Hi,

I found the following quite nonstandard inequality which somehow reminded me of the inequality mentioned in hartshorne( http://www.mathlinks.ro/Forum/viewtopic.php?t=22459 ):

Let $0\leq x_1,...,x_k\leq 1$ be reals and let $l>0$ be a positive real, such that $x_1^l + x_2^l + ... + x_k^l = n$ where $n$ is a positive integer, $n\leq k$.
Prove that there exists an $n$-element subset $S\subset \{1,...,k\}$ such that $\sum_{x\in S} x\geq x_1^{l+1}+x_2^{l+1}+...+x_k^{l+1}$.

(particular case: If some nonnegative reals add to 1, then there is at least one which is at least as large as the sum of the squares of all those numbers.)

i hope i didnt get something wrong, because my solution seems quite simple:

So let $\sum_{i=1}^k x^l = n$ and let $x_1 \ge x_2 \ge \dots \ge x_k$

now let us consider the cyclic sums $x_i + x_{i+1} + \dots + x_{i+n}$ (whereas if $i+j > k$, we take $i+j-k$ as index) \[ \sum x_i^l + x_{i+1}^l + \dots + x_{i+n}^l \le \sum x_1^l + x_2^l + \dots + x_n^l \] \[ \sum x_i^{l+1} + x_{i+1}^{l+1} + \dots + x_{i+n}^{l+1} \le \sum x_1^l\cdot x_i^l + x_2^l \cdot x_{i+1}^l+ \dots + x_n^l\cdot x_{i+n}^l \] \[ n \cdot \sum x_i^{l+1} \le x_1^l \cdot \sum x_i^l + x_2^l \cdot \sum x_i^l + \dots + x_n^l \cdot \sum x_i^l = n \cdot \sum_{i=1}^n x_i ^l \] \[ \sum x_i^{l+1} \le \sum_{i=1}^n x_i ^l \]

Ok Ok ... i actually proved the inequality only for $l= 1$ but i noticed that only later :( .... anyway its a weak generalization in a diffrent direction ...

[quote="Peter Scholze"]Well, actually I don't understand the second line of your proof. Indeed, it must be wrong as it does not make use of the condition $x_i\leq 1$ for all $i=1,...,k$. This condition is important as the example of $x_1=n^{\frac 1l}$, $x_2=...=x_k=0$ shows.
However, I am very interested in generalizations which weaken that last condition. E.g., to prove the hartshorne inequality nicely one would need a version of this inequality with some weaker condition.[/quote]

yeah there is a huge mistake in the second line. but if we change that mistake, we conclude at a different inequality without the restriction of $x_i\in[0;1]$.

[hide="Edited Proof"]
\[ \sum x_i^l + x_{i+1}^l + \dots + x_{i+n}^l \le \sum x_1^l + x_2^l + \dots + x_n^l \] \[ \sum x_i^{2l} + x_{i+1}^{2l} + \dots + x_{i+n}^{2l} \le \sum x_1^l\cdot x_i^l + x_2^l \cdot x_{i+1}^l+ \dots + x_n^l\cdot x_{i+n}^l \] \[ n \cdot \sum x_i^{2l} \le x_1^l \cdot \sum x_i^l + x_2^l \cdot \sum x_i^l + \dots + x_n^l \cdot \sum x_i^l = n \cdot \sum_{i=1}^n x_i ^l \] \[ \sum x_i^{2l} \le \sum_{i=1}^n x_i ^l \][/hide]

(i'll think about your inequality later ;) )


In a sequence of real numbers, the sum every 5 consecutive terms is positive, and the sum of every 7 consecutive terms is negative. How long may be the sequence?

Same idea [hide="IMO 1977 Problem 2"]http://www.mathlinks.ro/Forum/viewtopic.php?p=367411#367411[/hide]


geometry
circumcircle
cyclic quadrilateral
geometry unsolved

Let $ ABCD$ be a cyclic quadrilateral,$ E$-the midpoint of $ BD$ and $ C_1$, $ C_2$, $ C_3$,$ C_4$-the circumcircles of triangles
$ AEB$,$ BEC$,$ CED$ and $ DEA$,respectively.Show that if $ C_4$ is tangent to the line $ CD$,then $ C_1$, $ C_2$ ,$ C_3$ are
respectively tangent to the lines $ BC$,$ AB$,$ DA$.

$ ABCD$ is a harmonic quadrilateral, hence the conclusion.

Best regards,
sunken rock

what does it mean HARMONIC quadrilateral

Harmonic quadrilateral means:
a) the tangents to its circumcircle at two opposite vertices concur on a diagonal.
b) the bisectors of two opposite angles concur on a diagonal.

If a, b, c and d are the lengths of the consecutive sides, from each of the above properties we get:
$ a\cdot c \equal{} b \cdot d$ which, with [b]Ptolemy's[/b] gives: "[i][color=darkblue]the product of two opposite sides equals the one of a diagonal and half of the other one[/color][/i]".

Best regards,
sunken rock


If $ a,b,c$ are nonnegative real numbers, then

$ \sum a\sqrt{a^2\plus{}8bc} \le (a\plus{}b\plus{}c)^2\le \sum a\sqrt{4a^2\plus{}5bc}$.

The left follows from the Cauchy Schwarz Inequality.
And the right follows from the following known of yours :)
\[ \sum \sqrt{\frac{a^2}{4a^2\plus{}5bc}} \le 1\]
which is equivalent to
\[ \sum \sqrt{\frac{a}{4a\plus{}5b}} \le 1\]
for all $ a,b,c>0$.
This inequality, I remember is in your book. :)

Very nice your idea for the right inequality, can_hang2007. :lol:
A similar inequality is the following:

$ \sum a\sqrt{9a^2\plus{}7b^2}\ge (a\plus{}b\plus{}c\plus{}d)^2$,

where $ a,b,c,d$ are nonnegative real numbers.

