inequalities symmetry trigonometry AMC USA(J)MO USAMO quadratics

[quote="cn2_71828182846"]Can someone explain a solution to USAMO 2001 problem 3?  I am fine on the left inequality, but I'm a little hazy on the right-hand one.

Thanks for assistance.

Let a, b, c, be  :ge: 0 and satisfy

a :^2:  + b :^2:  + c :^2:  + abc = 4.

Show that

0  :le:  ab + bc + ca - abc  :le:  2.[/quote]

I'm not so sure but I'll give it a shot.

-abc = a :^2:  + b :^2:  + c :^2: - 4
0 :le: ab + bc + ca + a :^2:  + b :^2:  + c :^2: - 4 :le: 2
4 :le: ab + bc + ca + a :^2:  + b :^2:  + c :^2: :le: 6
4 :le: [(a+b):^2: + (b+c):^2: + (a+c):^2:]/2 :le: 6
8 :le: (a+b):^2: + (b+c):^2: + (a+c):^2: :le: 12

That's all I've gotten so far, I'll work on it later.

I think I got the first part
If one of a,b,c is 0, we are done. So assume a,b,c >0

By AM-HM(or cauchy):
[tex]\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}[/tex]

By the equation, a, b, c :le: 2. So a+b+c can be at most 6.
Therefore [tex]\frac{9}{a+b+c} \geq 1[/tex]
and
[tex]\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 1[/tex]
multiply both side by abc gives us the first inequality.

For the left side, just notice that not all of the variables can be >1, so WLOG assume a<=1. Then, bc(1-a)+a(b+c) >= 0 since each term is non-negative.

It's the right side that is difficult. I have a solution, but it isn't entirely mine so I feel guilty posting it.

following off beta's work (i'm not quite sure if this works):

The last inequality holds true if
[tex]
2abc \ge 2 \\
abc \ge 1 \\
\sqrt[3]{abc} \ge 1\\[/tex]

By GM-HM, we can substitute to get
[tex]
\displaystyle \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \ge 1 \\
3 \ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\\
3abc \ge ab+bc+ca [/tex]

From cauchy or AM-GM, we find
[tex]a^2+b^2+c^2 \ge ab+bc+ca \\[/tex]

So we substitute again
[tex]3abc \ge a^2+b^2+c^2 \\[/tex]

Sorta guessing, but I think it can be proved that as long as the condition is satisfied, this inequality holds true, but i dont know how to do that.

[quote="paladin8"][tex]3abc \ge ab+bc+ca [/tex]

From cauchy or AM-GM, we find
[tex]a^2+b^2+c^2 \ge ab+bc+ca \\[/tex]

So we substitute again
[tex]3abc \ge a^2+b^2+c^2 \\[/tex]
[/quote]
That' not true, you can't substitute like that.

Yes, it works. If this is proved:
[tex]3abc \ge a^2+b^2+c^2 \\[/tex]

and this is true:
[tex]a^2+b^2+c^2 \ge ab+bc+ca\\[/tex]

then this is true:
[tex]3abc \ge ab+bc+ca\\[/tex]

which is what we wish to prove.

ok, i see. I thought you proved
[tex]3abc \ge ab+bc+ca [/tex]
with these two steps, not realizing that you are working backwards
[tex]a^2+b^2+c^2 \ge ab+bc+ca \\[/tex]
[tex]3abc \ge a^2+b^2+c^2 \\[/tex]

[quote="beta"]All I have for the second inequality:

[tex]a^2+b^2 \geq 2ab[/tex]
[tex]b^2+c^2 \geq 2bc[/tex]
[tex]a^2+c^2 \geq 2ac[/tex]
add all together and divide by 2
[tex]a^2+b^2+c^2 \geq ab+bc+ac[/tex]
so
[tex]ab+bc+ac+abc \leq 4[/tex]
[tex]ab+bc+ac-abc \leq 4-2abc[/tex]

I will try to do figure it out later.[/quote]

I don't think my idea is going to lead anywhere. (It's too strong.) To prove this statement to be true, we need [tex]abc \ge 1[/tex]. However, by applying AM-GM to [tex]a^2 + b^2 + c^2 + abc = 4[/tex],

[tex]\displaystyle{{a^2 + b^2 + c^2 + abc}\over{4}} \ge \sqrt[4]{a^3b^3c^3}[/tex]

From which we obtain [tex]abc \le 1[/tex].

Hmm, thinking on then...

Paladin: your steps don't hold. Try a = 2, b = c = 0.

Some more work
One of a, b, c must be < or =1, one must be > or =1. Let a=1+x, let b=1-y.
, where [tex]x, y \leq 1[/tex]
 so
[tex](1+x)(1-y)c \leq 1[/tex]

[tex](1+x)^2+(1-y)^2+c^2+(1+x)(1-y)(c)=4[/tex]
since [tex]x^2+y^2+c^2-xyc \geq 0[/tex] because
[tex]xy+yc+xc-xyc=xy+yc+xc(1-y) \geq 0[/tex]
so we have
[tex](2+c)(1+x-y) \leq 4[/tex]

[quote]Paladin: your steps don't hold. Try a = 2, b = c = 0. [/quote]

I assumed a, b, c > 0 because it is trivial to prove if any are 0 as noted by beta earlier on.

Let a = 1.999, b = c = 0.01. It still doesn't work.

that doesnt satisfy the original condition, but i have also found a counterexample, (1.99, .1, .1)

The problem is where you do $2(x^2+y^2+z^2+xy+yz+zx)\le 4(x^2+y^2+z^2)$. This is true, but there's no reason to expect that the RHS will also be at most $3$. In fact you've just proven that $2(x^2+y^2+z^2+xy+yz+zx)\le 4(x^2+y^2+z^2)$ and $4(x^2+y^2+z^2) \ge 3$; perfectly true statements, just that they don't solve the problem.

[b]Another way to solve the problem :[/b]

We may take $a=\frac{2x}{\sqrt{(x+y)(x+z)}},b=\frac{2y}{\sqrt{(y+z)(y+x)}},c=\frac{2z}{\sqrt{(z+x)(z+y)}}$ where $x,y,z$ are positive real numbers.

For the first inequality, observe that by AM-GM it is enough to show that $ab+bc+ca \ge 3 \cdot (abc)^{\frac{2}{3}} \ge abc$. That is $27 \ge abc$, which is clear.

For the second inequality, write $a=2 \cos A, b=2 \cos B, c=2 \cos C$ where $A+B+C=\pi$, thanks to the hypothesis and the identity $\cos^2 A+\cos^2 B+\cos^2 C+2 \cos A \cos B \cos C=1$.
Rewrite the desired as \[ \sum_{cyc} 2 \cos A \cos B + \sum_{cyc} \cos^2 A - 1 = \sum_{cyc} (\cos A + \cos B)^2 - 1 \le 2 \]
Subtracting $\cos^2 A + \cos^2 B + \cos^2 C$ from both sides, this becomes \[ (\cos A + \cos B + \cos C)^2 \le \sin^2 A + \sin^2 B + \sin^2 C \]
Now we will prove two trig identities. \\
[b] Lemma 1:[/b] $\sum_{cyc} \cos A \sin B \sin C = \frac{1}{2} (\sin^2 A + \sin^2 B + \sin^2 C)$ where $A+B+C = \pi$. \\
[b]Proof:[/b] \begin{align*}
\cos A \sin B \sin C
&= \frac{1}{2} \cos A (\cos (B-C) + \cos A) \\
&= \frac{1}{2} \cos^2 A + \frac{1}{4} (\cos(A+B-C)+\cos(A+C-B)) \\
&= \frac{1}{2} \cos^2 A - \frac{1}{4} (\cos 2B + \cos 2C) \\
&= \frac{1}{2} \cos^2 A + \frac{1}{2} \sin^2 B + \frac{1}{2} \sin^2 C - \frac{1}{2} \\
&= \frac{1}{2} (\sin^2 B + \sin^2 C - \sin^2 A)
\end{align*}
Sum cyclically. \\
[b]Lemma 2:[/b] $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C$ where $A+B+C = \pi$. \\
[b]Proof: [/b] \begin{align*}
\sin 2A + \sin 2B + \sin 2C
&= 2 \sin (A+B) \cos (A-B) + \sin 2C \\
&= 2 \sin C (\cos (A-B) + \cos C) \\
&= 2 \sin C (\cos(A-B) - \cos (A+B)) \\
&= 4 \sin A \sin B \sin C
\end{align*}
As a simple consequence, we have that
\begin{align*}
\sum_{cyc} \frac{\cos A}{\sin B \sin C}
&= \frac{\sin A \cos A + \sin B \cos B + \sin C \cos C}{\sin A \sin B \sin C} \\
&= \frac{\sin 2A + \sin 2B + \sin 2C}{2 \sin A \sin B \sin C} \\
&= 2
\end{align*} \\
Now we apply Cauchy to finish the proof, combined with Lemmas 1 and 2:
\[ \sin^2 A + \sin^2 B + \sin^2 C = (\sum_{cyc} \cos A \sin B \sin C)(\sum_{cyc} \frac{\cos A}{\sin B \sin C}) \ge (\cos A + \cos B + \cos C)^2 \]
We are done.

[quote=v_Enhance]
[i]Case 3:[/i] Now suppose that $a=b \neq c$. That means $\lambda = \frac{1}{2-c}$
[/quote]
Since we have that
\[ (a-b)(\left[ 2\lambda - 1 \right] - \lambda c) = 0. \]
The result quoted seems to point to the fact that we need $(\left[ 2\lambda - 1 \right] - \lambda c)$ to equal zero, which confuses me. Isn't $(a-b)(\left[ 2\lambda - 1 \right] - \lambda c)$ already zero because $a=b$?
Also, how does this work:
[quote=v_Enhance]Now \[ 2a+2c = a+b+2c = \lambda (2c+ab)
	= \frac{1}{2-c} (2c+a^2) \] which implies \[ 4a+4c-2a^2-2ac = 2c+a^2 \] [/quote]
Should it be $\lambda=\frac{1}{2-a}$?

A final more general question about process, what was the motivation for letting having the condition for $U$ be $a^2+b^2+c^2<1000$? Why not let it be say $999$ or $1001$?. Is there anything specific about $1000$, or is it just a matter of choosing a number that is sufficiently large?

$\lambda = \frac{1}{2-a}$ was meant in both places. Thanks, I'll fix it.

There's nothing special about $1000$ of course: as you say any large number suffices. The point is to capture the constraint set in some compact set, for which the boundary ($a^2+b^2+c^2=1000$ in this case) is a non-issue.

If $a,b,c\ge0$ such that $a^2+b^2+c^2+abc=4$, [url=https://artofproblemsolving.com/community/c6h1847853p12451169]then[/url]
$$a+b+c\le\sqrt{8+abc}\qquad$$

[quote=v_Enhance]OK let's practice bashing.

The left-hand side of the inequality is trivial; just note that $\min \left\{ a,b,c \right\} \le 1$.  Hence, we focus on the right side.
We use Lagrange Multipliers.

Define \[ U = \left\{ (a,b,c) \mid a,b,c > 0 \text{ and } a^2+b^2+c^2 < 1000 \right\}. \] This is an intersection of open sets, so it is open. Its closure is \[ \overline U = \left\{ (a,b,c) \mid a,b,c \ge 0 \text{ and } a^2+b^2+c^2 \le 1000 \right\}. \] Hence the constraint set \[ \overline S = \left\{ \mathbf x \in \overline U : g(\mathbf x) = 4 \right\} \] is compact, where $g(a,b,c) = a^2+b^2+c^2+abc$.  Excellent.

Define \[ f(a,b,c) = a^2+b^2+c^2+ab+bc+ca. \] It's equivalent to show that $f \le 6$ subject to $g$. Over $\overline S$, it must achieve a global maximum.  Now we consider two cases.

If $\mathbf x$ lies on the boundary, that means one of the components is zero (since $a^2+b^2+c^2=1000$ is clearly impossible). WLOG $c=0$, then we wish to show $a^2+b^2+ab \le 6$ for $a^2+b^2=4$, which is trivial.

Now for the interior $U$, we may use the method of Lagrange Multipliers. Consider a local maximum $\mathbf x \in U$.  Compute \[ \nabla f = \left<2a+b+c, 2b+c+a, 2c+a+b \right> \] and \[ \nabla g = \left<2a+bc, 2b+ca, 2c+ab\right>. \] Of course, $\nabla g \neq \mathbf 0$ everywhere, so introducing our multiplier yields \[ \left<2a+b+c,a+2b+c,a+b+2c\right> = \lambda \left<2a+bc,2b+ca,2c+ab\right>. \] Note that $\lambda \neq 0$ since $a,b,c > 0$.  Subtracting $2a+b+c = \lambda(2a+bc)$ from $a+2b+c = \lambda(2b+ca)$ implies that \[ (a-b)(\left[ 2\lambda - 1 \right] - \lambda c) = 0. \] We can derive similar equations for the others.  Hence, we have three cases.

[i]Case 1:[/i] If $a=b=c$, then $a=b=c=1$, and this satisfies $f(1,1,1) \le 6$.
[i]Case 2:[/i] If $a$, $b$, $c$ are pairwise distinct, then we derive $a = b = c = 2 - \lambda^{-1}$, contradiction.
[i]Case 3:[/i] Now suppose that $a=b \neq c$. That means $\lambda = \frac{1}{2-a}$ (of course our conditions force $c < 2$).  Now \[ 2a+2c = a+b+2c = \lambda (2c+ab)
	= \frac{1}{2-a} (2c+a^2) \] which implies \[ 4a+4c-2a^2-2ac = 2c+a^2 \] meaning (with the additional note that $a \neq 1$) we have \[ c = \frac{3a^2-4a}{2-2a}. \] Note that at this point, $c > 0$ forces $1 < a < \frac 43$.

