For $ a,b,c$ positive reals prove that

$ (a^2\plus{}b^2\plus{}c^2)^2(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\plus{}27abc \geq 2(a\plus{}b\plus{}c)^3$

[quote="manlio"]For $ a,b,c$ positive reals prove that

$ (a^2 \plus{} b^2 \plus{} c^2)^2(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \plus{} 27abc \geq 2(a \plus{} b \plus{} c)^3$[/quote]
SOS kills it easy enough.  :wink:


inequalities proposed
inequalities
algebra

Let $a;b;c$ are real numbers satisfying $0 \leq a;b;c \leq 2$
Find the maximum value of the expression:
$E=\frac{a}{4+bc}+\frac{b}{4+ca}+\frac{c}{4+ab}$

[quote=sqing][hide]Let $a,b,c\in [0,2].$ Prove that$$\frac{a}{4+bc}+\frac{b}{4+ca}+\frac{c}{4+ab}\leq 1$$
$$\frac{a}{4+bc}+\frac{2b}{4+ca}+\frac{c}{4+ab}\leq \frac{3}{2}$$
$$\frac{a}{4+bc}+\frac{2b}{4+ca}+\frac{2c}{4+ab}\leq 2$$[/hide][/quote]
Let $f(a,b,c)=LHS$ is convex in all variables and done!!!


From a pack of 52 playing cards, one extracts the 26 red cards and pairs them up randomly. The back sides of each pair of cards are then glued together, resulting in 13 cards with both sides being "the front".
Prove that it is always possible to flip the cards so that the 13 sides facing upward are A,2,3,...,10,J,Q,K.

Pls explain to me with steps.. THZ A LOT!!

[hide="my attempt"]
Suppose this was true for n cards.
Now add in annother card. There are n-1 out of the original cards that are not paired with the new one. So you can make n-1 cards face up automatically. Now turn the two new cards so that the new card is bottom down. Becuase of the pigeonhole principal, one of these cards top faces must already be among the n-1 so you can flip it to reveal the new card.

The basis step with n=2 is obvious.
[/hide]


If $ A\sin(\alpha \plus{} \beta) \equal{} \sin \alpha$,prove that $ \tan(\alpha \plus{} \beta) \equal{} \frac {\sin\beta}{\cos\beta \minus{} A}$

[quote="Dr Sonnhard Graubner"]hello, what are $ \alpha$, $ \beta$ and $ A$ here?
Sonnhard.[/quote]

$ \alpha,\beta$ are angles.$ \alpha\plus{}\beta \not\equal{} k\pi ,k \in Z$

$ A\equal{}\frac{\sin\alpha}{\sin(\alpha\plus{}\beta)}\implies A$ is a parameter

hello, ok, thank you, i understood.
Sonnhard.


Suppose that $ x \in \mathbb{R}$ and $ x^n \plus{} a_{n\minus{}1}x^{n\minus{}1} \plus{} a_{n\minus{}2}x^{n\minus{}2} \plus{} \ldots \plus{} a_0 \equal{} 0$, where $ a_{n\minus{}1}, \ldots. a_0 \in \mathbb{Z}$. Prove that either $ x$ is an integer or it is irrational.

What relevant theorems are applicable in this problem? Any hints?  :)

Hint: The integers have unique prime factorization. Try substituting $ x \equal{} m/n$ where $ (m,n) \equal{} 1$, clear denominators, and see what happens.

I used $ k$ instead of $ n$ to avoid confusion:

Suppose $ x \equal{} \frac {k}{m}$, $ (k, m) \equal{} 1$. Then $ x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}a_{n\minus{}2}x^{n\minus{}2}\plus{}\ldots\plus{}a_{0}\equal{} 0$ becomes $ k^{n}\plus{}mk^{n\minus{}1}a_{n\minus{}1} \plus{}m^2k^{n\minus{}2}a_{n\minus{}2}\plus{}\ldots\plus{}m^{n\minus{}1}ka_1 \plus{} m^na_{0}\equal{} 0$. Where do we go from here?

$ m | 0 \implies  m | k^{n}\plus{}mk^{n\minus{}1}a_{n\minus{}1}\plus{}m^{2}k^{n\minus{}2}a_{n\minus{}2}\plus{}\ldots\plus{}m^{n\minus{}1}ka_{1}\plus{}m^{n}a_{0} \implies m|k^n \implies m |1$


i)Αν $ a^2 \plus{} x^2 \equal{} 1 , b^2 \plus{} y^2 \equal{} 1$ kai $ ay \minus{} bx \equal{} 1$ na dei3tei oti $ a^2 \plus{} b^2 \equal{} 1$

ii) $ A \equal{} 4 \plus{} x^2y^4 \plus{} x^4y^2 \minus{} 3x^2y^2$ na bre8ei to Amin (h elaxisth timh tis parastashs..)

[quote="KapioPulsar"]

ii) $ A \equal{} 4 \plus{} x^2y^4 \plus{} x^4y^2 \minus{} 3x^2y^2$ na bre8ei to Amin (h elaxisth timh tis parastashs..)[/quote]
$ A_{min} \equal{} 3$ apo AM-GM
Gia tin i)
isxyei $ (a\minus{}y)^2\plus{}(b\plus{}x)^2\equal{}0$ ara $ a\equal{}y$ kai $ b\equal{}\minus{}x$. antikathistontas stin (1) opou $ x$ to $ \minus{}b$ pairnoume to zitoumeno

[quote="Γιώργος"][quote="KapioPulsar"]

ii) $ A \equal{} 4 \plus{} x^2y^4 \plus{} x^4y^2 \minus{} 3x^2y^2$ na bre8ei to Amin (h elaxisth timh tis parastashs..)[/quote]
$ A_{min} \equal{} 3$ apo AM-GM
Gia tin i)
isxyei $ (a \minus{} y)^2 \plus{} (b \plus{} x)^2 \equal{} 0$ ara $ a \equal{} y$ kai $ b \equal{} \minus{} x$. antikathistontas stin (1) opou $ x$ to $ \minus{} b$ pairnoume to zitoumeno[/quote] :P tis ebala gia ligh e3askhsh opios 8elei sto am-gm enopsei tou efkleidh


geometry
algebra
polynomial
3D geometry
Putnam
college contests

A solid in Euclidean 3-space extends from $z = -\frac{h}{2}$ to $z = +\frac{h}{2}$ and the area of the section $z = k$ is a polynomial in $k$ of degree at most $3$ . Show that the volume of the solid is $h \frac{(B + 4M + T)}{6}$, where$B$ is the area of the bottom $(z = -h/2)$,$M$is the area of the middle section ($z = 0$), and $T$ is the area of the top $(z = h/2)$. Derive the formulae for the volumes of a cone and a sphere.

Shouldn't we try to get the actual wordings for these, as opposed to ripping off Kalva?

Do you know any place we can get the 'official' wordings??? Kalva seems to be the only place where old Putnam problems are stocked!

The books?  See if one of the other moderators have them.

I live in India where none of the books are available..... I've checked! I'll ask Valentin!


let $ x$ , $ y$ , $ z$ $ \in \mathbb R$ , prove that :

$ 3(x^2\minus{}x\plus{}1)(y^2\minus{}y\plus{}1)(z^2\minus{}z\plus{}1) \geq (xyz)^2\plus{}xyz\plus{}1$

[quote="y-a-s-s-i-n-e"]let $ x$ , $ y$ , $ z$ $ \in \mathbb R$ , prove that :

$ 3(x^2 \minus{} x \plus{} 1)(y^2 \minus{} y \plus{} 1)(z^2 \minus{} z \plus{} 1) \geq (xyz)^2 \plus{} xyz \plus{} 1$[/quote]
$ \prod_{cyc}(x^2 \minus{} x \plus{} 1)\geq\prod_{cyc}\sqrt {\frac {x^4 \plus{} 1}{2}}\geq\sqrt {\frac {\left(\sqrt [3]{x^4y^4z^4} \plus{} 1\right)^3}{8}}.$
Let $ \sqrt [3]{xyz} \equal{} t.$ Then it remains to prove that $ 9(t^4 \plus{} 1)^3\geq8(t^6 \plus{} t^3 \plus{} 1)^2.$
If $ t \equal{} 0$ it's obvious.
If $ t\neq0$ we can assume $ t \plus{} \frac {1}{t} \equal{} u.$
Id est, $ 9(t^4 \plus{} 1)^3\geq8(t^6 \plus{} t^3 \plus{} 1)^2\Leftrightarrow9(u^2 \minus{} 2)^3\geq8(u^3 \minus{} 3u \plus{} 1)^2\Leftrightarrow$
$ \Leftrightarrow u^6 \minus{} 6u^4 \minus{} 16u^3 \plus{} 36u^2 \plus{} 48u \minus{} 80\geq0\Leftrightarrow(u \minus{} 2)^2((u \plus{} 1)^4 \minus{} 12(u \plus{} 1) \minus{} 9)\geq0.$
If $ u \geq 2$ then $ (u \plus{} 1)^4 \minus{} 12(u \plus{} 1) \minus{} 9\geq27(u \plus{} 1) \minus{} 12(u \plus{} 1) \minus{} 9\geq0.$
If $ u\leq\minus{}2$ then $ (u \plus{} 1)^4 \minus{} 12(u \plus{} 1) \minus{} 9)\geq1 \minus{} 12( \minus{} 2 \plus{} 1) \minus{} 9\geq0.$ Done!

hi!
thank's arqady

did you use the $ 3$ ?

$ %Error. "blue" is a bad command.
{3}(x^2 - x + 1)(y^2 - y + 1)(z^2 - z + 1) \geq (xyz)^2 + xyz + 1$

[quote="y-a-s-s-i-n-e"]
did you use the $ 3$ ?
[/quote]
Yes! It's here:
[quote="arqady"]
Then it remains to prove that $ 9(t^4 \plus{} 1)^3\geq8(t^6 \plus{} t^3 \plus{} 1)^2.$[/quote]


Let $z=\frac{(3+4i)(\sqrt 2-\sqrt 2i)}{(\sqrt 3-i)\sqrt 5i}$. Find $|z|$.

Please hide your answer next time so that more people can try.

[hide="my idea of this problem"]We find $|z|$ for each term seperately (I hope you can do that). Since $\sqrt{a^2+b^2}=|z|$, we have

$\frac{(5)(2)}{(2)(\sqrt5)}=\frac{5}{\sqrt5}=\sqrt5$. [/hide]

236factorial, thank you very much because you make me know that $|z_1z_2|=|z_1||z_2|$.

[Let $z_1=r_1\exp i\theta_1$ and $z_2=r_2\exp i\theta_2$. Then $z_1z_2=r_1r_2\exp (\theta_1+\theta_2)i$. So $|z_1z_2|=r_1r_2=|z_1||z_2|$, and $\arg z_1z_2=\theta_1+\theta_2=\arg z_1+\arg z_2$.] :D

the idea of this problem is that you have to know the properties

$|w*z|=|w|*|z|$ and $\frac{|w|}{|z|}=\left|\frac{w}{z}\right|$

for complex numbers $w$ and $z$, with $z\ne 0$


LaTeX
geometry
algebra
system of equations
area of a triangle
Heron\u0027s formula

Solve system of equations
$ \left\{\begin{array}{c}x(y+z)=x^2+2y(z+x)=y^2+3z(x+y)=z^2+4\end{array}$

Here's a start...

Multiply the first equation by 3 and the second equation by 2 and subtract. Solve for z and plug that into the first and the third equations to get (after clearing denominators...possibly loosing solutions...)

$ 3x^3y\minus{}x^2y^2\minus{}2xy^3\equal{}18x^2\plus{}24xy\minus{}8y^2$
$ 2x^2y\minus{}2xy^2\equal{}3x\minus{}2y$

From here it follows that $ \frac{x}{y}\equal{}\minus{}2,0,\frac{2}{9}, 1$ and maybe $ \frac{2}{3}$.

Now there is a lot of casework...

You sure the first equation isn't x(y+z) on the LHS instead of (y+z)?  :maybe:

It is in the $ \text{\LaTeX}$.  I'm not sure why it's not rendering.

[quote="mathfunny"]Solve system of equations
$ \{\begin{array}{c} x(y + z) = x^2 + 2 \\
y(z + x) = y^2 + 3 \\
z(x + y) = z^2 + 4\end{array}$[/quote]
I corrected it.

[hide="Solution"]
[i]-- Partitioned because it looks messy otherwise --[/i]

[hide="1 System Rewritten"]
i. $ x(\minus{}x\plus{}y\plus{}z)\equal{}2$
ii. $ y(x\minus{}y\plus{}z)\equal{}3$
iii. $ z(x\plus{}y\minus{}z)\equal{}4$[/hide][hide="2 We can divide by x y and z"]First of all, easily notice that no variable can be zero, or else there will be one of the following contradictions: $ 0\equal{}2,3,4$. So, we can happily divide by x, y, or z.[/hide][hide="3a Elimination"]Adding $ (i)\plus{}\frac{x}{y}(ii)$ gives:
$ x(\minus{}x\plus{}y\plus{}z)\equal{}2$
$ x(x\minus{}y\plus{}z)\equal{}3\frac{x}{y}$

$ \implies 2xz \equal{} 2 \plus{} 3\frac{x}{y} \implies 2xyz \equal{} 2y \plus{} 3x$.[/hide][hide="3b More Elimination"]Performing a similar elimination with $ (ii)\plus{}\frac{y}{z}(iii)$ gives $ 2xyz\equal{}2z\plus{}4y$
Finally, a similar operation with $ (iii)\plus{}\frac{z}{x}(i)$ gives $ 2xyz\equal{}4x\plus{}2z$ [/hide][hide="4 Combination"]Hence, $ 3x\plus{}2y \equal{} 3z\plus{}4y \equal{} 4x\plus{}2z$[/hide][hide="5 Some Solving"]
$ 3x\plus{}2y \equal{} 3z\plus{}4y \iff 3(x\minus{}y) \equal{} 2y \iff y \equal{} \frac{3}{2} (x\minus{}z)$

$ 3x\plus{}2y \equal{} 4x\plus{}2z \implies 2(y\minus{}z) \equal{} x \implies 3(x\minus{}z)\minus{}2z \equal{} x \implies x \equal{} \frac{5}{2} z$

$ y \equal{} \frac{3}{2} (x\minus{}z) \implies y \equal{} \frac{9}{4}z$
[/hide][hide="6 Substitution"]
$ z(x\plus{}y)\equal{} \left( \frac{5}{2} \plus{} \frac{9}{4} \right) z^2 \equal{} z^2 \plus{} 4 \implies \frac{15}{4} z^2 \equal{} 4 \implies z \equal{} \pm \frac{4}{\sqrt{15}}$ [/hide][hide="7 Solution"]
$ (x,y,z) \equal{} \pm \left(\frac{2}{3} \sqrt{15} , \; \frac{3}{5} \sqrt{15} , \; \frac{4}{15} \sqrt{15} \right)$
[/hide][/hide]

[hide="Symmetrically"] First, we rewrite as

$ x( \minus{} x \plus{} y \plus{} z) \equal{} 2$
$ y(x \minus{} y \plus{} z) \equal{} 3$
$ z(x \plus{} y \minus{} z) \equal{} 4$

Next, let $ x \equal{} \frac {a \plus{} b}{2}, y \equal{} \frac {b \plus{} c}{2}, z \equal{} \frac {c \plus{} a}{2}$.  Now we rewrite as

$ bc \plus{} ca \equal{} 4$
$ ab \plus{} ca \equal{} 6$
$ ab \plus{} bc \equal{} 8$

Adding them all and dividing, we have $ ab \plus{} bc \plus{} ca \equal{} 9$, hence

$ ab \equal{} 5$
$ bc \equal{} 3$
$ ca \equal{} 1$

Multiplying them all and taking the square root, we have $ abc \equal{} \pm \sqrt {15}$, hence

$ a \equal{} \pm \frac {\sqrt {15} }{3}$
$ b \equal{} \pm \sqrt {15}$
$ c \equal{} \pm \frac {\sqrt {15} }{5}$

and we find the same solution set as The Zuton Force.  :) [/hide]
Edit:  I generally prefer algebraically symmetric solutions to substitution.  As to the motivation for the above manipulations, they are rooted in geometric considerations.  The tangent points of the incircle of a triangle with side lengths $ x, y, z$ divide those sides into lengths $ a, b, c$ so that $ x \equal{} a \plus{} b, y \equal{} b \plus{} c, z \equal{} c \plus{} a$.  Indeed, Heron's formula can be rewritten as

$ A \equal{} \sqrt{abc(a \plus{} b \plus{} c)}$


:)

Find positive integers $t, u, v, w, x$ such that $2^{u}\cdot 3^{v}\cdot 5^{w+2}\cdot 7^{x}= t!$

impossible, 5^3|5^k+2|t! so t has three factors of five so it is at least fifteen but then other primes like 13 divide t!.

