Find all the function $ f,g,h: \mathbb{Q}\rightarrow \mathbb{R}$ such that :

$ f(x)\plus{}f(y)\plus{}f(z)\equal{}g(x\plus{}y\plus{}z)\plus{}h(x\minus{}y)\plus{}h(y\minus{}z)\plus{}h(z\minus{}x)\ ,\ (\forall)x,y,z\in \mathbb{R}$

$ P(x,x,x): g(3x)\equal{}3f(x)\minus{}3h(0)$ (F1)
Let $ H(x)\equal{}h(x)\minus{}h(0)$, we have
$ f(x)\plus{}f(y)\plus{}f(z)\minus{}3f(\frac{x\plus{}y\plus{}z}3)\equal{}H(x\minus{}y)\plus{}H(y\minus{}z)\plus{}H(z\minus{}x)$. (F2)
WLOG, we can assume $ f(0)\equal{}0$, otherwise replace $ f(x)$ with $ f(x)\minus{}f(0)$.
Let $ x,y,z$ increase by $ c$, the RHS does not change. So we have
$ f(x)\plus{}f(y)\plus{}f(z)\minus{}3f(\frac{x\plus{}y\plus{}z}3)\\\equal{}f(x\plus{}c)\plus{}f(y\plus{}c)\plus{}f(z\plus{}c)\minus{}3f(\frac{x\plus{}y\plus{}z}3\plus{}c)$
Define $ F_c(x)\equal{}f(x\plus{}c)\minus{}f(x)$, we have
$ F_c(x)\plus{}F_c(y)\plus{}F_c(z)\equal{}3F_c(\frac{x\plus{}y\plus{}z}3)$
Replace $ y,z$ with $ y\plus{}z,0$ above, we get
$ F_c(y\plus{}z)\plus{}F_c(0)\equal{}F_c(y)\plus{}F_c(z)$.
So $ F_c(x)\minus{}F_c(0)$ satisfies Cauchy equation and we have
$ F_c(x)\minus{}F_c(0)\equal{}x(F_c(1)\minus{}F_c(0))$, or
$ f(x\plus{}c)\minus{}f(x)\minus{}f(c)\equal{}x(f(1\plus{}c)\minus{}f(1)\minus{}f(c))$ (F3)
Switch $ x,c$ above,
$ \frac{f(1\plus{}c)\minus{}f(1)\minus{}f(c)}c\equal{}\frac{f(1\plus{}x)\minus{}f(1)\minus{}f(x)}x,\forall c,x\ne0$.
Hence, $ \frac{f(1\plus{}x)\minus{}f(1)\minus{}f(x)}x\equal{}D,\forall x\ne0$, where $ D$ is constant.
or $ f(1\plus{}x)\minus{}f(1)\minus{}f(x)\equal{}Dx,\forall x$
Plug in (F3),
$ f(x\plus{}c)\minus{}f(x)\minus{}f(c)\equal{}Dxc,\forall x,c$.
Obviously $ f(x)\minus{}\frac D2x^2$ satisfies Cauchy equation and
$ f(x)\equal{}\frac D2x^2\plus{}Ex$
Plug in (F2)
$ H(x\minus{}y)\plus{}H(y\minus{}z)\plus{}H(z\minus{}x)\equal{}\frac D3(x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}yz\minus{}xz)$.
Let $ a\equal{}x\minus{}y$ and $ b\equal{}y\minus{}z$, we see
$ H(a)\plus{}H(b)\plus{}H(\minus{}a\minus{}b)\equal{}\frac D6(a^2\plus{}b^2\plus{}(a\plus{}b)^2)$
Let $ T(x)\equal{}H(x)\minus{}\frac D6x^2$, then
$ T(a)\plus{}T(b)\plus{}T(\minus{}a\minus{}b)\equal{}0$ and $ T(0)\equal{}H(0)\equal{}0$.
Let $ b\equal{}0$ we see $ T$ is odd and so it satisfies Cauchy equation.
$ T(x)\equal{}Fx$ or $ h(x)\equal{}\frac D6x^2\plus{}Fx\plus{}G$.

Conclusion:
$ f\equal{}\frac D2x^2\plus{}Ex\plus{}A$, $ h\equal{}\frac D6x^2\plus{}Fx\plus{}G$, $ g\equal{}\frac D6x^2\plus{}Ex\plus{}3A\minus{}3G$.


Find all subsets A of {1,2,3,...} with $ |A| \geq 2$ ,such that for all $ x,y \in A$,we have that $ \frac {x\plus{}y}{gcd(x,y)} \in A$

I guess a condition ,that $ A$ is finite, is missing.
Clearly $ x \equal{} 2$ is in the set. Assume that $ A$ contains odd numbers. Let $ y$ be the greatest among them. But then $ y \plus{} 2$ is in the set.
Contradiction.
So $ A$ contains only even number and let $ z\neq 2$ be the least of them, but this leads to contradiction, since $ \frac{z\plus{}2}{2}$ would be in the set. So there is not such subsets.


Determine all sequences of real numbers $a_1$, $a_2$, $\ldots$, $a_{1995}$ which satisfy:
\[ 2\sqrt{a_n - (n - 1)} \geq a_{n+1} - (n - 1), \ \mbox{for} \ n = 1, 2, \ldots 1994,  \] and \[ 2\sqrt{a_{1995} - 1994} \geq a_1 + 1.  \]


$ I(k) \equal{} \int_{0}^{\frac {\pi}{2}}log(sin^{2} \theta \plus{} k^{2}cos^{2} \theta)d\theta$
By Leibnitz Differntiation,

$ I'(k) \equal{} 2k \int_{0}^{\frac{\Pi}{2}} \frac {d \theta }{tan^2 \theta \plus{} k^2}$
Now Let $ tan \theta \equal{} t$
Split it into partial fractions and re-integrate with $ k$ as the integrating variable to get the answer.

Seri don't feel too much.

$ I'(k) \equal{} \frac{ \Pi}{1\plus{}k}$
$ I(k) \equal{} \Pi ln(1\plus{}k) \plus{} c$

A Clever substitution: We know that I(1) = 0 from the question. So
$ I(k) \equal{} \Pi ln(1\plus{}k/2)$


Let $ p\neq2$ prime such that $ p\not\equiv 1 (mod 7)$.

Prove:

$ 1^7 \plus{} 2^7 \plus{} 3^7 \plus{} .... \plus{} (p \minus{} 1)^7 \equiv 0 (mod p)$

Use the existence of a primitive root modulo $ p$.

Thanks.

The condition on $ p$ is wierd. I think you can replace $ 7$ by an integer $ n$ such that $ p\minus{}1 \nmid n$.

To use the given hint, consider the sets $ \{ 1, \cdots , p\minus{}1 \}$ and $ \{ r , r^2, \cdots, r^{p\minus{}1} \}$, where $ r$ is a primitive root.

Alternatively, consider the sets $ \{ 1, \cdots , p\minus{}1 \}$ and $ \{ a ,2a, \cdots, a(p\minus{}1) \}$, for an appropriate $ a$ (similar to Fermat's proof)


calculus
integration
function
quadratics
number theory unsolved
number theory

Find all non negative integral solutions of the equation :

$ a^3  \equal{} a(3b\plus{}1) \plus{} 3T(b)$

where, T(n) is the nth triangular number.

what do you mean?
n is fixed number or not

T(n) is the function for the nth triangle number.

In closed form, T(n)=(n²+n)/2

So the OP is asking for all non-negative integral solutions of the equation

a³=a(3b+1) + 1.5(b²+b)

Solving as quadratic towards $ b$ we get $ D\equal{}12a\left(a\plus{}1\right)\left(2a\plus{}1\right)\plus{}9$ which must be perfect square. Let $ 12a\left(a\plus{}1\right)\left(2a\plus{}1\right)\plus{}9\equal{}m^{2}$. Then we have $ 12a\left(a\plus{}1\right)\left(2a\plus{}1\right)\equal{}\left(m\plus{}3\right)\left(m\minus{}3\right)$ and since $ \left(a,a\plus{}1\right)\equal{}1$, $ \left(a,2a\plus{}1\right)\equal{}1$ and $ \left(a\plus{}1,2a\plus{}1\right)\equal{}1$ we have some systems to solve.

[quote="SAPOSTO"] we have some systems to solve.[/quote]

sorry but what kind of equations you are talking about.


superior algebra
superior algebra unsolved


There is no message board for Intermediate Algebra class I just enrolled.

Perhaps I should wait a little or ?

I'm asking this because there was one when I enrolled in AIME Problem Seminar...

There is one - they probably just forgot to put your name on? *shrugs*

There will be soon - we run a little script periodically to give new sign-ups message board access to their classes.


[quote="Ihatepie"]For example, say you have the fraction (sqrt5  +  7)/ sqrt 5

would that be simplified, or would it be 1 + 7sqrt5/5?[/quote]
You mean
$ \frac{7\plus{}\sqrt{5}}{\sqrt{5}}$?
Well you can't have a radical in the denominator, so it would just be
$ \frac{5\plus{}7\sqrt{5}}{5}$

[quote="Ihatepie"]For example, say you have the fraction (sqrt5  +  7)/ sqrt 5

would that be simplified, or would it be 1 + 7sqrt5/5?[/quote]

In general, you want to clear the radical from the denominator, so your second answer, \[ \frac{\sqrt{5}\plus{}7}{\sqrt{5}}\Rightarrow\frac{\sqrt{5}\plus{}7}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}}\Rightarrow\frac{5\plus{}7\sqrt{5}}{5} \Rightarrow \boxed{1 \plus{} \frac{7\sqrt{5}}{5}} \] is correct.

oops, I didn't mean that, srry

I meant something like (5 + sqrt3)/5

would that be simplified, or would 1 + sqrt3/5 be simplified? or would both work?

Ihatepie is just basically asking which is acceptable as a common fraction. I am honestly not sure either... :huh:

I think that both would work, unless they specifically ask that it be a "common fraction in simplest radical form", in which case you should use the first.

i think its usually $ \frac{something}{somethingelse}$ if otherwise, they might say express in teh form of a+bsqrtc where a is blah and b is blah.

Usually, you either want
$ \frac{a\plus{}b\sqrt{c}}{d}$
Or
$ a\plus{}b\sqrt{c}$
Personally, I wouldn't go with $ a\plus{}\frac{b\sqrt{c}}{d}$
But I think they're all acceptable.

look at what they are asking for
generally in mathcounts format, they will ask for common fraction
but if they ask for something obscure like "mixed number in simplest radical form" or something, just do what they say and you will be fine :)


I've read an inequalities book online recently.
There are many problems in it,but I don't have the solution
brothers and sisters,who can help me?
I WANT THE ANSWERS~

I think maybe someone can help me to look for the answers online?
The problems in this book seem very difficult~

[quote="Apollo_Mars"]
The problems in this book seem very difficult~[/quote]
Which problem? Give number of the problem.

[quote="Apollo_Mars"]Maybe all of them...[/quote]
For all I don't have time. I am sorry.

Could you post the next park of this chapter(wchich contains solutions of this problems) in pdf...?

sorry for my low quality of english... :maybe:

:|
I don't have the answers so I ask for help here~~


Is the one-point compactification of $ \left[0,1\right]$ connected?

Huh, no... in this case the set consisting of the added ideal point is both open and closed.


Ok, I know how to work these problems, in radians, but the directions say, "In each case convert radian measure to degree measure." However, the answers are going to be either x, y, 1/x, etc. (trig functions) of the 16 points around the Unit Circle, so would it even matter if it was in degress or radians? Because it is just the coordinates.

Ok, so, the first problem:

sin -pi / 6         (sin negative pi over six)

I get -1/2, but that is in radians, right? So how do I convert to degrees?

[quote="cooly291"]Ok, I know how to work these problems, in radians, but the directions say, "In each case convert radian measure to degree measure."

Ok, so, the first problem:

sin -pi / 6         (sin negative pi over six)

I get -1/2, but that is in radians, right? So how do I convert to degrees?[/quote]

The sin of a value is a ratio of two lengths.  It has no units.

Converting radians to degrees is done on the argument, not on the result.

In this case, $-\frac{\pi}{6}=-30^{\circ}$

Generally, $\pi = 180^{\circ}$.

So, what, does it end up being sin(-30) to solve?

BTW, my math teacher expressed very strongly to me that 180 deg. doesn't equal pi, pi lands on 180 degrees

If your math teacher was really cool you would already have stopped using degrees.   :D

We've only been in school for 3 days ^.^

Ok, I know what the radians equal in degrees, so, for this problem, I would simply do sin(-30) ? since -pi/6 in radians = -30 degress?

hmm, 1) that's hard to start school that early( i start on sept. 5)
        2) Just assume pi is 180 degrees
        3) Just tell your math teacher to just let it be and let you solve the math problems your way. (Make sure u have the right answer, it would be awefully degrading if you got it wrong.) :lol:

Besides, how would anyone ever know if you didn't do it the same exact way as it was told to do in the problem? :D  I do that a lot with my math teacher :lol:

Yes.  That happens too.

Talk with your teacher and try to compromise.

In my school we use radians.

In some cases they specify when degrees are used.


function
advanced fields
advanced fields unsolved

Let $ E^2$ be the closed unit disk on the plane. What is an example of a continuous function $ f: E^2\rightarrow E^2$ such that the only fixed points (we know there exists some by Brouwer's fixed pt. thm.) of $ f$ lie in the boundary of $ E^2$, i.e. lie in $ S^1$?

Any Mobius transformation $ z\mapsto\frac{z\minus{}a}{1\minus{}\bar a z}$ with $ a\ne 0$ will do.


For $a,b,c,d$ positive real numbers prove that

$ \frac{a^3+b^3+c^3}{a+b+c} +\frac{b^3+c^3+d^3}{b+c+d} + \frac{d^3+c^3 +a^3}{d+c+a} +\frac{a^3+b^3+d^3}{a+b+d} \geq a^2+b^2+c^2+d^2 $

It's obvious (for example by Chebyshef) that \[\frac{a^3 + b^3 + c^3}{a + b + c} \geq \frac{a^2 + b^2 + c^2}{3}.\]
Add those 4 inequalities.


Today I thought about how you could make an interesting number theory game. I'm not sure it this will be that interesting, but maybe it can give some ideas for better games.

In this game, I will pick a positive integer greater than or equal to 3. Suppose I pick 3. The next person to post must pick the smallest integer greater than 3 with exactly 3 divisors. That would be 4. Then the next person must pick the smallest integer greater than 4 with exactly 4 divisors. That happens to be 6. The game continues in the same manner.

So i'll start a new round with the number

4

[quote="Ignite168"]6

How is this a game?

And how is it fun? It involves thinking.[/quote]

12. a lot of people on here like thinking.

