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inequalities
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i)Let $ a,b,c>0$ . Prove that :
$ 2(a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1) \geq (a\plus{}1)(b\plus{}1)(c\plus{}1)(abc\plus{}1)$
ii)Let $ a,b,c \in [0;1]$ . Prove that :
$ \frac{1}{1\plus{}a\plus{}b}\plus{}\frac{1}{1\plus{}b\plus{}c}\plus{}\frac{1}{1\plus{}c\plus{}a} \leq \frac{3}{1\plus{}2\sqrt[3]{abc}}$
iii)Let $ a,b,c>0$ and $ ab\plus{}bc\plus{}ca\equal{}3$ . Prove that :
$ a^3\plus{}b^3\plus{}c^3\plus{}7abc \geq 10$
Can you help me ? :(

[hide="iii"]posted at least two times before:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1294329900&t=185868[/url]

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=162013[/url] (this link is in persian,but you can understand the whole idea.)[/hide]

[b]iii,[/b] For this kind of inequality, in my opinion we always can solve it by the "p,q,r technique"  :wink:
Above one is not hard. I have a solution for this one. See here:
http://www.mathlinks.ro/viewtopic.php?search_id=1852732440&t=156163

(sorry, I have posted 2 posts in recently topics, and I couldn't delete :D)

[b]ii,[/b] It was appeared in Vietnam's inequality Forum. I used to Mixing Varible method, it is not very hard  :wink:

My solution for the 2nd inequality:
First, we consider that when we decrease all of 3 variables a,b,c by the same amount of c=min(a,b,c),which means (a-c),(b-c),0, respectively we will have: LHS is increasing but RHS is decreasing.
Therefore,it's sufficient to prove this inequality in the case when one variable is equal to zero. Assume that c=0
We will have: 1/1+a+b +1/1+b +1/1+a=<3(0=<a,b<=1)
Clearing all denominators, we will have:(2+b+a)(1+a+b)+(1+b)(1+a)=<3(1+a+b)(1+a)(1+b)
It's equivalent to: a^2+b^2+4(a+b)+3ab=<3(1+2a+2b+3ab+a^2+b^2+a^2b+b^2a)
And we will have: 2(a^2+b^2)+3ab(a+b)+2(a+b)+3ab>=0, which is clearly true.
The equality happens when a=b=c=1 or a=b=c=0(Done)

[quote="conan_naruto236"]i)Let $ a,b,c > 0$ . Prove that :
$ 2(a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \geq (a \plus{} 1)(b \plus{} 1)(c \plus{} 1)(abc \plus{} 1)$
ii)Let $ a,b,c \in [0;1]$ . Prove that :
$ \frac {1}{1 \plus{} a \plus{} b} \plus{} \frac {1}{1 \plus{} b \plus{} c} \plus{} \frac {1}{1 \plus{} c \plus{} a} \leq \frac {3}{1 \plus{} 2\sqrt [3]{abc}}$
iii)Let $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$ . Prove that :
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 7abc \geq 10$
Can you help me ? :([/quote]
Can you help me i) and ii) ?  :)

For [b]ii[/b], you can see my solution at:
http://diendantoanhoc.net/forum/index.php?showtopic=37391

I don't think ghjk's solution was correct ;)

[quote="BaBaK Ghalebi"][quote="ghjk"]when we decrease all of 3 variables a,b,c by the same amount of c=min(a,b,c),which means (a-c),(b-c),0, respectively we will have: LHS is increasing but RHS is decreasing.
[/quote]
 :huh:
I'm not sure about this step of your solution,could you please explain it in more details.[/quote]
Zaizai was right! My solution is wrong :blush: .I will try to find another solution for this one and post it later :wink:

Finally, I got the solution! Hope it will be a correct one.
Here it is:
We will prove the inequality  in the case when there exists at least 1 variables>=1/2.W.L.O.G, assume that a>=1/2
Now we will have: LHS<=1/(3/2+b)+1/(3/2+c)+1/(1+b+c)
To prove the RHS =<3/1+2cube root(abc), we will prove that: f(a,b,c)<=f(a,square root(bc),square root(bc))(it's not hard to prove, so I'll leave it for you guys).
Now the rest is easy since we just consider the case when b=c and a>=1/2
Therefore, if there exists at least one variable>=1/2, then the inequality is proved!
Now, we will prove it when all 3 variables<1/2
We will prove the following inequality: 1/1+2a +1/1+2b<=2/[1+2square root(ab)]
It's equivalent to: 2(a+b)[1-2.square root(ab)]>=0, which is obviously true since a,b<=1/2
Now applying it twice we will have: 1/1+2a +1/1+2b +1/1+2c=<3/1+2.cube root(abc)(1)(it's called Jensen's inequality)
In addition we also have: 1/(1+2a)+ 1/(1+2b)>=2/(1+a+b) by AM-GM and similar to 2 more pairs[1/(1+2a) and 1/(1+2c), 1/(1+2b) +1/(1+2c)].(2)
So, from (1) and (2) we are done in this case!
From 2 cases above, we proved the inequality.

[quote="conan_naruto236"][quote="conan_naruto236"]i)Let $ a,b,c > 0$ . Prove that :
$ 2(a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1) \geq (a \plus{} 1)(b \plus{} 1)(c \plus{} 1)(abc \plus{} 1)$
ii)Let $ a,b,c \in [0;1]$ . Prove that :
$ \frac {1}{1 \plus{} a \plus{} b} \plus{} \frac {1}{1 \plus{} b \plus{} c} \plus{} \frac {1}{1 \plus{} c \plus{} a} \leq \frac {3}{1 \plus{} 2\sqrt [3]{abc}}$(1)
iii)Let $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$ . Prove that :
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 7abc \geq 10$
Can you help me ? :([/quote]
Can you help me i) and ii) ?  :)[/quote]
I have a nice solution .
+If $ abc \equal{} 0$ => The inequality is right
+If $ 0 < abc\leq 1$.
There fore exist $ k , a' , b' , c' > 0 : a\equal{}ka',b\equal{}kb',c\equal{}kc'\equal{}>abc \equal{} k^3.a'.b'.c'$ and $ k \equal{} \sqrt [3]{abc}$
=> $ a'b'c' \equal{} 1$
=> exist $ x,y,z : a' \equal{} \frac {x^2}{yz} , b' \equal{} \frac {y^2}{zx} , c' \equal{} \frac {z^2}{xy}$
$ (1) < \equal{} > \sum_{cyc} \frac {k(a' \plus{} b')}{1 \plus{} k(a' \plus{} b')} \geq \frac {6k}{1 \plus{} 2k}$
$ < \equal{} > \sum_{cyc} \frac {1}{\frac {xyz}{x^3 \plus{} y^3} \plus{} k} \geq \frac {6}{1 \plus{} 2k}(2)$
I have :
$ x^3 \plus{} y^3 \geq xy(x \plus{} y)$
$ \equal{} > \frac {1}{\frac {xyz}{x^3 \plus{} y^3} \plus{} k} \geq \frac {1}{\frac {z}{x \plus{} y} \plus{} k} \equal{} \frac {x \plus{} y}{z \plus{} k(x \plus{} y)}$
=>$ \sum_{cyc} \frac {1}{\frac {xyz}{x^3 \plus{} y^3} \plus{} k} \geq \sum_{cyc} \frac {x \plus{} y}{z \plus{} k(x \plus{} y)}$
Let $ z \plus{} k(x \plus{} y) \equal{} C,y \plus{} k(z \plus{} x) \equal{} B,x \plus{} k(y \plus{} z) \equal{} A$
+If $ abc \equal{} 1$ or $ k \equal{} 1 \equal{} >$ the inequality is right
+If $ 0 < k < 1$ .
$ \equal{} > x \plus{} y \equal{} \frac {A \plus{} B \minus{} 2kC}{(1 \minus{} k)(2k \plus{} 1)} , y \plus{} z \equal{} \frac {B \plus{} C \minus{} 2kA}{(1 \minus{} k)(2k \plus{} 1)} , z \plus{} x \equal{} \frac {C \plus{} A \minus{} 2kB}{(1 \minus{} k)(2k \plus{} 1)}$
$ (2) < \equal{} > \sum_{cyc} \frac {A \plus{} B \minus{} 2kC}{C(1 \minus{} k)} \geq 6$
$ < \equal{} > \frac {1}{1 \minus{} k}.\sum_{cyc} \frac {A \plus{} B}{C} \minus{} \frac {6k}{1 \minus{} k} \geq 6$
The inequality is right because :
$ \sum_{cyc} \frac {A \plus{} B}{C} \geq 6 \equal{} > \frac {1}{1 \minus{} k}.\sum_{cyc} \frac {A \plus{} B}{C} \minus{} \frac {6k}{1 \minus{} k} \geq 6$


How's can I solve it
x^3 + x + 57 is congruent to 0 (mod 53)

Hensel's Lemma?... I maybe wrong, number theory is my weakest subject by far.


Let $ n$ be a positive integer,and $ a,b,c > 0$ with $ a\plus{}b\plus{}c \equal{} 1$.Prove that
\[ \sum_{cyc}\frac{a(b^n\plus{}c^n)}{a^{n\plus{}1}\plus{}b^n\plus{}c^n} \leq \frac{6}{7}\]

[hide]It's equivalent to
\[ \sum_\text{cyc} \frac {a^{n \plus{} 2}}{a^{n \plus{} 1} \plus{} b^n \plus{} c^n}\geq \frac {1}{7}
\]
To do this, we have
\[ a^{n \plus{} 1} \plus{} b^n \plus{} c^n \equal{} a^{n \plus{} 1} \plus{} (b^n \plus{} c^n)(a \plus{} b \plus{} c)
\]

\[ \equal{} a^{n \plus{} 1} \plus{} b^{n \plus{} 1} \plus{} c^{n \plus{} 1} \plus{} ab^n \plus{} ac^n \plus{} b^nc \plus{} bc^n
\]

\[ \leq \frac {n \plus{} 3}{n \plus{} 1}\cdot a^{n \plus{} 1} \plus{} \frac {3n \plus{} 2}{n \plus{} 1}\cdot b^n \plus{} \frac {3n \plus{} 2}{n \plus{} 1}\cdot c^n
\]
(using AM-GM). So it remains to show
\[ \sum_\text{cyc} \frac {a^{n \plus{} 2}}{(n \plus{} 3)a^{n \plus{} 1} \plus{} (3n \plus{} 2)b^{n \plus{} 1} \plus{} (3n \plus{} 2)c^{n \plus{} 1}}\geq \frac {1}{7(n \plus{} 1)}
\]
WLOG let $ a\geq b\geq c$ and use Chebyshev on the sequence of denominators and the sequence of fractions to get
\[ \sum_\text{cyc} \frac {a^{n \plus{} 2}}{(n \plus{} 3)a^{n \plus{} 1} \plus{} (3n \plus{} 2)b^{n \plus{} 1} \plus{} (3n \plus{} 2)c^{n \plus{} 1}}
\]

\[ \geq \frac {3}{7(n \plus{} 1)}\cdot \frac {a^{n \plus{} 2} \plus{} b^{n \plus{} 2} \plus{} c^{n \plus{} 2}}{a^{n \plus{} 1} \plus{} b^{n \plus{} 1} \plus{} c^{n \plus{} 1}}
\]
It's easy to finish now.[/hide]


[quote="mathpk"]Find $x$ is a real number such that $x^{2}+2^{x}=1$[/quote]

Well,
here are my observations...

x=0 is the trivial solution

Also for x> 0 there are no solutions .

This is because 2x  > 1 for x> 0 .
Also x 2  > 0 for any real x

So  x 2 + 2x > 1 for all x >0

Need to check the possibility of any solution in the range x < 0.

U can  do a graphical analysis of the problem by plotting the graphs for  x 2 and 2x

It would be equal to 0. Plug 0 in for x and you will see that x squared=0 + 2 to the 0 power is equal to 1. So 0+1=1.[/b]

[quote="shobber"]is there an element algebra way to find the root?[/quote]

Nope..there are no std algebraic methods to determine the roots of such equations..

u need to use mathematical analysis to get to the solutions to such probs..U can also take a graphical approach to visualise the behaviour of the related fns

You can narrow it down a bit further however. Because if $x\leq-1$, $x^2=1$ and $2^{x}>0$, $-1<x<0$ (since we already have 0 as an answer).

Furthermore, I do not think that there is a rational answer to this problem (at least). This is because if we have $|x|<1$, then $2^{x}$ will never be rational (this follows from the proof that square root of 2 is irrational if we generalize it). This means that $x^2$ must also be irrational and of course $x$ is irrational also.


For 1, Squaring both sides of the given inequalities.
For 2, Use triangle inequality.

[quote="sidk47"]1. Solve:
(a) |2x-3|<=|x+2|
(b) |3x-2|<=|x-1|

2. Solve:
(a) Prove: |a-b|>=||a|-|b||
(b) Prove: |a-b|<==|a|+|b|

Detailed solutions would be most appreciated.[/quote]
1a) There are two ways to do this.
First way:
Case 1: $x\ge\frac{3}{2}$
Solving $2x-3\le x+2$ yields $\frac{3}{2}\le x \le 5$
Case 2: $-2< x< \frac{3}{2}$
Solving $-(2x-3)\le x+2$ yields $\frac{1}{3}\le x < \frac{3}{2}$
Case 3: $x\le-2$
Solving $-(2x-3)\le-(x+2)|$ yields nothing in that range.
Therefore, the answer is $\boxed{\frac{1}{3}\le x \le 5}$

The second way is the following:
$|2x-3|\le|x+2|$
$(2x-3)^{2}\le (x+2)^{2}$
$4x^{2}-12x+9 \le x^{2}+4x+4$
$3x^{2}-16x+5\le 0$
And again,
$\boxed{\frac{1}{3}\le x \le 5}$

Part (b) you can figure out yourself based on these two options.

