[b]1/[/b] Let $ a,b,c$ be positive real numbers. Prove that
$ \frac{(a\plus{}b\plus{}c)^2}{abc}\plus{}\frac{54}{\sqrt{3(a^2\plus{}b^2\plus{}c^2)}}\ge \frac{81}{a\plus{}b\plus{}c}$

[b]2/[/b] Let $ a,b,c$ be positive real numbers. Prove that
$ \frac{8(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2}\plus{}\frac{3(a\plus{}b)(b\plus{}c)(c\plus{}a)}{abc}\ge 48$

[b]3/[/b] Let $ a,b,c$ be positive real numbers. Prove that
$ \frac1{abc}\plus{}\frac{27}{(a\plus{}b\plus{}c)^3}\ge \frac{16}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$

[b]4/[/b] Let $ a,b,c$ be positive real numbers. Prove that
$ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{abc}\plus{}\frac{16(a\plus{}b\plus{}c)}{\sqrt{3(ab\plus{}bc\plus{}ca)}}\ge 24$

[b]5/[/b] Let $ a,b,c$ be positive real numbers. Prove that
$ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{abc}\plus{}\frac{4(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2}\ge 12$

[b]6/[/b] Let $ a,b,c$ be nonnegative real numbers. Prove that
$ \frac{a^3\plus{}b^3\plus{}c^3}{a^2\plus{}b^2\plus{}c^2}\plus{}\frac{18}{5}.\frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{(a\plus{}b\plus{}c)^2}\ge \frac{7}5(a\plus{}b\plus{}c)$

All above Inequalities are easy, but nice :)

number 2

$ a\plus{}b \ge 2\sqrt{ab}$
$ b\plus{}c \ge 2\sqrt{bc}$
$ c\plus{}a \ge 2\sqrt{ca}$

$ \rightarrow (a\plus{}b)(b\plus{}c)(c\plus{}a) \ge 8abc$

$ (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2 \ge 0$
$ a^2 \plus{} b^2 \plus{}c^2 \ge ab\plus{}bc\plus{}ca$
$ a^2 \plus{}b^2 \plus{}c^2 \plus{}2(ab\plus{}bc\plus{}ca) \ge 3(ab\plus{}bc\plus{}ca)$

$ \rightarrow (a\plus{}b\plus{}c)^2 \ge 3(ab\plus{}bc\plus{}ca)$

 :D  the net step is easy   :rotfl:

I don't think so. Sorry but indeed, your proof is so silly  :wink:
There are many inequalities similar above, but I think it isn't quite easy.
Example, with number 2, here is my proof :
By SOS, we are easily to prove: $ (a\plus{}b\plus{}c)^2(a\plus{}b)(b\plus{}c)(c\plus{}a) \geq\ 24abc(a^2\plus{}b^2\plus{}c^2)$
So, it's becomes: $ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{abc} \plus{} \frac{64abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \geq\ 16$ (trues with Am-Gm)
@ddlam:  Many people 'll be happy by your return. I am a man like this  :)

Hey, Raja Oktovin
I don't think your solution is right. Please detail your post carefully again.

ups... I misread the problem..  :blush:  sorry i think it is $ \frac {8(a \plus{} b \plus{} c)^2}{ab \plus{} bc \plus{} ca}$... haha.. yes, that's a silly one.. sorry...  :blush:

[quote="dduclam"]
[b]2/[/b] Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {8(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {3(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc}\ge 48$
[/quote]
$ \frac {8(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {3(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc}\ge 48\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(3a^4b\plus{}3a^3b^2\plus{}11a^2b^2c\minus{}17a^3bc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}b)^2(6c^3\minus{}11abc\plus{}3ab(a\plus{}b))\geq0.$
But $ 6c^3\minus{}11abc\plus{}3ab(a\plus{}b)\geq6c^3\plus{}3\sqrt{a^3b^3}\plus{}3\sqrt{a^3b^3}\minus{}11abc\geq\left(3\sqrt[3]{54}\minus{}11\right)abc\geq0.$

[quote="dduclam"]

[b]4/[/b] Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc} \plus{} \frac {16(a \plus{} b \plus{} c)}{\sqrt {3(ab \plus{} bc \plus{} ca)}}\ge 24$

[/quote]
We have $ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc} \ge\ 8$

Need prove $ \frac {16(a \plus{} b \plus{} c)}{\sqrt {3(ab \plus{} bc \plus{} ca)}}\ge 16$

$ \leftrightarrow (a\plus{}b\plus{}c)^2 \ge\ 3(ab\plus{}bc\plus{}ca)$ which is obvious

[quote="dduclam"][b]1/[/b] Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {(a \plus{} b \plus{} c)^2}{abc} \plus{} \frac {54}{\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}\ge \frac {81}{a \plus{} b \plus{} c}$
[/quote]
By AM-GM inequality we have
\[ \frac {{\left( {a \plus{} b \plus{} c} \right)^3 }}{{abc}} \plus{} \frac {{54\left( {a \plus{} b \plus{} c} \right)}}{{\sqrt {3\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)} }} \ge 27\sqrt [3]{{\frac {{\left( {a \plus{} b \plus{} c} \right)^5 }}{{3abc\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)}}}}
\]
It suffices us show that
\[ \left( {a \plus{} b \plus{} c} \right)^5 \ge 81abc\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)
\]
which is well known.  :P
[quote="dduclam"]
[b]4/[/b] Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc} \plus{} \frac {16(a \plus{} b \plus{} c)}{\sqrt {3(ab \plus{} bc \plus{} ca)}}\ge 24$
[/quote]
Using AM-GM inequality too we have
\[ \frac {1}{{abc}} \plus{} \frac {{27}}{{\left( {a \plus{} b \plus{} c} \right)^3 }} \ge 2\sqrt {\frac {{27}}{{abc\left( {a \plus{} b \plus{} c} \right)^3 }}}
\]
It reamains us to prove that
\[ 2\sqrt {\frac {{27}}{{abc\left( {a \plus{} b \plus{} c} \right)^3 }}} \ge \frac {{16}}{{\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}}
\]

\[ \Leftrightarrow 27\left[ {\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)} \right]^2 \ge 64abc\left( {a \plus{} b \plus{} c} \right)^3
\]
which is true since
\[ \left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right) \ge \frac {8}{9}\left( {a \plus{} b \plus{} c} \right)\left( {ab \plus{} bc \plus{} ca} \right)
\]
:)

[quote="onlylove_math"][quote="dduclam"]

[b]4/[/b] Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc} \plus{} \frac {16(a \plus{} b \plus{} c)}{\sqrt {3(ab \plus{} bc \plus{} ca)}}\ge 24$

[/quote]
We have $ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc} \ge\ 8$

Need prove $ \frac {16(a \plus{} b \plus{} c)}{\sqrt {3(ab \plus{} bc \plus{} ca)}}\ge 16$

$ \leftrightarrow (a \plus{} b \plus{} c)^2 \ge\ 3(ab \plus{} bc \plus{} ca)$ which is obvious[/quote]

I think Brother Lam posted wrong. The ineq he want to post is:
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc} \plus{} \frac {16(a \plus{} b \plus{} c)}{\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}\ge 24$
In general, problem $ 1, 2, 4, 5$ was created from:
$ (a \plus{} b \plus{} c)^2(a \plus{} b)(b \plus{} c)(c \plus{} a) \geq\ 24abc(a^2 \plus{} b^2 \plus{} c^2)$
Problem $ 3$ was created from:
$ (a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2 \geq\ \frac {64}{27}abc(a \plus{} b \plus{} c)^3$
And problem $ 6$ was created from...  :wink:
$ \frac{a^3\plus{}b^3\plus{}c^3}{a^2\plus{}b^2\plus{}c^2}\plus{}\frac{16}{5}.\frac{ab\plus{}bc\plus{}ca}{a\plus{}b\plus{}c}\ge \frac{7}5(a\plus{}b\plus{}c)$   :maybe:

[quote="Thinkers045"][b]Problem 5.[/b]

By AM - GM Inequality, the leftmost component of the left hand side of the inequality is at least 8. Moreover, the next component of the left hand side of the inequality is at least 4 since a^2+b^2+c^2 at least ab + ac + bc. That is,

(a+b)(b+c)(a+c)/abc >= 8 ..................................................(*)

and

4( ab + ac + bc)/(a^2+b^2+c^2) >= 4.................................................(**)

Adding (*) and (**) gives the desired inequality.[/quote]
[b]Problem 5.[/b]
Assum c=max{a;b;c}
$ Ineq \leftrightarrow (\frac {2}{ab} \minus{} \frac {2}{a^2 \plus{} b^2 \plus{} c^2})(a \minus{} b)^2 \plus{} (\frac {a \plus{} b}{abc} \minus{} \frac {2}{a^2 \plus{} b^2 \plus{} c^2})(a \minus{} c)(b \minus{} c) \ge\ 0$
Which is obvious  :)

[quote="Thinkers045"][b]Problem 5.[/b]

By AM - GM Inequality, the leftmost component of the left hand side of the inequality is at least 8. Moreover, the next component of the left hand side of the inequality is at least 4 since a^2+b^2+c^2 at least ab + ac + bc. That is,

(a+b)(b+c)(a+c)/abc >= 8 ..................................................(*)

and

4( ab + ac + bc)/(a^2+b^2+c^2) >= 4.................................................(**)

Adding (*) and (**) gives the desired inequality.[/quote]

The second component is[b] less or equal of 4[/b] because of $ a^2\plus{}b^2\plus{}c^2\ge ab\plus{}bc\plus{}ac$, so your solution isn't correct.

[quote="dduclam"][b]2/[/b] Let $ x,y,z$ be positive real numbers. Prove that

$ \frac {3(y \plus{} z)(z \plus{} x)(x \plus{} y)}{xyz} \plus{} \frac {8(x \plus{} y \plus{} z)^2}{x^2 \plus{} y^2 \plus{} z^2}\ge 48$.[/quote]We shall prove the stronger inequality :

$ \frac {(y \plus{} z)(z \plus{} x)(x \plus{} y)}{xyz} \plus{} \frac {2\sqrt {2}(x \plus{} y \plus{} z)^2}{x^2 \plus{} y^2 \plus{} z^2}\geq 8 \plus{} 6\sqrt {2},$

with equality if $ x \equal{} \sqrt {2},y \equal{} z \equal{} 1.$
[hide="Proof."]$ \frac {(y \plus{} z)(z \plus{} x)(x \plus{} y)}{xyz} \plus{} \frac {2\sqrt {2}(x \plus{} y \plus{} z)^2}{x^2 \plus{} y^2 \plus{} z^2} \minus{} 8 \minus{} 6\sqrt {2}\equiv\frac {F(x,y,z)}{xyz\left(x^2 \plus{} y^2 \plus{} z^2\right)}$,

$ F(x,y,z) \equal{} x^4(y \plus{} z) \plus{} x^3\left[y^2 \minus{} 2\left(3 \plus{} 2\sqrt {2}\right)yz \plus{} z^2\right]$

$ \plus{} x^2(y \plus{} z)\left[y^2 \plus{} \left(1 \plus{} 4\sqrt {2}\right)yz \plus{} z^2\right]$

$ \plus{} x\left[y^4 \minus{} 2\left(3 \plus{} 2\sqrt {2}\right)y^3z \plus{} 2\left(1 \plus{} 2\sqrt {2}\right)y^2z^2 \minus{} 2\left(3 \plus{} 2\sqrt {2}\right)yz^3 \plus{} z^4\right]$

$ \plus{} yz(y \plus{} z)\left(y^2 \plus{} z^2\right)$

$ \equal{} \frac {(y \plus{} z)}{8}(2x \minus{} y \minus{} z)^2\left(\sqrt {2}x \minus{} y \minus{} z\right)^2$

$ \plus{} \frac {(y \minus{} z)^2}{16}\Big\{2\left(2 \plus{} \sqrt {2}\right)(2x \minus{} y \minus{} z)^3 \plus{} \left(13 \plus{} 2\sqrt {2}\right)(y \plus{} z)(2x \minus{} y \minus{} z)^2$

$ \plus{} 2(2x \minus{} y \minus{} z)\left[13y^2 \minus{} 2\left(6\sqrt {2} \minus{} 5\right)yz \plus{} 13z^2\right]$

$ \plus{} 3(y \plus{} z)\left[5y^2 \minus{} 2\left(4\sqrt {2} \minus{} 1\right)yz \plus{} 5z^2\right]\Big\}\geq 0$,

which is clearly true for $ x \equal{} \max\{x,y,z\}$.[/hide]

[quote="dduclam"]
[b]6/[/b] Let $ a,b,c$ be nonnegative real numbers. Prove that
$ \frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {18}{5}.\frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \plus{} c)^2}\ge \frac {7}5(a \plus{} b \plus{} c)$
[/quote]
Dear dduclam,  :)

This problem is wrong with $ c\equal{}0$.  :)

[quote="Ji Chen"]
$ \frac {(y \plus{} z)(z \plus{} x)(x \plus{} y)}{xyz} \plus{} \frac {2\sqrt {2}(x \plus{} y \plus{} z)^2}{x^2 \plus{} y^2 \plus{} z^2} \minus{} 8 \minus{} 6\sqrt {2}\equiv\frac {F(x,y,z)}{xyz\left(x^2 \plus{} y^2 \plus{} z^2\right)}$,

$ F(x,y,z) \equal{} x^4(y \plus{} z) \plus{} x^3\left[y^2 \minus{} 2\left(3 \plus{} 2\sqrt {2}\right)yz \plus{} z^2\right]$