Was I the only one that used a calculator for this?

72

[quote="drunner2007"]In this game, I will pick a positive integer greater than or equal to 3. Suppose I pick 3. The next person to post must pick the smallest integer greater than 3 with exactly 3 divisors. That would be 4. Then the next person must pick the [b]smallest integer[/b] greater than 4 with exactly 4 divisors. That happens to be 6. The game continues in the same manner.
[/quote]

 :P

Elemennop, are you sure that's the smallest one?

So, should new rules be set?

Maybe like starting a pattern:

1,2,3

and telling people to write the next three terms based on their interpretation.

For example,

1,1/2,1

can be interpreted as a cycle of two terms or as something else maybe... yea...


(1=2) $ P(x)$ is a monic quartic such that $ P(1) \equal{} 7, P(2) \equal{} 8, P(3) \equal{} 9,$ and $ P(4) \equal{} 10$. Calculate $ P(5)$.

$ P(x)\equal{}x\plus{}6\plus{}(x\minus{}1)(x\minus{}2)(x\minus{}3)(x\minus{}4)\ \therefore P(5)\equal{}35$.

[hide="Solution?"]Solve a giant system of equations. Have fun. :P[/hide]
[hide="Solution."]Define $ Q(x) \equal{} (x \minus{} 1)(x \minus{} 2)(x \minus{} 3)(x \minus{} 4)$. You can tell that $ P(x) \equal{} Q(x) \plus{} x \plus{} 6$. Plug in for $ P(5)$: $ 4 \cdot 3 \cdot 2 \cdot 1 \plus{} 5 \plus{} 6 \equal{} \boxed{35}$.[/hide]
Edit: Ahh! People keep beating me!


can someone please answer my question?  :|

[quote="mihail911"]are all orthodiagonal trapezoids also isosceles trapezoids?[/quote]

Nope:

[img]http://i193.photobucket.com/albums/z278/Anaxerzia/Hello.png[/img]


[quote="espark52"]There are positive integers $k$, $n$, and $m$ such that $\frac1k+\frac1n+\frac1m < 1$. What's the smallest possible value of $k+n+m$?[/quote]

[hide]Let $k \le n \le m$. Then $\frac1k \ge \frac1n \ge \frac1m$.

1) $k=2$. If $n=3$, then $m>6$, which results in at least $12$.
If $n=4$, then $m>4$, which results in at least $11$.
Otherwise, $n=m=c>4$ and it results in at least $2+2c$ which is at least 12.

2) $k=3$. If $n=3$, then $m>3$, which results in at least $10$.
Otherwise, $n=m=d>4$ which results in at least $13$.

In other possibilities for $k$ where $k \ge 4$, the minimum possible value of $k+n+m$ is $3k$, which should be at least 12.

So, the minimum possible is $\boxed{10}$ when $k=3$, $n=3$ and $m=4$.[/hide]

[quote="7h3.D3m0n.117"][quote="#H34N1"][hide="Is it"]
$\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\boxed{11}$?[/hide][/quote]

No because

$\frac{1}{2}+\frac{1}{4}+\frac{1}{5}= \frac{19}{20}$ and the original problem says it must be less than[/quote]

Umm.. what were you trying to say? less than ______

[quote="7h3.D3m0n.117"][quote="#H34N1"][hide="Is it"]
$\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\boxed{11}$?[/hide][/quote]

No because

$\frac{1}{2}+\frac{1}{4}+\frac{1}{5}= \frac{19}{20}$ and the original problem says it must be less than[/quote]
No, because 3,3,4 is a smaller solution  :P
edit there was a recently posted problem that had the sum strictly greater than 19/20, that confused probably


For $ f(x)\equal{}x^3\plus{}ax^2\plus{}bx\plus{}b$ , all of the factors of $ f(x)$ are linear with integral coefficients. Find the sum of all $ a$ and $ b$ for which neither a nor b are zero or one.

[hide="solution"]Note that since the factors have integer coefficients, the three roots of the polynomial are all rational.  Also, because it is a monic polynomial, the roots are all integers.  We let these integers be $ \minus{}c, \minus{}d, \minus{}e$. (Why they're all negative will become clear in a moment.)

Thus the polynomial can be written as $ (x\plus{}c)(x\plus{}d)(x\plus{}e)\equal{}x^3\plus{}(c\plus{}d\plus{}e)x^2\plus{}(cd\plus{}de\plus{}ec)x\plus{}cde\equal{}x^3\plus{}ax^2\plus{}bx\plus{}b$.  This means that we need to find all integers so that $ cde\equal{}cd\plus{}de\plus{}ec$.

None of $ c, d, e$ are zero because otherwise $ b$ is 0, which isn't allowed, so we divide both sides by $ cde$ to get $ \frac1c\plus{}\frac1d\plus{}\frac1e\equal{}1$.  Note that if any of $ c, d, e$ is negative, say $ c$, then $ 1\plus{}\left(\frac{1}{\minus{}c}\right)\equal{}\frac1d\plus{}\frac1e$, and we can see that the left side is greater than 1, so one of the two on the right side must be greater than $ \frac12$, but that means that, say, $ d\equal{}1$ and $ e\equal{}\minus{}c$, so $ a\equal{}c\plus{}d\plus{}e\equal{}1$ which isn't allowed.

So we know that $ \frac1c\plus{}\frac1d\plus{}\frac1e\equal{}1$ where all $ c, d, e$ are positive.  It is well known that the only solutions to this equation are $ \{c, d, e\}\equal{}\{2, 3, 6\}, \{2, 4, 4\}, \{3, 3, 3\}$.  These give $ (a, b)\equal{}(11, 36), (10, 32), (9, 27)$ whose pairwise sum is $ \boxed{(30, 65)}$.[/hide]


calculus
integration
trigonometry
function
logarithms
real analysis
real analysis unsolved

Find:
1. $ \int \sqrt [x]{e}x^{ \minus{} 2}\,dx$
2. $ \int arcsin(cosx) \,dx$
3. $ \int \frac {xe^{2x}}{(2x \plus{} 1)^{2}}\,dx$
4. $ \int \frac {x^3e^{x^{2}}}{(x^{2} \plus{} 1)^{2}}\,dx$
5. $ \int ln(a \plus{} (x^{2} \plus{} a^{2} \plus{} b^{2})^{\frac {1}{2}})\,dx$
6. $ \int e^{acosx}cosbx\,dx$
You will spend good time with them. I found them in sosmath.com.

[hide="Number 5"]$ P \equal{} \int \ln(a \plus{} \sqrt{x^2 \plus{} a^2 \plus{} b^2})\,dx \equal{} \underbrace{x \ln(a \plus{} \sqrt{x^2 \plus{} a^2 \plus{} b^2})}_{\equal{} A} \minus{} \int \frac{x^2}{x^2 \plus{} a^2 \plus{} b^2 \plus{} a \sqrt{x^2 \plus{} a^2 \plus{} b^2}}\,dx$.

Now make the substitution $ x \equal{} \sqrt{a^2 \plus{} b^2} \tan t$, and so

$ P \equal{} A \minus{} (a^2 \plus{} b^2) \int \frac{\sin^2t}{\cos^2t(a \cos t \plus{} \sqrt{a^2 \plus{} b^2})}\,dt \equal{} A \minus{} k^2  \underbrace{\int \frac{\sin^2t}{\cos^2t(a \cos t \plus{} k)}}_{\equal{} P'}$.

To compute P' make the substitution $ \tan(t/2) \equal{} u$, and so

$ P' \equal{} 8 \int \frac{u^2}{(1\minus{}u^2)[(k\minus{}a)u^2 \plus{} k \plus{} a]}\,du$

and now this fraction can be "easily" decomposed.
[/hide]

We may rationalize the fraction for the integral after integration by parts for integral 5 and then to split it to easier integrals. Your substitutions are very interesting! Very good ideas!

P.S. -> EDITED -> We may rationalize the fraction for the integral after integration by parts for integral 5 and then to split it to easier integrals. - it is wrong that we will have the desired effect after rationalization...we shouldn't mutliply the two fractions with radicals - instead we must write them as a difference and then after lots of "focuses" of kind 0=1-1 and gymnastics with radicals we will have irrational integrals. Then they may be solved by substitution. It is interesting for me how Carcul discover such kind of trygonometric substitutions?!

[quote="borislav_mirchev"]$ 2\blacktriangleright F\equiv\int\arcsin\ (\cos x)\ \mathrm {dx}\ .$
[color=darkred]You will spend good time with them. I found them in http://www.sosmath.com .[/color][/quote]
[color=darkblue]The identity $ x\in [ \minus{} 1,1]\implies \arcsin x \plus{} \arccos x \equal{} \frac {\pi}{2}$ is well-known.

Particularly, $ x\in\mathrm R\implies$ $ \arcsin (\cos x) \equal{} \frac {\pi}{2} \minus{} \arccos (\cos x)\ .$ Prove easily that

$ \left\{\begin{array}{c} n\in\mathrm Z \\
 \\
n\pi\le x < (n \plus{} 1)\pi\end{array}\right\|$ $ \implies$ $ \arccos (\cos x) \equal{} ( \minus{} 1)^n\left(x \minus{} 2\left[\frac {n \plus{} 1}{2}\right]\pi\right)\ .$

$ \left\{\begin{array}{c} n\in\mathrm Z \\
 \\
n\pi\le x < (n \plus{} 1)\pi\end{array}\right\|$ $ \implies$ $ F(x) \equal{} \frac {\pi x}{2} \minus{} ( \minus{} 1)^n\left(\frac {x^2}{2} \minus{} 2\left[\frac {n \plus{} 1}{2}\right]\pi x\right) \plus{} K_n\ .$

Suppose $ K_0 \equal{} 0$ and must for any $ n\in\mathrm Z$ the condition $ F(n\pi \minus{} 0) \equal{} F(n\pi \plus{} 0)\ .$

Using the easy relation $ \left[\frac n2\right] \plus{} \left[\frac {n \plus{} 1}{2}\right] \equal{} n$ for any $ n\in \mathrm Z$

you can ascertain nicely the "brazing-constants" $ K_n$ for any $ n\in\mathrm Z\ .$

[hide="Brazing-constants."]
$ K_n \equal{} \pi^2\cdot\sum_{k \equal{} 1}^n( \minus{} 1)^{k \minus{} 1}k^2\implies$ $ K_n \equal{} ( \minus{} 1)^{n \minus{} 1}\cdot\frac {n(n \plus{} 1)}{2}\cdot\pi^2$ for any $ n\in \mathrm N\ .$[/hide]
[b]Success, Borislav_Mirchev & Co. ![/b][/color]

I cannot understand your idea in depth...
I saw integral in 2 ways solved.
[hide]  1. by parts with two substitutions.
  2. by some things what exactly is sin, cos, arcsin, arccos...[/hide]
I posted it here because I think it is not very easy and it is interesting.
Why did you use "Borislav_Mirchev & Co."? What does it mean?

[quote="borislav_mirchev"] ... Why did you use "Borislav_Mirchev & Co."? What does it mean?[/quote]
[color=darkred]Sorry, for me and hope and for you,

"[b]Borislav_Mirchev & Co.[/b]" means the all users of this section ![/color]  [color=darkblue]In the my opinion,

the determination of a primitive for $ f(x)\equal{}\arcsin \cos x$ over $ \boxed {\mathcal R}$ doesn't find through parts

because it isn't differentiable in the points $ x_n \equal{} n\pi$ for any $ n\in \mathrm Z\ !$

Prove easily that $ f'_l(n\pi )\equal{}(\minus{}1)^n$ and $ f'_r(n\pi )\equal{}(\minus{}1)^{n\plus{}1}$ for any $ n\in\mathrm Z\ .$[/color]

Virgil Nicula - see this: http://www.sosmath.com/CBB/viewtopic.php?t=1910&highlight=integral
What is your opinion?
"Borislav_Mirchev & Co" - thank you! All the best you too.

About number 6
kunny, I think it is not so difficult as it seems ;-). Try again but be more careful.

[quote="borislav_mirchev"]What is your opinion?[/quote]
[color=darkred][b]In the my opinion, there is wrong ![/b]

Determination of a primitive for a function is no only "[b]technics[/b]" ![/color]

It is very good opinion :-)...but it is the way when you check 500-600 topics in finding good integrals... anyway - it is very interesting integral...I'll check it carefully when I've more time. It is surprising how he is wrong by using two technicues, but ... it is human to err.

To Virgil Nicula:
  I'm no sure if they are correct but I tried to calculate this integral on the following site: http://www.quickmath.com/www02/pages/modules/calculus/index.shtml.
I entered the integral and took a primitive. Then entered the primitive and differentiate it and took the function under integral sign.
  So I think this integral may exist. If not exist we may report for a bug in this software. In the past I was a QA I found many issues in ICQ, in Microsoft Windows Calculator etc... (for example if you try to calculate 0 on power 0 you get 1 in MS Windows Calculator but it is undefined in the reality ;))
What do you think about this?

To all: What about problem 6?

Are you sure that number 6 is right? :maybe:

Yes, kunny I'm sure. I'll send you a link with the solution ;-)

Thank you for your prompt reply. I will solve again.

I have just done! :lol:

7. $ \int e^{a\cos x}\sin x\sin bx\ dx$.

Integral 6 - when I saw the solution seemed to be easier than 4 of the proposed here integrals. I even wondered if to post it. If it was good integral I'm felling happy.
  Kunny, if you use some different method - you may send it. I'm wondering how it is possible for you this integral to be unknown. Maybe you were tired or not in good condition when you started with it.
  To be complete... it is the solution of integral 6.
http://www.sosmath.com/CBB/viewtopic.php?t=5214&highlight=integral
  What do you think about proposed here integrals? Did you like it? Who is the best and where you have lots of troubles? ;-)
  You may try Virgil Nicula's integrals they are very good as it seems or to check some of my topics there are not solved integrals. Numbered ones are most interesting for me but ... who knows I may be wrong.

I have just looked your link, but on reading few lines, I'm dissapointed the solution.
This paper is worth while to 0 point. His paper can not understand the basic of Mathmatics.
Maybe few people could solve completely, that's why I have said this problem is difficult.

What is your solution, I'm interested? What do you think for Problem 2? Virgil Nicula wrote interesting arguments, but I'm not very good in Calculus  :( and I cannot understand them :(. As I said I never repeated myself  :) ... I like simple and genious things ... like me

P.S. kunny I saw a problem posted by you from an exam. It is very good, too.

6) is definitely terribly wrong. If we could do it for the general case, we could do it for $ a\equal{}i$, $ b\equal{}0$, which means we would get an elementary formula for the Bessel function.

Till now I cannot integrate 6. Software I'm using - also. I'm wondering what was kunny's solution, or you were wrong? There is nothing possible for me except to say - it is Bessel function even I don't know who is Bessel.

