Given $ a,b,c \ge 0$
provethat
$ (a^3b^3\plus{}b^3c^3\plus{}c^3a^3)[(a\plus{}b)(b\plus{}c)(c\plus{}a)\minus{}8abc] \ge abc(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2$

[quote="babyloverain"]Given $ a,b,c \ge 0$
provethat
$ (a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3)[(a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 8abc] \ge abc(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$[/quote]
Let $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2,$ $ abc\equal{}w^3$ and $ u^2\equal{}tv^2.$
Hence, your inequality is equivalent to  $ (27v^6\minus{}27uv^2w^3\plus{}3w^6)(9uv^2\minus{}9w^3)\geq$
$ \geq27w^3(3u^2v^4\minus{}4v^6\plus{}6uv^2w^3\minus{}4u^3w^3\minus{}w^6)\Leftrightarrow$
$ \Leftrightarrow(4u^3\plus{}4uv^2)w^6\minus{}(12u^2v^4\plus{}5v^6)w^3\plus{}9uv^8\geq0.$
$ (12u^2v^4\plus{}5v^6)^2\minus{}4\cdot4(u^3\plus{}uv^2)\cdot9uv^8\equal{}(25v^2\minus{}24u^2)v^{10}.$
Hence, for $ t\geq\frac{25}{24}$ our inequality is proven.
But for $ 1\leq t\leq\frac{25}{24}$ we'll prove that $ w^3\leq\frac{12u^2v^4\plus{}5v^6\minus{}v^5\sqrt{25v^2\minus{}24u^2}}{8(u^3\plus{}uv^2)}.$
$ (a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2\geq0$ gives $ w^3\leq3uv^2\minus{}2u^3\plus{}2\sqrt{(u^2\minus{}v^2)^3}.$
Id est, it remains to prove that
$ 3uv^2\minus{}2u^3\plus{}2\sqrt{(u^2\minus{}v^2)^3}\leq\frac{12u^2v^4\plus{}5v^6\minus{}v^5\sqrt{25v^2\minus{}24u^2}}{8(u^3\plus{}uv^2)}.$
But $ 3uv^2\minus{}2u^3\plus{}2\sqrt{(u^2\minus{}v^2)^3}\leq\frac{12u^2v^4\plus{}5v^6\minus{}v^5\sqrt{25v^2\minus{}24u^2}}{8(u^3\plus{}uv^2)}\Leftrightarrow$
$ \Leftrightarrow8(t\plus{}1)\left(3t\minus{}2t^2\plus{}2\sqrt{t(t\minus{}1)^3}\right)\leq12t\plus{}5\minus{}\sqrt{25\minus{}24t}\Leftrightarrow$
$ \Leftrightarrow16t^3\minus{}8t^2\minus{}12t\plus{}5\minus{}\sqrt{25\minus{}24t}\geq16(t\plus{}1)\sqrt{t(t\minus{}1)^3}\Leftrightarrow$
$ \Leftrightarrow4t^2\plus{}2t\minus{}1\plus{}\frac{6}{1\plus{}\sqrt{25\minus{}24t}}\geq4(t\plus{}1)\sqrt{t^2\minus{}t}\Leftrightarrow$
$ \Leftrightarrow\left(4t^2\plus{}2t\minus{}1\plus{}\frac{6}{1\plus{}\sqrt{25\minus{}24t}}\right)^2\geq\left(4(t\plus{}1)\sqrt{t^2\minus{}t}\right)^2\Leftrightarrow$
$ \Leftrightarrow12t^2\plus{}12t\plus{}1\plus{}\frac{12(4t^2\plus{}2t\minus{}1)}{1\plus{}\sqrt{25\minus{}24t}}\plus{}\frac{36}{\left(1\plus{}\sqrt{25\minus{}24t}\right)^2}\geq0,$ which is obvious.

We can use cauchy for problem,try it  :blush:

[quote="fjwxcsl"]
\[ (a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3)[(a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 8abc] \ge k abc(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2,
\]

\[ k_{max} \equal{} 9.
\]
[/quote]
Yes, you are right! But I don't see an easy proof for your nice result. :(
Let  $ ab \equal{} z,$ $ ac \equal{} y$ and $ bc \equal{} x.$
Hence, $ (a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3)((a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 8abc)\ge k abc(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\Leftrightarrow$
$ \Leftrightarrow(x^3 \plus{} y^3 \plus{} z^3)\sum_{cyc}(x^2y \plus{} x^2z \minus{} 2xyz)\geq k(x \minus{} y)^2(x \minus{} z)^2(y \minus{} z)^2.$
If $ z \equal{} 0$ and $ x^2 \plus{} y^2 \equal{} pxy$ then, should be $ (x^3 \plus{} y^3)(x \plus{} y)\geq k(x \minus{} y)^2xy$ and
 $ (x^3 \plus{} y^3)(x \plus{} y)\geq k(x \minus{} y)^2xy\Leftrightarrow$
$ \Leftrightarrow(x^2 \plus{} y^2 \minus{} xy)(x^2 \plus{} y^2 \plus{} 2xy)\geq k(x^2 \plus{} y^2 \minus{} 2xy)xy\Leftrightarrow$
$ \Leftrightarrow(p \minus{} 1)(p \plus{} 2)\geq k(p \minus{} 2)\Leftrightarrow p^2 \minus{} (k \minus{} 1)p \plus{} 2k \minus{} 2\geq0,$ which gives $ k\leq9.$
Thus, it remains to prove that
$ (x^3 \plus{} y^3 \plus{} z^3)\sum_{cyc}(x^2y \plus{} x^2z \minus{} 2xyz)\geq9(x \minus{} y)^2(x \minus{} z)^2(y \minus{} z)^2.$
Let $ x \plus{} y \plus{} z \equal{} 3u,$ $ xy \plus{} xz \plus{} yz \equal{} 3v^3,$ $ xyz \equal{} w^3$ and $ u^2 \equal{} tv^2.$
Hence, $ (x^3 \plus{} y^3 \plus{} z^3)\sum_{cyc}(x^2y \plus{} x^2z \minus{} 2xyz)\geq9(x \minus{} y)^2(x \minus{} z)^2(y \minus{} z)^2\Leftrightarrow$
$ \Leftrightarrow(27u^3 \minus{} 27uv^2 \plus{} 3w^3)(9uv^2 \minus{} 9w^3)\geq$
$ \geq9\cdot27(3u^2v^4 \minus{} 4v^6 \plus{} 6uv^2w^3 \minus{} 4u^3w^3 \minus{} w^6)\Leftrightarrow$
$ \Leftrightarrow8w^6 \plus{} (27u^2 \minus{} 44v^2)uw^3 \plus{} 9v^2(u^2 \minus{} 2v^2)^2\geq0.$
Id est, if $ t\geq\frac {44}{27}$ hence, our inequality is proven.
If $ 1\leq t\leq\frac {44}{27}$ and $ (27u^2 \minus{} 44v^2)^2u^2 \minus{} 4\cdot8\cdot9v^2(u^2 \minus{} 2v^2)^2\leq0,$ which is equivalent to
$ \Leftrightarrow729t^3 \minus{} 2664t^2 \plus{} 3088t \minus{} 1152\leq0,$ we obtain $ 1.01648...\leq t\leq\frac {44}{27}.$
Hence, it remains to prove our inequality for $ 1\leq t\leq1.01648...,$ for which we'll prove that
$ w^3\leq\frac {(44v^2 \minus{} 27u^2)u \minus{} \sqrt {729u^6 \minus{} 2664u^4v^2 \plus{} 3088u^2v^4 \minus{} 1152v^6}}{16}.$
$ (x \minus{} y)^2(x \minus{} z)^2(y \minus{} z)^2\geq0$ gives $ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$
Hence, enough to prove that
$ 3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\leq\frac {(44v^2 \minus{} 27u^2)u \minus{} \sqrt {729u^6 \minus{} 2664u^4v^2 \plus{} 3088u^2v^4 \minus{} 1152v^6}}{16},$
which is equivalent to
$ 5u^3 \minus{} 4uv^2 \minus{} \sqrt {729u^6 \minus{} 2664u^4v^2 \plus{} 3088u^2v^4 \minus{} 1152v^6}\geq32\sqrt {(u^2 \minus{} v^2)^3}.$
Easy to check that $ 5u^3 \minus{} 4uv^2 \minus{} \sqrt {729u^6 \minus{} 2664u^4v^2 \plus{} 3088u^2v^4 \minus{} 1152v^6}\geq0$ is true for $ 1\leq t\leq1.01648....$
But $ 5u^3 \minus{} 4uv^2 \minus{} \sqrt {729u^6 \minus{} 2664u^4v^2 \plus{} 3088u^2v^4 \minus{} 1152v^6}\geq32\sqrt {(u^2 \minus{} v^2)^3}\Leftrightarrow$
$ \Leftrightarrow \minus{} 135t^3 \plus{} 184t^2 \plus{} 16t \minus{} 64\geq(5t \minus{} 4)\sqrt {t(729t^3 \minus{} 2664t^2 \plus{} 3088t \minus{} 1152)},$ where
$ \minus{} 135t^3 \plus{} 184t^2 \plus{} 16t \minus{} 64\geq0$ is still true for $ 1\leq t\leq1.01648....$
In the end we obtain
$ \minus{} 135t^3 \plus{} 184t^2 \plus{} 16t \minus{} 64\geq(5t \minus{} 4)\sqrt {t(729t^3 \minus{} 2664t^2 \plus{} 3088t \minus{} 1152)}\Leftrightarrow$
$ \Leftrightarrow45t^5 \minus{} 162t^4 \plus{} 213t^3 \minus{} 116t^2 \plus{} 16t \plus{} 4\geq0\Leftrightarrow$
$ \Leftrightarrow(t \minus{} 1)^2(45t^3 \minus{} 72t^2 \plus{} 24t \plus{} 4)\geq0,$ which is obviously true.

With,k=1,we can use cauchy.
$ [c(a\minus{}b)^2\plus{}b(a\minus{}c)^2\plus{}a(b\minus{}c)^2][\frac{1}{c^3}\plus{} \frac{1}{b^3}\plus{} \frac{1}{a^3} ] \ge$
$ ( \frac{a\minus{}b}{c}\plus{} \frac{b\minus{}c}{a}\plus{} \frac{c\minus{}a}{b} )^2 \equal{} \frac{(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}{a^2b^2c^2}$
and we get
$ (a^3b^3\plus{}b^3c^3\plus{}c^3a^3)[(a\plus{}b)(b\plus{}c)(c\plus{}a)\minus{}8abc] \ge abc(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2$

Comment: in arqady's proof, he multiplies by abc in order to do the substitution $ x\equal{}bc$, etc. which has the problem if $ a\equal{}0$ then he hasn't verified the inequality... But those cases are trivial. Anyway, the inequality is equivalent to the one:

$ f(x,y,z)\equal{}(x^3\plus{}y^3\plus{}z^3)((x\plus{}y)(y\plus{}z)(z\plus{}x)\minus{}8xyz)\minus{} (x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2\ge 0$ where $ x\equal{}bc\ge 0$, etc

Now $ (x\plus{}y)(y\plus{}z)(z\plus{}x)\minus{}8xyz\equal{}x(x\minus{}y)^2\plus{}y(y\minus{}z)^2\plus{}z(z\minus{}x)^2$.

So it is easy to see that $ f(x,y,z)\ge f(x\minus{}d,y\minus{}d,z\minus{}d)$ where $ d\ge 0$. So $ f(x,y,z)\ge f(x\minus{}y,x\minus{}z,0)\equal{}f(p,q,0)$ where WLOG $ z$ is the minimum of $ x,y,z$ and $ p,q\ge 0$.

But $ f(p,q,0)\equal{}p,q(p^2\plus{}q^2)^2\ge 0$.

[quote="Altheman"]
$ f(x,y,z) \equal{} (x^3 \plus{} y^3 \plus{} z^3)((x \plus{} y)(y \plus{} z)(z \plus{} x) \minus{} 8xyz) \minus{} (x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2\ge 0$

So it is easy to see that $ f(x,y,z) \geq f(x \minus{} d,y \minus{} d,z \minus{} d)$ where $ d\ge 0$. [/quote]
I have obtained $ f(x,y,z) \minus{} f(x \minus{} d,y \minus{} d,z \minus{} d) \equal{} \minus{} 6\sum_{cyc}(x^2 \minus{} xy)d^4 \plus{}$
$ \plus{} 3\sum_{cyc}(x \plus{} y \plus{} 2z)(x \minus{} y)^2d^3 \minus{} 3\sum_{cyc}(x^2 \plus{} y^2 \plus{} 2z^2 \plus{} xz \plus{} yz)(x \minus{} y)^2d^2 \plus{}$
$ \plus{} \sum_{cyc}(x^3 \plus{} y^3 \plus{} 4z^3 \plus{} 3x^2z \plus{} 3y^2z)(x \minus{} y)^2d.$ :wink:

[quote="fjwxcsl"]
\[ (a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3)[(a \plus{} b)(b \plus{} c)(c \plus{} a) \minus{} 8abc] \ge k abc(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2,
\]

\[ k_{max} \equal{} 9.
\]
[/quote]
Replacing $ a,b,c$ by $ \frac{1}{a},\frac{1}{b},\frac{1}{c}$ respectively, then our inequality becomes
$ (a^3\plus{}b^3\plus{}c^3)[(a\plus{}b)(b\plus{}c)(c\plus{}a)\minus{}8abc] \ge 9(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2.$
Now, assume that $ c\equal{}\min\{a,b,c\}$ then we have
$ a^3\plus{}b^3\plus{}c^3 \ge a^3\plus{}b^3,$
$ (a\plus{}b)(b\plus{}c)(c\plus{}a)\minus{}8abc\equal{}2c(a\minus{}b)^2\plus{}(a\plus{}b)(a\minus{}c)(b\minus{}c) \ge (a\plus{}b)(a\minus{}c)(b\minus{}c),$
and
$ (a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2 \le ab(a\minus{}b)^2(a\minus{}c)(b\minus{}c).$
Therefore, we can reduce our inequality to
$ (a^3\plus{}b^3)(a\plus{}b)(a\minus{}c)(b\minus{}c) \ge 9ab(a\minus{}b)^2(a\minus{}c)(b\minus{}c),$
or
$ (a^3\plus{}b^3)(a\plus{}b) \ge 9ab(a\minus{}b)^2,$
which is equivalent to
$ (a^2\minus{}4ab\plus{}b^2)^2 \ge 0,$ which is trivial. :)

I should specify that min $ (x,y,z)\ge d\ge 0$.

