[size=150]Let $ a,b,c > 0$,
$ \sqrt {\frac {2a}{b \plus{} c}} \plus{} \sqrt {\frac {2b}{c \plus{} a}} \plus{} \sqrt {\frac {2c}{a \plus{} b}}\le \frac {(a \plus{} b \plus{} c)^2}{bc \plus{} ca \plus{} ab}$
BQ[/size]

[size=150]more Strong:
$ \sqrt{\frac{a}{b\plus{}c}}\plus{}\sqrt{\frac{b}{c\plus{}a}}\plus{}\sqrt{\frac{c}{a\plus{}b}} \le \frac{\sqrt{6}}{2}\frac{a\plus{}b\plus{}c}{\sqrt{ab\plus{}bc\plus{}ca}}$
BQ[/size]

[quote="xzlbq"][size=150]more Strong:
$ \sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{c \plus{} a}} \plus{} \sqrt {\frac {c}{a \plus{} b}} \le \frac {\sqrt {6}}{2}\frac {a \plus{} b \plus{} c}{\sqrt {ab \plus{} bc \plus{} ca}}$
BQ[/size][/quote]
It's obviously trues because by AM-GM, easily to see that:
$ \sum\ \frac{1}{a\plus{}b} \le \frac{3}{2}.\frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca}$
 :)

[size=150]thank you so much arqady!
you so say,show  it is not well.replace it.

Let $ x,y,z > 0$,prove that:
$ \frac {\sqrt {x(x + y)}}{y + z} + \frac {\sqrt {y(y + z)}}{z + x} + \frac {\sqrt {z(z + x)}}{x + y} \geq \frac {3}{4}\sqrt {10 - \frac {16xyz}{(y + z)(z + x)(x + y)}}$

please
BQ[/size]

[quote="xzlbq"]
Let $ x,y,z > 0$,prove that:
$ \frac {\sqrt {x(x + y)}}{y + z} + \frac {\sqrt {y(y + z)}}{z + x} + \frac {\sqrt {z(z + x)}}{x + y} \geq \frac {3}{4}\sqrt {10 - \frac {16xyz}{(y + z)(z + x)(x + y)}}$
[/size][/quote]
Dear my friend, did you check it carefully? Sorry but I don't think this ineq is trues  :maybe:

Dear nguoivn:this ineq holds,but I have not proof.
BQ

Wow, if it's trues, I think it's a very strong result. I'll try. Thank you for your check, my dear friend  :)

I jsut have nice proof for the little weaker  :) :
\[ \frac{\sqrt{a(a\plus{}b)}}{b\plus{}c}\plus{}\frac{\sqrt{b(b\plus{}c)}}{c\plus{}a}\plus{}\frac{\sqrt{c(c\plus{}a)}}{a\plus{}b} \ge \sqrt{\frac{16}{3}\minus{}\frac{20}{3}.\frac{abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}}\]
Indeed, with your result, it's very hard to find nice solution  :oops:

[size=150]Of your problem-solving ability , be very good. and your hard work, be very good too.

BQ[/size]


integration
logarithms
calculus
real analysis
real analysis unsolved

Well, integrate by parts to get

$\frac{2}{3}x^{\frac{3}{2}}\ln{(1+x^{2})}-\frac{4}{3}\int \frac{x^{\frac{5}{2}}}{1+x^{2}}dx$.

Unfortunately, that last integral appears to come out to something very nasty.

that's right paladin,first i tried to do that, but I didn't (cuz it got so difficult),
I said let $t=\sqrt{x}$ so $dt=\frac{dx}{2\sqrt{x}}$
and we have
$I=\int\sqrt{x}\ln(1+x^{2})dx=\int\frac{x\ln(1+x^{2})}{\sqrt{x}}=2\int t^{2}\ln(1+t^{4})dt$

and at last i came to this [b]nasty[/b] answer:

$I=2[ \frac{t^{3}}{3}\ln(1+t^{4})-\frac{4}{3}[\frac{t^{3}}{3}-\frac{1}{4\sqrt{2}}\ln|t^{2}+1-\sqrt{2}t|$
$-\frac{1}{4}\times\frac{2}{\sqrt{2}}\arctan\frac{t-\frac{1}{2}}{\frac{\sqrt{2}}{2}}+\frac{1}{4\sqrt{2}}\ln|t^{2}+1+\sqrt{2}t|$
$-\frac{1}{4}\times\frac{2}{\sqrt{2}}\arctan\frac{t+\frac{1}{2}}{\frac{\sqrt{2}}{2}}]]$+$c$

where you put $t=\sqrt{x}$.

I hope it's correct, because if it is correct i'll get full mark in my first Math exam in University :D


linear algebra
matrix
vector
algebra
system of equations

I have been wondering, why do you need, in general, at least n equations to find a solution for n variables?  It's something we all take for granted, but I don't know why it's true.[/i]

You need at least $n$ equations in $n$ unknowns to find a unique solution.  You need the rank theorem from linear algebra to really explain why (otherwise, you can find a free variable)

[quote="DPopov"]A linear system of equations is a set of $n$ linear equations in $k$ variables (sometimes called "unknowns"). Linear systems can be represented in matrix form as the matrix equation

\[Ax=b\]

where $A$ is the matrix of coefficients, $x$ is the column vector of variables, and $b$ is the column vector of solutions.

If $k < n$, then the system is (in general) overdetermined and there is no solution.[/quote]

But that doesn't explain why at all, which is what vinny wanted.

As blahblahblah said (I always have to count those blahs every time I quote your name..  ;) ), you really need a theorem from linear algebra to prove this properly. But just as an idea you can work on - lets say we have 3 equations in 5 variables. Choose any 2 of those variables and give them whatever values you want. Now you have 3 equations in 3 variables - so those first 2 variables could have any values you like; theres not a unique solution. (in general, of course.)

If there were more equations than variables, you could solve the first n equations (if there are n variables) to get a possible answer. Then in general that answer won't fit the rest of the equations; so theres no answer.

Indeed, it's most convenient if we work with homogeneous systems of equations, where we know that we have at least the zero solution.

Then, like TripleM said, in a system of m equations in n unknowns with n>m, we can set (n-m) unknowns arbitrary, and then we have a system of m equations in m unknowns.  Because it is homogeneous, there is the zero solution.  But since each of the (n-m) unknowns are arbitrary, changing them gives a new solution.  Thus there are infinitely many solutions.

FYI, a homogenous system of equations is a set of equations of the form

$\displaystyle \sum^n_{i=1} c_ix_i = 0$, where not all of the $c_i$'s are zero.

Consider:
[tex]\displaystyle x+y+z=2[/tex]
[tex]\displaystyle -x-y+z=4[/tex]

One can see, by quickly adding the equations, that [tex]\displaystyle 2z=6\rightarrow z=3[/tex]

Then:
[tex]\displaystyle x+y=-1[/tex]
[tex]\displaystyle -x-y=1[/tex]

And nothing further can be determined about the variables. One important thing to note, is that you do come up with a solution, but it is not unique. In the example, both [tex](3, -4)[/tex] and [tex](-4,3)[/tex] are solutions.

What you're essentially doing, is taking the system, getting rid of one variable per equation (I believe this is called Gaussian elimination), until you're left with a one variable statement. (i.e. [tex]\displaystyle x=3[/tex]). Without enough equations, after you reduce, you'll end up with an equation containing more than one variable. (In this case, [tex]\displaystyle x+y=-1[/tex])

Hope that helped.


By the Remainder Theorem we can write $Q(x)-n=(x+5)(x+2)P(x)$, where $P(x)$ is also a polynomial with integer coefficients.  If we write $Q(x)=q_0+q_1 x+q_2x^2+\cdots$, $P(x)=p_0+p_1 x+p_2x^2+\cdots$, then
\begin{eqnarray*} q_0-n&=&10p_0\\ q_1&=&10p_1+7p_0\\ q_i&=&10p_i+7p_{i-1}+p_{i-2} \text{ for }i\geq 2\\ \end{eqnarray*}
We can work down this list to determine all $p_m$ and $q_m$ uniquely: each $q_i$ is uniquely determined by $p_0,p_1,...p_{i-1}$ and the fact that $q_i$ is an integer between 0 and 9, and then each $p_i$ is uniquely determined by $p_0,p_1,...p_{i-1}$ and $q_i$.

So if $n=0$, that polynomial is $Q(x)=0$ for all $x$?

Yes. That brings up a point- why does this process terminate? You've got uniqueness, but you haven't proved existence yet.

This was on the first USAMO I took- I don't think I got anywhere on it at the time.

[quote="MithsApprentice"]Prove that for any integer $n$, there exists a unique polynomial $Q$ with coefficients in $\{0,1,\ldots,9\}$ such that $Q(-2) = Q(-5) = n$.[/quote]
[b] [color=blue]Algorithm for $Q(x): =\sum\limits_{k=0}^{m}a(k)x^k \; , \; \; m\ge 2 .$ [/color][/b]  Let  $n\in {\mathbb Z}$  and denote by   $\lfloor \cdot \rfloor$  the [i]floor-function[/i] ( integral  part).
[b][u]STEP 1.[/u][/b] Define
\[ \begin{array}{|lcl|} \hline &&\\ b(0)&=& -\; \left\lfloor \ds \frac{ n }{10}\right\rfloor \\ &&\\ b(1)&=& -\; \left\lfloor \frac{7b(0)}{10}\right\rfloor \\ &&\\ \hline \end{array}\; .  \]
[b][u]STEP 2.[/u][/b] Find the integers   $b(2) , b(3),...,b(\nu),b(\nu +1),b(\nu+2), ...$   using the rule
\[ (***)\; \; \; \; \; \; \begin{array}{|c|} \hline \\ b(k)= -\; \displaystyle \left\lfloor \frac{7b(k-1)+b(k-2)}{10}\right\rfloor \; ,\; 2\le k \\ \\ \hline \end{array}\; .  \]
 Therefore, using  (***)  it's possible to find  the numbers      $b(k) \; ,\; k\ge 2\; ,$ from the table
\[ \begin{array}{||c||c|c|c|c|c|c|c|c||} \hline\hline {\mathbf k}&2&3&4&\ldots & \nu &\nu+1& \nu+2&\ldots\\ \hline {\mathbf b(k)}&b(2)&b(3)&b(4)&\ldots & b(\nu) & {\mathbf 0}& {\mathbf 0}&\ldots\\ \hline\hline \end{array}\; .  \]
If a certain step two consecutive are zero, e.g.     $b(\nu+1)=0$ and  $b(\nu+2)=0$ , then STOP and    $\begin{array}{|c|} \hline m: =\nu: = \mbox{degree} [Q]\\ \hline \end{array} .\;$   A justification is that from      (***) we have  \[ b(\nu+3)=b(\nu+4)=\ldots =0 \; . \]
[b][u]STEP 3.[/u][/b] By means of the system of numbers  $\{b(0),b(1),b(2),\cdots,b(m-2)\}$ we compute numbers
$\{a(0),a(1),a(2),\cdots,a(m)\}$ where \[ \left\{\begin{array}{lclcl} a(0)&=&n+10b(0)&=& n-10\left\lfloor \frac{n}{10}\right\rfloor \\ a(1) &=&7b(0)+10b(1)&=& -7\left\lfloor \frac{n}{10}\right\rfloor-10\left\lfloor \frac{7b(0)}{10}\right\rfloor \end{array}\right.\; .  \]
\[ \begin{array}{|lcl|} \hline a(k) &=&10b(k)+7b(k-1)+b(k-2)\\ a(m-1)&=&b(m-3)+7b(m-2)\\ a(m)&=&b(m-2)\\ \hline \end{array}\; \; ,\; \; 2 \le k\le m-2  \]

 [b]Questions: [i]

1)  Find the polynomial $Q\in {\mathbb Z}[x]$ ,of minimal degree $m$ ,  having all coefficients in the set  $\{0,1,2,...,9\}$ such that
\[ P(-2)=P(-5)=2006\; .  \]
 [/i] [/b]

2) [b](Unsolved)  [color=green]To find  bounds for $m: {\mathbb Z} \to \{2,3,...\}$  , $m: =m(n) \;$  ,  $m$  being  the minimal degree in the posted problem . [/color] [/b]

Let $c_k=|b_k|$. It is obviosly that $c_k\le \frac 45 c_{k-1}<cb_{k-1}$. Therefore $c_m=0$ for $m>\frac{ln(n)}{ln(1.25)}$.

[quote="Rust"]Let $c_k=|b_k$. It is obviosly that $c_k\le \frac 45 c_{k-1}<cb_{k-1}$. Therefore $c_m=0$ for $m>\frac{ln(n)}{ln(1.25)}$.[/quote]
Nice, but a question !  What means $c_k=|b_k$ ? It's  $c_k=|b_k|$ ? It's possible to find an upper bound for $m$ ?

I'm very interested in this exercise, & i'm searching for a   solution with continued fraction...it seems to me the value have the particularity that division by 2 and/or 5 doesn't make a infinite periodic reminder different by 0

[hide="Solution"]Let $Q(x)=\sum_{i=0}^ka_ix^i$. If there exists $Q(x)$ such that $Q(-2)=Q(-5)=n$, taking $\mod 2, 5$ yields $a_0\equiv n\mod 10$; hence $a_0$ is unique. Inducting up to $a_k$, we conclude that $Q(x)$ is unique.

Now for existence. Define a sequence of integers $\{n_k\}_{k\ge 0}$ such that $n_0=n$ and $n_k=\frac{n_{k-1}-a_{k-1}}{-5}$. We claim that $n_k=0$ for some $k$. Indeed, if $|n_{k-1}|\le |n_k|$, we can readily determine that $-2\le n_{k-1}\le -1$ and verify that all such $n_{k-1}$ lead to $0$. Otherwise, if $|n_{k-1}|>|n_k|$, then $|n_k|$ is strictly decreasing. Hence, at some point $n_k$ must lie within the interval $[-2, -1]$ and we can finish with the above. It follows that $n_k=0$ for some $k$. Backwards induction yields our desired $Q(x)$, so we're done. $\blacksquare$[/hide]

This is sort of the mo sol, except I avoided the intricate stuff by reducing it to a casework bash.