[quote="spanferkel"][quote="can_hang2007"]
\[ \sum \sqrt {\frac {a^2}{4a^2 \plus{} 5bc}} \le 1
\]
which is equivalent to
\[ \sum \sqrt {\frac {a}{4a \plus{} 5b}} \le 1
\]
for all $ a,b,c > 0$.
[/quote]

Why?   :huh:[/quote]

Why not?          :)

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum a\sqrt {a^2 \plus{} 8bc} \le (a \plus{} b \plus{} c)^2\le \sum a\sqrt {4a^2 \plus{} 5bc}$.[/quote]

[quote="can_hang2007"]The left follows from the Cauchy Schwarz Inequality.
[/quote]

If $ a,b,c$ are nonnegative real numbers, then

$ \sum a\sqrt {a^2 \plus{} 8bc} \le (a \plus{} b \plus{} c)^2$

By Cauchy it is sufficient to prove

$ (\sum a)^4 \geq (\sum a^2)(\sum (a^2\plus{}8bc))$

Expanding I obtain

$ \sum ab \geq \sum a^2$

which is false.  Where am I wrong?  :blush:

[quote="can_hang2007"]And the right follows from the following known of yours :)
\[ \sum \sqrt {\frac {a^2}{4a^2 \plus{} 5bc}} \le 1
\]
which is equivalent to
\[ \sum \sqrt {\frac {a}{4a \plus{} 5b}} \le 1
\]
for all $ a,b,c > 0$.
This inequality, I remember is in your book. :)[/quote]

Why?  Sorry,but I don't understand. :blush:

Thank you very much.

[quote="manlio"][quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum a\sqrt {a^2 \plus{} 8bc} \le (a \plus{} b \plus{} c)^2\le \sum a\sqrt {4a^2 \plus{} 5bc}$.[/quote]

[quote="can_hang2007"]The left follows from the Cauchy Schwarz Inequality.
[/quote]

If $ a,b,c$ are nonnegative real numbers, then

$ \sum a\sqrt {a^2 \plus{} 8bc} \le (a \plus{} b \plus{} c)^2$

By Cauchy it is sufficient to prove

$ (\sum a)^4 \geq (\sum a^2)(\sum (a^2 \plus{} 8bc))$

Expanding I obtain

$ \sum ab \geq \sum a^2$

which is false.  Where am I wrong?  :blush:

[quote="can_hang2007"]And the right follows from the following known of yours :)
\[ \sum \sqrt {\frac {a^2}{4a^2 \plus{} 5bc}} \le 1
\]
which is equivalent to
\[ \sum \sqrt {\frac {a}{4a \plus{} 5b}} \le 1
\]
for all $ a,b,c > 0$.
This inequality, I remember is in your book. :)[/quote]

Why?  Sorry,but I don't understand. :blush:

Thank you very much.[/quote]
I think my style of writing makes you confused, so I will rewrite it as
\[ \sum \sqrt{\frac{a^2}{4a^2\plus{}5bc}} \le 1\]
for all $ a,b,c >0$.
which is equivalent to
\[ \sum \sqrt{\frac{x}{4x\plus{}5y}} \le 1\]
for all $ x,y,z>0$.
:)

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum a\sqrt {a^2 + 8bc} \le (a + b + c)^2\le \sum a\sqrt {4a^2 + 5bc}$.[/quote]
There is another nicer and simpler way to prove the right, we can rewrite it as
\[ 5abc \sum \frac{1}{2a+\sqrt{4a^2+5bc}} \ge 2\sum ab -\sum a^2\]
By CauchySchwarz,
\[{ \sum \frac{1}{2a+\sqrt{4a^2+5bc}} \ge \frac{9}{2\sum a +\sum \sqrt{4a^2+5bc}}} =\frac{9\sum a}{2(\sum a)^2 +\sum (a+b+c)\sqrt{4a^2+5bc}} \ge \frac{9\sum a}{2(\sum a)^2 +\sum \frac{(a+b+c)^2+4a^2+5bc}{2}} =\frac{18\sum a}{ 11\sum a^2+19\sum ab} \ge \frac{3\sum a}{2\sum a^2+3\sum ab}\]
It suffices to prove that
\[ 15abc \sum a \ge (2\sum ab -\sum a^2)(2\sum a^2+3\sum ab)\]
\[ \Leftrightarrow 2\sum a^2(a-b)(a-c) +\sum ab(a-b)^2 \ge 0\]
which is true.
We have done. :)

[quote="Vasc"]Very nice your idea for the right inequality, can_hang2007. :lol:
A similar inequality is the following:

$ \sum a\sqrt {9a^2 \plus{} 7b^2}\ge (a \plus{} b \plus{} c \plus{} d)^2$,

where $ a,b,c,d$ are nonnegative real numbers.[/quote]
I think it is easier than the previous, since by CauchySchwarz, we have
\[ 4\sum a\sqrt {9a^2 \plus{} 7b^2} \ge \sum a(9a \plus{} 7b)
\]
It suffices to prove that
\[ 9(a^2 \plus{} b^2 \plus{} c^2\plus{}d^2) \plus{} 7(a \plus{} c)(b \plus{} d) \ge 4(a \plus{} b \plus{} c \plus{} d)^2 \equal{} 4(a \plus{} c)^2 \plus{} 4(b \plus{} d)^2 \plus{} 8 (a \plus{} c)(b \plus{} d)
\]

\[ \Leftrightarrow 9(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \ge 4(a \plus{} c)^2 \plus{} 4(b \plus{} d)^2 \plus{} (a \plus{} c)(b \plus{} d)
\]
which is true because
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \ge (a \plus{} c)(b \plus{} d)
\]

\[ 2(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \equal{} 2(a^2 \plus{} c^2) \plus{} 2(b^2 \plus{} d^2) \ge (a \plus{} c)^2 \plus{} (b \plus{} d)^2
\]
:)

$ \displaystyle\left(\sum a\right)^2\ge \sum a\sqrt {a^2 \plus{} 8bc}$

$ \displaystyle\sum a\sqrt {a^2 \plus{} 8bc} \equal{} \sum \sqrt a\sqrt {a^3 \plus{} 8abc}\le \sqrt {\left(\sum a\right)\left(\sum a^3 \plus{} 24abc\right)}$. (1)

$ \displaystyle\left(\sum a\right)^3 \equal{} \sum a^3 \plus{} 6abc \plus{} 3\sum(a^2b \plus{} ab^2)\ge\sum a^3 \plus{} 6abc \plus{} 3\cdot 6\sqrt {a^6b^6c^6} \equal{}$
$ \displaystyle \equal{} \sum a^3 \plus{} 24abc$. (2)