[quote="borislav_mirchev"]$ 5\blacktriangleright$ $ I \equal{} \int\ln\left(a \plus{} \sqrt {x^{2} \plus{} a^{2} \plus{} b^2}\right)\ \mathrm {dx}\ .$
[color=darkred]You will spend good time with them. I found them in [b] http://www.sosmath.com [/b] .[/color]  [/quote]
[color=darkblue][b]Proof.[/b]I suppose w.l.o.g. $ a\ne 0$ and $ b > 0\ .$ Therefore,

$ I \equal{} \int\ln\frac {x^2 \plus{} b^2}{\sqrt {x^2 \plus{} a^2 \plus{} b^2} \minus{} a}\ \mathrm {dx} \equal{} L \minus{} J$, where

$ L \equal{} \int \ln (x^2 \plus{} b^2)\ \mathrm {dx} \equal{} x\ln (x^2 \plus{} b^2) \minus{} x \plus{} b\arctan\frac xb \plus{} \mathcal C$ and

$ J \equal{} \int\ln\left(\sqrt {x^2 \plus{} a^2 \plus{} b^2} \minus{} a\right)\ \mathrm {dx} \equal{} K\left(\sqrt {x^2 \plus{} a^2 \plus{} b^2} \minus{} a\right) \plus{} \mathcal C$, where

$ \boxed {\begin{array}{c} \boxed {\ t \equal{} \phi (x) \equal{} \sqrt {x^2 \plus{} a^2 \plus{} b^2} \minus{} a\ } \\
 \\
\left(x \equal{} \sqrt {t^2 \plus{} 2at \minus{} b^2}\right) \\
 \\
\phi '(x) \equal{} \frac {x}{\sqrt {x^2 \plus{} a^2 \plus{} b^2}}\end{array}}$ $ K \equal{} \int\frac {t \plus{} a}{\sqrt {t^2 \plus{} 2at \minus{} b^2}}\cdot\ln t\ \mathrm {dt}$ $ \implies$

$ \boxed {\begin{array}{ccc} u(t) \equal{} \ln t & \implies & u'(t) \equal{} \frac 1t \\
 \\
v'(t) \equal{} \frac {t \plus{} a}{\sqrt {t^2 \plus{} 2at \minus{} b^2}} & \implies & v(t) \equal{} \sqrt {t^2 \plus{} 2at \minus{} b^2}\end{array}}$

$ K \equal{} \sqrt {t^2 \plus{} 2at \minus{} b^2}\cdot\ln t \minus{} L$, where $ L \equal{} \int\frac {1}{t\sqrt {t^2 \plus{} 2at \minus{} b^2}}\ \mathrm {dt}$ $ \implies$

$ \boxed {\begin{array}{c} \boxed {\ y \equal{} \psi (t) \equal{} \frac 1t\ } \\
 \\
\left(t \equal{} \frac 1y\right) \\
 \\
\psi '(t) \equal{} \minus{} \frac {1}{t^2}\end{array}}$ $ L \equal{} M\left(\frac 1t\right) \plus{} \mathcal C$, where $ M \equal{} \minus{} \int\frac {1}{\sqrt {1 \plus{} 2ay \minus{} b^2y^2}}\ \mathrm {dy}$ $ \implies$

$ M \equal{} \minus{} \frac 1b\cdot\int\frac {\left(by \minus{} \frac ab\right)'}{\sqrt {\frac {a^2 \plus{} b^2}{b^2} \minus{} \left(by \minus{} \frac ab\right)^2}}\ \mathrm {dy} \equal{}$ $ \minus{} \frac 1b\cdot\arcsin\frac {by \minus{} \frac ab}{\frac {\sqrt {a^2 \plus{} b^2}}{b}} \plus{} \mathcal C$ $ \implies$

$ M \equal{} \minus{} \frac 1b\arcsin\frac {b^2y \minus{} a}{\sqrt {a^2 \plus{} b^2}} \plus{} \mathcal C\ .$

I hope that upper all is correcty ! Now you can return to $ I$ a.s.o.

[b]Remark.[/b] In the exercise $ \blacktriangleright 2\blacktriangleleft$ the blazing-constans are $ K_n \equal{} \pi^2\cdot\sum_{k \equal{} 1}^n( \minus{} 1)^{k \minus{} 1}k^2\implies$ $ K_n \equal{} ( \minus{} 1)^{n \minus{} 1}\cdot\frac {n(n \plus{} 1)}{2}\cdot\pi^2$ for any $ n\in \mathrm N$ (see the upper my reply).[/color]

Virgil Nicula, I'm understanding you are having lots of experience with integrals. I like your problems, posted recently and in particular - twin and district competition.
For me - problem two answer's is: $ \frac{x^{2}}{2} \plus{} x(arcsin(cos(x))) \plus{} C$
If you don't believe differentiate. I don't understand your argumets they may be correct but my theoretical knowledge is not enough.
Problem 6. - I'm sorry for the wrong problem. I'll try to solve it again, but it seems you are right with Bessel's functions.
Thank you for excellent solutions, opinions and interest to this topic. I'm proud I communicate with people like you.

[color=darkblue]Sorry, I don't like http://www.sosmath.com because there are mistakes and nobody correct them !

[b]Definition.[/b] A function $ f$ is [b]antiderivative[/b] ([u]admits a primitive[/u]) [b]on the interval [/b]$ I$

if exists a function $ F$ which is differentiable on $ I$ and for any $ x\in I$, $ F'(x)\equal{}f(x)$ !

The function $ F(x) \equal{} x^{2} \plus{} x(\arcsin (\cos x))$ isn't differentiable over $ \mathcal R$ !

For example, in $ \pi$ ! Try (with the definition or with the fundamental theorems) for to convince !

For any $ x\not\in \pi \mathrm Z$we have $ F'(x) \equal{} 2x \plus{} \arcsin (\cos x) \minus{} x\cdot\frac {\sin x}{|\sin x|}$ ...

[b]Good night and an easy sleep with nice dreams ![/b][/color]

You are right! Their integrals as a whole are at low level. But if you hardly find there are interesting integrals - there are many visitors. mathlinks.ro is my favorite site - there is almost every problem I'm interested is solved or if not - there is something similar who gives me an idea. People are nice and almost all my problems are solved - materials from IMO appeared even before these in official site. Other sites I like are http://www.kalva.demon.co.uk/, http://www.imocompendium.com, http://www.cut-the-knot.com, site with Math Excalibur and some American, Canadian, Russian, Chinese, Vietnamese, and Ukrainian sites. My latest discoveries are http://math10.com and http://sosmath.com. I'm sure there are plenty of good sites but I cannot know everything. I believe in interchange of ideas, not only concurrention.

Good night and have a nice dreams!

P.S. I don't remember this definition. It is said - it is very hard to remember something that you haven't ever ridden ;). Thank you for new knowledge. This even makes the problem better and challenging.


A 0.52 kg rubber ball has a speed of 1.5 m/s at point A and kinetic energy of .585 J.  At point B its speed is 5.299 m/s and kinetic energy of 7.3 J.  What is the total work done on the ball as it moves from A to B?

I need help on this problem.  I have no clue how to work it.
Please help.

I'm not sure about the name of the concepts in english but the work of non conservative forces is the difference between the final energy and the inicial energy
so, you calculate Kinetic energy=(1/2)mass.speed^2 but you dont need to calculate it because it is already done   :what?:  so you just need to write
delta energy=work=7.3-0.585=6.715 J

Off topic, but I really like your signature :).

(By the way you're quite right, it seems like you're interested in physics. It's hard to find too many mathematicians who like physics nowadays...)


The triangular numbers are the numbers 1, 3, 6, 10, ..., where the $ n$th triangular number is the sum of the first $ n$ positive integers. The square numbers are the numbers 1, 4, 9, 16, ..., where the $ k$th square number is the $ k$th positive integer squared. What is the sum of the first three positive numbers that are both triangular numbers and square numbers?

um the numbers are 1,36 and 1225 the first two are easy to find but 1225 takes some brute forcing.


In acute triangle ABC, AC=14 and the length of the altitude from B is 12. A square has two of its vertices on AC and its two other vertices on AB and BC respectively. Find the length of teh side of the sqaure.

[hide="Not sure if right"]Let the square be $ DEFG$ such that $ D$ and $ G$ are on $ \overline{AC}$, and $ E$ and $ F$ are on $ \overline{AB}$ and $ \overline{BC}$ respectively.  Let the side length of the square be $ x$.  Then:
$ [ABC]\equal{}\frac{1}{2}\times14\times12\equal{}84$, $ [BEF]\equal{}\frac{x}{2}(12\minus{}x)$
Since $ \overline{EF}\parallel \overline{AC}$, $ \triangle BEF \sim \triangle BAC$, which means:
$ \frac{[BAC]}{[BEF]}\equal{}\frac{168}{x(12\minus{}x)}\equal{}\left(\frac{AC}{EF}\right)^2\equal{}\frac{196}{x^2}$
Solving for $ x$ gives $ x\equal{}\frac{84}{13}$.
[/hide]

That's right
[hide="My solution"]
Let H be the foot of the altitude from B. Let $ AH \equal{} n$, then $ CH \equal{} 14 \minus{} n$. Let s be the side of the square DEFG, with D and G on AB. Then
$ \frac {n}{12} \equal{} \frac {AD}{s} \implies AD \equal{} \frac {sn}{12}$ and
$ \frac {14 \minus{} n}{12} \equal{} \frac {CG}{s} \implies GC \equal{} \frac {s(14 \minus{} n)}{12}$
We find $ s \equal{} DG \equal{} AC \minus{} AD \minus{} GC \equal{} 14 \minus{} \frac {sn}{12} \minus{} \frac {s(14 \minus{} n)}{12} \equal{} 14 \minus{} \frac {7}{6}s$. Solving, $ s \equal{} \frac {84}{13}$[/hide]
[hide="Better solution"]
Let s be the side of the square DEFG, with D and G on AB.
Since $ \triangle BEF \sim \triangle BAC$, their heights are in equal propotion to their base side, so $ \frac {12 \minus{} s}{s} \equal{} \frac {12}{14} \implies s \equal{} \frac {84}{13}$[/hide]


Hello all,

Anyone know how to make the gaps between \section and previous paragraph more narrow?

This is a sample show how big the gaps are. This is very annoyanced cuz I have to put lots of things on a A4 page  :)

http://www.hn.is.uec.ac.jp/~vuhung/tmp/test-sec.tex
http://www.hn.is.uec.ac.jp/~vuhung/tmp/test-sec.pdf

TIA

[quote="DPatrick"]You can, but it's not simple.
(I told you it wasn't simple. :) )[/quote]
 I will not dissapoint you Patrick :). Here is my solution without anyone's help except you :).

[code]
\renewcommand\section{\@startsection {section}{1}{\z@}%
{-0.3ex \@plus -0.3ex \@minus -.2ex}%
{0.9ex \@plus.4ex}%
{\normalfont\large\bfseries}}
[/code]

I have another question on the same thread( sorry if it vilolates the rules)

How can I replace the \maketitle command. I want \maketitle command without date. A nice link to read would be nice.

Thanks

I assume you could just do \renewcommand{\maketitle}{whatever you want here}

As far as reading, I do it the old fashioned way, with books:

[i]A Guide to LaTeX[/i] (I have the 3rd edition, but the most recent is the 4th I think)

[i]The LaTeX Companion[/i] (I have the new 2nd edition which is light-years better than the first)

[quote="DPatrick"]You can, but it's not simple.

[code]
\makeatletter
\renewcommand{\section}{\@startsection{section}{1}{0pt}%
  {pre-skip}{post-skip}{style}}
\makeatother
[/code]
where you replace "pre-skip" with the (rubber) length to leave between the previous paragraph and the section title, replace "post-skip" with the (rubber) length to leave between the section title and the next paragraph, and replace "style" with the style commands for the section title.

(I told you it wasn't simple. :) )[/quote]

Could someone explain this a bit more? E.g., what are \makeatletter and \makeatother, and what is with the @ in the middle of \@startsection? And in vuhung's solution, what are \z@, \@plus, and \@minus?

[quote="zabelman"]
Could someone explain this a bit more? E.g., what are \makeatletter and \makeatother, and what is with the @ in the middle of \@startsection? And in vuhung's solution, what are \z@, \@plus, and \@minus?[/quote]
He told me it wasn't simple and and it really is. That's why I am reading Latex Companion now :)

LaTeX has a lot of "internal" commands, like \@startsection, which have an @ in them.  Theoretically, these are only used in package or class files.  If you want to use them in a regular tex file, you first need to use the \makeatletter command, so that LaTeX will treat the @ just like another letter.  Then, when you're done using commands with @ in them, you use the \makeatother command to tell LaTeX to go back to treating @ like it usually does, as a special symbol.

Things like
{1ex plus 0.5ex minus 0.5ex}
are "rubber lengths".  These are used for lengths where you want TeX to have some flexibility in how much space it inserts.  1ex is the "optimal" length, but you're telling TeX that it's OK to add or subtract 0.5ex of additional space, for example in order to make pages break more nicely.

I don't know why vuhung used \@plus instead of just plus: all of the examples I have just use plus (I didn't even know that \@plus existed as a command).  Similarly I have no idea what \z@ is.

[quote="Nomendil"][quote="vuhung"]How can I replace the \maketitle command. I want \maketitle command without date. A nice link to read would be nice.
[/quote]

I'd guess simply putting \date{} before \maketitle should do the trick...[/quote]
It worked. Thank you


Let ABC be a triangle. Find the maximum value of
\[\frac{r}{s^{2}}\left [ (l_{a}+l_{b})\sqrt{\cot \frac{\gamma}{2}}+(l_{b}+l_{c})\sqrt{\cot \frac{\alpha}{2}}+(l_{c}+l_{a})\sqrt{\cot \frac{\beta}{2}}\right ]\]

[hide="solution"]

Disregard the $\frac{r}{s^{2}}$ term for now - we aim to work with $E$, the expression inside square brackets, first.

We have from Cauchy-Schwarz Inequality that $\left(\sum_{cyc}\left(l_{a}+l_{b}\right)^{2}\right) \cdot \left(\sum_{cyc}\cot{\frac{\alpha}{2}}\right) \ge E^{2}$

First the cotangents.  We have $\cot{\frac{\alpha}{2}}= \frac{s-a}{r}$, hence
$\sum_{cyc}\cot{\frac{\alpha}{2}}= \frac{3s-\left(a+b+c\right)}{r}= \frac{s}{r}$

Now the sum of squares: AM-GM gives $\sum_{cyc}\left(l_{a}+l_{b}\right)^{2}\le 4\left(a^{2}+b^{2}+c^{2}\right)$.  We claim that $a^{2}+b^{2}+c^{2}\ge 4K\sqrt{3}$ where $K$ is the area of the triangle.  this is a result from IMO 1961/2 that is available elsewhere. We leave it as an exercise to the reader.