We prove both uniqueness and existence by providing a construction:

By the remainder theorem, it follows that Q(x) = $(x^2+7x+10)(r_k \cdot x^k + r_{k-1} \cdot x^{k-1} + \dots + r_1 \cdot x^1 - r_0) + 10r_0+c$, where $r_i \in (0, 1, 2, \dots 9), \forall i \geq 1, c \in (0, 1, 2, \dots 9)$ and $r_0$ is an integer.

Since n = $10r_0 + c$, $r_0\ and\ c$ are uniquely determined. Specifically, c is the residue of n mod 10, and $r_0 = \frac{n-c}{10}$. We now prove that each $r_i$ is uniquely determined.

$\textbf{Lemma One}$: Each  $r_i$ is uniquely determined from $r_0$.

We have that $r_i = -\lfloor\frac{7r_{i-1} + r_{i-2}}{10}\rfloor$. Using the base cases of $r_{-1} = 0$ and $r_0$ to be defined as before, it follows that each $r_i$ is uniquely determined by induction.

The procedure terminates when $r_k$ = 1 or 2, since any other value would make the coefficient of the $x^{k+1}$ term or $x^{k+2}$ term not among the ten digits specified in the problem statement. Any following $r_i$ would have to equal zero (by the procedure given
above), which proves uniqueness.

Now we want to prove existence, or that repeating this procedure actually gives a finite polynomial.

For this, we need another lemma:

$\textbf{Lemma Two}$ The above procedure eventually yields $r_k$ = 1, 2 or a special case for all n.

For this, we note that $\mid r_i \mid < \mid r_{i-1} \mid$, with exceptions at $(r_{i-1}, r_{i-2}) = (-6,-9), (-6,-8), (-5,-9), (-5,-8), (-5,-7), \dots (3, 9) $. This follows from looking at the recursion formula for $r_i$. For these special cases, we can test by hand that the polynomial terminates using the recursion. For all other cases, trivial bounding suffices.

Since $\mid r_i \mid$ is monotonocally decreasing (with a few exceptions which we can compute and show that it terminates), $\mid r_i \mid$ eventually reaches one of our special cases, one or two. If $r_i$ is positive and not one of our special cases, it equals one or two and we're done.
If it's negative one or negative two, we can just test these by hand and we're done.
And we have already shown the special cases.

So the process terminates, hence proving existence.

If we let \[ Q(x) = \sum_{k \ge 0} a_k x^k \] then $a_k$ is uniquely determined by $n \pmod{2^k}$ and $n \pmod{5^k}$. Indeed, we can extract the coefficients of $Q$ exactly by the following algorithm: [list] 	[*]Define $b_0 = c_0 = n$. 	[*]For $i \ge 0$, let $a_i$ be the unique digit 	satisfying $a_i \equiv b_i \pmod 2$, $a_i \equiv c_i \pmod 5$. 	Then, define 	\[ b_{i+1} = \frac{b_i - a_i}{-2}, \qquad 		c_{i+1} = \frac{c_i - a_i}{-5}. \] [/list] The proof is automatic by Chinese remainder theorem, so this shows uniqueness already. The tricky part is to show that all $a_i$ are eventually zero (i.e.\ the ``existence'' step is nontrivial because a polynomial may only have finitely many nonzero terms).
In fact, we will prove the following claim:
[b][color=red]Claim:[/color][/b]  	Suppose $b_0$ and $c_0$ are any integers such that 	\[ b_0 \equiv c_0 \pmod 3.\] 	Then defining $b_i$ and $c_i$ as above, 	we have $b_i \equiv c_i \pmod 3$ for all $i$, 	and $b_N = c_N = 0$ for large enough $N$.
[i]Proof.[/i]  Dropping the subscripts for ease of notation, we are looking at the map \[ (b, c) \mapsto \left( \frac{b-a}{-2}, 		\frac{c-a}{-5} \right) \] for some $0 \le a \le 9$ (a function in $b$ and $c$).
The $b \equiv c \pmod 3$ is clearly preserved. Also, examining the size, [list] 	[*]If $|c| > 2$, 	we have $\left\lvert \frac{c-a}{-5} \right\rvert 	\le \frac{|c|+9}{5} < |c|$. 	Thus, we eventually reach a pair with $|c| \le 2$. 	[*]Similarly, if $|b| > 9$, 	we have $\left\lvert \frac{b-a}{-2} \right\rvert 	\le \frac{|b|+9}{2} < |b|$, 	so we eventually reach a pair with $|b| \le 9$. [/list] this leaves us with $5 \cdot 19 = 95$ ordered pairs to check (though only about one third have $b \equiv c \pmod 3$). This can be done by the following code:
[python]
import functools
@functools.lru_cache()
def f(x0, y0):
	if x0 == 0 and y0 == 0:
		return 0
	if x0 % 2 == (y0 % 5) % 2:
		d = y0 % 5
	else:
		d = (y0 % 5) + 5

	x1 = (x0 - d) // (-2)
	y1 = (y0 - d) // (-5)

	return 1 + f(x1, y1)

for x in range(-9, 10):
for y in range(-2, 3):
	if (x % 3 == y % 3):
		print(f"({x:2d}, {y:2d}) finished in {f(x,y)} moves")
[/python]
Running the program, we see it terminates, hence we are done. $\blacksquare$

We will use notations from the above solution. One can show that:
[list]
[*] $c_i \in [2,-1]$ for all sufficiently large $i$. [hide=Proof] Just consider the $c_k$ for which $|c_k|$ is minized and use $|c_{k+1},c_{k+2}| \ge |c_k|$.[/hide]
[*] $b_1 \in [5,-2]$ for all sufficiently large $i$. [hide=Proof]This is not that direct. If we consider the $b_k$ for which $|b_k|$ is minimized, then we only get $b_k \in [6,-3]$. Now if at any point, if $b_k$ atains a values $\ne 6,-3$, then we are done. So the only worry is when $b_i$ eventually becomes $6,-3,6,-3,\ldots$. But then $a_i$ would become $0,-9,0,-9, \ldots$. But then $a_i \equiv c_i \pmod{5}$ would be contradicted. Thus we are fine here too.[/hide]
[/list]
Moreover, $b_i \equiv c_i \pmod{3}$ is a invariant. So we can just work out these small cases for $(b_i,c_i)$ by hand, as follows:
[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(6cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -10.26, xmax = 12.7, ymin = -1.73, ymax = 11.47;  /* image dimensions */
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451);
 /* draw figures */
draw((-3,3)--(-1,1), linewidth(0.8) + wrwrwr,EndArrow(6));
draw((-1,3)--(-1,1), linewidth(0.8) + wrwrwr,EndArrow(6));
draw((1,3)--(-1,1), linewidth(0.8) + wrwrwr,EndArrow(6));
draw((-1,5)--(-1,3), linewidth(0.8) + wrwrwr,EndArrow(6));
draw((-3.98,4.85)--(-1,3), linewidth(0.8) + wrwrwr,EndArrow(6));
draw((0.3,4.77)--(-1,3), linewidth(0.8) + wrwrwr,EndArrow(6));
draw((2,4)--(-1,3), linewidth(0.8) + wrwrwr,EndArrow(6));
draw((-4,7)--(-3.98,4.85), linewidth(0.8) + wrwrwr,EndArrow(6));
draw((-2,7)--(-1,5), linewidth(0.8) + wrwrwr,EndArrow(6));
draw((0.26,6.99)--(-1,5), linewidth(0.8) + wrwrwr,EndArrow(6));
 /* dots and labels */
dot((-1,1),dotstyle);
label("$(0,0)$", (-1.18,0.45), NE * labelscalefactor);
dot((-3,3),dotstyle);
label("$(2,2)$", (-3.9,2.29), NE * labelscalefactor);
dot((-1,3),dotstyle);
label("$(1,1)$", (-1.06,2.47), NE * labelscalefactor);
dot((1,3),dotstyle);
label("$(-1,-1)$", (0.9,2.27), NE * labelscalefactor);
dot((-1,5),dotstyle);
label("$(4,1)$", (-2.1,4.73), NE * labelscalefactor);
dot((0.3,4.77),dotstyle);
label("$(3,0)$", (0.28,4.75), NE * labelscalefactor);
dot((2,4),dotstyle);
label("$(2,-1)$", (1.68,4.21), NE * labelscalefactor);
dot((-3.98,4.85),dotstyle);
label("$(5,2)$", (-4.98,4.55), NE * labelscalefactor);
dot((-2,7),dotstyle);
label("$(-1,2)$", (-2.62,7.23), NE * labelscalefactor);
dot((-4,7),dotstyle);
label("$(-1,-1)$", (-4.86,7.29), NE * labelscalefactor);
dot((0.26,6.99),dotstyle);
label("$(-2,1)$", (0.34,7.19), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */
[/asy]

Sketch:
Let $Q(x)=a_nx^n+\cdots$. More generally if $Q(-2)=a,Q(-5)=b$, and $a \equiv b \pmod{3}$, then there is a unique $Q$. We can find $Q$ as follows: pick $k \in \{0,\ldots,9\}$ such that $a \equiv k \pmod{2}$ and $b \equiv k \pmod{5}$, which is unique and exists by CRT. Then set $a_0=k$ and transition from $(a,b) \to (\tfrac{a-k}{-2},\tfrac{b-k}{-5})$, which preserves $a \equiv b \pmod{3}$. Uniqueness comes from the fact that $a_0=k$ is in fact necessary, so it suffices to show that this process terminates. This can be done by bounding $a$ and $b$ such that something like $|a+b|$ won't decrease from this operation (since otherwise we just apply it and induct down) and then just manually checking them


Hello

Do you know if the booklet RMC 2007 is available ?

I am interested to buy one.

thx

no, it isn't available yet, i think it will appear after a week or two after the BMO.


While reading a book on logic sometime ago, I came across the terms 'sufficient' and 'necessary', which is connected to "if... then..." statements, i.e. implications.

I don't quite understand the context and usage of 'sufficient' and 'necessary'.  :huh:  Can someone help explain? Thanks in advance! :P

Think about putting these in a sentence:

A is [i]sufficient[/i] for B.
In order for B to be true, it is [i]sufficient[/i] for A to be true.
A implies B.

A is [i]necessary[/i] for B.
If B to be true, it is [i]necessary[/i] that A is also true.
B implies A.

Each block consists of three equivalent statements. These words are functionally equivalent to implication, but come from somewhat different styles.

" sufficient " is like " enough " where the fact/statement given is enough for you to implies one thing . Example , if $k$ is any positive integer then it is " sufficient " that $2k$ is even number .

"neccesary" is only one among the conditions that need to satisy a fact/statement (or it is not enough to use this to conclude a statement but it is also needed .) Like , for $k$ to be a positive odd integer , it is " neccesary" for $k$ to be an integer . This statement($k$ has to be integer) however doesnt implies $k$ is a positive odd integer but only satisfying one among the conditions (which is $k$ has to be integer).

Thanks jmerry and shyong for explaining. I understand now :D


Hi,

I am looking for math materials for my 4th grader. I am looking mostly for books with word problems and all kinds of magazines of the sort of "Mathematical Gazette" that could be found in Europe.

Thank you,

:-)

You should visit

https://cemc.math.uwaterloo.ca/cgi-bin/WebObjects/sigma.woa/wa/orderPubs

These are books by the University Of Waterloo. Just click "browse catalogue". They have quite a few for 4th graders too.


Just an easy problem I found, does anyone has a nice and elegant solution?

A fair dice bearing number 1 through 6 on its faces is thrown repeatedly until the running total first exceeds 12. What is the most likely total that will be obtained?

If our last throw, x, took us to a total greater than 13, then there was exactly the same chance that we threw something that made us equal to 13 instead. So the chances of ending on 13 are at least as big as any other value. (for example, if we were on 9 and threw a 5, we had exactly the same chance of throwing a 4).

On the other hand, by throwing seven 1s followed by a 6 gives us 13, and changing that last throw can't get you more than 12. So 13 has the biggest chance.


Pascal\u0027s Triangle
number theory proposed
number theory

Let $p$ be prime, let $n$ be a positive integer, show that \[ \gcd\left({p - 1 \choose n - 1}, {p + 1 \choose n}, {p \choose n + 1}\right) = \gcd\left({p \choose n - 1}, {p - 1 \choose n}, {p + 1 \choose n + 1}\right). \]

Unless I have blundered, $p$ does not need to be prime. Sketch of proof: Let $q$ be a prime power. Because the multiples of $q$ in Pascal's triangle occupy "convex" triangular sets, if $q$ divides these three binomials
\[ {p - 1 \choose n - 1}, {p + 1 \choose n}, {p \choose n + 1} \]
then it divides the other three, and vice versa.


search
number theory unsolved
number theory

1) You don't have to use colour to be read :wink:
2) I'm quite sure that this problem appeared before on this forum, so use the 'search' command...

Pierre.

[b]
hello Pierre.

I tried search before I post it ,but I couldn't find it.

could you find it for me please .thanks :wink:  [/b]

[quote="ZetaX"]You also don't need to use bold text...[/quote]

[b]I just use bold and color to be nice form for problem .

Not to see problem .

I adjusted it.

[/b]
 :D

[quote="Hidden Scofield"]here is the problem I lost my mind :

if we write $ H \equal{} (\sqrt {2} \minus{} 1)^r$ where $ r$ is positive number .

show that $ \exists$ positive number k, such that :

$ \frac {H} {{\sqrt {k \minus{} 1} }} \equal{} \sqrt {\frac {k} {{k \minus{} 1}}} \minus{} 1$

 or equivalently to $ H \equal{} \sqrt {k} \minus{} \sqrt {k \minus{} 1}$ .

I'm waiting from someone to help me.

Thanks[/quote]

r=1,k=2
r=2,k=9
r=3,k=50
r=4,k=289
r=5,k=1682
r=6,k=9801
r=7,k=57122
r=8,k=332922
r=9,k=1940342
r=10,k=11308248

The problem is easy.
Proof:
Let $ r \equal{} 2k$,then if we denote
$ A \equal{} \sum_{i \equal{} 0}^{k}2^{i}\binom{2k}{2i}$.
$ B \equal{} \sqrt {2}(\sum_{i \equal{} 0}^{k \minus{} 1}2^{i}\binom{2k}{2i\plus{}1})$.
$ (\sqrt {2} \minus{} 1)^{2k} \equal{} A \minus{} B$.
So now we only need to prove that
$ A^2 \equal{} B^2 \plus{} 1$,and we will be done.
Which is obviously true because,
$ (A \minus{} B)(A \plus{} B) \equal{} ((\sqrt {2} \minus{} 1)(\sqrt {2} \plus{} 1))^{2k} \equal{} 1$.
The case when $ r$ is odd,is similar.