$ \plus{} x^2(y \plus{} z)\left[y^2 \plus{} \left(1 \plus{} 4\sqrt {2}\right)yz \plus{} z^2\right]$

$ \plus{} x\left[y^4 \minus{} 2\left(3 \plus{} 2\sqrt {2}\right)y^3z \plus{} 2\left(1 \plus{} 2\sqrt {2}\right)y^2z^2 \minus{} 2\left(3 \plus{} 2\sqrt {2}\right)yz^3 \plus{} z^4\right]$

$ \plus{} yz(y \plus{} z)\left(y^2 \plus{} z^2\right)$

$ \equal{} \frac {(y \plus{} z)}{8}(2x \minus{} y \minus{} z)^2\left(\sqrt {2}x \minus{} y \minus{} z\right)^2$

$ \plus{} \frac {(y \minus{} z)^2}{16}\Big\{2\left(2 \plus{} \sqrt {2}\right)(2x \minus{} y \minus{} z)^3 \plus{} \left(13 \plus{} 2\sqrt {2}\right)(y \plus{} z)(2x \minus{} y \minus{} z)^2$

$ \plus{} 2(2x \minus{} y \minus{} z)\left[13y^2 \minus{} 2\left(6\sqrt {2} \minus{} 5\right)yz \plus{} 13z^2\right]$

$ \plus{} 3(y \plus{} z)\left[5y^2 \minus{} 2\left(4\sqrt {2} \minus{} 1\right)yz \plus{} 5z^2\right]\Big\}\geq 0$,

which is clearly true for $ x \equal{} \max\{x,y,z\}$.[/hide][/quote]

Sorry but indeed, I don't like your proof because it's so hard and not nice with me  :wink:
SOS and Am-Gm can solve it.
$ Sa \equal{} \frac {1}{bc} \minus{} \frac {2\sqrt []{2}}{a^2 \plus{} b^2 \plus{} c^2}$...
We 'll prove $ Sb \geq\ 0 \Leftrightarrow f(c) \equal{} a^2 \plus{} b^2 \plus{} c^2 \minus{} 2\sqrt []{2}ca \geq\ 0$
$ f'(c) < 0$, so: $ f(c) \geq\ f(b) \equal{} (a \minus{} \sqrt []{2}b)^2 \geq\ 0$ (always trues)
We only neeed to prove: $ Sb \plus{} Sc \geq\ 0 \Leftrightarrow \frac {1}{ab} \plus{} \frac {1}{ca} \geq\ \frac {4\sqrt []{2}}{a^2 \plus{} b^2 \plus{} c^2}$
Using Am-Gm: $ LHS \geq\ \frac {2}{a\sqrt []{bc}}$ and $ a^2 \plus{} b^2 \plus{} c^2 \geq\ a^2 \plus{} 2bc \geq\ 2 \sqrt []{2}a \sqrt []{bc}.$
We have done. Equality occurs when $ a \equal{} b \equal{} c$ or $ a \equal{} \sqrt []{2}b \equal{} \sqrt []{2}c$
 :)
PS: I remember this problem was created by Pham Kim Hung and his solution used SOS and mixing variables.  :maybe:

[quote="nguoivn"]....In general, problem $ 1, 2, 4, 5$ was created from:
$ (a \plus{} b \plus{} c)^2(a \plus{} b)(b \plus{} c)(c \plus{} a) \geq\ 24abc(a^2 \plus{} b^2 \plus{} c^2)$
[/quote]

It can be rewritten as

$ \sum (ab(a\plus{}b)\plus{}4c^3\minus{}5abc)(a\minus{}b)^2 \geq 0$

but how to prove it by SOS theorems ?? :(

[quote="manlio"][quote="nguoivn"]....In general, problem $ 1, 2, 4, 5$ was created from:
$ (a \plus{} b \plus{} c)^2(a \plus{} b)(b \plus{} c)(c \plus{} a) \geq\ 24abc(a^2 \plus{} b^2 \plus{} c^2)$
[/quote]

It can be rewritten as

$ \sum (ab(a \plus{} b) \plus{} 4c^3 \minus{} 5abc)(a \minus{} b)^2 \geq 0$

but how to prove it by SOS theorems ?? :([/quote]
Here is my proof:
$ (a \plus{} b \plus{} c)^2(a \plus{} b)(b \plus{} c)(c \plus{} a) \geq\ 24abc(a^2 \plus{} b^2 \plus{} c^2)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}c(a \minus{} b)^2(a \plus{} b \minus{} 3c)^2\geq0.$
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62923

[quote="manlio"]Ohhhh....  Very nice!!!  :)

How did you find it?

Thank you very much.[/quote]
The equality occurs when $ a\equal{}2b\equal{}2c.$ :wink:
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62923

Thank you very much, Arqady, but I ask myself how to transform my form in your form in a simple way. Have you an answer?

Thank you very much.

[quote="manlio"]Arqady, but I ask myself how to transform my form in your form in a simple way. Have you an answer?
[/quote]
Through $ \sum (ab(a \plus{} b) \plus{} 4c^3 \minus{} 5abc)(a \minus{} b)^2 \geq 0\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(a^4b \plus{} 3a^3b^2 \minus{} 9a^3bc \plus{} 5a^2b^2c)\geq0\Leftrightarrow\sum_{cyc}c(a \minus{} b)^2(a \plus{} b \minus{} 3c)^2\geq0.$
But it's not easy.

Subtracting these two SOS form we have

$ \sum (a\minus{}b)^2 (a^2(c\minus{}b)\plus{}b^2(c\minus{}a) \plus{}5c^3 \plus{}7abc \minus{}6c^2(a\plus{}b)) \equal{}0$  :huh:

Very strange to imagine that!!!!

I think the magic trick is to use

$ A\equal{}\sum c(a \minus{} b)^2(c^2 \plus{} ab \minus{} ac \minus{} bc) \equal{} 0$

and

$ B\equal{}\sum (a\plus{}b)(a \minus{} b)^2(c^2 \plus{} ab \minus{} ac \minus{} bc) \equal{} 0$

very easy to prove

and to add at our SOS form the expression $ 5A\minus{}B\equal{}0$

:wink:

[quote="Maxim Bogdan"]Using the well-known identity:$ (a \plus{} b)(b \plus{} c)(c \plus{} a) \equal{} \frac {(a \plus{} b \plus{} c)^3 \minus{} a^3 \minus{} b^3 \minus{} c^3}{3}$ we obtain:

$ LHS \equal{} \frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {6}{5}(a \plus{} b \plus{} c) \minus{} \frac {6}{5}\frac {a^3 \plus{} b^3 \plus{} c^3}{(a \plus{} b \plus{} c)^2}\ge \frac {7}{5}(a \plus{} b \plus{} c)$

$ \Leftrightarrow$ $ (a^3 \plus{} b^3 \plus{} c^3)\frac {10(ab \plus{} bc \plus{} ca) \minus{} a^2 \minus{} b^2 \plus{} c^2}{(a \plus{} b \plus{} c)^2(a^2 \plus{} b^2 \plus{} c^2)}\ge (a \plus{} b \plus{} c) > 0$

But $ 10(ab \plus{} bc \plus{} ca) \minus{} a^2 \minus{} b^2 \plus{} c^2$ isn't always positive.For example $ a \equal{} 10^x,b \equal{} c \equal{} \frac {1}{10^x}$ where $ x\ge 1$.[/quote]

See
[quote="dduclam"]
@Honey_S: You're right. I'm sorry because I forgot the true form of problem. But I think the following result's true

$ \frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {9}{4}.\frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \plus{} c)^2}\ge a \plus{} b \plus{} c$[/quote]

[quote="manlio"][quote="nguoivn"]....In general, problem $ 1, 2, 4, 5$ was created from:
$ (a \plus{} b \plus{} c)^2(a \plus{} b)(b \plus{} c)(c \plus{} a) \geq\ 24abc(a^2 \plus{} b^2 \plus{} c^2)$
[/quote]

It can be rewritten as

$ \sum (ab(a \plus{} b) \plus{} 4c^3 \minus{} 5abc)(a \minus{} b)^2 \geq 0$

but how to prove it by SOS theorems ?? :([/quote]

Here is my solution by SOS ( I think it's short and easy to find)  :)

[quote="dduclam"]Thanks for care of everybody about that problems !

The problem 4 I posted incorrect, please read again as follows
[b]4/[/b] Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc} \plus{} \frac {16\sqrt {3(ab \plus{} bc \plus{} ca)}}{a \plus{} b \plus{} c}\ge 24$

@Honey_S: You're right. I'm sorry because I forgot the true form of problem. But I think the following result's true

$ \frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2} \plus{} \frac {9}{4}.\frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \plus{} c)^2}\ge a \plus{} b \plus{} c$[/quote]

Note that we have $ \frac{{a^3  \plus{} b^3  \plus{} c^3 }}{{a^2  \plus{} b^2  \plus{} c^2 }} \ge \frac{{a^2  \plus{} b^2  \plus{} c^2 }}{{a \plus{} b \plus{} c}}$

Thus it suffices to prove that
\[ \frac{{a^2  \plus{} b^2  \plus{} c^2 }}{{a \plus{} b \plus{} c}} \plus{} \frac{{9\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}}{{4\left( {a \plus{} b \plus{} c} \right)^2 }} \ge a \plus{} b \plus{} c
\]
\[ \Leftrightarrow \sum {ab\left( {a \plus{} b} \right)}  \ge 6abc
\]
true by AM-GM.  :)

Nice your SOS proof, nguoivn

But I like also

[quote="manlio"]I think the magic trick is to use

$ A \equal{} \sum c(a \minus{} b)^2(c^2 \plus{} ab \minus{} ac \minus{} bc) \equal{} 0$

and

$ B \equal{} \sum (a \plus{} b)(a \minus{} b)^2(c^2 \plus{} ab \minus{} ac \minus{} bc) \equal{} 0$

very easy to prove using cyclic expression $ (a\minus{}b)(b\minus{}c)(c\minus{}a)$

and to add at our SOS form the expression $ 5A \minus{} B \equal{} 0$

[/quote]

[quote="dduclam"][b]2/[/b] Let $ x,y,z$ be positive real numbers. Prove that

$ \frac {3(y \plus{} z)(z \plus{} x)(x \plus{} y)}{xyz} \plus{} \frac {8(x \plus{} y \plus{} z)^2}{x^2 \plus{} y^2 \plus{} z^2}\ge 48$.[/quote][quote="Ji Chen"]We shall prove the stronger inequality :

$ \frac {(y \plus{} z)(z \plus{} x)(x \plus{} y)}{xyz} \plus{} \frac {2\sqrt {2}(x \plus{} y \plus{} z)^2}{x^2 \plus{} y^2 \plus{} z^2}\geq 8 \plus{} 6\sqrt {2},$

with equality if $ x \equal{} \sqrt {2},y \equal{} z \equal{} 1.$[/quote][quote="nguoivn"]SOS can solve it.[/quote]Indeed, you are right.

$ \frac {(y \plus{} z)(z \plus{} x)(x \plus{} y)}{xyz} \plus{} \frac {2\sqrt {2}(x \plus{} y \plus{} z)^2}{x^2 \plus{} y^2 \plus{} z^2} \minus{} 8 \minus{} 6\sqrt {2}$

$ \equal{} \sum{\frac {(y \minus{} z)^2\left[y \plus{} z \minus{} \left(1 \plus{} \sqrt {2}\right)x\right]^2}{yz\left(x^2 \plus{} y^2 \plus{} z^2\right)}}\geq0.$


Let $ I$ be the incenter of $ \triangle{ABC}$. A line through $ I$ meets $ AC$ and $ AB$ at $ D$ and $ E$ respectively. Through $ D$ construct $ DF \parallel BC$ with $ F$ on line $ AI$.

Show that $ B$, $ C$, $ D$ and $ E$ are concyclic if and only if $ \angle{FBA}\equal{}\frac12 \angle{ACB}$.

[quote="shobber"]Let $ I$ be the incenter of $ \triangle{ABC}$. A line through $ I$ meets $ AC$ and $ AB$ at $ D$ and $ E$ respectively. Through $ D$ construct $ DF \parallel BC$ with $ F$ on line $ AI$.

Show that $ B$, $ C$, $ D$ and $ E$ are concyclic if and only if $ \angle{FBA} \equal{} \frac12 \angle{ACB}$.[/quote]

Let's $ DF \cap AC\equal{}M$

1. If $ B$, $ C$, $ D$ and $ E$ are concyclic then we have:
$ \frac{AB}{AC}\equal{}\frac{AE}{AD}$

$ \triangle ADF \sim  \triangle AEI$($ \angle DAF\equal{}\angle EAI$ and $ \angle ADF\equal{}\angle AEI$) so
$ \frac{AE}{AD}\equal{}\frac{AF}{AI}$
therefore
$ \frac{AB}{AC}\equal{}\frac{AF}{AI}$
therefore $ \triangle ACI \sim  \triangle ABF$(we have $ \angle CAI\equal{} \angle BAF$)
so $ \angle ABF\equal{}\angle ACI \equal{}\frac{1}{2}\angle ACB$

2. If $ \angle{FBA} \equal{} \frac12 \angle{ACB}$ then we have
$ \triangle ACI \sim  \triangle ABF$(becouse $ \angle CAI\equal{} \angle BAF$ and $ \angle ABF\equal{}\angle ACI$ )
so $ \frac{AF}{AI}\equal{}\frac{AB}{AC}\equal{}\frac{AD}{AM}$
$ \triangle AFM \sim  \triangle AID$($ \angle MAF\equal{}\angle DAI$) therefore
$ \angle ADE\equal{} \angle ADI\equal{}\angle AMF\equal{}\angle ACB \equal{}\angle ECB$
therefore $ B$, $ C$, $ D$ and $ E$ are concyclic.


linear algebra
matrix
linear algebra unsolved

Find all Symmetric, Positive Definite matrices in which every entry is either 0 or 1.