$ x^3\ge (x\minus{}d)^3$ so

$ (x^3\plus{}y^3\plus{}z^3)\ge (x\minus{}d)^3\plus{}(y\minus{}d)^3\plus{}(z\minus{}d)^3\ge 0$

also $ (x)(x\minus{}y)^2\ge (x\minus{}d)(x\minus{}y)^2\equal{}(x\minus{}d)((x\minus{}d)\minus{}(y\minus{}d))^2\ge 0$ so

$ \sum_{cyc} x(x\minus{}y)^2\ge \sum_{cyc}(x\minus{}d)((x\minus{}d)\minus{}(y\minus{}d))^2\ge 0$

Multiplying these two relations and adding $ (x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2$ gives us the result

Obviously, I did not communicate my solution well (or at all for that matter), I apologize. Thanks for checking my post carefully.

Very nice and easy your proof, Can! :lol:


In trigonometric form, is the complex number $z^{\frac{x}{y}}$ equal to:

$r^{\frac{x}{y}} cis \frac{x\theta}{y} + \frac{2\pi n}{y}$
(equivalent of raising to xth power, then taking yth root)

or

$r^{\frac{x}{y}} cis \frac{x\theta}{y} + \frac{2\pi xn}{y}$
(equivalent to taking yth power, then raising to xth power)

The key is to remember that $cis \theta=cis(\theta+2n\pi)$, where n is an integer. Thus if $z=rcis\theta$, then

$z^{\frac{x}{y}}=(rcis\theta)^{\frac{x}{y}}=(rcis(\theta+2n\pi))^{\frac{x}{y}}=r^{\frac{x}{y}}cis(\frac{(\theta+2n\pi)x}{y})$.

Shouldn't the first expression be equivalent to the second one?

i believe so, but only if x is an integer.


prove that doesnot exist the function f: R -> R satisfy f(f(x)) = x^2 - 2008

Prove that exist $ a,b$ different satisfy condition :
$ a\equal{}b^2\minus{}2008$
$ b\equal{}a^2\minus{}2008$
Then consider $ f(a)$ and $ f(b)$


inequalities
calculus
algebra
function
domain
inequalities unsolved

You think it is well known.Can you tell me where have you seen it,please?
  By the way do you have I proof for this?

Cezar, this is everything except new and important. I suggest you to browse the forum and you will see it in different forms and generalizations. Anyway, it is direct consequence of Holder inequality. And a little sugestion: please use more appropriate names for your topics.

Sorry, I thought that this one was interesting.  :oops:

[quote="keira_khtn"]The general inequal was posted a long time before:
[tex]\sum_{i=1}^n \frac{a_i^u}{b_i^v}\geq \frac{(\sum a_i)^u}{n^{u-v-1}(\sum b_i)^v}[/tex] with [tex]u,v[/tex] be positive s.t:[tex]u\geq v+1[/tex][/quote]
I think this is not the most general one. This is stronger. http://www.mathlinks.ro/Forum/viewtopic.php?p=111563#p111563
jaloweic's proof is amazing.  :lol:

cant we solve many inequalities in general with series expansions  :idea:

perhaps a silly idea  :roll:

but i was thinking f(x;y;z;..)>g(x;y;z;...)

turn them into series expansions like e.g. taylor and use some kind of calculus formula to decide if it is true or not ??

of course the taylor series must converge for all variables in the conjectured domain and integer conditions might make it "diophantic-like" but in general like algebraic inequalities with less than 5 variables just to say something...

just "dreaming" about that
guess someone is going to tell me nope it cant be done  :D
but if you do plz tell me why not , im not that smart  :oops:

greetings to all

[quote="siuhochung"]do you mind telling me the relationship between your post and the inequality of this thread???[/quote]

you have a point there  :blush:

but the inequation lead to that ( silly ? ) theory.
and i felt it was not worth starting a new topic for its since its pretty vague ...  :|


Anyone got any good ti-89 games to share. For a start, 68k mario on ticalc.org is extremely good (basically and abridged version of mario for the N.E.S.)

Block Man?  :lol:
Sudoku might be fun on 89.


analytic geometry
graphing lines
slope
trigonometry
conics
parabola
number theory

Source: 2005 AMC-12 B Problem 24

[b]I am not asking for a solution, only an explanation.[/b]

In an AMC-12 solution packet, one solution uses trig to define the slopes of a triangle:

The problem states "All three vertices of an equilateral triangle are on the parabola $y=x^{2}$, and one of its sides has a slope of $2$.  The $x$-coordinates of the three vertices have a sum of $\displaystyle\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers.  What is the value of $m+n$?"

Part of the solution states that "The slope of one side is $2=\tan\theta$, for some angle $\theta$, and the two remaining sides have slopes $\tan(\theta\pm\frac{\pi}{3})$."

Their statement makes sense, and it simplifies the problem greatly, but I have never seen anything relating the slopes of the sides of a triangle with trig before, and I had a difficult time proving it to myself.

Is there a simple relationship here?  Does it apply to equilateral triangles only, or any triangle?

Well if you draw a line, you'll see the slope is $\frac y x$ if you draw a right triangle to the x-axis, and letting y be the height and x the base of the right triangle.  Then, if $\theta$ is the angle the line makes with the x-axis, clearly $\tan \theta = \frac y x \implies \tan \theta = m$

I understand that part, but how do they get $(\theta\pm\frac{\pi}{3})$?

because you have the original slope defined by $\tan{\theta}$, then the next one is $60$ degrees more, so it's defined by $\tan(\theta+\frac{\pi}{3})$, and the next one is $120$ degrees more, or $60$ degree less (you can show this easily by drawing a parallel line line to that third side through the vertex opposite that side), so it's slope is defined by $\tan(\theta-\frac{\pi}{3})$.


calculus
integration
algebra unsolved
algebra

[b]1.[/b] Does there exists positive integral infinite sequences $\{a_n\}$ such that $a_{n+1}^2\geq 2a_{n+2}a_{n}$ for any positive integer $n$?

[b]2.[/b] Does there exists positive irrational infinite sequences $\{a_n\}$ such that $a_{n+1}^2\geq 2a_{n+2}a_{n}$ for any positive integer $n$?

We get $\frac{a_{n+1}}{a_n}\le\frac{a_2}{a_1}\cdot\frac 1{2^{n-1}}$. We may assume WLOG $a_2\le a_1$, because otherwise, we start from $n>1$ s.t. $a_{n+1}\le a_n$. Anyway, the sequence becomes strictly decreasing eventually, so there's no such sequence made up of positive integers.

Of course there has to be one made up of irrationals. Just take a transcendental number $\alpha$, put $a_1=\alpha,a_2=\alpha^2$, and then we can even construct the sequence inductively s.t. we have equality (not merely inequality). We get $a_n=r_n\cdot \alpha^n$, where $r_n$ are positive rationals.


Yes, T.O.'s a bad, baaaaaaad man, but the Patriots will pull this one out.  They'll be ahead at the half by a touchdown, and will be up by 17 by the end of the game.

Its been widely discussed that the advertising during the game will be "toned down", but there will certainly not be a lack of humor in the commercials.  Watch for the Lays ad that features the unceremonious return of MC Hammer...

Who do you think's going to win the big game??

edit:  i knew i thought i saw a previous thread about this...just now found it.  oops.  feel free to post here, though...can't have too many discussions about the super bowl!

Dynasty shall be continued

Go Patroits :lol:

Go Patriots! I agree. They are a DYNASTY!

I'm not very good at debating so I practice my debating skills against people who think the Eagles are gonna win.... and I actually win these debates! That's how good the Patriots are! If I can convince others that they are good (and I can't convince anyone of anything), then there's no way they can lose!


there are 16 games you have to get them all correct you choose home or visitor win in each box how many different combinations are there to ensure you get perfect

hi Bobby,

Please do not post a topic twice. All math belongs in the math forums.

This one goes to Getting Started unless I misinterpreted the problem and it's actually much more difficult than I imagine.

I'm not quite sure what your asking.....

Assuming that each game is unique:

[hide="If your looking for number of combos"]$2^16$=66536 possibilities[/hide]
[hide="If your looking for probability that you will pick the correct combo"]Let $x$ be the number of guesses you have, your probabiliby will be $x/(2^16)$ or $x/65536$[/hide]
[hide="If you want to know how many picks you need to ensure you will get the correct combo"]65536....[/hide]


function
calculus
derivative
algebra
polynomial
limit
calculus computations

how many critical points does the function provided have?

f(x)=(x+2)^2 * (x-3)^4

What is its derivative? Find and factor that.

i got 2(x-3)^3 * (x+2) * (3x + 1)

do i then set that equal to 0 and see my solutions...basically do i, in essence, to find the critical points take the deriv. and set equal to 0?

Yes. This function has critical points at $-2,-\frac13,$ and $3.$

Our next task is to classify them. There are three basics families of information we can use to classify critical points.

1. Information about the function values, or
2. Inforamtion about the sign of the first derivitative in between the critical points, or
3. Information about the sign of the second derivative at the critical points.

All three families of information have their uses. I'll show you #1 in this case.

Note that $f(x)$ is a sixth degree polynomial with a positive coefficient of $x^{6}.$ Here's what we know:

$\lim_{x\to\infty}f(x)=\infty$ and $\lim_{x\to-\infty}f(x)=\infty.$ (That is, both ends of the graph turn up.)
$f(-2)=0$ and $f(3)=0.$
But with the function being a second power times a fourth power, we have that $f(x)\ge0$ for all $x.$

That last item tells us what is happening at $-2$ and at $0:$ each point is local minimum - and these two points are tied for being the global mininum. Since the function is positive in between those two points, there must be a local maximum in that interval - and that tells us that the point at $-\frac13$ is that local maximum. But it can't be the global maximum because of the behavior of this polynomial for large $x.$ There is no global maximum.


geometry
perimeter
function
calculus
quadratics
area of a triangle
Heron\u0027s formula

An isosceles triangle $ABC$ has a perimeter of $90 cm$. Find the side lenghts that will maximize the area of the triangle. (I know the answer... but the function is coming out ridiculous for the expected difficulty of the problem.)

Intuitively it is clearly the equilateral triangle, which is what you're thinking of. Without getting into sin/cosine, the equation to maximize is: $y\sqrt{x+y}\sqrt{|x-y|}$ given $2x+y=90$. Not so bad, but it is definitely not high school basics forum! (hint... move to preolympiad...)

You can use Heron's formula and $\frac{a+b+c}{3}\geq \sqrt[3]{abc}$ for $a>0,\ b>0,\ c>0.$ This problem is beyond this forum, I think.

I don't see how this is too hard for HSB at all.

[hide]The equilateral triangle is always the maximum area of a triangle with a certain perimeter. So you plug in 30 into $\frac{s^{2}\sqrt{3}}{4}$ to get $\boxed{225\sqrt{3}}$.[/hide]

Ok guys... I'll be honest, this was a question a friend of mine had while I was tutoring him before his calculus test... I just figured that if you came up with a function for area expressed through one variable it would probably be quadratic and so the calculus part would just be a fancy way to find the vertex... It seems that the easiest way to do it however would be through Heron's and that function certainly isn't quadratic and so then for it to be easy it would require some calculus techniques.  :blush: Could an admin move this wherever they see fit or delete it please?

One can prove strictly geometrically that the equilateral triangle is optimal (with the assumption that an optimal triangle exists).  Note that the question is equivalent to showing that among triangles with a given area, the equilateral triangle has the least perimeter.  (If you can shrink the perimeter with the same area, then there is a triangle similar to the new own with the original perimeter but larger area.)  Rather than showing directly that the equilateral is optimal, we show that every other triangle is sub-optimal (which is why we need the assumption that an optimal triangle exists).  For any non-equilateral triangle, choose two unequal sides, and draw a line through their shared vertex parallel to the third side.  As we move the vertex along this line, the area of the triangle is unchanged.  The perimeter is minimized when we choose the point which lies on the perpendicular bisector of the third side.  (This is a particularly simple instance of the "man needs to walk to the river and then back to his cottage" problem.)  This leaves us with those two sides equal, so we have a different triangle with smaller perimeter and the same area, done.

Also, if calculus student doesn't know Heron's, easier to do by dropping an altitude then by teaching Heron's.  (It's isosceles, so that makes everything simpler.)

Yes, I showed him originally by dropping the altitude (he only new A= bh/2 so it's the formula I used) but the numbers get really dumb when you express A through the height or the base. Heron's makes it very much cleaner, but teaching it would be a pain.


calculus
integration
logarithms
trigonometry
calculus computations

Can anyone help me to show me how to calculate this integral:
$\int_{0}^{\frac{\pi}{2}}{\ln(\sin x)}dx$

[quote="sos440"]Or you can use the famous identity

$\prod_{k=1}^{n-1}\sin \left( \frac{k \pi}{n}\right) = \frac{n}{2^{n-1}}$

to calculate the Euler's integral. Simple manipulation gives

$\sum_{k=1}^{n-1}\ln \sin \left( \frac{k \pi}{n}\right) \frac{\pi}{n}= \frac{\pi}{n}(\ln n-(n-1) \ln 2)$

The left side converges to $\int_{0}^{\pi}\ln(\sin x) \, dx$ and the right side converges to $-\pi \ln 2$.[/quote]
Thank you very much!I thought about this but fogot how to calculate it.


analytic geometry
real analysis
integration
LaTeX
geometry
3D geometry
sphere

Let $ S_{k\minus{}1}$ be the unit circle in $ \mathbb{R}^{k}$, i.e the set of all $ u\in\mathbb{R}^{k}$ whose distance from the origin is 1 . It is known that $ \forall x\in\mathbb{R}^{k}$ then $ x\equal{}ru$ for some $ u\in S_{k\minus{}1}$ and  $ r\geq 0$. Thus $ \mathbb{R}^{k}\minus{}\{0\}$ may  be regarded as the Cartesian product of $ (0,\infty) x  S_{k\minus{}1}$
Let $ m_{k}$ be the Lebesgue measure on $ \mathbb{R}^{k}$ and define a measure $ \sigma_{k\minus{}1}$ on $ S_{k\minus{}1}$ as follow :  If $ A\subset S_{k\minus{}1}$ and $ A$ is a Borel set, let $ \bar{A}$ be the set of all the points $ ru$ where $ 0<r<1$ and $ u\in S_{k\minus{}1}$, and define \[ \sigma_{k\minus{}1}\equal{}k.m_{k}(A)\]

then ( the book says) we have  the formular \[ \int_{R^{k}}fdm_{k}\equal{}\int_{0}^{\infty} r^{k\minus{}1}dr\int_{S_{k\minus{}1}}f(ru)d\sigma_{k\minus{}1}(u)\]
I read and read the definition of $ \sigma_{k\minus{}1}$ but I still do not get it.
Can someone give me some (geometric) explanation for $ \sigma_{k\minus{}1}$ , what is the motivation to define such measure on $ S_{k\minus{}1}$ ? and  how to get the above formula? It would be great if you can give me some pictures about it.