[quote="sidk47"]1. Solve:
(a) |2x-3|<=|x+2|
(b) |3x-2|<=|x-1|

2. Solve:
(a) Prove: |a-b|>=||a|-|b||
(b) Prove: |a-b|<==|a|+|b|

Detailed solutions would be most appreciated.[/quote]
$\text{(A)}$

[hide]$|a-b|\geq \parallel a|-|b\parallel$; Square both sides.
$a^{2}-2ab+b^{2}\geq a^{2}-2|a| \cdot |b|+b^{2}$
$-2ab\geq-2|a| \cdot |b|$
$ab\leq |a| \cdot |b|$
$\frac{a \cdot b}{|a| \cdot |b|}\leq 1$
Which is most certainly true.  You could further square both sides if you wanted and get $1\leq 1$
$\boxed{True}$[/hide]

$\text{(B)}$
[hide]$|a-b|\leq |a|+|b|$
$a^{2}-2ab+b^{2}\leq a^{2}+2|a| \cdot |b|+b^{2}$
$-2ab\leq 2|a| \cdot |b|$
$ab\geq-|a| \cdot |b|$
$\frac{a \cdot b}{|a| \cdot |b|}\geq-1$
Which is most certainly true.  You could square both sides if you wanted and get $1\geq 1$
$\boxed{True}$[/hide]

Thank u so much 4 ur help, ur kindness, n ur patience. Thank u people.


Prove that the probability of exactly $ n$ heads in $ 2n$ tosses of a fair coin is given by the product of the odd numbers up to $ 2n \minus{} 1$ divided by the product of the even numbers up to $ 2n$.

I might be overestimating the difficulty of this problem, so please move it if necessary.

Nice solution!

And it seemed so difficult... :ninja:


i think that in order to have a solution to this problem you must add that f is even... :maybe:  in this case it is not hard to prove that f is a constant function  :wink:

Can you post a full proof, given that it is even? Note f(x)=0 is also a solution.

Only constant functions satisfy our condition.
Let $ P(x,y)$ be our condition
Let $ a\equal{}f(0)$

$ P(x,0)\minus{}P(x,\minus{}x)\Rightarrow f(f(\minus{}x))\equal{}f(x)\plus{}a$ so $ f(f(\minus{}0))\equal{}f(0)\plus{}a\Rightarrow f(a)\equal{}2a$
$ P(x,\minus{}x)\Rightarrow \ a\equal{}f(f(x))\plus{}f(f(\minus{}x))\equal{}f(f(x))\plus{}f(x)\plus{}a$ so $ \minus{}f(x)\equal{}f(f(x))$
Putting that into $ P(x,0)$ gives us $ (f(x)\equal{}\minus{}f(x)\plus{}f(a))\Rightarrow f(x)\equal{}a$ Q.E.D. :wink:

[quote="polskimisiek"]$ P(x,0) \minus{} P(x, \minus{} x)\Rightarrow f(f( \minus{} x)) \equal{} f(x) \plus{} a$ so $ f(f( \minus{} 0)) \equal{} f(0) \plus{} a\Rightarrow f(a) \equal{} 2a$[/quote]
I fail to see why $ f(f(\minus{}x)) \equal{} f(x) \plus{} a$ should hold.

[quote="polskimisiek"]Putting that into $ P(x,0)$ gives us $ (f(x) \equal{} \minus{} f(x) \plus{} f(a))\Rightarrow f(x) \equal{} a$ Q.E.D. :wink:[/quote]
Firstly, check your solutions. If $ f$ is constant, we should have $ a \equal{} f(x\plus{}y) \equal{} f(f(x))\plus{}f(f(y)) \equal{} f(a)\plus{}f(a) \equal{} 2a$, so $ f$ is identically zero.
Secondly, $ f(x) \equal{} x$ is a solution as well.

[quote="Kurt Gödel"]
Secondly, $ f(x) \equal{} x$ is a solution as well.[/quote]
Didn't we make an assumption that $ f$ is an even function,but $ f(x)\equal{}x$ is obviously not.

Well, iakovos1729 threw in the condition to have it even so it is a constant function. Without that condition, Kurt Gödel has a solution ><.

[quote="jedaihan"]Find all functions f:R→R satisfying
f(x+y)=f(f(x))+f(f(y))
for all x,y∈R[/quote]
Is it continous?I think set
$ f(f(x))\equal{}a,f(f(y))\equal{}b \minus{}>f(a\plus{}b)\equal{}f(a)\plus{}f(b)$
Any one can follow? :wink:

To follow it I think:
let $ y\equal{}0 \minus{}> f(x)\equal{}f(f(x))\plus{}f(f(0))$
easy to see that $ x\equal{}y <\minus{}> f(x)\equal{}f(y)$
so:
$ f(f(x))\plus{}f(f(y))\equal{}f(f(x\minus{}1))\plus{}f(f(y\plus{}1))$ for all x,y
$ \minus{}> G(x)\equal{}f(f(x))\minus{}f(f(x\minus{}1))\equal{}const$
 :wink:


Let be $ a,b,c\in (0,\infty)$ such that $ 3abc\equal{}1$ . Show that :

$ \frac{3a^5}{3a^5\plus{}2bc}\plus{}\frac{3b^5}{3b^5\plus{}2ca}\plus{}\frac{3c^5}{3c^5\plus{}2ab}\ge 1$

nice  :wink:
$ \sum \frac{3a^5}{3a^5+2bc}=\sum \frac{9a^4}{9a^4+2}\geq \frac{(a^2+b^2+c^2)}{3%Error. "suma" is a bad command.
^4+2}$
then it's suffice to prove that,$ \frac{3(a^2+b^2+c^2)}{3%Error. "suma" is a bad command.
^4+2}\geq1$
$ \Leftrightarrow \sum a^2b^2 \geq \frac{1}{3}$ wich is AM-GM :wink:

[quote="peine"]nice  :wink:
$ \sum \frac {3a^5}{3a^5 \plus{} 2bc} \equal{} \sum \frac {9a^4}{9a^4 \plus{} 2}$
[/quote]
you're wrong, because.
$ \sum \frac {3a^5}{3a^5 \plus{} 2bc} \equal{} \sum \frac {9a^6}{9a^6 \plus{} 2}$.

[quote="alex2008"]Let be $ a,b,c\in (0,\infty)$ such that $ 3abc \equal{} 1$ . Show that :

$ \frac {3a^5}{3a^5 \plus{} 2bc} \plus{} \frac {3b^5}{3b^5 \plus{} 2ca} \plus{} \frac {3c^5}{3c^5 \plus{} 2ab}\ge 1$[/quote]
$ \frac {3a^5}{3a^5 \plus{} 2bc}\equal{}\frac {a^4}{a^4 \plus{} 2b^2c^2}>\equal{}\frac {a^4}{a^4\plus{}b^4\plus{}c^4}$
similarly to $ b$ and $ c$.

[quote="aev5peru"]
$ \sum \frac {3a^5}{3a^5 \plus{} 2bc} \equal{} \sum \frac {9a^6}{9a^6 \plus{} 2}$.[/quote]

Well , to continue this proof , from Titu Andreescu we have that :

$ \sum_{cyc}\frac{9a^6}{9a^6\plus{}6abc}\ge \frac{(3a^3\plus{}\plus{}3b^3\plus{}3c^3)^2}{9a^6\plus{}9b^6\plus{}9c^6\plus{}6}\ge 1$

The last part of the inequality is equivalent with :

$ 18\sum_{cyc}a^3b^3\ge 6$ , which is $ MA\minus{}MG$.

So my proof is complete .


$\mbox{let L be the linear operation on } R^3 \mbox{ defined by}$

$L(x) = \left(\begin{array}{cc}x_2-x_1\\x_3-x_1\\x_3-x_1\end{array}\right)$

$\mbox{Determin the kernel of L.}$

Kernel = $t{(1,1,1)}$
where $t$ is a scalar..

$\mbox{Thank up so much, I was having trouble figuring out if the Ker}(L) = {0}$

$\mbox{or if the Ker(L) = Span}(e_1, e_2, e_3)$

The matrix of this linear tranformation is $\begin{\bmatrix}-1&1&0\\-1&0&1\\-1&0&1\end{bmatrix}.$

This is row equivalent to $\begin{\bmatrix}-1&1&0\\0&-1&1\\0&0&0\end{bmatrix}.$

The matrix has rank 2, hence nullity 1, so its kernel is one-dimensional. What Igor Castro said in comment #2 is correct; I don't know what he was trying to say in comment #4.

[quote]...I don't know what he was trying to say in comment #4.[/quote]
I just was trying to say that Kernel is the sub-space that when we apply L we get $0$(neutral element). But i used the davidbrw notation(i assume) for this when he said "$Ker(L)=0$"
I think is more correct to say $L(Ker)=0$...


calculus
integration
trigonometry
calculus computations

(This problem may be to "easy" for this forum, but I can't find where else it fits, so here it goes...)

In my Calc BC class, we have discovered something that no one has been able to explain yet.  The problem in question is the integral of (cos^3 x sin x) dx  (cosine cubed x times sine x).  Using u-substitution, with u = cos x, the integral becomes  - u^3 du which equals (-cos^4 x)/4 + C.  However, knowing (cos^2 x) = (1 - sin^2 x), if you write the integral as (1-sin^2 x)(sin x)(cos x) dx, then you use u-substitution with u = sin x, the integral becomes (u - u^3) du which equals (sin^2 x)/2 - (sin^4 x)/4 + C.  As far as we can tell, there is no way to configure (sin^2 x)/2 - (sin^4 x)/4 + C to equal (-cos^4 x)/4 + C, so it seems as if the two methods give different answers.  Is there an explanation for this?

[quote="Ravi B"]The two results are actually the same, provided you choose an appropriate constant $C$ for the two.  (The constant will be different for the two expressions.)  The equality of the expressions follows from the identity
\[\frac{\sin^{2}x}{2}-\frac{\sin^{4}x}{4}=-\frac{\cos^{4}x}{4}+\frac{1}{4}\, . \]
[/quote]

Thanks

[Edit - I figured out my other question]


algebra
polynomial
number theory proposed
number theory

Does it exist integer numbers [tex] a,b,c (ab \neq 0) [/tex] satisfying [tex] \forall n [/tex] integer,we can find at least one number $ x $ so that [tex] ax^2+bx+c=n! [/tex] :)

Wait....I must be misreading this....

Couldn't you just say that x=1, then find three integers that add up to n!?

Not really :): you have to find a [b]fixed[/b] triple $(a,b,c)$ which satisfies this for any $n$. After you've found such a triple you're only allowed to move $x$.

Hi, some questions:
i)Find all triple (a.b.c) in $Z^3$ such that:there exits $n_0$ satisfied for any n>=$n_0$ then exits an integer x satisfied:
                $ax^2+bx+c=n!$
in part i) I only prove c=0 if ab<>0.
ii)If we change $ax^2+bx+c$ to a polynomial in Z[x] whose degree >2 then what is the conclusion of the problem? :maybe:

I think I've reduced it to the question: do there exist $t,a$ s.t. $t^2+4an!$ is a perfect square for all $n$?

there are no such (a,b,c): just use the condition for n=1,2,3,4 (you don't need more) and use the fast $x-y|P(x)-P(y)$ for any polynomial P with integer coefficients. let $P(x)=ax^2+bx+c$. wlog we have $P(0)=1$ (else translate x) and $P(1)=2$ (actually, $P(x)=2$ is only possible for -1 and 1 by the fact, thus by sending -x to x we can assume $P(1)=2$). then we have some x such that $P(x)=6$. it follows that $x-1|4$ and $x-0|5$, hence $x=5$ or $x=-1$. thus, we have in the first case $c=1$, $a+b+c=2$ and $25a+5b+c=6$. these equations lead to $a+b=1$, thus $20a=0$, hence $a=0$, contradiction.
in the second case $c=1$, $a+b+c=2$ and $a-b+c=6$, thus $b=-2$ and $a=3$. now take the condition for n=4: the discriminant $4+4\cdot 23\cdot 3=280$ must be a square, contradiction.

Peter

Very nice and good solution.But what about [tex] x \in {\mathbb Q} [/tex]. :)


$\frac{n(n+1)}{2}$ coins are placed in a number of piles, with an abitrary number of coins in each pile. Each step, we remove one coin from each pile, and put the removed coins into a new pile.

Prove or disprove:
Eventually we will have $n$ piles all with a different number of coins.

i think the answer is yes, it's closely related to "coins on chessboard", another problem you posted. More generally, for any positive number N, if N = n(n+1)/2 + r where 0 <= r < n+1, any initial arrangement after finite actions will stay in a recursion among some arrangements containing n or n+1 piles with i-th pile containing i or i+1 coins if piles are sorted in increasing order.

This is a game known as Bulgarian Solitaire.  You can find more info in the post below, or through a Google search.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30537[/url]


pigeonhole principle
MIT
college
calculus
integration
algebra
function

I hate to ask such a simple question, but Im stuck.

Let $ R$ be a commutative ring with $ 1 \neq 0$.  If $ R$ has only finitely many maximal ideals and exactly one unit, then show that $ R$ is a finite ring.

$ R^\ast\equal{}{1}\implies (R/\mathfrak{m})^\ast\equal{}{1}$ is false in general. ($ \mathbb{F}_2[X]$)

[quote="Gockel"]$ R^\ast \equal{} {1}\implies (R/\mathfrak{m})^\ast \equal{} {1}$ is false in general. ($ \mathbb{F}_2[X]$)[/quote]
But here for $ R \cong \prod_{\mathfrak{m} \in Spm(R)} R/\mathfrak{m}$ we have $ R^\times \cong \prod_{\mathfrak{m} \in Spm(R)} (R/\mathfrak{m})^\times$.

