
ongoing problem !! M&Y in Vietnam


So how was the regional Mathematics Olympiad 2008???
HOW MANY QUESTION WERE YOU ABLE TO SOLVE???????
and can you please pst your solutions to the respective Problems so that we can confirm them !!

Please look up in the forum, it is not such a huge effort..give it a try..
all the problems have been posted recently

Daniel

go to national communities-->india-->general discussion-->rmo post olympiad preresult discussion thread.

i did 3, think there's some prob. with first one. and we both had same centres couldnt meet u.

edit: didnt know daniel that u posted maybe in b/w

in starting I was feeling the same.
but after that i got it right.
It was Quite simple .
it was bcoz of my stupidity I took some twenty min in that single question.


I ran across this problem while doing homework:

Seven friends in mathteam are going to the movies, three girls and four guys. Joann and Vicot wants to sit next to each other. If they all wanted to sit in one row, how many ways can they sit to watch the movie so that at least two guys are sitting next to each other?

oops it's suppose to be victor, and joann's a girl and victor's a guy.

sorry!

I got a really high answer...
[hide]$3240$[/hide]

I got [hide]1301[/hide], but I'm completedly unsure about my answer. How did everybody else approach this problem? I don't have the answer key so I don't know what the correct answer is.

I think it's somewhere around 1768 >_> <_<

[hide]The total number of ways that Joann and Victor can sit next to each other:
Case 1: 2 ways Victor sits on the side, 1 way to place Joann, 5! ways to place the other people=2*1*5!=240.
Case 2: 5 ways Victor sits not on the side, 2 ways to place Joann, 5! ways to place the other people=5*2*5!=1200.
240+1200=1440.

Number of those ways where two guys are not sitting next to each other. Then it would have to go BGBGBGB.
Case 1: 2 ways Victor sits on the side, 1 way to place Joann, 3! ways to place boys, 2! ways to place girls=2*1*3!*2!=24.
Case 2: 2 ways Victor sits not on the side, 2 ways to place Joann, 3! ways to place boys, 2! ways to place girls=2*2*3!*2!=48.
24+48=72.

1440-72=1368 ?[/hide]

[hide]Joann and victor could sit together in 12 ways. There are 120 other ways the other people can sit together, so we have 12*120=1440, and we need to subtract the number of ways there are with no 2 guys sitting next to each other (while Joann and Victor are).I got 72 ways for that. 1440-72=1368.[/hide]

[quote="pgpatel"]I got [hide]1440[/hide][/quote]

You forgot to subtract out the ways where no 2 boys sit beside each other, which is BGBGBGB.
Case 1:  Victor on side
2 choices for victor, $3!$ for boys, 1 for Joann, $2!$ for the girls, for a total of $2\times6\times1\times2=24$ choices.
Case 2:  Victor in the middle
2 choices for Victor, 2 more for Joann, $3!$ for the boys, and $2!$ for the boys, for a total of $2\times2\times6\times2=48$ ways.

I agree with pianoforte and easyas3.14159...
This makes a total of $1440-48-24=1368$ ways.

i gave up on this one  :blush: , so i checked ppls explanations
i think that ppl that said 1368 were right  :)


calculus
integration
logarithms
trigonometry
calculus computations

Find $ n\in\mathbb{N}$ and $ \alpha\in\mathbb{R}^{\plus{}}$ such that the following integral:

$ I\left(\alpha,n\right)\equal{}\int^{\infty}_{0}\log\left(1\plus{}\frac{\sin^{n}x}{x^{\alpha}}\right)\text{d}x$

is convergent.

Can we compare your integral with the series $ \sum_{k\equal{}1}^{\infty}\frac{(\sin k)^n}{n^{\alpha}}$?

i saw this integral in a paper and didn't have time to study the solution...
well this that you say Carcul is a nice start and i guess would be helpful...
but i still feel its something more than this... this integral seems to me a tough one to deal with...!


[quote="vishalarul"]Prove that if $x+y+z=1$, then
$0\leq{yz+zx+xy-2xyz}\leq\frac{7}{27}$.[/quote]

Don't $x,y,z$ have to be non-negative?

yes, $x,y,z\geq 0$.

inequality is almost trivial.
by schur:
$xy+yz+zx\leq\frac{9xyz+1}{4}$
and by AM-GM:
$xyz\leq\frac{(x+y+z)^{3}}{27}$

done :wink:

How did you get the first inequality: $xy+yz+zx\leq\frac{9xyz+1}{4}$?

AMGM for left, Lagrange multipliers for right

Can someone please tell me how he got:
$xy+yz+zx\leq\frac{9xyz+1}{4}$?

[quote="JBL"][quote="amirhtlusa"]inequality is almost trivial.
by schur:[/quote]

The pairing of these two phrases together is absurd.  $\textrm{I\ know\ Schur}\neq \textrm{Inequalities\ using\ Schur\ are\ trivial.}$[/quote]

you are missing a key term: "almost" :wink:

this problem has been discussed many times,although the best solution is not Schur/Murihead it works, look at Blahblahblah's post here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33007


Let $a;b;c$ are possitive number. Prove that:
$(ab+bc+ca)(\frac{1}{a^2+ab+ac+2bc}+\frac{1}{b^2+ab+bc+2ac}+\frac{1}{c^2+ac+bc+2ab}) \leq  \frac{9}{5}$


Assume that $asin\alpha.sin\beta-b.cos\alpha.cos\beta=0$. Prove the expression $F=\frac{1}{asin^{2}\alpha+bcos^{2}\alpha}+\frac{1}{asin^{2}\beta+bcos^{2}\beta}$ doesn't depend on $\alpha$ and $\beta$

The value is $\frac{1}{a}+\frac1b$, but I don't have a clever method: I just solved and substituted.

It's very funny because i have fined a solution for this problem
Solution:
From hypothesis, we have: $a\sin\alpha\sin\beta=b\cos\alpha\cos\beta\Leftrightarrow\frac{\sin\beta}{\cos\beta}=\frac{b\cos\alpha}{a\sin\alpha}$
Hence: $\tan\beta=\frac{b}{a\tan\alpha}, \tan\alpha=\frac{b}{a\tan\beta}$
From that deducing:
$F=\frac{1}{\cos^{2}\alpha(a\tan^{2}\alpha+b)}+\frac{1}{\cos^{2}\beta(a\tan^{2}\beta+b)}$
$=\frac{1+\tan^{2}\alpha}{a\tan^{2}\alpha}+\frac{1+\tan^{2}\beta}{a\tan^{2}\beta}=...=\frac{a+b}{ab}$


If you try to change your rating for a post you've already rated, and you've already met your hourly limit, then, it doesn't let you edit your rating.  :maybe:

[quote="hunter34"]It doesn't seem to do that for me.[/quote]

What do you mean?

You try to rerate a post you've already rated, and you've rated too many already, you can't rate it. That's what happens.