I don't see how there could be an answer (I probly missed the trick)

[hide]
As you can see, the only prime factors of $t!$ are 2, 3, 5 and 7. That means $t! < 11!$. On the other hand, we see that 5 appears at least two times, so we need a factorial smaller than 11! but which has twice the factor 5. That would be 10!.

$10! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 = 2 \cdot 3 \cdot 2^{2}\cdot 5 \cdot 2 \cdot 3 \cdot 7 \cdot 2^{3}\cdot 3^{2}\cdot 2 \cdot 5= \boxed{2^{8}\cdot 3^{4}\cdot 5^{2}\cdot 7}$

Sorry, the "hide" tag seems not to work :|
[/hide]

But then $w = 0$ and is not a positive integer.

This is a very old problem (1967), don't blame me. Anyway, i knew that $0$ cand be either positive or negative ;)

Hm? Postive means greater than zero. That's why the distinction positive and non-negative are made; the second includes zero...


[quote="jeez123"]Simplify:

$ \frac {(1 \minus{} x)(x^5 \plus{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1)}{(1 \minus{} x \plus{} x^2)(1 \plus{} x \plus{} x^2)}$[/quote]
[hide]
Note that $ x^5\plus{}x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1\equal{}\frac{x^6\minus{}1}{x\minus{}1}$, as it is a geometric series, as are $ x^2\plus{}x\plus{}1\equal{}\frac{x^3\minus{}1}{x\minus{}1}$ and $ x^2\minus{}x\plus{}1\equal{}\frac{x^3\plus{}1}{x\plus{}1}$. So the fraction becomes $ \frac{(1\minus{}x)(x^5 \plus{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1)}{(1 \minus{} x \plus{} x^2)(1 \plus{} x \plus{} x^2)}\equal{}\frac{(1\minus{}x)\frac{x^6\minus{}1}{x\minus{}1}}{(\frac{x^3\plus{}1}{x\plus{}1})(\frac{x^3\minus{}1}{x\minus{}1})}\equal{}\frac{1\minus{}x^6}{\frac{x^6\minus{}1}{x^2\minus{}1}}$, Which simplifies to $ 1\minus{}x^2$.[/hide]

[quote="Icy"]Is there some general method for factoring of this sort, or do you just have to memorize it?[/quote]

[[Geometric series]]:
$ x^n \plus{} x^{n \minus{} 1} \plus{} \cdots \plus{} x \plus{} 1 \equal{} \frac {x^n \minus{} 1}{x \minus{} 1}$

Thats pretty much all you needed to know here.

But then how do you go about factoring $ x^n \minus{} 1$?  ;)

The "general method" that you're looking for is the concept of [url=http://mathworld.wolfram.com/CyclotomicPolynomial.html]cyclotomic polynomials[/url].


inequalities
number theory unsolved
number theory

Find all positive integers $ x,n$ so that $ x^n \plus{} 2^n \plus{} 1|x^{n \plus{} 1} \plus{} 2^{n \plus{} 1} \plus{} 1$.

If $ n\equal{}1$ then $ x\equal{}4$ or $ x\equal{}11$.
If $ n\ge 2,x\ge 3$, then $ (x\minus{}1)(x^n\plus{}2^n\plus{}1)<x^{n\plus{}1}\plus{}2^{n\plus{}1}\plus{}1<x(x^n\plus{}2^n\plus{}1)$.
Check $ x\equal{}1,2$ (non solution).
Only $ n\equal{}1,x\equal{}4,11$.


for positive $a_i$ and $u_i$, for : $1\le i< j\le n$ we have: $a_i\geq a_j, u_i\geq u_j$
prove:
$\frac{(\displaystyle{\sum_{i=1}^n {a_i}})(\displaystyle{\sum_{i=1}^n {u_i}})}{n}\le\frac{n(\displaystyle{\sum_{i=1}^n {{a_i}^2}{{u_i}^2})}}{\displaystyle{\sum_{i=1}^n {a_i}{u_i}}}$

One simple aplication of Chebychev and Cauchy  :)


[hide]
guess and check yields e=7[/hide]

You should probably mention in the problem that $e$ is an integer.

[quote="Arvind_sn"]What is the highest possible $e$ in $2^{e}<200$?[/quote]

[hide]We got that $2^{7}=128$ and $2^{8}=256$. Then, $2^{7}<200<2^{8}$. So, the answer is $\boxed{7}$[/hide]

By using the guess and check method, we obtain 7 as our final answer!

[hide="if e is not an integer"]
if e is not an integer e is approximately 7.64[/hide]

[hide]2^8 is too high, but 2^7=128 works, so 7 is the answer[/hide]

[quote="nutz_for2.718281828"]Do you mean that e is 2.718281828... or e is an unknown?[/quote]

In this case, it's obviously an unknown. He should really use a different variable like $x$ or something.

Oh, well!You should really mention if it an integar otherwise 7 isn't the right solution :maybe:

[hide]$2^{7}$ is 128, and $2^{8}$ is 256, which is too high. Thus the highest possible value of $e$ is [b]7[/b].[/hide]

Because, if not, you can't find the highest possible value

I believe it is...

[hide]$x < \frac{log(200)}{log(2)}$

...or 7 if it is an integer.[/hide]

[quote="Arvind_sn"]What is the highest possible $e$ in $2^{e}<200$?[/quote]

[hide]I listed all the possible ones for e, and the closest one is e=7, which yields a result of 128. However, one could also use logs, but that is not Classroom math.[/hide]

[hide]$2^{e}<200$
$2^{e}<2^{3}*25$
$2^{e-3}<25$
$e-3<\frac{log(25)}{log(2)}$
$\frac{log(25)}{log(2)}$ is about 4.643856
$e<7.643856$[/hide]

[size=75][color=green]Mod edit: Try not to use such complicated methods in a middle school forum. :) [/color][/size]

[hide]2^1=2
2^2=4
2^3=8
continue
16
32
64
128
thats all then count so its 7!
ok so this way is lame, but it works on small numbers[/hide]

[hide]2^7 is the largest for of 2^e < 200, (2^7 = 128)[/hide]


How can we solve this 3rd degree equation where the unknown is $ x\ge0$ and a,y constants <br />
$ \frac{x^3}{9a^2}\minus{}\frac{2x^2}{3}\plus{}a^2x\minus{}y^2\equal{}0$

 The above equation comes from the main problem which is to  express this formula  $ y\equal{}(\frac{x}{3a}\minus{}a)\sqrt x$  as term of x

Is there any other way to  express the second formula    as term of x  except to solve the first 3rd grade equation??

Thanks anyone in advance

[quote="Immanuel Bonfils"]May be better make $ z \equal{} \sqrt x$ to get $ z^3 \minus{} 3a^2z \minus{} 3ay \equal{} 0$ that still is of the third degree, but now we can call $ z \equal{} \sum_1^2u_n$ (apud Cardano) an obtain a second degree equation in $ w \equal{} u^3$.
If you'll nedd more help just ask.[/quote]

What do you mean by this: but now we can call $ z \equal{} \sum_1^2u_n$ (apud Cardano) an obtain a second degree equation in $ w \equal{} u^3$.

Thanks for your reply

hello, one solution is
$ \displaystyle{x\equal{}2 a^2\plus{}\frac{\sqrt[3]{2} a^4}{\sqrt[3]{\minus{}2 a^6\plus{}9 a^2 y^2\plus{}3
   \sqrt{9 a^4 y^4\minus{}4 a^8 y^2}}}\plus{}\frac{\sqrt[3]{\minus{}2 a^6\plus{}9 a^2 y^2\plus{}3
   \sqrt{9 a^4 y^4\minus{}4 a^8 y^2}}}{\sqrt[3]{2}}}$
Sonnhard.

If we put  $ z \equal{} u_1 \plus{} u_2$ in equation $ \;\;\; (*)\;\;\;  z^3 \minus{} 3a^2z \minus{}3ay \equal{} 0$ we'll get

 $ u_1^3\plus{}u_2^3 \minus{} 3ay \plus{} 3(u_1\plus{}u_2)(u_1u_2\minus{}a^2) \equal{} 0$, leading to

$ u_1^3 \plus{} u_2^3 \equal{} 3ay$ and $ u_1u_2 \equal{} a^2$,

so, $ (u_1^3, u_2^3)$ are the roots of the second degree equation

$ (**)\;\;\;  w^2 \minus{} 3ayw \plus{} a^6 \equal{} 0$ where $ w \equal{} u^3$.


Which countries are there where
- the people speak English- i.e. an American can communicate with the people there easily
- there is electricity and the country is industrialized
- the country does not have frequent coup d’états or violence

wait all of them speak English? Would an English speaker be required to learn some of the language in those countries as part of a class or could one get by with just knowing a few phrases of those languages. Like "where's the bathroom" in Korean?

many do, especially at a university. I, for one, started studying English at age 6. But yeah it'd be nice if you tried to learn some of the language. :) However, I'm under the impression that Korean (and Japanese, too) is a hard language to learn for Westerners.

[quote="undefined117"]I know South Korea does (the Seoul National University is apparently extremely hard to get into?), and I'm guessing most of Europe..[/quote]
Are you coming from Korea,underfined117?How about KASIT,I don't remmember what exactly it means (may be ...Seoul Insitute of Technology :D )?My friends all say it is quite good to learn majors concering computer such as computer science,computer engineering,..
and it seems to be easier to get into.If I am not mistaken,KASIT official requires only SATI .Really ?? And I want to know more about scholarship at here  :lol:
Thank in advance!

KA[b]IS[/b]T stands for Korea Advanced Institute of Science and Technology, located in Taejon (another metropolitan city not far from Seoul). From what I've heard, KAIST and Seoul Natl. University used to be/are some of the hardest universities to get into in South Korea. My dad got his bachelor's and master's degrees at those two schools---now he's a nuclear engineer.

I don't know much about the admission processes to those universities. But sure, computer science is a great field. :lol: I'm considering it too. Good luck!

Assuming that you're American, why are you interested in studying abroad?  The answer is highly relevant to the question of which colleges would be right for you.


I can't remember the name for this form of reasoning and it's driving me crazy.  Someone (I forget who) used it to "prove" that two moving (in the same direction) objects seperated by a finite distinance would never meet even though the trailing object was moving faster.  It basically involved using extremely small measurements of time.  If anyone knows what I'm talking about, please let me know.
Thanks.

As ComplexZeta points out:
[url=http://mathworld.wolfram.com/ZenosParadoxes.html]mathworld.wolfram.com/ZenosParadoxes.html[/url]


Prove that $\left(\frac{1-i\sqrt{7}}{2}\right)^n +\left(\frac{1+i\sqrt{7}}{2}\right)^n$ (where $n$ is a positive integer) is odd.

Denote $a_n=\left(\frac{1-i\sqrt{7}}{2}\right)^n+\left(\frac{1+i\sqrt{7}}{2}\right)^n$

We have $(x-\frac{1-i\sqrt{7}}{2})(x-\frac{1+i\sqrt{7}}{2})=x^2-x+2$

Since this is the characteristic polynomial of the sequence $a_n$, we have $a_{n+2}-a_{n+1}+2a_n=0$, which is easily checked as true.

Since $a_1=1$ and $a_2=-3$, all $a_n$ are integer. Also, the difference $a_{n+2}-a_{n+1}$ is even, hence $a_{n+2}$ is of the same parity as $a_{n+1}$, meaning that all $a_n$ are of the same parity as $a_1$, and $a_1=1$


H dosmeni sxesi ginetai $2(xy+yz+xz)=3xyz \ \ (1)$

H pros apodeiksi anisotita grafetai $\frac{x^{2}}{x^{2}+xyz}+\frac{y^{2}}{y^{2}+xyz}+\frac{z^{2}}{z^{2}+xyz}\geq 1$

Omws apo tin anisotita Andreescu exoume oti

\begin{eqnarray}\frac{x^{2}}{x^{2}+xyz}+\frac{y^{2}}{y^{2}+xyz}+\frac{z^{2}}{z^{2}+xyz}&\geq&\frac{(x+y+z)^{2}}{x^{2}+y^{2}+z^{2}+3xyz}\nonumber \\ &=&\frac{x^{2}+y^{2}+z^{2}+2(xy+yz+xz)}{x^{2}+y^{2}+z^{2}+3xyz}\nonumber \\ &\stackrel{(1)}{=}& 1 \nonumber \end{eqnarray}

Alexandros

Σωστα!Μαλλον παραηταν ευκολη... :oops:

Μμμ! Δεν ήταν καθόλου εύκολη! Εύκολη φαίνεται σε όποιον τη λύσει ή τη δει λυμένη.Φυσικά δεν είναι επιπέδου ΙΜΟ ,  αλλά είναι πολύ ωραία και διδακτική. Μακάρι να μπορούσαν να τη λύσουν οι μισοί από τους μαθηματικούς και το 1/1000 των μαθητών του Λυκείου!!!