The constraint $a^2+b^2+c^2+abc=4 \iff c^2 + + a^2c + (2a^2-4) = 0$ now gives \[ \left( 3a^2-4a \right)^2 + a^2 \left( 3a^2-4a \right)\left( 2-2a \right) + \left( 2a^2-4 \right)\left( 2-2a \right)^2 = 0. \] Before expanding this, it is prudent to see if it has any rational roots. A quick inspection finds that $a=2$ is such a root (precisely, $16-32+16=0$).  Now, we can expand and try to factor:
\begin{align*}
	0 &= -6a^5 + 31a^4 - 48a^3 + 8a^2 + 32a - 16 \\
	&= (a-2)(-6a^4 + 19a^3 - 10a^2 - 12a + 8) \\
	&= (a-2)^2 \left( -6a^3+7a^2+4a-4 \right) \\
	\intertext{Seeing a cubic, we desperately try every rational root and miraculously discover}
	&= (a-2)^2 (2-3a)(2a^2-a-2).
\end{align*} The only root $a$ in the interval $\left( 1,\tfrac43 \right)$ is $a = \frac 14 \left( 1 + \sqrt{17} \right)$. You can guess what comes next -- write
\[ c = \frac{a(3a-4)}{2-2a} = \frac{1}{8} \left( 7 - \sqrt{17} \right) \] and \[ f(a,b,c)
	= 3a^2+2ac+c^2
	= \frac{1}{32} \left( 121 + 17\sqrt{17} \right)
	. \] This is the last critical point, so we're done once we check this is less than $6$.  This follows from the inequality $17^3 < (6 \cdot 32 - 121)^2$; in fact, we actually have \[ \frac{1}{32} \left( 121 + 17\sqrt{17}  \right) \approx 5.97165. \] This completes the solution.[/quote]

In case 3 , can anyone tell how c<2 is concluded?

$c\ge2 \Longrightarrow c^2 \ge 4 \Longrightarrow a^2 + b^2 + c^2 + abc >4,$ contradiction.

[quote=cn2_71828182846]Let $a, b, c \geq 0$ and satisfy  \[ a^2+b^2+c^2 +abc = 4 . \] Show that \[ 0 \le ab + bc + ca - abc \leq 2. \][/quote]
[url=https://artofproblemsolving.com/community/c6h2112266p15298270]2017 Preparation for juniors, Romania:[/url]
Let $a, b, c > 0$ and satisfy  \[ a^2+b^2+c^2 = 3. \] Show that \[ ab + bc + ca - abc \leq 2. \]

Let $a \leq 1$ and $b,c \geq 1$. Now, obviously $(b-1)(c-1) \geq 0$ and:
$ab+bc+ca-abc=a(b+c)+bc(1-a) \geq 0$
We will solve quadratic equation $a^2+b^2+c^2+abc-4=0$ for $a$, so:
$a=\frac{\sqrt{(4-b^2)(4-c^2)}-bc}{2}$. And we will use this in second equation:
$ab+bc+ca-abc=-abc+ab+ac-a+a+bc=-a(b-1)(c-1)+a+bc \leq a+bc=\frac{bc+\sqrt{(4-b^2)(4-c^2)}}{2}$
By Cauchy-Schwarz inequality we see that:
$\frac{bc+\sqrt{(4-b^2)(4-c^2)}}{2} \leq \frac{\sqrt{(4-b^2+b^2)(4-c^2+c^2)}}{2}=\frac{\sqrt{16}}{2}=2$
So, $0 \leq ab+bc+ca-abc \leq 2$ :)

It is well-known that we can substitute $a=2\cos A,b=2\cos B,c=2\cos C$ for a triangle $ABC$. Observe that
\[ab+bc+ca\ge 3\sqrt[3]{a^2b^2c^2}\ge abc\]
because $abc\le 27$, which shows the first part of the inequality. For the latter part, rewrite it as showing
\[3\ge \sum_{\mbox{cyc}}(\cos A+\cos B)^2.\]
Note since triangle $ABC$ is not obtuse, we can let $a^2=y+z,b^2=z+x,c^2=x+y$ for some nonnegative $x,y,z$, so the problem amounts to showing
\[3\ge \sum_{\mbox{cyc}}\left(\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}\right)^2 = 2\sum_{\mbox{cyc}}\frac{x^2}{(x+y)(x+z)}+2\sum_{\mbox{cyc}}\frac{yz}{(y+z)\sqrt{(x+z)(x+y)}}.\]
By rearranging, the problem amounts to checking
\[6xyz+3\sum_{\mbox{sym}}x^2y\ge 2\sum_{\mbox{cyc}}x^2(y+z)+2\sum_{\mbox{cyc}}yz\sqrt{(x+z)(x+y)}.\]
By cancelling, it is equivalent to show
\[6xyz+\sum_{\mbox{sym}}x^2y\ge 2\sum_{\mbox{cyc}}yz\sqrt{(x+z)(x+y)}.\]
But by AM-GM, the latter expression is at most
\[\sum_{\mbox{cyc}}(2xyz+y^2z+z^2y),\]
so done.

For the lower bound, write $0 \le ab + bc + ca - abc$ as
\[0\le -a((b-1)(c-1)-1)+bc.\] This is clearly true since $(b-1)(c-1)-1$ must be nonpositive.
Now we deal with the upper bound. WLOG, we can let $a-1$ and $b-1$ have the same sign. Then $(a-1)(b-1)=ab-b-a+1 \ge 0 \implies ab+1 \ge a+b$. Multiplying by $c$ and subtracting $abc$, we get $c \ge ac+bc-abc$. Adding $ab, $
\[c+ab \ge ac+bc+ab-abc.\] This means to prove the upper bound it suffices to prove $c+ab \le 2$. Assume FTSOC that $c+ab>2$. Then,

\begin{align*}
4
&=a^2+b^2+c(c+ab) \\
&\ge 2ab+c(c+ab) \\
&> 2ab+2c
\end{align*}

which implies that $2>ab+c,$ contradiction. Hence,  $c+ab \le 2$ and we are done.

[b][color=#00f]My solution:[/color][/b]
We know that $min\{a,b,c\}\leq1$ so wlog any two of $a,b,c$ can be $\geq1$ or $\leq1$\\
So let $a\leq 1$ and $b,c\geq 1$ so we have \[ab+bc+ca-abc=a(b+c)+bc(1-a)\geq 0\]So lower bound is proved.So now we need to prove upper bound so \[a^2+b^2+c^2+abc=4\]and we know that $b^2+c^2\geq 2bc$ by AM-GM so now put this up\[a^2+2bc\leq 4\implies bc(2+a)\leq 4-a^2=(2+a)(2-a)\\\implies bc\leq 2-a\]now lets see what we have until now,\\
$\bullet$ $(1-b)(1-c)\geq 0$ for both the cases\\
$\bullet$ $bc\leq 2-a$\\
So now begin the sol lets see what we have
\[ab+bc+ca-abc\leq ab+2-a+ca-abc=2-a(1-b-c+bc)=2-a(1-b)(1-c)\leq 2\]So we are done now !

Part 1: $0\leq ab+bc+ca.$
Notice $$\min (a,b,c) \leq 1 \implies ab+bc+ca \geq bc \geq abc.$$ Equality when $a=b=0, c=2.$ Q.E.D.

Part 2: $ab+bc+ca-abc\leq 2.$
WLOG, $a\leq 1, c\geq 1,$ then $$abc+b \geq ab+bc \implies b(a-1)(c-1)\geq 0.$$
We have to prove $2\geq ac+b.$
Notice: $$a^2+c^2+b(ac+b)=4 \implies 2ac+b(ac+b) \leq 4 \implies (b+2)(ac+b-2) \leq 0 \implies ac+b \leq 2.$$ Equality when $a=b=c=1.$ Q.E.D.


modular arithmetic
combinatorics unsolved
combinatorics

there`s a 4*4grid and one is allowed to fill in no.s from 0 to 3 and a no. can get repeated as many times how many ways are there to fill in no.s so that the no.s of each row and each column sum up to a multiple of 4 ?
any other such counting problems willl be appreciated.

the upper-left $ 3\times 3$ subtable $ T$ can be assigned its values in $ 4^9$ ways; the rest of the table is then determined. so the answer is $ 4^9$.

to see that the rest of the table is determined... clearly, elements $ a_{1,4},a_{2,4},a_{3,4}$ are uniquely determined because their values $ \pmod{4}$ are uniquely determined by the rows of $ T$; a similar thing goes for $ a_{4,1},a_{4,2},a_{4,3}$. and finally, to make sure $ a_{4,4}$ can be filled in without inconsistencies, we need to check that $ a_{1,4} \plus{} a_{2,4} \plus{} a_{3,4}\equiv a_{4,1} \plus{} a_{4,2} \plus{} a_{4,3} \pmod{4}$. and this is true because both are $ \equiv \minus{} S\pmod{4}$, where $ S$ is the sum of the elements of $ T$.


function
calculus
integration
complex analysis

http://en.wikipedia.org/wiki/Stieltjes_constants
For those who don't know the stieltjes constants are the coefficients of the laurent series for the zeta function around 1. I don't have the derivation of their exact value in my textbook, and I can't find it on the internet, but I'd like to know it if anyone can help. the exact value is in the article, but I'm not so good with tex so I won't write it here. thanks a lot!

What do you mean by "the exact value"? The formula with the limit, the integral representation, or something else? :?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=942796#942796]this post[/url] and the thread around it. I don't think I ever posted a full derivation, but it's a divide-and-conquer trick estimating an infinite sum for $ (s\minus{}1)\zeta(s)$ in two parts.


Points $ P$ and $ Q$ are midpoints of two sides of the square. What fraction of the interior of the square is shaded? Express your answer as a common fraction.

[asy]filldraw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)--cycle,gray,linewidth(1));
filldraw((0,1)--(1,2)--(2,2)--(0,1)--cycle,white,linewidth(1));
label("P",(0,1),W);
label("Q",(1,2),N);[/asy]

Let the side length be $ s$.  Then the area of the white triangle has height

 $ \frac{s}{2}$ and base $ \frac{s}{2}$, making the area of triangle $ \frac{s}{2}
\times\frac{s}{2} \times \frac{1}{2}= \frac{s^2}{8}$.  Therefore, the area of

the shaded part is $ s^2- \frac{s^2}{8}= \frac{7s^2}{8}.$  The desired fraction is

therefore $ \boxed{\frac{7}{8}}$.


For reals $ a_i, b_i\ (i\equal{}1,...,n)$ and $ 0\le x\le 1$, prove

\[ \left(\sum_{i\equal{}1}^na_i^2\plus{}2x\sum_{i< j}a_ia_j\right) \left(\sum_{i\equal{}1}^nb_i^2\plus{}2x\sum_{i< j}b_ib_j\right) \ge\left(\sum_{i\equal{}1}^na_ib_i\plus{}x\sum_{i\ne j}a_ib_j\right)^2\]

[quote="spanferkel"]For reals $ a_i, b_i\ (i \equal{} 1,...,n)$ and $ 0\le x\le 1$, prove
\[ \left(\sum_{i \equal{} 1}^na_i^2 \plus{} 2x\sum_{i < j}a_ia_j\right) \left(\sum_{i \equal{} 1}^nb_i^2 \plus{} 2x\sum_{i < j}b_ib_j\right) \ge\left(\sum_{i \equal{} 1}^na_ib_i \plus{} x\sum_{i\ne j}a_ib_j\right)^2\]
[/quote]
Notice that
$ \sum_{i \equal{} 1}^na_i^2 \plus{} 2x\sum_{i < j}a_ia_j \equal{} (1 \minus{} x)\sum^n_{i \equal{} 1} a^2_i \plus{} x\left(\sum_{i \equal{} 1}^n a_i\right)^2,$
$ \sum_{i \equal{} 1}^nb_i^2 \plus{} 2x\sum_{i < j}b_ib_j \equal{} (1 \minus{} x)\sum^n_{i \equal{} 1} b^2_i \plus{} x\left(\sum_{i \equal{} 1}^n b_i\right)^2,$
and
$ \sum_{i \equal{} 1}^na_ib_i \plus{} x\sum_{i\ne j}a_ib_j \equal{} (1 \minus{} x)\sum^n_{i \equal{} 1}a_ib_i \plus{} x\left(\sum^n_{i \equal{} 1}a_i\right) \left(\sum^n_{i \equal{} 1}b_i\right).$
Therefore, the original inequality can be written as
$ \begin{aligned} & \left[ (1 \minus{} x)\sum^n_{i \equal{} 1}a^2_i \plus{} x\left( \sum^n_{i \equal{} 1}a_i\right)^2\right] \left[ (1 \minus{} x)\sum^n_{i \equal{} 1}b^2_i \plus{} x\left( \sum^n_{i \equal{} 1}b_i\right)^2\right] \ge \\
& \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ge \left[ (1 \minus{} x)\sum^n_{i \equal{} 1}a_ib_i \plus{} x\left(\sum^n_{i \equal{} 1}a_i\right)\left(\sum^n_{i \equal{} 1}b_i\right)\right]^2.\end{aligned}$
By the Cauchy-Schwarz Inequality, we have
$ \begin{aligned} & \left[ (1 \minus{} x)\sum^n_{i \equal{} 1}a^2_i \plus{} x\left( \sum^n_{i \equal{} 1}a_i\right)^2\right] \left[ (1 \minus{} x)\sum^n_{i \equal{} 1}b^2_i \plus{} x\left( \sum^n_{i \equal{} 1}b_i\right)^2\right] \ge \\
& \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ge \left[ (1 \minus{} x)\sqrt {\left(\sum^n_{i \equal{} 1}a^2_i\right)\left(\sum^n_{i \equal{} 1}b^2_i\right)} \plus{} x\left|\left(\sum^n_{i \equal{} 1} a_i\right)\left(\sum^n_{i \equal{} 1}b_i\right)\right|\right]^2 \\
& \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ge \left[ (1 \minus{} x)\left| \sum^n_{i \equal{} 1}a_ib_i\right| \plus{} x\left|\left(\sum^n_{i \equal{} 1} a_i\right)\left(\sum^n_{i \equal{} 1}b_i\right)\right|\right]^2 \\
& \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ge \left[ (1 \minus{} x)\sum^n_{i \equal{} 1}a_ib_i \plus{} x\left(\sum^n_{i \equal{} 1}a_i\right)\left(\sum^n_{i \equal{} 1}b_i\right)\right]^2.\end{aligned}$
The proof is completed.


function
probability
geometry
3D geometry

Here are COPSD problems. These are only open to middle schoolers and in order for you to qualify, you have to get at least 7 problems correct with solution. I made this COPSD a lot harder than the previous two, hoping that harder problems will attract more middle schoolers. Enjoy! Problems are due 9/4/05. PM me the solutions.

Note: You can also do COPSD AND COP but that's optional.

[size=150]Problems[/size]

1. The tough sequence $t_n$ is defined as below:

$t_n = 3t_{n-1} - 2t_{n-2}$

where $n \geq 3$. If $t_1 = 3$ and $t_2 = 5$, compute the last digit of $t_{20}$.