So from (1) and (2) we get the inequality

[quote="can_hang2007"][quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum a\sqrt {a^2 + 8bc} \le (a + b + c)^2\le \sum a\sqrt {4a^2 + 5bc}$.[/quote]
There is another nicer and simpler way to prove the right, we can rewrite it as
\[ 5abc \sum \frac {1}{2a + \sqrt {4a^2 + 5bc}} \ge 2\sum ab - \sum a^2
\]
By CauchySchwarz,
\[{ \sum \frac {1}{2a + \sqrt {4a^2 + 5bc}} \ge \frac {9}{2\sum a + \sum \sqrt {4a^2 + 5bc}}} = \frac {9\sum a}{2(\sum a)^2 + \sum (a + b + c)\sqrt {4a^2 + 5bc}} \ge \frac {9\sum a}{2(\sum a)^2 + \sum \frac {(a + b + c)^2 + 4a^2 + 5bc}{2}} = \frac {18\sum a}{ 11\sum a^2 + 19\sum ab} \ge \frac {3\sum a}{2\sum a^2 + 3\sum ab}
\]
It suffices to prove that
\[ 15abc \sum a \ge (2\sum ab - \sum a^2)(2\sum a^2 + 3\sum ab)
\]

\[ \Leftrightarrow 2\sum a^2(a - b)(a - c) + \sum ab(a - b)^2 \ge 0
\]
which is true.
We have done. :)[/quote]

very nice your proof !!! :)

[quote="can_hang2007"]
I think my style of writing makes you confused, so I will rewrite it as
\[ \sum \sqrt {\frac {a^2}{4a^2 \plus{} 5bc}} \le 1
\]
for all $ a,b,c > 0$.
which is equivalent to
\[ \sum \sqrt {\frac {x}{4x \plus{} 5y}} \le 1
\]
for all $ x,y,z > 0$.
:)[/quote]

I still don't get it. If you mean a substitution, which one??  :maybe:  :maybe:

spanferkel, try $ x \equal{} \frac {a}{c},$ $ y \equal{} \frac {b}{a}$ and $ z \equal{} \frac {c}{b}$
or $ a\equal{}\sqrt[3]{\frac{x}{y}},$ $ b\equal{}\sqrt[3]{\frac{y}{z}}$ and $ c\equal{}\sqrt[3]{\frac{z}{x}}.$ :wink:

[quote="can_hang2007"][quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, then

$ \sum a\sqrt {a^2 + 8bc} \le (a + b + c)^2\le \sum a\sqrt {4a^2 + 5bc}$.[/quote]
There is another nicer and simpler way to prove the right, we can rewrite it as
\[ 5abc \sum \frac {1}{2a + \sqrt {4a^2 + 5bc}} \ge 2\sum ab - \sum a^2
\]
By CauchySchwarz,
\[{ \sum \frac {1}{2a + \sqrt {4a^2 + 5bc}} \ge \frac {9}{2\sum a + \sum \sqrt {4a^2 + 5bc}}} = \frac {9\sum a}{2(\sum a)^2 + \sum (a + b + c)\sqrt {4a^2 + 5bc}} \ge \frac {9\sum a}{2(\sum a)^2 + \sum \frac {(a + b + c)^2 + 4a^2 + 5bc}{2}} = \frac {18\sum a}{ 11\sum a^2 + 19\sum ab} \ge \frac {3\sum a}{2\sum a^2 + 3\sum ab}
\]
It suffices to prove that
\[ 15abc \sum a \ge (2\sum ab - \sum a^2)(2\sum a^2 + 3\sum ab)
\]

\[ \Leftrightarrow 2\sum a^2(a - b)(a - c) + \sum ab(a - b)^2 \ge 0
\]
which is true.
We have done. :)[/quote]
Nice, Can. :lol:
But, if we use the inequality

$ \frac {18\sum a}{ 11\sum a^2 + 19\sum ab} \ge \frac {6\sum a}{5(\sum a^2 + \sum ab)}$

then it suffices to prove that

$ 6abc\sum a \ge (\sum a^2+\sum ab)(2\sum ab-\sum a^2)$,

which is

$ \sum a^4+abc\sum a \ge \sum ab(a^2+b^2)$.

[quote="can_hang2007"]The left follows from the Cauchy Schwarz Inequality.
And the right follows from the following known of yours :)
\[ \sum \sqrt {\frac {a^2}{4a^2 \plus{} 5bc}} \le 1\]
which is equivalent to
\[ \sum \sqrt {\frac {a}{4a \plus{} 5b}} \le 1\]
for all $ a,b,c > 0$.
This inequality, I remember is in your book. :)[/quote]
See here: http://canhang2007.wordpress.com/2009/11/03/inequality-43-v-cirtoaje/


Find all triples of positive integers $ (a, b, c)$ such that

$ (1\plus{} \frac{1}{a})(1\plus{}\frac{1}{b})(1\plus{}\frac{1}{c}) \equal{}2$.

[quote="gaussintraining"]
By AM-GM,

$ \frac {ab \plus{} bc \plus{} ac}{3}\ge abc$
[/quote]

Are you sure about this?

[quote="AndrewTom"]Find all triples of positive integers $ (a, b, c)$ such that

$ (1 \plus{} \frac {1}{a})(1 \plus{} \frac {1}{b})(1 \plus{} \frac {1}{c}) \equal{} 2$.[/quote]

Wlog say $ a\ge b\ge c\ge 1$. We get $ 2\le(1 \plus{} \frac 1c)^3$ and so $ c\le 3$. Since $ c \equal{} 1$ is obviously not a solution, it remains two possibilities :

1) $ c \equal{} 3$ and so $ (1 \plus{} \frac 1a)(1 \plus{} \frac 1b) \equal{} \frac 32$ which implies $ \frac 32\le(1 \plus{} \frac 1b)^2$ and so $ b\le 4$ and so $ b\in\{2,3,4\}$
Checking these three values, we get two solutions : $ (3,4,5)$ and $ (3,3,8)$

2) $ c \equal{} 2$ and so $ (1 \plus{} \frac 1a)(1 \plus{} \frac 1b) \equal{} \frac 43$ which implies $ \frac 43\le(1 \plus{} \frac 1b)^2$ and so $ b\le 6$ and so $ b\in\{2,3,4,5,6\}$
Checking these five values, we get three solutions : $ (2,6,7)$ and $ (2,5,9)$ and $ (2,4,15)$

Hence the anwer : $ \boxed{(3,4,5),(3,3,8),(2,6,7),(2,5,9),(2,4,15)}$ (and permutations)


advanced fields
advanced fields unsolved

Let $\mathbb D$ be the unit circle and $f,g: \mathbb D\longrightarrow\mathbb D$ are continuous bijections that for each $s\in\partial\mathbb D$, $f(s)=g(s)$. Prove that there is a continuous mapping $F(s,t): \mathbb D\times [0,1]\longrightarrow\mathbb D$ that $F(s,0)=f(s)$ and $F(s,1)=g(s)$ and $F(s,t)=f(s)$ for each $s\in\partial\mathbb D$, and also $F(s,t)$ is a bijection for each $t$.