 Αυτά τα γράφω γιατί ποτέ δεν πρέπει να υποτιμάτε  τις ικανότητές σας.Στα μαθηματικά και η πιο απλή άσκηση κρύβει τη δυσκολία της.Αν λυθεί , αυτό είναι πάντα μεγάλη επιτυχία , και αυτός που τη λύνει πρέπει να απολαύσει το αποτέλεσμα της προσπάθειάς του , αναξάρτητα από το βαθμό δυσκολίας της άσκησης.
   Συμπέρασμα :
                         [color=red]  ΛΥΣΤΕ ΚΑΙ ΑΠΟΛΑΥΣΤΕ [/color]

 κάθε άσκηση. Κάθε άσκηση που δεν έχουμε λύσει είναι δύσκολη και κάθε άσκηση που λύσαμε  είναι τελικά εύκολη!

Κυριε Στεργιου εχετε δικιο!Καποιες ασκησεις σου δειχνουν απο μονες τους οτι πρεπει να σκεφτεις απλα,πραγμα που πολλες φορες το αγνοουμε..Και το απλο,ειναι πραγματι εντυπωσιακο...!

Συμφωνω φιλε!Καλοσηρθες! :lol:

Alternative:
$ \sum\frac{1}{1\plus{}\frac{xyz}{x^2}} \ge 1$
Έστω η $ f(x)\equal{}\frac{1}{1\plus{}\frac{a}{x^2}}$, me $ a\equal{}xyz$.
$ f''(x) \ge 0$
Ara
$ LHS \ge \frac{3}{1\plus{}\frac{a}{\left(\frac{x\plus{}y\plus{}z}{3} \right)^2}}$
Etsi exoume ndo:
$ (x\plus{}y\plus{}z)^2 \ge \frac{9xyz}{2}$
Alla apo ton periorismo:
$ xy\plus{}yz\plus{}zx \equal{}\frac{3}{2}xyz \Leftrightarrow 3\sum xy \equal{}\frac{9xyz}{2} \le (x\plus{}y\plus{}z)^2$.

[quote="Dimitris X"]
Έστω η $ f(x) \equal{} \frac {1}{1 \plus{} \frac {a}{x^2}}$, me $ a \equal{} xyz$.
$ f''(x) \ge 0$
Ara
[...][/quote]

Εδώ χάνει η προσέγγισή σου... δε σου δίνεται η συνθήκη $ xyz \equal{} \text{const.}$ για να μπορείς να μεταχειρισθείς το $ a$ ως σταθερά στην παραγώγιση.

Cheerio,

Durandal 1707

Νομιζω οτι η κατασταση στη λυση του Δημητρη σωζεται.

Πιστευω πως αυτο γινεται δουλευοντας με ολες τις συναρτησεις $ f(x)\equal{}\frac{1}{1\plus{}\frac{a}{x^{2}}}$, για καθε πραγματικη σταθερα $ a$

Για να γινει πιο ευκολα κατανοητο θα προσπαθησω να δωσω μια εξηγηση με αντιθετοαντιστροφη και ατοπο. Εστω μια σταθερη τριαδα $ (x_0, y_0, z_0)$ με $ x_0y_0z_0 \equal{} k \equal{} const$ που ικανοποιει τη συνθηκη αλλα οχι την ανισοτητα. Τοτε απο Jensen στην $ f(x) \equal{} \frac{1}{1\plus{}\frac{k}{x^{2}}}$ εχουμε την πρωτη ανισοτητα που διχνει ο Δημητρης για ολες τις τριαδες οποτε και για την συγκεκριμενη τριαδα $ (x_0, y_0, z_0)$. Οποτε για να μην ισχυει η ανισοτητα θα πρεπει να ειναι $ 3(x_0y_0 \plus{} y_0z_0 \plus{} z_0x_0) > (x_0 \plus{} y_0 \plus{} z_0)^2$ το οποιο ειναι αδυνατο.
Πειτε μου αν χανει καπου αυτο που λεω

[quote="it's me"]Αν \[x,y,z>0\] με \[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2}\] να δειξετε οτι \[\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}\geq 1\][/quote][quote="cretanman"]H dosmeni sxesi ginetai $2(xy+yz+xz)=3xyz \ \ (1)$

H pros apodeiksi anisotita grafetai $\frac{x^{2}}{x^{2}+xyz}+\frac{y^{2}}{y^{2}+xyz}+\frac{z^{2}}{z^{2}+xyz}\geq 1$

Omws apo tin anisotita Andreescu exoume oti

\begin{eqnarray}\frac{x^{2}}{x^{2}+xyz}+\frac{y^{2}}{y^{2}+xyz}+\frac{z^{2}}{z^{2}+xyz}&\geq&\frac{(x+y+z)^{2}}{x^{2}+y^{2}+z^{2}+3xyz}\nonumber \\ &=&\frac{x^{2}+y^{2}+z^{2}+2(xy+yz+xz)}{x^{2}+y^{2}+z^{2}+3xyz}\nonumber \\ &\stackrel{(1)}{=}& 1 \nonumber \end{eqnarray}
[/quote] μια συμπλήρωση από [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=780274#p780274]εδω[/url][quote="cretanman"]Gia na isxiei i isotita stin anisotita Andreescu, tha prepei na isxiei \[\frac{x}{x^{2}+xyz}=\frac{y}{y^{2}+xyz}=\frac{z}{z^{2}+xyz}\] 'h isodinama \[x+yz=y+xz=z+xy\] ap'opou linontas (fisika prepei to zeugos $(x,y,z)$ na epalitheuei kai tin $(1)$) pairnoume oti i isotita isxiei ean $x=y=z=2$[/quote]


superior algebra
superior algebra solved

Prove that the group morphisms $f: (\mathbb{C},+)\to(\mathbb{C},+)$ for which there exists a positive $\lambda$ such that $|f(z)| \leq \lambda |z|$ for all $z\in\mathbb{C}$, have the form
\[ f(z) = \alpha z + \beta \overline{z} \] for some complex $\alpha$, $\beta$.

[i]Cristinel Mortici[/i]


Here's a very interesting qn:

P is a point on the side BC and Q is a point on the side AC of a triangle ABC such that AB=BC=CA. AP and BQ meet at R and angleARB = 120°. CB is produced to a point S such that B is between S and C and SP = BQ. CA is produced to a point T such that A is between T and C and TQ = AP. AS and BT meet at U. Prove that angleSUB = 30°.

I've a hard time finding the congruent triangles needed to prove that SUB is 30°! :oops:

Dont know if it helps, but QBC is congruent to PAB, actually, which tells us AP=BQ!!! And the problem collapses to nothing!

Hi, thx for your reply! But how did you know that they are congruent? Which congruency test did you use here?

For proving QBC and PAB congruent,

[hide]We know BC=AB and <BCQ=<ABP(=60) (because you're given AB=BC=CA).

Let <PAB=x. Since triangle ABR has angles x & 120, the remaining angle, <RBA, is 60-x. Since we know that <CBA=60 because triangle ABC is equilateral, <QBC=60-(60-x)=x. So <PAB=<OBC.

So triangle QBC is congruent to triangle PAB by ASA.[/hide]

Thank you very much for your help! I really appreciate it lots!  :D


advanced fields
advanced fields unsolved

Forgive my ignorance if this is like a well known problem in mathematics, but is there any general method for solving $\ x^x=k$ where k is  a constant real number?

Dunno if it's well known, but you can simplify it to
x logx=logk
You could then put in k and solve the equation, I suppose...it might be difficult...
In this state, however, might be easier to deal with.  For instance, let k, whatever it is, be the base of the log.  Then you have log[k]x=1/x
You could probably solve that...

Here's an interesting one I once found a formula for, but it looked pretty disgusting..$a^b=6, b^a=7$ or any numbers, don't just have to be 6 and 7.  Solve for a and b.

I think there is no general method for solving this properly and exact, since most times the answer will be transcendental.
But for an aproximate solution you could use Newton-interpolation (start with some accurate $x_0$ and set $x_{n+1} = x_n- \frac{1 - \frac{k}{x_n^{x_n}}}{1+\ln(x_n)}$, because then $x_n$ will be very close to solution for big $n$)

K81o7 your method does not work, ZetaX I pretty much have no idea what you are talking about.


Let a triangle $ APQ$ and its incircle which touches the sides $ AQ,\ QP,\ PA$ at $ C,\ F, B$ .The parallel through $ P$ meets $ BC$ at $ R$ and $ RQ$ meets $ AP$ at $ M$ . Prove that the tangent to incircle which is parallel to $ AQ$ passes through $ M$ .

Denote $ F$ the point where the tangent from $ M$ w.r.t to the incircle cuts the circle and $ H \in MF \cap PQ$.

But $ X \in BD$ by the Newton theorem ($ D$ is the tangenty point of the incircle with $ PQ$), i.e $ X$ is on the polar of $ P$. Also $ X$ is on the polar of $ Y$, where $ Y \in MF \cap AQ$. $ (1)$

Thus $ PY$ is the polar of $ X$, so $ IX \perp PY$., But $ IX \perp AQ$, i.e $ IX \perp PR$, thus $ Y \in PR$.

But again, $ Y$ is on $ AQ$ from the definition in $ (1)$, thus $ Y \in AQ \cap PR \Rightarrow Y \equiv \infty$, so by replacing now in $ (1)$, we obtain $ MF \| AQ$.

Very nice ,great solution  :)  .Thank you for sharing it with us .
My solution is an algebraic one with trigonometry .


Prove the inequality for any point P and triangle ABC:

   $AB^2+BC^2+AC^2\leq 3(PA^2+PB^2+PC^2)$

To view all the problems of Uzbekistan Undergraduate Olympiad problems visit
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=81099[/url]

here is a more general one

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15558[/url] ;)

[quote="rustam"]IN your second solution you assumed that P is on the plane of triangle ABC.

My solution in the contest:
Using the Leibniz theorem I proved that RHS-LHS=9(PG)^2.

Another solution:
If you use the Vectors then its easy to find that RHS-LHS=([b]PA[/b]+[b]PB[/b]+[b]PC[/b])^2
(VECTORS IN BOLD)[/quote]

A very good generalization.
Can you tell the detail of the Leibniz theorem you used?
I only know the Leibniz theorem in Calculus. :)

[quote="luimichael"]
Can you tell the detail of the Leibniz theorem you used?
I only know the Leibniz theorem in Calculus. :)[/quote]

The Leibniz theorem is :
Let ABC be the triangle with the point M(centre of mass,usually you use it as G) and P arbitrary point(maybe in space too) then
$PA^2+PB^2+PC^2=\frac13(AB^2+BC^2+AC^2)+3PM^2$ if i'm not mistaken..........
During the contest i proved

[quote="luimichael"]curious, that is what I have shown in the post.
Besides, in the contest , do you score bonus marks if you can prove a stronger statement?[/quote]

No.
As i said I first prove the Leibnitz but official solution was: RHS-LHS=[b](PA+PB+PC)^2 [/b]
(VECTORS IN BOLD)
Only one contestant solved it like that and he get the  :first:

There is another general version of this inequality.

If a point $X$ coplanar to $\triangle ABC$ has areal co-ordinates $(\lambda, \mu, \nu)$ (ie. $\lambda + \mu + \nu = 1$) then for any other coplanar point $P$, \[ \lambda AP^2 + \mu BP^2 + \nu CP^2 = \lambda AX^2 + \mu BX^2 + \nu CX^2 + PX^2  \] This naturally leads to the inequality \[ \lambda AP^2 + \mu BP^2 + \nu CP^2 \geq \lambda AX^2 + \mu BX^2 + \nu CX^2  \]


calculus
integration
logarithms
calculus computations

Define $ I_{n} \equal{} \int_{1}^{e} x^{k}(\ln x)^{n}\ dx$, where $ k$ is any constant number other than $ \minus{}1$ and $ n$ is a positive integer or zero. Show that, for $ n\ge1$, $ (k\plus{}1)I_{n} \equal{} e^{k\plus{}1}\minus{} nI_{n\minus{}1}$.

Hence find $ \int_{0}^{e} \frac{(\ln x)^{2}}{\sqrt{x}}\ dx$.

What is the value of $ I_{n}$ when $ k\equal{}\minus{}1$?

$ I_n\equal{}\int_1^e x^k(\ln x)^n dx \equal{}\int_1^e x^{k\plus{}1} \frac{(\ln x)^n}{x}dx$

$ \equal{} \left[ x^{k\plus{}1} \frac{(\ln x)^{n\plus{}1}}{n\plus{}1}\right]_1^e \minus{} \int_1^e \frac{k\plus{}1}{n\plus{}1}x^k (\ln x)^{n\plus{}1}dx$

$ \equal{}\frac{e^{k\plus{}1} \minus{}(k\plus{}1)I_{n\plus{}1}}{n\plus{}1}$

which, when you plug in n-1 for n, gives you the desired reduction formula

for the exact value.. it seems you are looking for $ I_2$ when $ k\equal{}\minus{}\frac{1}{2}$ so

$ I_2\equal{}2\sqrt e\minus{}4I_1\equal{}2\sqrt e \minus{}4(2\sqrt e \minus{} 2I_0)\equal{}10\sqrt e\minus{}16$

when $ k\equal{}\minus{}1$ $ I_n\equal{}\int_1^e (\ln x)^n d(\ln x) \equal{} \int_0^1 u^n du \equal{} \frac{1}{n\plus{}1}$

hello, we can derive from above that
$ \int_{0}^{e}\frac{(\ln(x))^2}{\sqrt{x}}\,dx\equal{}10e$.
Sonnhard.


Hi,

I am working on a article and do not want a boring title. Is it possible to do things like here http://zoonek.free.fr/LaTeX/LaTeX_samples_title/0.html ?
More precisely I want something like sample number 23. I copied it to winedt and it works but when i want to view it as a pdf file, the line beneath the title disappears. Very strange.

If you'd looked at the Error that is reported:
File 'dessin' not found
you could have found the problem line which is
\fbox{\includegraphics[width=3em]{dessin}}%

So you need to draw your own line and save the image in a file called dessin.png.

Allright, but where i am supposed to draw my own line ? Paint ?

It doesn't matter as long as you save it as one of the file types recognized by LaTeX. Paint is completely fine for that task.

I've just realised that you mean sample 22 not 23 dessin is the portrait in sample 23. Did you compile via PS since it uses postscript in psboxit and pstricks?

No, but wait. To be clear I meant idd sample 22. But I do not where to button to compile via PS. There is only latex and tex.

edit: When I open it as a ps file I can see that black line.  What should I do no so that black line stays if I open it as a pdf file? I think change psboxit in pdfboxit etcetera ?