Hence we have $E^{2}\le 16K\sqrt{3}\cdot\frac{s}{r}= 16\sqrt{3}\cdot s^{2}\Leftrightarrow E \le 4\sqrt[4]{3}s$

Thus $\frac{rE}{s^{2}}\le 4\sqrt[4]{3}\frac{r}{s}= 4\sqrt[4]{3}\frac{K}{s^{2}}$

We want to find an upper bound on $\frac{K}{s^{2}}$.  Note that $K = \sqrt{s(s-a)(s-b)(s-c)}$.  Hence $\frac{K}{s^{2}}= \sqrt{\frac{(s-a)(s-b)(s-c)}{s^{3}}}$

From AM-GM we have $\frac{s}{3}= \frac{(s-a)+(s-b)+(s-c)}{3}\ge \sqrt[3]{(s-a)(s-b)(s-c)}$, hence rearranging gives us $\frac{(s-a)(s-b)(s-c)}{s^{3}}\le \frac{1}{27}$

thus $\frac{K}{s^{2}}\le \frac{\sqrt{3}}{9}$ and we get that the original expression has a maximum value of $4\cdot 3^{-\frac{5}{4}}$
[/hide]

edit: just caught a major but almost unnoticable error - see if you can find it!  :P

[quote="tjhance"]What are $l_{a},l_{b},l_{c}$?  :blush:[/quote]

angle bisectors.

AM-GM gives:
\[\sum_{cyc}\left(l_{a}+l_{b}\right)^{2}\le 4\left ({l_{a}}^{2}+{l_{b}}^{2}+{l_{c}}^{2}\right)\]
But why
\[\sum_{cyc}\left(l_{a}+l_{b}\right)^{2}\le 4\left (a^{2}+b^{2}+c^{2}\right )\] ?


The Date Game is here! The rules are simple. take the numbers of the date, add them together, divide by 2 and then square it. For example: today is 8/1/09. $ 8 \plus{} 1 \plus{} 0 \plus{} 9 \equal{} 18$,  $ 18 / 2 \equal{} 9$ and $ 9^2 \equal{} 81$. After you do this, add that number to the previous post's total.

We'll start with 81.

Have fun! :)

(As an extra bonus, you can try to find the date in the first ten thousand digits of pi. Try to see how many times you find it. You can then post the number of times you found it. If the date is 8/1/09, then try to find 8109.)

$ 8 \plus{} 3 \plus{} 0 \plus{} 9 \equal{} 20\Rightarrow\frac {20}{2} \equal{} 10\Rightarrow10^2 \equal{} 100\Rightarrow100 \plus{} 81 \equal{} 181$

[quote="meewhee009"][quote="Midnight Moon"]I am confused, if this isn't right then skip it!:

8+1+3+0+9=21 :arrow: 21/2=10.5  :arrow: [b]10.5^2= 110.25+110.25= 220.5[/b][/quote]

Squaring is multiplying the number by itself, example $ 2^2$ is $ 2 \times 2$ which is 4.

$ \[8 \plus{} 1 \plus{} 6 \plus{} 2 \plus{} 9 \equal{} 26$
$ \[\dfrac{26}{2} \equal{} 13$
$ \[13^2 \equal{} 169$
$ \[169 \plus{} 635.5 \equal{} \boxed{804.5}$

Adding 2009 in there :D[/quote]

didn't I do that?


Fie un triunghi ascutitunghic ABC. Construim in exteriorul acestuia triunghiurile dreptunghice BCM, ACN si ABP astfel incat AN=AP, BM=BP si CM=CN. Determinati masura unghiului NAP stiind ca AN=BM+CM.

Unde sunt unghiurile drepte ? banuiesc ca sunt in $M,N,P$ !

$===================================================$
Iata [u]o solutie (simpla)[/u] a acestei frumoase probleme. Fie punctele $U\in (AP)$, $V\in (AN)$ astfel incat $AU=y$ ($PU=z$) si $AV=z$ ($NV=y$). Din congruentele $PBU\equiv MBC\equiv NVC$ rezulta $UB=BC=CV$, $UV\perp UB$, $UV\perp VC$, adica patrulaterul $UBCV$ este un patrat si congruenta $AUV\equiv PBU\Longrightarrow AP\perp AN$.

[u]Observatie.[/u] Notam $AN=AP=x$, $BP=BM=y$, $CM=CN=z\Longrightarrow$
$y^2+z^2=a^2,\ z^2+x^2=b^2,\ x^2+y^2=c^2,\ x^2+y^2+z^2=\frac 12 (a^2+b^2+c^2)\Longrightarrow$
$x^2=\frac 12 (b^2+c^2-a^2),\ y^2=\frac 12 (c^2+a^2-b^2),\ z^2=\frac 12 (a^2+b^2-c^2).$
$x=y+z\Longrightarrow \sqrt{b^2+c^2-a^2}=\sqrt{a^2+c^2-b^2}+\sqrt{a^2+b^2-c^2}\Longleftrightarrow$
$b^2+c^2-3a^2=2\sqrt{(a^2+b^2-c^2)(a^2+c^2-b^2)}\Longleftrightarrow$
$3a^2\le b^2+c^2\ \wedge \ 5a^4-6(b^2+c^2)a^2+(b^2+c^2)^2+4(b^2-c^2)^2=0.$
$\Delta '=9(b^2+c^2)^2-5(b^2+c^2)^2-20(b^2-c^2)^2\ge0\Longleftrightarrow (b^2+c^2)^2\ge 5(b^2-c^2)^2\Longleftrightarrow$
$b^4+c^4\le 3b^2c^2\Longleftrightarrow (b^2+c^2)^2\le 5b^2c^2\Longleftrightarrow \overline {\underline {\left| \ 3a^2\le b^2+c^2\le bc\sqrt 5\ \right| }}$.
Aceasta relatie reprezinta o conditie de compatibilitate pentru existenta celor trei triunghiuri dreptunghice construite in exteriorul triunghiului $ABC$ si pentru care $x=y+z$.

Felicitari! Foarte frumoasa solutie! Cred ca am reusit sa generalizez problema si pentru cazul cand triunghiurile exterioare au unghiurile M, N si P egale cu un a. Dar unghiul NAP nu este tot de 90.

Eu iti sugerez sa o generalizezi la un romb si la un unghi $\alpha$ !

Asa ma si gandisem. In acest mod unghiul cautat va fi 180-a.

Foarte frumos ! Aranjeaz-o corect si posteaz-o !. Succes !


[color=blue][b]Problem.[/b]
Let $a, b, c$ three positive real numbers such that $a+b+c=1$. Prove that the following inequality holds true:

                                     \[ \frac{a+bc}{b+c}+\frac{b+ca}{c+a}+\frac{c+ab}{a+b}\geq 2. \]

 [/color]

First multiply both sides by the common denominator, or $(1-a)(1-b)(1-c)$. This gives you:

$(a+bc)(b-1)(c-1) + (b+ca)(a-1)(c-1) + (c+ab)(a-1)(b-1) \geq 2(a-1)(b-1)(c-1)$

expanding the last two terms of each unit, we get...

$(a+bc)(bc-b-c+1) + (b+ca)(ac-a-c+1) + (c+ab)(ab-a-b+1) \geq 2(a-1)(bc-b-c+1)$

seeing that $a+b+c = 1$, we can replace terms such as $-b-c+1$ with $a$, etc. thus...

$(a+bc)^2 + (b+ca)^2 + (c+ab)^2 \geq 2(a-1)(bc+a)$, because terms such as $(bc-b-c+1) = a+bc$ and so on.

Then, just expand...

$a^2b^2 + b^2c^2 + a^2c^2 + a^2 + b^2 + c^2 + 6abc \geq 2(ab+c)(c-1)$

Transfer everything on the right to the left, yielding...

$a^2b^2 + b^2c^2 + a^2c^2 + a^2 + b^2 + c^2 + 4abc + 2ab + 2c - 2c^2 \geq 0$

We're almost finished... all we need to prove is that the left side is always positive. When you look at $2c - 2c^2$, we can factor this as $2c (1-c)$, which, we can replace the $1 - c$ by:

$a + b + c = 1$
$1 - c = a + b$

Therefore,

$a^2b^2 + b^2c^2 + a^2c^2 + a^2 + b^2 + c^2 + 4abc + 2ab + 2c(1-c) \geq 0$

$a^2b^2 + b^2c^2 + a^2c^2 + a^2 + b^2 + c^2 + 4abc + 2ab + 2c(a + b) \geq 0$,

which is positive, because $a$, $b$, and $c$ are all positive, and there are no negative terms in the expression on the left. And by the way... I'm not quite sure as to how to insert LaTeX stuff in here (the funky pictures, I mean...) If someone could tell me how to do that, I'd really appreciate it  :)

EDIT: Made all text into LaTeX  :D

Put dollar signs at the beginning and end of the equation. More information about specifics of LaTex can be found by following the link on the left.

Put x = b+c etc.  It becomes $\sum \frac{xy}{z} \ge 2 \Leftrightarrow \sum x^2y^2 \ge \sum x^2yz$ which is trivial.


i recently helped tutor a girl who did pure math 30 in alberta through distance learning. In this program there is a project that every student in alberta must do -- second semester it was called student project:The Winter Sport Of Biathalon. She did great except for part C and i was wondering if anyone knew where we could gte answers or if anyone had the correct answers. thanks
 :D

I remember doing those silly lil assignments.  Though I took math 30 a while ago, good luck finding answers though.   Our teacher never even bothered to tell us what our marks were, let alone the answers!


I was deleting an image off of my [url=http://www.artofproblemsolving.com/Forum/galleries.php?f=435&u=23588]personal gallery[/url] and the Information page showed a typo (as well as the lack of a link).


BuLi abstracts the proton from the NH2 grp then nucleophilc attack on the carbpnyl grp.

hope this is correct.

You mean the methyl group.I second that.

yup . sorry . I wrote NH2 instead of CH3 . my mistake

BuLi is a super base.it abstracts H from CH3

at iit2009 ,have you also written INChO ?


Every other time that Cheri Linn saw a friend today, she gave away half of her flowers.  At the end of the day, she had five flowers left.  How many flowers did she start the day with if she saw eight friends?

If she had 8 friends, the she gave away flowers 4 times. $ 5 \times 2^{4}\equal{}80$

80 flowers

Since she saw 8 friends, she gave away flowers 8 times. That means 5 is $ \frac{1}{2^8}$ of the number of flowers she started with. Therefore, Cheri started with 1280 flowers. (I wonder how she can carry that much)

I think the answer is $ 80 \text{ flowers}$.

We can work backwards to solve this.

She had 5 flowers at the end. So, before her 8th friend, she had 10 flowers. Before her 6th friend, she had 20 flowers. Before her 4th friend, she had 40 flowers. And before her 2nd friend, she had 80 flowers.

5 flowers=8th friend
10 flowers=6th
20 flowers=4th
40 flowers=2nd

She had 40 flowers

But she didn't have 5 flowers BEFORE her 8th friend, she had it AFTER her 8th friend. I know for a fact the answer is $ \boxed{80}$. It's rather tricky, though.


calculus
integration
LaTeX
function
number theory
least common multiple
algebra

Hi all.  I'm trying to complete a problem for my college Thermo class, I have figured out the integral but I cannot solve it.
I need to solve the following integral:

 {  x * ((1-x)/3)^(1/7)  dx

where { represents the integral symbol b/c I don't know how to type that.

I've tried integration by parts and don't get anywhere.  None of the integration techiques I know seem to work and nothing I've found online says what to do when you have this kind of a situation.

Any help would be greatly appreciated. I'm not trying to cheat here, but I don't know where to go from here.   Thanks everyone!!

,Mike

Integration by parts should work fine, but you might have an easier time with a simple $ u$-substitution.  What do you think a natural choice for $ u$ would be?

Also, to learn about writing mathematics on this forum, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]this thread[/url] and the [[LaTeX]] guide.

This integral prettified:
$ \int x\cdot\left(\frac{1\minus{}x}{3}\right)^{1/7}\,dx$

You can quote me or click on the code to see how it works.

The integral $ {\displaystyle \int x^m(a+bx^n)^p\ \mathrm dx}$ (where $ m$, $ n$, $ p$ are rationals) can be expressed in terms of elementary functions only in three cases:
[b]Case I:[/b] $ p$ is an integer: Then if $ p>0$, the integrand can be expanded by using Newton's binomial formula, but if $ p<0$, then we can substitute $ x=t^k$ where $ k$ is the lcm of the denominators of $ m$ and $ n$.

[b]Case II:[/b] $ \dfrac{m+1}{n}$ is an integer: We can then substitute $ a+bx^n =t^\alpha$, where $ \alpha$ is the denominator of the fraction $ p$.

[b]Case III:[/b] $ \dfrac{m+1}{n}+p$ is an integer: We can then substitute $ a+bx^n =t^\alpha x^n$ where $ \alpha$ is the denominator of the fraction $ p$.

These substitutions will always reduce the given integral to that of integration of a rational function which can always be done (at least theoretically :)) However, in a given problem its not mandatory to use these; one can proceed by any other means if available.

In the given problem we have Case II. Set $ \dfrac{1-x}{3}=t^7$, then $ -\mathrm dx = 21 t^6\ \mathrm dt$ and $ x=1-3t^7$, so the given integeral
$ I=\int (1-3t^7)(-21t^6\ \mathrm dt)=21\int (3t^{13}-t^6)\ \mathrm dt$
I hope you can carry over from here.

$ \int x\cdot\left(\frac{1\minus{}x}{3}\right)^{1/7}\,dx$
Set $ u\equal{}\frac {1\minus{}x}{3}$
then $ \minus{}3du\equal{}dx$
Also, $ x\equal{}1\minus{}3u$

$ \int x\cdot\left(\frac{1\minus{}x}{3}\right)^{1/7}\,dx$
$ \equal{}\minus{}3\int (1\minus{}3u)u^{1/7}du$
and then the rest is trivial :)


function
algebra
functional equation
algebra unsolved

Find all functions $f: \mathbb N \to \mathbb N$ such that
\[f(m^{2}+f(n))=f(m)^{2}+n,\]
for all $m, n \in \mathbb N$.

Hello anuj kumar!
[quote="anuj kumar"]find all f:N->N such that
\[f(m^{2}+f(n))=f(m)^{2}+n.\ m,n belong to N \]
[/quote]

Let the proposal P(m,n) be : $f(m^{2}+f(n))=f(m)^{2}+n$

1) $f(x)$ is injective
$f(a) = f(b)$ ==> $f(m^{2}+f(a))=f(m^{2}+f(b))$ $\Rightarrow$ $f(m)^{2}+a=f(m)^{2}+b$ $\Rightarrow$ $a=b$
Q.E.D.