Many thanks

A $ \text{\LaTeX}$ remark first: $ X\times Y$ and $ a \cdot b$, look much nicer than $ XxY$ and $ a.b$ for products.

The dea behind this definition is that if $ U\subset S_{k-1}$ is a small (in the sense of diameter) set on the sphere, then it is almost flat and the set $ \{rx: x\in U,0\le x\le 1\}$ is something like a $ k$-dimensional pyramid with the apex at the origin, base $ U$ and height $ 1$, whose volume (according to elementary geometry) must be $ \frac 1k$ times $ 1$ times the area of $ U$. But we know the volume ($ k$-dimensional Lebesgue measure), so to have everything consistent, we are forced to define the surface measure on the unit sphere by the cited formula at least for small sets. But once it is defined so for small sets, it has to be defines so for all sets because of additivity.

As to the formula, it is easy for characteristic functions of "sectorial domains" $ \{rx: x\in U,r\in[a,b)\}$ (it just immediately follows from the definition then). From there you can pass to open sets, then measurable sets, then simple functions, then positive measurable functions, then integrable functions (well, you know the routine...) :).

I feel too lazy today to draw Asymptote pictures, but, I hope, my explanation is clear enough even without them :).

Thanks fedja
But I am still confused  :blush:

[quote="fedja"]

The dea behind this definition is that if $ U\subset S_{k \minus{} 1}$ is a small (in the sense of diameter) set on the sphere, then it is almost flat and the set $ \{rx: x\in U,0\le x\le 1\}$ is something like a $ k$-dimensional pyramid with the apex at the origin, base $ U$ and height $ 1$, whose volume (according to elementary geometry) must be $ \frac 1k$ times $ 1$ times the area of $ U$. [/quote]

why it is $ \frac{1}{k}\times A(U)$ ? ( where $ A(U)$ is the area of $ U$) could you please explain more?

[quote="fedja"]
But we know the volume ($ k$-dimensional Lebesgue measure), so to have everything consistent, we are forced to define the surface measure on the unit sphere by the cited formula at least for small sets. B
.[/quote]

What do you mean "cited formula at least for small sets"?

thanks for your time

Ok, let's draw the picture (in dimension $ k\equal{}2$):
[asy]size(200);
pair O=D("O",(0,20),N), A=D("A'",7,(-7,0)), B=D("B'",7,(7,0));
path P=D(CR(O,20,-200,20));
D("A",7,IP(D(A--O),P),N);
D("B",7,IP(D(B--O),P),N);
D(L(A,B),magenta);[/asy]
We want to assign something that we would like to call "circular length" to a small arc $ U\equal{}AB$. Since the corresponding part of the circle is almost flat, it should be close to the length ($ 1$-dimensional measure) of the interval $ A'B'$ and also additive. The length of the interval $ A'B'$ is twice the area of the triangle $ OA'B'$ (because the height is $ 1$) which is pretty much the same as the area ($ 2$-dimensional measure) of the circular sector $ OAB$ (which, in this case, is exactly that set $ \{rx: x\in U,0< r\le 1\}$). This explains the heuristic behind the formula for small arcs. Now just note that the formula gives you a $ \sigma$-additive quantity, so it can be used to define the circular length for arbitrary sets.

Is it clear so far? If yes, I'll continue; if not, I'd prefer to see what is confusing at this level first :).

Thanks for your time fedja
and thanks for your picture, it worths  thousand words

I know that the "Volume" of the triangle is $ \frac{1}{\bold{2}}\times \mbox{volume of the "base"}$
                 the "Volume" of the "[b]Pyramid[/b]" is $ \frac{1}{\bold{3}}\times \mbox{volume of the "base"}$
Is this formula true  in $ n\minus{}$dimensional spaces?  if possible could you please explain why? ( intuitively I understand it but how to prove it?)

Could you please keep going?

Let's talk about a generalized cone or pyramid in $ \mathbb{R}^n.$

We have a region $ U$ in $ \mathbb{R}^{n\minus{}1},$ such that $ m_{n\minus{}1}(U)\equal{}B.$ Let the base of our cone be the set $ U\times\{0\}.$ Let the apex of the cone be a point $ A$ whose coordinates we give as $ (x,h),$ where $ x\in \mathbb{R}^{n\minus{}1}$ and $ h\ne 0.$ WLOG, assume $ h>0.$

The cone $ C$ is then the convex hull of $ A$ and $ U\times\{0\}$ - that is, the union of all the line segments from an arbitrary point on the base to the apex.

Now take the section through this cone at height $ y,$ for some $ y\in[0,h].$ That is, slice through the cone along a "plane" (OK, hyperplane) parallel to the base. The cross section is similar to $ U,$ with a coefficient of similarity equal to $ \frac{h\minus{}y}{h}.$ That is, the cross section has the same shape as $ U$ but with all linear dimensions multiplied by $ \frac{h\minus{}y}{h}$ (and possibly also translated, but the translation doesn't matter.)

We do need to know how $ (n\minus{}1)$-dimensional measure behaves with respect to similarity. We need to know that the $ (n\minus{}1)$-dimensional measure of that cross section is $ \left(\frac{h\minus{}y}{h}\right)^{n\minus{}1}B.$

We then compute the $ n$-dimensional measure of the cone by integrating this in $ y.$ Call it Cavalieri's principle; call it Fubini's theorem. That $ n$-measure is:

$ V\equal{}\int_0^h\left(\frac{h\minus{}y}{h}\right)^{n\minus{}1}B\,dy\equal{}\left.\frac{B}{h^{n\minus{}1}}\cdot\left(\minus{}\frac1n(h\minus{}y)^n\right)\right|_0^h\equal{}\frac1n\,Bh.$

Is that what you were asking for, 1234567a?

Thanks prof. Kent, yes that is what I am asking for

Thanks a bunch

--------------------------------------

[b]For Fedja:[/b]
[quote]
But we know the volume (k-dimensional Lebesgue measure), so to have everything consistent, we are forced to define the surface measure on the unit sphere by the cited formula at least for small sets. B

[/quote]

I still do not understand " [b]the cited formula at least for small sets.[/b]", would you mind explaining ?

Oh, that was just the reference to the formula $ \sigma_{k\minus{}1}(A)\equal{}k\cdot m_{k}(\widetilde A)$ defining the measure. ("cited"="the one you rewrote from the book") :) I was just saying that this formula should hold (at least, approximately) when the diameter of $ A$ is small. That's all.


Let [$ a_n, b_n$], $ n \equal{} 1,2,.....$ be closed intervals with [$ a_n, b_n$] $ \cap$ [$ a_m, b_m$] $ \neq \emptyset$ for all $ n, m$. Prove that $ \bigcap_{n\equal{}1}^{\infty}[a_n, b_n] \neq \emptyset$

Thanks for answer, pleurestique, but what do you mean by "compactness"? And aren't there any other approches to this problem (since the idea of "compactness", I guess, is pretty general)?

i meant use the definition of compact...


number theory
relatively prime
calculus
calculus computations

here is another easy problem.   :D

the value of the  series

$ \sum _{n\equal{}2}^\infty \; \frac { x^4\plus{} x^3 \plus{}x^2 \minus{} x \plus{}1} {x^6\minus{}1}$

can be expressed as $ \frac{p}{q}$  where $ p$ and $ q$ are relatively prime $ \plus{}ve$ numbers.

[b]without[/b] explicitly solving for $ p$ and $ q$ , calculate $ p\plus{}q$

I am confused :(  , the index is n but it appears nowhere in the summation.  Would it not blow up to infinity then?

sorry, that was a typo  :(  , the summation should read as $ \sum_{x\equal{}2}^\infty$

Let's try to simplify our sum:

$ \textbf{S}\equal{}\sum _{x\equal{}2}^\infty \; \frac { x^4\plus{} x^3 \plus{}x^2 \minus{} x \plus{}1} {x^6\minus{}1}\equal{}\sum _{x\equal{}2}^\infty \; \frac { x^4\plus{} x^3 \plus{}x^2 \plus{} x \plus{} 1 \minus{} 2x} {x^6\minus{}1}\equal{}$

$ \equal{}\sum _{x\equal{}2}^\infty \; \frac {\frac{x^5\minus{}1}{x\minus{}1}\minus{}2x} {x^6\minus{}1}\equal{}\sum _{x\equal{}2}^\infty \; \frac {x^5\minus{}1} {(x\minus{}1)(x^6\minus{}1)}\minus{}2\sum _{x\equal{}2}^\infty \; \frac {x} {x^6\minus{}1}\equal{}$

$ \equal{}\sum _{x\equal{}2}^\infty \; \frac {x^6\minus{}1\minus{}x\plus{}1} {x(x\minus{}1)(x^6\minus{}1)}\minus{}2\sum _{x\equal{}2}^\infty \; \frac {x} {x^6\minus{}1}\equal{}$

$ \equal{}\sum _{x\equal{}2}^\infty \; \frac {1} {x(x\minus{}1)}\left(1\minus{}\frac{x\minus{}1}{x^6\minus{}1}\right)\minus{}2\sum _{x\equal{}2}^\infty \; \frac {x} {x^6\minus{}1}\equal{}$

$ \equal{}\sum _{x\equal{}2}^\infty \; \frac {1} {x(x\minus{}1)}\minus{}\sum _{x\equal{}2}^\infty \; \frac {1} {x(x^6\minus{}1)}\minus{}2\sum _{x\equal{}2}^\infty \; \frac {x} {x^6\minus{}1}\equal{}$

$ \equal{}1\minus{}\sum _{x\equal{}2}^\infty \; \frac {1} {(x^6\minus{}1)}\left(2x\plus{}\frac{1}{x}\right)\equal{}1\minus{}\textbf{G}$.

We can find that $ \textbf{G}\equal{}\frac{1}{12}$. This gives $ \textbf{S}\equal{}\frac{11}{12}$ and $ \boxed{p\plus{}q\equal{}23}$. But how not to use calculation of $ \textbf{G}$ find $ p\plus{}q$ I don't know.

You're working too hard with the summand.

I haven't figured out how to answer this question without explicitly evaluating the sum in question, though.

[hide]
We factor the denominator:

$ (x^6 \minus{} 1) \equal{} (x^3 \minus{} 1)(x^3 \plus{} 1) \equal{} (x \minus{} 1)(x^2 \plus{} x \plus{} 1)(x \plus{} 1)(x^2 \minus{} x \plus{} 1)$.

Then we wish to find coefficients $ A, B, C, D, E, F$ such that

$ \frac {A}{x \minus{} 1} \plus{} \frac {B}{x \plus{} 1} \plus{} \frac {Cx \plus{} D}{x^2 \minus{} x \plus{} 1} \plus{} \frac {Ex \plus{} F}{x^2 \plus{} x \plus{} 1} \equal{} \frac {x^4 \plus{} x^3 \plus{} x^2 \minus{} x \plus{} 1}{x^6 \minus{} 1}$.

Multiplying through and collecting like terms and solving the partial fraction decomposition gives

$ A \equal{} 1/2, B \equal{} \minus{} 1/2, C \equal{} 0, D \equal{} 1/2, E \equal{} 0, F \equal{} \minus{} 1/2$,

giving

$ \sum_{x \equal{} 2}^\infty \frac {x^4 \plus{} x^3 \plus{} x^2 \minus{} x \plus{} 1}{x^6 \minus{} 1} \equal{} \frac {1}{2}\sum_{x \equal{} 2}^\infty \frac {1}{x \minus{} 1} \minus{} \frac {1}{x \plus{} 1} \plus{} \frac {1}{1 \minus{} x \plus{} x^2} \minus{} \frac {1}{1 \plus{} x \plus{} x^2}$.

The first two terms clearly telescope, but it also turns out that

$ 1 \plus{} x \plus{} x^2 \equal{} \minus{} x \plus{} (x \plus{} 1)^2 \equal{} 1 \minus{} (x \plus{} 1) \plus{} (x \plus{} 1)^2$,

so the last two terms in the summand also telescope.  Hence the sum is equal to

$ \frac {1}{2}\left(\frac {1}{2 \minus{} 1} \plus{} \frac {1}{3 \minus{} 1} \plus{} \frac {1}{1 \minus{} 2 \plus{} 2^2}\right) \equal{} \frac {11}{12}$,

and since 11 and 12 are relatively prime, their sum is 23.
[/hide]


Let a=[m^(m+1)+n^(n+1)]/(m^m+n^n) where m,n are from N*. Prove that a^m+a^n>=m^m+n^n.

cheers!

Oh.. for a solution you could use the hint that if n>=m then m<=a<=n and the inequality to be proved is like so:

(a-m)(a^(m-1)+a^(m-2)*m+..+am^(m-2)+m^(m-1))>=(n-a)(n^(n-1)+n^(n-2)*a+..+na^(n-2)+a^(n-1)).

but I am looking for another solution...cheers!

If I am not mistaking.. this problem was given for a test in USA  :?


induction
algebra
binomial theorem
advanced fields
advanced fields unsolved

( The better presentation of the question with correct symbols is here.
http://www.geocities.com/johntyung/Q2.pdf )

*** For this problem the best way to solve it by just using the identity and manipulating it algebraically. Without using induction. ***

for n even, prove that

(n choose 0) + (n choose 2) + (n choose 4) +...+ (n choose n)=2^(n-1)

and

(n choose 1) + (n choose 3) +...+(n choose (n-1)) = 2^(n-1)

using mathematical induction and/or various identities involving binomial coefficients.
(Hint: first establish the identity,

((m+2) choose k) = (m choose k)+ 2(m choose (k-1)) +(m choose (k-2) )

Then, when doing the induction, do a simultaneous induction on both identities.