(This is a variation of a result from the paper "Acyclic Digraphs and Eigenvalues of (0,1) Matrices" by McKay, Oggier, Royle, Sloane, Wanless, and Wilf).

I could be getting this all wrong, but ....

I don't see that there are any other than the identity matrix. Plenty of positive semidefinite matrices, but not positive definite.

The idea: suppose we have a matrix with an off-diagonal element that is $1.$ Do a pemutation similarity to put $1$'s in the $(1,2)$ and $(2,1)$ positions. Then if the whole matrix is positive definite, then the upper-left $2\times 2$ matrix would need to be positive definite. But all of the $2\times 2$ symmetric zero-one matrices with $1$'s in the off-diagonal positions are either semidefinite or indefinite.


Let $ x,y,z$ be real numbers such that $ 0\le x,y,z \le 2$ and $ x \plus{} y \plus{} z \equal{} 3$.
Find the min and the max value of $ P \equal{} x^4 \plus{} y^4 \plus{} z^4 \plus{} 12(1 \minus{} x)(1 \minus{} y)(1 \minus{} z)$

$ x^2\plus{}y^2\plus{}z^2\equal{}(x\plus{}y\plus{}z)^2\minus{}2(xy\plus{}yz\plus{}zx)\equal{}9\minus{}2(xy\plus{}yz\plus{}zx),$

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

$ \equal{}(x^2\plus{}y^2\plus{}z^2)^2\minus{}2\{(xy\plus{}yz\plus{}zx)^2\minus{}2xyz(x\plus{}y\plus{}z)\}$

$ \equal{}\{9\minus{}2(xy\plus{}yz\plus{}zx)\}^2\minus{}2\{(xy\plus{}yz\plus{}zx)^2\minus{}6xyz\}$

$ \equal{}2(xy\plus{}yz\plus{}zx)^2\minus{}36(xy\plus{}yz\plus{}zx)\plus{}12xyz\plus{}81.$

On the other hand, we have $ (1\minus{}x)(1\minus{}y)(1\minus{}z)\equal{}1\minus{}(x\plus{}y\plus{}z)\plus{}xy\plus{}yz\plus{}zx\minus{}xyz\equal{}xy\plus{}yz\plus{}zx\minus{}xyz\minus{}2.$

$ \therefore P\equal{}2(xy\plus{}yz\plus{}zx)^2\minus{}36(xy\plus{}yz\plus{}zx)\plus{}12xyz\plus{}81\plus{}12(xy\plus{}yz\plus{}zx\minus{}xyz\minus{}2)$

$ \equal{}2(xy\plus{}yz\plus{}zx)^2\minus{}24(xy\plus{}yz\plus{}zx)\plus{}57$

$ \equal{}2(xy\plus{}yz\plus{}zx\minus{}6)^2\minus{}15$, the rest is using $ (x\plus{}y\plus{}z)^2\geq 3(xy\plus{}yz\plus{}zx)$ with $ x,\ y,\ z\in{[0,\ 2]}.$

[quote="kunny"]$ x^2 \plus{} y^2 \plus{} z^2 \equal{} (x \plus{} y \plus{} z)^2 \minus{} 2(xy \plus{} yz \plus{} zx) \equal{} 9 \minus{} 2(xy \plus{} yz \plus{} zx),$

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

$ \equal{} (x^2 \plus{} y^2 \plus{} z^2)^2 \minus{} 2\{(xy \plus{} yz \plus{} zx)^2 \minus{} 2xyz(x \plus{} y \plus{} z)\}$

$ \equal{} \{9 \minus{} 2(xy \plus{} yz \plus{} zx)\}^2 \minus{} 2\{(xy \plus{} yz \plus{} zx)^2 \minus{} 6xyz\}$

$ \equal{} 2(xy \plus{} yz \plus{} zx)^2 \minus{} 36(xy \plus{} yz \plus{} zx) \plus{} 12xyz \plus{} 81.$

On the other hand, we have $ (1 \minus{} x)(1 \minus{} y)(1 \minus{} z) \equal{} 1 \minus{} (x \plus{} y \plus{} z) \plus{} xy \plus{} yz \plus{} zx \minus{} xyz \equal{} xy \plus{} yz \plus{} zx \minus{} xyz \minus{} 2.$

$ \therefore P \equal{} 2(xy \plus{} yz \plus{} zx)^2 \minus{} 36(xy \plus{} yz \plus{} zx) \plus{} 12xyz \plus{} 81 \plus{} 12(xy \plus{} yz \plus{} zx \minus{} xyz \minus{} 2)$

$ \equal{} 2(xy \plus{} yz \plus{} zx)^2 \minus{} 24(xy \plus{} yz \plus{} zx) \plus{} 57$

$ \equal{} 2(xy \plus{} yz \plus{} zx \minus{} 6)^2 \minus{} 15$, the rest is using $ (x \plus{} y \plus{} z)^2\geq 3(xy \plus{} yz \plus{} zx)$ with $ x,\ y,\ z\in{[0,\ 2]}.$[/quote]
So $ P \equal{} 2(xy \plus{} yz \plus{} zx \minus{} 6)^2 \minus{} 15\ge 3$
Nice solution! :)
[hide="Hint"]Substituting $ x\equal{}a\plus{}1,y\equal{}b\plus{}1,z\equal{}c\plus{}1$, we have a nice solution, too[/hide]
And what about the max value?


Careful $m+n=33$ has only $\boxed{32}$ solutions in positive integers.

[quote="ch1n353ch3s54a1l"][quote="mdk"]How many ordered pairs of positive integers (m,n) for which mn>=0 and m^3+n^3+99mn=33^3.[/quote]

Hopefully I got this right...

[hide]$m^{3}+n^{3}+(-33)^{3}-3(33)mn=0$

$(m+n-33)(m^{2}+n^{2}+(33)^{2}-mn+33(m+n))=0$

First case: $m+n=33$, where there are $32$ solutions.

Second case: $(m^{2}+n^{2}+(33)^{2}-mn+33(m+n))=0$. Solving the quadratic in $m$, $\frac{n-33\pm \sqrt{-3n^{2}-198n-3267}}{2}$ and the discriminant is never positive, so no solutions in this case

In all, $\boxed{32}$ solutions.[/hide]

EDIT: answer changed...[/quote]

First of all, what is the nice factoring trick you use there? How did you just pop that out? Second, nice solution.

[quote="mna851"]$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(\frac{1}{2}(a-b)^{2}+\frac{1}{2}(a-c)^{2}+\frac{1}{2}(b-c)^{2})$.[/quote]

Ahh, yes the famous identity, how could I have forgotten? Lol.

[quote="t0rajir0u"][quote="mna851"]$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(\frac{1}{2}(a-b)^{2}+\frac{1}{2}(a-c)^{2}+\frac{1}{2}(b-c)^{2})$.[/quote]

Wow.  The last time I'd seen this problem I didn't even know about this identity.  Wish I had, haha.[/quote]

It was the first thing they talked about in Mathematical Olympiad Treasures I believe. Also they talked about it alot in the Russian Olympiad book (forgot name, it was like yellow cover). And it appears also in Chinese Olympiads and practice books too. I can't believe I forgot.


Let $ABC$ be a triangle, $E$ and $F$ the points where the incircle and $A$-excircle touch $AB$, and $D$ the point on $BC$ such that the triangles $ABD$ and $ACD$ have equal in-radii. The lines $DB$ and $DE$ intersect the circumcircle of triangle $ADF$ again in the points $X$ and $Y$.

Prove that $XY\parallel AB$ if and only if $AB=AC$.

One part of the problem is very easy.  :D

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32586]Note[/url] that $AD=\sqrt{p(p-a)}$, therefore:
\[AE\cdot{AF}=AD^{2}\Longleftrightarrow \angle{ADE}=\angle{AFD}. \]
It seems that we're done :lol: !

[quote="dragonfire"]How are we done? I got that it remains to prove $AB=AC\leftrightarrow AD=AY$. But how?[/quote]

If $XY\parallel AB$ then $\angle{XDF}=\angle{YDA}=\angle{AFD}$ so $D$ should be the tangency point of the ($A$) ex-circle with $BC$.
However, $AD\leq{m_{a}}$ thus $AB=AC$ (for (in any triangle) if $A^{\prime}$ is the point of tangency of the ex-circle with $BC$ then $M\in{[DA^{\prime}]}$, where $M$ is the midpoint of $BC$).

The converse is quite obvious.

[quote="Sailor"](for (in any triangle) if $A^{\prime}$ is the point of tangency of the ex-circle with $BC$ then $M\in{[DA^{\prime}]}$, where $M$ is the midpoint of $BC$).

[/quote]

How do you prove this?

[color=darkblue][b][u]Remark.[/u][/b] Generally, $pr^{2}+ar_{1}r_{2}=pr(r_{1}+r_{2})$ (see http://www.mathlinks.ro/Forum/viewtopic.php?p=637179#637279 ).
If $r_{1}=r_{2}=\rho$, then $\rho=\frac{r}{a}\cdot\left[p-\sqrt{p(p-a)}\right]$. From the obvious relation $\rho=\frac{pr}{p+AD}$ obtain :
$\frac{r}{a}\cdot\left[p-\sqrt{p(p-a)}\right]=\frac{pr}{p+AD}$ $\implies$ ${\frac{r}{a}\cdot\frac{ap}{p+\sqrt{p(p-a)}}=\frac{pr}{p+AD}}$ $\implies$ $p+\sqrt{p(p-a)}=p+AD$ $\implies$ $\boxed{AD=\sqrt{p(p-a)}}$.[/color]

can anyone post a complete solution please?


trigonometry quadratics algebra polynomial

Find all solutions of the equation

\[ \sin x + \tan x = \frac{1}{12} \]

Maybe not too hard  :D

But i tried alot and didn't get it. :mad:

First, we'll assume all solutions involve $\sin x, \tan x$ as neat fractions.  Then we can use some integer Pythagorean triple and write $\sin x = \frac{2mn}{m^2 + n^2}, \tan x = \frac{2mn}{m^2 - n^2}, \sin x + \tan x = \frac{2mn(2m^2)}{m^4 - n^4} = \frac{1}{12}$.

So, we wish to find integer solutions to

$m^4 - n^4 = 48m^3n$

And I'm stuck.   :)

Integer Schminteger

$\sin x + \tan x = \frac{1}{12}$

Multiply by $12\cos x$. ($\cos x \neq 0$ since tan exists)

$\Rightarrow 12\sin x \cos x + 12\sin x = \cos x$

$\Rightarrow \cos x = \frac{12\sin x}{1 - 12\sin x}$

$\Rightarrow \left(\frac{12\sin x}{1 - 12\sin x}\right)^2 = \cos^2 x = 1 - \sin^2 x$

$\sin x = M$

$\Rightarrow (12M)^2 = (1 - M^2)(1 - 12M)^2 = 1 - 24M + 144M^2 - M^2 + 24M^3 - 144M^4$

$\Rightarrow 144M^4 - 24M^3 +M^2 + 24M - 1 = 0$

I dunno solve that now and take the arcsin and check for extraneous ones.

[quote="t0rajir0u"]$\sin x = \frac{2mn}{m^2 + n^2}, \tan x = \frac{2mn}{m^2 - n^2}, \sin x + \tan x = \frac{2mn(2m^2)}{m^4 - n^4} = \frac{1}{12}$.

So, we wish to find integer solutions to

$m^4 - n^4 = 48m^3n$

And I'm stuck.   :)[/quote]

If $mn=0$ then our original equation isn't satisfied so we can assume it's not and divide by it to get:

$\frac{m^2}{n^2} - \frac{n^2}{m^2} = 48\frac{m}{n}$

Say $\frac{m}{n} = P \Rightarrow P^2 - \frac{1}{P^2} = 48P$

$\Rightarrow \sin x = \frac{2}{P + \frac{1}{P}}$

On the interval, $[0+2\pi k, \pi/2+2\pi k)$ both $\sin x$ and $\tan x$ are increasing, so on that interval $\sin x + \tan x$ is increasing.

Therefore there will be a solution in that interval (close to $0+2\pi k$) to the question.

On the interval, $(\pi/2+2 \pi k, \pi+2\pi k]$ if there was a solution in that interval then $\sin x + \tan x > 0$, but it isn't on that range.

Similarly, there will be no solution on the interval (3\pi/2+2\pi k, 2\pi + 2\pi k].

There will be another solution on the interval [\pi + 2\pi k, 3 \pi/2 + 2\pi k), although it will be farther from $\pi + 2\pi k$ than the first solution is from $2 \pi k$.

Note $k$ is an integer and $\sin x + \tan x$ is periodic about $2 \pi$.

In both cases the equation that describes the solutions is $\sin 2x + 2\sin x = \frac1{6}$

This leads to, $144M^4 - 24M^3 +M^2 + 24M - 1 = 0$ for $M = \sin x$

This factors to $(12M^2)^2 +(M^2-1)(1-24M) = 0$

NVM.  Tough problem.

[quote="FMako"]
In both cases the equation that describes the solutions is $\sin 2x + 2\sin x = \frac1{6}$

This leads to, $144M^4 - 24M^3 +M^2 + 24M - 1 = 0$ for $M = \sin x$

NVM.  Tough problem.[/quote]

Maybe [url=http://mathworld.wolfram.com/QuarticEquation.html]this[/url] help to finish.