2. Three functions $f(x), g(x)$ and $h(x)$ are defined in this way:

$f(x) = 2x$
$g(x) = x+3$
$h(x) = f(x) g(x)$

Find $h(4)$.

3. A three digit integer $abc$ has its "brother" number, which is in form $cba$. Compute the maximum sum of $abc$ and $cba$ if $b > a,c$ and $a \neq c$.

4. Set $T$ contains 6 two-digit numbers that have identical unit and tens digit in base 7. Set $R$ contains 6 numbers from set $T$, except in base 10 instead of 7. What is the probability to choose a perfect square or cube from set $R$?

5. In how many ways can we arrange 7 apples and 3 bananas so that no two bananas are next to each other? (They are arranged in single row)

6. What is the smallest number $n$ such that $n+n+1+n+2$ is divisible by 4?

7. How many four-digit numbers have 79 as its two middle digits and divisible by 15?

8. Find the remainder when $12^{12}$ is divided by 13.

9. Let $p$ and $q$ be rational numbers such that

\[p-q = pq =\frac{p}{q} = k\]

If $p,q,k \neq  0$, find $k$. Express your answer as a common fraction.

10. Given that $ABCD$ is an unit squre with diagonals intersecting in $E$, find the area of largest circle that can be inscribed in the square $ABCD$.

For 6, does n have to be a natural number? Does the diagonal in 10 have any revelance?

$n$ is a positive integer.

In number 10, do your best.  :)

Okay, then, what form should our answer be in number 10?

The diagonals can't pass through the circles, right?

Wait, so do we only PM you the answers, or do we have to show all of our work too?

I don't really understand what you mean by "solutions"

[quote="Drunken_Math"]
I don't really understand what you mean by "solutions"[/quote]

Explain your answers, like you always do.  :)

[quote="236factorial"][quote="Drunken_Math"]I don't understand number 4.

If set R contains 6 numbers from set T, and set T has 6 numbers in it... isn't set R the same as set T?  :?[/quote]

It's the same, except in different bases.[/quote]

Yes. The amount of numbers do not change but their base changes.

I only accepted one solution. I know these problems are quite challenging but solving these will definitely help you for upcoming Mathcounts.

So, just take a shot! :)

All solutions must be sumbitted by tomorrow 11:59:59 EST!

Please keep trying!

I submitted my answers and it said that you read them already, but I don't have a reply? What's up?  :(

I did receive it.

But I'm not giving the results away until all answers are submitted. This applies same to COP.


trigonometry
limit
function
calculus
calculus computations

I took AP calc in High School, but I didn't pass the Exam.  I am now taking Calc 1 in college, and there seems to be many new things that I don't understand.  Things like 0/0 =1 and when dividing by 0 the answer is infinite.  In high school I was always told it isn't possible to divide by zero so the problem was undefined.  That seems like a big difference.  If someone could explain, I would greatly appreciate it.

Secondly the problem I don't understand currently is:

"Evaluate the lim sin(3x)/tan(2x) as x approaches 0".  I would normally just plug in zero and get undefined, but I know that isn't right.  I tried changing tan to sin/cos and ending up with: sin(3x)cos(2x)/sin(2x) but I don't know where to go from there. :?:

$ \lim_{x\to0}\frac{\sin3x}{\tan2x}\equal{}\frac{3}2\lim_{x\to0}\frac{\sin3x}{3x}\cdot\frac{2x}{\tan2x}$

Does that make sense?

Limits have a formal definition not just plugging in (which I'd hope an AP Calc class would go over). It would be much better to think in terms of the definition. $ \frac{0}{0}$ is undefined, although a function approaching $ \frac{0}{0}$ may not be undefined since it only matters what happens near that point and not actually at that point.

Example:

$ f(x)\equal{}\begin{cases}1 ~\text{if}~ x\neq 0\\ 0 ~\text{if}~ x\equal{}0\\ \end{cases}$

$ \lim_{x\to 0}f(x)\equal{}1$, but $ f(0)\equal{}0$.

$ \frac{0}{0}$ is ... not a number (formally, [url=http://en.wikipedia.org/wiki/NaN]NaN[/url]). $ \frac{\text{near }0}{\text{near }0}$ could be anything. The interesting limits are really the second; the definition of a limit cares about the values nearby, but [i]not[/i] at the actual point. The only place either $ \sin$ or $ \tan$ is zero near the origin is zero itself, and that's the point we exclude. The quotient is defined everywhere that matters, in a punctured interval around the origin.

[quote]I tried changing tan to sin/cos and ending up with: sin(3x)cos(2x)/sin(2x) [/quote]
That is progress. For one thing, that cosine is "harmless" since $ \lim_{x\to0}\cos(2x)\equal{}1.$ Just factor that $ 1$ out.

Now we're looking at $ \frac{\sin(3x)}{\sin(2x)}.$

We have to know something. We have to know that $ \lim_{\theta\to0}\frac{\sin\theta}{\theta}\equal{}1.$

Another way to phrase this: the sine of a very small angle (in radians) is very nearly the same as the angle itself. In which case that $ \frac{\sin(3x)}{\sin(2x)}$ might as well be $ \frac{3x}{2x},$ or is at least getting closer to that as $ x$ gets closer to zero. (This is also some of the motivation behind what Hetidemek did.)

[quote]the sine of a very small angle (in radians) is very nearly the same as the angle itself[/quote]

I'm glad you explained that because that had been puzzling to me.

I think I finally understand this problem.  It is much easier to solve when the reasoning behind the procedure is explained.  Thanks for everyone's explanations.


(La  proba individuala) Aratati ca pentru oricare numere reale pozitive a,b,c cu $abc\geq2^{9}$ avem inegalitatea: \[\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}}\geq \frac{3}{\sqrt{1+(abc)^{\frac{1}{3}}}}\]
(La proba pe echipe) Aratati inegalitatea in triunghi :\[h_{a}+h_{b}+h_{c}\leq 9r+\frac{a^{2}+b^{2}+c^{2}-ab-bc-ca}{p}\]

Interesanta inegalitatea 2, e echivalenta cu $\sum{a}\cdot \sum{\frac{1}{a}}\leq  9+\frac{\sum{a^{2}}-\sum{ab}}{S}$, unde $a,b,c$ sunt laturile triunghiului.

Imi pare rau.Nu prea stiu clasamentele .Am plecat imediat dupa ce s-a terminat concursul (adik poba pe echipe).Iar despre ineg a doua, nu am aflat ink cum se face.O sa ma interesez si de subiectele de la alte clase. :D

[quote="Zamfirmihai"]stii cumva si clasamente, alte probleme, ceva?[/quote]

http://www.concursurigorj.ro/titeica/index.php

[quote="pohoatza"]Hmm, pai probabil o metoda brute-force dar sigura de reusita, este folosind identitatile:
$\sum{ab}=p^{2}+r^{2}+4Rr$
$\sum{a^{2}}=2p^{2}-2r^{2}-8Rr$
$abc=4pRr$, iar apoi folosind una din inegalitatile lui Gerretsen. ($p^{2}\geq 16Rr-5r^{2}$ sau $p^{2}\leq 4R^{2}+4Rr+3r^{2}$), depinde ce parte vrem sa majoram/minoram.[/quote]
nu cred ca merge asa, am incercat eu aseara :)

[quote="Vrajitoarea Andrei"](La  proba individuala) Aratati ca pentru oricare numere reale pozitive a,b,c cu $ abc\geq2^{9}$ avem inegalitatea:
\[ \frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}}\geq \frac{3}{\sqrt{1+(abc)^{\frac{1}{3}}}}\]
[/quote]
Fie $ a,b,c\geq$ a.i. $ \frac{a}{8}=x$,$ \frac{b}{8}=y$ si $ \frac{c}{8}=z$ ,astfel inegalitatea devine:
\[ \sum \frac{1}{\sqrt{1+8x}}\geq \frac{3}{\sqrt{1+8(xyz)^{\frac{1}{3}}}}, cu ,xyz\geq 1 \]
De asemenea fie $ x',y',z'\geq 0$ si $ t\geq 1$ a.i.$ x=t\cdot x'$ ,$ y=t\cdot y'$ si $ z=t\cdot z'$ si $ x'y'z'=1$
Deoarece $ x'y'z'=1$ $ \exists$ $ a',b',c'>0$ a.i. $ x'=\frac{b'c'}{a'^{2}}$,$ y'=\frac{c'a'}{b'^{2}}$ si $ z'=\frac{a'b'}{c'^{2}}$.
Astfel inegalitatea devine
\[ \sum \frac{a'}{\sqrt{a'^{2}+8\cdot t b'c'}}\geq \frac{3}{\sqrt{1+8t}}, \forall t\in [1,\infty) \]
(Obs: pentru t=1 este echiar inegalitatea de la IMO 2001)
Deoarece inegalitatea este omogena in $ a',b',c'$ presupunem $ a'+b'+c'=1$ si conform inegalitatii lui Jensen pt functia $ f(x)=\frac{1}{\sqrt{x}}$ (convexa) avem:
\[ \sum \frac{a'}{\sqrt{a'^{2}+8\cdot t b'c'}}\geq \frac{1}{\sqrt{\sum a'^{3}+8t \sum a'b'c'}}\geq \frac{3}{\sqrt{1+8t}}\]

\[ \Leftrightarrow \]

\[ (8t+1)(a'+b'+c')^{3}\geq 9(\sum a'^{3}+8t \sum a'b'c') \]

\[ \Leftrightarrow \]

\[ (8t+1)\left( \sum a'^{3}+3 \sum a^{2}(b+c)+6abc\right)\geq 9(\sum a'^{3}+8t \sum a'b'c'),(*) \]
Tinand cont de inegalitatea $ \sum a^{2}(b+c)\geq 6abc$
Inegalitatea (*) se reduce la
\[ (t-1)\sum a'^{3}\geq (t-1)\sum a'b'c' \]
.qed


superior algebra
superior algebra theorems

Do (1 2), (1 3), and (1 4) generate S_4? If not, what does?

I'm not certain about those three elements, but I am certain that S_4 = <(1234), (1243)>.  It's possible to get (1234) from the elements you have given, but I don't know about (1243).


Σε αυτό το χώρο θα μαζέψουμε ωραία θέματα από διαγωνισμούς, εθνικες ολυμπιάδες και SL - ΙΜΟ. Βέβαια , τέτοια θέματα μπορούμε να βάλουμε και στα επιμέρους κεφάλαια , θα πρότεινα όμως τα πιο δύσκολα και τα πιο περίεργα να τα βάλουμε εδώ. Έτσι δε θα χαλάει η συνοχή που πρέπει να έχουν οι επιμέρους ενότητες , αλλά και το βιβλίο θα καλύπτει όλα τα επίπεδα δυσκολίας.
    Για το χώρο αυτό [color=red]θα περιμένουμε και τα παιδιά που διακρίθηκαν στον Αρχιμήδη και τώρα διαβάζουν για τη μεσογειάδα[/color]. Αυτοί σίγουρα έχουν πολλές ασκήσεις να μας προτείνουν για εδώ.Ωστόσο βάλτε ο καθένας μια - δυό ασκήσεις για αρχή και συμπληρώνουν οι άλλοι λίγο αργότερα !

             [color=blue]Καλή Συνέχεια !!![/color]

[i]Λοιπόν , ας κάνω εγώ την αρχή με ένα εκπληκτικό θέμα.Προτείνω να το μελετήσετε όλοι και κυρίως αυτοί που πάνε για τη Β΄Λυκείου. Αλλά και σε αυτούς που ετοιμάζονται για τη Μεσογειάδα θα φανεί πολύ χρήσιμο. Σας το δίνω σε pdf.Τυπώστε το και  : [/i]
                                [color=red] ... απολαύστε το !!! [/color]

 [size=150][color=red]ΑΣΚΗΣΗ 1[/color] [/size]

[quote="silouan"]ΑΣΚΗΣΗ 1-3ος τρόπος  :wink:  . (Μπορείτε να τον ελέξτε για τυπογραφικά ?)[/quote]

 Εκτός και αν δεν βλέπω κάτι , αλλά είναι ο πρώτος τρόπος. Απλά παρουσιάζει άλματα στην απόδειξη και φαίνεταο πιο σύντομος.Κοιτάξτε τον και εσείς μήπως μου ξεφεύγει κάτι !

 Μπάμπης

Ναι είναι περίπου ο πρώτος τρόπος,,,

[size=150][color=red]ΑΣΚΗΣΗ 2 [/color][/size]

 Το παρακάτω θέμα είναι tst από το Ιράν 2006. Όχι πολύ δύσκολο για ΙΡΑΝ, αλλά αρκετά καλό. Τα προβλήματα με τον εγγεγραμμένο κύκλο έχουν πάντα ενδιαφέρον και κατά κανόνα είναι από τα πιο δύσκολα. Πάντα ήθελα να μαζέψω όλα τα προβλλήματα που αφορούν τον εγγεγραμμένο κύκλο και να τα παρουσιάσω σε μια ενιαία ενότητα ενός βιβλίου, αλλά ποτέ δε βρήκα το χρόνο. Ας δούμε τι επιφυλλάσσει το μέλλον.

 Παρακαλώ να τυπώσετε την άσκηση και να της ρίξετε μια ματιά.

  Ευχαριστώ !

[size=150][color=red]ΑΣΚΗΣΗ 3 [/color][/size]

  Στους μαθητές που ξεκινάνε να διαβάζουν γερά για τους διαγωνισμούς από την αρχή της Α΄Λυκείου , το επόμενο θέμα είναι πολύ καλό.

[size=150][color=red]ΑΣΚΗΣΗ 4[/color][/size]

 Ανισότητα , Πολωνία 2007

Επειδή βλέπω ότι αυτήν την κατηγορία του βιβλίου την έχουμε εγκαταλέιψει (ιδιαίτερα μετά τη συζήτησή μας για το αν θα έπρεπε να υπάρχει ένα ξεχωριστό κεφάλαιο γι' αυτήν) βάζω μια άσκηση που πραγματικά αρμόζει σε αυτήν την κατηγορία αφού δεν ταιριάζει πουθενά αλλού :wink: Προτείνω λοιπόν να μαζέψουμε θέματα όπως το παρακάτω σε αυτό το topic και να μεταφέρουμε τις ασκήσεις γεωμετρίας ή τις ανισότητες που υπάρχουν εδώ, στα ανίιστοιχα topic για την κάθε κατηγορία ώστε να μην πάνε και χαμένες σε περίπτωση που δεν συμπεριληφθεί το κεφάλαιο...