[quote="Erken"]The problem is easy.
Proof:
Let $ r \equal{} 2k$,then if we denote
$ A \equal{} \sum_{i \equal{} 0}^{k}2^{i}\binom{2i}{2k}$.
$ B \equal{} \sqrt {2}(\sum_{i \equal{} 0}^{k \minus{} 1}2^{i}\binom{2i \plus{} 1}{2k})$.
$ (\sqrt {2} \minus{} 1)^{2k} \equal{} A \minus{} B$.
So now we only need to prove that
$ A^2 \equal{} B^2 \plus{} 1$,and we will be done.
Which is obviously true because,
$ (A \minus{} B)(A \plus{} B) \equal{} ((\sqrt {2} \minus{} 1)(\sqrt {2} \plus{} 1))^{2k} \equal{} 1$.
The case when $ r$ is odd,is similar.[/quote]

Thank you .

but what is the value of $ \binom{2i}{2k}$ when $ 2i$ less than $ 2k$.

Thank you northwolves for the nice numbers.

[quote="Erken"][quote="Hidden Scofield"]
but what is the value of $ \binom{2i}{2k}$ when $ 2i$ less than $ 2k$.
 [/quote]
Sorry,i've already edited my solution.[/quote]

No problem dude , now it's very nice proof.

Thank you again.

[quote="Αρχιμήδης 6"]$ H \equal{} (\sqrt {2} \minus{} 1)^r$ We want to prove that $ \exists$ $ k$ so $ H \equal{} \sqrt {k} \minus{} \sqrt {k \minus{} 1}$.
We have $ (\sqrt {2} \minus{} 1)^r \equal{} a_n \minus{} b_n\sqrt {2}$ then $ (\sqrt {2} \plus{} 1)^r \equal{} a_n \plus{} b_n\sqrt {2}$
From sum we get $ a_n \equal{} [(\sqrt {2} \plus{} 1)^r \plus{} (\sqrt {2} \minus{} 1)^r]/2 \equal{} > k \equal{} [(\sqrt {2} \plus{} 1)^r \plus{} (\sqrt {2} \minus{} 1)^r]^2/4$ :wink:[/quote]

Wow ; it's really wonderful .

 :first:

thanks :wink:


number theory
least common multiple
inequalities
algorithm

Prove that there exist two infinite sequences $ \{a_n\}_{n\ge 1}$ and $ \{b_n\}_{n\ge 1}$ of positive integers such that the following conditions hold simultaneously:
$ (i)$ $ 0 < a_1 < a_2 < a_3 < \cdots$;
$ (ii)$ $ a_n < b_n < a_n^2$, for all $ n\ge 1$;
$ (iii)$ $ a_n \minus{} 1$ divides $ b_n \minus{} 1$, for all $ n\ge 1$
$ (iv)$ $ a_n^2 \minus{} 1$ divides $ b_n^2 \minus{} 1$, for all $ n\ge 1$

[19 points out of 100 for the 6 problems]

A One-liner:

Let $ a_n$ be odd for all n.

We define $ b_n\equal{}\frac {a_n^2\plus{}1}{2}$

 :wink:

P.S.
Not my solution but i tried along the same lines and kept getting stuck as i tried satisfying $ a_n\plus{}1|b_n\plus{}1$ and $ a_n\minus{}1|b_n\minus{}1$ simultaneously. :(

My solution:

i think this sum can be solved easily by taking a_n=2n+1 and b_n=2n^2+2n+1

herez my proof

(a_n^2)-1=2n*(2n+2)
So, 2n*(2n+2)/ (b_n^2)-1
But equating b_n to (2n+1)^2 does not work because given b_n < (a_n^2)
So i looked for the lcm of 2n*2n+2
which is 2n^2+2n
and added 1 to this because 2n*(2n+2) divides only (b_n^2)-1 and not (b_n^2)
hence proved

sry coz this is sorta ugly without latex

setting $ a_{m} \equal{} 2(m)(m \plus{} 1) \minus{} 1$  and $ b_{m} \equal{} (m \plus{} 1)(2m(m \plus{} 1) \minus{} 2) \plus{} 1$
also does the trick

Here is the solution in latex.
My solution:

i think this sum can be solved easily by taking $ a_n$=$ 2n\plus{}1$ and $ b_n$=$ 2n^2\plus{}2n\plus{}1$

herez my proof

$ (a_n^2)\minus{}1$=$ 2n*(2n\plus{}2)$
So, $ 2n*(2n\plus{}2)/ (b_n^2)\minus{}1$
But equating $ b_n to (2n\plus{}1)^2$ does not work because given $ b_n < (a_n^2)$
So i looked for the lcm of $ 2n*2n\plus{}2$
which is $ 2n^2\plus{}2n$
and added 1 to this because $ 2n*(2n\plus{}2)$ divides only $ (b_n^2)\minus{}1$ and not $ (b_n^2)$
hence proved

This is my hundreth post. :D

well i did this question only halfway through
so i am not supposed to get any marks

btw somebody told me that this question could be done by weirstrass inequality

Hell ever; I wrote what you told.
 :)

bn = (an^2 + 1) /2  with an always odd!

[quote="rt_08"]hey can this solved with any inequality[/quote][quote="rt_08"]btw somebody told me that this question could be done by weirstrass inequality[/quote]
I want to see this solution. If possible, would you pls post in here?

well ill post it as soon as possible(ill have to learn latex............. :blush:  :blush: )

[b]it is a rather detailed solution. please read patiently and carefully. think for a while before clicking on certain proofs.[/b]

[hide="CLICK TO VIEW SOLUTION"]we have $ a _n - 1 | b _n - 1$

also, $ {a _n}^2 - 1| {a _n}^2 - 1$

we will assume the simplest case when this happens i.e when
$ a _n + 1 | b _n + 1$

[hide="click below to see proof"]

now, $ {a_n} - 1|{b_n} - 1 \implies {b_n} - 1 = \lambda \left( {{a_n} - 1} \right)$ [b]AND[/b]

$ {b_n} + 1 = \mu \left( {{a_n} + 1} \right) \\
\& \implies \left( {{b_n} - 1} \right)\left( {{b_n} + 1} \right) = \lambda \mu \left( {{a_n} - 1} \right)\left( {{a_n} + 1} \right) \\
\& \implies \left({{a_{n}} - 1}\right)\left({{a_{n}} + 1}\right)|\left({{b_{n}} - 1}\right)\left({{b_{n}} + 1}\right) \\
\& \implies {a_n}^2 - 1|{b_n}^2 - 1$
[/hide]

$ {a_n} - 1|{b_n} - 1$
$ \implies {a_n} - 1|\left[ {\left( {{b_n} - 1} \right) - \left( {{a_n} - 1} \right)} \right]$ [By Euclids division algorithm]

$ {a_n} - 1|{b_n} - {a_n}$

similarly, $ {a_n} + 1|{b_n} - {a_n}$

[u][b]CASE I[/b][/u]

$ a_n$ is even $ \implies a_n \pm 1$ is odd
two odd numers separated by 2 are always coprime.

$ \because {a_n} + 1|{b_n} - {a_n}$ And
$ {a_n} - 1|{b_n} - {a_n}$

$ {a_n} - 1$ and $ {a_n} + 1$ are coprime, hence -
$ \left( {{a_n} - 1} \right)\left( {{a_n} + 1} \right)|{b_n} - {a_n}$

$ \implies {a_n}^2 - 1|{b_n} - {a_n}$

$ \implies {b_n} - {a_n} = \lambda \left( {{a_n}^2 - 1} \right)$ where $ \lambda \in N$
$ {b_n} = \lambda {a_n}^2 - \lambda + {a_n}$

now, $ {b_n} > {a_n} \implies \lambda {a_n}^2 - \lambda + {a_n} > {a_n}$
$ \implies {a_n}^2 - 1 > 0$

$ \implies{a_n} \in \left[ {\left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)} \right] \cap N \\
\& \implies {a_n} \in N$

Also $ {b_n} < {a _ n}^2$
$ \implies \lambda {a_n}^2 - \lambda + {a_n} < {a_n}^2$

$ \implies \left( {\lambda - 1} \right){a_n}^2 - \lambda + {a_n} < 0$
but for all$ \lambda \in N$ and $ \lambda - 1 \neq 0$, we have, $ \lambda - 1 > 0$ thus it is an upward parabola and we hence do not have $ \left( {\lambda - 1} \right){a_n}^2 - \lambda + {a_n} < 0$ for infinite integer solutions, against the requirement of the question. hence we see that [b]OUR CASE IS WRONG[/b]

[u][b]CASE 2[/b][/u]

$ \therefore {a_n}$  is odd

$ \therefore {a_n} \pm 1$ is even

following the logic in the previous case we have.
$ {a_n}^2 - 1|2\left( {{b_n} - {a_n}} \right)$
[hide="Click to see why"]
2 even numbers separated by 2 have only 2 as a common factor.
$ \implies a_n + 1 = 2m, a_n - 1 = 2n$ where m,n are coprime.

$ \because \left( {{a_n} - 1} \right)|\left( {{b_n} - {a_n}} \right)$ AND
$ \because \left( {{a_n} + 1} \right)|\left( {{b_n} - {a_n}} \right)$

$ \implies {b_n} - {a_n} = 2km \\
\& {b_n} - {a_n} = 2ln$

$ \because m,n$ are coprime hence $ b_n - a_n$ divides $ mn$
$ \implies {b_n} - {a_n} = 2pmn$

$ \& {a_n}^2 - 1 = \left( {{a_n} - 1} \right)\left( {{a_n} + 1} \right) = 4mn \\
\& 4mn|2\cdot 2pmn \\
\& \implies \boxed{{a_n}^2 - 1|2\left( {{b_n} - {a_n}} \right)}$

[/hide]

$ \& \implies 2\left( {{b_n} - {a_n}} \right) = \mu \left( {{a_n}^2 - 1} \right)$ where $ \mu \in N$
  $ \implies {b_n} = \frac {{\mu {a_n}^2 - \mu + 2{a_n}}}{2}$

now $ b_n > a_n \implies a_n \in N$ [working similar to that in previous case]

$ b_n < {a_n}^2 \\
\implies \frac {{\mu {a_n}^2 - \mu + 2{a_n}}}{2} < {a_n}^2 \\
\implies \left( {\mu - 2} \right){a_n}^2 + 2{a_n} - \mu < 0$

for there to be infinite values of $ a_n$ satisfying this condition then $ \left( {\mu - 2} \right){a_n}^2 + 2{a_n} - 2$ must represent a downward parabola.
$ \implies \mu - 2 < 0 \\
 \\
\because \mu \in N \implies \mu = 1$

resubstituting $ \mu$ we get

$ \& \implies - {a_n}^2 + 2{a_n} - 1 < 0 \\
\& \implies {a_n}^2 - 2{a_n} + 1 > 0 \\
\& \implies {\left( {{a_n} - 1} \right)^2} > 0$

$ \implies \boxed{a_n \in \left\{3,5,7,9,11..........\right\}}$ [$ \because a_n \in N$ from previous condition and $ a_n$is odd from case]

[i][b]for any sequence $ a_n$ as always an odd number, we can get $ b_n$ by plugging $ \mu = 1$ in $ {b_n} = \frac {{\mu {a_n}^2 - \mu + 2{a_n}}}{2}$[/b][/i]

EXAMPLES -

$ \& {a_n} = 2n + 1 \\
\& {b_n} = \frac {{{a_n}^2 - 1 + 2{a_n}}} {2} = \frac {{{{\left( {2n + 1} \right)}^2} - 1 + 2\left( {2n + 1} \right)}} {2} = 2{n^2} + 4n + 1$

$ \& {a_n} = 2n + 3 \\
\& {b_n} = \frac {{{a_n}^2 - 1 + 2{a_n}}} {2} = \frac {{{{\left( {2n + 3} \right)}^2} - 1 + 2\left( {2n + 3} \right)}} {2} = 2{n^2} + 7n + 7$

$ \& {a_n} = {2^n} + 1 \\
\& {b_n} = \frac {{{a_n}^2 - 1 + 2{a_n}}} {2} = \frac {{{{\left( {{2^n} + 1} \right)}^2} - 1 + 2\left( {{2^n} + 1} \right)}} {2} = {2^{2n - 1}} + {2^{n + 1}} + 1$

their properties in accordance to given conditions can be verified experimentally ,inductively or otherwise.

[u][i][b]HENCE WE SEE THAT THERE ARE AN INFINITE NUMBER OF SUCH SEQUENCES. [/u][/i][/b]

[u][i][b]Q.E.D[/b][/i][/u]

[NOTE :- The above sequences are not the only possibilities. i have only taken the case when $ a_n +1 | b_n +1$ there may be other cases also. however there are still infinite cases in this assumption itself. hence it is sufficient to prove the required statement]
[/hide]

just take
a_{n} = 2n+1 ,
b_{n}= 2n^2 + 4n +1.
Now just prove that this satisfies all the four conditions.


modular arithmetic
floor function
combinatorics unsolved
combinatorics

Let $k \in \mathbb{N},$ $S_k = \{(a, b) | a, b = 1, 2, \ldots, k \}.$ Any two elements $(a, b)$, $(c, d)$ $\in S_k$ are called "undistinguishing" in $S_k$ if $a - c \equiv 0$ or $\pm 1 \pmod{k}$ and $b - d \equiv 0$ or $\pm 1 \pmod{k}$; otherwise, we call them "distinguishing". For example, $(1, 1)$ and $(2, 5)$ are undistinguishing in $S_5$. Considering the subset $A$ of $S_k$ such that the elements of $A$ are pairwise distinguishing. Let $r_k$ be the maximum possible number of elements of $A$.

(i) Find $r_5$.
(ii) Find $r_7$.
(iii) Find $r_k$ for $k \in \mathbb{N}$.

Solution:When $k=1,2,3$, it's easy to know that $r_k=1$.
If $k\geq4$,we assume that $A_i=\{(i,b)|b=1,2,\ldots,k\}(i=1,2,\ldots,k)$,$B_i=A_i\cup A_{i+1}(i=1,2,\ldots,k,A_{k+1}=A_k$.