You mean that $\mathbb{D}$ is the unit disk- the circle is its boundary.

As stated, $F(s,t)=(1-t)f(s)+tg(s)$ works.

Excuse me, I had missd something in the problem. Now it's correct. Your solution does not work here.

Assuming $f$ is the identity map, define $F(s,t)=\begin{cases}tg(s/t),\quad & |s|\le t \\ s,\quad & |s|\ge t\end{cases}$


Consider a triangle $ ABC$ and let $ A_1\in (BC)$, $ B_1\in (AC)$, $ C_1\in (AB)$ s.t. $ AA_1,BB_1$ and $ CC_1$ are concurent at the point $ P$. Let $ K,L,M$ be the midpoints of $ [A_1B_1],[B_1C_1]$ and $ [C_1A_1]$, respectively. Prove that $ P$ is found in the interior of triangle $ \triangle KLM$.

[hide="Solution"]$ AA_1$ meets $ B_1C_1$ at $ A_2$. Because $ (APA_1A_2)\equal{}\minus{}1$, $ \frac{PA_1}{PA_2}\equal{}\frac{AA_1}{AA_2}>1$.
Thus, $ PA_1>\frac{A_1A_2}{2}>PA_2$, so P lies in the interior of quadrilateral $ KMB_1C_1$.
Similarly, P lies in the interior of quadrilaterals $ LKA_1C_1, MLB_1A_1$.
Hence P lies in the interior of triangle KLM. [/hide]


Find $ n\in N*$ such that
1) $ n^2\equal{}a^3\minus{}b^3$
2) $ 2n\plus{}79\equal{}c^2$
$ ( a,b,c\in N*)$

Hi all.
I have found those numbers just by computer(by programming in C++) :)


How much time it takes to read each book and how much time spent per day? Thanks.  :lol:

AoPS books are written to be like a supplement in my opinion when practicing challenging problems. IT teaches you ways how to solve some problems and introduces the basics of some subjects. At least in my volume 1 that is how it is. ANd it takes a long time to understand concepts. :)

^agrees^ I think it is best if you have the solutions manual close at hand!!! :wink:


trigonometry algebra polynomial inequalities

Prove that  $ sin\frac{\pi}{14}$  is a root of the equation $ 8x^3\minus{}4x^2\minus{}4x\plus{}1\equal{}0$

Here is what I have so far:
Let x=sin a. We need to show sin(7a)=1
From the given, we have
$ 4sin^2a\plus{}4sina\minus{}8sin^3a\equal{}1$
We need to show
$ 7sin a\minus{} 56sin^3a \plus{} 112sin^5a \minus{} 64sin^7a\equal{}1$
I ripped this from
[url]http://mathforum.org/library/drmath/view/54058.html[/url].
How can we do this?

An alternate approach is to [url=http://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html]take out the heavy machinery[/url].

Note $ \sin \pi/14 \equal{} \cos 3\pi/7$. The given polynomial is equal to $ U_3(x) \minus{} U_2(x)$. Substituting $ x \equal{} \cos \theta$ (so now we want $ 3\pi/7$ to be a root), we get $ U_3(\cos \theta) \minus{} U_2(\cos \theta)$, and now we use the trigonometric properties of Chebyshev polynomials to turn this into $ \frac {\sin 4 \theta}{\sin \theta} \minus{} \frac {\sin 3 \theta}{\sin \theta}$. (EDIT: a dumb mistake was here) This is $ 0$ at $ \theta \equal{} 3\pi/7$.

No, no mistake was made, it works out on my calculator.

Blarg, the calculator I used spit out an answer with an "e-39" tacked on at the end. In any case, I'm not sure what I was talking about, since that expression does actually evaluate to $ 0$ with $ \theta \equal{} 3\pi/7$. So let's just pretend that last line wasn't there. :D

Hint: Let $ \theta \equal{} \frac{\pi}{14}$ Then $ 7 \theta \equal{} \frac{\pi}{2}$ or $ 4 \theta \equal{} \frac{\pi}{2} \minus{} 3 \theta$
$ \Rightarrow \sin 4 \theta \equal{} \cos 3 \theta$ You get an equation in $ \sin \theta$ which is equivalent to the given equation and its obvious that hence $ \sin \frac{\pi}{14}$ is a root of this equation


Moadeleye zir ra dar adade sahih hal konid:
\[ x^2\plus{}(x\plus{}1)^2\equal{}y^4\]

belakhare yeki javab dad.see here:
http://www.mathlinks.ro/viewtopic.php?t=241526


There are 6 kids in a circle.  They all hold out their hands and randomly grabs two other hands other than their own.  It is possible for the kids to unravel theirselves without letting go of eachother.  what is the probability that everybody will be included in a chain?

hmm
this looks like we need to use [hide]graph theory[/hide] but i dont have time to do it right now/know how to do it right now :D
i'll think about it though

[hide]since every person is touching two other people, they will all be touching at least one person in common. Thus it is impossible to have a chain[/hide]

The chain he is referring to is a loop

haven't you ever played this game  :roll:

yeah I play it all the time, somewhere between recess and third period. lol.


Suppose that
\[ \frac {2x}{3} \minus{} \frac {x}{6}
\]is an integer. Which of the following statements must be true about $ x$?