[color=red][size=150](Ασκηση 5)[/size][/color]
[color=blue]Τέταρτο πρόβλημα στην BMO του 1985[/color]
Σε ένα διεθνές συνέδριο λαμβάνουν μέρος 1985 άτομα. Σε κάθε ομάδα τριών ατόμων υπάρχουν τουλάχιστον 2 άτομα που έχουν μια κοινή γλώσσα. Γνωρίζοντας ότι μιλάει το πολύ 5 γλώσσες, να αποδείξετε ότι υπάρχουν τουλάχιστον 2 άτομα που μιλούν την ίδια γλώσσα.

[color=red]Λύση[/color](by me)
Εστω ότι υπήρχε ένα ζευγάρι ατόμων (Α, Β) το οποίο δεν είχε καμία κοινή γλώσσα.Αυτά τα 2 άτομα συνολικά θα μιλούν το πολύ 10 γλώσες και όλα τα υπόλοιπα 1983 άτομα θα έχουν τουλάχιστον μια κοινή γλώσσα με τουλάχιστον έναν απ' τους δύο.
Αρα απ'την αρχή της περιστεροφωλιάς μια τουλάχιστον γλώσσα θα ομιλείται από τουλάχιστον $[\frac{1983}{10}]+1=199$ άτομα (όπου $[x]$ δηλώνει το ακέραιο μέρος του χ) και συμπεριλαμβανομένου του Α ή του Β η γλώσσα αυτή θα ομιλείται από τουλάχιστον 200 άτομα που είναι και το ζητούμενο.

Ας εξετάσουμε τώρα την περίπτωση όλα τα ζευγάρια να έχουν από τουλάχιστον μια κοινή γλώσσα.Τότε ο Α θα έχει από μια κοινή γλώσσα με τα άλλα 1984 άτομα. Δεδομένου όμως ότι ο Α μιλάει το πολύ 5 γλώσσες, μία τουλάχιστον γλώσσα θα ομιλείται από τουλάχιστον $[\frac{1984}{5}]+1=397$ άτομα και άρα η απόδειξη ολοκληρώθηκε.

[color=red]Άσκηση 6[/color]
Το θέμα αυτό είναι δύσκολο κατά τη γνώμη μου και η λύση είναι δικιά μου, γι' αυτό αν και δεν είναι θέμα Ολυμπιάδας σκέφτηκα να το προτείνω εδώ. Εξάλλου το περσινό 6ο χρησιμοποιούσε διανύσματα και δύσκολες τεχνικές στην επίσημη λύση του


I got an answer, but I'm not sure. does anyone know the answer? or can post a solution?  :lol:

[quote="now a ranger"]I got an answer, but I'm not sure. does anyone know the answer? or can post a solution?  [/quote]
Post your solution.  :)


geometry
geometric transformation
rotation
3D geometry
sphere
inequalities

1. Is momentum of some system always equal to its total mass times the velocity of its center of mass, even if the mass is changing or something?

2. Suppose you throw some object, such as a bowling ball, with sufficient initial center-of-mass velocity $v$ so that upon reaching the ground, its angular speed $\omega$ is not sufficient for rolling as $v > r \omega$; on the contrary, the object will skid for some distance. It eventually stops skidding due to friction: the kinetic frictional force applied by the ground both slows the center-of-mass velocity and exerts a torque to increase the angular speed until we eventually have $v = r \omega$ and the object then rolls. However, we know two things: (i) the object continues to roll, without skidding, at this point, and (ii) the object does slow down due to friction. However, if there were any further frictional force exerted on the object, this would cause both a decrease in $v$ and an increase in $\omega$ and we would then have $v < r \omega$, contradicting the $v = r \omega$ required for rolling. Thus, due to (i), we know that there can be no further frictional force. However, this is a blatant contradiction of (ii).

3. What is rolling resistance?

4. In rotation, how do we know what point to rotate about? For example, consider a cylinder resting on some surface, on which some horizontal force is applied. We set up the system so that the torques act with respect to a rotational axis that extends through the cylinder, passing through the center of mass. Why don't we take the parallel axis passing through the top lateral edge of the cylinder instead? If we did, then we'd find that first, the center of mass would still be accelerated as before (since there is a force), but that now, there would also be rotation about this new axis and the cylinder should rise into the air? (Ignore gravity and all that for now - it seems reasonable to suppose that the experiment could also be performed in space).

1. Yes.

2. No. Friction will not act when there is no relative motion. IE, when $v = r\omega$

3. No idea.

[quote="AndersoN"][quote="MysticTerminator"]If friction does not act after the object starts rolling, what slows it down? (I know that even rolling objects *do* slow down.) Is it air resistance?[/quote]
It sure is, but I think there is a certain friction of rolling... only I don't knwo the formula. I believe it's hard to get theoretically because to make the deduction simple, we use a perfect sphere and a plane. But it's quite clear that if everything's perfect, then the friction disappears. So it's as if we're neglecting the things that cause this friction: irregularities.

[quote="MysticTerminator"]4. In rotation, how do we know what point to rotate about? For example, consider a cylinder resting on some surface, on which some horizontal force is applied. We set up the system so that the torques act with respect to a rotational axis that extends through the cylinder, passing through the center of mass. Why don't we take the parallel axis passing through the top lateral edge of the cylinder instead? If we did, then we'd find that first, the center of mass would still be accelerated as before (since there is a force), but that now, there would also be rotation about this new axis and the cylinder should rise into the air? (Ignore gravity and all that for now - it seems reasonable to suppose that the experiment could also be performed in space).[/quote]

Your thinking is wrong. The cyllinder won't rise into the air because the axis that you chose (axis passing through the top lateral edge of the cylinder ) will move, change it's position... because of that, the torque formula will change (quite ugly). That is why it's a lot more convenient to choose the cyllinder's main axis.[/quote]

4. What? Why does the axis move? I didn't tell it to move... (assume the axis stays in the same vertical position).

2. Hm...I should now try to find the formula. That didn't really resolve the contradiction in the first post, however. As long as there's any force (due to irregularities or whatever it may be), the center-of-mass speed should slow (due to the force) and the angular speed should increase (due to the torque) and we consequently have $v < r \omega$, contradicting the $v = r \omega$ required for rolling.

[quote="oooooh ya"]3.

i was curious myself, so i looked in [url=http://en.wikipedia.org/wiki/Rolling_resistance]wikipedia[/url], it was pretty good.[/quote]

Thanks ... I should have thought of that. I keep forgetting that there are (some) physics resources on the web (usually I just do math, of course).

[quote="MysticTerminator"]
2. Hm...I should now try to find the formula. That didn't really resolve the contradiction in the first post, however. As long as there's any force (due to irregularities or whatever it may be), the center-of-mass speed should slow (due to the force) and the angular speed should increase (due to the torque) and we consequently have $v < r \omega$, contradicting the $v = r \omega$ required for rolling.
[/quote]

An interesting discussion... I'd like to point out that the condition of rolling without slipping is equivalent to a very large friction force (I don't understand why, but some IPhO solutions say exactly that). This makes matters even more complicated, because one would think that in the case of rolling without slipping there is [i]no[/i] friction...

[quote="psychicalchemist"]I dont think it makes sense for there to be a large coefficient of friction when the object is rolling without any net forces on it. if $v > r\omega$, friction works in opposite the perceived direction of motion for the bottom of the rolling object. Thus, $\omega$ decreases and $v$ increases. However, when $v = r\omega$, then the motion of a particle at the bottom of the object is $v - r\omega = 0$. Since the ground is not moving as well, there is no relative motion and friction ceases to act. (these are all my own conclusions, feel free to correct me if I'm wrong)

You can caculate the required coefficient of friction if an object is accelerating to keep it rolling under than acceleration with a similar method. Maybe this is what they mean by a "large" coefficient of friction acting if the object is rolling.[/quote]

? Does this mean to say that friction does not act on cars? Don't you have static friction (which is even [i]higher[/i] that kinetic friction) if there is no relative motion of ground and particle-at-the-bottom?

[quote="psychicalchemist"][quote="AndersoN"][quote="MysticTerminator"]If friction does not act after the object starts rolling, what slows it down? (I know that even rolling objects *do* slow down.) Is it air resistance?[/quote]

Your thinking is wrong. The cyllinder won't rise into the air because the axis that you chose (axis passing through the top lateral edge of the cylinder ) will move, change it's position... because of that, the torque formula will change (quite ugly). That is why it's a lot more convenient to choose the cyllinder's main axis.[/quote]

could you please post an example of such an equation. I already knew that it is best to use the center of mass in a situation where the object is accelerating, but I've never actually seen what happens when you dont. The closest I could get to such a situation would be to represent the net acceleration of the object as a "ghost" force acting in the opposte direction on the center of mass.[/quote]

yeah, exactly. We all use center of mass, but why? Also, there are some situations in which it is clearly best to not use center of mass: I remember in my physics book there's one (simple) problem which states that given a wheel that has been wheeled up to a corner, find the necessary force acting at the center to get it up over the corner. In this case, we clearly choose the axis of rotation through the corner of the obstacle, not the center of the wheel.

(I can't do diagrams, but if you're wondering, it looks something like [u]O[/u]---, except of course the obstacle is at less than the radius of the circle).

Ok, I've found it. The mathematical condition for the absence of slipping during rotation. I hope it's right, though: this is what I came up with after some hard thinking :D

[b]psychicalchemist[/b] was right: there is no force of friction acting on a ball which rolls on the ground. And I was wrong -- I said that non-slipping rotation is obtained when the force of friction is sufficiently large. Well, actually, the true condition for non-slipping is that the [i]coefficient of friction[/i] is sufficiently large. Let me explain:

A common mistake made in many hard mechanics problems is assuming that the friction force is always given by \[ F = \mu N \], where $\mu$ is the coefficient of friction and $N$ is the reaction of the ground. This is the expression for the force of kinetic friction. The force of static friction is \[ F \leq \mu N!!! \] (this means that the force of friction acting on a still object may not be equal to $\mu N$ -- this is really sensible if you think about it). I appreciate that a lot of you people knew this difference, but I wanted to point it out before I returned to our rolling without slipping. Here, the non-slipping condition is equivalent to $F\leq \mu N$, too! This means that in mechanics problems with rolling the non-slipping condition will be fulfilled for virtually any $N$ if the coefficient of friction is sufficiently large so that the above inequality is satisfied, and with this I've explained why I thought that large force of friction is equivalent to non-slipping.

Clearly, with $F=0$ we have that $F<\mu N$ for any $\mu$ and $N$ (except for the case $\mu = 0$ where there is clearly no need even to bother calculating friction, because it doesn't even exist). Furthermore, $F = 0$ implies that there will be no change in both the translational and the rotational velocity of the rolling object (deformations, air resistance etc. neglected), which is what some of you thought from the beginning.

[quote="demji"]Deformation of the wheel is the cause of the majority of the friction.  Although there is no friction at the point where a perfect wheel touches the ground we must realize that this is not a perfect world.  In the real world wheels touch the ground is several points, only one of which there is no friction at.  It is the other points which give the wheel its frictional tendencies.  All this friction over friction is realy great for the mind.[/quote]

This explaination hits the nail right on the head. I remember our high school physics teacher asked us this exact same question and we racking our brain over it for several days. In a theoretical world where wheels were perfectly round and undeformable and planes were perfect flat and undeformable, there would be no rolling resistance.


Columbia
function
pigeonhole principle
number theory
relatively prime
number theory unsolved

To each positive integer $ n$ it is assigned a non-negative integer $f(n)$ such that the following conditions are satisfied:

(1) $ f(rs) \equal{} f(r)\plus{}f(s)$
(2) $ f(n) \equal{} 0$, if the first digit (from right to left) of $ n$ is 3.
(3) $ f(10) \equal{} 0$.

Find $f(1985)$. Justify your answer.

$0=f(10)=f(2)+f(5)$; thus $f(5)=0$.
Thus $f(1985)=f(5)+f(397)=f(397)$.
Now $0=f(3573)=f(9)+f(397)=0$, and the result follows from $f(397)=0$.

It's not hard to show that the function is identically $0$ in that way.

I'm sorry for reviving this thread, but I'm not quite sure how $ f(2) \plus{} f(5) \equal{} 0 \implies f(5) \equal{} 0$.

It follows because $ f(n)$ is non-negative by the problem statement. Indeed, to show that $ f(n)$ is identically zero, simply consider primes ending in $ 2$, $ 5$, ...

Note that $ f(10k\plus{}r)\equal{}f(r)$ so we need only to prove that $ f(1)\equal{}f(2)\equal{}f(3)\equal{}\ldots\equal{}f(10)\equal{}0$.

$ f(10)\equal{}f(1)\plus{}f(10)\equal{}f(2)\plus{}f(5)\equal{}0\implies f(1)\equal{}f(2)\equal{}f(5)\equal{}f(10)\equal{}0$

$ f(4)\equal{}f(2)\plus{}f(2)\equal{}0, f(6)\equal{}f(2)\plus{}f(3)\equal{}0$

$ f(8)\equal{}f(2)\plus{}f(2)\plus{}f(2)\plus{}f(2)\equal{}0$

$ 0\equal{}f(35)\equal{}f(5)\plus{}f(7)\implies f(5)\equal{}f(7)\equal{}0$

$ f(9)\equal{}f(3)\plus{}f(3)\equal{}0$

Done. :D

[quote="Johan Gunardi"]Note that $ f(10k \plus{} r) \equal{} f(r)$ so we need only to prove that $ f(1) \equal{} f(2) \equal{} f(3) \equal{} \ldots \equal{} f(10) \equal{} 0$.