$\sum_{k\text{ is even}}2\binom nk=\sum_{k=0}^n(1+(-1)^k)\binom nk=(1+1)^n+(1-1)^n$
$\sum_{k\text{ is odd}}2\binom nk=\sum_{k=0}^n(1-(-1)^k)\binom nk=(1+1)^n-(1-1)^n$

The binomial theorem is powerful.

ok smarty pants LOL

how about explaining your steps....

thanks in advance any way.

From the binomial theorem, we know that

$\binom{n}{0}+\binom{n}{1}+\ldots+\binom{n}{n}=2^{k}$

We also know that

$\binom{n}{0}-\binom{n}{1}+\ldots\pm\binom{n}{n}=0$

Now add or subtract this to get what jmerry wrote (I'm assuming you are not comfortable with sigma notation)

OK.  can u show the steps of what u subtract from what?

U have to use the Identity to solve this.... Can u show how u use the identity?


inequalities
function
inequalities solved

I'll try:

Let there be a concave function f(x)=x/(x^2+1)
From Jensen we get

f(a)+f(b)+f(c)<=3f((a+b+c)/3)=3*1/3*9/10=9/10

Hope it's the corect solution!

Hey, yours soln is true if a,b,c>0. trie to solv :D

[quote="hxtung"]Comment:
 I found out (*), because I suppose  a/(a2+1) <= xa +y with a>-3/4. multiplying (a2+1)and change the member of an inequality, we get :xa3 + ya2 +a(x-1) +y>=0.Put f(a)=xa3 + ya2 +a(x-1) +y
 We suppose f(a)=(3a-1)2(ra+t).By indentity, we get:
 9r=x, 9t-6r=y, r-6t=x-1 and t=y. Solv this system, we have x=18/25, y=3/50.[/quote]

Nice solution! Since you are expecting f has a factor (3a-1) 2, we must have f(1/3)=f'(1/3) = 0. This gives us two linear equations in two variables x and y, then solve for them.

It's interesting to see how you applied the same technique  to Japan1997. I found your technique can also be applied to #5 of USAMO2003.

We can solv  this one with $a,b,c>-1;a+b+c=1$

[quote=hxtung]Prove that :
$\frac{a}{a^2+1} + \frac{b}{b^2+1} + \frac{c}{c^2+1} \leq \frac{9}{10}$
for all real numbers $a$, $b$, $c$ satisfying $a+b+c=1$ and $a,b,c>-\frac34$.[/quote]

$$\text{RHS -LHS} = \sum \frac{(2a-b-c)^2 (4a+3)}{50(a^2+1)} \geqq 0$$

[quote=hxtung]Prove that :
$\frac{a}{a^2+1} + \frac{b}{b^2+1} + \frac{c}{c^2+1} \leq \frac{9}{10}$
for all real numbers $a$, $b$, $c$ satisfying $a+b+c=1$ and $a,b,c>-\frac34$.[/quote]
$$\dfrac{x}{x^2+1}\le \dfrac{18}{25}x+\dfrac{3}{50}$$
[url=https://artofproblemsolving.com/community/c6h514132p3003386]for[/url] $x\ge -\dfrac{3}{4}$


Find $ a,b \in\mathbb{Z}_ \plus{} ^*$ such that $ (\sqrt [3]{a} \plus{} \sqrt [3]{b} \minus{} 1)^2 \equal{} 49 \plus{} 20\sqrt [3]{6}$.

Its in PEN

[url]http://www.mathlinks.ro/viewtopic.php?t=150693[/url]


calculus
integration
algebra
polynomial
real analysis
real analysis unsolved

Well as the integrand has no dependence on $ x$, you can pull it out, leaving you with $ C\int_{\minus{}0.5}^{0.5}\,dx\equal{}Cx\Bigg|_{\minus{}0.5}^{0.5}\equal{}C$, where $ C$ is that whole big mess.

First, make the obvious substitution $ t \equal{} y \plus{} \frac12$. The integral becomes $ 2^N\int_0^1 \frac {y^N}{p_N(y)}\,dy$, where $ p_N$ is the $ N$th Taylor polynomial of the exponential at the origin and $ N$ is $ 2009$.
Note that $ p_N'(y) \equal{} p_N(y) \minus{} \frac {y^N}{N!}$. Thus $ \frac {y^N}{p_N(y)} \equal{} N!\frac {p_N(y) \minus{} p_N'(y)}{p_N(y)} \equal{} N! \minus{} N!\frac {p_N'(y)}{p_N(y)}$
This is easy to integrate, and we have
$ 2^N\int_0^1 \frac {y^N}{p_N(y)}\,dy \equal{} 2^N\left[N!y \minus{} N!\ln(p_N(y))\right]_0^1$

$ \equal{} 2^N\cdot N!\cdot(1 \minus{} \ln(p_N(1)))$

That's an exact answer. If you want a reasonable approximation, the above calculation was basically a waste of time; the bad subtraction will kill you. To first order, it's approximately $ \frac {2^N}{N\cdot e}$.


Hi guys. I just started my own blog site. I'm going to be putting up some cool, basic effects that you can make in Photoshop or Photoshop Elements. Take a look! [url=http://www.wangsblog.com/andrew/]My Site[/url]


Find remainder when $19^{92}$ is divided by 92.

note that $\phi(92)=44$ and $(92,19)=1$ so you can reduce the exponent by multiples of $44$, and that is how t0rajirou got from $7^{92}$ to $7^4$, the theorem that he used is called euler's generalization of fermat's little theorem, if you don't know it, search first

Then shouldnt that be $19^{92} \equiv 361^{46}?$

Oops.  Did I write $361^{92}$?

 :blush:


Theorem: All numbers are equal.
Proof: Choose arbitrary a and b, and let t = a + b. Then
a + b = t
(a + b)(a - b) = t(a - b)
a^2 - b^2 = ta - tb
a^2 - ta = b^2 - tb
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
(a - t/2)^2 = (b - t/2)^2
a - t/2 = b - t/2
a = b


Hi, Can any body help solve the following problems in Angel's book?

What intergers m and n is valid for the equation
(5 + 3 sqrt 2) ^ m = ( 3 + 5 sqrt 2 ) ^n.

The solution in the book gives the following hint:
(5 + 3 sqrt 2) ^ m = ( 3 + 5 sqrt 2 ) ^m --> (5 - 3 sqrt 2) ^ m = ( 3  - 5 sqrt 2 ) ^n.
I can not figure  out the equivalence of the given equation to the second one.

Thank you,

[hide="One way to think about it"] When the expressions are expanded via the Binomial Theorem, the terms such that $ 3\sqrt{2}$ or $ 5\sqrt{2}$ have an even exponent are integer, and are the same whether we are expanding $ (5\minus{}3\sqrt{2})$ or $ (5\plus{}3\sqrt{2})$, and the terms such that $ 3\sqrt{2}$ or $ 5\sqrt{2}$ have an odd exponent are not (and are positive when we expand the positive version and negative when we expand the negative version).

In other words, the "integer part" remains unchanged when we flip the sign on the radical, while the "irrational part" flips signs.  Since the original statement implies that the "integer" and "irrational" parts are already equal, flipping the sign does not change equality. [/hide]

in other words
$ (5\plus{}3\sqrt{2})^{m}\equal{}A\plus{}B\sqrt{2}$
$ (3\plus{}5\sqrt{2})^{n}\equal{}C\plus{}D\sqrt{2}$
where A;B;C;D are some integer numbers
because $ (5\plus{}3\sqrt{2})^{m}\equal{}(3\plus{}5\sqrt{2})^{n}$ so $ A\equal{}C$ and $ B\equal{}D$
obviously $ A\minus{}B\sqrt{2}\equal{}C\minus{}D\sqrt{2}$
finally we get $ (5\minus{}3\sqrt{2})^{m}\equal{}(3\minus{}5\sqrt{2})^{n}$

If you mulitply those two expressions, you get $ 43^{m}\equal{}53^{n}$...which has the only solution when $ m\equal{}n\equal{}0$...

This works in the original equation, hence it is a solution.


geometry
circumcircle
perimeter
inequalities
inequalities proposed

For $m_a,m_b,m_c$ medians of a triangle of semiperimeter $s$ and circumradius $R$ prove

$m_a+m_b+m_c < 2R +s $

First I solve for acute triangles .

Denote p= semi perimeter
[b]Lemma 1[/b] For acute triangles:  $ p \geq\ 2R+r $

[b]Lemma 2[/b]: For acute triangles: $ m_a + m_b + m_c \leq\ 4R+r $

Proof for lemma 2
 Let A', B', C' be three midpoints of BC, CA, AB.
 We have  $ m_a \leq\ R + OA' = R + RcosA $ ( ABC is an acute triangle)
 Similarly, $ m_b \leq\ R + OB' = R + RcosB $
                $ m_c \leq\ R + OC' = R + RcosC $
Add them together we will get $ m_a + m_b + m_c \leq\ R(3+cosA+cosB+cosC) $
which is equivalent to lemma 2

By using two lemmas manlio's inequality is proved.

I haven't proved for obstute triangles yet. I still try.

For obtuse triangle  in angle A you must use  $2m_c<a+b$, $2m_b<a+c$,$m_a<R$,$a<2R$ to obtain

$m_a+m_b+m_c < R +s +\frac{a}{2} < R + R +s=2R +s$


In a 4x4 grid, place the numbers 2,3,4,..., N (with N as large as possible) such that if A and B share are placed in squares that share a side, then either A divides B or B divides A.

[quote="NoSoupForYou"]Do the numbers have to be $2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17$ or can they be any numbers?[/quote]Yes, they have to be consecutive numbers, and probably you won't be able to use them all!

Can't be done.  17 has to be next to something, and there is no X<17 such that X|17 or 17|X.

But you aren't required to place all of them, only as many as possible.
EDIT: Here's a solution with N=10. It's probably not even close to the maximum, but it's a start.
8|_|_|7

4|2|6|_

_|_|3|9

5|10|_| _

I think N=10 might be the maximum.

Suppose N=11, so there are 10 numbers. Then

11 cannot be next to any (automatically takes out 2 squares)
9 can only be next to 3 (takes out 1 square)
7 cannot be next to any (takes out 2 squares)
5 can only be next to 10 (takes out 1 square)

However, minimizing the removal of squares above requires all 4 of those numbers to be in the corners. Hence 10, which remains, is not in a corner. But 10 can only go next to 5 and 2, so it takes out another square.

So 7 squares are gone, leaving 16-7 = 9. But we have 10 numbers. Contradiction. Thus N=10 is the maximum, in an arrangement shown by LynnelleYe.

But we don't use 1. So N=10 implies 9 numbers, not 10. And I just got an arrangement using your logic for guidance which has N=12.
9|_|10|5
3|6 |2 |8
_|12|4|_
11|_|_|7

Err... $5\not|8$ and $8\not|5$. I think my logic is correct...

[quote="LynnelleYe"]Ooooops.  :blush: I can be an idiot sometimes.[/quote]Don worry, we all are sometimes.  :P


six more than 5 times a # is the same as the # is deacresed by 13. find the #

$ 6 \plus{} 5 \alpha \equal{} \alpha \minus{} 13$

$ 4 \alpha \equal{} \minus{}19$

$ \alpha \equal{} \minus{}\frac{19}{4}$


Show that $ 7^{2010}\minus{}2^{2010}$ is a multiple of $ 3^3 \cdot 5^2 \cdot 11\cdot 13\cdot 31\cdot 61\cdot 67$.


I was just wondering since i heard from some of friends that by attending a college's own summer programs it raises your chance of acceptance. is this true or somewhat true or completel false?

thanks!

I would not count on that at all. I don't plan to encourage my oldest son to go to any college's summer program of that kind. There are too many good math, science, and debate programs for the summers he has left anyway.

I'll throw in my speculation that IF you go to a particular college's summer program, AND you meet and get to know some of the faculty there, it COULD increase your chances of acceptance if you get a good letter of recommendation from one of them.  There's no reason for a college to favor students who have simply gone to a summer program sponsored by them.  However, if one of their own faculty recommends you for admission, I expect that that holds a lot of weight in the admissions process (especially because, if they have more questions, the admissions committee can go talk to that professor, and because presumably that professor knows pretty well what sort of student is qualified to go to to the school).  As a result, a recommendation letter from a professor at a college you're applying to is probably a boon to your application.

HOWEVER, never bank on this getting you in - I have friends who applied to prestigious colleges (e.g. Stanford) and were deferred even though they had a letter of rec from a professor at the school.  It's more important that you choose a program that really interests you and in which you can really excel and demonstrate intellectual talent and leadership.  Your demonstration of that in a non-college-sponsored summer program (and remember that you can always get a letter of rec from faculty at that program) will mean more than your participation in a college's summer program which you were less enthusiastic about.

Doing any type of summer program at a college will increase your chances of getting in [i]anywhere[/i].  It probably matters more of how prestigious that program is that you did rather than where that program is.  On the other hand, getting a recommendation from someone at the school you're applying to would definitely be a major plus.

Thanks guys. These advices are really helpful! :lol:


1) Let $n$ be an odd positive integer. Prove that
\[\binom{n}{0}, \binom{n}{1}, ... \binom{n}{\frac{n-1}{2}}\]
contains an even number of odd numbers.

2) Let $n$ be a positive integer. Determine the number of odd numbers in $\binom{n}{0}, \binom{n}{1}, ... \binom{n}{n}$.

[hide="1"]
It is not difficult to see

$\binom{n}{0}+\binom{n}{1}+\ldots+\binom{n}{\frac{n-1}{2}}=2^{n-1}$,

which is even.

This is only possible if all the numbers in the sequence are even, or if there are an even number of odd numbers in the sequence.
[/hide]

For 2) see here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=102805[/url].


Prove that tan(a)>a when 0<a<pi/2 and a is in radians

This can be proved easily with the help of calculus.
[hide]
Let $f(x)=tanx - x$
$f'(x)=sec^2x-1 \geq 0$
thus, f(x) is increasing in 0<x<pi/2
f(x)>f(0)=0
thus, tanx - x>0
tanx>x[/hide]

[quote="Gabriel(fr)"]We can prove it by the area of the $90^o$ arc[/quote]
can you elaborate?