[quote="-oo-"][quote="dallasboy"]Let $ R$ be a commutative ring with $ 1 \neq 0$.
If $ R$ has only finitely many maximal ideals and exactly one unit,
then show that $ R$ is a finite ring.[/quote]
we have $ Jac(R) \equal{} \cap_{\mathfrak{m} \in Spm(R)} \mathfrak{m} \equal{} \{x \in R : \forall r \in R : 1 \plus{} rx \in R^*\} \equal{} \{0\}$.
the chinese remainder theorem then yields $ R \cong \prod_{\mathfrak{m} \in Spm(R)} R/\mathfrak{m}$.
now $ R^* \equal{} \{1\}$ implies $ (R/\mathfrak{m})^* \equal{} \{1\}$ i.e. $ R/\mathfrak{m} \cong \mathbb{F}_2$ since it is a field.
hence $ R \equal{} \mathbb{F}_2^n$ and $ R$ is finite. conversely, every such ring satisfies the properties.[/quote]

$ \rm REMARK\quad$ Above is a special case of this ring-theoretic generalization of Euclid:
[b]infinite fewunit rings have infinitely many prime ideals[/b]. I quote from my post [1]:

Below, generalizing Euclid's classic argument, is a simple proof that an infinite ring
has infinitely many maximal (so prime) ideals if it has fewer units than elements
(i.e. smaller cardinality).  The key idea is that Euclid's construction of a new prime
generalizes from elements to ideals, i.e. given some maximal ideals $ \rm P_1,\ldots,P_k$
then a simple pigeonhole argument employing $ \rm CRT$ implies that $ \rm 1 \plus{} P_1\cdots P_k$
contains a nonunit, which lies in some maximal ideal $ \rm P$ which, by construction,
is comaximal (so distinct) from the prior max ideals $ \rm P_i$.

$ \rm THEOREM$ An infinite ring $ \rm R$ has infinitely many maximal ideals
if it has fewer units $ \rm U \equal{} U(R)$  than it has elements, i.e. $ \rm|U| < |R|$.

$ \rm PROOF\ \; R$ has a max ideal $ \rm P_1$, since nonunit $ \rm 0$ lies in some max ideal.
Inductively, suppose $ \rm P_1,\ldots,P_k$  are maximal ideals in $ \rm R$, with product $ \rm J.$

$ \rm Case\ 1: \; 1 \plus{} J \not\subset U$. So $ \rm 1 \plus{} J$ contains a nonunit $ \rm p,$ lying in some max
ideal $ \rm P.$ It's new: $ \rm P \neq P_i$  since  $ \rm P \plus{} P_i \equal{} 1$  via  $ \rm p \in P,\ 1 \minus{} p \in J \subset P_i$

$ \rm Case\ 2: \; 1 \plus{} J \subset U$ is impossible by the following pigeonhole argument.
$ \rm R/J \equal{} R_1 \times \cdots \times R_k,\ R_i \equal{} R/P_i$ by the Chinese Remainder Theorem.
We deduce that $ \rm|U(R/J)| \leq |U|$  because  $ \rm uv \in 1 \plus{} J \subset U \Rightarrow u \in U.$
Thus $ \rm|U(R_i)| \leq |U(R/J)| \leq |U|$  via injection $ \rm u \mapsto (1,1,\ldots,u,\ldots,1,1).$
$ \rm R_i$ field $ \rm\ \Rightarrow\ |R| > 1 \plus{} |U| \geq |R_i|,$ and also  $ \rm|J| \leq |U| < |R|$  via  $ \rm 1 \plus{} J \subset U.$
Therefore $ \rm|R| \equal{} |R/J|\ |J| \equal{} |R_1|\ \cdots |R_k|\ |J|$  yields the contradiction that
the infinite $ \rm|R|$ is a finite product of smaller cardinals. $ \rm QED$

The only piece of the proof that might be unfamiliar to a first-year
algebra student is the relationship between the units in $ \rm R$ and $ \rm R/J.$
We state this in more familiar terms using the Jacobson radical $ \rm J(R)$

$ \rm THEOREM\ \; TFAE$ in a ring $ \rm R$ with units $ \rm U$, ideals $ \rm I, J, J(R).$

$ \rm 1)\quad J \subset J(R)$ .................... i.e.  $ \rm J$ lies in every  max  ideal  of $ \rm R$

$ \rm 2)\quad 1 \plus{} J \subset U\,$  .................. i.e.  $ \rm 1 \plus{} j$ is a unit for every  $ \rm j \in J$

$ \rm 3)\quad I < 1 \Rightarrow I \plus{} J < 1\,$ .... i.e. proper ideals survive in $ \rm R/J$

This implies that all the nonunits in $ \rm R$ survive as nonunits in $ \rm R/J,$
so the only units in $ \rm R/J$ are those that are images of units in $ \rm R.$
The proof of this theorem is left as a short and elementary exercise.

I still recall the pleasure of discovering this and other related theorems
almost 3 decades ago while reading Kaplansky's book "Commutative Rings"
as an MIT undergrad. There Kaplansky presents a simpler integral domain
version as exercise 8 in Section 1-1, namely

....... (This exercise is offered as a modernization of Euclid's theorem on
....... the infinitude of primes.) Prove that an infinite integral domain with
....... with a finite number of units has an infinite number of maximal ideals.

I highly recommend Kap's classic textbook to everyone interested
in mastering commutative ring theory.  In fact I highly recommend
everything by Kaplansky - it is almost always very insightful and
elegant. Learn from the masters!  For more about Kaplansky see
the interesting article [0] which includes quotes from many eminent
mathematicians (Bass, Eisenbud, Kadison, Lam, Rotman, Swan, etc).

................................. I liked the algebraic way of looking at things.
................................. I'm additionally fascinated when the algebraic
................................................ method is applied to infinite objects.
............................................................................ --Irving Kaplansky

--Bill Dubuque

[0] Irving Kaplansky 1917-2006. Hyman Bass and T.Y. Lam.
Notices of the AMS, v.54, 11, 2007, 1477-1493.
http://www.ams.org/notices/200711/tx071101477p.pdf

[1] Bill Dubuque, sci.math, July 30 2008,
Infinite fewunit rings have infinitely many prime ideals (Euclid bis)
http://groups.google.com/groups?selm=y8zk5f3rn4e.fsf%40nestle.csail.mit.edu
http://groups.google.com/group/sci.math/msg/3d06e78d2ae5fa47


function
inequalities solved
inequalities

If $x,y,z \in [0,1]$ then $ \sum \frac{x}{yz+1} \leq 2$.

This prb was proposed in Kvant 20 year ago. In our forum it was solved. I remember that the main idea is to assume that x<=y<=z and then replace yz+1, xz+1 by xy+1.

Namdung

Another approach : The Lhs is a convex function in each of the variables, so it reaches its maximum at the endpoints of [0;1], that is when x,y,z are in {0;1}.
It is easy to check that, in these cases, the maximum is 2.

Pierre.

WLOG assume 0<x<=y<=z<=1
    we have xy+1<=yz+1
            xz+1<=yz+1
            xy+1<=xz+1
 hence : x/(yx+1)<=x/(xy+1) , y/(xz+1)<=y/(xy+1)
 so    : sum(x/(yz+1))<=(x+y+z)/(xy+1)
 and (1-x)(1-y)>=0 we have x+y<=2xy+1 and x+y+z<=2xy+2
 From that we have conclusion


[quote="DonkeyKong"][quote="alanchou"][quote="Fraenkel"]Two dogs run around a circular track 300 ft. long. One dog runs at a steady rate of 15 feet per second, and the other at a steady rate of 12 feet per second. Suppose they start at the same point and time. What is the least number of seconds that will elapse before they meet again?[/quote]

Are they running in the same direction?

@DonkeyKong - How did you get 12 secs?[/quote]

I was assuming that they are running opposite direction.. if that helps..
Actually I really don't have a clue how I got my answer.. And I'm probably wrong anyways..(I don't know why, but I think I did 300/27(15+12) to get my answer)
I'll figure it out though.[/quote]

I believe that $ \frac{300}{27}\equal{}11.\overline{1}$.

Which is very close to 12.

But if they run in the same direction than we get
[hide]
100
[/hide]

(by essentially the same idea)

[hide="same direction"]One gains 3 ft. each second, so they meet in 100 seconds.[/hide]
[hide="different"]They get 27 feet farther from each other every second, so  $ 11\frac{1}{9}$.[/hide]
So yeah, same.

[quote="alanchou"][/quote][quote="DonkeyKong"][quote="alanchou"][quote="Fraenkel"]Two dogs run around a circular track 300 ft. long. One dog runs at a steady rate of 15 feet per second, and the other at a steady rate of 12 feet per second. Suppose they start at the same point and time. What is the least number of seconds that will elapse before they meet again?[/quote]

Are they running in the same direction?

@DonkeyKong - How did you get 12 secs?[/quote]

I was assuming that they are running opposite direction.. if that helps..
Actually I really don't have a clue how I got my answer.. And I'm probably wrong anyways..(I don't know why, but I think I did 300/27(15+12) to get my answer)
I'll figure it out though.[/quote][quote="alanchou"]

I believe that $ \frac {300}{27} = 11.\overline{1}$.

Which is very close to 12.

But if they run in the same direction than we get
[hide]
100
[/hide]

(by essentially the same idea)[/quote]

Yeah, I rounded it up.

going in the same direction they'll meet in 100 seconds?

going the opposite I'll need to think about it

 :weightlift:

 :ddr:

Well, I agree with the answers above me.
And BTW, why doesn't my signature show up in my posts?

Above the button that says "Submit", there is a box that can be checked that says "Attach signature (signatures can be changed in profile)"


LaTeX
number theory unsolved
number theory

$n\in \mathbb{N}$
$\exists a \in \mathbb{N}, 3a^2+3a+1=n^2$

Prove:
 $\exists c\in \mathbb{N}, n=2c^2+2c+1$

They are my first posts, so I don't know if it's right...

[hide]

$n^2=3a^2+3a+1 \rightarrow a(a+1)=(2a+1-n)(2a+1+n)$

$gcd(a;a+1)=1 \implies a|(2a+1-n)$ and $a+1|(2a+1+n)$ or

$a+1|(2a+1-n)$ and $a|(2a+1+n)$) $\rightarrow$ impossible in $\mathbb{N}$

So, $a|(2a+1-n) \Longleftrightarrow n=ka+1$, $k \in \mathbb{Z}$

and it must be $a+1|(2a+2+ka) \Longleftrightarrow k=(a+1)b$, $b \in \mathbb {Z}$

then, $n=ab(a+1)+1=a^2b+ab+1$. With $b=2$ we have the proof.
[/hide]

[quote="M@re"][quote="hydro"]They are my first posts, so I don't know if it's right...

$gcd(a;a+1)=1 \implies a|(2a+1-n)$ and $a+1|(2a+1+n)$ or

$a+1|(2a+1-n)$ and $a|(2a+1+n)$) $\rightarrow$ impossible in $\mathbb{N}$

[/quote]

This is not correct, because $a$ can be $pq$ and $a+1$ can be $rs$ where numbers $p,q,r,s$ are coprime pairwise, and $pr|(2a+1-n)$ and $sq|(2a+1+n)$

bye[/quote]
sorry,I'm not clear about what you wrote.however, I think it's right.
I haven't learned to use Latex,so sorry that I can't post my proof.

He wrote that from $a(a+1)=(2a+n+1)(2a-n+1)$ and $qcd(a,a+1)=1$ it comes $a|(2a-n+1)$ and $a+1|(2a+n+1)$ and that is not correct! Simply, if that is correct then it must stand that $a\leq 2a-n+1$ and $a+1\leq 2a+n+1$ but $a(a+1)=(2a+n+1)(2a-n+1)$ so $a=2a-n+1$ and $a+1=2a+n+1$, in other words, product of some coprime numbers can be presented like product of two numbrs in only one way and that is faaaaar  from true.

you are right m@re my solution was wrong, excuse me!


Find all real all x-intercepts to the following functions.
Attempt to use period of the [i]oscillating[/i] function to your advantage.
[hide="#1"]This one is a warm up: $ f(x) \equal{} \displaystyle \cos \left(\frac {x}{2}\right)$ on the interval $ [0,2\pi]$
[b]Do not use any trig identities[/b].[/hide][hide="#2"]$ f(x) \equal{} \minus{} \frac {4}{x^2} \plus{} \frac {\pi}{8\cos^2(\frac {\pi}{8}x)}$ on the interval $ ( \minus{} \infty, \infty)$
[b]Solutions without any trig identities are preferred.[/b][hide="Hint"]Number of solutions : 2[/hide][/hide]
[b]Edit: [/b]Mistake in the hint.

[hide="#1"]$ \cos\left(\frac{x}2\right)\equal{}0$

$ \frac{x}2\equal{}\pm\frac{m\pi}2$

$ x\equal{}\pm m\pi$, where $ m$ is any odd integer.

In interval $ [0,2\pi]$, $ x\equal{}\pm\pi$.[/hide]

[hide="#2"]$ \minus{}\frac4{x^2}\plus{}\frac{\pi}{8\cos^2(\frac{\pi}8x)}\equal{}0$

$ \left(\sqrt{\frac{\pi}2}\frac1{2\cos(\frac{\pi}8x)}\plus{}\frac2{x}\right)\left(\sqrt{\frac{\pi}2}\frac1{2\cos(\frac{\pi}8x)}\minus{}\frac2{x}\right)\equal{}0$

$ \cos\left(\frac{\pi}8x\right)\equal{}\pm\frac14\sqrt{\frac{\pi}2}x$

That's all I can do so far without a calculator. The rest is just graphing and maybe consider the convergence of $ x_n\equal{}4\sqrt{\frac2{\pi}}\cos\left(\frac{\pi}8x_{n\minus{}1}\right)$. Yes, two solutions by your hint.[/hide]


Let A be a 3x3 matrix. It is known for that mtatrix that A[b]v[/b] = [b]0[/b] and A[b]w[/b] = [b]0[/b] for [b]v[/b] = [1,0,2]^T and [b]w[/b] = [2,2,1]^T. Calculate from this A[b]u[/b], where [b]u[/b] = [5,2,7]^T or say why it is not possible.

I figured out that it is possible, but I had to solve for a "A" to get this. Is there another reasonable way to answer the question?

Much thanks!

EDIT: I think I got the answer, but can someone confirm for me?