I don't use WinEdt so I don't know what you need, but the dvi file that is produced needs to be processed by something like dvips and Ghostscript's ps2pdf. So you are quite likely to need Ghostscript installed and then tell WinEdt where it is.

[quote="Amazigh"]edit: When I open it as a ps file I can see that black line.  What should I do no so that black line stays if I open it as a pdf file? I think change psboxit in pdfboxit etcetera ?[/quote]If you don't go via ps then psboxit is ignored, so no line - hence dvi -> ps -> pdf using dvips and ps2pdf.

Brilliant, it is ok now :)
Thank you very much.

I actually have another question;  where do I need to change te parameters if I want that black line to be taller ?
Right now, that black line stops in the middle of the page.

ps: On many articles you see at the top of each page the subject written, you know what I mean :) How can I do that ?

psboxit.sty is an ancient package made for an older version of LaTeX. To use it you must manually adjust the parameters inside the file. Try replacing
[code]        \number\dp0 \space 15800 div    % hand tuned for dvips
        \number\wd0 \space 15800 div    % hand tuned for dvips
        \number\ht0 \space -15800 div   % hand tuned for dvips [/code]by
[code]        \number\dp0 \space 595 div    % hand tuned for dvips
        \number\wd0 \space 595 div    % hand tuned for dvips
        \number\ht0 \space -595 div   % hand tuned for dvips[/code]adjsting the 595 to whatever suits you.

[quote="Amazigh"]ps: On many articles you see at the top of each page the subject written, you know what I mean :) How can I do that ?[/quote]Use [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Packages#fancyhdr]fancyhdr[/url]

[quote="stevem"]psboxit.sty is an ancient package made for an older version of LaTeX. To use it you must manually adjust the parameters inside the file. Try replacing ...[/quote]
But that code does not appear in my tex-file ?

I am sorry but I don't understand where I can do the change.

You need to edit the file psboxit.sty which is in your LaTeX tree [i]your_latex_directory/tex/latex/misc209/psboxit.sty[/i]
Or you could get yourself another copy from [url=http://tug.ctan.org/tex-archive/macros/latex209/contrib/misc/psboxit.sty]CTAN[/url] and put it it the same directory as your tex file and this version, which you edit, will take precedence over the one in the LaTeX tree.

Allright, I decided to do option 2.
Which program can open psboxit.sty ? I tried with notepad, but I get a chaotic text.

Notepad doesn't know how to deal with non-Windows files' line endings. Why not use WinEdt?


A very old one.
[hide]The rectangle with sides parallel to the angle bisectors of the two lines, and passing through $ P$. [/hide]


geometry
percent
ratio
algebra
linear equation

hi, Is there any easy way to find the radius without using the caculator?

Billy-Jo has two Frisbees.  One has an area of 120 square inches and the other has an area of 76.8 square inches.  What percent shorter is the radius of the smaller Frisbee compared to the radius of the larger Frisbee?

Dill, Nill, and Swill each were measuring some string. Dill wrote the number of inches in his string, Nill wrote the number of feet in his string and Swill wrote the numer of yards in his string. The number Dill wrote was ten times what Nill wrote, while Nill's number was ten more than Swill. THe total length of the three strings was 52 feet 2 inches. How long was each person's string?

the ratio of the radius is a/b with area of a^2/b^2

EDIT: oh i forgot second problem

are they measuring the same string?

Dill's number is $ 10$ times larger than Nill's, but his unit of measurement is $ 12$ times smaller, so the length of his string is really $ \frac{5}{6}$ that of Nill's.  Nill's number is $ 10$ times larger than Swill's, but his unit of measurement is $ 3$ times smaller, so the length of his string is really $ \frac{10}{3}$ that of Swill's.  Let the length of Dill's string be $ x$ inches.  Then Nill's string is $ \frac{6}{5}x$ inches, and Swill's string is $ \frac{9}{25}x$ inches.  We know the total is $ 12(52)+2=626$ inches, so we can write a linear equation:

\begin{align*}
x+\frac{6}{5}x+\frac{9}{25}x&=626\\
25x+30x+9x&=15650\\
64x&=15650\\
x&=\frac{7825}{32}
\end{align*}

From here, we can quickly solve for the length of Nill's string and Swill's string.  Sorry if I made any dumb arithmetic errors.

[hide="1"]The ratio of the areas of the $ 2$ frisbees is simply $ \frac {76.8}{120} \equal{} 0.64$.  The ratio of the radii of these frisbees would be the square root of the fraction we just attained: $ \sqrt {0.64} \equal{} \boxed {0.8}$.[/hide]

[hide="2"]Let the numbers that Dill, Nill, and Swill wrote be $ D,N,$ and $ S$ respectively.  By simple conversions from feet and yards to inches, we have:
$ N$ feet equals $ 12N$ inches, and $ S$ yards equals $ 36S$ inches.
From the following information, we know that:
(1) $ D \equal{} 10N$.  Therefore, Dill measured $ 10N$ inches.
(2) $ N \minus{} 10 \equal{} S$.  Therefore, Swill measured $ 36(N \minus{} 10) \equal{} 36N \minus{} 360$ inches
Nill measured 12N inches.

Since $ 52'2" \equal{} 626"$, we have that:
$ 10N \minus{} 360 \plus{} 36N \plus{} 12N \equal{} 626$, or $ \boxed N \equal{} 17$ feet.

Thus, Dill's number was $ 10(17) \equal{} \boxed {170}$ inches,
and Swill's number was $ 36(7) \equal{} 252$ inches, or $ \boxed {7}$ yards.
[/hide]

AIME15, Nill's number was $ 10$ MORE than Swill's, not ten times Swill's.

Meh, at least no dumb arithmetic errors :P

Anyway, if you can follow what I did you should be able to solve the problem.


If $ n^{n}$ has $ 10^{100}$ decimal digits, how many has $ n$?

[hide]
to have $ 10^{100}$ digits, the number must be such that:

$ 10^{99}\leq n^{n}< 10^{100}$

$ 99 \leq n\log{n}< 100$

just by sight here, n is going to be 2 digits with $ \log n$ being between 1 and 2

EDIT: more rigorously
at n = 10, $ n \log n = 10$
at n = 100, $ n \log n = 200$

since $ 10 < [99, 100) < 200$,
$ 10 < n < 100$ meaning that n has 2 digits
[/hide]


LaTeX
combinatorics unsolved
combinatorics

The set $M= \{1;2;3;\ldots ; 29;30\}$ is  divided in $k$
subsets such that if $a+b=n^2, (a,b \in M, a\neq b, n$ is an
integer number $)$, then $a$ and $b$ belong different subsets.
Determine the minimum value of $k$.

k can't be 1.
If k=2, we can devided the set M into 2 subsets A,B that satisfy the requirement above.
We can suppose 1 is in A, so 3 is in B, so 6 is in A, so 10 is in B ,so 15 is in A.
We now have contrast: 1 and 15 are in A.
so, k is larger or equal to 3.

We can easily observe that : that set M can be devided into 3 subsets that satisfy the requirement above:
M1= {1,2,4,6,11,13,16,17,18,22,26,28,29}
M2={3,5,7,8,10,12,14,19,21,23,25,27}
M3={9,15,20,24,30}

Conclude: min k=3.

Let me expand this problem:

find the least value of k that:
for each subset of M that consists k elements ,there are always a,b that: a>b and a+b is not a perfect square number.

M1 shows that k larger or equal to 14.

i have another solution to show that the min is 3
look at the three numbers 6,19,30
if with k=2 it is possible then two numbers are in the same subset .

My solution was the same, to show it is not possible for $k=2$ and give and example for $k=3$


[quote="tuanduycvp"]co ai bit cho down duoc cuon BAT DANG THUC vuc Pham Van Thuan ko mach em voi. em nho nguoi o Ha Noi mua ma ko duoc? Cam on nhe![/quote]

Bạn xem ở đây: http://www.mathlinks.ro/viewtopic.php?t=161263


[b]For which positive numbers $\{ w_{1}, w_{2},\ldots, w_{n}\}$ do we have
\[\left(\sum_{k=1}^{n}x_{k}y_{k}\right)^{2}\le \left(\sum_{k=1}^{n}w_{k}x_{k}^{2}\right)\left(\sum_{k=1}^{n}\frac{1}{w_{k}}y_{k}^{2}\right) \le \left(\sum_{k=1}^{n}x_{k}^{2}\right)\left(\sum_{k=1}^{n}y_{k}^{2}\right) \]
for all positive integer $n$ and all real $x_{k},y_{k}$ ? [/b]

By substituting $x_{j}=\sqrt{w_{j}}$ and $y_{j}=\sqrt{w_{j}}$ we will obtain
\[n\left(\sum_{j=1}^{n}w_{j}^{2}\right)\leq \left(\sum_{j=1}^{n}w_{j}\right)^{2}\]
thus $w_{j}=c.$

[quote="bodan"]...thus $w_{j}=c.$[/quote]
Hmm ? Only this solution (possibility) for $w_{j}$ [b] ?[/b]


What year in this century, when its digits are written in reverse
order, is a year in the next century?


Evaluate the following definite integral.

\[\int_{e^{e}}^{e^{e+1}}\left\{\frac{1}{\ln x \cdot\ln (\ln x)}+\ln (\ln (\ln x))\right\}dx\]

$\frac{d}{dx}\;\left( \; x\;\cdot\; \ln(\ln(\ln x)) \;\right) \;=\;\ln(\ln(\ln x)) \;+\;\frac{1}{\ln(\ln x)}\;\cdot\;\frac{1}{\ln x}$

so, your answer is $\boxed{ e^{e+1}\;\cdot\; \ln(\ln(e+1))}$ and thanks.  :D

That's right, misan. :)  I made this problem as follows.

$\int \frac{1}{\ln x \cdot \ln (\ln x)}dx=\int \frac{1}{x\ln x \cdot \ln (\ln x)}\cdot x \ dx$

As you know, you can use the integral by parts, but I like your solution because I myself don't almost use integral by parts  except special example.


If f(x) = x^2x and x = 1/x, what is f(x) multiplied by g(x) = 1/x + x^2x?


geometry
3D geometry
sphere
calculus
integration

Prove that if you have a point inside a sphere it is actually weightless.

Bomb

[i]Mod Note - Please do not Double-Post ----- even if you do, please delete them... there is a little x button by the posts. The other thread is deleted. [/i]

Double posts...

And for the question: Isn't a point weightless?

ok I will post my response here, and hopefully the other thread will be deleted.

What is "it"? What is "a point"?

Also, what would be the difference between a point in a sphere and a point outside of a sphere?

I still don't understand his representation of a point? A point is not a physical thing. A point cannot have mass or be massless...

perhaps weight was a specific term (I tend to disregard people's use of mass/weight when they are indeed different). Is the question asking, prove that inside a spherical SHELL, there is no gravitational field and thus there is no force and thus no "weight"?

assuming Spoon is right, use gauss's law applied to gravitation:

$\oint g \cdot dA = - 4 \pi G m_{enclosed}$

$m_{enclosed} = 0$ (inside a shell)

$\therefore g = 0$

(its the same sort of reasoning that is used to show that the $E = 0$ inside a charged conducting shell)

exactly, that is why I am betting that this is what he meant... though I would have put it in a hint :).

But the important part is a SHELL. This is not the case for a solid Sphere, because m_enclosed isn't 0.

Whoa, just wait a sec. How is it exactly in English? Isn't a "sphere" a set of points with distance R from a fixed centre, and a "ball" a set of points that has distance R or less from a fixed point?

I know that this is only a technical matter but I'd just like to know .. in Slovak language, we have two words for a circle and its interior (a disk?) but we use the same word for a sphere and a ball and it causes misinterpretations sometimes...

hmm... well in that case it's a "shell" anyway. But, is it possible to have a massless sphere? And if the sphere is massless, there is no gravity anywhere anyway, right? If you have something with mass (macroscopically speaking), it needs to be made up something, and therefore it's a shell.

If you're asking about the Sphere Theorem its actually physically false at close distances, because mass is not a true continuum of distribution.  There is a purely mathematical 'proof' which is rather irrelevant.  So in spite of its reputation for being true, the Sphere Theorem of Newton is actually a bust.  It is trivially true at distances which dwarf the radius of the sphere ( d >> r ) by orders of magnitude, but strictly speaking inside the hollow sphere there is no stable saddlepoint.
[b]
Newton's Sphere Theorem:[/b]
The gravitational net force on a test-particle (a point-like particle of any arbitrary mass) is zero, and the particle is free to float in any direction in the uniform field inside the hollow sphere of uniform density and negligible thickness.  The gravitational field inside a solid sphere is usually constructed by relying upon this theorem to build a solid sphere out of concentric hollow spheres.  But I warned you, it's not really true:

Here is the link to the integration ('proof') for the hollow sphere:


.  Chlorine reacts with ammonia and ammonia-like compounds that are formed from organic waste to form chloramines.  Chloramines do not sanitize; however, they block free chlorine molecules from oxidizing bacteria and result in the “chlorine” smell.  In addition, hypochlorite ion is destroyed by UV light from the sun to make chloride ion and O2.  Cyanuric acid is added to pools as a stabilizer to prevent the loss of chlorine.
Write a chemical equation(s) that describes the formation of chloramines.  Use reaction rate concepts to explain how chloramines block chlorine from oxidizing bacteria.  Write a chemical equation that shows how cyanuric acid prevents the loss of chlorine.

[quote="Afcweswarrior"]Write a chemical equation(s) that describes the formation of chloramines.[/quote]

$NH_{3}(aq)+Cl_{2}(aq) \longrightarrow NH_{2}Cl(aq)+H^{+}(aq)+Cl^{-}(aq)$.


Same logic as this problem I got from friend:

(x)/(y+z) + (y)/(z+x) + (z)/(x+y) >= 3/2

[b]runningking7[/b], yours is Nesbitt's inequality. :)[/b]

[quote="fuzzylogic"][quote="shyong"]for positive real $a,b,c$ satisfying $a+b+c=1$ , prove that

$(a+\frac{1}{a})(b+\frac{1}{b})(c+\frac{1}{c})\geq \frac{1000}{27}$[/quote]

For $0<x<1$, we have $x + \frac{1}{x} \ge \frac{50}{3}\cdot\frac{1}{1+12x}$, this is equiv. to $(3x-1)^2(4x+3) \ge 0$.