Hmmm, I'm going to, at least I hope I'm going to, take the AMC 10 or 12 in February for the first time, so I decided to read about it in the posts... I just heard about it this year... and I come across A and B versions of the tests... which have totally confuzzled me... what is the difference... what should I choose to take?  I am relatively new at this math competition stuff... I was on a different competition track in elementary and middle school... and I am currently enrolled in Honors Alg. 2/Trignometry with a satisfactorily high grade... not super advanced, like some people I know are, but I'll just try to learn it well... By February, which is when I have the concept that the test is offered, I'll probably have finished most of Alg. 2 and delved into some trig.... perhaps I should self-study the whole book... Only problem is... I can do really well in school mathematics... but most things outside of that... well, I have difficulties.  So, based on your knowledge of my mathematical ability, what do you think I should take?  I really appreciate your help in advance ^^.  thanks.  Oh yes, and I don't know if this helps... but I also took this Mathematical Logic class at John Hopkins' CTY summer programs two years ago... load of interesting proofs... anyhow, Happy New Years everyone!

You definitely need to cover geomtry a lot. Geometry is strong at AMC-10 and if you're taking AMC-12, make sure you understand the trig and other logarithrms and stuffs very much. And it might be too late but do LOTS of problems. I think I did more problems this vacation than any other time in my lifetime for just this AMC.

AMC A and B are basically same. They cover same topics, difficulty but some say A's easy, some say B's easy.

Hope this helps! :)


how do you write a program on a TI-84 that allows you to find the last $x$ digits of a number.

for example,

what are the last 2 digits of $2^{100}$?

Use modular arithmetic.

First, use Euler's totient theorem $\left( a^{\phi(n)}\equiv 1 \bmod n \right)$, where $\phi(n)$ represents the [url=http://mathworld.wolfram.com/TotientFunction.html]Totient function[/url], to reduce the exponent. In your case we would have $2^{\phi(100)}=2^{40}\implies 2^{100}\equiv 2^{20}\bmod 100$.

From there, you can have the calculator multiply that out, or you can use the [url=http://mathworld.wolfram.com/ChineseRemainderTheorem.html]Chinese Remainder Theorem[/url] to (I think) simplify it further.


group theory
combinatorics proposed
combinatorics

A polygon with 2n + 1 vertices is given. Show that it is possible to assign numbers 1, 2, …, 4n + 2 to the vertices and midpoints of the sides of the polygon so that for each side the sum of the three numbers assigned to it is the same.

let the vertices of the polygon be(clockwise,say)$A_1,A_2,\ldots,A_{2n+1}$.we label the vertices using the labels $1,2,\ldots,2n+1$ as follows:
   Consider the cyclic group $\mathbb{Z}_{2n+1}$.clearly $o(2)=2n+1$,so $2$ is a generator for the group.let the vertices of the polygon denote the group elements(i.e.$A_i\equiv i$ in $\mathbb{Z}_{2n+1}$).put the label $k$ on $A_{2k(mod\ 2n+1)}$. it is easy to check that the edge sums now read $n+2,n+3,\ldots,3n+2$.now insert the remaining numbers $2n+2,2n+3,\ldots,4n+2$ along the midpoints of the edges such that the sums all read $5n+4$.


find all integers positives $x ,y ,z$ with $x^2+y^2-2z=18$

Isn't there an infinite amount of these? As long as $x$ and $y$ are both even or both odd, and $x^2+y^2>18$, there will always be a $z$ value to satisfy the equation. Am I missing something?

Yes that's right,  the amount of solutions is infinite.


If you were awarded a prize of maths, Which mathematican do you want to issue you the cup(or certificate,etc.) the most?

I don't reallyy know any famous mathematicians :oops: By the way, A Beautiful Mind was a great movie :D

[quote="cezar lupu"]A Beautiful Mind is about a great mathematician: John Forbes Nash,isn't it? ;)[/quote]

Oh, I didn't know it was based on a real mathematician, haven't watched it in a while. Anyway, Cezar, you changed your avatar :(

Yes,I've changed it for a while. It is me there. ;)

Well I like A Beautiful Mind mostly because I like J.F.Nash a lot.
As for me, he or P.Erdös would be the best...

[quote="Skyward_Sea"]P.Erdös would be the best...[/quote]
if he was alive.

There r a ton of mathematicians that were simply great, just a few that could be names r:

Leonard Euler
Riemann
Gauss
Lagrange
Cauchy
Abel
Galois
Cayley
Polya
Wiles
Van Neumann
Kolmogorov
Chern

etc etc etc etc

does it have to be relatively recent? archimedes gets my vote then :surf:

Fermat - maybe he could tell me his proof for FLT while he was giving it to me.

i would have pythagoras if he was still alive :(


trigonometry
symmetry
calculus
algebra unsolved
algebra

solve in R² the system:
sin²x+cos²y=y²
sin²y+cos²x=x²

Add the two equation to obtain $x^{2}+y^{2}= 2$. Substituting for y in the first equation gives $\sin^{2}x+sin^{2}\sqrt{2-x^{2}}= 1-x^{2}$.

Clearly $x=1$ is a solution. If we let $f(x) = \sin^{2}x+sin^{2}\sqrt{2-x^{2}}+x^{2}-1$, we see that $f'(x) = 2\sin x \cos x+\frac{2x\sin \sqrt{2-x^{2}}\cos \sqrt{2-x^{2}}}{\sqrt{2-x^{2}}}+2x > 0 \ \forall x \in [0, \sqrt{2})$. Thus $(1, 1)$ is the only solution in the 1st quadrant. By symmetry, the solutions are $(\pm 1, \pm 1)$.

[b]Edit:[/b] Left out addition signs

An other one, whithout calculus ...
First of all it's easy to see that $|x|\leq \sqrt 2< \frac{\pi}2 ,|y|\leq \sqrt 2< \frac{\pi}2$
There are 2 cases :
$|x|\leq |y|\Longrightarrow y^{2}= sin^{2}x+cos^{2}y \leq sin^{2}y+cos^{2}x=x^{2}$ , so $|x|=|y|$
$|y|\leq |x|\Longrightarrow x^{2}= sin^{2}y+cos^{2}x \leq sin^{2}x+cos^{2}y=y^{2}$ , so $|x|=|y|$

Finally $|x|=|y|=1$
 :cool:

[quote="hien"][quote="mathisfun1"][quote="Diogene"]
There are 2 cases :
$|x|\leq |y|\Longrightarrow y^{2}= sin^{2}x+cos^{2}y \leq sin^{2}y+cos^{2}x=x^{2}$ , so $|x|=|y|$
$|y|\leq |x|\Longrightarrow x^{2}= sin^{2}y+cos^{2}x \leq sin^{2}x+cos^{2}y=y^{2}$ , so $|x|=|y|$

Finally $|x|=|y|=1$
 :cool:[/quote]

I'm sorry, but could you explain this? I'm not sure how you got the inequalities...[/quote]

No, It's wrong[/quote]
??? I see nothing wrong.

Assume $|x|\leq |y|$ by symmetry.
Then $y^{2}\leq x^{2}+y^{2}=2\Rightarrow |y|\leq \sqrt{2}<{\pi\over 2}$ and $0\leq |x|\leq |y|\leq{\pi\over 2}$.
Then $0\leq \sin |x|\leq \sin |y|$ and $\cos |x|\geq \cos |y|\geq 0$,
as $\sin$ is increasing positive on $[0,{\pi\over 2}]$
and $\cos$ is decreasing positive on $[0,{\pi\over 2}]$.

So
$y^{2}= sin^{2}x+cos^{2}y= sin^{2}|x|+cos^{2}|y|\\ \hspace*{0.85cm}\leq sin^{2}|y|+cos^{2}|x|=sin^{2}y+cos^{2}x=x^{2}$
and $|y|\leq |x|$. ($\sin$ is odd and $\cos$ is even)

Together with the assumption this give $|x|=|y|$ and $2=x^{2}+y^{2}=|x|^{2}+|y|^{2}=2|y|^{2}$,
and so $|x|=|y|=1$ and $(x,y)=(\pm 1,\pm 1)$.

Nice solution [b]Diogene[/b]!

Yes, i'm wrong. Sory of my stupid! :huh:

THanks Olorin. I wasn't thinking straight. :blush:

it is czech- slovakia regional round 2006


geometry
parallelogram
geometry proposed

Let $ BB_1$ and $ CC_1$ are heights of the triangle. Then triangles $ HBC_1$ and $ HCB_1$ are similar and the points $ U,V$ are midpoints of segments $ BC_1, CB_1$.

[quote="Virgil Nicula"][color=darkred]In the acute triangle $ ABC$ denote the orthocenter $ H$ , the midpoint $ M$ of the side $ [BC]$

and the projections $ U$ , $ V$ of the point $ M$ on the lines $ AB$ , $ AC$ .  Prove that $ \widehat {AUH}\equiv\widehat {AVH}$ .[/color][/quote]
I think that this is similar to [b]baysa[/b]'s solution:
[hide="Solution"]
Let $ CQ$ and $ BR$ be altitudes, let $ MU$ meet $ BR$ and $ X$, and let $ MV$ meet $ CQ$ at $ Y$.  We  have that $ MV\parallel BR$ and $ MU\parallel CQ$, so ${ \frac {BX}{BH} = \frac {BM}{MC} = \frac {1}{2}\implies BX = \frac {BH}{2}\implies XH} = \frac {BH}{2}$. Similarly, $ HY = \frac {CH}{2}$.  Also, we have that $ \frac {UX}{QH} = \frac {BM}{MC} = \frac {1}{2}\implies UX = \frac {QH}{2}$.  Simiarly, we have that $ YV = \frac {HR}{2}$.  Now, notice that $ \triangle QHB\sim \triangle RHC\implies \frac {QH}{BH} = \frac {HR}{HC}$.  Thus, $ \frac {\frac {QH}{2}}{\frac {HB}{2}} = \frac {\frac {HR}{2}}{\frac {CH}{2}}\implies \frac {UX}{XH} = \frac {YV}{YH}$.  Notice that $ XHYM$ is a parallelogram since $ XH\parallel MY$ and $ HY\parallel XM$.  Thus, $ \angle HXM = \angle HYM\implies \angle UXH = \angle HYV$.  Therefore, by SAS similarity, we have that $ \triangle HYV\sim \triangle HXU\implies \angle HUX = \angle HVY\implies \angle AVH = \angle AUH$, as desired. [/hide]


abstract algebra
modular arithmetic
function
superior algebra
superior algebra unsolved

Let $ G$ be a finite group of order $ pq$, where $ p,q$ are prime and $ p < q$. Suppose that $ q\not\equiv 1\pmod{p}$. Prove that $ G$ is abelian and cyclic.

It's obvious with Sylow's Theorem, but can be done only with Cauchy:

Let $ a$ be an element of order $ q$. Then for every element $ x$ of order $ q$, the cosets
\[ <a>,x<a>,x^2<a>,\cdots ,x^p<a>\]
are not all different since there are only $ p$ distinct cosets of $ <a>$. If $ x^j<a>\ \equal{}x^k<a>$ for $ p\geq j>k\geq 0$, then $ x^{j\minus{}k}\in <a>$ where $ q>p\geq j\minus{}k>0$. In particular, $ j\minus{}k$ has a multiplicative inverse modulo $ q$, so $ x\in <a>$.

Now that we know $ <a>$ contains precisely the identity and every element of order $ q$, it's normal. Let $ b$ be an element of order $ p$, so conjugation by $ b$ (i.e. the function $ \beta$ such that $ \beta (g)\equal{}bgb^{\minus{}1}$) is an automorphism of $ <a>$. The order of $ \beta$ divides both $ p$ (as $ b^p\equal{}1$) and $ q\minus{}1$ (as an automorphism of $ <a>$: $ <a>$'s automorphism group has $ q\minus{}1$ elements). But $ q\not\equiv 1\pmod{p}$, so the order of $ \beta$ is 1.

This means conjugation by $ b$ is the identity function on $ <a>$. In particular, $ bab^{\minus{}1}\equal{}a$. We get $ G\equal{}<ab>$ as a result.


function
real analysis
real analysis solved

Find n so that this limit is finite and it is not 0:

lim(x-->0) (x*e^(x^3)-sin(x)*e^(x^2)+sin(x)-x)/x^n.

Not hard, but I like it...


Hey, I was just curious if any of you could share some independent study stratagies with me. thanks

That's a much, much, much too vague question to lead to anything useful.  What aspect of independent study are you asking about?  Finding a project?  Actually carrying out mathematical research?  Something else entirely?

Finding a project. For instance, learn about a specific topic in linear algebra or analysis with a alot of depth, rather than just a superficial skim.


[quote="Tony2006"]un problema de mis entrenamientos en yucatán:

sean a, b, c enteros tales que b distinto de c. si $ax^{2}+bc+c$ y $(c-b)x^{2}+(c-a)x+(a+b)$ tienen una raiz en comun, pruebe que a+b+2c es multiplo de 3.[/quote]
tengo dos preguntas
1.- la raiz en comun es entera,racional, real?.
2.-es asi $ax^{2}+bc+c$ o es asi $ax^{2}+bx+c$.
chaufas espero ke respondas mis preguntas.


What did you all think.... I just finished a few minutes ago..

I thought the Mech was pretty standard, the last response question was hard for me... I thought the E&M was exceptionally hard, I am a bit scared I might not get a 5...

Don't forget, don't discuss any questions specifically.

How about don't discuss anything about it until tomorrow.

I am taking the Physics C next week, and I am wondering where I could find a simple explanation of some electricity concepts. I find it really difficult to attain an intuitive understanding of electricity.

I thought it was fairly difficult, but I did not take the class and did not study extensively. The last E&M FR was difficult - part of it, at least.


Let a,b,c be positive numbers
Prove that
 $ (a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}) \geq \frac{(a\plus{}2b)^2}{a^2\plus{}b(c\plus{}a)}\plus{}\frac{(b\plus{}2c)^2}{b^2\plus{}c(a\plus{}b)}\plus{}\frac{(c\plus{}2a)^2}{c^2\plus{}a(b\plus{}c)}$

[quote="8826"]Nobody's interested in it?[/quote]

No :blush: , I'm trying.