$ \textbf{(A)}\ \text{It is negative.} \qquad \textbf{(B)}\ \text{It is even, but not necessarily a multiple of }3\text{.}$
$ \textbf{(C)}\ \text{It is a multiple of }3\text{, but not necessarily even.}$
$ \textbf{(D)}\ \text{It is a multiple of }6\text{, but not necessarily a multiple of }12\text{.}$
$ \textbf{(E)}\ \text{It is a multiple of }12\text{.}$

$ \frac{2x}{3}\minus{}\frac{x}{6}\equal{}\frac{4x\minus{}x}{6}\equal{}\frac{3x}{6}\equal{}\frac{x}{2}$

If $ \frac{x}{2}$ is an integer, $ x$ must obviously be even, so the answer is $ boxed{B}$.

$ \frac{2}{3}\minus{}\frac{1}{6}\equal{}\frac{1}{2}$. So $ \frac{x}{2}\equal{}a\iff x\equal{}2a$ where $ a$ is an integer. Hence it is necessary and sufficient that $ x$ is even, so the answer is $ \boxed{B}$.

[hide=solution]$\cfrac{2x}{3}-\cfrac{x}{6}=\cfrac{1}{2}x$
If $\cfrac{1}{2}x$ is an integer, then $x$ is even, but not necessarily negative, or a multiple of $3, 6,$ or $12$.
The answer is $\boxed{\textbf{(B)}}$[/hide]


an ugly solution

First test: $n^2 \equiv 0,1\pmod 4$
We obviate $11\cdots 11, 22\cdots 22, 55\cdots 55, 66\cdots 66, 99\cdots 99$ and $33\cdots 33,44\cdots 44,77\cdots 77,88\cdots 88$ are left.

Second test : $n^2 \equiv 1,4,5,6,9\pmod {10}$
Only $44\cdots 44$ remains.

if is a square, $A^2 = 44\cdots 44$, thus $4|A^2$ and hence $2|A$, say $A=2B$, thus $B^2 = 11\cdots 11$ which is impossible due to the first test.

Let's look at the array vertically. It proceeds as follows:

n
2n
3n
...
9n

Since n>9, it always stops before n (a perfect square).

Now, let's look at the first horizontal line. It's made up entirely of 1s. It's mathematically impossible to multiply two identical integers or two identical fractions that do not  equal 1 and end up with such a number. I don't know how to prove this, though. :blush:

hun?? I din't get a word of that sorry :blush:

[quote="breez"]1st row is obviously all divisible by 11, thus they all have an even number of divisors.
[/quote]
Firstly, if you take a sting of the first row with an odd number of 1's, the number is not divisible by 11. Secondly, how would being divisible by 11 be proof that a number has an even number of divisors? If it were an even power of 11, it would have an odd number of divisors. In your method, it seems like you have to prove that $11^n$ is never a string of all 1's.

[hide]
All the numbers in the 1st row can be written in the form:
$\frac{10^k-1}{9}$ where $k$ is the number of integers.
The other numbers below can be written as multiples of this number.
We can straight away rule out that $2*\frac{10^k-1}{9}$ or $6*\frac{10^k-1}{9}$ or $8*\frac{10^k-1}{9}$ or $5*\frac{10^k-1}{9}$ can be perfect squares because $\frac{10^k-1}{9}$ is not a factor of $2$ or $5$.
Can't figure out the rest.
[/hide]

[quote="Peter VDD"]yes liyi, with mods it's easy. But since this was for 10th grade max, none had ever heard of mods.

there's a simpler solution that any school kid can come up with (if they have brains though :D)[/quote]

I'm looking forward to the other solution. :)

[quote="Meghrabi"][quote="Meghrabi"]111111... can't be a square because it is always 3 mod 4.
222.... can't be a square because it is 2 mod 4

3333... cant be a square ( notice that it is equal to 3*1111...)
4444.... cant be a square because 1111... isnt a square

and so on.

Is it correct ?[/quote]

Just look at that :)[/quote]

Well, maybe there are more people thinking it is not a proof... why does bein 3*11..11 guarantee it's not a square? what with further numbers as 77..77?

Okay then

Squares always end on 0,1,4,9,6,5

So we can already cancel 22..,333..,777...,8....

111... is always 3 mod 4
444..=4*11... so 1111.. has also to be a square and it isn't.
555... is always 3 mod 4
666... is always 2 mod 4
999... is always 3 mod 4

Is it now proved ?
If so can you give the other proof ? Thanks :)

yes, that's better. ;)

I do not remember exactly what solution I had in mind when writing that... there are several ways to solve it though. I think it had to do with looking at what number it would be the square of.

Prove, none of the  numbers are a  perfect power. (except   1st power, ofcourse)

[quote=Peter]yes liyi, with mods it's easy. But since this was for 10th grade max, none had ever heard of mods.

there's a simpler solution that any school kid can come up with (if they have brains though :D)[/quote]

huh? how come mods are teached in high-school only? In Romania the mods are teached in the 5th or 6th grade. Sorry for bad english

Modular arithmetic is often not taught at all in American schools. Mostly because it does not have anything to do with elementary calculus which is what the college track American curriculum is mainly designed to prepare students for.

A mod is essentially the remainder function. For example, $2=5\mod 3 $ since $\frac53=1\ \mathcal{R}\ 2$

$11=8+3$
$111=108+3$  Not expressible in form $8m+1$

$1111=1108+3$
.
And so for next sequence bcoz an even perfect square is of form $4k$
But they are not of form..... And so follows
And an odd perfect square is of form $8m+1$
Proof
$a^2=(2k+1)^2=4k(k+1)+1=4.2m+1=8m+1$

$a^2=(2k)^2=4k^2=4m$

It remains to prove that none of \[11 \ \ 111 \ \ \ 1111 \ldots\] are perfect squares. By induction, $11$ is not a perfect square, for a number,$a$ in the sequence with $k$ $1$'s that is not a perfect square, the next number $10a+1$ is also not a perfect square since if \[10a+1=j^2 \implies 10a=(j-1)(j+1)\] but $LHS$ is divisible by $2$ as $a$ is odd and $RHS$ is divisible by $4$ as $j$ is odd which is a contradiction.