3[b]v[/b]+[b]w[/b] = [b]u[/b]

Therefore, A[b]u[/b]=[b]0[/b]

Yes, that's it. You can find $Ax$ if and only if $x$ is a linear combination of $v$ and $w$.


modular arithmetic
number theory unsolved
number theory

Prove or disprove the following statements.
a) For every integer $ n \ge 3$, there exist $ n$ pairwise distinct positive integers such that the product of any two of them is divisible by the sum of the remaining $ n \minus{} 2$ numbers.
b) For some integer $ n \ge 3$, there exist $ n$ pairwise distinct positive integers, such that the sum of any $ n \minus{} 2$ of them is divisible by the product of the remaining two numbers.

a) n=3 we can take 2,3,6. So We can let $ n \ge 4$

Let P(i,j), i<j, be the smallest number s.t. $ n(n \plus{} 1)/2 \minus{} (i \plus{} j) | P(i,j)ij$. Then let Q = The product of all P(i,j) with i < j. Then Q, 2Q, ..., nQ works, since for any i,j we have$ Q(n(n \plus{} 1)/2 \minus{} (i \plus{} j))|Q^2ij$ which can be rephrased as $ n(n \plus{} 1)/2 \minus{} (i \plus{} j)|Qij$ which is true since $ n(n \plus{} 1)/2 \minus{} (i \plus{} j)| P(i,j)ij | Qij$

b)false.
let the numbers $ a_1,...,a_n$ have the property. suppose without loss of generality that they are ordered in increasing order
$ a_n|a_2\plus{}a_3\plus{}\cdots \plus{}a_{n\minus{}1}$ and $ a_n|a_1\plus{}a_3\plus{}\cdots \plus{} a_{n\minus{}1}$ so $ a_1\equiv a_2 \pmod{a_n}$ which is impossible as $ a_1\not\equal{} a_2$


As always, thanks for the great resources on this site. I am preparing for a parent meeting at which some new math teams will be organized in my town, and I was going to print out some AoPS articles, such as

[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]What is Problem Solving?[/url]

and

[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Contests.php]Pros and Cons of Math Competitions[/url].

But I realize now that those articles are formatted strictly for online reading. I have recommended them several times by email, and can certainly include their URLs on a page I pass out to parents on paper at the meeting, but it is regrettable that those articles aren't available in a printer-friendly format. That's my friendly suggestion: make it easier to spread the word about AoPS through old-fashioned media such as paper printouts as well as modern media such as emailed hyperlinks.

Thanks.

Good call - I'll put it on the todo list (which is terrifyingly long right now, but it'll happen).

At long last, I finally did this for a few of the longer articles.


vector
linear algebra
superior algebra
superior algebra solved

I found this question in one of the previous years exams in linear algebra, and I'm almost sure I missed something and my solution is somewhat wrong, as it seems too easy to be an exam question.

"Let U and W be subspaces of a vector space V, and V is the direct sum of U and W. let Bu={u1,...,un} be the basis for U, and Bw={w1,...,wk} be the basis for W. Prove that Bu :cup: Bw is a basis for V."

my solution:

V is the direct sum of U,W, so by definition for every v in V there exists vectors u in U and w in W such as v = u + w, and this decomposition is unique. Since Bu and Bw are bases for U and W, these u, w vectors can be written uniquely as a linear combination of vectors in U and W, respectivly. Adding these two up, every v in V can be written uniquely as a linear combination of {Vi,Wi}, and by definition its a base for V.

So, is there anything wrong with this?

That's perfectly right. Exam questions can be easy, especially if you understand the material.


Βρείτε όλους τους θετικούς ακεραίους $ a,b,c$ ώστε $ ab\plus{}bc\plus{}ca>abc$.

Η ανισότητα αυτή μπορεί να γραφεί:

\[ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}>1\]

Συνεπώς κάθε τριάδα που περιέχει έναν άσο ή δύο δυάρια ικανοποιεί την ανισότητα αυτή. Οι μόνες τριάδες  οι οποίες ικανοποιούν την ανισότητα χωρίς να έχουν μια από τις δυο προηγούμενες ιδιότητες είναι η $ (2,3,3)$, η $ (2,3,4)$ και οι μεταθέσεις τους.


geometry
inequalities
algebra
polynomial
number theory
functional equation
IMO

[quote="Davron"]Who can guess? In which order this years IMO problems will be?

for example IMO 2005:
q1-Geometry
q2-Number Theory
q3-Algebra (Inequality)
q4-Number Theory
q5-Geometry
q6-Combinatorics

for example IMO 2004:
q1-Geometry
q2-Algebra (Polynomials)
q3-Combinatorics
q4-Algebra (Inequalities)
q5-Geometry
q6-Number Theory

davron[/quote]
Hi Davron :wink: .
I can't say which order we will have this year,but I wonder,what do you think,what is the possibility that we will have two combinatorial problems this year :maybe: ?.

I hope there will be an inequality :jump:

I'll try..

1-Combinatorics
2-Number Theory
3-Geometry
4-Geometry
5-Algebra (functional equation)
6-Combinatorics

OK,let me try...
1.Algebra
2.Geometry
3.Combinatorics
4.Number Theory
5.Algebra(ineq.)
6.Geometry :D

I'll risk a nice ineq as problem 6, and 2 problems of geometry

Here's my wish list:

1. Number theory
2. Geometry
3. Combinatorics
4. Geometry
5. Algebra (functional equation!)
6. Number theory

I  thinks tiks's prediction is  correct,  but i  would definitely change alebra on number 1  for  number theory its really hard to
find a suitable algebra 1 problem for the imo (theyre normally very hard) and i think  were gonna have a number theory problem 1 because theyre difficulty is easy to measure and we dont want another hard imo 1 problem like last year. plus were probably gonna have a number theory six because in the last years the imo number theory  has  been   easy.  :lol:

[quote="vedran6"]here is my worst nightmare(everything would be better so i will say this :))
 1.Number Theory
 2.Combinatorics
 3.Geometry
 4.Combinatorics
 5.Geometry
 6.Algebra[/quote]
I think you like combinatorics :lol: .

[quote="Valentin Vornicu"]Algbera
Geometry
Combinatorics

Number Theory
Combinatorics
Geometry[/quote]
Is it reliable?

I think
1-Geometry
2-Combinatorics
3-Number Theory

4-Combinatorics or Number Theory
5-Geometry
6-Algebra

combinatorics
geometry
number theory

geometry
combinatorics
algebra

Valentin :yoda:  are you in Slovenia now :D

[quote="Albanian Eagle"]Valentin :yoda:  are you in Slovenia now :D[/quote]Still home :) I'm coming over tomorrow. But I won't be in Ljubljana  until at least Friday.

i think there will be two algebra problems but  not combinatorial problems (2) because they are hard then other problems for most of the teams and it will not be chosen by the leaders  ( it is only my guess )
hence we have remainded two geoetrzy one: number theory and combinatorics

but here is my guess ...
1. geometry
2. number t
3.combinations
4.algebra
5.geometry
6.number t

davron

I think it will be like this one

q1-geometry
q2-number theory
q3-combinatorics

q4-algebra or number theory
q5-geometry
q6-algebra

My bet:

1. Algebra
2. Number Theory
3. Geometry

4. Geometry
5. Algebra
6. Combinatoric.

for now it is going like this :

geometry
combinatorics
algebra (inequalities)

my predict for tomorrow is :

number theory
geometry
combinatorics

Here is my guess for tomorrow
4.Combinatorics
5.Number Theory
6.Geometry
We will see.
Good luck!!

Here is my guess:
4) Combinatorics
5) Geometry
6) Number Theory

We haven't had a killer Number Theory in the past two years. Last time was in 2003, and I died under it. I'm quite confident qn 6 is going to be Number Theory, and quite a number of other people at the IMO also share my opinion as well. History somehow repeats itself in weird ways.

Also, since Qn 3 was a hard algebra (inequality) question, chances are there will not be another algebra qn for qn 5 or 6.

We shall see who's right in over 10 hours time. :)

see dblues i was closer ...  :)

here were the problem of the second day :

number theory
algebra
combinatorial geometry

Haha.. My guess was completely wrong.  :P Anyway, this year's difficulty distribution of the questions is similar to IMO 2003. Qn 3 was 'crazy' for both 2003 and this year (2003 was crazier though), Qn 6 was very hard for both years, and Qn 5 is quite hard also, for both years. Qn 1 and 4 were very easy. Qn 2 was reasonable for a 2nd problem.

Overall, IMO 2004 and 2005 questions were easier than questions of IMO 2003 and 2006.

History repeats in strange ways. I think the cut-offs for the medals this year will be very similar to that of 2003. Recall that in 2003, the cut-offs for gold, silver and bronze were 29, 19 and 13 respectively. We shall wait and see.  :)


Let a, b and c be real numbers with |a*x^2+b*x+c|<=1;
Prove that |c*x^2+b*x+a|<=2.

Please do not answer to this topic . This problem is from a running contest .

[hide="@Ionut"]
ce faci ? hai sa incercam sa fim corecti si sa facem problemele pe bune[/hide] .


Show that for every nonzero number $ a$ and integer $ n>1$, $ x(x\minus{}a)$ is a factor of $ ax^n\minus{}a^nx$.

[hide]
$ ax^n\minus{}a^nx \equal{} ax(x^{n\minus{}1} \plus{} a^{n\minus{}1}) \equal{} ax(x\minus{}a).(x^{n\minus{}2} \plus{} x^{n\minus{}3}a...\plus{}a^{n\minus{}2})$

$ \Rightarrow ax^n\minus{}a^nx \equiv 0 \mod ax(x\minus{}a)$
[/hide]


(a) Apply DeMoivre's theorem to drive
\[ \cos5\theta=16cos^5\theta-20\cos^3\theta+5\cos\theta  \]
from
\[ \cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta  \]

(b) Prove that the solution set of $16x^4-20x^2+5=0$ is $\left\{\cos\frac{\pi}{10},\cos\frac{3\pi}{10},\cos\frac{7\pi}{10},\cos\frac{9\pi}{10}\right\}$

(c) Express $\cos\frac{7\pi}{10}$ as a root.

[quote="ifai"]
(b) Prove that the solution set of $16x^4-20x^2+5=0$ is $\left\{\cos\frac{\pi}{10},\cos\frac{3\pi}{10},\cos\frac{7\pi}{10},\cos\frac{9\pi}{10}\right\}$

(c) Express $\cos\frac{7\pi}{10}$ as a root.[/quote]

Please refer to my solution given earlier on
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=276276&highlight=#276276

This will be of some help.

[quote="scorpius119"]solution to (a):

[hide](cos 5x)+i (sin 5x)=((cos x)+i (sin x))^5=(cos^5 x)+5i (cos^4 x)(sin x)-10(cos^3 x)(sin^2 x)
-10i (cos^2 x)(sin^3 x)+5(cos x)(sin^4 x)+i (sin^5 x)

Set real parts equal: cos 5x=(cos^5 x)-10(cos^3 x)(sin^2 x)+5(cos x)(sin^4 x)
=(cos^5 x)-10(cos^3 x)(1-cos^2 x)+5(cos x)(1-2(cos^2 x)+cos^4 x)
=16(cos^5 x)-20(cos^3 x)+5(cos x)[/hide][/quote]
it right


In general, how would you approach problems that go like, given a function $f$ and it fulfils certain properties. Find all functions that fit this criteria?

I'd say I'm relatively good at those sort of problems.

So here are some things to consider:
Especially in the case where one side has $f(a-b)$ or similar, consider the case $a = b$, and see what you can deduce from there.

Try setting $x = 0$, with arbitrary $y$, and try separately $y = 0$, with arbitrary $x$.  Then, try $x = y = 0$.

Try solving for $f(0)$, and/or $f(1)$, depending on the problem.
(Think about what is useful).  The above alone may sometiems be enough for this.

Once you've gotten used to those and more familiar with these types of problems, you may find that in certain circumstances, the methods below are helpful:

If you find that $f(2x) = 2f(x)$ and $f(3x) = 3f(x)$ (or similar), whether for a specific $x$ or for a fixed value (ie. $f(0)$ or $f(1)$), then ask yourself: can you use induction to generalize it?

Can you show that $f(-x) =-f(x)$?
Generally, the easiest way to do so is by showing $f(x)+f(-x) = 0$.

Hmm, that's all of the "general" advice I can think of right now.

Here are some problems that I've made up (I'm not sure of the difficulty, as I'm new to these kinds of problems too. I made the first problem with a function in mind, and the second i just came up with. Lets try to solve these together:

1)   Find $f(x)$ such that $f(a+b)-f(a)-f(b)=2ab+1$.

2)   Find $F(x)$ such that $F^{2}(x+1) = F^{2}(x)+x$.

[quote="Franz Joseph"]They usually come with $R \rightarrow R$ or $Q \rightarrow Q$ or $Z \rightarrow Z$ etc.[/quote]

Ok, well the first is $Z \rightarrow Z$, and I have no idea for the second  :P .

[quote="boostage"][hide="Number 1"] $f(x)=x^{2}-1$. The right hand side looked like part of the expansion of a quadratic, so I played around with simple quadratic functions until I found one that worked.[/hide]

For the second question, by $F^{2}(x)$, do you mean $[F(x)]^{2}$ or $F(F(x))$?[/quote]

For the second, i meant $[F(x)]^{2}$.

And for the first, there's a larger set of functions that work.

[quote="lotrgreengrapes7926"][quote="cincodemayo5590"]2)   Find $F(x)$ such that $F^{2}(x+1) = F^{2}(x)+x$.[/quote]
[hide] For $\mathbb{Z}\to\mathbb{Z}$, I got $F(x)=\pm\sqrt{k^{2}+\frac{x(x-1)}{2}}$, where $F(0)=k$.  :maybe: [/hide][/quote]

That's what i got also, lol. Is there a way to find $F (0)$ from what's there, or do you think it needs to be given ?

[quote="Altheman"]two things:

1) you have to find all solutions...