AM-GM gives $(1+12a)(1+12b)(1+12c)\le 5^3$

Therefore, $(a+\frac{1}{a})(b+\frac{1}{b})(c+\frac{1}{c})\geq \big({\frac{50}{3}}\big)^3\frac{1}{5^3}$[/quote]

wow ! thats nice . But how do you know to use $x + \frac{1}{x} \ge \frac{50}{3}\cdot\frac{1}{1+12x}$ ? :roll:

[quote="arqady"]Let $x_1>0,x_2>0,...,x_n>0,x_1+x_2+...+x_n=1.$
Prove that $(x_1+\frac{1}{x_1})\cdot(x_2+\frac{1}{x_2})\cdot...\cdot(x_n+\frac{1}{x_n})\geq(n+\frac{1}{n})^n.$[/quote]

we can prove by Jensen by considering $f(x)=\ln (x+\frac{1}{x})$ over the interval $(0,1)$ since $f''(x)=\frac{2(1-x^2)}{(x^2+1)^2}+\frac{1}{x^2}>0$

so $\frac{1}{n}(\sum f(x))\geq \ln (\frac{\sum x_i}{n}+\frac{n}{\sum x_i})=\ln (n+\frac{1}{n})$ which simplify to our desired inequality.

[quote="arqady"]shyong, I assumed, that you will be consecutive:
$(x_1+\frac{1}{x_1})\cdot...\cdot(x_n+\frac{1}{x_n})\geq\frac{(n^2+1)^2}{n(n(n^2-1)x_1+2)}\cdot...\cdot\frac{(n^2+1)^2}{n(n(n^2-1)x_n+2)}=$
$=\frac{(n^2+1)^{2n}}{n^n(n(n^2-1)x_1+2)\cdot...\cdot(n(n^2-1)x_n+2)}\geq\frac{(n^2+1)^{2n}}{n^n}\cdot(\frac{n}{n(n^2-1)(x_1+...+x_n)+2n)})^n=$
$=\frac{(n^2+1)^{2n}}{n^n}\cdot\frac{n^n}{(n^3-n+2n)^n}=(n+\frac{1}{n})^n.$ ;)[/quote]

is it that you use

$a+\frac{1}{a}\geq \frac{(n^2+1)^2}{n(n(n^2-1)a+2)}\longleftrightarrow (na-1)^2((n^2-1)a+2n)\geq 0$ ? And the second inequality is AM-GM since

$\frac{\sum n(n^2-1)x+2}{n}\geq \left((n(n^2-1)x_1+2)\cdots (n(n^2-1)x_n+2)\right)^{\frac{1}{n}}$

but the first one is hard to see  :(  I did it when i try to compare it with fuzzylogic's given pattern above . Arqady , how you figure out the first inequality ?

[quote="arqady"][quote="shyong"]
 Arqady , how you figure out the first inequality ?[/quote]
Let $f(x)=x+\frac{1}{x}-\frac{a}{x+b}.$
Then I want that $f(\frac{1}{n})=0$ and $f'(\frac{1}{n})=0.$ ;)
Hence, $a=\frac{(n^2+1)^2}{n^2(n^2-1)},b=\frac{2}{n(n^2-1)}.$ :)[/quote]

cool  :ninja: I learn new thing :D

[quote="shyong"]
cool  :ninja: I learn new thing :D[/quote]

The same technique can be applied to many interesting problems. See:

http://www.mathlinks.ro/Forum/topic-3048.html

http://www.mathlinks.ro/Forum/topic-14088.html

http://www.mathlinks.ro/Forum/topic-180.html

etc.

Enjoy!


integration
calculus
derivative
inequalities
real analysis
real analysis unsolved

You don't mean $ f(x)\equal{}0$; that would make both sides zero, which is uninteresting. Also, you need at least one derivative for this to make sense; a continuous $ f$ is not enough.
If you mean $ f(0)\equal{}0$, the inequality is far from sharp, and we could put a $ \frac{1}2$ on the right-hand side. If you put the zero in the middle, we get an even stronger constant.

A proof of the stronger version:
First, $ \frac{1}{b\minus{}a}\int_{a}^{b}(f'(t))^{2}\,dt\ge\left(\frac{1}{b\minus{}a}\int_{a}^{b}|f'(t)|\,dt\right)^{2}$ by Cauchy-Schwarz, or the AM-QM inequality.
Now, we compare $ \int_{0}^{1}|f(t)f'(t)|\,dt$ to $ \frac{1}2\left(\int_{0}^{1}|f'(t)|\,dt\right)^{2}$. Let $ g(t)\equal{}\int_{0}^{t}|f'(s)|\,ds$; $ g(t)\ge |f(t)|$ for all $ t\in [0,1]$, and we no longer have to worry about signs.
$ |f(t)f'(t)|\le g(t)g'(t)$, so $ \int_{0}^{1}|f(t)f'(t)|\,dt\le\int_{0}^{1}g(t)g'(t)\,dt\equal{}\frac{1}2g(1)\equal{}\frac{1}2\left(\int_{0}^{1}g'(t)\,dt\right)^{2}\equal{}\frac{1}2\left(\int_{0}^{1}|f'(t)|\,dt\right)^{2}$
Applying the previously stated case of Cauchy-Schwarz, we get
$ \int_{0}^{1}|f(t)f'(t)|\,dt\le\frac{1}2\int_{0}^{1}(f'(t))^{2}\,dt$.


calculus
derivative
integration
limit
advanced fields
advanced fields unsolved

For $ u\in W^{1,p(x)}(R^n),p(x)\ge 1,$

$ J(u)\equal{}\int_{\Omega}\frac{1}{p(x)}|\nabla u|^{p(x)}dx$. Prove

\[ J'(u)v\equal{}\int_{\Omega\subset R^n}|\nabla u|^{p(x)\minus{}2} \nabla u \nabla vdx.\]

The derivative is here with respect to Frechet mean.

Thanks

If we assume or already know that the frechet differential exists, we can calculate it straight foreward:

$ J'(u)v\equal{}\lim_{t\to 0} \frac{J(u\plus{}tv)\minus{}J(u)}{t} \equal{} \left.\frac{d}{dt} J(u\plus{}tv)\right|_{t\equal{}0}$. Now a simple differentiation under the integral sign yields the result. (Of course we have to makle sure that this is legal)

[quote="Gockel"]If we assume or already know that the frechet differential exists, we can calculate it straight foreward:

$ J'(u)v \equal{} \lim_{t\to 0} \frac {J(u \plus{} tv) \minus{} J(u)}{t} \equal{} \left.\frac {d}{dt} J(u \plus{} tv)\right|_{t \equal{} 0}$. Now a simple differentiation under the integral sign yields the result. (Of course we have to makle sure that this is legal)[/quote]

Hi,

Thank you! But I still confuse with the exponent $ p(x)$. Would you please give more details?

$ J(u+tv)=\int_\Omega \frac{1}{p} \langle \nabla u+t\nabla v,\nabla u+t\nabla v\rangle^{p/2} \,dx = \int_\Omega \frac{1}{p} \left(\langle \nabla u,\nabla u\rangle +2 t\langle \nabla u,\nabla v\rangle + t^2\langle \nabla v,\nabla v\rangle\right)^{p/2} \,dx$

Now differentiate with respect to $ t$:
$ \frac{d}{dt} J(u+tv) = \int_\Omega \frac{1}{p} \cdot \frac{p}{2}\cdot \left(\langle \nabla u,\nabla u\rangle +2 t\langle \nabla u,\nabla v\rangle + t^2\langle \nabla v,\nabla v\rangle\right)^{p/2-1} \cdot \left(2\langle \nabla u,\nabla v\rangle+2t\langle \nabla v,\nabla v\rangle\right) \,dx$

Now plug in $ t=0$ and everything's fine.

[quote="Gockel"]$ J(u + tv) = \int_\Omega \frac {1}{p} \langle \nabla u + t\nabla v,\nabla u + t\nabla v\rangle^{p/2} \,dx = \int_\Omega \frac {1}{p} \left(\langle \nabla u,\nabla u\rangle + 2 t\langle \nabla u,\nabla v\rangle + t^2\langle \nabla v,\nabla v\rangle\right)^{p/2} \,dx$

Now differentiate with respect to $ t$:
$ \frac {d}{dt} J(u + tv) = \int_\Omega \frac {1}{p} \cdot \frac {p}{2}\cdot \left(\langle \nabla u,\nabla u\rangle + 2 t\langle \nabla u,\nabla v\rangle + t^2\langle \nabla v,\nabla v\rangle\right)^{p/2 - 1} \cdot \left(2\langle \nabla u,\nabla v\rangle + 2t\langle \nabla v,\nabla v\rangle\right) \,dx$

Now plug in $ t = 0$ and everything's fine.[/quote]

Thanks. I also obtained the result by a different way. Let me see


At least $ n - 1$ numbers are removed from the set $\{1, 2, \ldots, 2n - 1\}$ according to the following rules:
(i) If $ a$ is removed, so is $ 2a$;
(ii) If $ a$ and $ b$ are removed, so is $ a \plus{} b$.
Find the way of removing numbers such that the sum of the remaining numbers is maximum possible.

is :$n^2$and the number that we removed are the even number!the proof for maximum of it is not hard :D

[quote="April"]At least $ n \minus{} 1$ numbers are removed from the set $ A \equal{} \{1, 2, \ldots, 2n \minus{} 1\}$ according to the following rules:
(i) If $ a$ is removed, so is $ 2a$;
(ii) If $ a$ and $ b$ are removed, so is $ a \plus{} b$.
Find the way of removing numbers such that the sum of the remaining numbers is maximum possible.[/quote]
We have to remove $n-1$ even numbers $2,4,...,2n-2$ so the sum of the remaining is maximum!
The solution is from these Lemma:
Lemma1: If we remove 1 we have to remove $2,3,...$ so we remove all numbers from the set A
Lemma2: If we remove 2, we remove $n-1$ numbers $2,4,..,2n-2$ so the sum of remaining numbers is $n^2$
Lemma3: If we remove p number $a_1<a_2<a_3<...<a_p$ so we must remove the number $a_1+a_p$ but this number greater than $a_p$ so $a_1+a_p > 2n-1 $
A similar way to $a_i$ and $a_{p+1-i}$. So the sum of the number which are removed is $\ge n(n-1)$ So the sume of the remaining numbers is $< n^2$.


no don't read a proof. try to work it out yourself. the second formula is not that hard to see, intuitively, and neither is the first if you think along the right lines.

please read the given formula and then respond

[quote="MathFiend"]I don't understand the formula because I'm stupid so I'm trying to figure it out!!! :evil:  :evil:  :evil:[/quote]
Woah! Cool down. Where specifically do you not understand?

what the letters stand for, what the underscore is for, when you use what power, everything...

allright, then, say so instead of just guessing. Let's start at the beginning, do you know the principle of prime factorization? i.e. writing every number as a product of primes. so for example, 48 would be 2^4*3^1, and 198 would be 2^1*3^2*11. Do you follow me so far?

yes I know what prime factorization is...I'm not an idiot!

OK, calm down. Remember, I'm the comoderator, and I can calm you down by force :) , which I don't want to do. I mean, I could change all your posts into "lalala! I'm a happy boy in a happy place! Smiles all around :) " but that would be abuse of my power and inappropriate, not to mention inpersonation (but it would be funny :) )

Did my post help? I mean, my prime job is to help people.

Please don't "split-post", because it's banned for reasons I don't know :)

because then I have to merge them together and it's such a pain, I mean it takes like 10 clicks! :)

no problem! :twisted: ...I mean, you weren't being a jerk :)

Yeah, I'm a jerk too, but I thought only boys could be called a "jerk"...

I got angry for no apparent reason, and I rephrase what I said, I wasn't a jerk because as you said tare, only boys can be jerks, I was being immature

lol, you WASNT a jerk, just confused, but all is cleared up

What's the girl equivalent word of "jerk"? :twisted:

but seriously, I agree with Syntax on this one, which happens rarely :twisted: </kidding!>


If you have a problem like
"How many ways can you make change for $\$$18 using 1,2,5, and 10 dollar bills?" (note: I'm not asking for the answer... I already know the answer. This is just an example to clarify what I'm saying)
Is there any pattern or way to solve without listing, besides using the Subset Sum Equation? How can this be applied to different sums and denominations?

[quote="neelnal"]This is partitioning.[/quote]
I'm pretty sure partitioning is only good for breaking it into any positive integers not specific ones. Then again, maybe I'm wrong (probably)...

[quote="Fraxia"]You can use generating functions for this problem. The generating function for the number of 1 dollar bills is
\[1+x^{2}+x^{3}+\cdots = \frac{1}{1-x}\]
The generating function for the number of 2 dollar bills is
\[1+x^{2}+x^{4}+\cdots = \frac{1}{1-x^{2}}\]
and so forth. So to find the number of ways, you can find the coefficient of $x^{18}$ in the expression
\[\frac{1}{(1-x)(1-x^{2})(1-x^{5})(1-x^{10})}\]
[/quote]
Even so, for a no calculators allowed contest, I'd guess that listing would still be faster (seeing as you're allotted 2 or less minutes per problem)


let  $ x_o\equal{}0$  and  sequence such that : $ x_{n\plus{}1} \equal{} \frac{1}{4^{x_n}}.$
 determine limit of sequence  where  $ n \to  \infty$

half of one.

(sequence is clearly bounded.)

i know
 we have   :
 let$ x_{2k\plus{}1}$   and$ x_{2k}$ the both have limit$ \equal{}1/2$.
   but i don't know  how do bound  it?

each x(n) is in the closed interval [0,1] since x(n) is a nonnegative sequence.

[quote="Hong Quy"]i know
 we have   :
 let$ x_{2k \plus{} 1}$   and$ x_{2k}$ the both have limit$ \equal{} 1/2$.
   but i don't know  how do bound  it?[/quote]

We show that :

[b]i)[/b] $ x_{_n} \in \left[ 0; 1 \right] , \forall n \in \aleph$ ... (1)

[b]ii)[/b] $ \left( x_{_{2k}} \right)$ is increasing sequence.
(1)$ \Rightarrow \exists \lim_{}x_{_{2k}}$ and $ \lim_{}x_{_{2k}} \equal{} \frac{1}{2}$ ... (2)

[b]iii)[/b] $ \left( x_{_{2k\plus{}1}} \right)$ is decreasing sequence.
(1)$ \Rightarrow \exists \lim_{}x_{_{2k\plus{}1}}$ and $ \lim_{}x_{_{2k\plus{}1}} \equal{} \frac{1}{2}$ ... (3)

(2)(3) $ \Rightarrow  \lim_{}x_{_n} \equal{} \frac {1}{2}$

Pro 2: let $ u_1\equal{}1$ and $ u_{n}\equal{}\frac{1}{1\plus{}u_{n\minus{}1}}$.
Caculate $ \lim\limits_{ n \to \infty} u_n$

Proving the existence of a limit is the one that requires work [Monotone Convergence Theorem is always my first line of attack], but if we assume the limit exists and are asked to find it, then we just do some calculation and decide later. In the case of the $ x_{n \plus{} 1} \equal{} (\frac {1}{4})^{x_{n}}$, it is clear that the numbers are positive. Then assume the limit exists, say $ s$.
Then we must have the relation:
$ s \equal{} (\frac {1}{4})^s \iff log{_\frac {1}{4}}s \equal{} s \iff \frac {log_{\frac {1}{2}}s}{log_{\frac {1}{2}}\frac {1}{4}} \equal{} s \implies s \equal{} \frac {1}{2}$.