First we show that $|A\cap B_i|\leq\lfloor\frac{k}{2}\rfloor$
Just consider $B_1$,It's obvious that $(1,b),(2,b),(1,b+1),(2,b+1)$ are pairwise undistinguishing,so the elements of $A$ have at most $\lfloor\frac{k}{2}\rfloor$ different $b$,for each $b$,there is at most one $a=1 \mbox{or} 2$.So $|A\cap B_i|\leq\lfloor\frac{k}{2}\rfloor$

Then if we have $|A|=r_k$,then consider
$\sum_{i=1}^k|A\cap B_i|=\sum_{i=1}^k|A\cap A_i|+|A\cap A_{i+1}|=2\sum_{i=1}^k|A\cap A_i|=2|A\cap S_k|=2r_k\leq k\lfloor\frac{k}{2}\rfloor$
So we have $r_k\leq\lfloor\frac{k}{2}\lfloor\frac{k}{2}\rfloor\rfloor$

Next we show that $r_k=\lfloor\frac{k}{2}\lfloor\frac{k}{2}\rfloor\rfloor$
In the next part,all number is in $\mod k$.$|b-d|>2$ means $b-d\not\equiv0\mbox{or}\pm1\pmod{k}$.

If $k$ is even.Suppose $k=2m(m\geq2)$,then $\lfloor\frac{k}{2}\lfloor\frac{k}{2}\rfloor\rfloor=m^2$.
Let $A=\{(a,b)|a,b\equiv1\pmod{2},(a,b)\in S_k\}$,then it's easy to see $|A|=m^2$,
and for different $(a,b),(c,d)$ at least one of $|a-c|,|b-d|>2$,so they are distinguishing.

If $k\equiv1\pmod{4}$,$k=4t+1(t\geq1)$,$\lfloor\frac{k}{2}\lfloor\frac{k}{2}\rfloor\rfloor=t(4t+1)$
Let $A\cap A_{2p+1}=\{(2p+1,b)|b=p+1,p+3,\ldots,2t+p-1\}(p=0,1,\ldots,4t)$
It's easy to know $|A\cap A_{2p+1}|=t$,so $|A|=t(4t+1)$.
And $A\cap A_{2p}=A_{2(p+2t)+1}=\{(2p,b)|b=p+2t+1,p+2t+3,\ldots,p+4t-1=p-2\}$.
And it's east to see for any two elements $(a,b),(c,d)$ in $A\cap A_{2p+1}$,$|b-d|>2$;
and $(a,b)\in A\cap A_{2p+1},(c,d)\in A\cap A_{2p}$,$|b-d|>2$.
So any different elements $(a,b),(c,d)\in A$ are distinguishing.

If $k\equiv3\pmod{4}$,$k=4t+3(t\geq1)$,$\lfloor\frac{k}{2}\lfloor\frac{k}{2}\rfloor\rfloor=4t^2+5t+1$.
Let $A\cap A_{2p+1}=\{(2p+1,b)|b=p+1,p+3,\ldots,2t+p-1\}(p=0,1,\ldots,2t+1)$
and $A\cap A_{2p}=\{(2p+1,b)|b=p+2t+1,p+2t+3,\ldots,p+4t+1=p-2\}(p=1,2,\ldots,2t+1)$
Likewise we can show $|A|=t(2t+2)+(t+1)(2t+1)=4t^2+5t+1$,
and any different elements $(a,b),(c,d)\in A$ are distinguishing.

So $r_k\leq\lfloor\frac{k}{2}\lfloor\frac{k}{2}\rfloor\rfloor$
And $r_5=5,r_7=10$.

$\sum_{i=1}^k|A\cap A_i|+|A\cap A_{i+1}|=2\sum_{i=1}^k|A\cap A_i|$
i think it is incorrect because you have only one A intersection A1 and only one intersection A with Ak.


linear algebra
matrix
inequalities
linear algebra unsolved

Assume $ A,B \in M_n(\mathbb{R})$

If $ AB\equal{}BA\equal{}0$ and $ rank(A^2)\equal{}rank(A)$

can we use "elementary transformation of block matrix"  to show that: $ \begin{pmatrix}
A\plus{}B &0 \\
0 & 0
\end{pmatrix}\sim \begin{pmatrix}
A &0 \\
0 & B
\end{pmatrix}$
In other words: to show: $ rank\begin{pmatrix}
A\plus{}B &0 \\
0 & 0
\end{pmatrix}\equal{}rank\begin{pmatrix}
A &0 \\
0 & B
\end{pmatrix}$

Suppose that it is true, i.e. $ \mathrm{rk}(A \plus{} B) \equal{} \mathrm{rk} A \plus{} \mathrm{rk} B$ holds if $ A^2 \equal{} A, AB \equal{} BA \equal{} 0$. We will prove that $ \ker(A \plus{} B) \equal{} \ker A\cap \ker B$. The inclusion $ \ker(A \plus{} B)\supset\ker A\cap \ker B$ is obvious. Let $ v\in \ker (A \plus{} B)$. Then $ A^{2}v \equal{} A(A \plus{} B)v \minus{} ABv \equal{} 0$. Thus $ Av \equal{} 0$ (since $ |\ker A^2| \equal{} |\ker A|$) and $ Bv \equal{} (A \plus{} B)v \minus{} Av \equal{} 0$, which proves $ \ker(A \plus{} B)\subset\ker A\cap \ker B$. Therefore $ |\ker A| \plus{} |\ker B| \equal{} 2n \minus{} (\mathrm{rk} A \plus{} \mathrm{rk} B) \equal{} 2n \minus{} \mathrm{rk}(A \plus{} B) \equal{} |\ker (A \plus{} B)| \equal{} | \ker A\cap \ker B|$, meaning that in $ | \ker A\cap \ker B|\leq |\ker A|\leq |\ker A| \plus{} |\ker B|$ the second inequality is an equality, i.e. $ |\ker A| \equal{} 0$. Analogously $ |\ker B| \equal{} 0$, which contradicts with $ AB \equal{} 0$.


based on general relativity, gravity is not instantaneous, meaning that it can not act faster than the speed of light.  if you were to remove the sun as you are saying, then that would be an instantaneous actions.  not that the sun would ever dissapear, but if it did, it could not occur instantaneously.  even if the time interval was extremely tiny, it would be a time interval in which the speed of light is not surpassed.  therefore, gravity would not act faster than the speed of light.  if what you said was true, then you would be dealing with some form of ideal physics (this relates to motion in that an object can not accelerate over an instant, on a graph there would always be curvature).

[quote="stupidkid"]I've read this problem somewhere but don't remember where I saw it, but anyways...
So Earth revolves around the sun, and let's hypothetically remove the sun (make it disappear) would Earth just suddenly go off its orbit? Because if it does, then wouldn't that mean gravity travels faster than light? Or is there a time difference between when the sun dissappears and when Earth goes off its orbit?

I would greatly appreciate your opinions! :D[/quote]

The Special Theory of Relativity, of which the General Theory of Relativity is an extention that includes gravitational effects, is founded upon two assumptions:

1. The laws of physics are the same in all reference frames.

2. The speed of light is constant in all reference frames.

Given these two postulates, were the sun to disappear as you suggest then its disappearance would be manifest in the conversion of the sun's mass into energy [i]vis-a-vis[/i] $E=M_{\bigodot}c^2$.

This "solar" energy would then propagate through space as a wave traveling at the speed of light, $c$.  We know that it would travel at $c$ because $c$ is determined by physical properties of space itself, namely $\mu_0$ and $\epsilon_0$;  $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$.

At the speed of light it would take the energy wave approximately 8 minutes to reach the Earth, and it would take a minimum time of $\frac{\hbar}{M_{\bigodot}c^2}$ for the "disappearance" to occur in the first place, according to the Heisenberg Uncertainty Principle. Upon arrival of the wave the Earth would be obliterated, so no need to worry whether Earth would be knocked out of its orbit.


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

How to prove using the class equations that a subgroup of order p^(n-1) in a group of order p^n is normal.

in general if $G$ is a group of order $n$ and $p$ is the smallest prime dividing $n$ then every subgroup of index $p$ is normal

Another related question, how to prove a group G of order p^n has subgroup of order p^x for all 0<=x<=n. My idea is to use the conclusion from the class equations that the group has a non-trivial centre and by induction the centre has a subgroup K of order p which will be of course normal in G as it is a subgroup of the centre. Construct the quotient group G/K. Now i want to use the elements of G which form the cosets of G/K to form a subgroup of order p^(n-1). How should i proceed. Any other way to make use of the class equations?

That's Sylow's First Theorem.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147186

[quote="unixguru"]Another related question, how to prove a group $G$ of order $p^{x}$ has subgroup of order p^x for all 0\leqx\leqn[/quote]

Let $G$ be a $p$-group, i.e. $|G|=p^{n}$. We use induction on $n$. The case $n=0$ is trivial. According to [url=http://planetmath.org/?op=getobj&from=objects&id=2186]Cauchy's theorem[/url] there exists an element of order $p$, say $x$. Then $G/\langle x \rangle$ has order $p^{n-1}$. Apply the induction hypothesis, take the inverse image of every subgroup in $G/\langle x \rangle$ and check that these give you subgroups in $G$ of order $p^{x}$, where $0\leq x \leq n$. The normality follows directly from the [url=http://en.wikipedia.org/wiki/Lattice_theorem]Lattice theorem[/url].

here is my sulution:
easy seen $|G: H|=p$ and we know
$|G|=p^{n}$ and $H=p^{n-1}$ and $G\geq H$.
next step:
$|G: H|=p$
next step:
use theorem Kelle.
next step:
because $|G: H|=p$ and $p$ is minimal prime number wich $|$ ord of group $G$ thereofer
$H$ is normal.
Really by the theorem Kelle $|G: H_{G}|$ $|$ $p!$
we know that $p$ is minimal prime number wich $|$ $|G|$
then $|G: H_{G}|=1$ or $|G: H_{G}|=1$ but $H_{G}\leq H$ therefore first case can't be. therefore we have seccond wich means that $H_{G}=H$ Done.

Here are my proofs, can someone please verify them for me:-

1. We'll apply induction on 'n'. It suffices to prove that a group G of order p^n has a subgroup of order p^(n-1). We know that G has a normal subgroup K of order p. Consider G/K, the quotient group of order p^(n-1). List this group as
{Ke, Ka1, Ka2.....}(where each entry must be distinct). Now traverse the list and upon encountering an element Kai, replace the symbol of its inverse by K(ai)^-1 and move the element next to Kai and also corresondingly change the symbols of all the products which can be formed by the elements traversed so far and move them all forward. After doing this the symbols {e, a1, a2,..........} will form a subgroup :?:

2. Given any subgroup H != G  where order of G is p^n, it can be proved that there exists an element 'x' such that x is not in H and xH(x)^-1 = H. Therefore in our problem N(H) is forced to be G and hence H is normal in G.

Another proof for the second result is from herstein's second proof os sylow's theorem using the class equations. He gives a proof of the existence of p-sylow subgroups using induction whre, if i am not mistaken, the hypothesis that p^(m+1) does not divide G is not at all used and hence the proof can be trivially modified to prove sylow's thorem a special case of which is our problem.

[quote="Jacobson"]Then $G/\langle x \rangle$ has order $p^{n-1}$. [/quote]
This isn't a group, in other words $\langle x \rangle$ is not necessary normal.

why? isn't $\langle x \rangle \cap Z(P) \neq 1$

$G= D_{8}$ (dihedral group with $8$ elements, represented by cycles of permutations on $1,2,3,4$), set $x=(1,2)(3,4)$.

But if you prove that the center of a $p$-group is nontrivial (which is a well known fact), you can take $x$ from it.

[quote="unixguru"]What is theorem kelle?[/quote]
Common case of theorem Kelle:
the Kernel of the Actions $G$ on $G/H$ is equal to $H_{G}$. and $G/H_{G}$ is isomorphic to symetric subgroup of  group $S_{G/H}$.


Prove that the equation $x^n+y^n=z^n$ with  $n$ natural $n>1$ has no solution from integers with $0<x\leq n$ and $0<y\leq n$

The problem may seem strange, but it's quite easy. We can even show more:

[color=blue]If n is a natural number such that n > 1, then the equation $x^n+y^n=z^n$ has no integer solutions x, y, z satisfying $0<x\leq n$ and 0 < y.[/color]

[i]Proof.[/i] We assume that there are integers x, y, z satisfying the equation $x^n+y^n=z^n$ and the conditions $0<x\leq n$ and 0 < y. We can WLOG assume $x\leq y$; if this is not so, we can just interchange x and y (in fact, if we have $x\geq y$, $0<x\leq n$ and 0 < y, then $0<y\leq n$, so the condition $0<x\leq n$ is still satisfied after the interchanging of x and y).

Since 0 < x, we have $z^n=x^n+y^n>y^n$, so that z > y, and thus, due to integrality, $z\geq y+1$. Hence,

$x^n=z^n-y^n\geq\left(y+1\right)^n-y^n=\left(y^n+ny^{n-1}+\underbrace{...}_{\text{positive terms}}\right)-y^n>ny^{n-1}$.

Since $x\leq n$, we have $ny^{n-1}\geq xy^{n-1}$. Thus, $x^n>ny^{n-1}\geq xy^{n-1}$, so that $x^{n-1}>y^{n-1}$, and thus x > y. This contradicts $x\leq y$. So we have obtained a contradiction, and our proof is complete.

  darij


Two participants A and B wrote a 3-hour long, 5-problem MO (guess which one.... ;) ). They both spent the full 3 hours and attempted all 5 problems. If for each problem, A spent a different amount of time working on it than B (so they can't both spend the same amount of time on the same problem), but they are allowed to work on the problems in any order. Knowing they can only work on one problem at a time, will there always be a moment such that both of them are working on the same problem?  :)

How about the following schedule:

A  |*1*|**3**|**5**|**2**|**4**|
B  |**2**|**4**|**1**|**3**|*5*|

Clearly, all intervals can be made of different length  :)

[quote="fedja"]Clearly, all intervals can be made of different length  :)[/quote]

What does that mean?? :?


N contestants answered an exam with n questions. ai points are awarded for a correct answer to question i and nil for an incorrect answer. After the questions had been marked it was noticed that by a suitable choice of positive numbers ai any desired ranking of the contestants could be achieved. What is the largest possible value of N?

shoma in soal o 2 bar inja; va ham too ghesmate a Olympiad Combinarotics post kardid.