2) f(0) = 0
Let $f(0)=a > 0$ Then $P(0,n) \Rightarrow f(f(n)) = n+a^{2}$ $\Rightarrow$ $f(n+a^{2}) = f(n)+a^{2}$ $\Rightarrow$ $f(n)=n+f(mod(n,a^{2}))-mod(n,a^{2})) = n+d(n)$ with $d(n)$ bounded
Then $f(m^{2}+f(n))=f(m^{2}+n+d(n))=m^{2}+n+d(n)+d(m^{2}+n+d(n))$ and $f(m)^{2}+n=(m+d(m))^{2}+n=m^{2}+n+2md(m)+d(m)^{2}$
And $d(n)+d(m^{2}+n+d(n))=2md(m)+d(m)^{2}$
But L.H.S is bounded and R.H.S is not (except if $d(m)=0$ $\forall m$, which implies $f(n)=n$ which is in contradiction with $f(0)=a > 0$
So $f(0)=0$
Q.E.D.

3)f(n) is bijective
$P(0,n) \Rightarrow f(f(n))=n$ $\Rightarrow$ f is surjective $\Rightarrow$ f is bijective (with point 1))
Q.E.D.

4) $f(n)=n$ is the only solution
$P(n,0) \Rightarrow f(n^{2})=(f(n))^{2}$
Since f is bijective, for a given n I can find an integer p > 0 such that $f(p)=2n+1$
Then $P(n,p) \Rightarrow f(n^{2}+2n+1)=(f(n))^{2}+p$ $\Rightarrow$ $f((n+1)^{2})=(f(n))^{2}+p$ $\Rightarrow$ $(f(n+1))^{2}=(f(n))^{2}+p$ $\Rightarrow$ $f(n+1) > f(n)$
So f is a strictly increasing bijection from N in N ==> f(n)=n
Q.E.D.

I'm affraid my demonstration is quite long an I think there exist some simpler one.

--
Patrick

[quote="me@home"]Sorry, I'm not sure what you mean by $d(n)$ is bounded? Do you mean linear?[/quote]

No I mean it exist reals A and B such that $A < d(n) < B$ $\forall n$

And this is easy to see since $d(n)=f(mod(n,a^{2}))-mod(n,a^{2})$ : d(n) can only take at most $a^{2}$ different values ==> a finite set of values is always "bounded" (I dont know if it is the good english word : french one is "borné")

--
Patrick

[quote="anuj kumar"]find all f:N->N such that
\[f(m^{2}+f(n))=f(m)^{2}+n.\ m,n belong to N \]
[/quote]
[quote="pco"]f(0) = 0
[/quote]
since the function is only defined for natural no.s how can you prove that $f(0)=0$

ok pco i read your other post   so it's the problem with your country but you should seriously like to revisit the proof.

[quote="pardesi"]ok pco i read your other post   so it's the problem with your country but you should seriously like to revisit the proof.[/quote]

OK, let's go :

1) f is injective :
$f(a)=f(b)$ $\Rightarrow $ $f(m+f(a))=f(m+f(b))$ $\Rightarrow $ $f(m)^{2}+a=f(m)^{2}+b$ $\Rightarrow $ $a=b$ Q.E.D.

2) f is strictly increasing :
If $f(a)>f(b)$, then $f(a)^{2}+n=f(b)^{2}+(n+f(a)^{2}-f(b)^{2})$ and then $f(a^{2}+f(n))=f(b^{2}+f(n+f(a)^{2}-f(b)^{2}))$ and, since f is injective :
$a^{2}+f(n)=b^{2}+f(n+f(a)^{2}-f(b)^{2})$ which implies $f(n+f(a)^{2}-f(b)^{2})=f(n)+(a^{2}-b^{2})$
$\Rightarrow $  $f(n+k(f(a)^{2}-f(b)^{2}))=f(n)+k(a^{2}-b^{2})$ $\Rightarrow $ $a>b$.
$f(a)>f(b)$ $\Rightarrow $ $a>b$ shows that f is monotonous increasing. Hence, since f is also injective, f is strictly increasing. QED

3) $f(n)=n$
Since f is strictly increasing, $f(a)=f(b)+1$ $\Rightarrow $ $a=b+1$
But we have $f(m^{2}+f(n+1))=f(m)^{2}+n+1 = f(m^{2}+f(n))+1$ and so $m^{2}+f(n+1)=m^{2}+f(n)+1$ and $f(n+1)=f(n)+1$
So $f(n)=n+f(1)-1$
When entering this expression in the functional equation, we conclude $f(1)=1$ and $f(n)=n$
Q.E.D.

--
Patrick

:coolspeak: though i haven't gone through the proof (in no mood to do so :sleep2: )

[quote="shfdfzhjj"][quote="pardesi"][quote="lovejrz"]what's QED mean?[/quote]

http://en.wikipedia.org/wiki/Q.E.D. :wink:[/quote]

but i can not open it[/quote]
Why can't you i can but any ways this thing is quite not related to topic please google for Q.E.D


What is mathematics?

Cheesy answers absolutely welcome if you mean it.

[hide]And we will all be very amused if you answer "what mathematicians do" and say that "a mathematician is one who does mathematics". It will be hilarious and well will fall off our seats and bonk our heads on the wall in unstoppable laughter.[/hide]

mathematics is what mathematicians do and mathematicians do mathematics

Math is what makes the world exist.

This is because of quantum physics that determined the formation of the universe, that came out of math.

Mathematics is the only creation of humans that has meaning outside of the context of the world.

Math is a pastime such that someday everyone will be doing it.

Math is the manipulation of numbers and diagrams


On Line 9 of Page 43 of Allen Hatcher's "Algebraic Topology", it says

[quote]the kernel of Ø contains all the elements
of the form $ i_{\alpha\beta}(\omega)i_{\alpha\beta}(\omega)^{ \minus{} 1}$[/quote]

Can someone explain to me why?

By definition, $ \Phi: *_{\alpha} A_{\alpha} \to X$ is the unique extension of the homomorphisms $ j_{\alpha}: \pi_1\left(A_{\alpha}\right) \to \pi_1(X)$. Therefore,

$ \Phi\left( i_{\alpha\beta}(\omega)i_{\beta\alpha}(\omega)^{ \minus{} 1}\right) \equal{} \left(j_{\alpha}i_{\alpha\beta}(\omega)\right)\left(j_{\beta}i_{\beta\alpha}(\omega)\right)^{ \minus{} 1}$

and by Hatcher's observation earlier in that sentence $ j_{\alpha}i_{\alpha\beta}(\omega) \equal{} j_{\beta}i_{\beta\alpha}(\omega)$, which puts $ i_{\alpha\beta}(\omega)i_{\beta\alpha}(\omega)^{ \minus{} 1}$ in the kernel of $ \Phi$.


It has been calculated that a large swarm of locusts can eat 3,000 tons of food in a single day. How many days would it take a restaurant to make enough quarter-pound hamburger patties (no buns) to feed the swarm of locusts for one day if they can make 100 hamburger patties per minute and they work around the clock? Express your answer to the nearest whole number. Note: 1 ton is equal to 2000 pounds.

There are 2000 pounds in a ton, so the locusts can eat $ 3000 \times 2000$, or 6,000,000 pounds of food per day. Since each patty is 1/4 pound, that would be 24,000,000 patties. The restaurant can make 100 patties per minute, or 6000 patties an hour. $ \frac{24,000,000}{6,000}$ would be 4,000 hours or $ 166\frac{2}{3}$ days?


Find the smallest number with 28 divisors.

[hide="Solution"]The number of divisors of a number is the product of one more than the exponents in its prime factorization, where the same primes are combined with exponents.

For example, the prime factorization of $ 24$ is $ 2^3 \cdot 3^1$. Thus, it has $ (3 \plus{} 1)(1 \plus{} 1) \equal{} 4 \cdot 2 \equal{} 8$ factors.

Now, since $ 28 \equal{} 2 \cdot 2 \cdot 7$, there are three cases. The exponents of the primes can be: 6 and 1 and 1, 13 and 1, 6 and 3, or just 27.

To minimize the number, we raise the smallest prime to the biggest exponent, the next smallest to the next biggest, and so on.

So for each case, the minimum is:

$ 2^6 \cdot 3^1 \cdot 5^1 \equal{} \mathbf{960}$

$ 2^{13} \cdot 3^1 \equal{} \mathbf{24576}$

$ 2^6 \cdot 3^3 \equal{} \mathbf{1728}$

$ 2^{27} \equal{} \mathbf{134217728}$

Clearly, the smallest of these is $ \boxed{960}$.[/hide]

@PowerofPi: make sure you type 2^{13}, not 2^13 when dealing with multi digit exponents in $ \text{\LaTeX}$  :)

@prezcoin: next time just use the "quote" button for that type of reply


In the figure shown, R is a long smooth fixed non-conducting rod of uniform linear charge density $ \lambda$ and radius of cross section r. A is a non-conducting uncharged smooth ring of mass m which fits completely on the rod and can move horizontally on the rod. B is a non-conducting charged small particle of mass 2m and charge Q and connected to A by an inextensible light string of length l. B is released from rest from the position shown in figure. Find velocity of B and tension in the string when the string becomes vertical. Assume gravitational accelaration g and $ \lambda Q$ >0

[hide="Ans"] $ v \equal{} \sqrt{\frac{2gl}{3} \plus{} \frac{\lambda q}{6 \pi \epsilon_o m}. \ln \frac{l\plus{}r}{r}}$

 $ T\equal{} 2mg\plus{} \frac{9mv^2}{l}\plus{}\frac{2 \lambda k Q}{r\plus{}l}$
 [/hide]

Electric field due to the rod is given by $ E(r)\equal{}\frac{\lambda}{2\pi\epsilon_0 r}$ and hence, potential difference can be written as $ V(r_2)\minus{}V(r_1)\equal{}\frac{\lambda}{2\pi\epsilon_0}\ln \frac{r_2}{r_1}$

Thus, change in electric potential energy is $ \frac{Q\lambda}{2\pi\epsilon_0}\ln \frac{l\plus{}r}{r}$ and change in total potential energy is $ \frac{Q\lambda}{2\pi\epsilon_0}\ln \frac{l\plus{}r}{r}\plus{}2mgl$

Hence, kinetic energy $ \tfrac{1}{2}mv_1^2\plus{}mv_2^2\equal{}\frac{Q\lambda}{2\pi\epsilon_0}\ln \frac{l\plus{}r}{r}\plus{}2mgl$ (here $ v_1,v_2$ are velocities of ring and particle respectively)

By conservation of linear momentum, $ v_1\equal{}\minus{}2v_2$ and hence,
$ v_2\equal{}\sqrt{\frac{Q\lambda}{6\pi\epsilon_0}\ln \frac{l\plus{}r}{r}\plus{}\frac{2}{3}mgl}$

In reference frame of the ring, velocity of particle $ v\equal{}3v_2$

Now, tension in string is given by $ T\minus{}2mgl\minus{}\frac{Q\lambda}{2\pi\epsilon_0 (l\plus{}r)}\equal{}\frac{mv^2}{l}$, and hence,

$ T\equal{}2mgl\plus{}\frac{Q\lambda}{2\pi\epsilon_0 (l\plus{}r)}\plus{}\frac{mv^2}{l}$


A person pie costs 8 dollars and weighs three pounds and four ounces. A dog pie costs 5 dollars and weights two pounds. How many dog pies can be bought for the cost of $48\frac{3}{4}$ pounds of person pie?

[hide]First we convert everything to ounces

Person pie: 52 ounces
Dog pie: 32 ounces
$48\frac{3}{4}=780$ ounces

the answer is $\frac{(780)(8)}{(52)(5)}=\boxed{24}$ dog pies[/hide]

[hide]3 pounds and 4 ounces is the same thing as 52 ounces.  $48\frac34$ pounds is 780 ounces.  Therefore, you can buy 15 person pies.  This would cost $8\times15=\$120$, which could be spent on $\frac{120}{5}=24$ dog pies.[/hide]

[hide]We convert everything to ounces to make our life easier.  Then I just did $\frac{(780)(8)}{(52)(5)}=24$.

$24$[/hide]


AMC
AIME
geometry
3D geometry
icosahedron
AMC 10

I've seen this question asked before, but never answered:

I received an AMC medal for winning the AMC 10 two years in a row this year.  It was bronze.  How exactly does the medal system work?  How many consecutive school wins do you need for a silver medal?  Gold?

It seems hard to win your school more than 3 times, considering most people only take the AMC 10 in 8th, 9th, and 10th grade.

Also, I've heard the consecutivity starts over when you go from the AMC 10 to the AMC 12 in 11th grade.

I guess gold and silver are for amc 12 participant winners and stuff.

and amc 10 just gives bronze?  :blush:

hypothetically speaking, what happens if you were top school scorer multiple years, but at different schools: say you moved and changed schools?

[quote="jmerry"]It always has been tough to win 4 or more times in a row, even when there was only the AHSME. The results book included a list of everyone who was school winner for the fourth time or more, and it wasn't a very long list. Many of the names were students who also did very well on the USAMO.
I pulled it off myself, winning every year from 1997 to 2000. (AHSME the first three years, AMC 12 in 2000)[/quote]
Not when you're the first person in your city to qualify for the AIME in 10+ years.

[quote="AMCDirector"]> I received an AMC medal for winning the AMC 10 two years in a
> row this year. It was bronze. How exactly does the medal system > work? How many consecutive school wins do you need for a silver > medal? Gold?

In addition to the Web reference already posted,
See the AMC 10/12 Teachers' Manaul, available at our website, page 8, Section XI, item 1, under Intramural and National Awards.  In particular there, it says:
"Details of how medals are awarded will be included with your school's results."
Consult with your school contest manager, the results for the A-date contests should already be there.

> hypothetically speaking, what happens if you were top school
> scorer multiple years, but at different schools: say you moved and > changed schools?

No medal.  We do not track individual students across multiple schools.

Steve Dunbar
Director, American Mathematics Competitions[/quote]
So if we want a gold medal, it would be to our advantage to take the AMC 12 in 9th grade?

In 1971, the three-time school AHSME leader award was a pin - a little silvery pin in showing an icosahedron (the MAA symbol). Not that I have any idea where mine would have gone.

this is what they have for School winner every year on the AMC 8

As well as the 10 last year. I don't know if it's a pin for the 12, but if so, I'll have collected all three pins by the end of this year.  :P

What happens in the tricky case of somebody being school winner for 5 years?

The medals are bronze (two years), silver (three years), and gold (four years).  The students who win gold medals also have their names listed in the Summary of Results that goes out to all the schools, coaches, and other folks.  It's quite nicely done.