[quote="tuandokim"]$ LHS \equal{} \sum_{cyc}(1 \plus{} \frac {b}{c} \plus{} \frac {b}{a}) \equal{} \sum_{cyc}\frac {a^2}{a^2} \plus{} \frac {4b^2}{bc \plus{} ba}\geq \sum_{cyc}\frac {{a \plus{} 2b}^2}{a^2 \plus{} bc \plus{} ba} \equal{} RHS$
Q>E>D :)[/quote]

Thank you, very nice solution. You have some mistake, but no problem


Each interior point of an equilateral triangle of side 1 lies in one of six congruent circles of radius [i]r.[/i] Prove that
[i]r[/i] >=  :sqrt: 3 /10

:blush: ...any help? ;) Hints are welcome, too :D
Artiona

Suppose $r<\dsp\frac {\sqrt 3}{10}$ and then consider the six centers $C_1,...,C_6$ of our circles. In order for our property to hold, any disk of radius $r$ centered in a point in the triangle has to contain at least one of the 6 centers. So construct 3 circles of radius $r$ centered at the vertices, and 6 more circles centered on the sides, at distance $2r$ from any of the vertices. Now, it's easy to see that these 9 circles do not intersect, except for the three pairs of circles centered on each of the sides, which intersect in a really small region (they intersect on a length of 0.04 of each side). In order for the 6 centers $C_1,...,C_6$ to lie in these 9 circles, three of them must lie in the circles that correspond to the vertices, and the other three in the common regions of the circles centered on the sides. But then it's easy to see that the centroid of the triangle is at a distance $>r$ from $C_1,...,C_6$ and thus we are done.

Actually, you can considerably improve the constant $\dsp \frac {\sqrt 3}{10}$, but who wants to carry out the computation?

[quote="RHS"]

Actually, you can considerably improve the constant $\dsp \frac {\sqrt 3}{10}$, but who wants to carry out the computation?[/quote]

That's the most difficult part! :P
Could you make an effort?

Stronger result: \[ \displaystyle r \geq  \frac 1 {2(1+ \sqrt 3)} \]


geometry
circumcircle
geometric transformation
reflection
incenter
geometry proposed

Let ABC be a triangle with circumcentre O and Nagel point N.
If ON intersects the circumcircle(O) at P.
Prove that the simson line wrt. P is parallel to NH.

Take anticomplementary triangle $ \triangle A_0B_0C_0,$ such that $ A, B, C$ are midpoints of $ B_0C_0, C_0A_0, A_0B_0.$ $ H$ is its circumcenter, $ O$ its 9-point circle center, $ N$ its incenter. Reflections of the diacentral line $ HN$ of $ \triangle A_0B_0C_0$ in the sidelines $ BC, CA, AB$ concur at the orthopole $ P \in (O)$ of $ HN$ WRT $ \triangle A_0B_0C_0$ $ \Longrightarrow$ $ HN$ is Steiner line of $ \triangle ABC$ with the pole $ P,$ parallel to Simson line $ p$ of $ \triangle ABC$ with the pole $ P.$ By 2nd Fontene theorem, circumcircle of pedal triangle (pedal circle), of $ \triangle A_0B_0C_0$ WRT any point $ Q \in HN$ goes through $ P \in (O).$ When $ Q$ coincides with the incenter $ N,$ pedal circle coincides with the incircle, which is tangent to the 9-point circle $ (O)$ at the Feuerbach point $ \Longrightarrow$ the orthopole $ P$ is identical with the Feuerbach point, which is identical with intersection of the ray $ \overrightarrow{ON}$ with $ (O).$


[hide="Partial solution"]
Well, using the change of base formula, we find that:

$\log_{35}28 = \frac{\log_{14}28}{\log_{14}35}$

$= \frac{\log_{14}(4*7)}{\log_{14}(7*5)}$

From log properties, this equals

$\frac{\log_{14}4+\log_{14}7}{\log_{14}7+\log_{14}5}$

From the values for $a$ and $b$, we see that this equals:

$\frac{\log_{14}4+a}{a+b}$

From here, though, I'm stuck on evaluating $\log_{14}4$.[/hide]

[hide]Solution?
The beginning is similar to I Am Me's:

$log_{35}28=\frac{log_{14}28}{log_{14}35}$

$=\frac{log_{14}(14\times2)}{log_{14}(7\times5)}$

$=\frac{log_{14}14+log_{14}2}{log_{14}7+log_{14}5}$

So the only thing left to figure out is $log_{14}2$:

$log_{14}7 = log_{14}(\frac{14}{2}) = a$
$log_{14}(\frac{14}{2}) = log_{14}14-log_{14}2 = a$
Therefore:
$log_{14}2 = 1-a$ since $log_{14}14 = 1$

Back to our original expression which becomes:
$\frac{1+(1-a)}{a+b}$
or
$\frac{2-a}{a+b}$
[/hide]


logarithms
real analysis
real analysis unsolved

Solve Cauchy's problem: \[ tx'-x-\ln x'=0, \] where $x(1)=1.$

notez x'=p. obtinem tp-x-lnp=0, care prin derivare conduce la $(t-\frac{1}{p})\frac{dp}{dt}=0$. se obtine $t=\frac{1}{p}$ si $x=1-lnp$. solutia singulara este $x=1+lnt$, iar cea generala $x=tc+lnc$ care impreuna cu conditia initiala devine $x=t$


inequalities
algebra
polynomial
binomial theorem
inequalities proposed

Hi!

Let $x_1,x_2,x_3$ be three real numbers such that $x_2x_3+x_3x_1\geq 1+x_1x_2+x_3^2$. What is the minimal value of $c$ in the inequality:
\[
(x_1-x_2)^{2n+1}+(x_2-x_3)^{2n+1}+(x_3-x_1)^{2n+1}\geq
c(x_1-x_2)(x_2-x_3)(x_3-x_1)
\]
where $n\in{\Bbb N}$.

I assume $c$ has to satisfy the inequality for all real $x_1,x_2,x_3$, otherwise I've done a completely different question.

Just one thing: suppose we have the inequality for some $x_1,x_2,x_3,c$. Then if we replace $x_i$ by $-x_i$, the requirement is preserved... but doesn't the inequality change direction?

Anyway: let $a=x_3-x_2, b=x_1-x_3$. The requirement becomes $ab \geq 1$, while the inequality is
\[(a+b)^{2n+1} - a^{2n+1} - b^{2n+1} \geq cab(a+b)\] Here we have the problem I outlined above: $a$ and $b$ must have the same sign, but the direction of the inequality depends on whether they are both positive or both negative.

Assuming $a,b > 0$, we can let $a=b=1$ to get $2^{2n+1} - 2 \geq 2c$. Hence $c \leq 2^{2n} -1$. To prove the inequality for this value of c, expand $(a+b)^{2n+1}$ by the binomial theorem, cancel and divide across by ab. The LHS becomes a polynomial with $2^{2n+1} -2$ terms, the RHS simply $(2^{2n}-1)(a+b)$, and since $ab \geq 1$, the inequality is trivial (just a case of repeated AM-GMs).


Let $ a,b,c,d>0$,prove that:
     $ \frac{a\minus{}b}{a\plus{}2b\plus{}c}\plus{}\frac{b\minus{}c}{b\plus{}2c\plus{}d}\plus{}\frac{c\minus{}d}{c\plus{}2d\plus{}a}\plus{}\frac{d\minus{}a}{d\plus{}2a\plus{}b}\geq 0$

[quote="super_math"]where are you read this inequality?[/quote]
My friend told me about this inequality.why did you ask about it?

[quote="super_math"]where are you read this inequality?[/quote]
If I don't wrong
This ineq is in Algebraic inequalities

[quote="super_math"]Can you write solution?[/quote]
If i know solution,i would post it in "Proposed and Own" problems section,isn't it?

I think I've seen this before on the forum....anyway,

This is the type of problem for which it's quite difficult to find an elegant solution (olympiad type) if you don't know the idea and it's quite easy to if you do know it.
The idea is to "get rid" of the differences $ a\minus{}b,\ldots,d\minus{}a$ as they may be negative and restrict our choices. We add $ \frac{1}{2}$ to each term of the [i]LHS[/i].
The inequality rewrites:
$ \frac{3a\plus{}c}{a\plus{}2b\plus{}c}\plus{}\frac{3b\plus{}d}{b\plus{}2c\plus{}d}\plus{}\frac{3c\plus{}a}{c\plus{}2d\plus{}a}\plus{}\frac{3d\plus{}b}{d\plus{}2a\plus{}b}\ge4$.
After amplifying each by its numerator, and applying Cauchy we actually get equality, as $ (3a\plus{}c\plus{}3b\plus{}d\plus{}3c\plus{}a\plus{}3d\plus{}b)^{2}\equal{} 16(a\plus{}b\plus{}c\plus{}d)^{2}$ and
$ 4(a\plus{}b\plus{}c\plus{}d)^{2}\equal{} (3a\plus{}c)(a\plus{}2b\plus{}c)\plus{}$ $ (3b\plus{}d)(b\plus{}2c\plus{}d)\plus{}(3c\plus{}a)(c\plus{}2d\plus{}a)\plus{}(3d\plus{}b)(d\plus{}2a\plus{}b)$.

I've firstly seen this idea in a problem from [i]Old & New Inequalities (GIL Publishing House)[/i]. The problem was proposed by [i][b]Vasc[/b]:
If $ a,b,c,d$ are positive real numbers, then:
$ \frac{a\minus{}b}{b\plus{}c}\plus{}\frac{b\minus{}c}{c\plus{}d}\plus{}\frac{c\minus{}d}{d\plus{}a}\plus{}\frac{d\minus{}a}{a\plus{}b}\ge0$
[/i]

[quote="freemind"]I've firstly seen this idea in a problem from [i]Old & New Inequalities (GIL Publishing House)[/i]. The problem was proposed by [i][b]Vasc[/b]:
If $ a,b,c,d$ are positive real numbers, then:
$ \frac {a \minus{} b}{b \plus{} c} \plus{} \frac {b \minus{} c}{c \plus{} d} \plus{} \frac {c \minus{} d}{d \plus{} a} \plus{} \frac {d \minus{} a}{a \plus{} b}\ge0$
[/i][/quote]
We have:
$ \frac{a\minus{}b}{b\plus{}c}\plus{}\frac{c\minus{}d}{d\plus{}a}\equal{}\frac{a\plus{}c}{b\plus{}c}\plus{}\frac{a\plus{}c}{d\plus{}a}\minus{}2\equal{}(a\plus{}c)(\frac{1}{b\plus{}c}\plus{}\frac{1}{d\plus{}c})\minus{}2$
$ \geq \frac{4(a\plus{}c)}{a\plus{}b\plus{}c\plus{}d}\minus{}2$
Similar,we have:
$ \frac{b\minus{}c}{c\plus{}d}\plus{}\frac{d\minus{}a}{a\plus{}b} \geq \frac{4(b\plus{}d)}{a\plus{}b\plus{}c\plus{}d}\minus{}2$
Done!

[quote="freemind"]The problem was proposed by [i][b]Vasc[/b]:
If $ a,b,c,d$ are positive real numbers, then:
$ \frac {a \minus{} b}{b \plus{} c} \plus{} \frac {b \minus{} c}{c \plus{} d} \plus{} \frac {c \minus{} d}{d \plus{} a} \plus{} \frac {d \minus{} a}{a \plus{} b}\ge0$
[/i][/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=102924

[quote="chien than"][quote="silouan"]In fact the stonger one holds (and it can be proved by Cauchy Swartz )
Let $ a,b,c,d > 0$,prove that:
$ \frac {a}{a \plus{} 2b \plus{} c} \plus{} \frac {b }{b \plus{} 2c \plus{} d} \plus{} \frac {c }{c \plus{} 2d \plus{} a} \plus{} \frac {d }{d \plus{} 2a \plus{} b}\geq 1\geq\frac {b}{a \plus{} 2b \plus{} c} \plus{} \frac {c }{b \plus{} 2c \plus{} d} \plus{} \frac {d }{c \plus{} 2d \plus{} a} \plus{} \frac {a}{d \plus{} 2a \plus{} b}$[/quote]

We have:
$ \frac {a}{a \plus{} 2b \plus{} c} \plus{} \frac {b }{b \plus{} 2c \plus{} d} \plus{} \frac {c }{c \plus{} 2d \plus{} a} \plus{} \frac {d }{d \plus{} 2a \plus{} b} \geq$
$ \geq \frac {(a \plus{} b \plus{} c \plus{} d)^2}{\sum a^2 \plus{} 2(ab \plus{} ac \plus{} bc \plus{} cd \plus{} ad \plus{} bd)} \geq 1$
This inequality is true!
$ \frac {b}{a \plus{} b \plus{} b \plus{} c} \leq \frac {b}{4(a \plus{} b)} \plus{} \frac {b}{4(b \plus{} c)}$
Similar,we have:
$ \frac {c}{b \plus{} c \plus{} c \plus{} d} \leq \frac {c}{4(b \plus{} c)} \plus{} \frac {c}{4(c \plus{} d)}$
$ \frac {d}{c \plus{} d \plus{} d \plus{} a} \leq \frac {d}{4(c \plus{} d)} \plus{} \frac {d}{4(d \plus{} a)}$
$ \frac {a}{d \plus{} a \plus{} a \plus{} b} \leq \frac {a}{4(d \plus{} a)} \plus{} \frac {a}{4(a \plus{} b)}$
So we have $ 1\geq\frac {b}{a \plus{} 2b \plus{} c} \plus{} \frac {c }{b \plus{} 2c \plus{} d} \plus{} \frac {d }{c \plus{} 2d \plus{} a} \plus{} \frac {a}{d \plus{} 2a \plus{} b}$
Done![/quote]

Very nice :lol:

Similar problem:
[b]Problem 1(Pham Kim Hung)[/b] Let $ a;b;c$ be positive real numbers.Prove that:
$ \frac {a \minus{} b}{\sqrt {b \plus{} c}} \plus{} \frac {b \minus{} c}{\sqrt {c \plus{} a}} \plus{} \frac {c \minus{} a}{\sqrt {a \plus{} b}} \geq 0$
[b]Problem 2 [/b]  Prove that:
$ \frac {a^2 \minus{} b^2}{\sqrt {b \plus{} c}} \plus{} \frac {b^2 \minus{} c^2}{\sqrt {c \plus{} a}} \plus{} \frac {c^2 \minus{} a^2}{\sqrt {a \plus{} b}} \geq 0$

[quote="chien than"]Similar problem:
[b]Problem 1(Pham Kim Hung)[/b] Let $ a;b;c$ be positive real numbers.Prove that:
$ \frac {a - b}{\sqrt {b + c}} + \frac {b - c}{\sqrt {c + a}} + \frac {c - a}{\sqrt {a + b}} \geq 0$
  [/quote]
Pham Kim Hung's first inequality is equivalent to this one:
${{ \frac{a+c}{\sqrt{b+c}}+\frac{b+a}{\sqrt{c+a}}+\frac{c+b}{\sqrt{a+b}}\geq \sqrt{b+c}+\sqrt{b+a}}+\sqrt{a+c}}$,which is true
from CBS inequality.