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=exam+answered+contestants&t=137750

this thread will be locked.


WHAT??????????

Are we gonna leave AoPS now :( ?

it is finally back tot the good old days

er yea...i'll just stay here...it's easier checking one math forum instead of 2 anyways

Nope, I need team forums.  Oh well.  Bye AOPS (even though I'll probably still check here daily from now on).  You other people just want to raise your AOPS post count.

I AM NOW NO LONGER A CHARMANDER, BUT A [one step below charizard].

One step below Charizard is Charmeleon, right?  OMG, I am still discussing Pokemon...I thought that was cleansed from my body years ago  :D  :P

Yea...I'm kinda embarrassed I still remembered it was charmeleon. Oh well...those good old 4th grade days. *sigh*

Haha, yeah. I spent half a minute trying to remember the name. And then I was like "YES! I GOT IT!" and felt like a dork.

Man, I couldn't remember that for the life of me.

Bulbasaur, Ivysaur?, Venusaur
Squirtle (that one took me a good 2 mins), Wartortoise (5 mins), Blastoise

The middle one is always sofa king hard to remember.

Nidoran (male)->Nidorhino->Nidoking
Nidoran (female)->Nidorhina->Nidoqueen
Pidgey->Pidgeyotto? (I thought there were 3 levels for this one)
...

Yea, I agree Steffy. And I think the 3rd one there is like Pidgeot or something like that...

why are we discussing this?

Swamp - Tap. Dark Ritual. Dark Ritual. Dark Ritual. Mox Ruby - Sac. Raging Goblin. Hatred for 19. Attack with goblin. You die!

Nice go MAGIC

although i don't play anymore....

Haha, magic sucks. I have a collection of magic cards I find in library boks though. So far, I've got 6.

I now have just 2 M:TG cards left that I use as bookmarks (a Stronghold Contemplation and an Ice Age Leviathan).

i always scrounged my cards -- crap that my friends didnt want. then my mom threw them all away so i stopped playing

haha...i still have some old magic cards...wow...*tosses into garbage* that solves that...


Let $ ABCDEF$ be a regular hexagon whose length of a side is $ 1$,and $ P,Q$ be points on the segments $ BC,DE$ respectively.
Find the minimum value of $ AP\plus{}PQ\plus{}QA$.


Triangle $ ABC$ has $ BC\equal{}20.$ The incircle of the triangle evenly trisects the median $ AD.$ If the area of the triangle is $ m \sqrt{n}$ where $ m$ and $ n$ are integers and $ n$ is not divisible by the perfect square of a prime, find $ m\plus{}n.$

Is it so hard to search?
Use Contest(AOPS) or Resources(ML).
http://www.mathlinks.ro/viewtopic.php?p=365372#365372


geometry
circumcircle
radical axis
geometry unsolved

Nice problem!  Let $ N$ be the midpoint of arc $ BC$, and let $ F$ be the intersection of $ AM$ with $ BC$.  Notice that $ E$ is the orthocenter of $ \triangle{FMN}$, so we get that $ F, D,$ and $ N$ are collinear, and that $ MD \perp FN$.  So then $ AEDF$ is cyclic.  Now it's easy to prove that the intersection of the tangent to $ \omega$ at $ D$ with $ BC$ is equidistant from $ D$ and $ E$ by a brief angle chase, and then we're done, because the circumcenter of $ AEDF$ must lie on $ BC$.

[quote="outback"]A scalene triangle $ ABC$ is inscribed in a circle $ w$. The bisector of angle $ A$ meets $ BC$ at $ E$. Let $ M$ be the midpoint of the arc $ BAC$. The line $ ME$ intersects $ w$ again at $ D$. Show that the circumcenter of triangle $ AED$ coincides with the intersection point of the tangent to $ w$ at $ D$ and the line $ BC$.[/quote]
[hide="Solution"]
Let the intersection between the tangent to $ \omega$ at $ D$ and $ BC$ be $ X$.  We consider the circle $ T$, that is tangent to $ \omega$ at $ D$ and tangent to $ BC$ as well.  We also consider the circle $ S$ that is tangent to $ \omega$ at $ A$ and tangent to $ BC$.  Letting $ E'$ be the point of tangency between $ T$ and $ BC$, we have that it is well-known that $ DE'$ passes through the midpoint of arc $ BAC$, which is $ M$.  Hence, $ DM$ meets $ CB$ at $ E'$, so $ E'\equal{}E$.  Similarly, $ S$ is tangent to $ BC$ at $ E$ since $ AE$ passes through the midpoint of arc $ BC$.  Then, since $ X$ is the intersection between the tangent at $ D$ to $ \omega$ and $ BC$ (which is tangent to both $ T$ and $ S$ at the same point, $ E$) we see that $ X$ is the radical center between $ S$, $ T$, and $ \omega$, which means that $ XA\equal{}XE\equal{}XD$. [/hide]

It can be easily shown that $ \frac{AB}{BC}\equal{} \frac{AD}{CD}$, which means $ ABCD$ is a harmonic quadrilateral and $ BC$, tangent to $ w$ at $ A$ and $ D$ are concurrent  :)

$ M$ is the midpoint of arc $ BAC$, so we have $ \angle MDC\equal{}\angle MDB$, and

$ \frac{BD}{DC}\equal{}\frac{BE}{EC}\equal{}\frac{BA}{AC}$

so $ AB\cdot CD \equal{}AC\cdot BD$

$ \Rightarrow ABCD$ is harmonic

$ \Rightarrow$ the tangent line of $ \odot \omega$ at $ A$ and $ D$ meet $ BC$ at the same point, denote the intersection point as $ O$

we have $ OA\equal{}OD$

$ \because \frac{OC}{OB}\equal{}(\frac{CD}{BD})^2$, and $ OC\cdot OB\equal{}OD^2$

$ \Rightarrow OB\equal{}\frac{OD\cdot BD}{CD}$

and $ OC\equal{}\frac{OD\cdot CD}{BD}$

but $ \frac{EC}{EB}\equal{}\frac{CD}{BD} \Leftrightarrow \frac{EC}{BC}\equal{}\frac{CD}{CD\plus{}BD}$

$ \Rightarrow OE\equal{}OC\plus{}EC\equal{}OC\plus{}\frac{CD}{CD\plus{}BD}\cdot BC$
$ \equal{}OC\plus{}\frac{CD}{CD\plus{}BD}\cdot (OB\minus{}OC)$
$ \equal{}OC\plus{}\frac{CD}{CD\plus{}BD}\cdot \Big(\frac{BD^2\minus{}CD^2}{CD\cdot BD}\Big)\cdot OD$
$ \equal{}\frac{OD}{BD}\cdot (BD\minus{}CD)\plus{}\frac{OD\cdot CD}{BD}\equal{}OD$

$ \Rightarrow OD\equal{}OE\equal{}OA$

$ \Rightarrow$ $ O$ is the circumcenter of $ \triangle AED$

QED.


Let a, b, c >0

Prove that:

If a+b+c>= 1/a+1/b+1/c     ----->             a+b+c>= 3/abc

[quote="conejita"]Let a, b, c >0

Prove that:

If $a+b+c \geq \frac{1}{ }a+\frac{1}{ }b+\frac{1}{ }c$    ----->             a+b+c>= 3/abc[/quote]

note that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \sqrt{3(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc})}$.....

very nice! given in romanian JTST 2004...

 :D

From the condition
$abc(a+b+c)\geq ab+bc+ca\geq \sqrt{3abc(a+b+c)}$

so the result   :)


I was doing my workbook just now and i started a new chapter on circles. in that chapter, it mentions radians. what are radians? it tells me it's another way to measure angles. How does it work? I have used degrees all my life :?

and if you can can you please show me how you would figure out the arc and area of a sector using radians? I have the  formula here no idea what i'm supposed to do with it.

[quote="akimatsu"]I was doing my workbook just now and i started a new chapter on circles. in that chapter, it mentions radians. what are radians? it tells me it's another way to measure angles. How does it work?[/quote]

An angle of 1 radian is defined such the arc length of such an angle is equal to the radius.

Because the circumference of the circle is $2\pi$ times the radius, there are $2\pi$ radians in a whole circle. This means that a degree is equal to $\pi /180$ radians.

Computing arc length is easy with radians: just multiply the angle by the radius and you're done.

The area of a sector is given by (area of whole circle)*(angle measure of sector)/(angle measure of whole circle). If $\theta$ is the angle and $r$ is the radius, this reduces to $\frac{1}{2}\theta r^2$.


Loki, Moe, Nick and Ott are good friends.  Ott had no money, but the
others did.  Moe gave Ott one-fifth of his money, Loki gave Ott
one-fourth of his money and Nick gave Ott one-third of his money.
Each gave Ott the same amount of money.  What fractional part of the
group's money does Ott now have?

Let the amount of money with Loki, Moe and Nick be represented by [b]L, M [/b]and [b]N[/b] respectively.

Amount of money they give to Ott are equal

So, $ \frac{1}{5}M \equal{} \frac{1}{4}L \equal{} \frac{1}{3}N \equal{} x$

$ M \equal{} 5x, L \equal{} 4x, N \equal{} 3x$

Total money among the group = [b]M + L + N = 12x [/b]

Money they give Ott = [b]3x[/b]

Fractional part of the group's money that Ott now has = $ \frac{3x}{12x} \equal{} \frac{1}{4}$.


Let me just give an example. The article "Chebyshevs inequality" is now a redirect to "Chebyshev's inequality", which, in its turn, is a redirect to "Chebyshev's Inequality".  If you go to the "Inequalities" article and click on "Chebyshev's inequality", you actually go to "Chebyshevs inequality" (that's how the link was set), from where you are automatically redirected to "Chebyshev's inequality", but it stops there and the second redirect to "Chebyshev's Inequality" has to be made by hand. Is it a bug or multiple redirects are just not supported because in a well-organized database they are never really needed?

Double redirects do not work, if you click on something and it brings you to a redirect page, click on "What links here" and change everything on there to link to the correct article (preferably not even a redirect), I already had to fix this with Amc, (Amc redirected to American mathematics competition which redirected to American Mathematics Competition)

I'm sure there are a few more of these to fix, particularly since we went through the process of determining which articles should be named as proper nouns and changed many article titles.  Just help fix them when you find them.


There is a one acre field in the shape of a right triangle, with sides of length $x$ and $y$. At the midpoint of each side there is a post. Tethered to the posts on each side is a sheep. Thethered to the post on the hypotenuse is a dog. Each animal has a rope just long enough to reach the two adjacent vertices of the triangle. How much area outside of the field do the sheep have to themselves?

Easy question :rotfl:

[quote="guile"]If you will draw this you will form 3 semicircles (area of the animals can graze) with diameters, x, y and sqrt(x^2 + y^2). So the radii of the circles are x/2, y/2 and sqrt(x^2 + y^2)/2.  Therefore finding the area of three semi circles:

With side x

    (pi(x/2)^2)/2 = pi x^2 / 8
    (pi(y/2)^2)/2 = pi y^2 / 8
    (pi(sqrt(x^2 + y^2)/2)^2)/2 = pi (x^2 + y^2 )/ 8

Therefore the area of the circles are  (2pi x^2 + 2 pi y^2)/8[/quote]

Oops... Sorry, i made a mistake. its should be (3pi x^2 + 3 pi y^2)/8

First of all, use the edit button. Second of all, the answer shouldn't have a variable because the area was given to you.

But I don't know how to do it either  :blush:  Mostly because the question asked only for the sheeps' space for only themselves. The dog's space conflicts...

the answer is in figures not in variables. :D  :rotfl:
Should I put this in Intermediate? :lol:

See how the problem goes for a while, and if no one can solve it, move it up. Does this problem involve calculus or something?

:rotfl:  :rotfl:  :rotfl:  Only basic geometry required.
The figure is the interesting thing. :D

We are only given the area of the triangle, we do not have the side lengths. All we know is that $xy=1$, but his equation is using the sides $x$ and $y$ separately.

Ill say it again the diagram is the main thing. After that its really easy.
 :jump:

[quote="JesusFreak197"]This is not an easy question, I have no idea what you're talking about. Drawing the diagram does absolutely nothing.[/quote] :rotfl:  :rotfl:  :rotfl:  :rotfl:  it is extremely easy
okay listen, draw me the diagram and then I can judge if it is right or wrong.
 ;)

maybe still if nobody gets it Ill give a hint. :lol:

My only guess is that the question is not worded correctly. Is it supposed to be space that both sheep have in common, outside of the field, that the dog does not have?

It is worded perfectly,
okay a hint
[hide="hint"]
The midpoint of the hypotenuse is the circumcente of the triangle,
after this it is tooooooooooooooooooo easy sorry I have to give the hint.
 :mad:  everyone can get it now.
[/hide]

Well, you know the area of the field is $1$, couldn't you simply pick sides that would give the area of $1$. I picked the sides as $2$ and $1$, thus the area that sheep $x$ has outside the field is $\frac{\pi}{2}$, and the other sheep $y$ has area outside the field of $\frac{\pi}{4}$, so add the two together and you get $\frac{3\pi}{4}$.

EDIT: But then wouldn't there be an infinite number of possibilities because you can pick different sides???

[quote="236factorial"]Simply adding the two areas together wouldn't work because the dog who is poled on the hypotenuse has some space conflicts with the sheep. It wants to know the area the sheep have to them selves.[/quote]

Ooh..now that makes it harder :P Well, then you can simply find the area of the entire region occupied by all the animals plus the area of the triangular field, then subtract the area occupied by the dog, which is a full circle and the tiny portion of the area of the triangle that's left. But that takes work :|

So then the total area would be $\frac{13\pi+4}{4}$ and then subtract the area occupied by the dog, $\frac{5\pi}{2}$ and now we need to find the area of the little part of the traingle that's left:

Here's way I might have made a mistake:

So you need to find the little area of the triangle, well, you know that the circle covers up all of it, and so the little area is left, well draw the radii, which are $\frac{\sqrt{5}}{2}$ and that would be an equilateral triangle so the area of that would be $\frac{5\sqrt{3}}{16}$ and subtract it from the area of the triangle in which the triangle is present, which is $\frac{5\pi}{16}$ so the little part of the circle outside of the equilateral triangle is $\frac{5\pi-5\sqrt{3}}{16}$. So then you have the other small right triangle which is similar to the larger one, which is the field, and you know that the area of that is $\frac{1}{4}$ and you subtract the small area of the circle to get the little piece of the triangle we needed all along, so you get $\frac{4-5\pi-5\sqrt{4}}{16}$ and then you got the area of the small part of the triangle.