I will try later  :)

There is no need for equations with degrees higher than two.
$\\\sin x+\cos x=\frac{1}{12}\\ \cos x\tan x+\cos x=\cos x(\tan x+1)=\frac{1}{12}$
This is because $\tan x=\frac{\sin x}{\cos x}$
Squaring both sides, we get
$\\\cos^2 x(\tan^2 x+2\tan x+1)=\frac{1}{144}\\ \cos^2 x(\sec^2+2\tan x)=2\cos x\sin x +1=\frac{1}{144}\\ \cos x\sin x=\frac{-143}{288}$
From the first equation, $\sin x+\cos x=\frac{1}{12}$, we have
$\cos x(\frac{1}{12}-\cos x)=\frac{-143}{288}\\ \frac{1}{12}\cos x-\cos^2 x+\frac{143}{288}=0\\ \cos^2 x-\frac{1}{12}\cos x-\frac{143}{288}=0$
Let $u=\cos x$.

Hence,
$u^2-\frac{1}{12}u-\frac{143}{288}=0\\ u=\frac{\frac{1}{12}\pm\sqrt{\frac{1}{144}+4\cdot\frac{143}{288}}}{2}\\ u=\frac{\frac{1}{12}\pm\sqrt{\frac{287}{144}}}{2}\\ x=\cos^{-1}\frac{\frac{1}{12}\pm\sqrt{\frac{287}{144}}}{2}$

By extraneous roots, only $x=\cos^{-1}\frac{\frac{1}{12}-\sqrt{\frac{287}{144}}}{2}$ works.

The goal of this was to substitute the first equation into another, and solve a quadratic equation.

Masoud Zargar

Unfortunately, the original equation was not $\\\sin x+\cos x=\frac{1}{12}$.

nooooo :(  I'll try and solve that one :P

I hope everyone realizes that "all solutions" means [u]all[/u] solutions.
The three small (quasi-??)circles are all solutions a+bi.

[quote="FMako"]
In both cases the equation that describes the solutions is $\sin 2x + 2\sin x = \frac1{6}$

This leads to, $144M^4 - 24M^3 +M^2 + 24M - 1 = 0$ for $M = \sin x$

NVM.  Tough problem.[/quote]

Strange ...

[quote="scorpius119"][hide="there are solutions, but not nice ones"]

Isolate $\sin x$ so that $\sin x=\frac 1{12}-\tan x$ and square both sides. $\sin^2 x=\frac 1{144}-\frac 16\tan x+\tan^2 x$.

Now if $\tan x=y$, then $\cot x=\frac 1y\Rightarrow \csc^2 x=1+\frac 1{y^2}\Rightarrow \sin^2 x=\frac{y^2}{y^2+1}$.

After multiplying by $144(y^2+1)$, this becomes $144y^2=(y^2+1)(1-24y+144y^2)\Rightarrow 144y^4-24y^3+y^2-24y+1$, so $\tan x$ is a root of this polynomial.

(after not being able to find rational roots, i plugged this into mathematica. The results... [b]YUCK!!!!!!!!!!!!!!!![/b][/hide][/quote]

I got same equation that [b]FMako[/b] got. But using [b]scorpius119[/b] approach i didn't found any mistakes. For sure one of them are wrong!  :mad:

So where are the mistake? :help:

[quote="kar"]
$t^4+48t-1=0$
Then solve this equation. I haven't found the solution yet. Anyone wants to try?[/quote]

The symbolic results are [u]not[/u] pretty (according to Mathcad).

[quote="Altheman"]one uses sin x and the other uses tan x as the variable that they substituted for...[/quote]

And ...  :huh:

[quote="kar"]$\sin x + \tan x = \frac{1}{12}$
$\tan x = \frac{2\tan {\frac{x}{2}}}{1-\tan^2 {\frac{x}{2}}}$ and $\sin x = \frac{2\tan {\frac{x}{2}}}{1+\tan^2 {\frac{x}{2}}}$
then $\frac{2\tan {\frac{x}{2}}}{1+\tan^2 {\frac{x}{2}}} + \frac{2\tan {\frac{x}{2}}}{1-\tan^2 {\frac{x}{2}}}=\frac{1}{12}$
Take $t=\tan {\frac{x}{2}}$ then it becomes $\frac{2t}{1+t^2}+\frac{2t}{1-t^2}=\frac{1}{12}$ which becomes
$t^4+48t-1=0$
Then solve this equation. I haven't found the solution yet. Anyone wants to try?[/quote]

It is getting more and more ugly... Now we have three different equations. Which is correct? :ewpu:

They are all correct. The reason that the three polynomials are different is simply because they are being used to find different roots.

In the first two, if $r$ is a root of the polynomial in $M$ that FMako used, then $\frac{1}{12}-r$ is a root in the polynomial in $y$ that I used. (This is similar to something like $y^2-2y+1=x^2+2x+1$ if $y=x+2$.)

Thanks scorpius.

Sorry about that  :blush:


function
trigonometry
geometry
geometric transformation

Find all the functions $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that $ f(x\minus{}\frac{\pi}{2})\equal{}\sin x\plus{}2f(\frac{\pi}{3})\ ,\ (\forall)x\in \mathbb{R}$ . How would the graph look ?

[hide]Plugging in $ x \plus{} \frac {\pi}{2}$, we have
\[ f(x) \equal{} \sin \left(x \plus{} \frac {\pi}{2}\right) \plus{} 2f\left(\frac {\pi}{3}\right) \equal{} \cos x \plus{} c.
\]
The functions are simply vertical translations of the function
\[ f(x) \equal{} \cos x.
\]
[/hide]

Not quite: the constant $ c$ isn't arbitrary here  :wink:

Sorry about that.

Considering
\[ f(x)\equal{}\cos x \plus{} 2f\left(\frac{\pi}{3}\right)\]
when $ x\equal{}\frac{\pi}{3}$, we have
\[ f\left(\frac{\pi}{3}\right)\equal{}\cos \frac{\pi}{3}\plus{}2 f\left(\frac{\pi}{3}\right).\]
Solving, we have
\[ 2f\left(\frac{\pi}{3}\right)\equal{}\minus{}1,\]
so our final function is
\[ f(x)\equal{}\cos x \minus{} 1.\]


I am having trouble finding how to map certain regions to other regions.

Is there a good place that collects how to map regions to other regions using the exponential map, log map or mobius functions?

for example I was able to derive that $(z-1)/(z+1)$ maps the line x=0 to the upper half of the unit circle. but now I am trying to map the upper half plane onto the unit disc centered at $-i$.

thanks for any help

I think this might work for mapping $\left\{ \Im z > 1 \right\}$ to $\left\{ |z+i| < 1 \right\}$:

$\displaystyle f(z) = \frac{ (1-i)z - 2 }{z}$

does that look right?

well, i see it as this:

invert on a circle of radius 2 with center on $-2i$: $f(z) = \frac{1}{z+i}$. that maps the upper half plane into the unit circle centered on $-3i$.
it's now only needed to add $2i$ in order to get the image within the unit circle centered on $-1i$


There are 10 tests, each test contains 4 problems. Within any two tests there are not more than 1 problem in common. At least how many problems are there?

Can someone help me with this problem??

Do you have the answer? Because i keep getting 15 as the solution.

You've already asked this question in another thread:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244256


I don't believe this was posted before, and I hope it isn't too hard.

Prove that the sum of two rational numbers is a rational number.

Say $x$ and $y$ are two rational numbers, i.e. $x=\frac{a}{b}$ and $y=\frac{c}{d}$.  Then $x+y=\frac{ad+bc}{bd}$, so their sum is also rational.  The only thing you'd need to do is prove (if you want to be [i]really[/i] rigorous) is that the product of two integers is an integer and the sum of two integers is an integer, but that is in essence the proof.

Well, if we have 2 numbers $\frac{p}{q}$ and $\frac{r}{s}\in\mathbb{Q}$, then their sum would be $\frac{p}{q}+\frac{r}{s}=\frac{ps+rq}{qs}$.  Since $p,q,r,s\in\mathbb{Z}$, the numerator and dinominator are also integers, thus an element of $\mathbb{Q}$.

EDIT: I know I was late $\leq 1\text{ minute}$. :(
Masoud Zargar

By the way, is the sum of two irrational numbers always irrational?

Yes, because if you have an irrational number with infinite digits and add it to another number, no matter what set it belongs to, it will still have infinite digits with no pattern.

EDIT: $(1-\sqrt{2})+\sqrt{2}=1\neq 0$.  This is not a case, since we are cancelling out teh irrational numbers.
Masoud Zargar

Hmm, I don't think so.  $1-\sqrt{2}$ is irrational and so is $\sqrt{2}$, but their sum is rational.  It's not even known if $e+\pi$ is rational or irrational.

[quote="kokosai"] The only thing you'd need to do is prove (if you want to be [i]really[/i] rigorous) is that the product of two integers is an integer and the sum of two integers is an integer, but that is in essence the proof.[/quote]

I got to where you were too. I was wondering how to be rigorous (not the kind of person I am).  :P

Two postulates for the set of integers is that the set is closed under addition and multiplication (I'm getting this from 2.1, [u]Elements of Abstract Algebra[/u] by Gilbert), so I guess that's about as rigorous as we can possibly be.


[hide]Let Amy's age be x. So Jamie's age would be x+5. So $2x+5=19$. Solving for this gets 7 which is Amy's age. So Jamie is 12.[/hide]

[quote="Interval"]Jamie is 5 years older than her sister Amy. If the sum of their ages is 19 , how old is Jamie?[/quote]
[hide]$j=a+5;\,j+a=19$

Adding the two equations, we get $2j+a=a+24$, which leads to $j=\boxed{12}$.[/hide]


algebra
difference of squares
special factorizations

If $p>3$ is a prime, prove that $p^2 \equiv 1 \text{ (mod 24)}$.

Just felt like posting something.  :)

If you have trouble:
[hide="Hint"]All primes greater than 3 can be written in the form $6k \pm 1$.[/hide]

This is what I did:

[hide]$p = 6k \pm 1$ for $p>3$, so $p^2 = 36k^2 \pm 12k + 1$

Thus

$p^2 = 12k(3k \pm 1) + 1$

If $k$ is even then $24|12k$ and if $k$ is odd then $24|12(3k \pm 1)$. So

$p^2 \equiv 1 \text{ (mod 24)}$

Same thing as yours but with a little less work.  :) [/hide]

The other more laborious solution. :D

[hide]

  We are asked to show $\frac{p^2-1}{24}$.  When factored as a difference of squares:

  $\frac{(p+1)(p-1)}{24}$

  Since $p$ is a prime larger than 3, it follows that $p$ is odd, and consequently, $p+1$ and $p-1$ are both even and produce a product divisible by 4.  We know that among any three consecutive integers exactly one is divisible by 3.  Since the problem states $p$ is both prime and larger than 3, it follows that either $p+1$ or $p-1$ is the multiple of 3, and consequently, produce a product divisible by 3.  Since $p$ is neither even nor a multiple of 4, it follows that p is congruent to either 1 or 3 mod 4.  In both cases either $p-1$ or $p+1$ would be congruent to 0 mod 4, or rather, divisible by 4.  Hence, it follows that $(p-1)(p+1)$ is divisible by: $4 \cdot 3 \cdot 2 = 24$ ( Note: the other even number is accounted for by the multiple of 4.) [/hide]

[hide="Solution"]$p^2-1=(p-1)(p+1)$
Since  p>3, and prime.
Hence p is odd.
p is of the form 2m+1
Simplifying we get (p-1)(p+1) as a multiple of 8.
Now p is not of the form 3m, since then it is not a prime.
Thus p is either of the form 3m+1 or 3m-1, in either way
(p-1)(p+1) is a multiple of 3. Thus, (p-1)(p+1) is a multiple of 24.
Hence proved.[/hide]

Not used much of the terminology. :oops:

[quote="Kurt Gödel"][hide="Ultimate  :D"]
We prove a stronger congruency: for every odd integer $n>0$ that isn't a multiple of $3$: $n^2 \equiv 1 \mod 24$.
(So, e.g.: $35^2 = 1225 \equiv 1 \mod 24$)

[i]Proof[/i]
$n \equiv -1, 1 \mod 3 \ \ \Rightarrow \ \ n^2 \equiv 1 \mod 3$
Remark that if $n$ is odd, $n\equiv 1,3,5,7 \mod 8 \ \ \Rightarrow \ \ n^2 \equiv 1^2, 3^2, 5^2, 7^2 \equiv 1, 9, 25, 49 \equiv 1 \mod 8$.
As $\gcd(3,8)=1$, these two congruencies lead to $n^2 \equiv 1 \mod 24$. $\square$
[/hide][/quote]

  Also note we can prove $n^2=1$ (mod 8) for any odd $n$, a more elegant way:

  $n=2k-1$ (mod 8)  For any integer $k$.

  $n^2=(2k-1)^2=4k^2 - 4k + 1=4(k^2-k) + 1=4(k)(k-1) + 1$ (mod 8)

  Since the product of any two consecutive integers is divisible by 2, $4(k)(k-1)$ must be divisible by 8 and reduces to 0, so we are left with: "1"

  Thus,  $n^2=1$ (mod 8)


Given a pentagon all of whose sides are equal.  Prove that the discs whose diameters are the sides of the pentagon cannot cover the pentagon entirely.

The pentagon needs to be convex, right? :?