[quote=Peter]Prove that there are no perfect squares in the array below:
\[\begin{array}{cccc}11&111&1111&...\\22&222&2222&...\\33&333&3333&...\\44&444&4444&...\\55&555&5555&... \\66&666&6666&...\\77&777&7777&...\\88&888&8888&...\\99&999&9999&...\end{array}\][/quote]

You can use Modular Arithmetic. For example start with mod 4 and you know all perfect squares are 0,1. mod 32 might be optimal though.

Let $S_n$ denote the infinite set of all positive integers greater than 10 whose only distinct digit is $n$ for all positive integers $n$ from 1 to 9, inclusive. Then note that $S_4$ and $S_9$ contain perfect squares iff $S_1$ contains a perfect square. Similarly, $S_8$ contains a perfect square iff $S_2$ contains a perfect square. Therefore, we just have to prove that none of $S_1, S_2, S_3, S_5, S_6,$ or $S_7$ contain a perfect square. Since all perfect squares end in one of 1, 4, 5, 6, 9, or 0, we now just have to show that none of $S_1, S_5,$ or $S_6,$ contain a perfect square. It's not hard to show that all perfect squares ending in 5 have tens digit 2, so we know that none of the elements of $S_5$ is a perfect square, since all elements of $S_5$ have tens digit 5. Also, we know that each element in $S_6$ has exactly one 2 in its prime factorization, which means that these are also not perfect squares. This means that we just have to prove none of the elements of $S_1$ is a perfect square. We can do this because all of the elements in $S_1$ end in 11, meaning that they are 3 more than a multiple of 4. All squares are either multiples of 4 or are 1 more than a multiple of 4, meaning that we have proven that none of the elements in $S_1$ are squares. Thus, we have proven the original statement. $QED$

[quote=Peter]Prove that there are no perfect squares in the array below:
\[\begin{array}{cccc}11&111&1111&...\\22&222&2222&...\\33&333&3333&...\\44&444&4444&...\\55&555&5555&... \\66&666&6666&...\\77&777&7777&...\\88&888&8888&...\\99&999&9999&...\end{array}\][/quote]

firstly we put 2s 3s 7s and 8s out of the game since no square has last number a 2,3,7,8
Then we work with $\frac{A(10^n-1)}{9} = X^2$ which suffices $\frac{A(10^n-1)}{9}$ to be a square there we check the case
 i) $A=1,4,9$ where $10^n-1$ is a square
$10^n= k^2+1$ so for $2\geq n$ no solution, so by $n\geq 3$      $8$ should divide $k^2+1$ which can't happen for any k which is a positive integer
ii) we now observe the cases of 5 and 6
For 5
if $5/X^2$ then $5^2/X^2$ and so $5/10^n-1$ impossible since 5 has to divide 1 which is not true
For 6
2 should divide $10^n-1$ which is also impossible and so we are done

[hide = "short and neat"]
We begin by proving that no square number can end with $11, 22, 33 \hdots 99$.

Let a square number end with $ab$ (the first digits don't contribute). Thus, $(10a + b)^2 = 100a^2 + 20ab + b^2 \rightarrow 20ab + b^2$. Clearly, since $k^2$ ends with $1,4,5,6,9$, we only need to consider $11,44,55,66,99$. Checking these, we find that only $44$ work. However, we can always write it as $4(11111\hdots 1111)$. Since $11111\hdots 1111$ is not a square by the earlier proof (ending w/11), $44$ doesn't work either and thus there are no squares in the number array.
[/hide]

[quote=breez]1st row is obviously all divisible by 11, thus they all have an even number of divisors.
2nd row's sum of digits all equal a multiple of three, thus they are all divisible by 3 and have an even number of divisors.
3rd row's alternating digits are all 0, so it is divisible by 11, once again, an even number divisors.

Perfect squares have an odd number of divisors.[/quote]

no
111 is not divisible by 11

We note that $2,3$ are non-quadratic residues of $4$ as well as $5$.

First row can't be a square as it is $3\pmod4$.

Second row can't be a square as it is $2\pmod4$.

Third row can't be a square as it is $3\pmod5$.

Fourth row can't be a square as it is $4$ times a non-perfect square.

Fifth row can't be a perfect square as it is $3\pmod4$.

Sixth row can't be a perfect square as it is $2\pmod4$.

Seventh row can't be a perfect square as it is $2\pmod5$.

Eighth row can't be a perfect square as it is $3\pmod5$.

Ninth row can't be a perfect square as it is $3\pmod4$.

[quote=mattjunior]there just aren't
believe me
i'm in 6th grade and i came across this problem and i kept grinding
i never found a perfect square[/quote]

why the 6th grade part?
also no offense but we really don't need your input on this if you're just going to grind
to be honesty modular arithmetic is the absolute go-to solution to this :)

mathmax12's solution (which he told me to post here):
 The first row can't be a square because it is $3\pmod{4}$,the second row can't be a square because it is $2\pmod{4}$, the third row can't be a square because it is $3\pmod{5}$, the fourth row can't  be a square because it is $4$ times a non perfect square, the fifth row can't be a square because it is $3\pmod{4}$, the sixth row can't be a square because it is $2\pmod{4}$, the seventh row can't be a square because it is $2\pmod{5}$, the eithg row can't be a square because it is $3\pmod{5}$, the ninth row can't be a square because it is $3\pmod{4}$, hence, proved.


algebra
polynomial
quadratics
function
Finite Differences

Compute the value of the expression
\[ 2009^4 \minus{} 4 \times 2007^4 \plus{} 6 \times 2005^4 \minus{} 4 \times 2003^4 \plus{} 2001^4 \, .\]

[hide="Brute Force"]We can write the given expression as:

$ \sum\limits_{i \equal{} 0}^4 (2009 \minus{} 2n)^4 \binom{4}{n} ( \minus{} 1)^n$

Expanding $ (2009 \minus{} 2n)^4$ especially and adding and noticing a lot of stuff cancels (line it up in order of coefficients) will give the answer (but it takes a while).[/hide]

Is there a non-brute force way to do it?

[quote="Ravi B"]Kouichi's method is pretty reasonable.  His key idea is to center on 2005, not 2009.