2) if you make up the problem randomly, it is unlikely that you can solve it

3) if the functions are not continuous, then you have many random solutions (i.e. a piecewise equation)[/quote]

Thats three things, alex  :D


geometry
circumcircle
perpendicular bisector

$\triangle{AOB}$ is a right triangle with $\angle{AOB}=90^{o}$. $C$ and $D$ are moving on $AO$ and $BO$ respectively such that $AC=BD$. Show that there is a fixed point $P$ through which the perpendicular bisector of $CD$ always passes.

Haha, you know, one of the contestants got a special prize that year for an outstanding solution. His name is Frits van Zyl. All other solutions given at that contest were analytical, I believe :)

Try to find the "beautiful" solution, it's really cool.

Did you went to that year's contest Arne?  :)

No, but I have been trainer of the South African IMO and PAMO (Pan African) teams this year, that's why I know the story. I went to IMO 2003 and 2004 myself.

What medal did you get at the IMO, Arne?  :)

And this problem. I never know how to solve the " fixed point " type of problems. They really confuses me...so can anyone please tell me some general ways of proving "line pass a fixed point" and "circle passes a fixed point" kinds of prolems?

For this problem, I constructed the perpendicular bisector of AB, since when C and D are very close to A and B, we will include AB as one of the lines. And then I found out that it seemed $\angle{APB}=90^{o}$. The geometer's sketchpad helped me to confirm it. Thus this problem becomes much easies.

Construct as I show in the figure: The perpendicular line meet the circumcircle of $\triangle{OAB}$ at $P$.
Then it is well-known that $PC=PD$. Also $\angle{PAC}=\angle{PBD}$ is trivial. Since we are gived $AC=BD$, so $\triangle{PAC} \cong \triangle{PBD}$, which gives $PA=PB$. Since $\angle{APB}=\angle{AOB}=90^{o}$, so $P$ is the intersection of the perpendicular bisector of $AB$ and the circumcircle of $\triangle{AOB}$, which is obviously "fixed".

Oh I made a little mistake in my proof: My three conditions of congruence only holds if $\angle{ACP}, \angle{BDP} > 90^{o}$.
Maybe this one is better: $P, C, D, O$ are cyclic because of $\angle{CPD}=\angle{COD}=90^{o}$. So $\angle{PCO}=\angle{PDO}$. Hence: $\angle{PCA}=\angle{PDB}$. This time it should be perfect.

Take $C,D$ on $AO$ and $BO$ such that $AC=BD$,now let the perpendicular bisector of $AB$ and $CD$ intersect at $P$.now take two points $X,Y$ at $AO$ and $BO$ such that $AX=BY$,from triangles $PAX$ and $PBY$ we get $AP=BP$,$AX=BY$ and ${\angle}PAX={\angle}PBY$,hence we get $XP=YP$,so $P$ is a point on perpendicular bisector of $XY$,hence we get the result.

is that necessary for AOB to be a right triangle?!!
i think the bisector of CD passes through a fixed point anyway...(which is the midpoint of the arc AOB in circumcircle of the triangle)

I believe even Cartesian coordinates can do it !!  :-D

There is a fixed center of rotation mapping $A\mapsto B$ and mapping line $AO$ to line $BO$. as this rotation also maps $C\mapsto D$, it lies on perpendicular bisector of $CD$. $\square$


geometry
rectangle
AMC
AIME
2006 AIME II

In convex hexagon $ABCDEF$, all six sides are congruent, $\angle A$ and $\angle D$ are right angles, and $\angle B$, $\angle C$, $\angle E$, and $\angle F$ are congruent. The area of the hexagonal region is $2116(\sqrt{2}+1)$. Find $AB$.

[hide]Call the side $x$.  There are two isocsceles right triangles with side $x$ and a rectangle sides $x$ and $x\sqrt2$. Total area: $x^2(\sqrt2+1)=2116(\sqrt2+1)$. $x=\boxed{046}$[/hide]

[quote="pkothari13"]but it says 2006??[/quote]
That's weird, lol  :P  Its supposed to be 2116

I know this is real old, but I found the area using trapezoids. This was easy, but then again most AIME 1s are  :blush:

[hide]first draw the hexagon
draw altitudes to form 2 isoceles right triangles and a big rectangle
say 1 side is x
the area would be:
(x^2)/2 + (x^2)/2 + x^(1/2)=2116(1+root2)
that simplifies to x^2(1+2^(1/2))=2116(1+root2)
simplify
x^2=2116
x=46
I think its 2116 because its a square number? [/hide]

Hey, just wanted to point out that the OP includes $\angle D$ in the list of the congruent angles whereas it doesn't appear there in the actual problem (as that would mean all the angles are right, which makes no sense). If someone could fix that, it would be great!

[color=#f00]Fixed; thanks for pointing that out!  ~dj[/color]

And just so I don't revive a 10 year thread just because of a typo:
[hide=Alternate Solution With Diagram]
Extend $DE$ and $AF$ so that they meet at $M$, and extend $AB$ and $DC$ so that they meet at $N$.
[asy]
draw((0,0) -- (0,2) -- ( sqrt(2),2+sqrt(2)) -- (2+sqrt(2),2+sqrt(2)) -- (2+sqrt(2), sqrt(2)) -- (2,0) -- cycle);
draw((0,2) -- (0, 2+sqrt(2)) -- (sqrt(2), 2+sqrt(2)), dashed);
draw((2,0) --( 2+sqrt(2),0) -- (2+sqrt(2), sqrt(2)), dashed); label("$A$",(0,0), SW); label("$B$",(0,2), W); label("$C$",(sqrt(2),2+sqrt(2)), N); label("$D$",(2+sqrt(2),2+sqrt(2)), NE); label("$E$",(2+sqrt(2),1.4), dir(0));  label("$F$",(2,0), S); label("$N$",(0,2+sqrt(2)),NW); label("$M$",(2+sqrt(2),0),SE); label("$x$",(0,1),dir(0)); label("$x$",(.707,2.707),SE); label("$\frac{x}{\sqrt{2}}$",(0,2.907),W);

 [/asy]
Now, because $\triangle BNC$ and $\triangle FME$ are isosceles right triangles, we know $FM=MF=BN=NC = \frac{x}{\sqrt{2}}$. Thus, the area of the hexagon is the same as $[AMDN] - [FME] - [BNC]$. But, using the formulas for area of a square and area of a right triangle, we see that this is $\left ( \frac{2x+x\sqrt{2}}{2} \right )^2 - \frac{x^2}{4} - \frac{x^2}{4} = x^2 ( 1+\sqrt{2})$. But, if this equals $2116 ( \sqrt{2} + 1)$, we need $x^2 = 2116 \implies x=46$. Thus, since $AB=x$, our answer is $\boxed{046}$. [/hide]


Can someone post inequalities that utilizes AM-GM or the Cauchy-Schwarz Inequalities?

THanks A lot


Hi, I'm having trouble solving these problems with variables on both sides

Example:

2z - 9 < 3z + 9

Can someone please post a solution? The math book doesnt explain it.

$2z-9<3z+9$
Firstly we substract $2z$ from both sides in order to get rid of $2z$ on the left.
$2z-9-2z<3z+9-2z$
$-9<z+9$
now we substract $9$ from both sides to left $z$ alone
$-18<z$ and we have the answer :)

alright, so...

4x + 5 [u]>[/u] 9x - 2

is the solution:

1.4 [u]>[/u] x

Yes.  You can solve inequalities just like you solve equations (do the same thing to both sides).  One note: Be careful, because if you multiply/divide both sides by a negative you must flip the sign.


[hide]$ a=(x+9)^{\frac{1}{3}}, b=(x-9)^{\frac{1}{3}}$
$ a-b=3$
$ a^{3}-b^{3}=18$
$ \implies a^{2}+ab+b^{2}=6$
$ 9+3ab=(a-b)^{2}+3ab=6$
$ \implies ab=-1$
$ a+\frac{1}{a}=3$
$ a^{2}-3a+1=0$
$ a=\frac{3 \pm \sqrt{5}}{2}$

So
$ a=\frac{3\pm\sqrt{5}}{2}, b=\frac{-3\pm\sqrt{5}}{2}$

So $ x=\left(\frac{3\pm\sqrt{5}}{2}\right)^{3}-9=\pm4\sqrt{5}$[/hide]

Did I mess something up? (I never checked the consistency of the solutions, but I assume the algebra works...)

[quote="kunny"]You are right.

$ \sqrt[3]{x+9}+\sqrt[3]{9-x}+(-3)=0$
$ \Longrightarrow (x+9)+(9-x)+(-27)=3\sqrt[3]{(x+9)(9-x)}\cdot (-3)$,
yielding $ (x+9)(9-x)=1\Longleftrightarrow x=\pm 4\sqrt{5}$.[/quote]

Nice, using identity $ a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$...
That IS more elegant...


calculus
integration
logarithms
calculus computations

Find
$I=\int_0^ \infty \frac{ e ^ { -5x } - e ^ { -6x }} x \ dx$

I used my TI 200 to aproximate some values and that is the constant it kept spitting at me!
I am workin on a real solution though!

for some reason I got $\ln{\frac{5}{6}}$ i wonder were i made the mistake. The way I solved it was through the followinf substitutions:
1.$x = \ln{y}$
2.${y = e^{z}}$
3.$5z = \ln{w}$
4.$w = e^{u}$
5.$u = \ln{f}$

this was for the forst part of the integral(the second part works ou the same)

and I got the following:

$\int_{0}^{\infty}({\ln{5x}-\ln{6x}})dx = \ln{\frac{5}{6}}$


function
floor function
algebra proposed
algebra

[quote="alexchang"]Let $ K = {\lfloor a - \frac {1}{2}\rfloor + \lfloor a + \frac {1}{2}\rfloor}$, where $ a\in \mathbb{R}$
Prove that  $ K = 0 (mod 2)$.[/quote]
$ x-1<[x]\le x$ so

\[ a-\frac32<\left[a-\frac12\right]\le a-\frac12\]
\[ a-\frac12<\left[a+\frac12\right]\le a+\frac12\]

So
\[ 2a-1<K\le 2a\] If $ a\not\in \mathbb{Z}$ de conclusion is nat valid.
If $ a=0$ then $ K=1\neq 0\mod 2$

[quote="alexchang"]Let $ K = {\lfloor a - \frac {1}{2}\rfloor + \lfloor a + \frac {1}{2}\rfloor}$, where $ a\in \mathbb{R^*}$
Prove that  $ K = 1 (mod 2)$.[/quote]
\[ \frac{k+1}{2}>a\ge\frac{k}{2}\]
If $ k=2p$ then $ p>a-\frac12\ge\frac{2p-1}{2}$ so $ [a-\frac12]=p-1$ but $ [a+\frac12]=p$ so we got $ k=2p-1$ imposible.

[quote="cckek"][quote="alexchang"]Let $ K = {\lfloor a - \frac {1}{2}\rfloor + \lfloor a + \frac {1}{2}\rfloor}$, where $ a\in \mathbb{R}$
Prove that  $ K = 0 (mod 2)$.[/quote]
$ x - 1 < [x]\le x$ so
\[ a - \frac32 < \left[a - \frac12\right]\le a - \frac12
\]

\[ a - \frac12 < \left[a + \frac12\right]\le a + \frac12
\]
So
\[ 2a - 1 < K\le 2a
\]
[/quote]

It is wrong. It should be $ 2a - 2 < K\leq 2a$. How can u prove it now?

[quote="alexchang"]Let $ K = {\lfloor a - \frac {1}{2}\rfloor + \lfloor a + \frac {1}{2}\rfloor}$, where $ a\in \mathbb{R^*}$
Prove that  $ K = 1 (mod 2)$.[/quote]
Let $ t=a-\frac{1}{2}$ then
$ [a+\frac{1}{2}]+[a-\frac{1}{2}]=[t]+[t+1]=2[t]+1$


If Pick's Theorem, $ I\plus{}\frac{B}{2}\minus{}1$, relies on counting the lattice points on the boundary and interior of a polygon, then is it useful when I have an accurate diagram on a coordinate grid?

Thanks in advance!

It is useful when you have an accurate diagram. It is also useful if you have the area and the coordinates. Then you can quickly calculate the points on the boundary and find the lattice points inside. There are probably many more applications of this.

Bump

So on contest math where there's no accurate diagram, Pick's Theorem can't be used?

Pick's Theorem isn't only useful for trying to find the area of a [i]particular [/i]polygon. Here are just a few examples of problems you can solve with Pick's Theorem:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175008]Show that a convex pentagon with lattice point vertices has area at least 5/2[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11019&start=17]Show that an equilateral triangle cannot have all three vertices on lattice points.[/url]


geometry
parallelogram
geometric transformation
trigonometry
vector
analytic geometry
complex numbers

Let ABCD paralelogram. Take a point  P inside satisfying:
angle PBA= 2(angle PDA)       and       angle PCD= 2(angle PAD).

Prove that AB=BP=PC=CD

We use a translation $T(\overrightarrow{BA})$, then $P \to P'$, hence $AP'PB$ and $DP'PC$ are both parallelograms. Therefore $\angle{AP'P}=\angle{ABP}=2\angle{ADP}$ and $\angle{DP'P}=\angle{DCP}=2\angle{DAP}$. Hence $P'$ is the circumcentre of triangle $APD$, therefore $BA=CD=PP'=P'A=P'D=PB=PC$.

Thank you, is very interesting your solution!!
But... any solucion using inversion???