For the second problem: $ u_{1} \equal{} 1, u_{n \plus{} 1} \equal{} \frac {1}{1 \plus{} u_{n}}$.
We have: $ \lim_{n \to \infty}u_{n} \equal{} \frac { \minus{} 1 \plus{} \sqrt {5}}{2}$, the other value which was negative was discarded.


logarithms
inequalities unsolved
inequalities

Let  $ x,y,z\in(1;\plus{}\infty)$
Prove that  $ (\log_x\dfrac{y\plus{}z}{2})(\log_y\dfrac{z\plus{}x}{2})(\log_z\dfrac{x\plus{}y}{2})\ge 1$

________________________________
$ \textbf{Azerbaijan Land of Fire }$ :ninja:  :rotfl:

[quote="Pirkuliyev Rovsen"]Let  $ x,y,z\in(1; \plus{} \infty)$
Prove that  $ (\log_x\dfrac{y \plus{} z}{2})(\log_y\dfrac{z \plus{} x}{2})(\log_z\dfrac{x \plus{} y}{2})\ge 1$[/quote]
First notice that $ \log_b a \equal{} \frac{\log a}{\log b}$
So we have to prove:
$ \log \frac{x\plus{}y}{2} \cdot \log \frac{y\plus{}z}{2} \cdot \log \frac{z\plus{}x}{2} \ge \log x \cdot \log y \cdot \log z$. (Since $ x,y,z > 1$)
Let $ f(x) \equal{} \log \circ \log x$
Then we have to prove $ f \left ( \frac{x\plus{}y}{2} \right ) \plus{} f \left ( \frac{y\plus{}z}{2} \right ) \plus{} f \left ( \frac{z\plus{}x}{2} \right ) \ge f(x) \plus{} f(y) \plus{} f(z)$. And hence it suffices to prove:
$ \frac{f(x) \plus{} f(y)}{2} \le f \left ( \frac{x\plus{}y}{2} \right )$
And since $ f''(x) \equal{} \frac{\minus{}(\log x \plus{} x)}{(x \log x)^2} < 0$ we see that $ f$ is concave and the result follows from Jensen's.

$ x;y;z > 1\Rightarrow \log_xy > 0;\log_yx > 0;\log_xz > 0;\log_zx > 0;\dots \\
\mbox{and: } \\
\left(\log_x{\frac {y \plus{} z}{2}}\right).\left(\log_y{\frac {x \plus{} z}{2}}\right).\left(\log_z{\frac {x \plus{} y}{2}}\right)\ge \\
\ge \left(\log_x\sqrt {yz}\right). \left(\log_y\sqrt {xz}\right). \left(\log_z\sqrt {xy}\right) \equal{} \\
\equal{} \frac {1}{8}.\left(\log_xy \plus{} \log_xz\right).\left(\log_yx \plus{} \log_yz\right).\left(\log_zx \plus{} \log_zy\right) \equal{} \\
\equal{} \frac {1}{8}.\left(1 \plus{} \log_xz \plus{} \log_yz \plus{} \log_xz.\log_yz\right).\left(\log_zx \plus{} \log_zy\right) \equal{} \\
\equal{} \frac {1}{8}.\left(\log_zx \plus{} \log_zy \plus{} 1 \plus{} \log_xy \plus{} \log_yx \plus{} 1 \plus{} log_yz \plus{} \log_xz\right) \equal{} \\
\equal{} \frac {1}{4}.\left(\frac {\log_xy \plus{} \log_yx}{2} \plus{} \frac {\log_xz \plus{} \log_zx}{2} \plus{} \frac {\log_yz \plus{} \log_zy}{2} \plus{} 1\right)\ge 1; \\
\mbox{beucause: } \\
 \\
\frac {\log_xy \plus{} \log_yx}{2}\ge \sqrt {\log_xy.\log_yx} \equal{} 1;\dots$


WAY too easy for Pre-Olympiad.
[hide]
$10=\log x\Rightarrow x=10^{10}$, assuming this is a common logarithm.
[/hide]

john, I hope this isn't HW..... :wink:

Anyways, moved to Basics.

the answer is 10 to the tenth power if you are using a commom log.

[hide][quote="john lai"]y = log x
If y = 10, then what is x?[/quote]

logx=10
x=10^10
x=10000000000[/hide]


I have a little question for y'all :P

I am to know some things about circuit solving using matrices. I understood Kirchhoff, nodal analysis and mesh analysis.
But here's what I don't get: it's something called "essential incidences matrix"; one considers the graph of the circuit, then one selects a spanning tree contains preferably all E's and builds a matrix of coefficients. Does anybody know what I'm talking about? Like how one find those coefficients :?:

An incidence matrix is one whose elements are all equal either to 1 or 0.

An essential matrix is the product  of a 3x3 orthogonal matrix and a 3x3 skew-symmetric matrix.

I don't think that helps me very much :(

The normal incidence matrix (which I understood) is like this: it's a $ (n\minus{}1) \times e$ matrix ($ n$ vertices, $ e$ edges), where the element $ (i,j)$ tells me whether the edge $ j$ goes in ($ \minus{}1)$, out ($ 1$), or neither ($ 0$) into node ($ i$). Multiply this matrix by $ I_1, \ldots, I_{n\minus{}1}$ and you get Kirchhoff I.

However, there seems to be another method, with spanning trees, that I do not understand, and I was wondering if any of you might have heard about it....


1)calculate according to m and r
$ \sum_{i\equal{}0}^r \binom {m\plus{}i} {i}$
2)$ \sum_{i\equal{}0}^m \binom {m} {i} \binom {m} {r\plus{}i}$
3)calculate according to m
$ \sum_{i\equal{}0}^m {\binom {m} {i}}^2$

hello, for the first sum i have got $ \sum_{i\equal{}0}^r\binom{m\plus{}i}{i}\equal{}\frac{(r\plus{}1)(m\plus{}r\plus{}1)!}{(m\plus{}1)(r\plus{}1)!m!}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, for the first sum i have got $ \sum_{i \equal{} 0}^r\binom{m \plus{} i}{i} \equal{} \frac {(r \plus{} 1)(m \plus{} r \plus{} 1)!}{(m \plus{} 1)(r \plus{} 1)!m!}$.
Sonnhard.[/quote] Yes, indeed! $ \sum_{i\equal{}0}^r\binom{m\plus{}i}{i}\equal{}$ $ \binom{m\plus{}1}{0}\plus{}\binom{m\plus{}1}{1}\plus{}\binom{m\plus{}2}{2}\plus{}\binom{m\plus{}3}{3}\plus{}...\plus{}\binom{m\plus{}r}{r}\equal{}$ $ \binom{m\plus{}2}{1}\plus{}\binom{m\plus{}2}{2}\plus{}\binom{m\plus{}3}{3}\plus{}...\plus{}\binom{m\plus{}r}{r}\equal{}$ $ \binom{m\plus{}3}{2}\plus{}\binom{m\plus{}3}{3}\plus{}...\plus{}\binom{m\plus{}r}{r}\equal{}$ $ ...\equal{}\binom{m\plus{}r}{r\minus{}1}\plus{}\binom{m\plus{}r}{r}\equal{}\binom{m\plus{}r\plus{}1}{r}$ (we use the identity $ \binom{n}{k}\plus{}\binom{n}{k\plus{}1}\equal{}\binom{n\plus{}1}{k\plus{}1}$)

hello, for the second sum i have found
$ \sum_{i\equal{}0}^m\binom{m}{i}\binom{m}{r\plus{}i}\equal{}\frac{2m)!}{(m\plus{}r)!(m\minus{}r)!}$
Sonnhard.

hello doctor.but how I can prove your finding?


Let $ d_1<d_2<....<d_n\equal{}n$ be all diffirent divisors of positive interger $ n$ written in ascending order.Dertermine all n such that    $ d_7^2\plus{}d_{10}^2\equal{}(\frac{n}{d_{22}})^2$

I have seen a similar problem before...
The idea is to obvserve taking congruences $ mod 3,4 and 5$ that $ n$ must be divisible by $ 3,4 and 5$ ( Obviously if $ d_i=0 (p)$ then $ n=0 (p)$)
Squares $ mod 3,4$ are $ 0,1$ and squares $ mod 5$ are $ 0,1,4$.
Then you get that the first six divisors are $ 1,2,3,4,5,6$ and also $ 10$ is a divisor. So $ d_7=7,8,9 or 10$ You have to check these cases and make the same things I have done above. I'm too lazy too finish it but I think the main idea is clear enough for you too solve it.
Please if you don't understand anything ask.

Daniel

Hello, i've never seen something like that , so i just can write what it hink about it:
[hide]If $ 1 \equal{} d_1 < d_2 < ... < d_n \equal{} n$ are all divisors of n, the number $ n/d_1,n/d_2,...,n/d_n$ are too all divisors of n  and see that
$ n/d_1 \equal{} d_n$, $ n/d_2 \equal{} d_{n \minus{} 1}$ , so $ n/d_{22} \equal{} d_{n \minus{} 21}$
we have
$ d_7 ^2 \plus{} d_{10} ^2 \equal{} d_{n \minus{} 21} ^2$ and we know that

$ d_7 \equal{} a^2 \minus{} b^2$
$ d_{10} \equal{} 2ab$
$ d_{n \minus{} 21} \equal{} a^2 \plus{} b^2$
notice that from the equations above we discover some divisors:
$ d_1 \equal{} 1$
$ d_2 \equal{} 2$
$ d_3 \equal{} 3$
$ d_4 \equal{} 4$
$ d_5 \equal{} 5$
$ d_6 \equal{} 6$,
[/hide]
ok i've psoted the whole solution but i found some mistakes so i deleted it, if you wanna i can complete it later, greetings ;)
but Daniel wrote how to do that so I think you'll do it easly


algebra
polynomial
superior algebra
superior algebra unsolved

Hello, I found this in my assignment

Fix a field $ \mathbb F$. Consider the polynomial $ f\equal{}x^4\plus{}1_{\mathbb F}$. Prove that is reducible for all fields F.

I think there's something wrong because if I pick $ \mathbb F \equal{} \mathbb Q$ I get that $ f$ has no rational roots, so I look for some divisor of degree 2. I write

\[ x^4\plus{}1 \equal{} (x^2\plus{}ax\plus{}1)(x^2\plus{}bx\plus{}1)\]

which gives $ a\equal{}\minus{}b$ and $ a$ such that $ a^2\equal{}2$ but there is no such rationals which verify this.
what do you think?

The correct problem is: Prove that is is reducible in any [b]finite[/b] field.

It is indeed irreducible in $ \mathbb Q$.

Any finite field? that poly is irreducible in $ \mathbb F_3[x]$ too.

I agree it is reducible on every field of char 2 but is there more?

In $ \mathbb F_3[x]$ we have $ x^4 \plus{} 1 \equal{} (x^2 \plus{} x \minus{} 1)(x^2 \minus{} x \minus{} 1)$.

right, but how do I prove the fact for $ char(\mathbb F) > 2$ ?

[quote="Αρχιμήδης 6"]Its not true....[/quote]

what's false?

Hint:
Try to factor it over $ \mathbb Q[\sqrt{\minus{}1}]$, $ \mathbb Q[\sqrt{2}]$ and $ \mathbb Q[\sqrt{\minus{}2}]$. Then show that at least one of $ \minus{}1, 2, \minus{}2$ is a square in $ \mathbb F_q$ and use the "same" factorisation.


inequalities quadratics function conics parabola

Let $ a,$ $ b$ and $ c$ are non-negative numbers and
$ u=\frac{a+b+c}{3},$ $ v=\sqrt{\frac{ab+ac+bc}{3}},$ $ w=\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}.$
Prove that for all ${ \alpha\geq1+\sqrt3}$ the following inequality holds:
$ \alpha w+v\geq(\alpha+1)u.$

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers and
$ u = \frac {a + b + c}{3},$ $ v = \sqrt {\frac {ab + ac + bc}{3}},$ $ w = \sqrt {\frac {a^{2} + b^{2} + c^{2}}{3}}.$
Prove that for all ${ \alpha\geq1 + \sqrt3}$ the following inequality holds:
$ \alpha w + v\geq(\alpha + 1)u.$[/quote]

$ \alpha w + v\geq(\alpha + 1)u.$
<=>$ ((a - b)^2 + (b - c)^2 + (c - a)^2)(\frac {\alpha}{\sqrt {a^2 + b^2 + c^2} + a + b + c} - \frac {1}{\sqrt {ab + bc + ca} + a + b + c}) \geq 0$
This inequality is true! :P

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers and
$ u = \frac {a + b + c}{3},$ $ v = \sqrt {\frac {ab + ac + bc}{3}},$ $ w = \sqrt {\frac {a^{2} + b^{2} + c^{2}}{3}}.$
Prove that for all ${ \alpha\geq1 + \sqrt3}$ the following inequality holds:
$ \alpha w + v\geq(\alpha + 1)u.$[/quote]
It can be $ \alpha\geq\frac{1 + \sqrt3}{2}.$

First note that $ u \equal{} \frac {\sqrt {3w^2 \plus{} 6v^2}} 3$ and our enequality is equivalent to the following: $ 3 \alpha w \plus{} 3v \geq (\alpha \plus{} 1) \sqrt {3w^2 \plus{} 6v^2}$.After obvious transformations we have: $ w^2 \cdot (2 {\alpha}^2 \minus{} 2 \alpha \minus{} 1) \plus{} 6\alpha wv \plus{} v^2\cdot (1 \minus{} 2\alpha^2 \minus{} 4 \alpha) \geq 0$(*). From our conditions we can derive that $ 2 {\alpha}^2 \minus{} 2 \alpha \minus{} 1 \geq0$. Let's look at the left side of the (*) by dividing on $ v^2$ and substituting $ t \equal{} w/v$ we can investigate the quadratic function of t $ f(t)$, where $ t\geq1$ (it's an easy consequence of the inequality of 3 squares). $ f(1) \equal{} 0$ and it's easy to see that function increases when $ t\geq1$ therefore the coffecient at the $ t^2$ is positive and the peak of parabola lies leftwars of 1.
The enequality is prooved. :roll:


winrar.exe, IE web browser, xunlei downloader, "straight flush" poker game, "ping" in command window, these are all you need! :rotfl:
No site is safe!

How could they be so CNN?

[sarcasm]
Great, all terrorists need now is a straight flush poker game. Thanks a lot.
[/sarcasm]

Maybe I should turn off the internet now...