I'll just sketch it. Let $ABCDE$ be our pentagon, and let $M,N,P,Q,R$ be the midpoints of $AC,BD,CE,DA,EB$ respectively.  We can show that $M$ lies outside the circle $(DE)$ (analog stuff holds for $N,P,Q,R$, and this is where we use the convexity; call this $(*)$). Since the sum of the angles of the pentagon is $3\cdot \pi$, it has at most one angle $\le\frac\pi 3$, so we may assume WLOG that $\angle CDE,\angle DEA>\frac\pi 3$. This shows that $M$ also lies outside $(CD),(EA)$, and, since $M$ lies on $(AB),(BC)$, it means that we can find a point close to $M$ which lies outside all the circles.

Sorry for the incomplete solution, but my proof for $(*)$ is really boring to write (so I won't do that). Just some basic estimates. I'm sure there's a nice, short proof out there.

yeah, convex; i guess i should have included that


induction
number theory
relatively prime

What is the third smallest triangular number that is also a perfect square?

Not counting $ 0$, correct?

[hide="Solution"]Triangular numbers are of the form $ n(n\plus{}1)/2$. Since $ n$ and $ n\plus{}1$ are relatively prime, for this to be a square number one of the two must be a square and the other must be double a square.
So let's consider doubled squares and whether them plus one or them minus one is also a square.
[list]
[*]$ 2$ is double a square, as is $ 1$. This gives the $ 1$st triangular number $ 1$ as being a square number. $ 3$ is not a square, though.
[*]$ 8$ is double a square, as is $ 9$. This gives the $ 8$th triangular number $ 36$ as being a square number. $ 8$ is not a square, though.
[*]$ 18$ is double a square, but neither $ 17$ nor $ 19$ are squares.
[*]$ 32$ is double a square, but neither $ 31$ nor $ 33$ are squares.
[*]$ 50$ is double a square, as is $ 49$. This gives the $ 49$th triangular number $ 1225$ as being a square number ($ 35^2$). This is the third lowest square number, so we stop here.[/list]
Answer: $ \boxed{1225}$[/hide]

[hide="Discussion"] We want to find positive integers $ m, n$ such that

$ \frac{n(n\plus{}1)}{2} \equal{} m^2 \Leftrightarrow$
$ 4n^2 \plus{} 4n \plus{} 1 \equal{} 8m^2 \plus{} 1 \Leftrightarrow$
$ (2n \plus{} 1)^2 \minus{} 2(2m)^2 \equal{} 1$

This is the Pell's equation $ x^2 \minus{} 2y^2 \equal{} 1$, with solutions

$ x_k \plus{} y_k \sqrt{2} \equal{} (3 \plus{} 2 \sqrt{2})^k$

It's provable by induction (or directly) that $ x_k$ is always odd and $ y_k$ always even.  Letting $ k \equal{} 3$,

$ x_3 \plus{} y_3 \sqrt{2} \equal{} 99 \plus{} 70 \sqrt{2}$

so $ 2m \equal{} 70 \implies m \equal{} 35$ and $ 2n\plus{}1 \equal{} 99 \implies n \equal{} 49$.  We therefore have

$ \boxed{ T_{49} \equal{} 35^2 \equal{} 1225 }$ [/hide]


LaTeX
number theory unsolved
number theory

Let X be a set with n elements, and 0  k  n. We denote by an;k (bn;k) the maximum
number of permutations of the set X such that every two of them has at least (at most) k
common components. (An element x 2 X is called a common component of permutations f
and g if f(x) = g(x).)
(a) Show that an;k  bn;k  n!.
(b) Find the exact value of ap;2 where p is a prime number.

[quote="bomb"]Anyone.....[/quote]

If you are too lazy too learn $\LaTeX$ nobody will read your postings!  ;)

hey,
sori for (a) it is meant to say a_(n,k) * b(n,k-1) <= n! for all n,k
and btw, u shudnt b posting these problems on mathlinks. u r meant to solve them urself.

[quote="vinoth_90_2004"] and btw vinh, u shudnt b posting these problems on mathlinks. u r meant to solve them urself.[/quote]

What are you talking about? Vin you also should learn some $\LaTeX$!  :D

After many many guesses, I conjecture this problem might be like this:

Let $X$ be a set with $n$ elements, and $0 \leq k \leq n$. We denote by $a_{n,k}$ the maximum number of permutations of the set $X$ such that every two of them has at least $k$ common components. And denote by $b_{n,k}$ the maximum number of permutations of the set $X$ such that every two of them has at most $k$ common components. (An element $x \in X$ is called a common component of permutations $f$ and $g$ if $f(x) = g(x)$.)
(a) Show that $a_{n,k} \cdot b_{n,k} \leq n!$ for all $n, k$.
(b) Find the exact value of $a_{p,2}$, where $p$ is a prime number.


Hi,

Can someone explain me what's the "Viete jumping's method" please (with exemple if it's possible) ?

Thanks.

Can someone explain the whole method?

Tanvir

[quote="Moonmathpi496"]for example u can have a look at IMO 1988 and 2007/5.[/quote]

Ok, but what's precisly the method ?

It's a technique wherein we write a number theoretic problem as an expression and treat it as a quadratic equation in suitable variables. We then find the minimum root of this quadratic i.e we assume one root is minimum and oh I can't really explain this well without an example - just check out those or Iran 1995 .
PS: @mod shouldn't this be in Algebra Theorems and Formulae?

you can refer to this thread
http://www.mathlinks.ro/viewtopic.php?t=165529 :wink:


[quote="FOURRIER"]Let ABCD a convex quadrilateral
M is the intersection of AC 'n BD
The Bisector of [img]http://www.gnux.be/latex/data/21e348d32acf9fd28e572ec835443fb0.png[/img]'meets' AB in K
And we have :  [img]http://www.gnux.be/latex/data/87237823645f1519233e0648780aa872.png[/img]

Prove : ..........[/quote]

... qed

  darij

[quote="mathmanman"]With those conditions, a correct conclusion is the following (probably posted before) :
$\angle BKC = \angle CDB.$[/quote]

Demonstration please


What will happen with Cuba now that Fidel Castro is not at the power??

For me, it will remain the same, because now is his brother Raúl.

However, I think that there will be slightly more freedom now. It's the beginning of the end of communism in Cuba. Please, post your opinion.

I'm not sure what will happen, but I don't think Cuba will be friendly to the United States for a while- there's just too much animosity built up. Much of Castro's power has come from his popularity, and his defiance of the USA is a large part of that. Before Castro's rule, the USA dominated the Cuban economy and propped up a corrupt dictator. In the fifty years since, it's been a policy of economic isolation, and the Soviet Union was the only place Castro could go for help (until they collapsed).

I'm not sure Castro was a communist until he had to be for protection. There wasn't a decent alternative at the time.

[quote="José"]It's the beginning of the end of communism in Cuba.[/quote]I thought that was (about) 1990. The Soviet supports disappeared, the Cuban economy collapsed, and survival trumped all in the recovery.

I do hope the USA takes this opportunity to restore at least some relations with Cuba.

I think that Cuba will stay on some similar goberment until something changes, like a falling of the populism in Venezuela and other latinoamerican countries -Or that Cuba start it- But I can't really tell anything.

By the way, this is a little bit out of topic but, Does anyone knows why Cuba has stopped from importing mexican shoes? This is really important to me and to my city, because my city produces 40% of the shoes of Mexico.

There are plans for Cuba to become the 51st state, but these plans ask that the southern tip of FL be part of Havami.

Some people think that Castro is dead, I don't think he is dead, atleast not yet.

But once he dies, there might be a revolution in Cuba.

Officially Castro's brother is just has interim power now.

[quote="Dave Vather"][quote="Xantos C. Guin"]There are plans for Cuba to become the 51st state, but these plans ask that the southern tip of FL be part of Havami.
[/quote]
My god! Let me see if i get it right... There are plans to steal a country sobereignity... What do you want next? Mexico? Central America? The moon?[/quote]

Yes! stop with the imperialism please, let us live!!!!!!!!!!

My guess is that Havami will not become the 51st state.

So far Sicily, Peuto Rico, New York City, and maybe a few others  have all tried and failed to become a state.

[quote="Dave Vather"][quote="Xantos C. Guin"]There are plans for Cuba to become the 51st state, but these plans ask that the southern tip of FL be part of Havami.
[/quote]
My god! Let me see if i get it right... There are plans to steal a country sobereignity... What do you want next? Mexico? Central America? The moon?[/quote]

Xantos' construction is a little bit odd ... "There are plans..." obscures the obvious question of whose plans they might be.  Frankly, I don't think he has any idea what he's talking about in this instance.

Also, the only serious pushes that I know of for New York City independence from New York State came well before there were 50 states; thus, NY was not in fact trying to become the 51st state.

[quote="Xantos C. Guin"]On CNN they covered the celebration in the street, and showed people reading a booklet that advertised a proposal for Havami to become the 51st state. Also, a few people that were interviewed said that they were in favor of this.

Some time in the last year or two, NYC had a cash surplus while NY  had a deficit. Some guy in NYC was trying to get NYC split from NY. This failed miserablely, but it got on the news, so I counted it as an attempt.[/quote]

When you make a statement like, "There are plans ..." there's an implication that those plans are being made by someone who is relevant to a decision-making process that could implement those plans, not by "one guy" or "a few people" (especially when those people are Miami Cubans talking about annexing Cuba).  The fact that some Miami Cubans think that it would be kind of cool if we annexed Cuba (or the fact that some CNN reporter thought that this might be an interesting question to ask) is not the same thing as there actually being plans for it to happen.

Well I think he is gonna get through this and live for another decade or something, his brother will probably do about the same.
But I don't know much of course  :o ?

On forums like IMDB I have already seen stuff like 'I hope the US invades now and supports some anti communism movement!  Cuba will crumble!"
People like them only understand the language of violence .....  :huh:  and I actually  [u]mean[/u] it, they are probably US'ers who don't speak any language as they write things like ' there country', 'dose not' and 'recieve'.

I once read that in general Castro was pro-USSR, and Che Guevara was more pro-China.  Does  Castro get any support from any of the 'official' four other communist countries : Laos, Vietnam, China and North Korea?

[quote="akalra1"]if castro dies, something outrageous will happen

i know it[/quote]

yes, outrageous, but surely good. Communism and amry government is always bad.

José, I am sorry but what country were you from again?  :oops:
Cuba or Argentina?

[quote="fredbel6"]For some reason I suspected Cuba too...
But now I remember topics about buying books in Argentina etc... :wink:

Aah, I hope one day the flags return.
Why don't you write your location in your profile?[/quote]

I will....

EDITED: Made.

Is Castro dead? Why, if not,  is all so misterious?


Let $ a;b;c$ be positive real numbers.Prove that:
$ \frac{1}{\sqrt{2b\minus{}a}}\plus{}\frac{1}{\sqrt{2c\minus{}b}}\plus{}\frac{1}{\sqrt{2a\minus{}c}}\geq\frac{\sqrt{3(a\plus{}b\plus{}c)^3}}{ab\plus{}bc\plus{}ca}$

[quote="chien than"]Let $ a;b;c$ be positive real numbers.Prove that:
$ \frac {1}{\sqrt {2b \minus{} a}} \plus{} \frac {1}{\sqrt {2c \minus{} b}} \plus{} \frac {1}{\sqrt {2a \minus{} c}}\geq\frac {\sqrt {3(a \plus{} b \plus{} c)^3}}{ab \plus{} bc \plus{} ca}$[/quote]
It's wrong! Try $ a\equal{}3$ and $ b\equal{}1.$ :wink:

can't be $ a\equal{}3$, $ b\equal{}1$, since $ 2b\minus{}a>0$

[quote="EUCLA"]can't be $ a \equal{} 3$, $ b \equal{} 1$, since $ 2b \minus{} a > 0$[/quote]
We need to prove that
\[ \frac {1}{\sqrt {2b \minus{} a}} \plus{} \frac {1}{\sqrt {2c \minus{} b}} \plus{} \frac {1}{\sqrt {2a \minus{} c}}\geq\frac {\sqrt {3(a \plus{} b \plus{} c)^3}}{ab \plus{} bc \plus{} ca}
\]
for reals $ a,$ $ b$ and $ c.$ ( chien than wrote it  :wink: )
Hence, we need to prove particularly that $ 2b \minus{} a > 0,$ which wrong. :wink:

Can you prove it if $ 2b\minus{}a>0;2c\minus{}b.0;2c\minus{}a>0$?

[quote="FOURRIER"][quote="arqady"]
[u]Hence, we need to prove particularly[/u] that $ 2b \minus{} a > 0,$ which wrong. :wink:[/quote]

I dont agree with you,
we have $ 2b \minus{} a > 0$[/quote]
No! We have not it.
[quote="chien than"]
Let $ a;b;c$ be positive real numbers.[/quote]
The rest we need to prove! :wink:
[quote="chien than"]Can you prove it if $ 2b \minus{} a > 0;2c \minus{} b.0;2c \minus{} a > 0$?[/quote]
Let $ 2b \minus{} a \equal{} x,$ $ 2c \minus{} b \equal{} y$ and $ 2a \minus{} c \equal{} z$ and use the following inequality:
$ \sum_{cyc}\frac {1}{\sqrt {x}}\geq\frac {3\sqrt [4]3}{\sqrt [4] {xy \plus{} xz \plus{} yz}}.$ :wink:


Prove that there is no set of infinitely many distinct powers of $ 2$, s.t. they have a common digit sum (base $ 10$).

It seems that this question is out of reach, however it is equivalent to one on a problems paper (and most surely there is a solution).


geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

Use the Mayer Vietoris sequence to compute the homology groups of a sphere with n handles.