[quote="chien than"]
[b]Problem 2 [/b]  Prove that:
$ \frac {a^2 \minus{} b^2}{\sqrt {b \plus{} c}} \plus{} \frac {b^2 \minus{} c^2}{\sqrt {c \plus{} a}} \plus{} \frac {c^2 \minus{} a^2}{\sqrt {a \plus{} b}} \geq 0$[/quote]
Very nice too. :)
Proof:
This inequality is equivalent to this one:
$ \frac{(a\plus{}b)(a\plus{}c)}{\sqrt{b\plus{}c}}\plus{}\frac{(a\plus{}b)(b\plus{}c)}{\sqrt{a\plus{}c}}\plus{}\frac{(c\plus{}b)(a\plus{}c)}{\sqrt{b\plus{}a}}\geq$ $ (a\plus{}b)\sqrt{b\plus{}c}\plus{}(c\plus{}b)\sqrt{a\plus{}c}\plus{}(a\plus{}c)\sqrt{b\plus{}a}$,which is obviously true from CBS too.

[quote="Erken"]which is true from CBS inequality.[/quote]

What is this?

[quote="FOURRIER"][quote="freemind"]
After amplifying each by its numerator, and applying Cauchy [/quote]
Do you mean apply CS like this:
$ (\frac {3a \plus{} c}{a \plus{} 2b \plus{} c} \plus{} \frac {3b \plus{} d}{b \plus{} 2c \plus{} d} \plus{} \frac {3c \plus{} a}{c \plus{} 2d \plus{} a} \plus{} \frac {3d \plus{} b}{d \plus{} 2a \plus{} b})(\sum (3a \plus{} c)(a \plus{} 2b \plus{} c))\ge...$??[/quote]
Yes,he meant to apply CBS this way :wink:


Find the volume in cubic inches of a cube whose surface area is $ 96$ square inches.

Each face has an area of 96/6=16, so each side is 4. Thus 4^3=64.


Can anyone provide me with easy to understand links to "rough set" and "Fuzzy set" ?

Fuzzy set :
It's hard to explain in a few words here.
I've got a good book about fuzzy sets, but it's in Polish :)
Maybe google knows something...


geometry
3D geometry
sphere
geometry unsolved

A minimum distance of a finite set points is defined as follows: Consider the length of all distances between each two points of this set. The shortest distance is denoted as minimum distance.

a.) Prove that on a sphere's surface K with radius R one can find 8 points whose minmum distance is greater than 1.15 R.
b.) Prove that on a sphere's surface K with radius R one can find 8 points whose minmum distance is greater than 1.2 R.
c.) Prove that the limit for the minimum distance for the situation described in a.) and b.) is 2/7* \sqrt 7 *  \sqrt (4 - \sqrt 2)

Are the distances mesured on the surface of the sphere or is it the usual euclidean distance?

Pierre.

To begin interrapted discussion...

I have just managed to find construction with the minimal distance $d=\frac{2R}{\sqrt 3}\approx 1.155$.
It looks like the following. Choose two opposite points $A$ and $B$. Construct now three edges from $A$ and $B$ or length $d$ which form equal angles. And the figure should be "antisymmetrical", it is if we watch at it from the top of the sphere, where $A$ is placed, then we should see six radial edges.


Can I ask where are all the other posts from the old forum? I hope we will have access to them.

look carefully: the inequalites are all in the inequalities.

only the topics that were in the proposed problems section (in advanced and math for college) I moved them all to algebra. from there the moderators are supposed to relocate them to their places (solved, unsolved, geometry etc.).

Upps, sorry. I noticed that earlier, but I forgot where I posted this. I hope I will acomodate with the new format (better). Interesting... Why exactly inequalities? I really like other things, too.

[quote]Why exactly inequalities? I really like other things, too.[/quote]
just gave a quick example :)

there was a section includes many math resource links.
where is it now?

Hm, will all the posts "outside" the subforums be moved into the subforums eventually?  It's a bit strange to leave a pile of problems "outside" and just continue the rest "inside", as it might get messy if people alternately post "outside" and "inside the subforums".

Actually I think it might be better if the "Post New Topic" option inside the forum, but outside the subforums could be removed...


If there is $ 3$ mouses eating $ 3$ cheeses in $ 3$ minutes. What is the time required if there is $ 100$ mouses eating $ 100$ cheeses?

I think its $ 3$ minutes also? But I don't know a good mathematical method to do it. So its might be wrong  :(

You are right.
$ 1$ mouse eats $ 1$ cheese for $ 3$ minuts
then $ 100$ mouses eats$ 100$ cheeses for $ 3$ minuts.

You should just go step by step
$ 5$ mouses eats $ 7$ cheeses in $ 9$ minutes , then
$ 1$ mouse eats $ \frac {7}{5}$ cheeses in $ 9$ minutes
$ 1$ mouse eats $ \frac {7}{5 \cdot 9}$ cheeses in $ 1$ minute

         * * *
Go on this way -step by step .I hope this is clear for you !

[quote="hasan4444"]Thanks a lot.
But how I can make it mathematically. For example if I face different numbers like:

If there is $ 5$  mouses eating $ 7$  cheeses in $ 9$  minutes. What is the time required if there is $ 91$ mouses eating $ 28$ cheeses?

Now I can't solve this  :([/quote]

Then in 9 minutes, 1 mouse eats $ \frac{7}{5}$ cheeses. Thus in 9 minutes, 20 mice eat 28 cheeses.

Since there are 91 mice, the time is $ \frac{91}{20} \cdot 9\equal{}\frac{819}{20}\equal{}40.95$ minutes.

This can be generalized using quite a few variables.

Thanks a lot "enndb0x" and "BOGTRO".
LOL I was mouses and its mice  :rotfl:

[quote="aguoman"]We could also use a ratio: mice/cheeses/minutes, which is equivalent to (mice*minutes)/(cheese). Basically, this just makes the common sense (more mice=more cheese, more mice=less time) a bit more rigorous :lol:[/quote]
Interesting
Good thinking anyway.  :lol:

[quote="person1133"]How about something like: 5 mice eat x slices of cheese in 9 minutes and x mice eat 10 slices of cheese in 2 minutes. Find x.[/quote]
Is $ x$ is the same in both places?

[quote]Then in 9 minutes, 1 mouse eats 7/5 cheeses. Thus in 9 minutes, 20 mice eat 28 cheeses.

Since there are 91 mice, the time is 91/20 * 9 = 819/20 = 40.95  minutes[/quote]

That doesn't make sense. If 20 mice eat 28 cheeses in 9 minutes, then 91 mice should be able to eat 28 cheeses in less than 9 minutes, not more.

Let there be $ m$ mice, $ c$ pieces of cheese that are eaten, and $ n$ minutes.
As $ n$ is kept the same, as $ m$ gets greater, so does $ c$.  However,
as $ c$ is kept the same, as $ m$ gets gets bigger, $ n$ gets smaller.
Therefore, the relationship between $ c,m$ and $ n$ is $ \frac{mn}{c}$, a
fraction which is always constant.
So, $ \frac{3*3}{3}\equal{}3$ is our constant.  Therefore, we have the equation:
$ \frac{100n}{100}\equal{}3 \rightharpoonup \boxed{n\equal{}3}$

[quote="hasan4444"]But how I can make it mathematically. For example if I face different numbers like:

If there is $ 5$  mouses eating $ 7$  cheeses in $ 9$  minutes. What is the time required if there is $ 91$ mouses eating $ 28$ cheeses?

Now I can't solve this  :([/quote]You can look at that type of problem in terms of "direct variation" and "inverse variation."
If doubling one variable makes the other double (for example, twice as much time, means twice as much cheese eaten), you have a direct variation. If doubling one variable makes the other half as much (for example, twice the number of workers will finish the job in half the time), you have an inverse variation.
For a direct variation, you'll make the dependent variable a factor in the function.
For an inverse variation, you make the inverse of the variable a factor. In other words, you put the variable in the denominator.
If the volume, $ V$ of a balloon varies directly with the temperature, $ T$, and inversely with the pressure, $ P$, you would write:
$ V\equal{} \frac{kT}{P}$ $ (k$ is a constant to be determined$ )$ to show the "joint variation" of $ V$ with $ P$ and $ T$.
The number of cheeses eaten, $ c$, varies directly with the time in minutes, $ t$, and also directly with the number of mice, $ m$. You can write that as:
$ c\equal{}k t m$ where $ k$ is a constant to be determined.
With the data given, you can find $ k\equal{} \frac{7}{45}$, so $ c\equal{} \frac{7tm}{45}$ in your problem.
Now you have $ c$ as a function of $ t$ and $ m$.
You can solve for $ t$ or for $ m$ if needed, to get
$ t\equal{} \frac{45c}{7m}$  and  $ m\equal{} \frac{45c}{7t}$
If they give you any two of the variables, you can find the other one.

All of these problems revolve around finding a constant $ k$, and then setting up an equation with the information you know.


[color=darkblue]Slove this equation: $ x^3\minus{}3x^2\plus{}3x\minus{}16\sqrt{x}\minus{}9\equal{}0$[/color]

Square it, and we get that as $ x$ divides every part of $ LHS$ except for $ 81$(we don't know) and it divides $ RHS$ than $ x\mid 81$, so try all cases(I assumed that $ x\in \mathbb{Z}$).

YOU CAN JUST REPRESENT THE SQUARE ROOT OF THE UNKNOWN(X) BY A VARIABLE(P) SO THAT (X)=P^2.
SO FOLLOWING THAT ARGUMENT,(X)^2=P^4.(X)^3=P^6.

hello, with $ \sqrt{x}\equal{}t$ you have to solve
$ t^6\minus{}3t^4\plus{}3t^2\minus{}16t\minus{}9\equal{}0$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, with $ \sqrt {x} \equal{} t$ you have to solve
$ t^6 \minus{} 3t^4 \plus{} 3t^2 \minus{} 16t \minus{} 9 \equal{} 0$.
Sonnhard.[/quote]

Yes it right. Equation $ t^6 \minus{} 3t^4 \plus{} 3t^2 \minus{} 16t \minus{} 9 \equal{} 0$ has a root $ t\equal{}3$?

[hide]and finally,(x)=9 :D [quote]all laws of nature are the same in all uniformly moving frames of refrence[/quote][/hide]


Hola,

Tal vez una forma de subir el número de personas que participan en este foro sea si nos conocemos un poco. Pues por que no nos presentamos.

Bueno, allí voy yo. Mi nombre es David Jiménez (el tipo de la foto de al lado), tengo 24 años, lo que me convierte en el mas viejo de los que participan en este foro con frecuencia. Lastimosamente, entre en el mundo de olimpiadas algo tarde como para hacer mucho como participante, pero Olimpiadas tuvo dos efectos positivos en mí: Me motivó a estudiar Matemática, y me hizo participar durante algunos años en el entrenamiento de los equipos que participan en la Centro y en la Ibero por Costa Rica. Actualmente vivo en Atlanta, Georgia, USA, donde curso mis estudios doctorales, de momento orientados hacia Teoría de Códigos. Pues al menos que alguien tenga una pregunta, creo que eso es todo lo que aplica decir por el momento. ¿Que tal ustedes?

Hola!, mi nombre es Leonardo Urbina y tengo 17 años (pronto 18 :)) . Me gradue de Bachiller en Julio de 2004, pero me quede un año fuera de la Universidad para poder ir por segunda vez a la IMO. Las Olimpiadas en las que he participado son la Centroamericana de 2003, la Ibero de 2003 y la IMO de 2004. Me gusta mucho la matematica, aunque no tengo un area preferida en particular, y ademas me gusta la musica, durante mi vida he tocado varios instrumentos: Violin, Piano, Cuatro y Guitarra :D. Quien mas se presenta??

Hola yo soy Pascual Restrepo de Colombia tengo 18 años y me gustan las matematicas mucho desde hace aproximadamente 3 años en los que he podido representar a colombia en varios eventos....Sin embargo quienes me conocen saben que soy un tipo tranquilo, alegre y amante de los placeres de la vida...jaja :D  :D . Tambien acabo de terminar el colegio, espero participar en esta IMO (2005) y luego estudiar en Colombia que es para mi el paraiso! Suerte!!!

Yo soy Manuel Rivera de Puerto Rico (el mejor pais del mundo) y tengo 16 anos y pues llevo estudiando matematicas desde hace poco (1 ano) pero cada dia aprendo mas y me estoy fajando bastante en este tiempo. Me interesa mucho las matematicas en especial la combinatoria y grafos, y lo menos que me gusta es la geometria jaj pero necesito aprender y mejorar mas ahi. Fui a Atenas a la IMO2004 y espero ir a Mexico en el 2005 y a Eslovenia en el 2006. (y a las dos proximas iberos). Ademas de la mate me interesan los deportes, la musica (reggaeton puertorro, vdd pascual?jaja) y pasarla bien por ahi. Y pues nada espero aprender aki en mathlinks mas de mate cn ustedes que ya llevan mas tiempo!

Hola
Soy David Torres de México. Vivo en León, Guanajuato y Me gustan mucho las mátematicas. A la fecha he representado a México en la Centro del 2004 y espero este año ir a la Internacional en Mérida (lejísimos por cierto (aunque puede ser que alguien de Puerto Rico o Cuba le quede más cerca que a mí)) y con suerte a la Ibero

Hola a todos, mi nombre es José Luis y soy de El Salvador, he participado en la centro 2003 y en la imo2005 y tambien soy parte del equipo de mi país para la ibero 2005, me gusta mucho la matemática y me perece "chivo" que existan sitios como este... :lol:

Bueno para extender la informacion de los miembros de este forum..

aqui pongo la direccion de un post
con la informacioin de los peruanos que estan participando en esta comunidad
espero les interese conocer sobre este grupo !!!:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56422

CarlosBrav :)
Lima- PERU

me parece genial este foro, y mejor aún haber encontrado gente que habla español...  :)

mi nombre es Francisco Campos, tengo 16 años, y vivo en Santiago de Chile

me encanta todo esto relacionado a olimpíadas matemáticas, fundamentalmente las internacionales

encontré el foro por casualidad (qué raro haha), y me gustó mucho saber que hay gente que se dedica a esto como una comunidad...

currículum... mmm... sólo una Olimpíada del Cono Sur  :blush:
pero espero ir este año a la Ibero...  :lol:

bien... eso sería, amigos... saludos!!