Now, finally...

Subtract everything from the total area and you get $\frac{7\pi+12+5\sqrt{3}}{4}$.

Xantos, so the lunes of the legs add up to the area of the triangle? Wow, then I was completely off :?

It be nice if I knew that lunes theorem... then the problem would be easy.

[quote="Xantos C. Guin"]Here is a diagram I made.

[hide]Those 2 moon shapped areas are Lunes of Hippocracies. They add up to the area of the triangle. (Lunes of Hippocraces Theorem)[/hide]

So the answer is [hide]$1$ acre[/hide][/quote]
Finally someone got it !!!!!!!!!! It was sooo easy and Xantos, it can be done without Lunes of Hippocraces Theorem :rotfl:  :rotfl:  :rotfl:  :rotfl:

If that's the case, then show us how we can do it without the theory. If the theory is the only way to do it, then that's pretty stupid; it's not an easy problem.

Give up so soon???  :rotfl:  :rotfl: I can give the anwer to you but I want you to try whats the need of giving it then?? Xantos got it by a theorem but there is no theorem required.
IT is extremely easy just put ur head to it and you'll get it.
Now its even more easy as Xantos has given a diagram.

Does this rule exist?

In a right triangle, the length from the midpoint of the hypotenuse to the vertex of the right angle is equal to the length of half the hypotenuse.

It sure does for a 45-45-90 and a 30-60-90, so let's [i]assume[/i] (I hate that word) that it works for all right triangles.

Let x=2 and y=1 in a right triangle. (area is 1). the hypotenuse is $\sqrt5$. The area of a semicircle the hypotenuse vertexes make is $(\frac{1}{2}\sqrt5)^2\pi$, or $\frac{5}{4}\pi$. Since we are assuming the rule above is true, the 90 angle vertex is on this semicircle. The area that it spills into the the sheep's space is $\frac{5}{4}\pi-1$, where 1 is the area of the triangle. The sheep's area outside the triangle is $\frac{5}{4}\pi$. So the space the sheep have to themselves is $\frac{5}{4}\pi-(\frac{5}{4}\pi-1)=1$.
Seriously that is easy?

no need for that you dont need to find the sides.
[hide="hint"]
use only areas.
[/hide]

[quote="JesusFreak197"]Umm... I'll try something here...

Given sides $a$, $b$, and $c$, with $c$ the hypotenuse. Drawing the circles through any two of these points, we have $\frac{a^2\pi}{4}$, $\frac{b^2\pi}{4}$, and $\frac{c^2\pi}{4}$. We can also equate these through Pythagorean Theorem: $\frac{a^2\pi}{4}+\frac{b^2\pi}{4}=\frac{c^2\pi}{4}$.

...After that, I have no idea.[/quote]
Carry on youll get there, but can you answer what Ive asked.

How can I get there when there's nothing more I can do from there?

[quote="riddler"][quote]
In a right triangle, the length from the midpoint of the hypotenuse to the vertex of the right angle is equal to the length of half the hypotenuse.
[/quote]
Can anybody explain this and why it works for ALL triangles because it works for only 45 one. this is really weird. and I know for sure that the Circumcentre of a right triangle is located on the midpoint of the hypotenuse. Im correct right????[/quote]

I didn't say it was true, I just said it was potential. It works in a 30-60-90 triangle.
What's circumcentre?

Sorry Ive understood my query :blush:
circumcentre is the centre of the circle which has all the points of the triangle on the circumference

But, I've used the 236factorial triangle postulate (just kidding  :P ) there, so there must be another way, as you said, using basic geometry.

So they all give up then except Xantos eh?    :)  :D  :lol:

[quote="Xantos C. Guin"]$a^{2}+b^{2}=c^{2}$

By mupliplying both sides by $\frac{1}{2}\cdot\frac{\pi}{2^{2}}$ We get:

$\pi\cdot(\frac{a}{2})^{2}+\pi\cdot(\frac{b}{2})^{2}=\pi\cdot(\frac{c}{2})^{2}$

This means that the sum of the 2 semi circles add up to the hypotenuse's semi circle.

The area that the sheep have is equal to:

A(semicircle1)+A(semicircle2)-[A(semicircle3)-1]

By substitution:

A(semicircle1)+A(semicircle2)-A(semicircle1)-A(semicircle2)+1=sheep land

So the sheep have 1 acre of land.[/quote]

yep that looks right. good job.

[quote="236factorial"][quote="Xantos C. Guin"][quote="236factorial"]Does this rule exist?

In a right triangle, the length from the midpoint of the hypotenuse to the vertex of the right angle is equal to the length of half the hypotenuse.
[/quote]

[hide]The proof for this is alot easier with a circle.

Inscribe the triangle in a circle. The hypotenuse is the diameter or 2 radii and the midpt of the hypotenuse to the opposite vertex is another radii. So it works for all rt. triangles.[/hide][/quote]

Nice proof  :D  much easier.
Is there a name for this theorem?[/quote]

Not that I know of.

[quote="Xantos C. Guin"][quote="Treething"]It's the Treething Theorem.[/quote]

lol[/quote]

Hey Treething, you didn't make it.  :( Are you trying to take my credit? :mad:
 :lol:


geometry circumcircle concurrency ratio romania

Let $O$ be a point in the plane of the triangle $ABC$. A circle $\mathcal{C}$ which passes through $O$ intersects the second time the lines $OA,OB,OC$ in $P,Q,R$ respectively. The circle $\mathcal{C}$ also intersects for the second time the circumcircles of the triangles $BOC$, $COA$ and $AOB$ respectively in $K,L,M$.

Prove that the lines $PK,QL$ and $RM$ are concurrent.

It's a very nice problem.

We invert wrt $O$ and get this equivalent problem:

A line $\ell$ cuts $OA,OB,OC$ in $P,Q,R$ respectively. The same line cuts the lines $BC,CA,AB$ in $K,L,M$ respectively. Show that the circles $(OPK),(OQL),(ORM)$ also concur in a point different from $O$.

From Desargue's theorem we find that the mapping $P\leftrightarrow K,\ Q\leftrightarrow L,\ R\leftrightarrow M$ is an involution on the line $\ell$. At the same time, we know that a pencil of circles (the set of circles passing through two fixed points determines an involution on any line (which cuts them, of course; if a line cuts two of them then it cuts all of them-this is easy to prove). Assume the circles $(OPK),(OQL)$ also intersect at $X$. The circle $(OXR)$ must cut $\ell$ at $M$ because the involution is already determined by the pairs $P\leftrightarrow K,\ Q\leftrightarrow L$, so the circle $(OXR)$ is the same as the circle $(ORM)$, so the circle $(ORM)$ passes through $X$, Q.E.D.

Do you happen to know who proposed it?

Dear Grobber can you explain more??????
Wht's involution?
What's Desargue theorem?
I will be thankful if you answer me

An involution is a homographic transformation (which conserves the cross-ratio of four points) from the set of points of a line to the same set, such that $f(f(P))=P$ for all points $P$ of the line. Desargue's theorem (not the one that's so well-known, but another one): A line $\ell$ intersects opposite pairs of sides of a complete quadrilateral in pairs of points which correspond to each-other in an involution (on the line $\ell$).

Let O be a point in the plane of the triangle ABC
A circle T which passes throught O intersects the second time the lines OA,OB,OC in P,Q,R and also  intersects the second time the circles (BOC),(COA),(AOB) in K,L,M.
Prove that PK,QL,RM are concurrent

I have  solution:
Let H be the center of T
E,F,G are centers of (BOC),(COA),(AOB)
We have
PK,QL,RM  are concurrent <=> LP.QM.RK=PM.QK.RL <=> sinAOMsinBOKsinCOL=sinAOLsinBOMsinCOK <=> sinHGFsinHEGsinHFE=sinHFGsinHGEsinHEF
Because EH,FH,GH are concurrent so applying the Ceva sin we have done

Dear Mathlinkers,
an article concerning the “’Du cercle des huit points au cercle des six pieds or the theorem on the six pedals’’ has been put on my website with interesting examples.
http://perso.orange.fr/jl.ayme    vol. 9    p. 8
You can use Google translator
Sincerely
Jean-Louis

Can someone explain to me why we have this "PK,QL,RM are concurrent <=> LP.QM.RK=PM.QK.RL"


I played piano for five or six years and the cello for three years. The first was hate at first sight and the latter was only a little better. I quit them two years ago and one year ago, respectively. I feel so sad when I think of all the hours of my life I wasted on them...

This topic's already been discussed, I think.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19417

[quote="miraculouspostmaster"]I've heard somewhere that skill in math is sometimes related to skill in music.  Any other musicians out there?  What do you play/do?

I'm a pianist/percussionist (love the steel pan).[/quote]

I hope you don't spect it to be a norm. I have known a lot of mathematicians with skills in music, but most of the ones I know (and I swear, I know many) don't have any (and sadly, I include myself).

I said piano in the poll but i also play the violin. You should make choices for people that play more than one instrument.

I used to play piano, and I play cello right now.

I play bassoon, mainly, but I also play piano (classical and jazz), and percussion (mostly marimba), and a little bit of trumpet.  I LOVE music!!!!!!!!! :lol:

ive played cello for almost ten years and  i can play a couple songs on the guitar :D

I play piano (7 years) and viola (3 or 4 years)

ok ok i admit it. at my school i'm the singing mathnerd.  :blush:

i play the piano.

as for a correlation between math people and musicians, my friends can play some sweet music but they aren't as talanted at math :D

I said woodwind to this poll.  I play oboe(YAY), saxophone(jazz, mostly) and piano...

I agree you should let people pick twice because I play a lot of instruments....  :P

OBOE ROCKS!!!

I've been playing piano for something like 9-10 years and violin for around 5-6.

I've been playing piano for 9 years, quit flute after 3 years, and quit cello after...uh...about 5 lessons :?

I play the violin, but I've missed many concerts for math competitions. My orchestra teacher hates me. I have to miss a banquet for testing out of chemistry and precalc.


There are 8 red, 8 blue, and 8 yellow marbles in a jar. What is the fewest marbles you can remove from the jar so that the ratio of red to non red marbles is 3 to 7 and so that the ratio of yellow to non yellow marbles is also 3 to 7?

$ r: b\plus{}y\equal{}3: 7, 3b\plus{}3y\equal{}7r...A$
$ y: r\plus{}b\equal{}3: 7, 3r\plus{}3b\equal{}7y...B$

$ A\minus{}B\equal{}>3y\minus{}3r\equal{}7r\minus{}7y, 3(y\minus{}r)\equal{}\minus{}7(y\minus{}r)$. Therefore, y=r.

put that into A.

$ r: b\plus{}r\equal{}3: 7, 7r\equal{}3b\plus{}3r, 4r\equal{}3b. b\equal{}8, r\equal{}6, y\equal{}6$

8-8+8-6+8-6=4.

answer : 4

I remember taking that problem on the sprint at regionals or something like that...

 :blush: I got it wrong :blush:


modular arithmetic
number theory unsolved
number theory

[quote="al.M.V."]Does there exist an infinite set A of positive integers such that the sum of elements of every subset in A is composite but any pair of elements in A is coprime :?:[/quote]
How about this?
$2^1+1$, $2^2+1$, $2^4+1$, $\cdots$, $2^{2^n}+1$, $\cdots$

[quote="al.M.V."]Does there exist an infinite set A of positive integers such that the sum of elements of every subset in A is composite but any pair of elements in A is coprime :?:[/quote]

Just choose the elements one by one, by using the Chinese Remainder Theorem:

Choose any $a_1$ that you want. Now pick $a_2$ s.t. for some prime $p$ which does not divide $a_1$ we have $p^2|a_1+a_2$, and $a_2$ is congruent to $1\pmod a_2$. This is done by choosing the residues of $a_2$ modulo $p^2$ and $a_1$, which can be done as we please, by employing the above-mentioned theorem. Then, choose $a_3$ which is congruent to $1\pmod{a_2},\pmod{a_3}$ and s.t. for each non-empty subset $S$ of $\{a_1,a_2,a_3\}$ we have a large prime $p_S$ s.t. $p_S^2$ divides the sum of the elements in the respective subset. We choose the primes $p_S$ among the primes which do not divide the numbers $a_1,a_2,a_3$. Then, we choose $a_4$, and so on. I think you get the idea.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ K$ be a perfect field. Prove that the following are equivalent:

1. $ K$ has at most one extension of given degree.
2. Every finite extension $ L/K$ is cyclic.
3. $ \operatorname{Gal}(\tilde{K}/K)$ is procyclic, where $ \tilde{K}$ is the algebraic closure of $ K$.

1=>2: Apply "if a finite group has at most one subgroup of every order, then it is cyclic" to the Galois closure of $ L$.
2=>3: obvious
3=>1: obvious


inequalities
linear algebra
linear algebra unsolved

$n \geq 2$
1/ Give an exemple of matrices $A,B \in M_n(C)$  such that
$rg(AB)-rg(BA) = E(n/2)$
2/Prove that for any $X,Y \in M_n(C)$
$rg(XY) - rg(YX) \leq E(n/2)$

$E(n/2)$ is the integer part of $n/2$

[url= https://artofproblemsolving.com/community/c7h2576955_matrix_and_rank] see [/url]
https://artofproblemsolving.com/community/c7h2576955_matrix_and_rank

I didn't think too hard about $a$ and I'm sure an example isn't hard to give. As for $b$, we use the only inequalities involving rank which I'm familiar with: $r(AB)\le r(A),r(B),\ r(AB)\ge r(A)+r(B)-n$ (the last one is Sylvester's inequality).

We thus have $r(XY)-r(YX)\le r(X)-(r(x)+r(Y)-n),\ r(Y)-(r(X)+r(Y)-n)\Rightarrow r(XY)-r(YX)\le n-r(X),n-r(Y)\ (*)$. If one of $r(X),r(Y)$ is $\le [\frac n2]$, then $r(XY)-r(YX)\le r(XY)\le \min(r(X),r(Y))\le [\frac n2]$. On the other hand, if $r(X),r(Y)>[\frac n2]$, then by using $(*)$ we get the conclusion.