Oops, forgot the link.

http://www.youtube.com/watch?v=yAUKfdNwhco

[quote="ZetaX"]Whatever you are talking about...

Someone please delete it or move it to Games&Fun Factory.[/quote]

What makes you think this is kinda game&fun topic?
Follow the link, CNN will tell you it is a hundred percent serious, if not more.

That's not reasonable at all. When in the video was any of that suggested as a method?

[quote="SimonM"]Firstly, most of what is said is true. There is a huge amount of bias, but its rubbish that the Chinese government are paying off hackers.

Why would you bother paying most of these people, many do it out of interest and curiosity, there is no reason to suspect that they are actually serious.

With regards to the interviewed hackers, I have my doubts about the seriousness of the claim. Everyone is going to "big" themselves up for the camera.[/quote]

We are talking about the seriousness of this topic. It seems like you agree with me. So the topic does not belong to the game&fun category.

As for the authenticity, I will leave it to the readers, including you.
However, personally I would think, a cheap lie can not be used to support any conclusions. Rather, it will make people doubt the IQ of the reporter and the media. If somehow the audiences believe this cheap lie, I would doubt their IQ too.

No, a cheap lie may not be used to support a conclusion. But misinformation provided by the American government (not that the Chinese government is any better) can lead to news stations and the general public to have a bad idea of what is going on.

Clearly that reporter knew nothing about hacking, so he made some mistakes. Does that make him stupid? No.

[quote="Buffalo"]A very reasonable hypothesis on the right method used by professional CNN style hackers:

step by step
1. Click "Start", then "run", type in "cmd", press Enter.

2. In the command line, type "[b]ping[/b] Pentagon/or something else", press Enter. You can get the IP address of your prey.

3. Turn on you [b]IE browser[/b], input the IP address you get in step 2, press Enter.

4. You have now entered the database of Pentagon, find all files you want, download them by [b]xunlei[/b] or BT or something else. IE is not recommended, because it is slow and unable to resume broken downloads.

5. We know that this kind of data-files are extremely large, it will take hours to download. So you can play some games like [b]straight flush poker[/b] for amusement while waitting.

6. Downloading finishes. Some files are compressed, so you need to extract them with [b]winrar[/b].

7. Bring the data to the nearest Chinese government for you reward. Good luck.[/quote]

This is approximately as high-tech and professional as requesting that the local bank kindly give you the contents of their vaults.

Doubtlessly the Pentagon - as a matter of fact, any organization that takes itself seriously - does not open unnecessary exploit points which curious teenagers with limited computer literacy can easily access.

@Lazarus - I think the hi-tech thing was what he was getting at. My point is that that is not the case shown in the movie.

Generally, it [i]is[/i] curious teenagers who develop a vast amount of computer knowledge who access these places

So you do think CNN is just a joke for fun?


calculus
integration
trigonometry
derivative
function
calculus computations

integral of  (xsin[x])/(1+(cos[x])^2) dx

i'm sorry

i cant use latex....

thx before...

hello, did you meant
$ \int\frac{x\sin(x)}{1\plus{}\cos^2(x)}dx$
Sonnhard.

Let me make this nice and neat.
$ \int\frac{x\sin{x}}{1\plus{}\cos^{2}{x}}\ \mathrm{d}x$

Use integration by parts:

$ uv \equal{} \int u dv \plus{} \int v du$

The idea is to consider $ \frac{x\sin(x)dx}{1 \plus{} \cos^2(x)} \equal{} u dv$. Can you think what would be the good choice for $ u$ and what for $ dv$? We're hoping that $ vdu$ will be a simpler expression, so you want to pick $ u$ so that its derivative is simpler and $ dv$ so that its antiderivative is simpler.

(If you already know about integration by parts, then I'm sorry for talking too much about it...)

Well, good luck with this one because Mathematica goes ballistic with it. The idea Xevarion wrote is a good one and it would give you:

$ \int\frac{x \sin x}{1\plus{}\cos^2 x}\,dx\equal{}\minus{}x \tan^{\minus{}1}\left(\cos x\right)\plus{}\int \tan^{\minus{}1}\left(\cos x\right) \,dx$

Form here you can try to write the Taylor series for $ \tan^{\minus{}1}\left(\cos x\right)$ and that will give you odd powers of $ \cos$ that will need to be integrated. That, on the other hand can be done using the substitution $ t\equal{}\sin x$ and rewriting this using binomial formula and undoing the substitution. In the end, you should end up with something like:

$ \int \frac{x \sin x}{1\plus{}\cos^2 x}\,dx\equal{}\minus{}x \tan^{\minus{}1}\left(\cos
x\right)\plus{}\sum^{\infty}_{i\equal{}0}\frac{(\minus{}1)^{i}}{2i\plus{}1}\sum^{i}_{j\equal{}0}\frac{(\minus{}1)^j C^i_j}{2j\plus{}1} \sin^{2j\plus{}1}x$

This is the best I can do using basic functions.

Funny, the whole time I was thinking as if the denominator was $ (1 \plus{} \cos(x))^2$< in which case it's easy after integrating by parts... That's why I assumed the OP didn't know that method...

There is probably no closed form expression for the antiderivative of that function as stated.

[quote="milin"]There is but posting it could be seen as spam![/quote]
A finite combination of elementary functions? Maple gave me an expression involving dilogs. I certainly believe it can be written pretty concisely if you just use hypergeometric series notation! Pretty much anything can be written that way!

I think (probably) that is the word about a definite integral, for example $ \int_0^{\pi}\frac {x\sin x}{1\plus{}\cos^2x}\ \mathrm {dx}$ ...

If so, this would change the situation drastically. Using integration by parts we would get the first part equal to $ \frac{\pi^2}{4}$, and the second (troubling us at the moment) would equal $ 0$. This is because $ \tan^{\minus{}1}\left(\cos \left(x\minus{}\frac{\pi}{2}\right)\right)$ is odd on the interval $ \left[\minus{}\frac{\pi}{2},\frac{\pi}{2}\right]$. So, we would have:

$ \int^\pi_0 \frac{x \sin x}{1\plus{}\cos^2 x}\,dx\equal{}\frac{\pi^2}{4}$

[b]Indeed[/b], $ I \equal{} \int_0^{\pi}\frac {(\pi \minus{} x)\sin x}{1 \plus{} \cos^{2}x}\ \mathrm{dx\ \ }$ $ \stackrel {\begin{array}{c} (x: \equal{} \pi \minus{} x) \\
\end{array}}{\ \ \implies\ \ }\mathrm {\ \ }$ $ I \equal{} \frac {\pi}{2}\cdot\int_0^{\pi}\frac {\sin x}{1 \plus{} \cos^2x}\ \mathrm {dx} \equal{} \left\|\frac {\pi}{2}\cdot \arctan (\cos x)\right\|_{\pi}^0$ $ \implies$ $ I \equal{} \frac {\pi^2}{4}$ .

[quote="milin"]Yes, dilogs are involved. Still, it is a closed form!  :)[/quote]
Then anything is closed form because I can just introduce a function with whatever values


There are six points on the plane, and no three points are collinear. Let $G_1$ be the centroid of a triangle formed by three points that are randomly chosen from the six points, and let $G_2$ be the centroid of a triangle formed by the other points. Prove that a line connecting $G_1$ and $G_2$ goes through a fix point regardless of how we choose the three points.

We can pick 2 different lines, so that there will be at-most one point.

Let the points be $P_i = (x_i, y_i)$ for $i \in \{1,2,3,4,5,6\}$.  We guess the "fixed point" we want is $P = (\frac{\sum_{i=1}^6 x_i}{6} , \frac{\sum_{i=1}^6 y_i}{6})$.

The easiest way is to look at the area $G_1G_2P$ for any choice of triangle that will give $G_1, G_2$.

Wlog, $P_1, P_2, P_3$ has centroid $G_1 = (x_1, y_1)$ and likewise, $G_2 = (x_2, y_2)$.

Then we only need to verify $\frac{x_1 + x_2}{2}y_1 + x_1y_2 + x_2\frac{y_1 + y_2}{2} = \frac{x_1 + x_2}{2}y_2 + x_1\frac{y_1 + y_2}{2} + x_2y_1$.  It works.

So we have found the unique point P.


The first problem from BMO 2005 Iasi, Romania
Let $ABC$ be a triangle whose inscribed circle touches $AB$ and $AC$ at $D$ and $E$, respectively. Let $X$ and $Y$ be the points of intersection of the bisectors of the angles $\angle ACB$ and $\angle ABC$ with the line $DE$ and let $Z$ be the midpoint of $BC$. Prove that the triangle $XYZ$ is equilateral if and only if $A$ is $60^o$.

They've already been posted, man :). I'll post the links in a second.

Edit: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35934[/url]


For the first one , we need to use Cauchy-Schwarz only:

$ \left(\sum_{cyc}ab^2 \plus{} abc \plus{} ac^2 \right)\left(\sum_{cyc}\frac {a}{b^2 \plus{} bc \plus{} c^2}\right)\geq (a \plus{} b \plus{} c)^2$

But $ \sum_{cyc}ab^2 \plus{} abc \plus{} ac^2 \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$

Therefore:

$ \sum_{cyc}\frac {a}{b^2 \plus{} bc \plus{} c^2} \geq \frac {(a \plus{} b \plus{} c)^2}{(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)} \equal{} \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}$

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
$ (b) \ \ \ \ \frac {b \plus{} c}{b^2 \plus{} 3bc \plus{} c^2} \plus{} \frac {c \plus{} a}{c^2 \plus{} 3ca \plus{} a^2} \plus{} \frac {a \plus{} b}{a^2 \plus{} 3ab \plus{} b^2} \le \frac {6(a \plus{} b \plus{} c)}{5(ab \plus{} bc \plus{} ca)}$.[/quote]
A hint: Rewriting the inequality in the simple form:
\[ \sum \frac{(b\minus{}c)^2(b\plus{}c\minus{}a)}{b^2\plus{}3bc\plus{}c^2} \ge 0,\]
which can be easily proved if we assme that $ a \ge b \ge c$. :)

[quote="can_hang2007"][quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
$ (b) \ \ \ \ \frac {b \plus{} c}{b^2 \plus{} 3bc \plus{} c^2} \plus{} \frac {c \plus{} a}{c^2 \plus{} 3ca \plus{} a^2} \plus{} \frac {a \plus{} b}{a^2 \plus{} 3ab \plus{} b^2} \le \frac {6(a \plus{} b \plus{} c)}{5(ab \plus{} bc \plus{} ca)}$.[/quote]
A hint: Rewriting the inequality in the simple form:
\[ \sum \frac {(b \minus{} c)^2(b \plus{} c \minus{} a)}{b^2 \plus{} 3bc \plus{} c^2} \ge 0,
\]
which can be easily proved if we assme that $ a \ge b \ge c$. :)[/quote]
Nice, Can_hang 2007. :lol:

[quote="encyclopedia"][quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that

$ (a) \ \ \ \ \frac a{b^2 \plus{} bc \plus{} c^2} \plus{} \frac b{c^2 \plus{} ca \plus{} a^2} \plus{} \frac c{a^2 \plus{} ab \plus{} b^2} \ge \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}$;

$ (b) \ \ \ \ \frac {b \plus{} c}{b^2 \plus{} 3bc \plus{} c^2} \plus{} \frac {c \plus{} a}{c^2 \plus{} 3ca \plus{} a^2} \plus{} \frac {a \plus{} b}{a^2 \plus{} 3ab \plus{} b^2} \le \frac {6(a \plus{} b \plus{} c)}{5(ab \plus{} bc \plus{} ca)}$.[/quote]

Some others

$ \sum\frac {(k \minus{} 1)(b \plus{} c) \plus{} 2a}{b^2 \plus{} kbc \plus{} c^2}\ge\frac {6k(a \plus{} b \plus{} c)}{(k \plus{} 2)(ab \plus{} bc \plus{} ca)}$

holds for all $ k > \minus{} 2$

$ \sum\frac {(k \minus{} 3)a \plus{} b \plus{} c}{b^2 \plus{} kbc \plus{} c^2}\le\frac {3(k \minus{} 1)(a \plus{} b \plus{} c)}{(k \plus{} 2)(ab \plus{} bc \plus{} ca)}$

holds for all $ 0\le k\le 3$[/quote]
Nice, Encyclopedia.  :lol:

But what you say about the following nice inequality.

Let $ a,b,c$ be nonnegative real numbers such that $ ab\plus{}bc\plus{}ca\equal{}3$. Prove that

$ 18(\frac 1{a^2 \plus{} b^2}\plus{}\frac 1{b^2 \plus{} c^2}\plus{}\frac 1{c^2 \plus{} c^2}) \plus{}5(a^2\plus{}b^2\plus{}c^2)\ge 42$,

with equality for $ a\equal{}b\equal{}c$ and again for $ 3a\equal{}b\equal{}c$ or any cyclic permutation.

It can be
$ (3x^{2} \plus{} 3y^{2} \plus{} 3z^{2} \plus{} xy \plus{} yz \plus{} zx)(xy \plus{} yz \plus{} zx)^{2} \ge12xyz(x \plus{} y \plus{} z)(x^{2} \plus{} y^{2} \plus{} z^{2})$
Is it hold?

[quote="can_hang2007"][quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
$ (b) \ \ \ \ \frac {b \plus{} c}{b^2 \plus{} 3bc \plus{} c^2} \plus{} \frac {c \plus{} a}{c^2 \plus{} 3ca \plus{} a^2} \plus{} \frac {a \plus{} b}{a^2 \plus{} 3ab \plus{} b^2} \le \frac {6(a \plus{} b \plus{} c)}{5(ab \plus{} bc \plus{} ca)}$.[/quote]
A hint: Rewriting the inequality in the simple form:
\[ \sum \frac {(b \minus{} c)^2(b \plus{} c \minus{} a)}{b^2 \plus{} 3bc \plus{} c^2} \ge 0,
\]
which can be easily proved if we assme that $ a \ge b \ge c$. :)[/quote]
Can you explain in details how you get this pls. I'm very curious about it

Flakky, write the inequality as

$ \sum{[\frac {b\plus{}c}{b^2\plus{}3bc\plus{}c^2}\minus{}\frac {4a\plus{}b\plus{}c}{5(ab\plus{}bc\plus{}ca)}]} \le 0$.

[quote="encyclopedia"] Some others

$ \sum\frac {(k \minus{} 1)(b \plus{} c) \plus{} 2a}{b^2 \plus{} kbc \plus{} c^2}\ge\frac {6k(a \plus{} b \plus{} c)}{(k \plus{} 2)(ab \plus{} bc \plus{} ca)}$

holds for all $ k > \minus{} 2$[/quote]
Actually, the following inequality is stronger since equality holds for $ a\equal{}b\equal{}c$ and again for $ a\equal{}0$ and $ b\equal{}c$:

$ \sum\frac {(3\minus{}k)a\plus{}(k \minus{} 1)(b \plus{} c)}{b^2 \plus{} kbc \plus{} c^2}\ge\frac {3(k\plus{}1)(a \plus{} b \plus{} c)}{(k \plus{} 2)(ab \plus{} bc \plus{} ca)}$.