$ H^0 \equal{} \mathbb{Z}$ trivial, $ H^1 \equal{} \mathbb{Z}^{2g}$, $ H^2 \equal{} \mathbb{Z}$, $ H^i \equal{} 0$, $ i > 2$, $ g$ = genus of the surface. Prove it by induction.

Alternatively: $ \pi_1(F_g) \equal{} \langle{x_1,\ldots,x_g,y_1,\ldots,y_g \mid \prod[x_i,y_i] \equal{} 1}\rangle$ by Seifert-van Kampen, then abelianize. You can write the surface as a quotient of a $ 2g$-gon with the above relation identified. From this, you can calculate $ H^2$.

You could also use cellular homology.


im stuck on level 13 and whats with the xof5

I don't know if there is a relation between 1of5, 2of5, etc with the answer, but notice that

<!-- he's been a bad boy --> (viewing source)

...And you should find where does he live.

Also, check the title

 ;)


I don't need the general term formula to deduce the limit.

Let's just start from the recurrent relation :
$F_n = F_{n-1} + F_{n-2}$.

We first consider another problem:
\[x_1 = 1, x_2=\frac{1}{2}, x_{n+1} = \frac{1}{1+x_n}\]
Find the limit : $\lim\limits_{n\to\infty} x_n$.

Suppose the limit does exist, say $A$. thus $A=1/(1+A)$ we get $A=\frac{-1+\sqrt5}{2}$. Now study $x_n$,

if $x_n < A$, then $x_{n+1} = 1/(1+x_n) > 1/(1+A) = A$
if $x_n > A$, then $x_{n+1} = 1/(1+x_n) < 1/(1+A) = A$

this tells us that $x_n$ jumps around $A$. and $x_1>A$, we know
\[x_1,x_3,x_5,\dots,x_{2n+1},\dots > A\]
\[x_2,x_4,x_6,\dots,x_2n,\dots < A\]

We are going to prove that $x_{2n}\nearrow A$ and $x_{2n+1}\searrow A$. For this, we consider $x_{n+2}-x_n$:

\[x_{n+2}-x_n = \frac{1}{1+x_{n+1}}-x_n = \frac{1}{1+\frac{1}{1+x_n}}-x_n = \frac{1-x_n-x_n^2}{2+x_n}\]

From $1-x-x^2 = (\frac{\sqrt5+1}{2}+x)(\frac{\sqrt5-1}{2}-x)$, we have

\[
x_{n+2}-x_n = \frac{1}{1+x_{n+1}}-x_n = \frac{1}{1+\frac{1}{1+x_n}}-x_n = \frac{(\frac{\sqrt5+1}{2}+x_n)(\frac{\sqrt5-1}{2}-x_n)}{2+x_n}
\left\{
\begin{array}{cc}
>0, & \fbox{if }x_n<A\\
<0, & \fbox{if }x_n>A
\end{array}
\right.
\]

Therefore $\{x_{2n}\}$ is increasing and $A$ is its upper bound, $\{x_{2n+1}\}$ is decreasing and $A$ is its lower bound. Thus they are convergent. Suppose $x_{2n}\to \alpha$ and $x_{2n+1}\to\beta$, take the limit in $\displaystyle x_{2n}=\frac{1}{1+x_{2n-1}}$ and$\displaystyle x_{2n+1}=\frac{1}{1+x_{2n}}$, we have

\begin{eqnarray*}
\alpha &=& \frac{1}{1+\beta}\\
\beta &=& \frac{1}{1+\alpha}
\end{eqnarray*}

Solving it, we get $\alpha = \beta = A$, hence $\lim x_n = A$.

Now we go back to Fibonacci.

From \[\frac{F_{n+2}}{F_{n+1}} = \frac{F_{n+1}+F_n}{F_{n+1}} = 1 + \frac{F_n}{F_{n+1}}\]
Let $y_n = F_{n+1}/F_n$, we get \[y_{n+1} = \frac{1}{1+y_n}\] and $y_1 = 1$. This is the problem we solved above. $\displaystyle y_n \to \frac{\sqrt5-1}{2}$ and hence, $\displaystyle \lim \frac{F_{n+1}}{F_n} = \lim \frac{1}{y_n} = \frac{\sqrt5+1}{2}$

liyi, while it's nice to have generalised proofs in the advanced section, this is intermediate section, and your proof is too complicated for here ;).

[quote="Pork_Chop8"]Given $F_{n} = \displaystyle\frac{(\phi)^{n}-(1-\phi)^{n}}{\phi-(1-\phi)}$, prove $\displaystyle\lim_{n\to\infty}\displaystyle\frac{F_{n+1}}{F_{n}}=\phi$.[/quote]Considering highest degrees:$\displaystyle\lim_{n\to\infty}\frac{F_{n+1}}{F_{n}}=\lim_{n\to\infty}\frac{(\phi)^{n+1}-(1-\phi)^{n+1}}{(\phi)^{n}-(1-\phi)^{n}}=\lim_{n\to\infty}\frac{2\phi^{n+1}}{2\phi^n}=\phi$

Thank you Peter. Also, that was a great proof liyi and I hadn't seen it before. Thank you.


The possible squares are $ 4 \times 4$, $ 5 \times 5$, $ 6 \times 6$, $ 7 \times 7$, $ 8 \times 8$, and a "special" $ 3 \times 3$.

For the squares with side length $ 4$ through $ 8$, you simply have $ 25\plus{}16\plus{}9\plus{}4\plus{}1$, or $ 55$.

The special $ 3 \times 3$ looks like this:

[asy]draw((0,0)--(3,0)--(3,3)--(0,3)--cycle);
fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,black);
fill((2,0)--(3,0)--(3,1)--(2,1)--cycle,black);
fill((0,2)--(1,2)--(1,3)--(0,3)--cycle,black);
fill((2,2)--(2,3)--(3,3)--(3,2)--cycle,black);
fill((1,1)--(2,1)--(2,2)--(1,2)--cycle,black);[/asy]

There are $ 4$ of these per row.  Multiply this by $ 8\minus{}3\plus{}1$ for the number of columns, to get $ 24$ squares.  Thus, the answer is $ 55\plus{}24$, or $ \boxed{79}$.

How?

This way clearly contradicts that.

there are 36 3x3 squares but half of them have 5 white squares. so it's 55+18=73. Check your work ernie

Check your grammar :wink:

But he is right. You forgot to include the case in which there are 5 white squares, and there are 6 of those. 79-6=73

There are 3 3x3 squares with 5 black squares per "row". its 73

[quote=DSL13]Wait but since it says "distinct", why wouldn't it be like this?

there is a 1 x 1, 2 x 2, 3 x 3, etc??? I am confused because I thought distinct meant different?[/quote]

Same as me. I think they should've re worded it.

[hide]
I used complementary counting on a casework problem  :o

Basically, first find all possible squares, so:
1x1:8^2
2x2:Notice that in this case, you get 7x7, because you increased the height and width by 1, and so decrease the height and width of the big square by 1
3x3:6x6 and so forth
64+49+36+25+16+9+4+1 = 8*9*(17)/6 = 12*17 = 204 total squares

Now count the ones with less than 5 squares
This is obvious, 1x1 and 2x2 are counted out immediately
3x3 has 3 in each row and the height is the same so 3x6
Ans: 204-64-49-18=73
[/hide]

[hide="simple sol if yk how to count squares in a chessboard"]
Since there's an "at least", complementary counting is probably the best.

The only cases with less than 5 square are the "white-centered 3x3", all 2x2s, and all 1x1s (squares).

Then there are $I64 + 49 + 36 + 25 + 16 + 9 + 4 + 1) - 1 - 4 - 18 = \boxed{73}$ squares we are looking for.

[i]Note: For those who don't know how to count the squares, if we have an nxn square, there are $8-n+1$ ways to shift it both horizontally and vertically, for a total of $(9-n)^2$ squares. The sum above is for all squares: 1x1 through 8x8.[/i]
[/hide]


For the nonzero numbers $a$, $b$, and $c$, define
\[J(a,b,c)=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\]'

Find $J(2,12,9)$.


i think we can start some sub-forums here, like the olympiad sections do.
What's your opinion? Any other good advice?

For now, no subforums. when the activity is going to be so intense that we cannot take anymore crowdness, we will create subforums and move all the topics where they would belong in the new hierarchy :)

well, as a moderator, i think we should have sub-forum in the future, because when i tried to gather resources, it was hard to categorize them. Just another opinion ;)


$x,y,z$ are positive real number such that $x+y+z=1$. Prove that $(xy)^{\frac{5}{4}}+(yz)^{\frac{5}{4}}+(zx)^{\frac{5}{4}}<\frac{1}{4}$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=2070

I can never forget this thread :) .

That's an easy solution, but we have discussed after that lot more difficult facts on this inequality. For example, try to amuse by finding the maximum of that ugly sum.

I suppose Harazi refers to the more general and interesting problem:

For $a,b,c$ positive reals which add up to one find the max of

$(ab)^r+(bc)^r+(ca)^r$ where $r$ is any real positive number

I remember Vasc solved this general problem but he didn't post his solution  :(


These two problems are sure to have you wrecking your brain

Find the value of cos 105 degrees
Find the value of tan 15 degrees

Not if you notice that $105=90+15$ and if you did your homework on half-angle formulas :)

Just note that $105^\circ=60^\circ+45^\circ$ and $15^\circ=60^\circ-45^\circ$.
You can then use the sum and difference formula :roll:

[quote="warmachine10"]These two problems are sure to have you wrecking your brain[/quote] about how much you need to review your trig formulas :)


vector
LaTeX
linear algebra
linear algebra unsolved

Supposed $ U$ is a subspace of $ {F}^{n.n}$ such that for all $ A$ in $ U$ we have $ A^T\equal{}A$. Calculate the dimension of $ U$. Thank you for the solution before.

I am somewhat new to Linear Algebra so I apologize if this is a basic question :)
I am assuming $ F^{n.n}$ is a vector space of all square matrices in $ \mathbb{R}^n$
Would the dimension be 1 since the subspace requires the $ zero$ $ vector$?
So $ U \equal{}\{0\}$ ?

Or would it be of $ dimension$ $ n$ and the subspace $ U$ will just be the $ zero$ $ vector$ for all $ n x n$ matrices?
Also what is the preferred "x" symbol for Latex ?

The symbol you're looking for is \times, as in $ n\times n.$

As a piece of abstract language: the space (or subspace) $ \{0\}$ is zero dimensional. The dimension of a vector space is the cardinality of a basis. Since $ \{0\}$ is the span of the empty set, a basis for it has zero elements.

That has nothing to do with this problem. If $ U$ is the largest possible such subspace - that is, the set of all possible symmetric matrices - then its dimension is $ \frac{n(n\plus{}1)}2.$ I'll leave to the reader the task of describing a basis for this set.


[b]question 1[/b]find the solutions to $a!+b!=c!$
[b]question 2[/b] difficulter $a!+b!+c!=d!$
are there any generalization here?

[size=75][color=red]b-flat: please use either bold font, or \text{stuff} for text to make it stand out.[/color][/size]

[quote="srulikbd"]$question 1$find the solutions to $a!+b!=c!$
$question 2$ difficulter $a!+b!+c!=d!$
are there any generalization here?[/quote]
[hide="ones I found"]$1!+1!=2!$
$2!+2!+2!=3!$
[/hide]

[quote="srulikbd"][b]question 1[/b]find the solutions to $a!+b!=c!$
[b]question 2[/b] difficulter $a!+b!+c!=d!$
are there any generalization here?

[/quote]

[hide="answers"]$\text{for question 1: }$
1!+1!=2!

$\text{for question 2: }$
2!+2!+2!=3!
[hide]
[/hide][/hide]

Please avoid latexing texts, its pointless and requires graphics loading (bad for dial-ups like me). If you want to make it stand out, just use bold or italics or underline or change font color or font size but no latexing text. THANK YOU.


Can someone figure out why this program won't work properly?

#Author:<mdk>
#Description: This program asks for a name to be input, and an age,
 then prints out the person's age in 25 years

#Ask the user for their name
def getName():
    name=raw_input("What is your name>")

    #Send that value to whoever called this function
    return name

def getAge():
    #Ask the user for their age
    age+raw_input("What is your age>")

    #Convert the age variable to numbers, an "integer", so we can do
 math
    #Instead of letters (a String) which is returned from raw_input
    ageInt=int(age)

    #Send that number to whoever called this function
    return ageInt

def main():
    name=getName()
    age=getAge()
    futureAge=age+25 #Add 25 to the age

    #Print out the information to the screen
    print "Your name is:", name, "in 25 years you will be", futureAge

    if (futureAge>=59):
        # If the future age is greater than or equal to 59, print the
 Clinton Message
        print "and older than Hillary Clinton"
    elif (futureAge>=46):
        # "else if" (we are not greater than or equal to 59, are we not
 greater than or eual to 46?)
        # If so print out the Barack message
        print "and older than Barack Obama"
    else:
        #Okay, we're neither >=59 or 46, just print out the message
 below then.
        print " Which is younger than Hillary Clinton and Barack Obama"

 # Run the "main" function above when this file is loaded

 main():

Well, it would help if you clarified what you mean by 'won't work properly'.

I can see one character that is obviously wrong, though. Just read through every line carefully and it will be obvious.

If that was just a typo and you meant something else, please provide some more information..

you have a typo:

age[b]+[/b]raw_input("What is your age>")

should probably be

age[b]=[/b]raw_input("What is your age>")

Not quite sure what language you're using, but this seems to be simple enough for logic to work.