I have also solved that on http://www.mathlinks.ro/Forum/viewtopic.php?t=59339 in quite the same way as Shobber here... I don't know of a solution using inversion; inversion actually seems to make the problem more difficult.

  darij

[quote="conejita"]Let $ABCD$ paralelogram. Take a point  $P$ inside satisfying :
$m(\widehat{PBA})=2\cdot m(\widehat{PDA}$ and $m(\widehat{PCD})=2\cdot m(\widehat{PAD})$. Prove that $AB=BP=PC=CD\ .$[/quote]
[color=darkblue][b]Trigonometrical proof.[/b] Denote : the points $R\in AB$, $S\in CD$ so that $P\in RS$ and $RS\perp AD$ ;

$m(\widehat{BAD})=\phi$ ; $m(\widehat{PAD})=x$ , $m(\widehat{PDA})=y$ . Thus, $m(\widehat{ABP})=2y$ , $m(\widehat{DCP})=2x\ .$ Therefore,

$\frac{RA}{RB}=\frac{PA}{PB}\cdot\frac{\sin x}{\sin (\phi+2y)}=\frac{\sin 2y}{\sin(\phi-x)}\cdot\frac{\sin x}{\sin (\phi+2y)}\ .$ ; $\frac{SD}{SC}=\frac{PD}{PC}\cdot \frac{\sin y}{\sin (\phi-2x)}=\frac{\sin 2x}{\sin (\phi+y)}\cdot\frac{\sin y}{\sin (\phi-2x)}\ .$

$\frac{RA}{RB}=\frac{SD}{SC}$ $\Longrightarrow$ $\frac{\sin 2y}{\sin (\phi-x)}\cdot\frac{\sin x}{\sin (\phi+2y)}=\frac{\sin 2x}{\sin (\phi+y)}\cdot\frac{\sin y}{\sin (\phi-2x)}$ $\Longrightarrow$

$\cos y\sin (\phi+y)\sin (\phi-2x)=\cos x\sin (\phi-x)\sin (\phi+2y)$ $\Longrightarrow$

$\cos y[\cos (2x+y)-\cos (2\phi+y-2x)]=\cos x[\cos (x+2y)-\cos (2\phi+2y-x)]$ $\Longrightarrow$

$\cos 2(x+y)+\cos 2x-\cos (2\phi+2y-2x)-\cos (2\phi-2x)=$

$\cos (2x+2y)+\cos 2y-\cos (2\phi+2y)-\cos (2\phi+2y-2x)$ $\Longrightarrow$

$\cos 2x-\cos (2\phi-2x)=$ $\cos 2y-\cos (2\phi+2y)$ $\Longrightarrow$ $\sin\phi\sin (\phi-2x)=$ $\sin (\phi+2y)\sin \phi$ $\Longrightarrow$

$(\phi-2x)+(\phi+2y)=180^{\circ}$ $\Longrightarrow$ $\phi+y=90^{\circ}+x$ $\Longrightarrow$ $\widehat{BAP}\equiv\widehat{BPA}$ and $\widehat{CPD}\equiv\widehat{CDP}$ $\Longrightarrow$ $BA=BP$ and $CP=CD\ .$

[b]Remark.[/b] See http://www.mathlinks.ro/Forum/viewtopic.php?t=107342 [/color]

yes yes, inversion really make it much harder.

some kinds of problems have nice ideas, which we can find them by

inversion.

but here there is not any hidden idea.every thing is clear.

i f we use inversion, we should just do alot of angle chasing :!:

we should use inversion for hard and powerful problems, not easy ones.

solving easy problems by inversion is like killing a sparrow with dynamite!

because u can just use stone to kill it. :D

[quote="Ashegh"]we should use inversion for hard and powerful problems, not easy ones.

solving easy problems by inversion is like killing a sparrow with dynamite! [/quote]

Yes, we have to use inversion for hard and powerful problems like "donkey=hemar" that say in triangle we have $a+b<c\ ,a+c<b\ ,b+c<a$

:rotfl:  that is really exception.

donkey theorem, will be solve very nice by inversion,isnt it?

[color=darkblue][b]You are right presently, Amir. S ! Sorry, Ashegh ![/b] If you have more proofs for the same problem, then you must choose the best (simply, shortly, quickly or nicely). In the contrary case, when you are only a solution, then is trivially if it is ugly, intricate or more long. I am wont to say " ... is like killing a butterfly with atom bomb " or " ... is like killing a fly with truck ". [u]Any method is good if it solves correctly the given problem.[/u]

[size=134] [b]The geometry (generally, all mathematics) is a monstrous tautology.[/b][/size]

[u]Example.[/u] A point $P$ from (lifeless) plane $\Longleftrightarrow$ A point $P(x,y)$ from a (animated) plane $\Longleftrightarrow$ a point $P(z)$ from a complex plane,
where $z=x+i\cdot y$ - [b]afix[/b] (complex number) $\Longleftrightarrow$ a position vector $P\equiv\overline{OP}$ from the vectorial plane a.s.o. (in the future).
It means " Synthetical geometry $\Longleftrightarrow$ Analytical geometry $\Longleftrightarrow$ Complex numbers $\Longleftrightarrow$ Vectorial geometry $\Longleftrightarrow$ Barycentrical coordinates, a.s.o."[/color]

[quote="Virgil Nicula"][color=darkblue][b]You are right presently, Amir. S ! Sorry, Ashegh ![/b] If you have more proofs for the same problem, then you must choose the best (simply, shortly, quickly or nicely). In the contrary case, when you are only a solution, then is trivially if it is ugly, intricate or more long. I am wont to say " ... is like killing a butterfly with atom bomb " or " ... is like killing a fly with truck ". [u]Any method is good if it solves correctly the given problem.[/u]
color][/quote]

but it is not a wonderfull work if u solve easy problem, by power full tools.

u are right. any method which solves the problem corectly, is good.

but which method is the most variable?

for example solving a similarity problem by similarity,

or solve the same problem by inversion.

my goal isnt to only prove problems.

i want to prove them nice and nice.this is my goal. :wink:


I've heard varying assertions on this, and I can't find anything official...

Do the teams in ARML have to be confined to a region? If yes, are there any specific limits? In my local area I can only think of one person that might be willing to do it, and $ 2 \neq 15$, so I was hoping I would be able to join a team from somewhere else. (Edit: To poster below, yes)

I believe the only team in Virginia that competes is TJ, and I almost certainly can't join them since I don't go to TJ.

Is the 2nd person me, or is there actually another person in this area who might do ARML? :P

Anyways, I've been also curious if there is any way to be on a ARML team in Virginia if you don't go to TJ.

same thoughts, but i probably wouldn't get to go anyway, because of an undescribable discrimination against middle schoolers :)

Forum member Setofallsets tried (unsuccessfully, I believe) to do this several years ago.  The related thread is http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8538 .  You could probably PM or email him to see if he has any helpful suggestions.

[quote="ARMLEXEC"]As the Executive Director of ARML I think I can help you Virginia mathletes out a bit. First of all, I do believe that TJ's team has some non-TG students. Also you can form a team from the rest of Virginia. I'm sure that there are 13 other qualified students in the rest of Virginia. Richmond, Charlottesville, and Norfolk/Portsmouth come to mind as places to start looking for members. You or a teacher in your school can get a copy of the AIME scores and contact high scorers. I would also look for a teacher or college professor to try and help you organize and handle coaching. Again, contact with some of the Universities in Virginia is a good way to start trying to get some help. That is the way several of our teams got started. We at ARML would love to see the rest of Virginia represented. If I can be of further help please contact me.
ARML EXECUTIVE DIRECTOR[/quote]

Hmm, TJ has not had non-TJ students on its ARML teams for at least two years, if ever.

I have to question if ARMLEXEC is actually the executive director of ARML...

Well, it has the right e-mail address.  (Compare w/ http://www.arml.com/members.htm )

Yes, that is Steve Adrian. I had emailed him calling attention to this particular topic.

Single-school teams at ARML do lead to the issue that MellowMelon raised: what about everyone else? In my own case, the team I coach started as a response to the existence of single-school teams, and may have eventually led to the disappearance of those teams - but there there are no state-level magnet schools in my area, so it's a little different. I do note that AAST seems to coexist with other teams representing portions of New Jersey, and that schools like Stuyvesant and IMSA do not have their own teams but whose students join the NYC or Chicago teams.

All those are possible models. My best guess is that the New Jersey model might be the one that can happen: TJ stays TJ, but a separate Virginia team gets organized.


[hide="Answer"]Let $x$ be the number of days that it takes $m+r$ men to finish the job. $m\cdot d=md$ man-days, and $(m+r)\cdot x=md$ man-days, so $x=\frac{md}{m+r}$ days, and the answer is $C$.[/hide]

Diffferent way of doing it:

Call $x$ the amount of work they have to do. $x=md$ because it is equivalent to rate x time.  So $m+r$ men could do $md$ work in $\frac{md}{m+r}$ days (much like t=d/r)

[quote="joejia"]If $m$ men can do a job in $d$ days, then $m+r$ men can do the job in:
\[\textrm{(A)} d+r\qquad \textrm{(B)} d-r\qquad  \textrm{(C)} \frac{md}{m+r}\qquad \textrm{(D)} \frac{d}{m+r}\qquad \textrm{(E) none of these}\]
days.[/quote]

[hide]  $ md = x $  where $x$ is the number of man-powered-days required to complete the task.  To finish the task with $ m+r $ men working, the task will be completed in $ \frac{x}{m+4} = \frac{md}{m+r} $.  Meaning the answer is [b] C [/b] [/hide]


function
algebra
functional equation
algebra unsolved

Find all continuous functions $f$ on $\mathbb{R}$ satifies following sondition: $f(xyz)=f(x)f(y)f(z)$, for all $x,y,z\in\mathbb{R}$

x=y=z=1 yields

f(1)=f(1)f(1)f(1)

and f(1)=1 or 0 [Edit. just realised f(1)=-1 works here as well...]

If f(1)=0
x=1 yields
f(yz)=0 for all y,z so f(x)=0 for all x

If f(1)=1 :
z=1 yields
f(xy)=f(x)f(y)

This is a fairly well known functional equation, with solutions $f(x)=0,{|x|}^{c}$or$sgnx \cdot{|x|}^{c}$


inequalities
function
inequalities unsolved

[quote="∫FaiLurE∮"]If $ x,y,z\ge\ 0$ and $ x^2 + y^2 + z^2 = 1$, prove:
$ 1 \le\ \sum_{cyc} \frac {x}^{1 + yz}$  $ \le\ \sqrt {2}$[/quote]
Let $ x+y+z=3u,$ $ xy+xz+yz=3v^2$ and $ xyz=w^3.$
Hence, $ 1 \le \sum_{cyc} \frac {x}{1 + yz}\Leftrightarrow f(w^3)\leq0,$ where $ f$ is a convex function;
$ \sum_{cyc} \frac {x}{1 + yz}\leq\sqrt2\Leftrightarrow g(w^3)\geq0,$ where $ g$ is a monotone function.
Thus, $ f$ ($ g$) gets a maximal (minimal) value, when $ w^3$ gets an extremal value, which happens when two numbers from $ \{x,y,z\}$ are equal or maybe $ w^3=0.$
Hence, for the proof enough to check only two cases:
1) $ y=z$;
2) $ z=0,$ which easy to do.

For the $\sum_{cyc} \frac {x}{1 + yz}\ge 1$,
we have $\frac {x}{1 + yz}\ge x^2\Leftrightarrow x+xyz\le 1\Leftrightarrow x+\frac {x-x^3}{2}\le1$ and it's just $(x-1)^2(x+2)\ge 0$.

For the $\sum_{cyc} \frac {x}{1 + yz}\le\sqrt2$,
WLOG $x\ge y\ge z$ then $\sum_{cyc} \frac {x}{1 + yz}\le \frac{x+y+z}{1+yz}$.
Let $y+z=p,yz=q$ and we just have to prove that $\sqrt{1-p^2+2q}\le \sqrt2(1+q)-p$
which is equivalent to $1-p^2+2q\le 2(1+q)^2+p^2-2\sqrt2p(1+q)$ and it's just $(\sqrt2p-q-1)^2+q^2\ge0$.


What is the reason for locking my thread?
I was conviced that some moderator thought this topic is about fun only.
But why is it locked?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=196383

Because you insulted another member and the moderator did not like that.
Why post this in Games and Fun?

[quote="serialk11r"]Because you insulted another member and the moderator did not like that.[/quote]
Please show me the evidence for your accusation, whom and how did I insult.

[quote="serialk11r"]Why post this in Games and Fun?[/quote]

I did not post it here. I posted it in the round table. Somehow ZetaX thought the CNN topic is a joke for fun, it seems he/she had successfully persuaded the moderator to move the thread here. This is why.

You might not have insulted anyone, but there might have been a flame war had the thread continued.

Your thread was moved to Games and Fun because it was presented in a jokingly manner. You made it seem as a joke. Now had it been presented in a serious manner, where the Youtube link had  been present in the original post, then your thread might have been taken more seriously.

[quote="7h3.D3m0n.117"]You might not have insulted anyone, but there might have been a flame war had the thread continued.
[/quote]

So every man should be castrated, for otherwise he might rape someone.


calculus
integration
real analysis
real analysis unsolved

find
$ \prod_{k \equal{} 0}^{\infty} \frac {k^{k} \minus{} 1}{k^{k} \plus{} 1}$  - ?
 :huh:
when I was solving this one I was playing with  integrals like this $ \int_{0}^{1}x^{ \minus{} x}$ ..
but I can't get any closed answer.

hello, i think it gives no closed answer.
Sonnhard.


algebra
polynomial
quadratics
number theory
greatest common divisor
calculus
trigonometry

1)If $n>1$, show that $(x+1)^{n}-x^{n}-1=0$ has a multiple root if and only if $n-1$ is divisible by 6.

2)Let $P(x)$ be a polynomial with real coefficients, and assume that $P(x) \geq 0$ for all $x$. Prove that $P(x)$ can be expressed in the form $(Q_{1}(x))^{2}+Q_{2}(x))^{2}+...+Q_{n}(x))^{2}$ where $Q_{1}, Q_{2}, ...Q_{n}$ are polynomials with real coefficients.

A solution to #2 can be found [url=http://wangsblog.com/jeffrey/?p=60]here[/url].

[quote="Phelpedo"]Nice solution, paladin8. I just have one comment: I don't think that $\sqrt{a}$ is a polynomial.[/quote]

It's not $P(x) = \sqrt{x}$; $a$ is a constant. So it is a polynomial.

Maybe this is useful:
A polynomial has a multiple root if it has a common root with its derivative. That is: the gcd of P(x) and P'(x) has degree at least 1.