1. There is a game which is popular in Asia called 24. In it you are given 4 numbers and you are to make these numbers equal 24 using only parentheses and +, -, /, and *. Use 5, 5, 5, and 1 each exactly 1 time to make 24.

2. A marching band is on the field. The director notes that of the shortest people in each column, Al is the tallest. Also, of the tallest in each row, Bob is the shortest. Is Al taller than Bob?

3. If x, y are positive reals such that $x+y=2$, prove $xy(x^2+y^2)\le 2$.

4. 4 points are chosen from the boundary of a square such that the figure formed by connecting the points is a rhombus. Prove the rhombus is actually a square.

5. Forgetful me. (Can ya'll remember?)

4. something about a rhombus being a square, if it touches the boundary points in a square

5. dont remember

[hide="1"]Well, I don't know how you really write a solution for a 24 problem, but here we go :
To form 24, we must first form 25, 24, 23, 60, 29, 21, or $\frac{25}{4}$.  We notice that $\frac{25}{4}=5-\frac{1}{5}$ and we have our answer :  $5\times (5-1\div 5)$.[/hide]

[hide="3"]Since $x+y=2$, $x^2+y^2=-2xy+4$.  Using this, we have
\begin{eqnarray*}4xy-2(xy)^2&\le &2\\ 2(xy)^2-4xy+2&\ge &0\end{eqnarray*}
which is clearly true by Trivial Inequality.

Q.E.D.
(Of course, we should have done that backwards, but it doesn't matter here.)[/hide]
[hide="4"]Let us call the square $ABCD$ and its side length $s$, and let us call the rhombus $PQRS$, with $P$ on $AB$, $Q$ on $BC$, $R$ on $CD$, $S$ on $DA$.  Now let us draw $PR$.  Clearly, the triangles into which is partitions $PQRS$ are congruent; therefore $PBQ\equiv RDS$.  We can then use side lengths to prove that $PBQ\equiv QDR\equiv RDS\equiv SAP$, and then we use sums of angles to prove that one of the angles in $PQRS$ is a square.  (That's much more clear when you draw the diagram.)

Q.E.D.
[/hide]

[quote="solafidefarms"][quote="Boy Soprano II"][hide="2"]Let us say that that Al is in column A, and Bob is in row 1.  If Al is in row 1, then Bob is the tallest person in that row; therefore Bob is taller.  If Bob is in column A, then Al is the shortest person in that column; therfore Bob is taller.  If neither is true, let us say that Charles is the person in row 1, column A.  Since Al is the shortest one in that column, Charles is the taller than Al, and since Bob is the tallest in his row, he is taller than Charles.  Therefore, Bob is always taller than Al.

Q.E.D.[/hide][/quote]

Not so; you can arrange the people so that Al is taller or Bob is taller.[/quote]

Really?  How?  What about the person in Bob's row and Al's column, then?  They are arranged in a rectangle, aren't they?

I think I see.  We can arrange the people such that we can [i]argue[/i] that Al is taller (e.g., say that all the tallest people in a row are arranged in the same column), but we could then use the argument I outlined above to prove that Bob must also be taller and therefore obtain a contradiction (from which we would conclude that, for example, the tallest people row-wise are not in the same column).

Answer to #3

[hide]Lemma 1: $xy \le 1$

By the AM-GM Inequality,

$\frac{x+y}{2} \le \sqrt{xy}$

Simplifying, we have $\sqrt{xy} \le 1$

End Lemma 1

Lemma 2: $x^2 + y^2 \le 2$

$(x+y)^2 - 2xy = x^2 + y^2$, so we can substitue known values to yield $x^2 + y^2$.

We have $x+y=2$ originally, and by lemma 1, we have $xy \le 1$. Thus, we have

$(x+y)^2 - 2xy = x^2 + y^2 \le 2$

End Lemma 2

Multiplying $xy$ by $(x^2 + y^2)$ yields $xy(x^2 + y^2) \le 2$, due to lemmae 1 and 2. Thus, our proof is complete.[/hide]


[quote="thanhnam2902"][b]Let $ a > 0, b > 0,c > 0$ and $ abc \equal{} 1$.
Prove that: $ \frac {1}{x^3 \plus{} y^3 \plus{} 1} \plus{} \frac {1}{y^3 \plus{} z^3 \plus{} 1} \plus{} \frac {1}{z^3 \plus{} x^3 \plus{} 1} \le 1$.[/b][/quote]

it has already posted before

$ a > 0, b > 0,c > 0$ and $ abc \equal{} 1$ ? --> $ x > 0, y > 0,z > 0$ and $ xyz \equal{} 1$
$ x^{3}\plus{}y^{3}\geq xy(x\plus{}y)$ ---> $ \frac {1}{x^3 \plus{} y^3 \plus{} 1} \plus{} \frac {1}{y^3 \plus{} z^3 \plus{} 1} \plus{} \frac {1}{z^3 \plus{} x^3 \plus{} 1}\leq \sum\frac{1}{xy(x\plus{}y)\plus{}1}\equal{}\sum\frac{z}{x\plus{}y\plus{}z}\equal{}1$


Ok so I am deeply troubled. There was this question on the preliminary exam this year, concerning two cylinders with equal mass distributions, that are brought together and in the end they both come to rest. So the question was which of the following has to be true. The choices basically narrowed down to (w_1)(r_1^3)=(w_2)(r_2^3) and (w_1)(r_1^4)=(w_2)(r_2^4). When I try to solve the problem by using the fact that the two cylinders exert equal and opposite forces on each other and then manipulate the stuff I get the first answer. But then when I try using conservation of angular momentum, I get the second answer is correct. Which one is actually right? Please explain to the best of your abilities.

Could you please clarify how you got the first option.We get the second one by using conservation of angular momentum which is correct beyond doubt,as there is no external torque acting.So I think second option is correct

Well so by newton's third law, the frictional force of cylinder 1 on cylinder 2 is equal in magnitude and opposite to the forse of 2 on 1. Then dL_1 = (F)(r_1)dt and dL_2 = (F)(r_2)(dt). Then you intergrate both of the things and divide the first by r_1 and the second by r_2 and set them equal since they're both equal to the integral of F with respect to time. What I think i forgot to mention last time: both of the cylinders have an equal and uniform density. After posting yesterday, I thought about this problem some more and concluded that there must be exteernal torques on the system, because if there weren't then the two cylinders would start moving in opposite directions. Is this correct? If not, why so?

Yes... there are forces acting on the axes of the cylinders to keep them in place. Think about it. Each cylinder is acted on by the other with a frictional force originating from the point of contact, but both cylinders are stationary. Thus, there must be support forces acting on the axes of the cylinders. These forces generate a torque which decreases the angular momentum of the system.

Ok, but don't these forces have a moment arm of lenght zero so there's no external torque anyway?

There are four forces involved here: the frictional forces acting on cylinder 1 and on cylinder two (maybe that's just one), and the two support forces acting on the cylinder axes. Depending on where you set your origin, some of these forces may produce zero torque. But since they do not all act on the same point, the net torque is never 0.

For example, if you choose your origin to be at the contact point between the two plates, the torque due to the frictional force is 0. However, then the support forces will generate torques. In fact, since the frictional force is within the system, it will never generate any torque, and since the support forces form a couple, they will generate the same torque about any point.


Let $ a,b,c$ be nonnegative real numbers, not all are zero. Prove that
\[ \frac{a}{a\plus{}b\plus{}7c}\plus{}\frac{b}{b\plus{}c\plus{}7a}\plus{}\frac{c}{c\plus{}a\plus{}7b} \plus{}\frac{2}{3}\cdot \frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \le 1\]
Equality holds if and only if $ (a,b,c) \sim (1,1,1), (2,1,0), (1,0,0).$
:)

[quote="vanhoadh"]My solution is pqr.  :)

With $ a\leq c\leq b$:
$ LHS \equal{} \frac {2}{3}\frac {q}{p^2 \minus{} 2q} \plus{} \\
\plus{} \frac {7(p^3 \minus{} 3pq \plus{} 3r) \plus{} 33(pq \minus{} 3r) \plus{} 24r \plus{} 18\sqrt { \minus{} 4q^3 \plus{} p^2q^2 \minus{} 27r^2 \minus{} r(4p^3 \minus{} 18pq)}}{7(p^3 \minus{} 3pq \plus{} 3r) \plus{} 57(pq \minus{} 3r) \plus{} 366r}$[/quote]
I am looking for another solution. ;)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, not all are zero. Prove that
\[ \frac {a}{a \plus{} b \plus{} 7c} \plus{} \frac {b}{b \plus{} c \plus{} 7a} \plus{} \frac {c}{c \plus{} a \plus{} 7b} \plus{} \frac {2}{3}\cdot \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2} \le 1
\]
Equality holds if and only if $ (a,b,c) \sim (1,1,1), (2,1,0), (1,0,0).$
:)[/quote]
Any nice solutions?  :(

I'm sure you won't like this solution either  :oops: but at least it's simple.

We set $ a\equal{}\min\{a,b,c\}$, we have $ b\equal{}a\plus{}x,c\equal{}a\plus{}y$ with $ x,y\ge0$ and after clearing denominators we find the equivalent one \[ x^4y\plus{}3x^3y^2\minus{}24x^2y^3\plus{}28xy^4\plus{}270a^3(x^2\minus{}xy\plus{}y^2)\plus{}3a^2(97x^3\minus{}24x^2y\plus{}3xy^2\plus{}97y^3)\plus{}a(29x^4\plus{}206x^2y\minus{}267x^2y^2\plus{}260xy^3\plus{}29y^4).\] The funny thing is that each coefficient is non-negative:
$ x^4y\plus{}3x^3y^2\minus{}24x^2y^3\plus{}28xy^4\equal{}xy(x\minus{}2y)^2(x\plus{}7y),$
$ x^2\minus{}xy\plus{}y^2\equal{}(x\minus{}y)^2\plus{}xy,$ $ 97x^3\minus{}24x^2y\plus{}3xy^2\plus{}97y^3\equal{}73(x^3\plus{}y^3)\plus{}27xy^2\plus{}24(x\plus{}y)(x\minus{}y)^2,$ and $ 29x^4\plus{}206x^2y\minus{}267x^2y^2\plus{}260xy^3\plus{}29y^4\equal{}$
$ \equal{}29(x^4\plus{}y^4)\plus{}\frac{267}2xy(x\minus{}y)^2\plus{}\frac{145}2x^3y\plus{}\frac{253}2xy^3.$
So that's it. Of course the equality cases are the ones you mention.

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, not all are zero. Prove that
\[ \frac {a}{a \plus{} b \plus{} 7c} \plus{} \frac {b}{b \plus{} c \plus{} 7a} \plus{} \frac {c}{c \plus{} a \plus{} 7b} \plus{} \frac {2}{3}\cdot \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2} \le 1
\]
Equality holds if and only if $ (a,b,c) \sim (1,1,1), (2,1,0), (1,0,0).$
:)[/quote]
Although I have not solved completely the problem with three equality, I would like to discuss about it.  :)

The inequality can rewrite  as follow with the denoting $ a\plus{}b\plus{}c\equal{}3$ for more simple
\[ \Leftrightarrow \frac{{ab}}{{1 \plus{} 2b}} \plus{} \frac{{bc}}{{1 \plus{} 2c}} \plus{} \frac{{ca}}{{1 \plus{} 2a}} \ge \frac{{ab \plus{} bc \plus{} ca}}{{a^2  \plus{} b^2  \plus{} c^2 }}
\]
The three equalty case I think it is base on the triple $ \left( {ab,bc,ca} \right)$.

Of cause I think the problem is true and follow this, after expanding then actually we can get a very nice inequalty of the nice cyclic expression $ a^2 b \plus{} b^2 c \plus{} c^2 a$ is that
\[ 2\left( {a^2 b \plus{} b^2 c \plus{} c^2 a} \right) \ge \frac{{q\left( {4q \plus{} 8r \plus{} 7} \right)}}{{9 \minus{} 2q}} \minus{} \left( {q \plus{} 18r} \right)
\]
 :)
Hope we will have a nice solution for it soon, actually not from the author.  :oops:

[quote="Honey_S"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, not all are zero. Prove that
\[ \frac {a}{a \plus{} b \plus{} 7c} \plus{} \frac {b}{b \plus{} c \plus{} 7a} \plus{} \frac {c}{c \plus{} a \plus{} 7b} \plus{} \frac {2}{3}\cdot \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2} \le 1
\]
Equality holds if and only if $ (a,b,c) \sim (1,1,1), (2,1,0), (1,0,0).$
:)[/quote]
Although I have not solved completely the problem with three equality, I would like to discuss about it.  :)

The inequality can rewrite  as follow with the denoting $ a \plus{} b \plus{} c \equal{} 3$ for more simple
\[ \Leftrightarrow \frac {{ab}}{{1 \plus{} 2b}} \plus{} \frac {{bc}}{{1 \plus{} 2c}} \plus{} \frac {{ca}}{{1 \plus{} 2a}} \ge \frac {{ab \plus{} bc \plus{} ca}}{{a^2 \plus{} b^2 \plus{} c^2 }}
\]
The three equalty case I think it is base on the triple $ \left( {ab,bc,ca} \right)$.