$ f(10) \equal{} f(1) \plus{} f(10) \equal{} f(2) \plus{} f(5) \equal{} 0\implies f(1) \equal{} f(2) \equal{} f(5) \equal{} f(10) \equal{} 0$

$ f(4) \equal{} f(2) \plus{} f(2) \equal{} 0, f(6) \equal{} f(2) \plus{} f(3) \equal{} 0$

$ f(8) \equal{} f(2) \plus{} f(2) \plus{} f(2) \plus{} f(2) \equal{} 0$

$ 0 \equal{} f(35) \equal{} f(5) \plus{} f(7)\implies f(5) \equal{} f(7) \equal{} 0$

$ f(9) \equal{} f(3) \plus{} f(3) \equal{} 0$

Done. :D[/quote]

Why $ f(10k\plus{}r)\equal{}f(r)$? It's $ f(rs)\equal{}f(r)\plus{}f(s)$, not $ f(r\plus{}s)\equal{}f(r)\plus{}f(s)$

$ f(n)\equal{}0$ for all $ n$ because WLOG $ n$ is not divisible by $ 2, 5$
$ \Rightarrow$ we can find a positive integer $ x \le 9$ so that the last digit of $ f(nx)$ is $ 3$
$ \Rightarrow f(n)\equal{}0$

[quote="Johan Gunardi"]Oopss. Sorry, I was wrong.

Let me try again :D

It suffices to prove that every positive integer has a multiple which begins with 3. But it is well known (and easy to prove by pigeonhole) that every positive integer has a multiple of the form 111...1000...0, so it also has a multiple of the form 333...3000...0. QED?[/quote]
But the first digit is from right to left.

Uh, just consider $ f(2)\equal{}f(5)\equal{}0$ and $ f(10n\plus{}3)\equal{}0$ and $ f((10n\plus{}1)*3)\equal{}0\equal{}f(10n\plus{}1)$ and $ f((10n\plus{}7)*9)\equal{}0\equal{}f(10n\plus{}7)$ and $ f((10n\plus{}9)*7)\equal{}0\equal{}f(10n\plus{}9)$ and the result follows.

An alternative approach is that $ f(ab) \equal{} 0 \Rightarrow f(a) \equal{} 0 \equal{} f(b).$ Therefore immediate that $ f(2) \equal{} 0 \equal{} f(5),$ and $ f(2^a5^bx) \equal{} f(x).$ But $ x$ relatively prime to both $ 2$ and $ 5 \Rightarrow \exists k \ni xk$ ends in $ 3 \Rightarrow f(xk) \equal{} 0 \Rightarrow f(x) \equal{} 0 \Rightarrow f(2^a5^bx) \equal{} 0.$


How many distinct isosceles trangles having sides of integral length and perimeter 113 are possible?

[hide]I get 28 as first 2x is greater than y.  So we divide 113 by 2 to get 56.5 so 2x is greater than 56.5, now for x to be an integer, 2x has to be even, so we count the evens from 56.5 to 113, thats 58 to 112.. thats 28 numbers[/hide]


function
algebra
polynomial
calculus
integration
inequalities
inequalities solved

The following problem was used for Vietnamese team selection test 1996. I have a solution, but it is quite complicated.

Find the minimal value of function
f(a, b, c) = (a+b)^4 + (b+c)^4 + (c+a)^4 - (4/7)(a^4+b^4+c^4), where a, b, c run over reals.

[i]Vietnamese Team selection 1996 problem *[/i]

It is a very hard problem and the knowing of your solution, though complicated, will give to the very clever solvers of the forum (I am not among them) new spurs to work for finding a simple solution.

I would like to know your solution, Namdung.

Thank you, very much.

Well, I will briefly describe the solution

We will show that f(a,b,c)>=0 for all a, b, c. From that, the minimum equal 0.

Among a, b, c, there exist at least one number which has the same sign as a+b+c, say a. By long, but easy computation, we have
f(a,b,c) - f(a,(b+c)/2,(b+c)/2) = 3a(a+b+c)(b-c)^2+(3/56)(7b^2+10bc+7c^2)(b-c)^2>=0. (Main idea! To fuse variables)

So, it sufficient (and necessary) to show that f(a, t, t) >=0 for all a, t. Is equivalent to f(0,t,t) >=0 and f(1,t,t)>=0 (due homogenusness). The first is trivial, the second is not, but is an exercise.

Namdung

Hi, it is secret of my cooking  :) . But I will share with you.

Well, we "want" h(a,b,c) = f(a,b,c) - f(a,(b+c)/2,(b+c)/2) >=0 (is well-known approach).
The first thing we must have is h(0,b,c) >= 0. h(0,b,c) is symmetric homogenus polynomial of b, c and it's easily to find that h(0,b,c) = +(3/56)(7b^2+10bc+7c^2)(b-c)^2. Now, take h(a,b,c) - h(0,b,c) and factor, we will get  h(a,b,c) - h(0,b,c) = 3a(a+b+c)(b-c)^2. (we can guess that factor (b-c)^2 will appear in this expression).

The last secret: I use Maple!  :)

Namdung

Thanks! You learn something new every day... :D  :D  :D

Ok it's clear that f(0,0,0)=0. We prove that f(a,b,c)>=0.

f(a,b,c) = 10/7  \sum a^4 + 4 \sum ab(a^2+b^2) + 6 \sum a^2b^2
f(a,b,c) >= 0 <=> 5/7 \sum a^4 + 2 \sum ab(a^2+b^2) + 3 \sum a^2 b^2>=0
<=> 5 \sum a^4 + 14 \sum ab(a^2+b^2) + 21 \sum a^2 b^2 \geq 0

We prove that 5/2 (a^4+b^4) + 14 ab(a^2+b^2) + 21 a^2 b^2 \geq 0.
Let x=ab , y=a^2+b^2 =>
5/2 (y^2-2x^2) + 14xy + 21 x^2 >=0 <=> 16x^2 + 14xy + 5/2 y^2 >= 0

if x=0 then a=b=0 and we will treat this case later. if not , we want  32 + 28 (y/x) + 5 (y/x)^2 >= 0 . If y/x = t with -1/2 <= t <= 1/2 , we must prove

32 + 28 t + 5t^2 >= 0.  The latter second degree function has roots
r_1 = (-28 + 12)/10 , r_2 = (-28 - 12)/10 . It's clear that -1/2 > r_1 and so 32 + 28t + 5t^2 >= 0.

All we have to do now is study the case a=b=0.
f(0,0,c) = 2c^4 - 4/7 c^4 >= 0.

I hope it's correct, I am recovering from a very nasty flu and I don't know how badly my brain was damaged  :)

For Kuba: maple is a maths software, like Mathematica, Mathlab, and others. You can graph functions, calculate limits, integrals and much more (you can do all this with Mathematica, I have never actually used Maple  :D )

I think... : Let a+b=2x , b+c =2y , c+a=2z
We can solve the problem

Could you please post the entire solution? Thank you.

[quote="grobber"]For Kuba: maple is a maths software, like Mathematica, Mathlab, and others. [/quote]
I always use Mathematica. It is designed for solving diff. eqn. at first.
Maple is upgrading every year... so it has much more packages.

How about replacing $\frac{4}{7}$ by k
And the question is find the maximum value of k such that f(a,b,c) > 0 for all real numbers a,b,c
Is it a good question  :D  :D

Good question, but I'm afraid that we dont have good answer.

If my guess is not wrong then k = 0.5778461359...

And 4/7 = 0.57142857 ...

So 4/7 is best "good" constant.

[quote="Namdung"]So, it sufficient (and necessary) to show that f(a, t, t) >=0 for all a, t. Is equivalent to f(0,t,t) >=0 and f(1,t,t)>=0 (due homogenusness). The first is trivial, the second is not, but is an exercise.

Namdung[/quote]

The inequality $f(1,t,t)\geq0$ can be proved in the following way:

$f(1,t,t)=59t^{4}+28t^{3}+42t^{2}+28t+5=$
$=\frac{6}{59}(20t+7)^{2}+\left(\sqrt{59}t^{2}+\frac{14t}{\sqrt{59}}-\frac{1}{\sqrt{59}}\right)^{2}> 0$

[quote="Stun"]How about replacing $\frac{4}{7}$ by k
And the question is find the maximum value of k such that f(a,b,c) > 0 for all real numbers a,b,c
[/quote]
The best $k$ is $f(t)$, where $f(x)=\frac{2(x+1)^4+16}{x^4+2}$ and $t$ is a real root of the equation
$x^5+4x^4+7x^3+14x^2+8x+2=0$.
The following inequality is also true.
Prove that:
\[(a+b)^4 + (b+c)^4 + (c+a)^4 \geq\frac{26}{45}(a^4+b^4+c^4)\]


[quote="tpulak"]heres the problem:

"What is the area of hexagon ABCDEF if AB=2 units? Express your answer in simplest radical form."

Note: this is a regular hexagon, so all side lenghts are the same[/quote]

[hide]
The area formula is $ 3s^{2}\sqrt{3}$, all that divided by 2, so we take $ 3(2^{2})\sqrt{3}\equal{}12\sqrt{3}$, then divide by 2 and we get $ 6\sqrt{3}$.
Try to derive the area formula for a regular hexagon used above
[/hide]

There are 6 equilateral triangles that fit in the hexagon.
6*(sqrt 3)/4*2^2=6*sqrt(3)

[hide="Another easy way"]You can also get the height of apothem, and letting $ a\equal{}$apothem and $ p\equal{}$perimeter, you can use the formula $ a\times (\frac12p)$ and works perfectly fine,too.[/hide]

[quote="tpulak"]heres the problem:

"What is the area of hexagon ABCDEF if AB=2 units? Express your answer in simplest radical form."

Note: this is a regular hexagon, so all side lenghts are the same[/quote]
Formula for the area of a regular hexagon:
$ \frac{3s^2\sqrt3}{2}$
Just plug in the numbers

[quote="alanchou"]@DK=It only really helps if the apothem is given or you know the circumradius of a regular polygon and the side length of it.  Here, the best way is the formula, since it uses the side length which was given[/quote]

Well, I was just proposing another way to look at it. But you are right. Formulas always work the best.

Thanks donkeykong , and alanchou for the formulas. I could solve this problem, but it would take me a long time to solve it. I guess it won't take long now :)


i was browsing the internet about information on traders and this came up: http://www.ftsweb.com/texts.htm

it seems to have lessons about some things you must learn to be a trader.  does anyone who knows what traders actually learn and use in their jobs (as this seems to be a very secretive topic in the financial world) know if this is something valid or just junk?

Trading stocks takes a good general knowledge of finance if you want to get serious about it.  Trading stocks is not gambling, it is a science that requires background knowledge.  However, you do not need a college degree to be successful, just ask my friend's dad!

I think there should be an investing forum in the Arts and Sciences section :).

It depends on what you mean by a trader, there are many types. Anyone who signs up for an account and makes a trade is a "trader". Do you mean doing it for a living? Investing your own money or someone elses? Do you mean be a stock broker?

A one minute look around that site says to me that it's valid information, but not particularly educational.  I can't imagine that many people could learn step-by-step from that site.  Almost anyone would have questions along the way that require significant thought or attention.

[quote="WarpedKlown1335"]Would trading on the market floor get me six figures?  Because that's half of what I care about.[/quote]Well that depends on if you're trading something worth 6 figures or not. If you put up a Bently I bet you'd get 6 figures on the floor.... ;)

Lol, on a more serious note, there is no market floor ;).

Ok, really on a serious note:

Where you stand when you trade does not determine your earnings in any way shape or form.... Having most of your trades go in your favor (consistantly) makes you 6 (or 7) figures, and that is the ONLY thing that will make you money.

[quote="ffdbzathf"]on the topic of different types of traders, what types of trading do companies such as DE Shaw and Jane Street Capital do?  I know of a few types of trading such as options, bonds, and stocks.  Anyone know about any of this?[/quote]

I would imagine they do all three of those... but I do not really know a lot about either company. There are some former employees here that could probably answer that for you.

[quote="mysmartmouth"]A serious investor can sit in fornt of his computer and tv for 10 hours a day from the comfort of his own home, and still make a six-figure profit.  But it takes a certain amount of experience to get to that level. :)[/quote]

It can certainly be done, but it's a shaky way to get started before you have a significant pool of capital and experience.  I traded stock options from my home for a little less than a year and made a bit more than had been my previous salary.  However, it was during the tech crash and the environment changed so quickly that I had to change what I was trading and my profits went from quick and fast to difficult to find and after some time of losing a bit of what I made I went out and found another trading job.

I wonder how many members on this forum invest and manage their own (or somebody else's) portfolio.

In 2000, my parents set up an account for me to trade, and I had put my allowance for the last 8 years into that: around 2,500 dollars. Now it is 84 dollars after the tech crash :( .

But some people lost their million dollar nest egg in the stock market, and a friend of my dad went bankrupt by buying on margin.

[quote="mysmartmouth"]A serious investor can sit in fornt of his computer and tv for 10 hours a day from the comfort of his own home, and still make a six-figure profit.  But it takes a certain amount of experience to get to that level. :)[/quote]

Met a few. Buncha druggies though.

I have a keen eye for politics and the sorts. London won the olympic thing, I was thinking as I read it, "I wonder if they'll get bombed..." And what do you know, they did. Wasn't during the olympics but they did.

I thought about becoming and investor, but I'm poor. Neurosurgery seems more my style.

[quote="mysmartmouth"]A serious investor can sit in fornt of his computer and tv for 10 hours a day from the comfort of his own home, and still make a six-figure profit.  But it takes a certain amount of experience to get to that level. :)[/quote]
If the experienced investor can sit around in front of his computer and tv all day, may I ask, what does a non-experienced investor do, walk around the block for 10 hours a day or maybe sit and knit...?

I sold Yahoo yesterday before they posted earnings, make my day a little bit better :). Did anyone else have any recent close calls?


Don't remember what year or what question it was, but on a certain state sprint, at the top of one of the pages ( :lol: ), there was a question about the orbits of planets... the answer was 2030. I think.

Something like this:

4 planets orbit around a sun. They take 40, 45, 20, and 30 months respectively to make a complete orbit around the sun (where [i]months[/i] is a consistent measure of time). The four planets are all aligned in the year 2000. When is the next year when they will all be aligned?

Not sure if I got the numbers right... but anyways I kept getting 2060 over and over with the lcms. The solutions say that it's 2030 because the planets will align on the other side of the sun as they did in 2000. So you have to divide 60/2.

But then, how would you know if the planets don't line up some other time? I would also appreciate it if someone found the actual problem.[/i]

I believe this is a 2000 state/chapter sprint round

how did you guys get anything before 2001?

anyways, the lcm of those three numbers is 360, and 360/12=30 years..

It wasn't those numbers, Ihatepie.

And, nope Quickster, it's not from 2000..

really?

I thought it was 2000

I remember this problem

maybe 2001 1998 or 2003?

This problem is from the 1999 State Sprint #10.  Here is the original wording.