We had a student (Shadow Hunter) win a gold medal for being the school winner on the AMC-10 four years in a row.  He won in fourth, fifth, sixth, and seventh grade.  Since then, he has been doing the AMC-12 (and AIME).

His last year, when he hoped to win the gold, he took the AMC-10 problem solving course here at AoPS.  He says it helped a great deal.

The medals look cool. :-).  MathDancer got one last year on the AMC-12, and I think she and E^(pi*i) will get them this year, on the 12 and 10, respectively.

Best,
Dr. Asaro

[quote="rnwang2"][quote="indianamath"]lol i don't think  someone who won the school for an AMC would be held back. :P[/quote]
Except for our school is really weird and has 5 grade levels.  :ninja:[/quote]
Or if someone is a winner in 8th grade :ninja:

Getting a gold medal is not necessarily easy, but with so much fluctuation between schools, are medals seen as significant? The average school (see AMC website) has a winner of around 110, and in other schools, (like mine) you need >140 to have a decent chance of winning. I think that the AMC should, in addition to intramural awards, give some kind of national recognition for getting a certain score for several years (i.e. >130 or 140 for four years).

Yeah, that's a good idea. At my school, which is rather bad at math, I was the school winner(!) with only a 112.5, with second place at 97.5. While at my best friend's school, the school winner had a 130+. Bear in mind that these two schools are less than ten miles apart.

oh man.. I took the AMC 10 freshman year.. got a 150.. It starts over again :-(... I think I ma just get the silver bleh..
This year, being a sophomore, I am the school winner with 138,

aawww. my school wouldn't let me take it cuz i was in 6th grade.


inequalities
geometry
incenter
inequalities proposed

[quote="flip2004"]In a triangle hold the inequalities   $ 2 r\le R \le 3.125 r \;  . $
[b]Prove (or disprove) that the orthocentre does not belongs to the incircle.[/b][/quote]
I am not sure about the problem " belongs" means IH=r or IH<r.
Consider equilateral triangle we have the orthocenter is the incenter so orthocenter is inside the incircle.

After my opinion, in case of my proposed problem, the term ,,belongs" is

equivalent to  IH=r .


In this addition problem, each letter stands for a different digit.
[asy]label("T", (1,1));
label("W", (2, 1));
label("O", (3, 1));
label("+", (0,.6));
label("T", (1,.6));
label("W", (2, .6));
label("O", (3, .6));
draw((-.5, .3)--(3.5,.3));
label("F", (0, 0));
label("O", (1, 0));
label("U", (2, 0));
label("R", (3, 0));[/asy]
If T = 7 and the letter O represents an even number, what is
the only possible value for W?

ok we have

7WO+7WO=FOUR

since O is even, tehres no carrying, and O=4, F is obviously 1, and R is 8

so 7W4*2=14U8

2W<10; W<5, and W isnt 1, 2, or 4

so W=3

If T is 7, and O is even, its means that O must be 4, and 2W<10.

This means W=1,2,3,or 4. O is already 4, so W can't be 4. W can't be 1 because F already is. So W must be 2 or 3. However, if W were 2, U would be 4, and O is already 4, so that doesn't work.

So the only solution for W is 3.

EDIT: Sorry mz I didn't see your post.  :blush:


1. (Modified from 1986 Individual Round Question Number 7)
Define $ f(n)$ as below:

$ f(n+1) = (-1)^n (n+1) - 2 f(n), \ n \geq 1$

If $ f(1) = f(2007),$ find $ f(1) + f(2) + \cdots + f(2006)$.

2. (Modified from 2007 AIME Question Number 14)
The sequence $ z_n$ is defined with $ z_{n+2} z_{n-2} = z^2_n + 2007$ for all $ n \geq 4$. It is given that $ z_2 = 1$ and $ z_4 = 3$. Find the largest integer less than or equal to

$ \displaystyle \frac{3z^2_{2008} + 3z^2_{2006} - z_{2008}z_{2006}}{3z_{2008}z_{2006}}$

3. (Modified from 2004 ARML Individual Round Question Number 4)

If ${ \displaystyle \sqrt[3] {7} = a + \frac{1}{b + \frac{1}{c + \frac{1}{d + \cdots }}}}$ where $ a,b,c,d, \cdots$ are positive integers, find $ b-a$.

[hide="Number 1"]$ 2f(n) \plus{} f(n \plus{} 1) \equal{} ( \minus{} 1)^n (n \plus{} 1)$

Letting $ n$ range from 1 to 2006 and summing, we have that:

$ 2f(1) \plus{} 3f(2) \plus{} 3f(3) \plus{} \ldots \plus{} 3f(2006) \plus{} f(2007) \equal{} \minus{} 2 \plus{} 3 \minus{} 4 \plus{} 5 \ldots \plus{} 2007$

$ 3 \left( f(1) \plus{} f(2) \plus{} \ldots \plus{} f(2006) \right) \plus{} (f(2007) \minus{} f(1)) \equal{}$ $ (3 \minus{} 2) \plus{} (5 \minus{} 4) \plus{} \ldots \plus{} (2007 \minus{} 2006)$

$ 3S \plus{} 0 \equal{} 1003$

$ \boxed{ S \equal{} \frac {1003}{3} }$[/hide]

[hide="Number 3"]
$ 1 \leq 7 < 8 \implies 1 \leq \sqrt [3]{7} < 2 \implies a \equal{} 1$

Now $ b$ is the floor of $ \frac {1}{\sqrt [3]{7} \minus{} 1}$

Clearly $ b$ is at least $ 1$. If $ b$ were 2, then $ \sqrt [3]{7} \minus{} 1 \equal{} \frac {1}{2} \implies \sqrt [3]{7} \equal{} \frac {3}{2}$.

Upon cubing, we see that $ \left( \frac {3}{2} \right) ^3 \equal{} \frac {27}{8} < 3.5 < 7$, so $ b \equal{} 2$ is too high, and $ b \equal{} 1$.

Hence, $ \boxed{b \minus{} a \equal{} 0}$
[/hide]


calculus
inequalities proposed
inequalities

What's the maximum of the expression $a^2+b^2+c^2$ can be take if $a+b+c=5$ and

(i) $a,b,c$ are three lengths of a triangle.

(ii) $ab,bc,ca$ are three lengths of a triangle.

(i) let $f(a,b,c)=a^2+b^2+c^2$. then,

$f\left(\frac{a+b+c}{2},\frac{a+b+c}{2},0\right)-f(a,b,c)=\frac{a(b+c-a)+b(c+a-b)+c(a+b-c)}{2} \\\\ \Rightarrow f(a,b,c)\leq f\left(\frac{a+b+c}{2},\frac{a+b+c}{2},0\right)= 2\left(\frac{5}{2}\right)^2=\frac{25}{2}$

with equality iff two of $a,b,c$ are equal and the other one equals $0$

cool  :D i tried to approach the other one in a similar way, but i didn't succeed  :(

i'll give it a try!

yes i solved the first one too :
$25=(a+b+c)^2=a^2+b^2+c^2+a(b+c)+b(c+a)+c(a+b)\geq 2(a^2+b^2+c^2) \rightarrow \frac{25}{2}\geq a^2+b^2+c^2$

[quote="hungkhtn"]Yes. Sure, this one is true. But anything for the other?[/quote]
Well, $a\to5,b=c\to0$ is as high as we can get... :) :)

[quote]Well,  is as high as we can get...  [/quote]

The second is more difficult very much.

Why? I don't understand
If $a,b,c\ge0$, we have $a^2+b^2+c^2\le (a+b+c)^2=25$ and this is the attained maximum.:pilot: If $a=5-2\epsilon,b=c=\epsilon$, the trianlge has sides $\epsilon^2, \epsilon(5-2\epsilon), \epsilon(5-2\epsilon)$ and is very acute if $\epsilon\to0$.

If we admit negative numbers, they cannot all be negative because of $a+b+c=5$. But if some are positive and some negative, your 'triangle' will have negative sides.

After some thinking: Maybe you mean both conditions [b]simultaneously[/b]? But then why did you write a full stop at the end of (i)?? ;)

In this case, suppose wlog $a\ge b\ge c$. So $ab\ge ac\ge bc$. I would [i]guess [/i](maybe some, but not much, calculus needed to show monotony) that in the extremal case we must have simultaneously $a=b+c$ and $ab=ac+bc$. This yields $a=\frac52, b= \frac54(\sqrt{5}-1), c=\frac54(3-\sqrt{5})$ and the maximum $\frac{25}2(3-\sqrt{5})\approx 9.549$. Is that your solution?  :P


I made a typo typing in my ID for Round 1.  I have my real name and user-name correct.  Will it make a difference?

Probably not.  Most likely it will be fine.


function
calculus
derivative
modular arithmetic
real analysis
real analysis unsolved

Let $f(x)$ be the $2\pi$-periodic function that is given by $f(x) = x$ for
$0\leq x < 2\pi$. Compute the Fourier series of $f(x)$. For which $x$ does the series converge? What is the
sum? For which $x$ does the series converge to $f(x)$?

Working with weak derivatives and the complex Fourier series to simplify my life:

$f'(x)=1-2\pi\delta(x).$ (That's the Dirac delta.)

For $k\ne0,\ \widehat{f'}(k)=-2\pi.$

Hence for $k\ne0,\ \widehat{f}(k)=\frac{-2\pi}{2\pi ik}=\frac{-1}{ik}.$

For $k=0,\ \widehat{f}(0)$ is the average value of the function, namely $\widehat{f}(0)=\pi.$

Hence $f(x)$ has Fourier series $\pi-\sum_{k\in\mathbb{Z}\setminus\{0\}}\frac1{ik}e^{ikx}.$

We can rewrite this series in real terms as

$\pi-\sum_{k=1}^{\infty}\frac2k\sin(kx).$

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121138]here[/url] for an argument that shows that this series converges for $x\in(0,2\pi)$ to $f(x).$

If $x\equiv 0\pmod{2\pi},$ then the series converges to $\pi,$ which, rather than being the value of the function, is actually the average of the right and left-hand limits.


Given the real number a.Find all the function$f: R\to R$satisfly:
          $f(f(a+y)-f(a))=y \forall y \in R$

Let \[ y = 0 :  \]

	\[ f(f(a)-f(a)) = 0 \Rightarrow f(0) = 0 \]

	\[ f(y) = 0 \Leftrightarrow y = 0  \]
	Proof:
	Let \[ b \in \Re : f(b) = 0  \]

\[ y = b-a \Rightarrow f(-f(a)) = b-a  \]
\[ y = -a \Rightarrow f(-f(a)) = -a \Rightarrow b-a = -a \Leftrightarrow b = 0 \]

One solution is obviously: \[ f(x) = x \,\, \forall x \in \Re  \]
because it satisfies all ipothesis.

Let \[ f(x)= x + c  \] with \[ c \in \Re  \]

\[ f(f(a+y) - f(a)) = y \Rightarrow f(a+y+c -a-c) = y \Rightarrow f(y)=y \Rightarrow y + c = y \Leftrightarrow c = 0  \]

Let \[ f(x)= cx  \] with \[ c \in \Re  \]
\[ f(f(a+y) - f(a)) = y \Rightarrow f(ac + cy -ac) = y \Rightarrow c^2y=y \Leftrightarrow c = \pm 1  \]

So, the solution is \[ f(x) = \pm x \,\, \forall x \in \Re  \]

[quote="Gauss_87"] ...
So, the solution is \[ f(x) = x \,\, \forall x \in \Re  \][/quote]

No, that's not so easy ! There are at least $2^{card(R)}$ of such functions  :D
 :cool:

briefly, we have:

$f(f(a+y) - f(a)) = y$

$f(f(a+y) - f(a)) - f(a) = y - f(a)$

$f(f(f(a+y) - f(a)) - f(a)) = f(y - f(a))$

but

$f(f(f(a+y) - f(a)) - f(a)) = f(a+y) - f(a) - a$

so

$f(a+y) - f(a) - a = f(y - f(a))$

putting  $x = a+y , k = - f(a) - a$

we obtain

$f(x) - k = f(x-k)$.

Diogene is in the right :).


How many positive perfect squares less than 10 to the power of 6 are multiples of 24?


(1) Prove that $ e^x\geq 1\plus{}\frac{1}{2}x^2$ for $ x\geq 0$.

(2) For positive integers, let $ M_n$ be a maximum value of the function $ f_n(x)\equal{}n^2(x\minus{}1)e^{\minus{}nx}\ (x\geq 0)$.

Find $ \sum_{n\equal{}1}^{\infty} M_n$.

[hide="Number (1)"]Let $ f(x) \equal{} e^x \minus{} 1 \minus{} \frac12 x^2$, and note that f(0) = 0. Now $ f'(x) \equal{} e^x \minus{} x$ and, for $ x \geq 0$, we have $ f'(x) > 0$. But this means that f(x) is an increasing function for $ x \geq 0$, i.e.,

$ f(x) \geq f(0)$

$ \equal{} > e^x \minus{} 1 \minus{} \frac12 x^2 \geq 0$

$ \equal{} > e^x \geq 1 \plus{} \frac12 x^2$

as we wanted.
[/hide]

[hide="Number (2)"]$ f_n'(x) \equal{} n^2 e^{ \minus{} nx} \minus{} n^3(x \minus{} 1) e^{ \minus{} nx} \equal{} (n^3 \plus{} n^2 \minus{} n^3 x) e^{ \minus{} nx}$, and $ f_n'(x) \equal{} 0$ for $ x \equal{} \frac {n^3 \plus{} n^2}{n^3} \equal{} 1 \plus{} \frac1n$. Now $ f_n''(x) \equal{} \minus{} n^3 e^{ \minus{} nx} \minus{} n(n^3 \plus{} n^2 \minus{} n^3 x) e^{ \minus{} nx}$ and $ f_n''\left(1 \plus{} \frac1n\right) \equal{} \minus{} n^3e^{ \minus{} n \minus{} 1} < 0$ for every $ n \in \mathbb{N}$. Therefore, $ x \equal{} 1 \plus{} \frac1n$ is a maximizing value and the corresponding maximum is $ f_n \left(1 \plus{} \frac1n\right) \equal{} M_n \equal{} n e^{ \minus{} n \minus{} 1}$.

Now $ \sum_{n \equal{} 1}^{\infty} M_n \equal{} \sum_{n \equal{} 1}^{\infty} \frac {n}{e^{n \plus{} 1}}$. But, for |x| < 1,

$ \sum_{n \equal{} 1}^{\infty}x^n \equal{} \frac {x}{1 \minus{} x}$, and differentiating both sides with respect to x gives

$ \sum_{n \equal{} 1}^{\infty}nx^{n \minus{} 1} \equal{} \frac {1}{(1 \minus{} x)^2}$

$ \equal{} > \sum_{n \equal{} 1}^{\infty}nx^{n \plus{} 1} \equal{} \left(\frac {x}{1 \minus{} x}\right)^2$.