Hola soy nuevo aquí  :blush:

Me llamo Sebastián Illanes, tengo 14 años y soy de Chile, amigo de Guía Rojo... encontré este foro gracias a otro amigo , y me pareció espectacular... este es mi primer año serio en las Olimpiadas y espero llegar a la Cono Sur el próximo año  :) . Bueno, que más decir, soy principiante en esto y espero subir mi nivel, que por lo cierto es bajo  :wink:  :blush:

Bueno, eso sería

Saludos :lol:

Bueno, nunca me presenté, así que acá vamos:

Me llamo José, tengo 17 años, y soy de Argentina. Vivo en una ciudad de 23.000 personas, que se dedica al turismo, por su patrimonio histórico. Participo en olimpíadas desde los 10 u 11 años, todos los años. Juego al básquet, me gusta estar en Internet, escuchar música, y estar con mis amigos y familiares. ¿Logros en olimpíadas? Ninguno. Me gusta practicar matemática, y charlar (hablar de modo informal) con otras personas sobre matemática (de cualquier cosa, no solo de problemas y soluciones exactamente). Bueno, eso es todo, cualquier cosa me preguntan. Saludos a la Comunidad!

Hola...Me llamo César Núñez, tengo 19 años y soy de Santiago, Chile. Estudio Ingeniería en la Universidad de Chile. Desarfotunadamente, me enteré muy viejo de estas competencias, por lo que apenas tengo un logro en una comptencia local; sin embargo, eso no implica que no quiera seguir aprendiendo cada vez más. Un saludo a toda la comunidad.

Un amigo mío también se llama así, y tiene también 19 años! Linda casualidad

[quote="mysmartmouth"]A great time to practice my Spanish!!

Hola.  Me llamo Sean.  Tengo 19 años.  Vivo en Carolina del Sur en el EE.UU, pero yo soy Indian.  Mi asignaturas favoritas son matematicas y ciencias.  No me gusta musica y Ingles.  Adios amigos!  Corriges mis errores, por favor!![/quote]

Soy indio.
No me gustan ni la música y ni el inglés.
Corrige mis errores...

Tenés un buen español! (You have a good spanish!)

[quote="José"]
Corrige mi errores...
[/quote]

So I assume we are using the imperative tense of corrigar, which is corrige?  And we don't use mis, even though we have a plural noun?

corrige mis errores / corregí mis errores

(it was just a typo). Yes, you use the imperative (usás el imperativo)

Hola a todos, no habia visto este topico de presentarse. A ver, yo soy Daniel Hernandez, competi por Cuba en las IMO, ahora estoy en Madrid haciendo el doctorado en física teórica. Pero de vez en cuando entro y me quedo con algun problemita que me guste para no perder la costumbre. 24 años.
La cosa más divertida que hay es coger una casa de campaña y una mochila e irse a la montaña o a cualquier lugar que incite a tu curiosidad por descubrir. Por suerte eso todavía lo podemos hacer en los paises latinoamericanos, en Europa ya poco monte queda inexplorado.
La segunda cosa mas divertida es la fiesta con tus amigos. Bailar o tomar el te, da igual mientras estes compartiendo con la gente de tu bando
La tercera cosa mas divertida son los placeres intelectuales, leer, en el tema cientifico aparte de la fisica, me encantan desde luego las matematicas y disfruto programando.
Y ya! que vaya pedazo de presentacion!

Daniel (La cosa blanca detras de mi Avatar es un rapido en el  Nilo :))

[color=red] Ah se me olvidaba preguntar, alguien de Cuba aqui?? alguien de España??[/color]

[color=green]Hola a todos!!!  :) Mi nombre es Gustavo Espínola, tengo 17 años, vivo en Capiatá, a 21 kilómetros de la capital, Asunción. Si bien estoy registrado desde hace un buen tiempo, estaba sólo como oyente... Mas me gusta el grado de respeto con el que se trabaja, lo que hace ideal el ambiente para compartir mis experiencias. Es una satisfacción para mí estar con ustedes, y espero aprender un poco más y ayudar en lo que pueda...

Como experiencia en olimpiadas matemáticas, además de la Nacional, sólo Cono Sur en el 2005, mi objetivo será la IBERO de Portugal... ya que me voy a la universidad desde el próximo año (chau IMO...).

Creo que es todo... Saludos!!![/color]

[color=blue]Soy Rafael Arias. Había creado, sin saber que existía este tópico, uno similar, que es el siguiente:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=110616[/url]

Discúlpenme por mi error.  :blush:  Aún era nuevo en el foro cuando sucedió eso.[/color]

Soy Jorge Castillo. No soy olimpico, sino que universitario  :blush: . Estudio ingenieria civil matemática en la universidad de concepcion, en 2 ° año, tgo 19 años, y eso..saludos

Hola, soy Roland Hablutzel, venezolano, y participé en la OMCC y en la OIM en el 2004, y en el 2005 en la IMO y en la OIM, y con premios en todas ellas (aunque nada importante :blush: ). A partir de ahí empezé la universidad en física, ya que conociendo tan profundamente la matemática no quise estudiarla, me volvió loco, jajaja, sin embargo me ha sido útil todo lo que he aprendido en ellas. En mathlinks entré un poco tarde... pero me sirve para no perder el ritmo de resolver problemas. Me da gusto esta reunión de hispanohablantes ya que hay pocos sitios así.

Hola!

Soy Esteban Arreaga, tengo 17 años, vivo en la ciudad de Guatemala, Guatemala. Hasta ahora solo he participado en una centro y una ibero, espero estar al menos en otra ibero. Me encanta la matemática, en especial las cosas algebraicas, soy malo en geometría aunque tambien la disfruto. Entro a la universidad en el 2009, así que talvez todavía tengo chance.

Saludos a todos!!

 :)


function
geometry
pigeonhole principle
combinatorics proposed
combinatorics

We have a closed path on a vertices of a $ n$×$ n$ square which pass from each vertice exactly once . prove that we have two adjacent vertices such that if we cut the path from these points then length of each pieces is not less than quarter of total path .

A really nice one, it is hard to explain without pictures but I will try to give a sketch of the proof.
Initially take the adjacent squares with the maximum lenght of the smaller path between them and assume that this small path's lenght is lesser than one quarter of the total path. Now, divide into two parts, first if at least one of these two squares is not on an edge and second both of them are on an edge. First case is really simple, a casework using the maximality of these two squares leads directly to a contradiction. (Yet, it's hard to explain it without too many pictures so i'm leaving this part now, if a problem occurs, tell me and i will answer it.) The main part is the second part. Let's look at the picture, the squares we have selected are the ones with a red and black edge connected to them. (since there is no edge between them, the edges that pass from them must be like red and black edges) Start from one of the black edges and go along with the path without passing through our selected squares. We can't reach a red edge before we reach the other black edge. So divide the path into two parts, the black edges and the path connects them without passing through our selected squares; and the red edges and the path connects them without passing through our selected squares. Say the first part black path and the second part red path. We can say from the maximality that the red path is the smaller one. Otherwise we could find a bigger smaller part by taking the squares that are on the just right of our selected squares.The squares that are on the black path are black squares and the others are red squares, two selected squares are also black.
Claim 1: Red squares are a connected set and there is no black squares inside red squares. Moreover there is no red square that is adjacent to the edge of the big square.
Proof: Easy properties that can be proved by using the black squares are also connected. So I'm again leaving the proof.
Therefore, the red squares has a shape like the grey colored squares in the picture. Say P to the polygon that formed of the edges of the shape of red squares. Now let's define a f function for the edges of lenght 1 on P (like the yellow edge on picture) A be an edge like that, one side of A is a red square and one side is a black square from the claim 1. Start from the edge next to the upside selected square say this edge A_1, go along P from the direction that doesn't pass through the edge of downside selected square and tell the i th reached edge lenght 1 A_i. (For example in the picture the yellow edge is A_7, while the downside selected squares' edge is A_24)
Claim 2: If i>j, the black path between the upside black edge (the one that drawn on picture) and the black square that contains A_i is longer than the one with the square that contains A_j.
Proof: Again start from the upside black edge and go along the black path, assume that we first reached A_i. Since A_i and A_j are on the edge of P and the black path can't go inside P, it is clear that it's impossible to connect A_j and the downside black edge. Contradiction. So claim 2 is true. Moreover the same for the red path is also true, I mean If i>j, the red path between the upside black edge (the one that drawn on picture) and the red square that contains A_i is longer than the one with the square that contains A_j.
f(A_i) is equal to the lenght of the path between the two squares that contain A_i as an edge and passing through the upside red and black edges (on the picture again). From claim 2, f(A_i) is a non decreasing function. Because of the maximality of the first selected squares, f(A_i) must be either lesser than one quarter or more than 3 quarter of the total path.
Claim 3: There is an i such that f(A_{i+1})-f(A_i) is bigger than the half of the total path.
Proof: f(A_1)=1 and f(the downside selected squares' edge)= Total path-1. So f(A_i) starts to be more than 3 quarter of the total path at some i, take the smallest such i. Then, f(A_{i-1}) is lesser than 1 quarter of the total path. Result follows.
Now we have a little to do remaining. Assume that for i, f(A_{i+1})-f(A_i) is bigger than the half of the total path. There are 3 cases
i) A_i and A_{i+1} are not perpendicular to each other. The red path between the red squares that contains A_i and A_{i+1} is lesser than one quarter of the total path, so the black path between the black squares that contain A_i and A_{i+1} must be bigger than one quarter of the total path. Moreover the other path between the black squares that contain A_i and A_{i+1} includes the whole red path, so it can't be the smaller path between the black squares that contain A_i and A_{i+1} because of the maximality of the first selected squares. So it's again a contradiction!
ii) A_i and A_{i+1} are perpendicular and has an angle 90 degrees in P. The path between the black squares that contain A_i and A_{i+1} as an edge is longer than the half of the total path. Let's look at the black square that is a neighbour to the both black squares that contain A_i and A_{i+1} as an edge. Say this square X, an argument like in claim 2 shows us that the black path between the black squares that contain A_i and A_{i+1} as an edge passes through X. So the black path between X and one of the black squares that contain A_i and A_{i+1} as an edge must be bigger than one quarter of the total path. So it's again a contradiction as in i)
iii) A_i and A_{i+1} are perpendicular and has an angle 270 degrees in P. since the red path is lesser than one half, this is absurd.
Finally we are done, if it is unclear, I am sorry. I can answer your questions if you ask.

very very nice problem :). here's my solution:

it's easy to see that $n$ must be even for such a path to exist, so from now for comfort we write it as $2n$.
this closed path divides the area of the square into two parts, inside and outside. if we consider each square in the inside part as a vertice of a graph and connect two vertices if their corresponding squares are adjacent, we get a tree. (one can easily show that this graph is connected and does not have any cycles). by pick's theorem we get that this tree has $2n^2-1$ vertices. the maximum degree of this tree is $4$, since each square can be adjacent to at most $4$ squares.

[i]Lemma[/i]: if the maximum degree of a tree is $4$, then there is an edge in it that divides the graph into two parts, each having at least $\frac{n-1}{4}$ vertices.

[i]Proof[/i]:if an edge has the above property, we call it good. let $v$ be an arbitrary vertex and make it the root of the tree. let it's neighbors be $v_1$ (possibly $v_2,v_3,v_4$). let the subgraph of our tree with the root $v_i$ be $V_i$. now, by pigeonhole principle, one of the $V_i$'s, has at least $\frac{n-1}{4}$ vertices, namely $V_1$. if  the edge $vv_1$ is not good, we conclude that $V_1$ must have at least $\frac{3(n-1)}{4}$ vertices except $v_1$. now do the same, again by pigeonhole principle, we get that in one of the subgraphs with roots neighbours of $v_1$, there is at least $\frac{n-1}{4}$ vertices. if the edge $v_1v_2$ is not good, again we conclude that this subgraph must have at least $\frac{3(n-1)}{4}$ vertices except $v_2$ and so on. this process must stop somewhere and we must get a good edge, since if not, we get a graph with infinite vertices. so the proof is complete.

now back to the main problem, there is an edge $ab$ dividing the graph into two parts each having at least $\frac{n^2-1}{2}$ vertices. this edge cuts the lattice edge between two vertices, namely $t$ and $s$. we claim that $t$ and $s$ have the desired property. delete the edge $ab$ of the graph. now we have two trees, each having at least $\frac{n^2-1}{2}$ vertices. again using pick's theorem, we easily find that each of the two paths connecting $t$ and $s$ have at least $n^2$ edges, so the proof is complete. :)

This beautiful problem has a solution using this problem as its main idea : https://artofproblemsolving.com/community/c6h85771p499536


geometry
parallelogram
vector
geometric transformation
rotation
complex numbers

Four squares are constructed externally on the sides of a parallelogram. Show that the centers of these squares form a square.

I know there's a simple vector solution, but what about a purely Euclidean solution?

Let the intersection of the diagonals of the parallelogram be $O$, the origin of the complex plane $\Pi$.  Let the vertices of the parallelogram be $a,b,-a,-b$.  Also, let the centers of the squares be $s_{1},s_{2},s_{3},s_{4}$.  Then note that the diagonals are orthogonal since upon division of the complex numbers $s_{i}-s_{j}$ and $s_{m}-s_{n}$, you obtain $\pm i$.  The conclusion follows.

I haven't thought of the Classical Geometry proof, yet.

Masoud Zargar

hmm... seems like there should be a solution involving rotation. Maybe you could show that if $X$ is the intersection of the diagonals of the parallelogram, then a $90^\circ$ rotation around $X$ would result in each center being sent to another center. While I can't figure out how I could prove this, I know that a $180^\circ$ rotation clearly does this (because it leaves the parallelogram unchanged), which at least provides an easy proof that the centers form a parallelogram.