[quote="MATHCOUNTS"]A solar system contains four planets and a sun as shown.  Planet A completes an orbit around the sun in 20 months; planet B completes an orbit in 36 months; planet C completes an orbit in 40 months; and planet D completes an orbit in 48 months.  If the four planets lie along a straight line in May of the year 2000, what is the next year that they will again lie on the same line?  Assume that the orbits are circular and that a year has exactly 12 months.[/quote]
Because the problem asks for the next year that the planets will "again lie on the same line", we only need to look at the case when the planets are at the same angle they began with or at the opposite angle.

The problem is imprecise in one way: It says "the four planets lie along a straight line", but it should say "the four planets and the sun lie on a straight line".  Otherwise the four planets could be on some other crazy line, which makes the problem a lot harder; in fact, we don't have enough information to solve that problem, because we weren't given the initial positions.

[quote="t0rajir0u"][quote="leoxnlin"]how would you know if the planets don't line up some other time?[/quote]

I'm going to take the problem as it's written.  The rigorous way to do this problem is to write out the position of each planet as follows, measuring the times in months and the positions in degrees (with the original aligned position being $ 0^{\circ}$):

$ p_1(t) \equal{} 9t$
$ p_2(t) \equal{} 8t$
$ p_3(t) \equal{} 18t$
$ p_4(t) \equal{} 12t$.

The rookie mistake here is then to solve for when $ p_1(t) \equiv p_2(t) \equiv p_3(t) \equiv p_4(t) \equiv 0 \bmod 360$, but you'll notice that the last condition is not necessary.  Working without the last condition,

$ 9t \equiv 8t \equiv 18t \equiv 12t \bmod 360 \implies$
$ t \equiv 0 \bmod 360$

so the first time the planets will be aligned is in $ 360$ months, or $ 30$ years.

I doubt, however, that this is the original problem.  You'll note that the fact that $ 9t \equiv 8t \bmod 360$ means that the answer is exactly as expected in the rookie solution.  Suppose we change that $ 9$, however, to a $ 10$, and consider

$ 10t \equiv 8t \equiv 18t \equiv 12t \bmod 360$;

then the solution is $ 2t \equiv 0 \bmod 360 \implies t \equiv 0, 180 \bmod 360$.  (These numbers correspond to month periods of $ 36, 45, 20, 30$.)  The $ 180^{\circ}$ solution corresponds to an alignment on the other side.  If we numbers with a greater $ \gcd$, for example

$ 9t \equiv 6t \equiv 18t \equiv 12t \bmod 360$,

then our solutions are $ 3t \equiv 0 \bmod 360 \implies t \equiv 0, 120, 240 \bmod 360$.  (These numbers correspond to month periods of $ 40, 60, 20, 30$.)  The $ 120^{\circ}$ and $ 240^{\circ}$ solutions are [b]not[/b] alignments on the other side in this case.[/quote]

Dude, you have a knack of making things extremely complicated

[quote="t0rajir0u"][quote="PI-Dimension"]It needs to be 180 degrees to be aligned. No  90 degrees, 45 degrees, 64.239479632 degrees...

(or a multiple)

so 180 degrees it the earliest[/quote]

I just gave an example of planets that align at $ 120^{\circ}$.

[quote="Ravi B"]This problem is from the 1999 State Sprint #10.  Here is the original wording.

[quote="MATHCOUNTS"]A solar system contains four planets and a sun as shown.  Planet A completes an orbit around the sun in 20 months; planet B completes an orbit in 36 months; planet C completes an orbit in 40 months; and planet D completes an orbit in 48 months.  If the four planets lie along a straight line in May of the year 2000, what is the next year that they will again lie on the same line?  Assume that the orbits are circular and that a year has exactly 12 months.[/quote][/quote]

This sounds ambiguous to me.  I read it as "what is the next year that they will again lie on the same line [b]as each other[/b]?"[/quote]

I think the question also wants it to be aligned with the sun too. Anyone got the 'official' problem?

[quote="Poincare"]Dude, you have a knack of making things extremely complicated[/quote]

First, I'm going to ask you not to quote the entire thing; you could have just said "t0rajir0u.

Secondly...this really isn't complex - it's just well written and rigorous.  I assume all of you want to do math after 8th grade; well, in Olympiads, you will learn that there's much more to the question than finding the answer...the RIGOROUS justification is so much more important.

Thanks everyone. Not very familiar with modular arithmetic yet though. I will be tomorrow, though  :D !

so about making it rigorous, how do we know that the planets have to align with the sun, as opposed to making some other crazy line(as ravi b said)

or, if it doesnt state that, how can we show that the "official" answer is correct?

[quote="mathking123"]so about making it rigorous, how do we know that the planets have to align with the sun, as opposed to making some other crazy line(as ravi b said)[/quote]

You don't.  You have to assume it.  You have no information about how the planets line up independent of the sun unless you also know the radii of their orbits.


geometry
geometric transformation
homothety
geometry unsolved

An acute angle with the verticex $O$ and the rays $Op_1$ and $Op_2$ is given in a plane . Let $k_1$ be a circle with the center on $Op_1$ which is tangent to $Op_2$. Let $k_2$ be the circle which is tangent to both rays $Op_1$ and $Op_2$ and to the circle $k_1$ frome outside. Find the locus of tangency points of $k_1$ and $k_2$ when center of $k_1$ moves along the ray $Op_1$.

When the center of $k_1$ moves on $Op_1$, the figured is simply transformed through a homothety centered at $O$, so the locus is a line through $O$, if we always make the same choice for the circle $k_2$ (note that there are two possible choices, since there are two circles satisfying the required conditions for $k_2$: we can take the one between $O$ and $k_1$, or the one farther away from $O$ than $k_2$).

More interestingly, however, the locus is the [b]same[/b] line, even if we choose the other possibility for $k_2$. In other words:

Let $k_1$ be as above, and take the two circles, $k_2,k'_2$ tangent to both $Op_1,Op_2$ in $T_1,T_2$ respectively, and externally to $k_1$. Then $O,T_1,T_2$ are collinear.

This can be shown immediately by performing an inversion of pole $O$ which invariates $k_1$. $k_2,k_2'$ are turned into one another, so the points where they touch $k_1$ are also turned into one another. In particular, they are collinear with the pole of inversion, $O$.


Where'd the alcumus discussion area go? :huh:

It has its own section now, just below the Online Classes section.

It doesn't show up anymore on my computer...


induction
number theory solved
number theory

Problem 3

i)Find a,b,c positive integers (a,b,c >= 0) such that

1/4 = 1/a^2 + 1/b^2 + 1/c^2

ii)Find all positive integer n (n>= 0) such that there exist n positives integers x_1,x_2,...x_n with

1 = 1/(x_1)^2 + ....+1/(x_n)^2

About ii:

We have a soln for n=1, for n=4 we have (2, 2, 2, 2). There are no solns for n=2, 3 or 5 (by a way similar to the soln for i). For n=6 we have the soln (2, 2, 2, 3, 3, 6), for n=7 we have (2, 2, 2, 4, 4, 4, 4) and for n=8 we have (2, 2, 3, 3, 3, 3, 6, 6). If we have soln (x1,.., xn) for n we also have soln          (x1,.., x(n-1), 2xn, 2xn, 2xn, 2xn) for n+3. Because we have solns for n=6, 7, 8, by using induction n->n+3 we get that we have solns for ecery n>=6. So we have a soln for n=1, n=4, n>=6.


geometry
geometric transformation
rotation

YOu have five different keys to be placed around a keyring....how many ways are there of arranging the 5 different keys??

[hide]5 for placing the first one, 4 for the next, 3 for the next, 2 for the next, 1 for the last, so it's 5!. But then there are 5 rotations of the same combination, so 5!/5 = 4! = 24, so I think he had the right answer. If not that, then 12, if you count flipping one arrangement over as identical to another.[/hide]

[hide="here is what i got"]
so u have 5 keys. let one of them stay in place. so we have 4 keys which can be ordered in $4!$ ways. however, if we use this we count every situation twice, since it reads the same foward and backwards on the keyring. so we really have $\frac{4!}{2}=12$ ways to arrange the keys.

a general formula for this is,

given $n$ key they can be arranges on a keyring $\frac{(n-1)!}{2}$

[/hide]

yehh that's right guys....very smart...how did the idea come to you about flipping??

well when i first did this kind of problem i didnt think about the flipping. so i would have gotten 24. but i remeber my algebra2 teacher saying something about it being the same backwards and fowards, and i never really thought about it untill a few months ago.

If you liked this one, you might also like that one :) :
http://www.mathlinks.ro/Forum/viewtopic.php?t=59927

When you say flipping it over, do you mean actually flipping the entire key chain over?  :huh:

Yes, I think so. (If you think about it, flipping the entire keychain over gives 1/2 as many combinations, right?)

Oh okay... I thought flipping over meant something else... never mind  :P


Hi! This is a very simple question...I would like to put in a document a word that links to another part of the same document. Any ideas???
thanks in advance.

you can do something similar with the package hyperref.  To learn how to use it, look in "the not so short introduction to Latex2e".  Simply look it up on google, it's free.

Ledurt


analytic geometry
vector
trigonometry
calculus
complex numbers
complex analysis

Hello again guys,

I have the following problem:

Problem:

a) Plot the following complex variables on an Argand diagram

(i) 2 + j5
(ii) 3 - j6
(iii) - 4 + j10
(iv) - 6 - j3

b) Convert the (i) and (ii) above into their Polar Equivalents, show ALL working.

c) Add (i) and (ii) above and express the result in rectangular form.

d) Subtract (iii) from (ii) above and express the result in rectangular form.

e) Multiply (i) and (iii) above by first converting to Polar form and express the result in Polar form.

f) Divide (iii) by (iv) above by first converting to Polar form and express the result in Polar form.

IT'S PRETTY HARD I KNOW! can anbody help as i'm totally lost.

To plot the points, just draw a cartesian plane, and instead of labling the axii $x$ and $y$, label them real and imaginary.

Here is a general method to convert a complex number in rectangular coordinates to polar coordinates:

$z=a+bi = |z|\cdot e^{i\arg{z}}$, where $\arg{z}$ is the angle formed by the vector from the origin to $z$ relative to the positive real axis.

Example: $2+5i = \sqrt{2^{2}+5^{2}}\cdot e^{i\left(\tan^{-1}\frac{5}{2}\right)}= \sqrt{29}\cdot e^{i\left(\tan^{-1}\frac{5}{2}\right)}$

As far as adding complex numbers, just add the correspondig real and imaginary components:

$(a+bi)+(c+di) = a+c+(b+d)i$

Multiplication is easiest in polar form:

$\left(r_{1}\cdot e^{i\theta_{1}}\right)\left(r_{2}\cdot e^{i\theta_{2}}\right) = r_{1}r_{2}e^{i\left(\theta_{1}+\theta_{2}\right)}$

On a side note, are you're an engineering major since you used $j$ instead of $i$? Also, this probably belongs in the intermediate forum, even though you're most likely taking a first course in complex analysis, and when you start doing calculus with complex variables, there is a complex analysis forum.


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

[quote="Dr Sonnhard Graubner"]hello, the online integrator says:
Mathematica could not find a formula for your integral. Most likely this means that no formula exists. [More information]
Sonnhard.[/quote]

But there are methods to find definite integrals without finding the antiderivative.
(double integration, contour integration, other complex methods, etc.)

Lovely integral, I have to say. It is probably the consequence of the general form which also yields Ahmed's integral ($ \int_{0}^{1}\frac {\tan^{ \minus{} 1}\sqrt {x^2 \plus{} 2}}{\sqrt {x^2 \plus{} 2}(x^2 \plus{} 1)}dx$), but I've never seen the general form.

(To be more precise, I have seen a generalized form, but not suitable for your integral - it was $ \int_{0}^{1}\frac {\tan^{ \minus{} 1}\sqrt {x^2 \plus{} a}}{\sqrt {x^2 \plus{} a}(x^2 \plus{} \frac{a}{2})}dx$).

That's why I'm very interested to see the solution, and the generalization, if possible.

$ \sum _{n\equal{}0}^{\infty } \frac{(\minus{}1)^n \left(x^2\plus{}3\right)^n}{(2 n\plus{}1) \left(x^2\plus{}1\right)} \equal{} \frac{\tan ^{\minus{}1}\left(\sqrt{x^2\plus{}3}\right)}{\left(x^2\plus{}1\right) \sqrt{x^2\plus{}3}}$

And... nothing came. I tried 3 different aproachs to this and ended all with hypergeometric series.

I wonder if there is a way without hgs, since this integral is a lot harder...

[quote="EulerIntegral"]$ \sum _{n \equal{} 0}^{\infty } \frac {( \minus{} 1)^n \left(x^2 \plus{} 3\right)^n}{(2 n \plus{} 1) \left(x^2 \plus{} 1\right)} \equal{} \frac {\tan ^{ \minus{} 1}\left(\sqrt {x^2 \plus{} 3}\right)}{\left(x^2 \plus{} 1\right) \sqrt {x^2 \plus{} 3}}$

And... nothing came. I tried 3 different aproachs to this and ended all with hypergeometric series.

I wonder if there is a way without hgs, since this integral is a lot harder...[/quote]

This identity holds only if $ \minus{}1 < \sqrt{x^2 \plus{} 3} \leq 1$, which is impossible in $ \mathbb{R}$.  :lol:

Actually, I have tried a similar approach using the identity $ \tan^{\minus{}1} \sqrt{x^2 \plus{} 3} \equal{} \frac{\pi}{2} \minus{} \tan^{\minus{}1} \frac{1}{\sqrt{x^2 \plus{} 3}}$, which turned out to be not a good idea. [img]http://myhome.shinbiro.com/~sosQED/images/m_0503.gif[/img]

$ \frac {1}{p} \plus{} \frac {1}{q} \equal{} \frac {p \plus{} q}{p\, q} \Rightarrow \frac {1}{p\, (p \plus{} q)} \plus{} \frac {1}{q\, (p \plus{} q)} \equal{} \frac {1}{p\, q}$

Set $ p \equal{} a^2 \plus{} x^2$ and $ q \equal{} a^2 \plus{} y^2$ and integrate with respect to $ x$ and $ y$ each time from $ 0$ to $ 1$ .