The quickest way I know of is to use finite differences, which lets you solve the problem in your head.  Anyone want to give it a try?[/quote]
I'm interested in this particular solution. Anyone care to elaborate a bit? :)

EDIT: [hide="My guess example"]$9^4-4\cdot7^4+6\cdot5^4-4\cdot3^4+1^4=(9^4-3\cdot7^4+3\cdot5^4-3^4)-(7^4-3\cdot5^4+3\cdot3^4-1^4)?$[/hide]

Sonnhard is implicitly assuming that the polynomial on the left-hand side of his equation is a constant (which is true).  With that approach, we may as well set $a = 0$, which simplifies the calculation to
\[
  4^4  - 4 \cdot 2^4 + 0 - 4 \cdot (-2)^4 + (-4)^4
    = 256 - 64 + 0 - 64 + 256
    = \boxed{384}.
\]

Here is the finite-differences solution I had in mind:
[hide="Finite-Differences Solution"]
Let $f(x) = {(2x + 1)}^4$.  Then our problem is to evaluate
\[
  f(1004) - 4 f(1003) + 6 f(1002) - 4 f(1001) + f(1000).
\]
That expression is the fourth finite difference of $f$ evaluated at $1000$.

Note that $f$ itself is a quartic polynomial with leading coefficient $16$.  So the first difference of $f$ is a cubic with leading coefficient $64$.  The second difference is a quadratic with leading coefficient $192$.  The third difference is a linear function with leading coefficient $384$.  The fourth difference is the constant $\boxed{384}$.
[/hide]

Finite differences also proves that Sonnhard's polynomial is a constant, and so then we can substitute $a = 0$.

[hide="Solution"]We let $x=2005$. Thus we can write,
\begin{align*}(x+4)^4 - 4(x+2)^4 + 6x^4 -4(x-2)^4 + (x-4)^4\end{align*}
Simplification of the first and last terms leaves us with $2x^4+192x^2+512$. Similarily, simplification of the second and fourth terms gives $-8x^4 - 192x^2-128$. The answer is
\[2x^4+192x^2+512 - 8x^4 - 192x^2-128 +6x^4 = \boxed{384}\]  [/hide]


For $a,b,c$ positive reals prove that

$\sum \frac{a}{(1+b)(1+c)}\geq\frac{3abc}{(1+abc)^{2}}$

Sorry, but just now I remember it is like USA inequality I proposed many year ago.

As $\sum \frac{a}{(1+b)(1+c)}\geq \sum \frac{a}{(1+a)(1+b)}$

it follows from the solution of USA ineq you can find in the forum.


modular arithmetic
number theory unsolved
number theory

Given $ r\geq 3$, find the minimum of positive integer d such that
$ 7^{d}\equiv 1(\mod 2^{r})$

If $ (a, m) = 1$ then $ a^{\phi(m)}\equiv 1 \pmod m$, and we are looking for the multiplicative order.

As a well-known fact, it's $ d = 2^{r-2}$ (remember we're admitting $ r \ge 3$). This is because $ v_{2}(7-1) = 1$, $ v_{2}(7^{2}-1) = 3$ and $ v_{2}(7^{2^{n+1}q}-1) = v_{2}(7^{2^{n+1}}-1) = v_{2}(7^{2^{n}}-1)+1$, whenever $ n, q \in \mathbb{N}$ and $ q$ is odd. Here as in fashion, $ v_{2}(\cdot)$ is a dyadic valuation and $ \mathbb{N}= \{1, 2, \ldots\}$.


Prove that there are infinitely many triplets $ (a,b,c)$ that satisfy the following equalities:

$ \dfrac{2a\minus{}b\plus{}6}{4a\plus{}c\plus{}2}\equal{}\dfrac{b\minus{}2c}{a\minus{}c}\equal{}\dfrac{2a\plus{}b\plus{}2c\minus{}2}{6a\plus{}2c\minus{}2}$

[hide="Solution"]A first attempt to solve this problem would be to set all those fractions equal to a number $ k\in\mathbb{R}$ and solving the system of equations. One would get $ 3(5k\minus{}2)(k\minus{}2)a \equal{} (5\minus{}3k)(k\minus{}2)$. The first observation is that if $ k \equal{} 2$, this equation is always satisfied and so we might suspect that in the case $ k\equal{}2$ there are infinitely many solutions. Solving with $ k \equal{} 2$, indeed, gives the solutions $ (a,b,c) \equal{} (x,2x,1\minus{}4x)$ for all $ x\in\mathbb{R}$.[/hide]


Let's just say (   ) is a combination symbol. sry no latex

I know that  ( n ) =  n(n-1)/2    using pascal's identity
                      2

but what do you do it is like ( n )    ( n ) .........  ??????
                                               3         4

and is there any shortcuts so i dont need to go through all the calculations?

sorry the letter should be like

( n )
( 2 )
and
( n )
( 3 )

( n )
( 4 )

$\binom{n}{k}=\frac{n!}{k!(n-k)!}$

For example, $\binom{n}{4}=\frac{n!}{4!(n-4)!}=\frac{n(n-1)(n-2)(n-3)}{4!}$ (make sure you understand that last step).

Please learn to use $\LaTeX$.  It is a heck of a lot easier.  Or use common notation such as C(n,k) for binomial coefficients.


projective geometry
ARML
Duke
college
trigonometry

yeah but whatever we have UNC so we pwn at bball, it's all good

[quote="mysmartmouth"]It seems official that NC is the best state in the country at math:

East Chapel Hill
Enloe
CMC
etc...

How many teams from NC were in the Top 10?  Do we know?

(They still are the worst at football though) (besides Hawaii)[/quote]

Yes, because whatever you say must be official  :moose:  You'll also have to define what you mean by the "best".  Many states aren't able to draw their full talent to ARML for various reasons.  For example, if AAST let their students join the other students from NJ for ARML the team would be even better.  Probably the best counter-example to what you said is California.  They have a ton of teams and if they put together their best team they also would have done better.

He meant because we won ARML I think. You can certainly take that as official since it is the only competition between states.

Not to say we definitely are the best state :maybe: .

[quote]Arnav, which problem(s) did you miss at Duke? We used the 9-point circle on one of the proofs (4c?) did you guys?[/quote]

Hey I solved 3 and 4 for East.  How'd you use the 9-point circle... It didn't really seem relevant?  To prove that the 6 points are cyclic I just noted that there are 3 really obvious cyclic quads due to the right angles.  If the circles surrounding those quads aren't all the same then the 3 radical axes are the sides of the triangle, which don't concur ---><---.   That sort of knocked out 3+4 at once.   You could also do 3b by Desargues but then you don't knock out 4 immediately.