Python? If so, maybe you should post stuff in [ code][/code] next time so we can see the indentation. Anyway it is probably the +/= swap.

Another note: the joy of Python is you don't need parenthesis around conditions unless you need a specific evaluation order. Don't do if (blah): do if blah:.

And another: why put stuff in main() ? Get rid of a level of indentation and a layer in the call stack and eliminate main.

[Sorry, I love Python and I love C and I like C++, but I don't like Python mixed with C/C++. ;)]


function
geometry
algebra proposed
algebra

OK, my wife is taking some college and she tested for Intermediate Algebra. Now, I pride myself on being very good in math all the way up and through at least beginning Calculus. Now my wife has been given a question and her professor and I have a disagreement on the proper way of doing the problem. As I was taught the order of operations it was Please Excuse My Dear Aunt Sally (PEMDAS). Now it is called the Order of Operations Agreement which, when I read it, seems to be pretty much the same thing. The only difference is that it specifically states that Multiplication and division are to be done as they occur from left to right and then it says the same for addition and subtraction. Here is the problem though: given the question 15/3[7+2(8-9)]-5 I come up with the solution of -4. The professor insists that the answer is 20 because you need to divide 15 by 3. Where I say that you must first eliminate your parenthesis by multiplying the 3 by 5 (the result of the parenthesis). I would appreciate some insight into this.

It it quite hard to establish what is truly the right answer because normal text isn't unambiguous when writing math. Still, it is "normal" practice that an entry of the form:

a/b(c+d)

to be treated as:

$ \frac {a}{b}(c \plus{} d)$

rather then

$ \frac {a}{b(c \plus{} d)}$.

It is the same thing with functions: does sin xn means:

$ n \sin(x)$ or $ \sin(x n)$.

So, the best way to be sure is to use parentheses. Also, I'd like to mention this is from my personal experience and it might not be true for all.

[quote="khar76"]15/3[7+2(8-9)]-5 I come up with the solution of -4. The professor insists that the answer is 20 because you need to divide 15 by 3. Where I say that you must first eliminate your parenthesis by multiplying the 3 by 5 (the result of the parenthesis).[/quote]
The professor is correct.  As you said, the expression inside the brackets evaluates to 5.  So the whole expression simplifies to 15/3[5] - 5.  The subexpression "3[5]" stands for "3 * [5]", so the whole expression becomes 15 / 3 * [5] - 5.  At this point, the brackets are no longer needed, so we get 15 / 3 * 5 - 5.  Evaluating from left to right, we get 5 * 5 - 5, which is 20.

The "P" in PEMDAS means to evaluate what is inside the parentheses first.  It doesn't mean evaluate things near the parentheses.

Ravi B, I thank you for your response however, I would like to ask if anyone else has another opinion before I consider this matter closed? It's not that I don't want to believe you, but rather that I have learned in all areas of Mathematics, there is always room for interpretation (what intrigues me most about the subject, I guess).


Find the largest positive integer $n$ such that the following equations have integer solutions in $x$, $y_1, y_2, ... , y_n$ :
$(x + 1)^2 + y_1^2 = (x + 2)^2 + y_2^2 = ... = (x + n)^2 + y_n^2$


[quote]The electric company provides continuous 240 V.A.C to your fuse panel. You have several lights on in the house. You plug in a 15 amp grinding wheel motor to a wall socket. What happens?

a) No lights dim.

b) All the lights in the house dim temporarily.

c) Some lights dim strongly, then come back up but are still dimmer than before.[/quote]You need Ohm's Law and a little extra for this interesting problem.

[hide](b) is the wrong answer.

[b]The correct answer is (c).[/b]

[b](1) The wiring and Connections have significant resistance.[/b]
In the case of ordinary house wiring, the resistance of the wire cannot be ignored. (The lines are 14 gauge copper 2-wire+gnd bundles, or 12 gauge copper 3-wire+gnd lines to carry dual fused circuits with standard 15 amp breakers or fuses at the breaker panel. The thickness of wire is chosen just large enough to safely carry 15 amperes of current.  Typical runs (lengths of wire from panel to outlet/light are 50-200 ft.  At junction boxes, the connections are typically either just twisted contacts -marrette connectors, or clamps built into outlets, act as high resistance points in the circuit, and must be housed in metal boxes for fire safety as they are point-of-failure risks.)

[b](2) The Voltage Drop is dependant upon the load to wire resistance ratio.[/b]
The voltage supplied at the fuse-panel is a constant 120 v.a.c. (rms).  The outlets and lights are in parallel, but the house wiring and junctions are in series with the load.  At 15 amperes of draw the effective resistance of the running motor load is about  R = V/I  = 120/15 = 8 ohms.  The resistance of the line in series might typically be 1 ohm.  The actual current is then I = V/R = 120/9 = 13.3 amperes.  The voltage drop across the load then V = IR = 13.3 x 8 = 106 volts.  This explains the lower running amperage of the motor and a [b]continuous dimming[/b] of the lights in parallel with it.

[b](3) The Surge Current causes a huge temporary voltage drop.[/b]
Typically in a motor or heavy heating appliance the 'cold' resistance is almost negligible, perhaps only 1 or 2 ohms, either caused by the simple resistance of the motor coils or the cold resistance of the material in the heating element. This results in a huge initial surge current:  I = V/R = 120 / 2 = 60 amperes!  However this is almost instantly strangled as the motor turning or the element heating up rapidly increases the resistance of the load to 'normal'.  For circuits or outlets that will routinely have such reactive loads plugged into them, special 'slow-blow' fuses or timed breakers are used, to allow for the surge current.  This presents no hazard as long as the current is very brief.  Sometimes a 'mini' circuit or component is added to the appliance to minimize this effect.

Any lights in parallel with the load but hooked up 'post' motor outlet physically (and in series with the line resistance) will experience the voltage drop, and [b]lights will exhibit both a large temporary dimming and a lasting weaker dimming[/b] during the duty cycle of the motor.  Lights on other lines experience relatively little effect, (mostly from the surge alone) as the resistance from hydro-pole to house is much lower than the house lines (but the secondary effects are dependant upon total draw from the panel and even local neighbourhood usage).

The slight constant dimming of some lights is often masked by the dramatic dimming from the surge current.

 [b](c) is the correct answer because it correctly explains the actual observed facts,[/b] which are noticable and significant.[/hide]

Umm... Don't you think your solution is contingent on assumptions that weren't clearly stated such as the 15A limit on household lines? How is one supposed to know that? Was I sleeping that day in my circuits class? Damn I should have taken the class "'Ordinary' Home Wiring 1101". I must have missed the signup. In any event, I had some doubts before, but now i know you're a complete idiot if you expected someone to produce that solution out of thin air and make all of those assumptions without them being specified in the problem.

It seems reasonable to expect most people over the age of 14 to know about the existance of their house-panel and fuses, and the idea that electricity from a wall socket comes from hidden wires running to the panel, and out to the street.  In any case, the fuses are not really germaine to the problem.

But the introduction of the simplification about wire having no resistance in basic circuits is precisely what leads to a possible difficulty for a student in grasping why a motor dims lights.

One of first things a student of basic electricity is confronted with in the real world is his own house wiring, and he naturally wants to apply his knowledge to these common circuits.
[hide]
The problem only required a knowledge of Ohm's law, series resistance and voltage dividers, along with a little wonder about commonly observed effects in the home and an apparent 'paradox' that should tweak anybody's curiosity who has an interest in electronics.

Any attempt to wrestle with the issues here should give partial marks for effort, and the highest marks to those who penetrated the (it turns out) rather simple causes behind the effects.  The knowledge of an electrician about wiring specs isn't needed.

As shown a student could have reasoned out the wire resistance without any specific knowledge other than the fact that RMS is chosen as a standard so that Ohm's Law can be used with A.C. voltages and currents, and that the house A.C. is a constant voltage source in which multiple loads are connected in parallel through long runs of wire.

Even this could have been deduced from printing on lightbulbs giving a voltage rating rather than a current rating, and the fact that single fuses control multiple lights and outlets.

Only a simple multimeter might be required to investigate the draw of the motor if it wasn't printed right on the housing.  The student could either simply measure the current of the motor or the voltage drop across it to begin his investigation and calculations.  Conversely, he could measure the voltage drop across the lightbulb and start reasoning from there.

A lazy student could just start making guesstimates without even getting out his chair, like I did in the explanation.

For instance, he could sketch himself this:
[img]http://lentils.imagineis.com/rouge/AC-Circuit1.jpg[/img]
Now he can either read off the amp draw of the motor or simply measure it, and then measure the voltage across it twice (on and off) to figure out the internal resistance of the A.C. power supply.  Possibly a grade 11 or 12 physics problem in the  Introduction to A.C. ciruits section: applying Ohm's Law to find the internal resistance of a power supply.
This kind of problem teaches first of all, a general method of approach to all basic problems in a.c. circuits where you need to find one or two unknowns, and prepares a student for Norton and Thevein theorems. [/hide]

[quote]I had some doubts before, but now i know you're a complete idiot if you expected someone to produce that solution out of thin air and make all of those assumptions without them being specified in the problem.[/quote]
Lets have a look at one possible step-by-step solution by a student:
[hide]
(1) Hmm.  I have experienced lights dimming before when somebody turns on some heavy appliance.  So the answer must be (b) or (c).

(2) But (c) seems complicated. I don't remember noticing that.   Should I just guess?

(3) No. (if yes give yourself a 50/50 probability of getting the mark.)

(4) The question is either offering some dumb trick answer, or suggesting there's more to it than I thought.  I'd know which if I knew what caused the dimming.

(5) I thought the A.C. outlet was a constant voltage:  But it can't be if the lightbulb dims.  And how can the motor change the voltage to the lamp?

(6) Something in my mind about voltage drops:  what was it?  resistors in series.  But aren't the motor and light in parallel?

(7) Maybe there is a resistor in series with *both* of them:  would that work?  yes. I can treat the light and motor as one resister by using the parallel resistors rule.

(8) Okay so what?  I'll never figure out all this crap.  I barely know Ohms law.

(9) Wait: I don't need to!  If there is a resistor in series with the motor and light, it must still be there all the time, so if the motor makes a voltage drop it must always be there when the motor is on!

(10) But what about when it first turns on?  The motor has to start up.  That must take extra juice or something.  That's it: If there's a resistor in the circuit, the answer must be (c).  Because I'm not doing anything to change the circuit afterward.

(11) I guess I just never noticed the lights actually stayed a bit dim.  I'm not sure it could be... I just can't remember.

(12) Wait, maybe the motor is a variable resistor.  No, still doesn't make sense.  There can't be a voltage drop unless the resistor is somewhere else.

(13) I have to go with (c)..  Hey, I think I almost understood why too.  And I didn't have to do any math at all.  I love multiple choice questions.

(14) ...still, are fuses resistors?  I guess they must be.  Hey this is a rip.  There's probably some resistor built into the fusebox stopping us from getting full power.  But they're probably charging us for it anyway!  Wait, yeah a transformer is a resistor too, right?  So some of my juice is going back to them, and they're ripping us off!  I got to ask my teacher about that!

(full marks for creative thinking, warts and all.)
[/hide]


Question:

"Find a recurrence relation and initial conditions for the sequence {a sub n} if a sub n is the number of bit strings of length n that contain three consecutive 0's."

So here's what I have so far...
n > 3
n = 4, 1000, 0001
n = 5, 10000, 00001, 00010, 01000, 10001
n = 6, 100000, 000001, 010000, 000010, 001000, 00100, 111000, 000111, 110001, 100011, 110000, 000011, 100001, 101000, 000101, 100010, 010001, ... I might have missed a few...

Anyways here's what I see... As n increases the number of bit strings increases (almost exponentially), and the maximum number of 1's in the string is n-3. In n-6, you have 3x1's.

That's all I have so far... And I don't know where else to go from here. So I would like some help with this.

Ask yourself how the strings can end / where the last 1 is.


If your average score on your first six mathematics tests was $ 84$ and your average score on your first seven mathematics tests was $ 85$, then your score on the seventh test was

\[ \textbf{(A)}\ 86 \qquad
\textbf{(B)}\ 88 \qquad
\textbf{(C)}\ 90 \qquad
\textbf{(D)}\ 91 \qquad
\textbf{(E)}\ 92
\]

[hide="Or..."]
This is equivalent to starting with $ 84$ for all $ 7$ numbers, then adding one to each, so we add a total of $ 7$.  However, this all goes to the last number, so we do $ 84\plus{}7\equal{}91$, or answer choice $ \boxed{\textbf{(D)}}$.
[/hide]

ummm why are you posting all these questions that you seem to already know the answer to?


If $ xy \equal{} 6$ and $ x^2 y \plus{} xy^2 \plus{} x \plus{} y \equal{} 63,$ find $ x^2\plus{}y^2$.