$ \int_0^1 \int_0^1 \frac {dx \, dy}{(a^2 \plus{} x^2)(2a^2 \plus{} x^2 \plus{} y^2)} \plus{} \int_0^1 \int_0^1 \frac {dy \, dx}{(a^2 \plus{} y^2)(2a^2 \plus{} x^2 \plus{} y^2)}$ $ \equal{} \int_0^1 \frac {dx}{a^2 \plus{} x^2}\cdot \int_0^1 \frac {dy}{a^2 \plus{} y^2}$

By reversing the roles of $ x$ and $ y$ we see that both integrals on the LHS are the same and both integrals on the RHS are the same.

So $ 2\int_0^1 \int_0^1 \frac {dx}{a^2 \plus{} x^2}\, \frac {dy}{\sqrt {2a^2 \plus{} x^2}^2 \plus{} y^2} \equal{} \left(\frac {1}{a} \arctan\frac {1}{a}\right)^2$ $ \Rightarrow \int_0^1 \frac {2a^2}{a^2 \plus{} x^2}\, \left.\frac {\arctan\frac {y}{\sqrt {2a^2 \plus{} x^2}}}{\sqrt {2a^2 \plus{} x^2}}\right|_0^1\, dx \equal{} \left(\arctan \frac {1}{a}\right)^2$ .

Now write $ \arctan\frac {1}{\sqrt {2a^2 \plus{} x^2}}$ as $ \frac {\pi}{2} \minus{} \arctan\sqrt {2a^2 \plus{} x^2}$ .

$ \underbrace{\int_0^1 \frac {2a^2}{a^2 \plus{} x^2}\, \frac {\frac {\pi}{2}}{\sqrt {2a^2 \plus{} x^2}}\, dx}_{\pi \left.\arctan \frac {x}{\sqrt {2a^2 \plus{} x^2}}\right|_0^1} \minus{} \int_0^1 \frac {2a^2}{a^2 \plus{} x^2}\, \frac {\arctan\sqrt {2a^2 \plus{} x^2}}{\sqrt {2a^2 \plus{} x^2}}\, dx \equal{} \left(\arctan \frac {1}{a}\right)^2$ .

Hence $ \int_0^1 \frac {2a^2}{a^2 \plus{} x^2}\, \frac {\arctan \sqrt {2a^2 \plus{} x^2}}{\sqrt {2a^2 \plus{} x^2}}\, dx \equal{} \pi \,\arctan\frac {1}{\sqrt {2a^2 \plus{} 1}} \minus{} \left(\arctan\frac {1}{a}\right)^2$ .

If I'm not mistaken, that is the form analogous to the generalization I gave in my post (without solution, though), but I still don't see its application in this particular case (maybe I'm overseeing something obvious, but..)

This was only a proof of the generalization in hsiljak's post.

It didn't refer to misan's question at the start.

This, I think, can be integrated in a very simple way, although it will be a bit lengthy.
First  put (3+x^2)^(1/2) as t
Then substitute (tan inverse t) as z
We get integration of  ((z)Sec^2(z) dz)/((tan^2(z)-1)((tan^2(z)-3)^(1/2)))
This can be integrated by parts with z as the first function and the rest as the second function.

Am I wrong???


[color=brown]Let be given a triangle $ ABC,$ prove that [/color] \[ 8\cdot\left(\frac{r+4R}{p}\right)^{3}+27\cdot\frac{r}{p}\ge 27\cdot \frac{r+4R}{p}\]

The inequality can be written as $ 2\left(\frac{r+4R}{p}\right)^{3}\geq 27\cdot\frac{R}{p}$
We have $ \frac{r+4R}{p}=\frac{4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}+4R}{R(\sin A+\sin B+\sin C)}=\frac{\cos A+\cos B+\cos C+3}{\sin A+\sin B+\sin C}$.
The inequality becomes $ 8\left(\frac{\cos A+\cos B+\cos C+3}{\sin A+\sin B+\sin C}\right)^{3}\geq 27\cdot\frac{4}{\sin A+\sin B+\sin C}\Leftrightarrow$
$ \Leftrightarrow 2(\cos A+\cos B+\cos C+3)^{3}\geq 27(\sin A+\sin B+\sin C)^{2}\Leftrightarrow$
$ \Leftrightarrow 2\left(2\cos^{2}\frac{A}{2}+2\cos^{2}\frac{B}{2}+2\cos^{2}\frac{C}{2}\right)^{3}\geq 27\left(4\cos\frac{A}{2}+\cos\frac{B}{2}\cos\frac{C}{2}\right)^{2}\Leftrightarrow$
$ \Leftrightarrow \left(\cos^{2}\frac{A}{2}+\cos^{2}\frac{B}{2}+\cos^{2}\frac{C}{2}\right)^{3}\geq 27\cos^{2}\frac{A}{2}\cos^{2}\frac{B}{2}\cos^{2}\frac{C}{2}\Leftrightarrow$
$ \Leftrightarrow \cos^{2}\frac{A}{2}+\cos^{2}\frac{B}{2}+\cos^{2}\frac{C}{2}\geq 3\sqrt[3]{\cos^{2}\frac{A}{2}\cos^{2}\frac{B}{2}\cos^{2}\frac{C}{2}}$

[color=darkblue]Is well-known or prove easily that $ \{\begin{array}{c}ab+bc+ca=p^{2}+r^{2}+4Rr\\\\ (p-a)(p-b)(p-c)=pr^{2}\\\\ \boxed{a(p-a)+b(p-b)+c(p-c)=2r(4R+r)}\end{array}$.

Apply AM-GM inequality :  $ [\sum a(p-a)]^{3}\ge 27\prod [a(p-a)]$ $ \Longleftrightarrow$

$ 8r^{3}(4R+r)^{3}\ge 27\cdot 4Rpr\cdot pr^{2}$ $ \Longleftrightarrow$ $ \boxed{2(4R+r)^{3}\ge 27p^{2}R}$.[/color]

[quote][color=darkred][b]Other similar inequalities:[/b][/color] $ \{\begin{array}{c}\boxed{2(2R-r)^{3}\ge 27Rr^{2}}\\\\ \boxed{\frac{p^{2}+r^{2}}{2Rr}\ge 4+3\cdot\sqrt [3]{\frac{2r}{R}}}\\\\ \boxed{\frac{R}{2r}+\frac{3}{4}\cdot(\frac{4R+r}{p\sqrt 3})^{2}\ge \frac{7}{4}}\end{array}$.[/quote]
[hide="Hint."][color=darkblue]You can use the well-known identities $ \{\begin{array}{c}\boxed{\sum a(p-b)(p-c)=2pr(2R-r)}\\\\ \boxed{\sum bc(p-a)=p(p^{2}+r^{2}-8Rr)}\\\\ \boxed{\sum a^{2}(p-b)(p-c)=4p^{2}r(R-r)}\end{array}$

and apply AM-GM inequality : $ \{\begin{array}{c}[\sum a(p-b)(p-c)]^{3}\ge 27\prod a(p-b)(p-c)\\\\ [\sum bc(p-a)]^{3}\ge 27\prod bc(p-a)\end{array}$

or the Cebasev's inequality $ \sum a^{2}\cdot\sum (p-b)(p-c)\le 3\cdot\sum a^{2}(p-b)(p-c)$ a.s.o.[/color][/hide]
[color=darkblue][b]Remark.[/b] The second inequality is weakly because $ p^{2}+r^{2}\ge 14Rr$ $ \implies$ $ \frac{p^{2}+r^{2}}{2Rr}\ge 7\ge 4+3\sqrt [3]{\frac{2r}{R}}$.[/color]

$ S=a+b+c,\,P=ab+bc+ca,\,Q=abc\implies\boxed{8\cdot S^{3}+27\cdot Q\ge 27\cdot PS}$

$ \left\{\begin{array}{c}a: =\tan\frac{A}{2},\,b: =\tan\frac{B}{2},\,c: =\tan\frac{C}{2}\\ \tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}=\frac{r+4R}{p}\\ \tan\frac{A}{2}\cdot\tan\frac{B}{2}+\tan\frac{B}{2}\cdot\tan\frac{C}{2}+\tan\frac{C}{2}\cdot\tan\frac{A}{2}=1\\ \tan\frac{A}{2}\cdot\tan\frac{A}{2}\cdot\tan\frac{A}{2}=\frac{p}{r}\end{array}\right\|$ $ \implies\boxed{8\cdot\left(\frac{r+4R}{p}\right)^{3}+27\cdot\frac{r}{p}\ge 27\cdot\frac{r+4R}{p}}$


Να αποδειχθεί ότι:

1) Αν ισχύει $a+c=a+b$ τότε $b=c$

2) $b-a=b+(-a)$

3) $-(-a)=a$

4) $a(b-c)=ab-ac$

5) $0\cdot a=0$

 :D

Απλά καταπληκτικά

[quote="ambros"]Απλά καταπληκτικά[/quote]

Οι αποδείξεις είναι ακόμη καλύτερες. Το κακό με αυτά είναι ότι είναι "παρεξηγημένα", μιας και τα μαθαίνουμε ως αξιώματα, ενώ στην πραγματικότητα δεν είναι  :wink:

να βαλουμε δικες αποδηξεις Στελιο ? εννοω οχι απο καπιοθιο βιβλιο

lolol γιατί μόνο για το RAKAR? δεν κάνουν συνολοθεωρία!
να πω, να πω?

για πες νεαρε ? :P

1)Για α=0 ισχύει
έστω ότι ισχύει για α=κ
τότε για α=κ+1 είναι
κ+1+β=κ+1+γ<=>κ+β=κ+γ<=>β=γ οεδ

2)β-α=β+(-α)<=>β=β+(-α)+α<=>0=(-α)+α<=>0-α=(-α) ισχύει

3) β=0, α=-α

4)προφανές

5) (/)=κενό
0*α=((/)XA)#=(/)#=0

(/)XA=(/)
γιατί αν χ=(β,α) ανήκει (/)XA τότε β ανήκει (/) άτοπο

and the angel og maths sings sets in my head... lolol
@kostas  :wink:

για το πρωτο : πρεπει να δειξεις οτι ισχυει για α=1
για το δευτερο πρεπει να δειξεις το πρωτο

@thatin:

Αυτά στα οποία βασίζεσαι για να αποδείξεις τις παραπάνω ισότητες θέλουν να αποδειχθούν και αυτά.

[b]Edit[/b]

Για να γίνω πιο σαφής, τα αξιώματα που θα χρειαστούν είναι τα εξής:

1) [u]Communicative laws[/u]: $x+y=y+x$, $xy=yx$
2) [u]Associative laws[/u]: $x+(y+z)=(x+y)+z$, $x(yz)=(xy)z$
3) [u]Distributive law[/u]: $x(y+z)=xy+xz$
4) [u]Identity element[/u]: Υπάρχουν δύο συγκεκριμένοι πραγματικοί αριθμοι, το $0$ και το $1$, τέτοιο ώστε για κάθε πραγματικό $x$ να ισχύει: $x+0=x$ και $1\cdot x=x$
5) [u]Existence of negatives[/u]: Για κάθε πραγματικό $x$ υπάρχει κάποιος πραγματικός $y$ τέτοιος ώστε $x+y=0$
6) [u]Existence of reciprocals[/u]: Για κάθε πραγματικό $x\neq0$ υπάρχει ένας πραγματικός $y$ τέτοιος ώστε $xy=1$

[quote="kostas"]για το πρωτο : πρεπει να δειξεις οτι ισχυει για α=1
[/quote]

έχει αποδειχθεί στη διπλανή σελίδα! (hm.. mallon den eprepe na to pw auto... :rotfl: )
α+1=β+1<=>(ΑU{0})#=(BU{1})#
δείχνω ότι υπάρχει αμφιμορφισμός μεταξύ Α,B

@mostel έτσι δε θα τελείωσουμε ποτέ! είναι αδύνατο να μη δεχθούμε κάποιες πρωταρχικές έννοιες...

μπερδευτικα και νομιζα οτι μιλαμε μονο ακεραιους.
(μολις περασαμε το Ρουμανικο φορουμ :clap2: )

για το πρωτο υποθετουμε οτι b>c ή b<c προσθετουμε το α στα δυο μελη και ατοπο.
για το τριτο απο το ποστ του mostel παιρνουμε απλα τα 1)communivative laws,2)associative laws για τον πολλαπλασιασμο

[quote="kostas"]...
(μολις περασαμε το Ρουμανικο φορουμ :clap2: )[/quote]

Άντε και καλορίζικοι στη 1η θέση ευρωπαϊκά και 2η παγκόσμια. Αν και θέλει χρόνο, νομίζω ότι μπορούμε να περάσουμε και τους Καναδούς...

Κώστας Βήττας.

$\bullet$[quote]topic is exclusively dedicated to RAKAR[/quote]
thanks gia tin afierosi.... :P

$\bullet$ ante +stin proti 8esi...!!! :)

$\bullet$ontos mostel...xriazonte akrivos ayta pu egrayes...
ine i idiotites pu isixun se enan antimeta8etiko daktilio me monada...

$\bullet$
[u]@thatin[/u]
[hide]dn ine anagki an kanume sinolo8eoria...[/hide]

 :D

Για το 5 έχουμε ότι:
0*α=(β-β)*α=α*(β-β)=αβ-αβ=0

Να σας ρωτήσω και κάτι που είμαι καινούργιος.
Πως μπορώ να γράφω με σωστά μαθηματικά σύμβολα??
Από το word δεν μπόρεσα να κάνω κάτι..

@RAKAR
[hide]Αυτό προσπαθούσα να πω τόσο καιρό ! Όχι, δεν είναι θέματα από την IMO του 1821  :P [/hide]

@Bouzoukman
[hide]Για να γράψεις μαθηματικά σύμβολα να χρησιμοποιήσεις latex, αν δεν ξέρεις τι είναι, ρίξε μία ματιά εδώ --> http://www.latex-project.org .

Ένα μικρό παράδειγμα..