[quote="t0rajir0u"]I was trying to remember that solution, but I"d forgotten that great use of induction.  :)

How does the alternate solution work, though?  There's certainly no guarantee that $P(x)$ can be factored into linear and quadratic polynomials.[/quote]

There is a solution like that for polynomials over the complex numbers.

[quote="me@home"][hide="1)"]
$\sum_{k=1}^{n-1}{n \choose k}x^{k}=0$ factor out the $x=0$ root (there is only one of this because we we factor out this root we are left with a polynomial with a constant term, namely) $\sum_{k=1}^{n-1}{n \choose k}x^{k-1}=0$ \[{n \choose n-1}x^{n-2}+{n \choose n-2}x^{n-3}+\cdots+n=0\]  \[x^{n-2}+\frac{1}{(n-2)!(2!)}x^{n-3}+\cdots+1=0\] now its time for an amazing multiple root trick!! ...  :o
[/hide][/quote]
What amazing multiple root trick? Did you stop writing the solution in the middle? :maybe:

[quote="Phelpedo"]The one that B23 refers to.[/quote]
Ah, ok, just checking. I forgot Larson uses calculus...

Sorry, i got to that part in my solution in realized that I didn't know what to do..

[quote="Phelpedo"]1)If $n>1$, show that $(x+1)^{n}-x^{n}-1=0$ has a multiple root if and only if $n-1$ is divisible by 6.
[/quote]

First note that 'derivative of a polynomial' is hardly calculus, because you can define it formally without the use of limits.

Case n=7: $(x+1)^{7}-x^{7}-1=7x(x+1)(x^{2}+x+1)^{2}$ motivates to investigate whether the roots of $x^{2}+x+1$ are of multiplicity >1.
The roots of $x^{2}+x+1$ are $\theta=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$ and $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ .
Writing $\theta$ in goniometric form:
$\theta=\cos(\frac{4\pi}{3})+i\sin(\frac{4\pi}{3})$
De Moivre tells us that $\theta^{6m+1}$ is the same(=$\theta$).
On the other hand,
$\theta+1=\cos(\frac{-\pi}{3})+i\sin(\frac{-\pi}{3})$
Again, with De Moivre, we have $\theta^{6m+1}=\theta+1$.
Whence $(\theta+1)^{6m+1}-\theta^{6m+1}-1=\theta+1-\theta-1=0$.
To prove $\theta$ is a multiple root just check it is also a root of the derivative $P'(x)=(6m+1)(x+1)^{6m}-(6m+1)x^{6m}$.
This is easily done since $(\theta+1)^{6m}=\theta^{6m}=1$

Yay! Nice polynomial tricks with DeMoivre's... whenever a problem reads (divisible by 6) and its talking about powers of roots, it is always good to look for something nice using trig/ck and solutions to $x^{2}+x+1$... is this just my impression or from experienced problem solvers, is this always the case?


Let $ k$ be the smallest positive integer with the following property:

There are distinct integers $ m_{1},m_{2},m_{3},m_{4},m_{5}$ such that the polynomial \[p(x)=(x-m_{1})(x-m_{2})(x-m_{3})(x-m_{4})(x-m_{5})\] has exactly $k$ non-zero integers coefficients.

Find a set of integers $\{m_{1},m_{2},m_{3},m_{4},m_{5}\}$ for which this minimum $k$ is achieved

[quote="Yosh..."]Let $k$ be the smallest positive integer with the following property:

There are distinct positive integers $m_{1},m_{2},m_{3},m_{4},m_{5}$
such that the polynomial $p(x)=(x-m_{1})(x-m_{2})(x-m_{3})(x-m_{4})(x-m_{5})$ has exactly $k$ non-zero integers coefficients.

Find a set of integers $m_{1},m_{2},m_{3},m_{4},m_{5}$ for which this minimum $k$ is achieved[/quote]

I don't understand : if all $m_{i}>0$, none of the coefficients of $p(x)=(x-m_{1})(x-m_{2})(x-m_{3})(x-m_{4})(x-m_{5})$ can be zero : coeff of $x^{5},x^{3},x$ are all $>0$ and coeff of $x^{4},x^{2},1$ are all $<0$.

So $k=6$ and any polynomial verify the property.

[quote="Yosh..."]Let $ k$ be the smallest positive integer with the following property:

There are distinct integers $ m_{1},m_{2},m_{3},m_{4},m_{5}$
such that the polynomial $ p(x)=(x-m_{1})(x-m_{2})(x-m_{3})(x-m_{4})(x-m_{5})$ has exactly $ k$ non-zero integers coefficients.

Find a set of integers $ m_{1},m_{2},m_{3},m_{4},m_{5}$ for which this minimum $ k$ is achieved[/quote]

Hello !
With this correction, I claim $ k=3$.

1) $ k=0$ is impossible : it would mean $ p(x)=0$
2) $ k=1$ is impossible : it would mean $ p(x)=x^{5}$ and all $ m_{i}$ would be equal (to 0)
3) $ k=2$ is impossible : it would mean $ p(x)=x^{5}+ax^{4}$ or $ p(x)=x^{5}+ax^{3}$ or $ p(x)=x^{5}+ax^{2}$ or $ p(x)=x^{5}+ax$ or $ p(x)=x^{5}+a$ :
$ p(x)=x^{5}+ax^{4}$ implies four identical roots 0, which is impossible
$ p(x)=x^{5}+ax^{3}$ implies three identical roots 0, which is impossible
$ p(x)=x^{5}+ax^{2}$ implies two identical roots 0, which is impossible
$ p(x)=x^{5}+ax$ implies a root 0 and the four roots of $ x^{4}=-a$. But this equation can't have four real roots, so impossible
$ p(x)=x^{5}+a$ implies $ m_{i}$ are the five roots of $ x^{5}+a=0$. But this equation can't have five real roots, so impossible

3) $ k=3$ is possible :
Take $ m_{1}=0,m_{2}=a,m_{3}=-a,m_{4}=b,m_{5}=-b$ with $ a\neq b$ and $ ab\neq 0$. Then :
$ p(x)=x^{5}-(a^{2}+b^{2})x^{3}+a^{2}b^{2}x$ and $ p(x)$ has exactly 3 non-zero integer coefficients.


function
real analysis
real analysis solved

Let f:R->R be a function which has a limit in any real point and has no local extreme points. Show that:

a)f is continuous.
b)f is strictly monotonous.

1) Let a be a real number, and L = limf(x) as x tends to a.
Suppose that L =/= f(a) :
- If L < f(a), it follows that there exist e > 0 such that if |x-a| < e then f(x) < (L + f(a))/2 < f(a). Thus a is a local maximum for f.
- We use a simila reasoning if L > f(a).
Then L = f(a) and f is continuous at a. Since a is arbitray chosen, we are done.

2) Let a,b,c be three real numbers such that a < b < c.
Suppose, for a contradiction, that f(b) > f(a),f(c). Since f is continuous on the compact set [a,c], it has a maximum which cannot be f(a) nor f(c). Thus f has a local maximum. A contradiction.
In the same way, we deduce that we cannot have f(b) < f(a),f(c).
Thus for all real numbers a,b,c such that a < b < c we have f(a)  \leq  f(b)  \leq  f(c) or f(c)  \leq  f(b)  \leq  f(a).

Now suppose that f(0)  \leq  f(1) (otherwise consider -f). Then, from above, f(0)  \leq  f(x)  \leq  f(1) for each x in [0,1].
Moreover f(0)  \leq  f(1)  \leq  f(x) for each x  \geq  1 (for a = 0, b= 1, c= x), and f(x)  \leq  f(0)  \leq  f(1) for each x  \leq  0.

Let x <y be in [0,1].
As above (for a = 0, b = x, c = y), we then have f(0)  \leq  f(x)  \leq  f(y),  from which we deduce that f is nondecreasing on [0,1].

Let x<y \leq 0 :
As above, for a = x, b = y and c = 0, we deduce that f(x)  \leq  f(y)  \leq  f(0), which gives that f is nondecreasing on ]-oo;0].

Let 1  \leq  x < y :
As above, for a =1, b = x, c = y, we deduce that f(1)  \leq  f(x)  \leq  f(y), which gives that f is nondecreasing on [1, +oo[.

And then, f is nondecreasing on |R.
From the choice made that f(0)  \leq  f(1), we deduce that f is monotonic on |R.

Pierre.

Ooops, i misread the question. We have to prove that f is strictly monotonic. But this easily follows from the fact that f is monotonic and that it cannot be constant on any interval [a,b] with a < b since otherwise it would have a local extreme point.

Pierre.

Good job Pierre, I move it to solved problems


If the rotational inertia of a sphere about an axis through the center of the sphere is I, what is the rotational inertia of another sphere that has the same density, but has twice the radius?

$ I \propto MR^2$.
$ \frac {I}{I'} \equal{} \frac {MR^2}{M'(2R)^2}$ where $ M' \equal{} V \rho \equal{} \frac{4}{3} \pi (2R)^3 \rho \equal{} 8M$
Hence, $ I'\equal{}32I$.


How many number have $ 4$ digits in which each digits $ 0,1,...,9$ is use $ \leq 2$ times in $ 2$ ways of thinking:
a) Number shape $ aabb$ is OK! (May have $ 0,1,2$ pair digits)
b) Number shape $ aabb$ isn't OK! (Only $ 0,1$ pair digits)

Could you please explain with some examples the two cases?

Example:
a) Number shape: $ 4$ digits are different ($ 1234$); $ 1$ pair digits ($ 1123$); $ 2$ pairs digits ($ 1122$).
b) Number shape: $ 4$ digits are different ($ 1234$); $ 1$ pair digits ($ 1123$).

My solution:
a)-) Number shape $ aaaa$: $ 9$ numbers.
-) Number shape $ 1$ digit $ a$ and $ 3$ digits $ b$ ($ a\neq b$):
+) $ a,b\neq 0$: $ 2.C^2_9.4\equal{}288$ numbers.
+) $ a\equal{}0$: $ 9.3\equal{}27$ numbers.
+) $ b\equal{}0$: $ 9$ numbers.
Then we have $ 9000\minus{}9\minus{}288\minus{}27\minus{}9\equal{}8667$ numbers.
b)-) Number shape $ 2$ digits $ a$ and $ 2$ digits $ b$ ($ a\neq b$):
+) $ a,b\neq 0$: $ C^2_9.6\equal{}216$ numbers.
+) $ a\equal{}0$ or $ b\equal{}0$: $ 9.3\equal{}27$ numbers.
Then we have $ 8667\minus{}216\minus{}27\equal{}8424$ numbers.


How many numbers are in the list $ 1, 4, 7, \ldots, 2005, 2008 ?$

[hide="Hint"]It's an arithmetic series with common difference 3, first term 1 and $ a_{n}\equal{}2008$ (I don't want to do the calculations myself-I'm lazy :lol:) [/hide]

We subtract every term by $ 1$, and we get $ 0,3,6,\dots,2004,2007$.  Then, we see, as JRav said, we have an arithmetic sequence with a common difference of $ 3$, so we divide every number by $ 3$, and get $ 0,1,2,\dots,668,669$.  Then, adding $ 1$ to all the numbers, we get $ 1,2,3,\dots,669,670$, and it is clear that we have $ \boxed{670}$ numbers.

Alternatively, (what I actually had in mind when I wrote the hint), $ 2008\equal{}1\plus{}3(n\minus{}1)\equal{}1\plus{}3n\minus{}3\equal{}3n\minus{}2\equal{}2008\Rightarrow 3n\equal{}2010\Rightarrow n\equal{}670$.


Points $ A$ and $ C$ lie on a circle centered at $ O$, each of $ \overline{BA}$ and $ \overline{BC}$ are tangent to the circle, and $ \triangle ABC$ is equilateral. The circle intersects $ \overline{BO}$ at $ D$. What is $ \frac {BD}{BO}$?

$ \textbf{(A)}\ \frac {\sqrt2}{3} \qquad \textbf{(B)}\ \frac {1}{2} \qquad \textbf{(C)}\ \frac {\sqrt3}{3} \qquad \textbf{(D)}\ \frac {\sqrt2}{2} \qquad \textbf{(E)}\ \frac {\sqrt3}{2}$

I got B

We're just finding the ratio.
If we caculate by using the angles, we get 1/2

How exactly do you do that?

This is one of the two problems that I couldn't figure out by myself. (probably because I forgot everything concerning circles other than their area.

I don't have the figure.
Anyway..
Let AB=BC=CA=a(for convenience)
Also, I will make E the intersection of BO and AC

We know that OD is the radius of the circle and OB is OE+EB
Angle BAO=90 Angle CAO=30
Using the trigonometric ratio at triangle AOE we know that AO=a/root3. So, OD=a/root3 (radius)

Then, BO=BE + OE
We can get BE because it is the height of an equilateral triangle.
We can get OE by trigonometric ratio.
BO is 2a*root/2.

So, BD/BO=1/2=A

Sorry for my complicated solution.
I don't know how to write mathematical symbols at my computer.
Also sorry for the English, I'm Korean.

Lines AB and BC are tangent to circle O.
Draw lines OA and OB and you will have right angles.
Triangle ABC is equilateral, therefore angle B is 60 degrees.
Angle ABO is half, or 30 degrees.
Angle OAB=90 angle AB0=30 therefore angle AOB=60
IF OA=X then OB=2X because it is a 30,60,90 triangle
OA=OD=X
BD=OB-OD=2X-X=X
BD/OB=X/2X=1/2


cos(-2t) + sin(-2t)

[cos(t)]^2-2sin(t)cos(t)-[sin(t)]^2 is the answer but can anyone show me how to get this answer??

This is just an application of the double angle formulas:

$\cos(-2t)+\sin(-2t) = \cos(2t)-\sin(2t) = \cos^{2}t-2\sin t\cos t-\sin^{2}t$

You should probably post this in the high school section.


Can someone attach a pdf file of All Russian Mathematical
olympiad, and also All Russian Geometry olympiad in English, I saw
that there are all the problems on this forum but not in
download-able shape.

Thanks in advance

Don't mind, see http://www.kalva.demon.co.uk/.

Thank you very much!! Now, only geometry olympiad left...