[hide]Factoring, we have:
\[ (xy\plus{}1)(x\plus{}y)\equal{}63\]Since $ xy\equal{}6$, $ xy\plus{}1\equal{}7$, and $ x\plus{}y\equal{}9$. Then:
\[ x^2\plus{}y^2\equal{}(x\plus{}y)^2\minus{}2xy\equal{}9^2\minus{}2 \cdot 6\equal{} 81\minus{}12\equal{}\boxed{69}\][/hide]

$ x^2 y \plus{} xy^2 \plus{} x \plus{} y \equal{} (xy \plus{} 1)(x \plus{} y) \equal{} 63$

$ xy \plus{} 1 \equal{} 7 \rightarrow x \plus{} y \equal{} 9$

$ (x \plus{} y)^2 \equal{} x^2 \plus{} 2xy \plus{} y^2 \equal{} x^2 \plus{} 12 \plus{} x^2 \equal{} 81 \rightarrow x^2 \plus{} y^2 \equal{} 69$


modular arithmetic
combinatorics proposed
combinatorics

$ n$ numbers in a row which are either $ 0$ or $ 1$ are given.we can choose $ s \in N$ consecutive numbers of this row with the property that their sum is even and then replace each $ 0$ by $ 1$ and each $ 1$ by $ 0$,
see an example:
the given sequence is
$ 1,0,0,0,1,1,1,1,0,1,0,1,0,1,1,0,1,0,1,0$
we can choose the third,fourth,...,and nineth number (because we have $ 0\plus{}0\plus{}1\plus{}1\plus{}1\plus{}1\plus{}0\equal{}4 \equiv 0 \pmod {2}$)
and then transform the sequence into :
$ 1,0,1,1,0,0,0,0,1,1,0,1,0,1,1,0,1,0,1,0$.
for which $ n$ can we obtain any permutations of $ 1$ and $ 0$  from any other permutations existing by doing finite steps as mentioned above?

From any sequence we can pick the 0s one by one to get the all-ones sequence.  From the all-ones sequence, if $ n > 1$, we either
* switch everything (if $ n$ is even), or
* switch all but the first entry to 0, then switch the second entry to 1, then switch the first and second entries together to 0 (if $ n$ is odd),
after which we have the all-zeroes sequence.  From here, we may change 0s to 1s one by one to get any desired sequence.

If $ n \equal{} 1$, we can get from 0 to 1 but not back.

(It seems like maybe there should be some way to make this a more interesting question ... forbid changing just a single entry, perhaps?)


algebra
polynomial
geometric sequence
algebra unsolved

$ (ab \plus{} ac \plus{} bc)^3 \minus{} abc(a \plus{} b \plus{} c)^3 \equal{} (a^2\minus{}bc)(b^2\minus{}ca)(c^2\minus{}ab)$

$ (ab \plus{} ac \plus{} bc)^3 \equal{} a^3b^3 \plus{} a^3c^3 \plus{} b^3c^3 \plus{} 3(ab \plus{} ac \plus{} bc)(abc)(a \plus{} b \plus{} c) \minus{} 3a^2b^2c^2$

$ (ab \plus{} ac \plus{} bc)^3 \minus{} 3(ab \plus{} ac \plus{} bc)(abc)(a \plus{} b \plus{} c) \plus{} 3a^2b^2c^2 \equal{} a^3b^3 \plus{} a^3c^3 \plus{} b^3c^3$

$ abc(a \plus{} b \plus{} c)^3 \minus{} 3(ab \plus{} ac \plus{} bc)(abc)(a \plus{} b \plus{} c) \plus{} 3a^2b^2c^2 \equal{} a^3b^3 \plus{} a^3c^3 \plus{} b^3c^3$

$ (a \plus{} b \plus{} c)^3 \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc) \minus{} 3abc$ then

$ abc\bigg\{a^3 \plus{} b^3 \plus{} c^3 \plus{} 3(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc) \minus{} 3abc\bigg\} \minus{} 3(ab \plus{} ac \plus{} bc)(abc)(a \plus{} b \plus{} c) \plus{} 3a^2b^2c^2 \equal{} a^3b^3 \plus{} a^3c^3 \plus{} b^3c^3$

$ abc(a^3 \plus{} b^3 \plus{} c^3) \plus{} abc3(a \plus{} b \plus{} c)(ab \plus{} ac \plus{} bc) \minus{} 3a^2b^2c^2 \minus{} 3(ab \plus{} ac \plus{} bc)(abc)(a \plus{} b \plus{} c) \plus{} 3a^2b^2c^2 \equal{} a^3b^3 \plus{} a^3c^3 \plus{} b^3c^3$

$ a^4bc \plus{} ab^4c \plus{} abc^4 \equal{} a^3b^3 \plus{} a^3c^3 \plus{} b^3c^3$

$ a^3(b^3 \plus{} c^3) \plus{} b^3c^3 \equal{} a^4bc \plus{} abc(b^3 \plus{} c^3)$

$ (b^3 \plus{} c^3)a(a^2 \minus{} bc) \equal{} bc(a^2 \minus{} bc)(a^2 \plus{} bc)$ :)

[b]a.[/b]  $ a^2 \equal{} bc \Rightarrow (ab \plus{} ac \plus{} a^2)^3 \equal{} aa^2(a \plus{} b \plus{} c)^3$

$ a^3(a \plus{} b \plus{} c)^3 \equal{} a^3(a \plus{} b \plus{} c)^3$ $ \Rightarrow$ $ b,a,c$ geometric series. $ a^2/c , a , c$ and  $ r \equal{} c/a$

[b]b.[/b] $ a^2\neq bc\Rightarrow (b^3 \plus{} c^3)a \equal{} bc(a^2 \plus{} bc)$

  $ ab^3 \plus{} ac^3 \equal{} a^2bc \plus{} b^2c^2$

  $ ab(b^2 \minus{} ac) \equal{} c^2(b^2 \minus{} ac)$

    [b]i.[/b] $ b^2 \equal{} ac \Rightarrow$ the same
    [b]ii.[/b] $ b^2\neq ac \Rightarrow ab \equal{} c^2$  :lol:

[quote="ertanrock"]$ a,b,c \in R \minus{} \{0\}$ ;If $ (ab \plus{} ac \plus{} bc)^3 \equal{} abc(a \plus{} b \plus{} c)^3$ then show $ \{a,b,c\}$ has a geometric sequence order.[/quote]
Assume the opposite.
We have to cases:
Case 1: $ a,b,c$ is all positive or all negative. Wlog $ a,b,c > 0$ and $ a \le b \le c$. If $ b^3 > abc$ then $ (ab\plus{}bc\plus{}ca)^3 \equal{} abc(a\plus{}b\plus{}c)^3 > (ab\plus{}b^2\plus{}bc)^3$
$ \iff ab\plus{}bc\plus{}ca > ab\plus{}b^2\plus{}bc \iff abc > b^3$. Contradiction. The case $ b^3 < abc$ is analogous.
Case 2: Exactly one of $ a,b,c$ has another sign than the rest. Wlog $ b > 0 > a,c$. $ abc > 0$ so $ \frac{ab\plus{}bc\plus{}ca}{a\plus{}b\plus{}c} > 0$. Assume $ b^3 > abc$, then $ \frac{ab\plus{}bc\plus{}ca}{a\plus{}b\plus{}c} \equal{} \sqrt[3]{abc} < b \iff ab\plus{}bc\plus{}ca > ab\plus{}b^2\plus{}bc$ $ \iff abc > b^3$. Contradiction. The case $ b^3 < abc$ goes the same way.

And we are done :)


algebra
polynomial
geometry solved
geometry

:?:    why we can't  make a quadrature of the cercle
en francais pourkoi on arrivera jamis a faire la quadrature de cercle ????
0.0000000000000001 THANKS :

Because $\sqrt{\pi}$ is transcedental  :P

When you construct some point in the complex plane as intersection of lines and/or circles, this point fulfills always an equation of at most second degree with coefficients the ones already constructed before.
The main problem is now to show that by this every point we construct fulfills some polynomial with the numbers we started with, so with rational ones.
Then we can exclude $\pi$ and al related numbers cause of transcedence.
Something related is http://www.mathlinks.ro/Forum/viewtopic.php?t=99227 .

Please use better titles from now :)


IMO Shortlist
combinatorics proposed
combinatorics

Two players play alternately on a 5 x 5 board. The first player always enters a 1 into an empty square and the second player always enters a 0 into an empty square. When the board is full, the sum of the numbers in each of the nine 3 x 3 squares is calculated and the first player's score is the largest such sum. What is the largest score the first player can make (if the second player plays so as to minimise this score)?

Edit: Sorry, I misread the problem statement.

I think it's not so easy... :huh:
Check again everything...
[hide="Hint"]Let $ B_1,B_2,.....B_8,B_9$ be a sum of the numbers in each squares 3 x 3, then let $ B$=max{$ B_1,B_2,.....B_8,B_9$}. So $ B$ - is score of first player. and problem is to find maximum $ B$ [/hide]

This is an IMO Shortlist problem. See this (still no solution): http://www.mathlinks.ro/viewtopic.php?p=352814#352814

Edit: official solution posted there


Hi. How many inequalities are known ? Can we find a list whith them ?

I'm intrested about the named inequalities only and not about exersizes...

For example  Cauchy - Swartz   --->     ( type )

e.c.

Thanks  :)

there a lot of known inequalities

you should read a book about them and i advice you to read this
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=88087[/url]

thanks all of you    :)

I'm newcomer in this section so i need your help...


group theory
superior algebra
superior algebra unsolved

Let $ G$ be anon-commutative group. Consider all the one-to-one mappings $ a\rightarrow a'$ of $ G$ onto itself such that $ (ab)'\equal{}b'a'$ (i.e. the anti-automorphisms of $ G$). Prove that this mappings together with the automorphisms of $ G$ constitute a group which contains the group of the automorphisms of $ G$ as direct factor.

this is also true for commutative groups, but then it's, of course, boring ;-).

so let $ G$ be non-commutative. then there is no antiautomorphism, which is also an automorphism. there is a canonical antiautomorphism, namely the inversion $ i : x \mapsto x^{ \minus{} 1}$ which satisfies $ i^2 \equal{} 1,i \neq 1$ and commutes with all automorphisms. furthermore, $ f \mapsto fi$ defines a bijection between the automorphisms and the antiautomorphisms, the inverse is also given by $ f \to fi$. from this we see that

$ Aut(G) \times \langle i \rangle \to Sym(G) , (f,g) \mapsto fg$

is an injective group homomorphism, whose image consists exactly of the automorphisms and the antiautomorphisms. in particular, these form a group isomorphic to $ Aut(G) \times \mathbb{Z}/2$.


when and how does AAPT announce the results for the preliminary round of USAPhO?

the results will be announced this week on aapt's website.  the exact day will depend on when shurtz and company finish grading the exam :)

i think your coaches will also get a letter in the mail if you make semis.

did anyone happen to get a letter informing your qualification as a semi-finalist?


Jenkas are a self-replicating species. In one generation a "healthy" jenka can produce 2 more healthy ones, but then the original becomes "sick". A sick one can produce one more healthy jenka before dying. A population that starts with one healthy jenka, free of mutations, will grow like this:
(g=generation)
g   h   s
0   1   0
1   2   1
2   5   2
3  12  5
4  29 12
5  70 29
6 169 70
etc.

Healthy jenkas have 5 body segments, and sick ones have 2. After 7 generations, a certain population of jenkas has a total of 10089 body segments. What is the smallest initial population possible?

Your whole question is far too vague.  First off, do jenkas HAVE to reproduce every generation?  That itself changes the question completely, but going by your chart, I assume they do.  Also, in the question, you don't specify what the starting jenkas have to be.  Does it have to start with x amount of healthy jenkas, or any kind of jenkas?  Well, I assumed that the colony started with a certain amount of only healthy ones and they had to reproduce each generation.  I solved it... but got an irrational answer.  Here's what I did
[hide]
I set up a recursive equation to find the amount of healthy jenkas in a colony at a specific generation.  The amount of sick ones is merely the amount of healthy ones two generations previous.
$H_x = 2H_{x-1} + H_{x-2}$
Of course, $x_0$ and $x_1$ would be given to find out the rest.  Once I could solve for how many healthy jenkas there were in the 7th generation, I could find out how many body segments there were, so I set up a chart the same way the poster did.  But since we don't know how many healthy ones we start with, the value is $y$.
g  h       s
0  y       0
1  2y     y
2  5y     2y
3  12y   5y
4  29y   12y
5  70y   29y
6  169y 70y
7  408y 169y
Since each healthy jenka has five parts and each sick has 2:
$S = 5h + 2s$
$10089 = 5(408y) + 2(169y)$
$10089 = 2378y$
$y = \frac{10089}{2378} = 4.2426$

Since I doubt there can be a fraction of a jenka reproducing, I probably did it wrong or understood it wrong... however, my work looks good to me.
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Hmm.. That looks right to me.  Hexadiditom, can you start out with some jenkas being healthy and some being sick, not just all healthy?  That might solve it.

Well, assuming that, I can expand my previous table with z being the starting number of sick jenkas.  Figuring out that the forums doesn't allow more than one space made it a lot easier to do because there was no point in making the spacing right...

g h s
0 y z
1 2y+z y
2 5y+2z 2y+z
3 12y+5z 5y+2z
4 29y+12z 12y+5z
5 70y+29z 29y+12z
6 169y+70z 70y+29z
7 408y+169z 169y+70z

Now, I already know I can't really do anything with it, but for the heck of it I'll at least get the equation into lowest terms.
$10089 = 5(408y+169z)+2(169y+70z)$
$10089 = 2378y + 985z$

I just made a program that goes thru all reasonable integer values of y and z.  It turns out (after finding out I forgot an extra "End" in the code and getting 0 answers...) that starting out with 3 sick and 3 healthy works.  So I assume that's the answer the poster wanted.  I guess we could have gotten the same answer by going thru reasonable integers the same way.  It would be relatively simple to rule out most of them - just make sure the units digit of the sum is 9, then fully multiply the ones with a units digit of 9.

:blush: Yes that was it. And since the last alien problem didn't work out so well, I made sure that this one had an answer. I wasn't sure there was an easy way to solve it though. Good job.