This inequality holds for $ \frac {\minus{}3}{2}\le k \le \frac 5{2}$. For $ k\equal{}2$ we get again Iran96 Inequality.

Moreover, for $ m\equal{}\frac {2\minus{}\sqrt 2}{4}$, the inequality holds

$ \sum\frac {ma\plus{}b \plus{} c}{b^2 \plus{} 3bc \plus{} c^2}\ge\frac {3(m\plus{}2)(a \plus{} b \plus{} c)}{5(ab \plus{} bc \plus{} ca)}$,

with equality for $ a\equal{}b\equal{}c$, and again for $ \frac a{\sqrt 2}\equal{}b\equal{}c$.


trigonometry
analytic geometry
function
ratio

Ok, first off I'd like to admit something. I'm no high-schooler. However, in high school I never had a trig class.
I'm a programmer by trade and most of the stuff I deal with is pulling and manipulating data. However, generating a pie chart is a bit out of my range right now. Bar graphs, line graphs, scatter plots, and trend lines are all easy. But the most I know about finding the size of a pie slice is it involves right traingles, the radius and the data set somehow. I've decided to do the pie chart in SVG, for various reasons.

I know that I need to do something with sine and cosine. I think I need to find the hypotenuse of said triangle, but I don't know exactly where it comes from and why.

Here is a sample data set, the question doesn't really matter:
Yes	144
No	261
N/A	35

The first thing I guess I would do is decide the size of the radius and model the data around that.
I have decided of the radius size 1. Respectively rounded, that would make the dataset:

Yes 0.33
No 0.59
N/A 0.08

I don't need to know the size of any of these slices, just the method used to obtain the arc end point.

Beggin' ya. :)

EDIT: Forgot to mention, I'm using cartesian coordinates with the center at 0, if that matters at all.

First, go here, and read about the sine, cosine, and tangent functions:

http://www.horacemann.pvt.k12.ny.us/academics/math/pcbch/trig/sohcahtoa.html

Then, go here, and read about the unit circle:

http://www.humboldt.edu/~dlj1/PreCalculus/Images/UnitCircle.html

Once you understand those basic concepts:

Any cartesian point (x,y) can also be written $(r\cos{\theta},r\sin{\theta})$ where r is the magnitude of the point (that is, $\sqrt{x^{2}+y^{2}}=r$) and $\theta$ is the angle from the positive x axis. That angle can be found by $\tan^{-1}{\frac{y}{x}}=\theta$, the inverse of the tangent function, where y is the "opposite side" of the right triangle and x is the "adjacent side." (If you don't follow, you'll need to review the two above links.)

The process also works in reverse, which is what is useful to you. There are 360 degrees in a circle. The "yes" responses need to occupy 33% of the central angle, or $(.33)(360)=118.8$ degrees. You've already decided the radius is 1 (good, that makes it easier) so if we start "yes" at the positive x axis and go counterclockwise (which conventionally is the positive angle direction):

Start point for "yes": $(x,y)=(\cos{0},\sin{0})=(1,0)$
End point for "yes": $(x,y)=(\cos{118.8},\sin{118.8})=(-0.482, 0.876)$

By the same reasoning, "no" needs to take up $(.59)(360)=212.4$ degrees. But we have to start at the end of "yes." So:

Start point for "no": $(x,y)=(\cos{118.8},\sin{118.8})=(-0.482, 0.876)$
End point for "no": $(x,y)=(\cos{(118.8+212.4)},\sin{(118.8+212.4)})=(\cos{331.2},\sin{331.2})=(0.876,-0.482)$

And then clearly for "N/A" the start/end points are (0.876, -0.482) and (1,0).

Hope that helps!

[quote="bthings_2000"]Ok, you made a common mistake: you forgot to put your calculator in "degree" mode instead of "radian" mode. When considering angles in degrees, make sure you're in degree mode or nothing makes sense.

All you $\sin$ and $\cos$ angles are right, but they did not evaluate correctly because your calculator is in the wrong mode.[/quote]
For all your help, I thank you kindly.

/me performs a medium ojigi

I finally feel like I'm catching up to what I should know by now.

Quick question relating to the topic at hand though.

After I have all the coordinates, can I scale pie graph by multiplying all the coordinates by the same number (say oh, 100)

Yep! Recall that we can express $(x,y)=(r\cos{\theta},r\sin{\theta})$. In the previous case we had r=1. Just make r anything you want, but make sure it's the same throughout the graph.

Arg! I've encountered something quite frustrating.

The programming language I used only outputs radians for cosine, sine, and tangent.

I then tried to convert this number to degrees with rad*(180/pi), it's way off. If I divide this number by 100, it gives a number similar to what the calculator outputs.

Example:
Angle: .67*360=241.2
Calculator Degree mode: $(\cos 241.2)=-0.481$
Programming Language(Rad): $(\cos 241.2)=-0.763$
Conversion: $(-0.763*(180/pi)/100)=-0.437$

That's a difference of -0.044. With my radius as 1, couldn't this effect my results?

Or maybe I'm just doing it wrong.

One way or the other, I've got to get radians to degrees.

I'm confused. You say the sine, cosine, and tangent only OUTPUT in radians? Those three functions accept angle measures as INPUT, but OUTPUT a ratio between sides...

[quote="bthings_2000"]You have input/output confusion.

A circle can be measured in 360 degrees or $2\pi$ radians. Thus the way to convert from radians R to degrees D is, like you said:

$D=\frac{180}{\pi}R$

and,

$R=\frac{\pi}{180}D$

So in the example you gave: you have 241.2 degrees. To convert to radians:

$R=\frac{\pi}{180}(241.2)=4.2097=R$

And indeed, $\cos{4.2097}=-.4818$ (did you round correctly before?)

Like tjhance said, the INPUT of sine, cosine, and tangent is degrees or radians. The OUTPUT of those functions is a ratio between side lengths, without units.

(And the INPUT of sine inverse, cosine inverse, and tangent inverse is a ratio between sides, and the output of those is an angle measure in degrees or radians.)

Dividing by 100 didn't do anything except produce an almost-coincidence. Note that measuring a circle in $2\pi\approx{6.283}$ radians means that the sine and cosine functions are periodic with period $2\pi$, which is only on the order $10^{0}$ (not very big) so there is a decent chance for coincidences if you move some numbers around.[/quote]

I was in fact very confused. I found out from someone in IRC what I really wanted was: $\cos(0.67*(2\pi))$

Sorry. -_-; I'm an imbecile.

yep! that goes directly from ratio to radians, instead of ratio to degrees to radians. So you're set.


[quote="iptgut"]help me !!![/quote]

Your solution is corect and nice, iptgut :)

[quote="iptgut"]with equality if and only if a=?,b=?, c = ?.[/quote]

Equality does not occur, iptgut.


What are the last three digits of $ 3^{210}$?


algebra
polynomial
function
number theory
least common multiple
ceiling function
logarithms

let a integer positive number
Let F(N) be the number of binary string which has length is N (INCLUDING SUCH BINARY STRINGS THAT HAS BEGINNING BITS IS 0) and NOT CONTAIN 101 OR 111
for examples with N=3, there are 8 binary strings
but only 6 are satisfying 000 001 010 011 100 110 , not 101 and 111
member Teak on ioicamp has proved that F(N)=F(N-1)+F(N-3)+F(N-4)
and he also find out integer number k so that F(N)=F(N+k)(mod 2005)
(for all N) .. and that k is 200 (I don't know whether it is true or not)
I want to generalize it that :let m a integer number, is there existence of integer number k  so that for all N : F(N)=F(N+k)(mod m) , and dind out that number k

First off, it is easy to verify that the minimal polynomial of the sequence F(N) is $f(x)=x^{4}-x^{3}-x-1.$
From the recurrence F(N)=F(N-1)+F(N-3)+F(N-4) and initial values F(0)=1, F(1)=2, F(2)=4, F(3)=6 it can be also obtained a generating function of the sequence F(N), that is $\frac{x^{3}+2x^{2}+x+1}{1-x-x^{3}-x^{4}}.$

The period of F(N) modulo m equals LCM of the periods modulo each power of prime in the prime factorization of m.

Let $q=p^{k}$ be the maximum power of prime $p$ dividing $m$. The period of F(N) modulo q equals $T(f(x))$ (i.e., the period  of its minimal polynomial $f(x)$) over the field $GF(q).$

There is a theorem stating that if $f(x)=x^{l}f_{1}(x)^{k_{1}}\dots f_{s}(x)^{k_{s}}$ is a factorization over $GF(q)$ into powers of irreducible polynomials, then
$T(f(x)) = p^{c}\cdot LCM(T(f_{1}(x)),\dots,T(f_{s}(x)))$ where $c=\lceil \log_{p}\max\{k_{1},\dots,k_{s}\}\rceil.$
And the period of each irreducible polynomial $f_{i}(x)$ is a divisor of $q^{\deg(f_{i})}-1$ (but not of $q^{m}-1$ for $m<\deg(f_{i})$).
Therefore, it is enough to find the smallest divisor $d$ of $q^\deg(f_{i})-1$ such that $x^{d}\equiv 1\pmod{f_{i}(x)}.$

[b]Example[/b]. $m=2005=5\cdot 401.$

Over GF(5) we have $f(x)=(x+2)^{3}(x+3),$
$T(x+2)=T(x+3)=4,\ c=1.$
Therefore, $T(f(x))$ modulo 5 is $4\cdot 5=20.$

Over GF(401) we have $f(x)=(x+20)(x+111)(x+289)(x+381),$
$T(x+20)=T(x+381)=4,\ T(x+111)=T(x+289)=200,\ \ c=0.$
Therefore, $T(f(x))$ modulo 401 is 200.

Combining results, we have $T(f(x))$ modulo 2005 is LCM(200,20)=200.

thank you very much but I don't understand 3 lines : from 1 to 3
about generating the sequence F(n) :mad:  :blush:

and why do we have "The period of F(N) modulo m equals LCM of the periods modulo each power of prime in the prime factorization of m."
i think it is Chinese Theorem ??? :huh:

[quote="nttjuvwamsncc"]thank you very much but I don't understand 3 lines : from 1 to 3
about generating the sequence F(n) :mad:  :blush: [/quote]
See http://mathworld.wolfram.com/GeneratingFunction.html
[quote="nttjuvwamsncc"]and why do we have "The period of F(N) modulo m equals LCM of the periods modulo each power of prime in the prime factorization of m."
i think it is Chinese Theorem ??? :huh:[/quote]
First notice that the period moludo m is divisible by the period modulo each prime power in the prime factorization of m and, hence, by LCM of these periods. On the other hand, using Chinese Theorem, it is easy to check that this LCM *is* actually the period modulo m.


The German IMO team 2006 is to be announced in the next few minutes.

Interested spectators may take a look at the current distribution of scores ([url=http://noname.schlussda.com/imo/a06.html]list[/url]).

The team members are:

Peter Scholze (139, silver '04, gold '05)
Christian Sattler (117, silver '04, '05)
Darij Grinberg (87, silver '04, '05)
Daniel Harrer (74, silver '05)
Matthias Ohst (72, honourable mention '04)
Georg Schönherr (69, bronze '05)

The following students will act as replacement in case of one of the above being prevented from participation:

Paul Jonas Hamacher (67)
Moritz Rodenhausen (65)

The scores inside the brackets indicate the scores achieved in the selection tests.

[quote="NoName"]The team members are:

Peter Scholze (139, silver '04, gold '05)
Christian Sattler (117, silver '04, '05)
Darij Grinberg (87, silver '04, '05)
Daniel Harrer (74, silver '05)
Matthias Ohst (72, honorable mention '04)
Georg Schönherr (69, bronce '05)

The following pupils will act as replacement in case of one of the above being prevented from participation:

Paul Jonas Hamacher (87)
Moritz Rodenhausen (85)

The scores inside the brackets indicate the scores achieved in the selection tests.[/quote]

The 7º and 8º students have a bigger score than 4º, 5º and 6º??? How's that???

Congratulations to all of the team members :)

I see that we only have team members who have already been to an IMO before. Seems to me like the German team could get quite a good result this year.  :lol:

congrats to all of you!

especially to Peter Scholze, who got 139(!!) of possible 147 points during the TSTs.
did any german have any comparable result in the TSTs?

and of course a lot of success in slowenia!!  (wishing good luck, always seems, that it is very unpredictable ... )

Glückwunsch an alle die es geschafft haben und viel Erfolg in Slowenien:)
Ich hoffe, daß ihr vor noch mehr anderen Ländern landet als letztes Jahr ;) :D

[quote="darij grimberg"](@Valentin: why did you lock http://www.mathlinks.ro/Forum/viewtopic.php?t=88832 ? :(  it was so much fun)[/quote]indeed!

[quote="darij grinberg"][quote="darij grimberg"]Congratulations to the real team  :D

(@Valentin: why did you lock http://www.mathlinks.ro/Forum/viewtopic.php?t=88832 ? :(  it was so much fun)[/quote]

congratulations to the unreal darij  :D

  dg[/quote]

thank you  :D

dg

[quote="Idril"]Gratulations to all of you.

must be a little mistake if you look at the startistics http://www.noname.schlussda.com/imo/a06.html
you'll see they have about 66 and 63 points[/quote]

Thanks.

congratulations to the team members, are there problems from the TST anywhere?

Some are in the resources ( http://www.mathlinks.ro/Forum/resources.php?c=64&cid=54&year=2006 ), and the others will be posted in the future (there are some reasons not to do this now...). Two or three are also already posted, so I think darij will add them soon.

@Peeta: yes, similar results were reached before.

[quote="darij grimberg"]Congratulations to the real team  :D

(@Valentin: why did you lock http://www.mathlinks.ro/Forum/viewtopic.php?t=88832 ? :(  it was so much fun)[/quote]Because I got 10 PMs asking me what the hell is going on :) It was fun but well all fun things come to an end :D Congrats to all german team and see ya all in Slovenia!


Let ABC be triangle and its incircle tangent to BC,CA,AB at D,E,F respectively.A',B'and C' are points on BC,CA and AB recpectively such that AA',BB' and CC' are

cocurrent.A" is a point on incircle such that A'A" is tangent to incircle (A" is distinct from D).We define B"and C" in the same way.Prove that AA",BB" and CC"

are cocurrent.

[hide]See for a solution at the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=140191]Interesting and hard[/url], of [b][size=100]cvix[/size][/b].[/hide]

Kostas Vittas.