Is there russia national olympiad 2002 and 2003 with solution?(Not the one in kalva demon). I'd like to know the distribute for three different grades exactly.


probability
FTW
number theory
prime numbers

First, let's count how many prime numbers or multiples of three there are in the grid:

2, 3, 5, 7, 9, 11, 12, 13, 15, 17, 18, 19, 21, 23, 24

There are 15 of them.
However, there are 25 spaces on the grid.
So, $ 15/25 \equal{} 3/5 \equal{} \boxed{60}$%

the solution was right but i think ftw counted 1 as a prime number so it would be $\frac{16}{25}$or $ \boxed {64 } $ %

Actually, Rubik's solution did not include the number $6$, which is a multiple of $3$, so there are indeed $16$ numbers that are either prime or a multiple of $3$, and that is without the inclusion of $1$.


summer program
PROMYS
Mathcamp
HCSSiM
Ross Mathematics Program
search
MIT

er.........can anyone inform me about the existence of summer camps for math in chennai ,I mean math summer camps like how US has Promys ,MathCamp,HCSSim etc

Only good quality summer camps in India I know of are the Bhaskaracharya Pratisthan camp and KVPY. Why don't you apply to some US camps like SSP, RSI, Ross, mathcamp or even something like ISSYP. Most of them are free for internationals. Moreover you are much more likely to be selected as a southpaw :P Its a sort of tradition(And my excuse for failing :P )

[quote="Scary math"]@nuclear
failing in kg u mean?  :huh:[/quote]
Eh? Maybe....but here I was referring to failure in getting into the international summer camps(SSP and ISSYP) which I applied to.

[quote="srinath.r"]Are mathcamp,SSP free for internationals.??[/quote]
Assuming you are not rich, they do offer up to 100% aid. Go check their sites for details.

[quote="Scary math"]http://www.artofproblemsolving.com/Forum/search.php?search_author=nuclear_alchemist&ppage=20&start=580
[hide]Dat's kind of an extraordinarily strict skool,I officially failed in kg,but they promoted me to next class somehow. :P I remember horrifying Miss Paxie :stink: She nearly boiled me alive dat day for colouring the Indian flag in pink and purple :rotfl:  :rotfl: [/hide][/quote]
I did NOT fail KG

[quote="Scary math"]Oops!Thot u were serious :P
But I did  :blush: And I'll take revenge on Miss paxie for failing me someday :bomb:[/quote]
How?

lemme think
The only 1 thing i remember abt her...............
She used to wear really scary heels[hide]still remember the sound of it[/hide]
wud have surely pierced my skull had she run after me someday with those shoes in hand:(
Those shoes even left pug marks :rotfl:  :rotfl:
But no idea on how to take revenge,I'll think of something when time comes :D
[hide]the campus has got pretty snakes,lizards,spiders[hide]I even have a wonderful collection of colourful house flies:evilgrin:[/hide] ..............what abt leaving a bunch of them at her doorstep :diablo: ??????????She'll definetely be scared to death at the sight of those :rotfl: [/hide]

Are there are any indians who have attended the camps like the mathcamp

Possibly.
Quoted from their site-
Home countries of past Mathcampers have included Australia, Bulgaria, China, the Czech Republic, Egypt, England, France, Germany, Greece, Hong Kong, Hungary, India, Japan, Korea, Latvia, Lithuania, Luxembourg, Macedonia, Mexico, Poland, Romania, Serbia, Singapore, Spain, Switzerland, Taiwan, Thailand, and Turkey. In 2007, 13% of the students at the camp were from overseas.

    Mathcampers are an amazingly creative group of people. The games here were absolutely hilarious. People here are not only incredibly smart but incredibly crazy (in a good way).   -- Rohan Agarwal

    Mathcamp reaffirmed my desire to be a teacher--with the great teaching provided by some of the guest lecturers and the staff. It reminded me of the reason I want to be a teacher: I want other people to see the beauty and elegance and feel the awesomeness of math (just like some of the lecturers and staff made me feel)... Everything seems to fall into the framework created by Mathcamp!

-- Sameer Shah, MIT

er.....do you know anyone who has attended it in the recent years or will be attending the following year from India.

Lol, I didn't even apply to Mathcamp. But if you want to get in touch, just search the indian participants on facebook or something....

@srinath: Contact "Brutus" on this forum.....I do know that he was going for the math camp this year and
 was at SSP last summer.

Yes. Never talked to him much here, but he's one brillliant guy from what I've heard :P

"Brutus" is very much an Indian((from Pune).....was with me at the  Astro camp....Is part of the senior Astro  Team this year.

anyone of you have his email id ,he seems inactive in mathlinks .

I have PM'ed you his gmail id....Check it out.

can anyone tell me where  mathcamp will be held?


I would appreciate any help on this question I cannot solve.

" In a particular town, 70% of the students are bused to school. If a statistician randomly sampled 1000 students from the town, then the probability that at least 720 of these students are bused to school, correct to the nearest hundredth, would be _____? "

The answer key says the answer is 0.09, but the closest I can get to that is 0.90 using some not-so-mathematical math techniques. Any help?

Edit: sorry, I guess I have to post the source? heres the address to the test:
http://education.alberta.ca/media/739560/pure_ma30_jan01_de.pdf

Yes, there lies the problem. Even using binompdf on a calculator isn't much faster at all, and there surely must be a better way than typing in over 200 numbers.


greatest common divisor
number theory
induction
algorithm
Euclidean algorithm

[quote="radio"]The sequence of words, each of which consists of two letters $ A$ and $ B$, is defined as follows:
The first word is $ A$ and the next one is constructed from the previous one by substituting each $ A$ with $ BAA$ and substituting $ B$ with $ AB$. Is it possible for this sequence to obtain a periodic word, that is the word of the form $ http://WWW...W$, for some word $ W$?[/quote]

[hide="small things"]
dunno if it's possible (i'm thinking it isn't), but here's something i noticed
the number of a's after the nth iteration is the n+2th fibonacci number, while the number of b's is the n+1th fibonacci number (if we count 0 as the first, 1 as the second, etc.)
i also noticed that the numbers of a and b must have a GCF larger than 1 for the repeating word to be possible.
last thing i noticed: the string will always end in an A.

can anyone magically make something out of this?

[/hide]

[quote="CircleSquared"]the number of a's after the nth iteration is the n+2th fibonacci number, while the number of b's is the n+1th fibonacci number (if we count 0 as the first, 1 as the second, etc.)
i also noticed that the numbers of a and b must have a GCF larger than 1 for the repeating word to be possible.[/quote]

Keep going.  What's $ \gcd(F_{n\plus{}2}, F_{n\plus{}1})$?  :)

[quote="t0rajir0u"][quote="CircleSquared"]the number of a's after the nth iteration is the n+2th fibonacci number, while the number of b's is the n+1th fibonacci number (if we count 0 as the first, 1 as the second, etc.)
i also noticed that the numbers of a and b must have a GCF larger than 1 for the repeating word to be possible.[/quote]

Keep going.  What's $ \gcd(F_{n \plus{} 2}, F_{n \plus{} 1})$?  :)[/quote]

wait what's that method where gcd(a,b)=gcd(a-b,b) called?
i'm thinking that will probably help in an inductive proof or something

edit: found it in the wiki.
$ GCD(a_1,\dots,a_n)\equal{}GCD(GCD(a_1,\dots,a_{n\minus{}1}),a_n)$
we know $ GCD(F(1),F(2))\equal{}1$
so $ GCD(F(1),F(2),F(3))\equal{}GCD(GCD(F(1),F(2)),F(3))\equal{}1$
and so on

and because we need the GCD of the a and b values to be greater than 1
there is no string with repeating word.

is that right?

Define $ a_i = \gcd{(F_{i+1},F_i)}$.  You just show that $ a_{n+1} = \gcd{(F_{n+2},F_{n+1}) = \gcd{(F_{n+2}-F_{n+1},F_{n+1})} = \gcd{(F_n,F_{n+1})} = a_n}$.  We know that $ a_0 = 1$, hence $ a_i = 1$ for all nonnegative integers $ i$.  It follows that any two consecutive Fibonacci numbers are coprime.

To show that $ \gcd{(a,b)} = \gcd{(a-b,b)}$ (with $ a > b$), let $ d = \gcd{(a,b)}$.  Then $ a = md$ and $ b = nd$, so $ a-b = (m-n)d$ which is divisible by $ d$.  Hence $ \gcd{(a-b,b)} \ge d$.  Now assume for contradiction that $ \gcd{(a-b,b)} > \gcd{(a,b)}$.  Set $ e = \gcd{(a-b,b)}$.  Then $ a-b = pe$ and $ b = qe$, so $ a = (a-b)+b = (p+q)e$.  We have a contradiction, hence $ \gcd{(a,b)} = \gcd{(a-b,b)}$.

You don't even need to use Fibbonacci numbers:

Let $ a_{n}$ and $ b_{n}$ be the numbers of $ A$'s and $ B$'s, respectively, in the $ n$th word of the sequence.  Then we have $ a_{n\plus{}1}\equal{}2a_{n}\plus{}b_{n}$ and $ b_{n\plus{}1}\equal{}a_{n}\plus{}b_{n}$, since each $ A$ generates two $ A$'s and a $ B$, and each $ B$ generates one of each.  Since the first word in the sequence is $ A$, we have $ a_{1}\equal{}1$ and $ b_{1}\equal{}0$.

Suppose for the sake of contradiction that the $ k$th word in the sequence is periodic and can be divided into $ m$ identical words, where $ m>1$.  Then $ (a_{k},b_{k})\equal{}m>1$.  Therefore it is sufficient to prove that $ (a_{k},b_{k})\equal{}1$ for all $ k$.

We use induction on $ k$.  The base case is true since $ (a_{1},b_{1})\equal{}(1,0)\equal{}1$.  Now assume $ (a_{k},b_{k})\equal{}1$.  Then $ (a_{k\plus{}1},b_{k\plus{}1})\equal{}(2a_{k}\plus{}b_{k}, a_{k}\plus{}b_{k})\equal{}(a_{k},a_{k}\plus{}b_{k})\equal{}(a_{k},b_{k})\equal{}1$ using the Euclidean Algorithm, and we're done.


(1)
$ (1 \minus{} x)^{1 \minus{} \frac {1}{x}} < (1 \plus{} x)^{\frac {1}{x}}$
$ \Longleftrightarrow(1 \minus{} \frac {1}{x})\ln(1 \minus{} x) < \frac {1}{x}\ln(1 \plus{} x)$
$ \Longleftrightarrow x\left(\ln(1 \minus{} x^2) \minus{} x\ln(1 \minus{} x)\right) > 0$,

hence it suffices to prove $ f(x) > 0$ on positive $ x$ and $ f(x) < 0$ on negative $ x$,here $ f(x) \equal{} \ln(1 \minus{} x^2) \minus{} x\ln(1 \minus{} x)$.

$ f'(x) \equal{} \frac { \minus{} 2x}{1 \minus{} x^2} \minus{} \ln(1 \minus{} x) \plus{} \frac {x}{1 \minus{} x} \equal{} \minus{} 1 \plus{} \frac {1}{1 \plus{} x} \minus{} \ln(1 \minus{} x)$ ,
$ f''(x) \equal{} \minus{} \frac {1}{(1 \plus{} x)^2} \plus{} \frac {1}{1 \minus{} x} \equal{} \frac {x(x \plus{} 3)}{(1 \plus{} x)^2(1 \minus{} x)}$ ,
so we have
$ f''(x) > 0$ $ (x > 0)$ and $ f''(x) < 0$ $ (x < 0)$,
hence $ f'(x) > f'(0) \equal{} 0$ $ (x > 0)$ and $ f'(x) >f'(0) \equal{} 0$ $ (x < 0)$, so we have
$ f(x) > f(0) \equal{} 0$ $ (x > 0)$ and $ f(x) < f(0) \equal{} 0$ $ (x < 0)$

(2)Multiply $ (1 \minus{} x)^{\frac {1}{x}} > 0$ both sides we have
$ (1 \minus{} x) < (1 \minus{} x^2)^{\frac {1}{x}}$,and put $ x \equal{} 0,01$ then we have $ 0,99 < 0,9999^{100}$

multiply $ (1 \plus{} x)^{1 \minus{} \frac {1}{x}} > 0$ both sides we have
$ (1 \minus{} x^2)^{1 \minus{} \frac {1}{x}} < 1 \plus{} x$ and put $ x \equal{} \minus{} 0,01$ then we have $ 0,9999^{101} < 0,99$,so we are done.


[If possible, without using a PC] , to find the best  positive (integers) constants $A, B$ such that  $A\cdot \pi <B+e^{\pi}$.

[quote="Ravi B"]What is the definition of "best"?[/quote]
$A\cdot \pi < B+e^{\pi}$
$B=$the least positive (integer)
$A=$the greatest positive (integer)

[quote="Ravi B"]Hmm, I'm afraid I still don't understand.  How do you optimize two conditions?  For example, is it better to choose $(A, B)$ to be $(7, 1)$ or $(8, 2)$?  The first pair has a smaller $B$, while the second pair has a larger $A$.[/quote]
Indeed! ........ Regards,Alex

But my example was when $(A, B)$ was $(8, 2)$, not $(2, 8)$.  So do you mean that $(7, 1)$ has won?

[quote="flip2004"]Hi Ravi !
Consider two pairs $(A,B)$ and $(A^{*},B^{*})$ of positive integers.
Let us say that $(A^{*},B^{*})$ is [b]better[/b] than $(A,B)$, iff
\[A\pi-B < A^{*}\pi-B^{*}< e^\pi \; . \]
[/quote]
Then there doesn't exist the best pair. It's very easy  to prove that for any $\alpha\in\mathbb{R}\setminus\mathbb{Q}$, $\beta\in\mathbb{R}$ and $\varepsilon>0$ there exist integers $x>0$ and $y$ such that $|\alpha x+y+\beta|<\varepsilon$.

[quote="RIP"] ...Then there doesn't exist the best pair. ...
[/quote]
Right, the proposer/Alex