Για να γράψεις: χ στο τετράγωνο + 2 = 5 θα γράψεις το εξής:

[code]$x^2 + 2 = 5 $[/code]

Και θα βγει:

$x^{2}+2 = 5$

:) [/hide]

οταν λεμε αξιωματα δεν δεν χρειαζεται αποδειξη.δεν εχουμε ορισει που μιλαμε,πχ στο Ζ  το R .σε καποιους δακτυλιους δεν ισχυει η αντιμεταθετικοτητα (κατι που κανει την αποδειξη του Bouzoukman λαθος) μιλαμε  στο R

[quote="kostas"]σε καποιους δακτυλιους δεν ισχυει η αντιμεταθετικοτητα (κατι που κανει την αποδειξη του Bouzoukman λαθος) μιλαμε  στο R[/quote]
sostos kosta...
px. i 2x2 pinakes me stixia pragmatikus ari8imus den ine antimeta8etikos daktilios...(+dn ine +pedio akereotitas)gt exi dieretes tu midenos...
px(1,0,0,0)epi (0,0,0,1)=(0,0,0,0) diladi ab=0 opu a=(1,0,0,0) kai b=(0,0,0,1)
ala kanenas apo ta a,b den ine miden...(diladi iparxun dieretes tu midenos..)

 :D

bouzoukman... :welcomeani:  +apo m1....

Έχετε δίκιο παιδιά για το γεγονός ότι δεν ισχυεί σε όλους τους δακτύλιους η αντιμεταθετική.
Εγώ για το σύνολο R αναφερόμουν.....
Γεια και σε σένα Rakar!!!!!
Ευχαριστώ Mostel γιατί δεν ήξερα τι να κάνω....

$1)$ $a+c=a+b \Leftrightarrow (-a)+(a+c)=(-a)+(a+b) \Leftrightarrow$
$\Leftrightarrow ((-a)+a)+c=((-a)+a)+b \Leftrightarrow 0+c=0+b \Leftrightarrow c=b$.

$3)$ Ισχύει $(-a)+(-(-a))=0 \Leftrightarrow a+((-a)+(-(-a)))=a+0 \Leftrightarrow$
$\Leftrightarrow (a+(-a))+(-(-a))=a \Leftrightarrow 0+(-(-a))=a \Leftrightarrow (-(-a))=a$.

$5)$ $0 \cdot a=(0+0) \cdot a=0 \cdot a+0 \cdot a \Leftrightarrow$
$\Leftrightarrow 0 \cdot a+0=0 \cdot a+0 \cdot a \Leftrightarrow 0=0 \cdot a$, λόγω του $(1)$.

$4)$ Λήμμα: $(-1) \cdot a=-a$ .
Απόδειξη λήμματος: ισχύει $a+(-1) \cdot a=a \cdot 1+a \cdot (-1)=a \cdot (1+(-1))=a \cdot 0=0 \cdot a=0$ (λόγω του $(5)$) $\Leftrightarrow$
$\Leftrightarrow (-a)+(a+(-1)\cdot a)=(-a)+0 \Leftrightarrow ((-a)+a)+(-1)\cdot a=-a \Leftrightarrow$
$\Leftrightarrow 0+(-1)\cdot a=-a \Leftrightarrow (-1) \cdot a=-a$.
Οπότε $a(b-c)=a(b+(-c))=ab+a(-c)=ab+a((-1)c)=ab+(a(-1))c=$
$= ab+((-1)a)c=ab+(-1)(ac)=ab-ac$.

To $(2)$ δε το πολυκατάλαβα. Αυτός δεν είναι ο ορισμός της αφαίρεσης; Με βάση τί πρέπει να το αποδείξουμε;

Btw bouzoukman  :welcomeani: . [hide]Αλήθεια,τί σε έκανε να διαλέξεις αυτό το  username;  :P[/hide]

Για το $2$...

Έστω $x=b-a$ και έστω $y=b+(-a)$. Θέλουμε να δείξουμε ότι $x=y$.

Όμως $x+a=b$ (από ορισμό του $b-a$ -μπορώ να βάλλω την απόδειξη αν το θελήσετε-)

Επίσης:

$y+a=[b+(-a)]+a=b+[(-a)+a]=b+0=b$

(Γιατί το $a+(-a)=0$ μπορώ να βάλλω επίσης απόδειξη)

Δηλαδή καταλήξαμε στο ότι $x+a=y+a$ όμως από το θεώρημα $1$ που έδωσα για απόδειξη θα έχουμε $x=y$, όπου και προκύπτει το ζητούμενο.

@stedes
[hide]Δεν είχα χρόνο και δεν είδα αν αυτά που έγραψες είναι σωστά. Θα τα δω αύριο και θα σου στείλω pm αν είναι !  :) [/hide]

Παίζω από μικρός μπουζούκι και σε όλα τα ουφατσίδικα που πήγαινα
έβαζα αυτό για παρατσουκλι και τώρα το έχω καθιερώσει........ :lol:
Παλιές συνήθειες Stedes.....


It is given that $2^{333}$ is a $101$-digit number whose first digit is $1$. How many of the numbers $2^k$, $1 \le k \le 332$, have first digit $4$?

Check problem 7 from [url]http://www.math.toronto.edu/oz/turgor/archives/TT2001F_JAsolutions.pdf[/url]


probability
geometry
geometric transformation
reflection
combinatorics unsolved
combinatorics

Do there exist such a triangle $T$, that for any coloring of a plane in two colors one may found a triangle $T'$, equal to $T$, such that all vertices of $T'$ have the same color.
[b]
proposed by S. Berlov[/b]

I don't know whether this probabilistic method can be accepted as a solution:
We prove that the triangle with angles $120,30,30$ and sides $\sqrt 3,1,1$ is good.
Firstly note that the center of an equilateral triangle of side $\sqrt 3$ has color opposite to the color that most vertices of the triangle are colored (or else we get our monochromatic triangle).(1)
That means if $O$ is an arbitrary point and $ABC$ is an equilateral with side $\sqrt 3$ with center $O$ then at least $2$ of segments $OA,OB,OC$ have endpoints of opposite colors.(2)
Now "integrating" that's what we see:
Denote by $r$ the probability that from a white point one reaches a black point by a $\sqrt 3$ translation(the point and the translation ar taken arbitrarily). Analogously define $s$ for black points.
"Integrating" the previous argument (2) we get $r,s\geq \frac 23$.
Now if $t$is the probability that from a given white point one reaches a black one by a translation by magnitude $2\sqrt 3$ is as easily seen $r(1-s)+s(1-r)=r+s-2rs \geq \frac 23+\frac 23-(\frac 23)^2=\frac 89>\frac 56$.
Now this means particularly that one can find a regular hexagon
$ A B C D E F$ with side $2\sqrt 3$ and center $O$ is white, while at least $5$ of $A,B,C,D,E,F$ are black.
WLOG suppose $A,B,C,D$ are black.
Then the centers $X,Y$ of $OAB,OCD$ are white by (1).
If we take $Z$ the center of $OEF$ then the triangle $XYZ$ with center $O$ contradicts (1)
QED????

Probability is not a magic wand. You cannot use it until you define probabilistic space.

Well no wonder I had such a hard time trying to show that no such triangle exists! :)

I think you're right about that triangle, iura, but we don't need such extreme measures (like probability spaces we're not at all sure about :)) in order to prove it.

If a $\sqrt 3$ equilateral triangle is monochromatic, (say it has color $r$), then its center and the reflections of its center in two sides must have color $b$, and we have a $1,1,\sqrt 3$ monochromatic triangle. This means that if no $1,1,\sqrt 3$ is monochromatic, then no $\sqrt 3,\sqrt 3,\sqrt 3$ is either.

Now construct a network of $\sqrt 3$ equilateral triangles (a tesselation of the plane)+their centers. You'll easily find that all pairs of centers distance $2$ apart are monochromatic. Any two points of the plane distance $2$  apart can be seen as two centers in such a network, so pairs of points distance $2$ apart are monochromatic. From here we easily get a contradiction (we find the entire plane to have the same color).

I hope it's correct, but if it's not, forgive me: I just woke up and I'm a bit confused :D.

[in the above, "an $a,b,c$" means a triangle with sides $a,b,c$, and "an $\ell$ equilateral triangle" means equilateral triangle with side $\ell$]

[quote="grobber"]I hope it's correct, but if it's not, forgive me: I just woke up and I'm a bit confused :D.[/quote]
Just woke up? What was your local time? :lol:

[quote="grobber"]I don't know.. About 10 AM. But I went to sleep at about 4 AM :).[/quote]
Analogously  :)

Can it be solved in a such way ?

    First we always can find a segment with vertices of the same
color with a given length (just construct equilateral triangle,
 from three vertices one pair will satisfy condition).

    Let  $A_{1}A_{2}A_{3}A_{4}A_{5}A_{6} $ be a regular hexagon.
We choose triangle $A_{1}A_{2}A_{4}$ , then
if we find a segment $A_{1}A_{4}$ with two vertices of the same color,
assuming that contradiction exists we get $A_{2},A_{3},A_{5},A_{6}$
are of another color so $A_{2}A_{3}A_{5}$ = $A_{1}A_{2}A_{4}$ satisfyies condition. q.e.d.

I can't understand sentence there three "then" are in row without any punctuation marks. Is there some meaning? I don't know.  ;)

I think, prowler means the following: without loss of generality,
$A_1$ and $A_4$ are red, hence all other vertices are blue and three of them form desired blue triangle. It's true. The other possible solution: some 4 of 7 vertices of regular 7-gon are of the same clour. Then they form a triangle with angles $\pi/7,\,2\pi/7,\,4\pi/7$.


In a scoring system, 6 points are given for each correct answer, 2.5 points are given for each unanswered question, and no points are given for an incorrect answer. Some of the possible scores between 0 and 150 can be obtained in only one way, for example, the only way to obtain a score of 146.5 is to have 24 correct answers and one unanswered question. Some scores can be obtained in exactly two ways, for example, a score of 104.5 can be obtained with 17 correct answers, 1 unanswered question, and 7 incorrect, and also with 12 correct answers and 13 unanswered questions. There are three scores that can be obtained in exactly three ways. What is their sum? Thanks to anyone for their assistance.
- and if you are wondering this is the AMC grading system

Let r be the number of right answers and w be the number of wrong answers. These two nonnegative integers are subject to the constraint that r+w:le:25. The grading system is (12r+5w)/2. You can temporarily ignore the division by 2 as that will not change the number of ways that a certain score is attainable. So you have 12r+5w. Note that this is equal to 12(r-5k)+4(w+12k), where k is a positive integer. Think about how we get new solutions as k varies.

I am not sure I get this...
I drew up an equation from what you explained. 12(r-5k) + 4(w+12k)=12r+5w
However, when I substitute in a positive number for k, I get w = negative number. How can there be negative solutions? So, I am a bit lost. I did something [i]really [/i]wrong.

A chef is required to serve 10 kg of meat. By experience the chef estimates that 30% of the meat is wasted during preparation and cooking. How much meat should be ordered  (Answer to the nearest 0.1 kg)

Out of a delivery of 12.0 kg of green pepper, 750g was wastage. Express the wastage in percentage. (Answer to the neares 0.1%)

Answer:  %

Out of a delivery of 25 kg of meat, there was a wastage of 12 kg, express the wastage in percentage. (Give answer to the nearest 0.1%)

Answer: %

The total food cost to produce 25 main course main meals was $49.27 . Calculate the cost per meal:
a) round off to three decimal places
b) round off to two decimal places
If the cost of labor is $0.20 for an item which sells for $0.76 the labor cost is said to be: (Answer: Percent to two decimal place)  %
If an operation is operating on a 30.0% Food Cost the correct selling price of an item whose raw food cost is
$0.75 would be:

What should the selling price of an item be if its cost of food is $4.00 The operation is seeking a 43% food cost.
Answer (rounded to two decimal places)
If the cost of food for a given period is
$8200, the labor cost is $8290, other costs are $1210 and 8000 meals were served, calculate the average cost per meal served (rounded to two decimals) $
50kg of beef was purchased. There was a 15% loss during cooking. Express the loss in kg.
What would be the wage cost expressed in percentage if sales per month
$120000 and wages were $78000
Answer (round off to two decimal places) =
A hundred and twelve 125 g portions of meat are required. Assuming that 30% has to be allowed for wastage, how many kg of meat has to be ordered _____.
Standardized recipe ensures precise _________ and _________
Answer
A) quality, quantity
B) time, accuracy
C) measurements, portion size
D) product, taste
Portion cost =
Answer
A) AP/EP
B) hidden costs - yield cost
C) cost of ingredients / number of portions
D) total cooked weight x serving portion weight
To convert a recipe from its original number of portions to a different number of portions, you must first calculate the:
Answer
A) new quantity
B) old total yield
C) new total yield
D) conversion factor
Food cost is
$1.80. Food cost desired is 30%. Menu price is:

Answer
a) $8.00
b) $5.00
c) $6.00
d) $4.00

Food cost percent is:

Answer
a) Menu price of an item divided by its food cost
b) Food cost of an item divided by its menu price
c) Food cost of an item divided by its EP
d) EP of a recipe divided by its total cost

If the total sales for a given period are $400,000 and the cost of food purchased is $100,000, the food cost is said to be:

Answer
a) 37%
b) 29%
c) 33%
d) 25%
e) 40%

Assume that you plan to serve 500 portions of beef stew, using 150 g of raw meat for each portion. To provide this you would need to order:

Answer
a) 7.5 kg of beef
b) 75 kg of beef
c) 150 kg of beef
d) 29 kg of beef

If a steak costs $1.25 and the food cost is said to be set at 25% the selling price should be:
Answer
a) exactly the same
b) at least five times as much
c) three times the cost
d) at least four dollars
e) four times the cost
The recipe cost is
$36.00 for 20 portions. If the cost increases by 10%, the portion cost will be:

Answer
a) $1.98
b) $1.81
c) $2.20
d) $2.12

Please do not revive old topics  :!:

Thanks,

[color=red]b-flat[/color]

Locked.


geometry
3D geometry
sphere
quadratics
algebra

(Assume that your eyes are at the top of your head)
(Assume that the Earth is a perfect sphere)

How tall would a person have to be if he/she could see for exactly a mile?

This is similiar to [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=152767[/url] (It also has a nice picture to help illustrate)

[hide="Solution"]
Using the radius of the Earth as 3960 miles, we solve for the triangle with legs 3960, 1, and hypotenuse 3960 + h. The 3960 is the radius of the Earth, the 1 is the mile the person can see, and h is the height of the person.

$3960^{2}+1^{2}= (3960+h)^{2}$
Forms this quadratic:
$0 = h^{2}+7920h-1$
Quadratic equation gives us:
$h = \frac{7920 \pm \sqrt{7920^{2}+4}}{2}$
$h \approx .000126,-7920$
Ignoring the negative, the answer is for h is
$h \approx .000126$ miles or
$h \approx .667$ feet

[/hide]