In our case, $ x \equal{} \frac1e$, and so $ \sum_{n \equal{} 1}^{\infty} M_n \equal{} \left(\frac {1/e}{1 \minus{} 1/e}\right)^2 \equal{} \frac {1}{(e \minus{} 1)^2}$.[/hide]

For (1) Why $ f'(x)>0$?

For (2), your solution is smart because you neglected the intention of problem maker. :lol:

It is easier to write

$ e^x \equal{} \sum_{k\equal{}0}^\infty \frac{x^k}{k!} \equal{} 1 \plus{} x \plus{} \frac{x^2}{2} \plus{} \frac{x^3}{6} \plus{} \cdots$,

and for $ x \ge 0$ all the terms are clearly nonnegative.  Therefore $ e^x \ge 1 \plus{} \frac{x^2}{2}$.

Your solution is right in mathematicaly, but Japanese HighSchool students don't study Taylor expansion or Maclaurin Series.

[quote="kunny"]For (1) Why $ f'(x) > 0$?

For (2), your solution is smart because you neglected the intention of problem maker. :lol:[/quote]
 $ f'(x) > 0$    because of:
 We have : $ f''(x) \equal{} e^x\minus{}1 \geq  0  , \forall x \geq  0$ hence :  $ f'(x)  \geq f'(0) \equal{}0$

That's right, or $ y \equal{} e^x$ is an increasing and downwards convex, so the tangent line at $ x \equal{} 0$ on the curve $ y \equal{} e^x$; $ y \equal{} x \plus{} 1$ is located below $ y \equal{} e^x$ i.e.
$ e^x\geq x \plus{} 1 > x$. I like the solution $ f''(x) > 0$ as you have shown.


[quote="Mutant"]Question says:

find the points $(r, \theta)$ where the curve $r = 1 + sin(\theta)$ has horizontal or vertical tangents.

How can I prove it?

Thank you :)[/quote]
I have sols here. Hope it is correct:

$x=rcos(\theta)=cos(\theta)+sin(\theta)cos(\theta)$
$\frac{dx}{d\theta}=-sin(\theta)+cos^2(\theta)-sin^2(\theta)$

$y=rsin(\theta)= sin(\theta)+sin^2(\theta)$
$\frac{dy}{d\theta}=cos(\theta)+2sin(\theta)cos(\theta)$

Using chain rule we have:

$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{cos(\theta)+2sin(\theta)cos(\theta)}{-sin(\theta)+cos^2(\theta)-sin^2(\theta)}$

It has vertical tangent when $-sin(\theta)+cos^2(\theta)-sin^2(\theta)=0$ --> $(2sin(\theta)-1)(sin(\theta)+1)=0$ --> $\theta=-\frac{\pi}{2} ,\frac{\pi}{6}, \frac{5\pi}{6}$

It has horizontal when  $cos(\theta)+2sin(\theta)cos(\theta)=0$ --> $\theta=\frac{\pi}{2}, \frac{-\pi}{6}, \frac{-5\pi}{6}$

Hope it help.


Hallo, Mathlinkers, the following problem i have solved but not sure if ans is correct. please check it and solve the problem:

An urn contains 6 balls of different colors. Probability of saying truth for P is $ p$ and for Q ,it is $ q$. A ball is picked up from urn and both P & Q assert it is red. Find the PROBABILITY of the ball being actually red.

I think it is [hide]$ \frac{pq}{pq \plus{} \frac{(1\minus{}p)(1\minus{}q)}{5}}$[/hide]

Hmm, I'm not good at this but this is what I got:

[hide="Solution."]
=====
Let $ P,Q,R$ be the events of $ P$ saying red, $ Q$ saying red, and drawing red, respectively. We have:

$ P(R) \equal{} \frac {1}{6}$
$ P(P|R) \equal{} p$
$ P(Q|R) \equal{} q$
Now:
$ P(P|\overline{R}) \equal{} 1 \minus{} p$
$ P(Q|\overline{R}) \equal{} 1 \minus{} q$
$ \implies P(P\cap Q|R) \equal{} pq$
$ \implies P(P\cap Q|\overline{R}) \equal{} (1 \minus{} p)(1 \minus{} q) \equal{} pq \minus{} p \minus{} q \plus{} 1$
$ \implies P(P\cap Q) \equal{} \frac {pq}{6} \plus{} \frac {5(pq \minus{} p \minus{} q \plus{} 1)}{6} \equal{} \frac {6pq \minus{} 5p \minus{} 5q \plus{} 5}{6}$
Thus:
$ P(R|P\cap Q) \equal{} \frac {P(R\cap (P\cap Q))}{P(P\cap Q)}$
$ \equal{} \frac {P((P\cap Q)|R)P(R)}{P(P\cap Q)}$
$ \equal{} \frac {\frac {pq}{6}}{\frac {6pq \minus{} 5p \minus{} 5q \plus{} 5}{6}}$
$ \equal{} \boxed{\frac {pq}{6pq \minus{} 5p \minus{} 5q \plus{} 5}}$
=====
[/hide]

[hide]

There is a $ \dfrac{1}{6}$ chance of the ball being red.  If this happens, the probability that $ P$ and $ Q$ both assert its redness is $ p*q$, for a total of $ \dfrac{1}{6}pq$

There is a $ \dfrac{5}{6}$ chance of the ball not being red.  In this case, the probability of $ P$ and $ Q$ both asserting its redness is $ (1 \minus{} p)(1 \minus{} q)$, for a total of $ \dfrac{5}{6}(1 \minus{} p)(1 \minus{} q)$

Thus, the probability of the ball actually being red is $ \dfrac{\dfrac{1}{6}pq}{\dfrac{1}{6}pq \plus{} \dfrac{5}{6}(1 \minus{} p)(1 \minus{} q)} \equal{} \dfrac{pq}{pq \plus{} 5(1 \minus{} p)(1 \minus{} q)} \equal{} \dfrac{pq}{6pq \minus{} 5p \minus{} 5q \plus{} 5}$
[/hide]

[hide="side note"]
What's Baye's Theorem?  Isn't this just conditional probability?
[/hide]

It is conditional probability. Here, [url=http://en.wikipedia.org/wiki/Bayes%27_theorem]Bayes' theorem[/url].

dear jvenezuela716 and feuxfollets it is wrong. because we dont know,if there exists a ball of color red in the urn or not. so, probability of not picking up a red ball from urn is not 5/6. :)  :) .

[hide]isn't the problem a twist? :D [/hide]

re. thanks for moving the post, i was thinking of requesting moderators today,but saw that its already moved. however i see that no answer given here is still seeming to be correct.

[quote="MellowMelon"]That makes this problem silly. You need the chance that the ball is not red to produce a solution, but we have no way of knowing it if we aren't guaranteed to have a red ball. (What's the chance of having a red ball?)[/quote]
How about following argument??

consider two events; P and Q say truth(event A), P and Q lie(event B).
if A happens we are sure there[b] is[/b] a red ball. so prob=pq/6.
if B happens one of two events are there:
   the ball is in urn.then probability=(1-p)(1-q)[hide="(5/6)"]since red isn't picked up in 5 ways out of 6[/hide]*[hide="(1/5)"]since according to P there is ared ball in the remaining 5[/hide]*[hide="(1/5)"]same for q[/hide]
   the ball isn't in urn.then probability=(1-p)(1-q)*(1/5)(1/5)*[hide="(1)"]since there are 6 ways of not picking up a red ball out of 6 balls[/hide]*[hide="(0)"][color=red]since red ball isn't in urn, probability for having a red ball outside urn is $ \frac {1}{infinity}$[/color][/hide]

So req. prob= $ \frac {\frac {pq}{6}}{\frac {pq}{6} \plus{} \frac {(1 \minus{} p)(1 \minus{} q)}{30}}$ :wink:  :wink:

So, [b]Daan was correct[/b] :lol:

[quote="ith_power"]
   the ball is in urn.then probability=(1-p)(1-q)[hide="(5/6)"]since red isn't picked up in 5 ways out of 6[/hide]*[hide="(1/5)"]since according to P there is ared ball in the remaining 5[/hide]*[hide="(1/5)"]same for q[/hide][/quote]
 :huh: What do you mean "according to $ P$ there is a red ball"? The probability that the red ball remains in the urn and $ P,Q$ both lie is $ \frac{5}{6}\cdot (1\minus{}p)(1\minus{}q)$: the chosen ball is not red with probability $ \frac{5}{6}$ and the two lie with probability $ (1\minus{}p)(1\minus{}q)$.

[quote="ith_power"][hide="(0)"][color=red]since red ball isn't in urn, probability for having a red ball outside urn is $ \frac {1}{infinity}$[/color][/hide][/quote]
This makes no sense.

so if B happens as you say, then how do you know the probability of there being a red ball in the urn?

See , i wanted to include the "1/5" terms in denominator, because it is the answer. :P   anyway, if my reasoning seems wrong ,please ask DAAN how he got the answer,i would like to know to. :maybe:

edit: [quote="jvenezuela716"][quote="ith_power"]
   the ball is in urn.then probability=(1-p)(1-q)[hide="(5/6)"]since red isn't picked up in 5 ways out of 6[/hide]*[hide="(1/5)"]since according to P there is ared ball in the remaining 5[/hide]*[hide="(1/5)"]same for q[/hide][/quote]
 :huh: What do you mean "according to $ P$ there is a red ball"? The probability that the red ball remains in the urn and $ P,Q$ both lie is $ \frac {5}{6}\cdot (1 \minus{} p)(1 \minus{} q)$: the chosen ball is not red with probability $ \frac {5}{6}$ and the two lie with probability $ (1 \minus{} p)(1 \minus{} q)$.[/quote]
its because we aren't sure if there is a red ball in the 6 or not

[quote="ith_power"]See , i wanted to include the "1/5" terms in denominator, because it is the answer. :P   anyway, if my reasoning seems wrong ,please ask DAAN how he got the answer,i would like to know to. :maybe:[/quote]

You cannot justify your logic by saying that you are trying to match someone else's answer.  :o

I don't think he's trying to justify his logic; he's admitting he's not sure of his reasoning.  There's nothing wrong with that.

Actually, I meant [hide]$ \frac{pq}{pq \plus{} 5(1\minus{}p)(1\minus{}q)}$[/hide]  :blush:

[hide]Of all the possibilities (P speaks the truth / P lies, Q speaks the truth / Q lies, the ball is red / not red in any combination), we have that

$ P(\textrm{ball is red, P speaks truth, Q speaks truth}) \equal{} \frac15 p q$ and
$ P(\textrm{ball is not red, P lies, Q lies}) \equal{} \frac56 (1\minus{}p) (1\minus{}q)$

Since P and Q both assert the ball is red, these are the only possible situations in our problem. With this information, we can now easily calculate the chance of the ball being red. It is:
$ P(\textrm{ball is red in our situation}) \equal{} \frac{ \frac15 p q }{\frac15 p q \plus{} \frac56 (1\minus{}p) (1\minus{}q)} \equal{} \frac{pq}{pq \plus{} 5(1\minus{}p)(1\minus{}q)}$[/hide]

If $ p\equal{}q\equal{}1$, that is both speak the truth, then the ball [b]is[/b] red. You can verify your answers like that!  :lol:

[quote="Rajiv"]If $ p \equal{} q \equal{} 1$, that is both speak the truth, then the ball [b]is[/b] red. You can verify your answers like that!  :lol:[/quote]

I coudn't understand what difference that would make?? :huh:    :ewpu:  :huuh:


Se dau in spatiu 6 puncte astfel incat distantele intre ele sunt diferite. Se construiesc toate triunghiurile. Sa se demonstreze ca putem gasi o latura care este cea mai mare intr-un triunghi si cea mai mica in altul.

Nimeni? :P Hint:mai stiu eu o problema cu 6 puncte si niste culori din folclor.....

m-am mai gandit si eu unpic la ea, dar m-am lasat, deci sunt $20$ triunghiuri formate, si $15$ drepte, deci trasand toate triunghiurile  fiecare segment "apare" de $4$ ori. si acum cred ca merge facut urmatorul lucru : luam un $2$ puncte $B$ si $C$, si acum luam un cerc care contine coarda $(BC)$, si luam al treilea punct pe cercul asta, acum alt cerc, si al 4-lea pe cercul asta, si etc. si facand asa analog , ne ducem la unghiuri : o latura $XY$ e max cand unghiul opus din triunghiul ce o contine e max, e la mijloc cand unghiu e la mijloc si etc. si daka o luam cu unghiuri tot asa, si presupunand ca sunt numai `max-uri` si `medium-uri`, da o contradictie, dar tre sa ma mai gandesc.

Ar fi interesanta o solutie pe aceasta cale, desi par sa apara foarte multe cazuri..... :maybe: Totusi poate se foloseste cineva si de hint :)

negatia problemei ar spune ca orice latura intr-un triunghi ori e maxima
ori e minima,insa asta s-ar putea intampla doar intr-un triunghi
echilateral,deci toate triunghiurile ar trebui sa fie echilaterale.
insa, deoarece putem construi maxim 4 puncte care sa aiba aceleasi
distante intre ele (tetraedru regulat),al cincilea punct neputand fi gasit
astfel incat sa indeplineasca in continuare conditia de a pastra toate triunghiurile echilaterale.
deci negatia ei este falsa,ramane ca propozitia initiala sa fie adevarata.

Nu prea e aia negatia propozitiei :wink:  Acum depinde si de terminologie dar dupa cum ai continuat nu cred ca "echilateral" era metafora la tine (paradoxal asta trebuia sa fie...)

din pacate nu vad nici un paradox si nici nu-mi pot da seama de ce este
gresita negatia.
te rog lamureste-ma

Negatia este urmatoarea: nu exista o latura care sa fie si minima si maxima in acelasi timp( in triunghiuri diferite$!!!$). Aceasta nu implica faput ca toate triunghiurile sunt echilaterale deoarece latura este minima [u]intr-un triunghi [/u]si maxima [u]in altul[/u].

Ok vad ca nu vrea nimeni sa rezolve problema :P  O sa postez solutia in curand dar poate pana atunci mai vine cineva cu niste idei :)

Având in vedere celălalt topic al tău, presupun că rezolvarea se bazează pe următoarea lemă:

[color=blue]Oricum am colora cu roşu/albastru muchiile unui graf complet cu 6 vârfuri,[/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=triangle&t=5820]există două triunghiuri monocromatice.[/url]

(În topicul ăla e demonstrat rezultatul general.)