Of cause I think the problem is true and follow this, after expanding then actually we can get a very nice inequalty of the nice cyclic expression $ a^2 b \plus{} b^2 c \plus{} c^2 a$ is that
\[ 2\left( {a^2 b \plus{} b^2 c \plus{} c^2 a} \right) \ge \frac {{q\left( {4q \plus{} 8r \plus{} 7} \right)}}{{9 \minus{} 2q}} \minus{} \left( {q \plus{} 18r} \right)
\]
:)
Hope we will have a nice solution for it soon, actually not from the author.  :oops:[/quote]
Who can prove this one ? :wink:
\[ 2\left( {a^2 b \plus{} b^2 c \plus{} c^2 a} \right) \ge \frac {{q\left( {4q \plus{} 8r \plus{} 7} \right)}}{{9 \minus{} 2q}} \minus{} \left( {q \plus{} 18r} \right)\]
The idea is let find the minimum value of $ a^2 b \plus{} b^2 c \plus{} c^2 a\equal{}f(r)$  :wink:

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, not all are zero. Prove that
\[ \frac {a}{a + b + 7c} + \frac {b}{b + c + 7a} + \frac {c}{c + a + 7b} + \frac {2}{3}\cdot \frac {ab + bc + ca}{a^2 + b^2 + c^2} \le 1
\]
Equality holds if and only if $ (a,b,c) \sim (1,1,1), (2,1,0), (1,0,0).$
:)[/quote]
Now, I will post my solution for this inequality.
Because
$ \frac{a}{a+b+c}-\frac{a}{a+b+7c} =\frac{6ca}{(a+b+c)(a+b+7c)},$
it suffices to prove that
$ \sum \frac{ca}{a+b+7c} \ge \frac{(a+b+c)(ab+bc+ca)}{9(a^2+b^2+c^2)}.$
If $ a=b=0$ or $ b=c=0$ or $ c=a=0$, the inequality becomes equality. For $ a+b>0,b+c>0,c+a>0,$ applying the Cauchy Schwarz Inequality, we get
$ \sum \frac{ca}{a+b+7c} \ge \frac{(ab+bc+ca)^2}{\sum ca(a+b+7c)}.$
The inequality is reduced to
$ 9(ab+bc+ca)(a^2+b^2+c^2) \ge (a+b+c)(7 \sum a^2b+\sum ab^2+3abc),$
or
$ \sum a^3b+ 4\sum ab^3 -4\sum a^2b^2 \ge abc\sum a,$
that is
$ \sum ab(a-2b)^2 \ge abc\sum a.$
Applying the Cauchy Schwarz Inequality again, we get
$ [ \sum ab(a-2b)^2] ( \sum c) \ge [ \sum \sqrt{abc} (a-2b)]^2 =abc (a+b+c)^2,$
hence
$ \sum ab(a-2b)^2 \ge abc\sum a,$
and our proof is completed. :)

Thank you very much everyone for spending your time on reading this proof.
Regards
V. Q. B. Can


[quote="sam3.14"]How would you graph the function
\[
f(x)=
\left\{
\begin{array}{ll}
1 & x\text{ is rational}\\
-1 & x\text{ is irrational}\\
\end{array}
\right.
\]

Would it be two dotted lines on 1 and -1? That seems logical, but I thought a dotted line meant that the graph wasn't actually there, like an asymptote. Some help would be greatly appreciated.

Thanks[/quote]There's a few things to understand here..

- It's not physically possible to graph this. The best you could do would be to draw straight lines at $y=\pm1$, but this isn't accurate since there are no actual lines in the set of points $(x,f(x))$ graphed on the plane. They're very much like lines though.

- It wouldn't look like dotted lines, since this implies that there are gaps on the lines $y=\pm1$, which is not the case since the sets of rationals and irrationals are dense.

- For your last question, we do draw dotted lines to represent the asymptotes on graphs, but the use of "dotted lines" is arbitrary. We could draw wavy lines or two parallel lines really close together or whatever. You have to keep in mind that what we draw on paper is only a representation of the mathematical ideas. So if you draw a graph of a function that will actually have dotted lines in it, that doesn't make it an asymptote.


Is there any way I can get copies of the pre and post tests for Intermediate Trig, Intermediate Counting, and Olympiad Geometry? I want to try to take some of these and independent study in my school, but my school needs some more info on the classes.

Thanks.

If you send me your email address I will send you copies of them on Monday.  My email is crawford@artofproblemsolving.com.


[quote="splatyango"]lol who doesn't take school?

and who would be K-3?[/quote]

Some people are homeschooled because the school is too far, too easy, or something like that.

oh.

[quote]The message is too small. Please make the message longer before submitting.[/quote]

Oh well. blah.

going to be a freshman next year.
and even if your homeschoold you technically still go to school.

Going to 9th...

What happens if you are in the middle of skipping a grade (like 5th before summer, 7th after)? What would they put?

[quote="kookamunga"][quote][quote]
lol who doesn't take school?

and who would be K-3?[/quote]

Some people are homeschooled because the school is too far, too easy, or something like that.[/quote]

Being homeschooled is still taking school...[/quote]

yeah. not taking school is probably like the beyond category. or ur some genius 4yr. old that isn't even IN school yet :D

Going into 10th! Woohoo! (see siggie =D)

[quote]umm shouldn't this be under polls? anyway. freshman in high school next year =][/quote]

Polls got merged into G&FF.

EDIT: Ackkkkkkkkk why doesn't quotes with names work anymore? WAAAAAAAAAAAAAAAAAAAH!

[quote="dreambold22"]6th. i feel small... :([/quote]

Yeah, but you have more of a chance to study for mathcounts! 2 years!

Goin into 8th! Last year fo' mathcounts.

[quote]5th... But I should have put "Don't take school" because I'm Homeschooled.[/quote]

lol as we said before being homeschooled still counts as taking school.

[quote="dragon96"]Some people are homeschooled because the school is too far, too easy, or something like that.[/quote]
 :o Someone actually understands a reason why people homeschool! I feel so invigorated!

I am in 8th grade(As of this fall), and guess which school I go to.

I'm a freshman in Fall 2008 and I get a $ \$$17,000 scholarship to cover my first year at this awesome boarding school.  :D

[quote="Nexmus"][quote="splatyango"]lol who doesn't take school?

[/quote]

Me.

[quote="splatyango"]and who would be K-3?

[/quote]

Me.[/quote]

How can you not take school and be K-3 at the same time? :wink:


Fie $a_{1},a_{2},...,a_{n}>0$ pentru care suma dintre suma si produsul lor este $n+1$. Demonstrati ca \[\sum_{i=1}^{n} a_{i} >= \sum_{i=1}^{n} \sqrt{a_{i}}. \]

[quote="harazi"]Hmmm... nu cumva ai postat-o si pe proposed inequalities forum??? ;) . Oricum, AM-GM+Cauchy=solutie.  :P[/quote]
Da, dar oricum era adresata unei anumit domeniu de oameni ... (clasa a 8-9).
Nu trebuie sa-i uitam.

[quote="Cristi Sav"]Fie $a_{1},a_{2},...,a_{n}>0$ pentru care suma dintre suma si produsul lor este $n+1$. Demonstrati ca \[ \sum_{i=1}^{n} a_{i} \geq  \sum_{i=1}^{n} \sqrt{a_{i}}.  \][/quote]
   Cred ca merge si cu Cebisev.


Find the value of the infinite product

$ \prod^{\infty}_{n\equal{}1} \frac{2^n \plus{} 1}{2^n \plus{} 2} \equal{} \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{9}{10} \cdots$

[hide="Answer"]
$ \prod_{n\equal{}1}^{k} \frac{2^n\plus{}1}{2(2^{n\minus{}1}\plus{}1)}$ telescopes to $ \frac{3}{4} \cdot \frac{5}{2(3)} \cdot \frac{9}{2(5)} \cdots \equal{} \frac{2^{k}\plus{}1}{2^{k\plus{}1}}$. As $ k \rightarrow \infty$ this approaches $ \frac{1}{2}$.
[/hide]

Thanks! I tried to this problem but I couldn't see how it telescoped (I got to the factoring on the first line but then I got stuck).


I need a way to solve this equation:
[img]http://i36.tinypic.com/s9yb.jpg[/img]

hello, after many elementary calculations i've got the solution $ x\equal{}\minus{}3$.
Sonnhard.

i did not understand how you got this equation.can you please write the full way?

Hello !

first of all, remark that $ (3x^2 \plus{} 8x \plus{} 4) \minus{} (3x^2 \plus{} 6x \minus{} 2) \equal{} 2(x \plus{} 3)$ and [b]magically[/b] :wink: $ (x^2 \minus{} 1) \minus{} (x^2 \minus{} 3x \minus{} 10) \equal{} 3(x \plus{} 3)$....so I write like that:

$ \frac {\sqrt {3x^2 \plus{} 6x \minus{} 2} \minus{} \sqrt {3x^2 \plus{} 8x \plus{} 4}}{1} \equal{} \frac {\sqrt {x^2 \minus{} 1} \minus{} \sqrt {x^2 \minus{} 3x \minus{} 10}}{1}$

And now, to use my remark, you must multiply left and right by the "conjuguée"(sorry, I don't know the english term)[hide]$ \frac {\sqrt {a} \minus{} \sqrt {b}}{1} \equal{} \frac {\left(\sqrt {a} \minus{} \sqrt {b}\right)\left(\sqrt {a} \plus{} \sqrt {b}\right)}{1\times\left(\sqrt {a} \plus{} \sqrt {b}\right)} \equal{} \frac {a \minus{} b}{\sqrt {a} \plus{} \sqrt {b}}$ and so on, with appropriate $ a$ and $ b$ ...The rest is like in the solution of mitdac123[/hide]


Outline a method for the synthesis of p-nitrophenol [i]with good yield[/i].

Can you tell me what reagents can one use in an organic synthesis?? I am new to this.....pls guide me  :(

Can we start with phenol??  :maybe:

[hide]first acetylating phenol ( i.e treating with $ (CH_3CO)_2 O$ in Pyridine) will give phenyl acetate.......nitrating this will give us para-nitro phenyl acetate.........acid catalysed hydrolysis of this will give para-nitro phenol!   :maybe: [/hide]

wouldn't nitrating phenyl acetate give both ortho- and para-nitro phenyl acetate?
also, esterification is an equilibrium process. would that result in a [i]good[/i] enough yield?

[quote="jdlkem"]wouldn't nitrating phenyl acetate give both ortho- and para-nitro phenyl acetate?
[/quote]

i think steric factor comes into picture ..  :wink:
so i feel para is the major product

[quote="Jdlkem"]also, esterification is an equilibrium process.[/quote]

That's true, but only with Fischer esterification, i.e., acid catalized reaction of an alcohol with an acid. On this case, we have essentially the reaction of a base with an anhydride, which is more reactive than an acid.

[quote="Da Vinci"]so i feel para is the major product[/quote]

That's right, but the ortho will always form in some amount.

Its acceptable, but other possibilities could be considered.

From phenol,
Protect OH using benzene sulfonate chloride and nitrate to get para product chiefly followed by hydrolysis..

That is roughly the same as the acetate based route.

i think going via  nitroso formation will help us,coz the nitroso compound formed will be predominantly para. and then oxidise it to nitro.

Finally. That was the route I had in mind. The nitrosyl cation, being only weakly electrophilic, reacts almost only at the para position. Then carefull oxidation yields p-nitrophenol.


Find all trinom  f(x) = x^2 + px + q with nonnegative integer coefficient, such that if GCD(f(a),f(b)) = 1 then GCD(a,b) = 1.

I think the condition is equivalent to $q=0$. Let's assume $q\neq 0$. We take $b>1$ s.t. $(b,q)=1$ and $a=kb$ s.t. $(k-1,f(b))=((k+1)b+p,f(b))=1$ and we have found $a,b$ s.t. $(a,b)=b<>1$ and $(f(a),f(b))=1$, which is a contradiction. This shows that the condition $q=0$ is necessary. It's obvious that it's also sufficient, so the trinomials we're looking for are $x^2+px,\ p\in N$.

Is it right?

You have forgot to explain the most important step of the solution: how can you be sure that b and k can be chosen with so many properties?

I think you're right to some extent..

First of all, since $q<>0$, we can definitely can choose $b$ s.t. $(b,q)=(b,f(b))=1$. Now forget about $b$ (we may consider it fixed).

If we take $\alpha$ to be a prime dividing $f(b)$, all we need to do is find a $k$ s.t. $\alpha$ doesn't divide $k-1$ or $(k+1)b+p$. This can obviously be done if $\alpha >2$, because the equations $k-1=0\ (mod\ \alpha )$ and $(k+1)b+p=0\ (mod\ \alpha )$ have at most two solns in total (in $Z_{\alpha }$), and we can choose $k$ to have a residue different from the two when divided to $\alpha$.

The only problem arises when $\alpha =2$, and we're only forced to accept this when $2|f(b)$ for all $b$, and this only happens when $p$ is odd and $q$ is even. In this case $b$ must be odd, $k$ must be even, so $(k+1)b+p$ will be even, and we didn't want this.

I guess (at least) this reduces it to the case when $p$ is odd and $q$ is even..

Sorry  :blush:

Stupid me! If $p$ is odd and $q$ is even, then we can never have $(f(a),f(b))=1$ (because $f(a)$ is even for all $a\in Z$), so $(f(a),f(b))=1$ implies anything we want, because it's a false sentence and false $\Rightarrow$ anything we want, so $(f(a),f(b))=1\Rightarrow(a,b)=1$, so this case works as well.

I think the final answer is:

$q=0,p\in N$
$q\in 2N,p\in 2N+1$.

Is this wrong as well? :)


Prove that there is a number such that its decimal represantation ends with 1994 and it can be written as $1994\cdot 1993^{n}$ ($n\in{Z^{+}}$)

It sufficies to prove that the exists $n\in\mathbb{Z}^{+}$ such that $1993^{n}\equiv 1\pmod{10000}$. Of course $1993\bot 10000$ and $\phi(10000)= 4000$, so $1993^{4000}\equiv 1\pmod{10000}$ and $4000$ is the desired $n$ which ends the proof.

i know this is a very easy question but the other ones aren't easy like this.

[quote="Gyan"][quote="Rzeszut"]......  such that $1993^{n}\equiv 1\pmod{1000}$.  [/quote] you may like to edit out a typo - should be 10000 instead of 1000.[/quote]
You're right  :blush: . I corrected it.

[quote="D.P.L"]But if $1994 \times 1993^{n}\cong 1994 (mod 10000)$, but gcd(1994,10000) is not equal to 1. Sorry if this sounds stupid :ninja:[/quote]

That means $1994 \cdot 1993^{n}\equiv 1994 \pmod{10000}\not\Rightarrow 1993^{n}\equiv 1 \pmod{10000}$ but we still have $1993^{n}\equiv 1 \pmod{10000}\Rightarrow 1994 \cdot 1993^{n}\equiv 1994 \pmod{10000}$.


superior algebra
superior algebra unsolved

1.
let $ k,m\in\mathbb{N}$ and define
\[ S(m,k) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {1}{n(mn \plus{} k)}
\]
prove that S(m,k) rational iff $ m|k$.

2.
my friend solved it :)

3.
$ S(n,k)$ = sum of digits of $ k$ written in the system having base$ n$.
is
\[ \sum_{k \equal{} 1}^{\infty}\frac {S(n,k)}{k(k \plus{} 1)}
\]
a rational number or not ?
Can you count it (give an explicit value) ?