Acuma mai rămâne de găsit modalitatea de colorare  :bruce:

se pare ca ai dreptate ciprian.
 :blush:

Ok...perfect_radio a fost cel mai aproape de solutie :)  Cam asta e:
In fiecare triunghi coloram cea mai mica latura cu negru. Unele muchii pot aveam mai multe "straturi de negru" dar asta nu prea ne intereseaza ca tot negre sunt :rotfl: Acum coloram restul muchiilor cu alb. Va exista un triunghi monocolor. Dar acesta nu poate fi alb deoarece are o cea mai mica latura. Rezulta ca este negru. De aici avem imediat ca cea mai mare latura in acest triunghi gasit este neagra, deci este si cea mai mica in alt triunghi q.e.d.


trigonometry
logarithms
calculus
quadratics
algebra
quadratic formula

Find all values of $t$ such that there exists an angle $A$, $0\le A<2\pi$, where $4\csc A=t^{-1}$ and $\cos^{2}A=6t$.

[hide]
$4\csc{A}= \frac{4}{\sin{A}}= \frac{1}{t}\Rightarrow 4t = \sin{a}\Rightarrow 16t^{2}= \sin^{2}A$
$\cos^{2}{A}= 6t$

we know that $\sin^{2}{A}+\cos^{2}{A}= 1$, thus:

$16t^{2}+6t = 1$
solving, we find t to be $\frac{1}{8}$ and $-\frac{1}{2}$

Case1 $t = \frac{1}{8}$:
$cos^{2}{A}= \frac{6}{8}\Rightarrow \cos{A}= \sqrt{\frac{3}{4}}= \frac{\sqrt{3}}{2}$

Case2 $t =-\frac{1}{2}$
$cos^{2}{A}=-3 \Rightarrow \cos{A}= \sqrt{-3}$ which is impossible

Thus the only value of t that fits the stipulations is: $\boxed{\frac{1}{8}}$
[/hide]

[quote="GlenthemanN"]$cos^{2}{A}=-3 \Rightarrow \cos{A}= \sqrt{-3}$ which is impossible[/quote]

[hide="If you are interested..."] This is, in fact, possible if $A$ is allowed to be complex.  We will make use of the identity $\cos A = \frac{u+\frac{1}{u}}{2}$ where $u = e^{i A}$.  Then

$u^{2}\mp 2 \sqrt{-3}u+1 = 0 \implies$
$u = \frac{ \pm 2 \sqrt{-3}\pm \sqrt{-16}}{2}= i \left( \pm \sqrt{3}\pm \sqrt{4}\right)$

Now, let $A = a+bi$.  Then

$e^{a}e^{bi}= i \left( \pm \sqrt{3}\pm \sqrt{4}\right)$

This part is somewhat subtle.  If we pick $+$ for $\sqrt{4}$ then we have the solutions

$e^{a}= \left( \sqrt{4}\pm \sqrt{3}\right), e^{bi}= i \implies$
$A = \boxed{ \ln \left( \sqrt{4}\pm \sqrt{3}\right)+i \frac{\pi}{2}}$

But if we pick $-$ for $\sqrt{4}$ then instead we have

$e^{a}= \left( \sqrt{4}\pm \sqrt{3}\right), e^{bi}=-i \implies$
$A = \boxed{ \ln \left( \sqrt{4}\pm \sqrt{3}\right)+i \frac{3\pi}{2}}$ [/hide]

[hide="that is definatly interesting.. but explain one thing"]
you say that $u = e^{i A}$ and then we find the value for u.
Then you say 'let $A = a+bi$'

then $e^{i A}$ becomes $e^{a}e^{bi}$ but shoudlnt it be

$e^{iA}= e^{i(a+bi)}= e^{ai-b}$

thanks for the clarification :)
[/hide]

Oops.  Yes, we should indeed have that instead.  My apologies.

I generally don't think of an "angle" being complex, just as I don't generally think of complex lengths.

Yeah, true, though they are interesting. It's like binominal coefficients with negative or non-integers; they don't make any real sense either. (by real, I mean applicable to real life)
Am I correct in assuming that the equation:
$e^{iy}=\cos{y}+i\sin{y}$ can only be proven with calculus?

Use the taylor series expansions of cosx and sinx, so basically yeah. It's probably on MathWorld.

I'm probably wrong but I got that $t=\frac{1}{8}$.

[hide="Solution"]

You know that $t=\frac{\sin{A}}{4}$ so you can substitute in the other equation you have to get $2\cos^{2}{A}=3\sin{A}$.

From that $4\cos^{4}{A}=9\sin^{2}{A}$
$4\cos^{4}{A}+9\cos^{2}{A}-9=0$
$\cos^{2}{A}=\frac{3}{4}$

You already know $\cos^{2}{A}=6t$ therefore you just solve for $t$ now since you know the value of $\cos^{2}{A}$ and thus you get $\boxed{t=\frac{1}{8}}$.[/hide]

Edit: Added Solution

[hide]$4\csc A=t^{-1}$ and $\cos^{2}A=6t$
$\implies \sin A/4=t=\cos^{2}A /6$
$\implies \sin 3/2=1-\sin^{2}A$.
From quadratic formula, $\sin A=1/2,-2$. -2 is extraneous.
So $t=\sin A/4=1/8$.[/hide]


In a tournament of $ n$ contestants, each contestant plays every other contestant exactly once. Let $ x_i$ be the number of wins of the $ i$th player, and $ y_i$ the number of losses. Prove

$ \displaystyle \sum_{i \equal{} 1}^n x_i^2 \equal{} \sum_{j \equal{} 1}^n y_j^2$

Does this belong in the Basics forum?

[quote="leoxnlin"]Does this belong in the Basics forum?[/quote]  Definitely not.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10992 is an old topic on the same question.  (This is from the 1965 Putnam exam.)

[hide="Solution"]
$ x_i \plus{} y_i \equal{} n\minus{}1$

$ \sum x_i \equal{} \sum y_i \equal{} \frac{n(n\minus{}1)}{2}$

$ \sum x_i^2 \equal{} \sum y_i^2 \Leftrightarrow \sum_{i,j, i\ne j} x_i x_j \equal{} \sum_{i,j, i\ne j} y_i y_j$

$ \Leftrightarrow \sum_{i,j, i\ne j} x_i x_j \equal{} \sum_{i,j, i\ne j} (n\minus{}1\minus{}x_i)(n\minus{}1 \minus{} x_j)$

A bit of simplification yields,

$ \Leftrightarrow \sum_{i,j, i\ne j} (n\minus{}1)^2 \equal{} (n\minus{}1) \sum_{i,j, i\ne j} (y_i \plus{} y_j)$

Which is trivial, as required.
[/hide]

$ \sum x_i^2\minus{}y_i^2\equal{}\sum (x_i\plus{}y_i)(x_i\minus{}y_i)\equal{}n\sum x_i\minus{}y_i\equal{}0$ because for each win a team has, another has a loss.


Determine all positive integers k :le: 2000 for which x4 + k can be factored into two distinct trinomial factors with integer coefficients.

Or you could recognize that the only way you can factor x^4 + k is when the Sophie-Germain identity applies to it.

*[i]x^4 + y^4/4 = (x^2 + y^2/2)^2 - (xy)^2 = (x^2 + xy + y^2/2)(x^2 - xy + y^2/2)[/i]*

[quote="Leonardoty"]Wow thank you this was very helpful.[/quote]

In the future, please don't bring up threads that have been dead for a while (e.g. this one is from more than a year ago).   :)

[hide] By the Sophie-Germain identity, we have that $x^4 + 4y^4 = (x^2 + 2y^2 + 2xy)(x^2 + 2y^2 - 2xy)$ In this case, $4y^4 = k$. We can only have 4 values of $y = 1, 2, 3, 4$, resulting in $k= 4, 64, 324, 1024$ respectively.  [/hide]

lol 10 years without a post this was so dead


I applied to Exeter, Andover, St.Pauls, Deerfield, and choate and got in to Exeter and Choate so far. The rest of confirmations etc are coming later. Anyone else get in/apply to these schools? these are high schools btw.

and what school do you think I should pick. feedback is welcome.

i so far think exeter or andover, though I am uncertain to which one is better.

[quote="mysmartmouth"]I would think Exeter is the better school if you are into math.  Just look at the teams they constantly put up.[/quote]

I agree. Dr. Feng's an awesome math coach!

Just wondering, were you at state MC last weekend, InsanelyBored?

Are these schools similar/same as the one in the movie Dead Poets' Society? Basically schools concentrated with rich kids who, for lots of them, make it to the Ivies?

[quote="Treething"][quote="Ubemaya"]How much does it cost to get into schools like these? What are the requirements?[/quote]

$\$$40,000+ for both questions[/quote]

Uh, not true at all. Plus, there's always financial aid. Yes, Exeter is better if your only interest is to focus on competitive math, water polo, or squash.

In my junior year i probably will make it to the Oklahoma School of Science and Math.  As you can guess, they're very science and math oriented.  Their whole curriculum is built around the ACT, the Math Olympiad, and th ephysics bowl.  I have to test in there, but i think I'll make it. i just have to make the top 80 in the state that apply for it, and ill make it.

Anyway, they placed third at the TEAMS contest last year, which is really good, since most of the rest of the country thinks we're hillbillies.  that rumor is just not true.  Only [u]most[/u] of us are hillbillies.

Plus, it's absolutely free!  It's paid for with grants and tax money! ha ha!

I got about 20 letters from schools asking me to join. about 5 were from schools like community colleges.  one was from the military. the rest were okay, i guess.  i was considering going to Baylor Something Or Other, but to be accepted, i would have to take my whole family of six on a plane to Tennessee to be interviewed. the whole thing would cost about $4000 not including$120,000 tuition for four years. :o
so we didn't do that.

answering easy as's question, I would have to tell you no on that one; I had to attend this science org so I decided not to go. our team from our school placed third; it sux. Did your school place?

anyways back to the answerers, it seems like most people concur that exeter can be the better school for me. As for me, Exeter carries more familiarity so I am glad that people here prefer Exeter.

In response to some answers I have heard:
Exeter, Andover and other such schools are not the schools that can be disregarded using ignorant stereotypes such as: "rich kids' school (when financial support is clearly available)", "fanatics' institution", or "school for kids who need extra help". In order to achieve admission, one must through an arduous process where many strong candidates are filtered in varying degrees. It is not merely the prestige that allures us all; instead, we come to meet others similar to us and learn from each other. An Exonian experience is something to be treasured; it cannot be degraded by mere words. Perhaps the unsavory rumors are the byproduct of the seemingly pervasive inaptitude of today's students.

ah end of tirade, I sometimes go into my nerd mode when I am bored. And of course by the inaptitude that is concurrent among people I am also referring to myself, as I am a pretty stupid individual.

[quote="InsanelyBored"]Exeter, Andover and other such schools are not the schools that can be disregarded using ignorant stereotypes such as: "rich kids' school (when financial support is clearly available)"[/quote]

This is an obvious logical fallacy.  If you want to know if Exeter is a "rich kids' school," you should look at the distribution of income/wealth of families of Exeter students.  It is undoubtedly skewed heavily towards the upper end of the spectrum.

EDIT: additional information: unlike at the university level, where applications are need-blind, Exeter considers the need for financial aid in its application process.  [quote="[url=http://www.exeter.edu/admissions/147_351.aspx]Exeter admissions website[/url]"]Being a candidate for financial aid places a student in a more competitive situation for admission to the Academy. [/quote]

Also, totally unrelatedly, does anyone care to explain to me what the following statement means? [quote="[url=http://www.exeter.edu/admissions/147_162.aspx]Exeter admissions website[/url]"]38 percent of us are ethnically diverse[/quote] This is like advertising bread as "100% multigrain."

Please read this page: [url]http://www.exeter.edu/news_and_events/news_events_9873.aspx[/url]

You might come across this sentence:
For the past two years, the school has been able to meet the full, demonstrated financial aid needs of all admitted students, making the Academy effectively “need blind.”

Also, I think the whole "ethnically diverse" thing just indicates the percentage of students that aren't Caucasian.

Please don't revive year-old threads...

Although I wana go to exeter too :coolspeak:


topology
real analysis
real analysis unsolved

Show that for every locally compact but not compact topological space X there is a compact topological space Y such that X is contained in Y and Y\X consists of a single point, and the space X is a subspace of the space Y.

I don't see how the construction will satisfy the compactness.  :(

The space being locally compact is irrelevant. You actually want one of the separation properties- the one thing we need is for compact sets to be closed. Depending on your exact definition, local compactness might not imply this.

Let the new point be $ p$. The definition of open sets in $ Y$: $ S$ is open in $ Y$ if (1) $ S$ is open in $ X$ (and doesn't contain $ p$) or (2) $ S$ is the union of $ p$ and the complement of a compact set in $ X$.
Since compact sets are closed in $ X$, the induced subspace topology on $ X$ is the same as the original topology.
Now, compactness. An open cover of $ Y$ contains at least one set of type (2)...

Your proof satisfies everything, including compactness. If only I had remembered that closures are closed sets.  :blush: It's hard to remember these things when most properties don't carry across all topological spaces. Could you clarify this question for me:

What happens when compact sets are not closed? Also in terms of ZF what does it mean to add the point p?

Local compactness of a topological space means every point of that space has a neighborhood whose closure is compact.

When we don't have that compact sets are closed, we use a slightly different construction: the type (2) open sets are complements of [i]closed[/i] compact sets. Local compactness is still irrelevant.

Jmerry, local compactness is not irrelevant. If $ X$ is not locally compact then $ Y$ is not Hausdorff and so it is useless

Useless or not, there is still a one-point compactification. The construction works for anything.

One example: the one-point compactification of a set $ X$ with the cofinite topology is a set with one more element and the cofinite topology.


Αυτό είναι ευκολάκι

Απο Chebychev $ \frac{a^5\plus{}b^5}{a^3\plus{}b^3}\geq\frac{a^2\plus{}b^2}{2}$

Άρα $ \sum\frac{a^5\plus{}b^5}{a^3\plus{}b^3}\geq a^2\plus{}b^2\plus{}c^2\geq\frac{(a\plus{}b\plus{}c)^2}{3}\equal{}\frac{1}{3}$


function
calculus
derivative
algebra
polynomial
superior algebra
superior algebra solved

Compute the "n^3" derivate of the function f:R-->R
f(x)=(2x+1)^2(3x^2+2x+1)^3*..*[(n+1)x^n+nx^(n-1)+..+2x+1]^(n+1) where n is a natural number.

Hi Lagrangia, is the problem asking for the third derivative f(3)(x)?

Rewriting the function in a more compact form:

f(x) =  \pi {i=1 to n+1} ( \sum {j=1 to i} (jxj-1)i)

hello 3X.lich !

it asks for f^(n^3) (x)

cheers!