Uh I'm pretty sure every single person on the East team put 0 for #7 because we are all idiots.  Other than that Arnav didn't miss anything...

Yeah, I put 0 for 7 too...

NOTE THAT POSITIVE+POSITIVE+POSITIVE IS POSITIVE!!!!!

It was crappy...

And on 8, which I should've got, I thought way too long before I found the right way of doing it. Waaaaaaa

How do you do #6 though?

we are not idiots

I was simply being logical and working in $\mathbb{R}/\pi\mathbb{Z}$, - it

I took the tangent and got x = arctan(tan(arctan(-3)+arctan(3))) = arctan(tan(0)) and I put 0... (Actually, I forgot about the arctan but that's ok)

[quote="MysticTerminator"]dude, no[/quote]

i got someone else to tell me it  :wink:

I7:

Find $\arctan 1+\arctan 2+\arctan 3$

If I recall correctly, Arnav missed the trig question last year, too.

he does

I AM SO MAD I DIDN'T GET IT

Also, [size=200]how do you do #6?????[/size]

Powers of 3... 3, 9, 27, 81 (5 possibilities)
Powers of 5... 5, 25, 125 (4 possibilities)
Powers of 7... 7, 49 (3 possibilities)
Powers of 11... 11, 121 (3 possibilities)
26 other primes ($2^{26}$ possibilities)

$2^{26}3^{2}2^{2}5=2^{28}3^{2}5^{1}$, so the answer is 31.

You obviously can't have 2s and they can be in increasing order in one way. Apparently, 1 isn't a power of a prime (I did it that way anyway but it's quite vague).

For the happy sequence problem...

[hide="HINT"]For each prime p such that 2<p<128 we may select either 0 or 1 elements of {p^k|p^k<128} to be in the sequence.[/hide]

[quote="not_trig"]the "happy sequence" crap

and he does... you missed #8 last year[/quote]

what the -?

WTFFFFFFFF.

I missed that problem because 1 is a - positive power of a prime.  Watch this:

$2^{0}= 1$

QEFD

I'm still curious as to what the 9-point circle has to do with anything

dude I do not believe them at all

they are wily south carolinians with their wily ways

I thought that in general, the test was a good test.

I was surprised at how well Winston-Salem did, when we got 8th (21 power, 16 team, 15 relay, 13 individual).

I personally only got a 5 because I decided that one was a positive power of a prime so I missed 6, then I did the same thing as not_trig on 7 (and evidently the East team).

However, the contest ran very smoothly and was quite fun  :lol:  :lol:

I guess I was like, the only one who compeltely misinterpreted #6

how do you find out your team's score spread

You must have the password for your team (your coach will have it) and then go to the website.

Then, you have access to everything.

haha for the happy sequence i misread the problem and thought it was the number of sets you can make from the set of 31 primes which is $2^{31}$  :rotfl:

[quote="Elemennop"]I guess I was like, the only one who compeltely misinterpreted #6
[/quote]

No I did that too... if I had read it right, it would have been TRIVIAL!!!!!

 :cursing:

NYEH

And #8 - was it 34

Heh... I guessed it right on this one. I knew that 31 had to do with something, so I put 31.  :P


parameterization
algebra unsolved
algebra

For a real parameter $a$, solve the equation $x^{4}-2ax^{2}+x+a^{2}-a=0$.

Find all $a$ for which all solutions are real.


Suppose that several number of lines, any three of which don't intersect at a point, are drawn on a paper. If we know that the total number of intersection points is 17, then what is the minimum number of lines that makes these intersections possible? Why? Is there a good, working procedure that allows us to solve general questions of this type?

[quote="OHO"][hide]I got one possible situation, but I don't know what the maximum.
7 lines which can divide into 4 groups of 3,2,1,1 lines such that the lines in each group are parallel to each other.[/hide][/quote]
[hide]Yeah, that's the minimum. For a procedure, you could count the maximum number of intersection points for each number of lines, and choose the first one that is bigger than your number. For example, we want 17 intersection points for this problem. The maximum for 1 line is 0 intersection points, 2 lines is 1, 3 lines is 3, 4 lines is 6, and so on. For 6 lines it is 15 and for 7 lines it is 21, so 7 lines is the minimum number of lines needed.[/hide]

Oh! I thought that we have to find the maximun. :oops:
Anyway, what's the maximun number of lines that can cause 17 intersetion? :huh:

18. 17 parallel lines, and one line crossing them all

Unless we get into 3 dimensions :rotfl:

Then, the answer is......infinity!!!!!!!!![/i]


abstract algebra
induction
superior algebra
superior algebra solved

let M,R-Module is noetherian . prove R/Ann(M)is noetherian .

$M$ is finitely generated, hence there exist $x_1,\ldots,x_n\in M$ that generate $M$ as an $R$ module.
For all $i=\overline{1,n}$, let $J_i=(0: x_i)$. $J_i$ is an ideal of $R$.
We have $AnnM=(0: M)=(0: \sum_{i=1}^nRx_i)=\cap_{i=1}^n(0: Rx_i)=\cap_{i=1}^n(0: x_i)=\cap_{i=1}^nJ_i$.
We have to prove that $\frac{R}{\cap_{i=1}^nJ_i}$ is noetherian.

For all $x_i$, we have an $R$-module homomorphism $f_i: R\to M$ such that $f_i(r)=rx_i$. Its kernel is $(0: x_i)=J_i$, hence there exist an injective $R$-module homomorphism $\varphi_i: R/J_i\to M$ sendind $r+J_i$ to $rx_i$ for all $r\in R$. This means that $R/J_i$ identifies with a submodule of $M$. Since $M$ is noetherian, every submodule is noetherian, hence $R/J_i$ is noetherian for all $i=\overline{1,n}$.

Therefore it is enough to solve this problem:
Let $R$ be a ring and let $I_i$ be ideals of $R$, $i=\overline{1,n}$, such that $R/I_i$ is noetherian for all $i$. Prove that $\frac{R}{\cap_{i=1}^nI_i}$ is noetherian.

We proced by induction on $n$. For $n=1$, there is nothing to prove.
For $n=2$, we have \[ \frac{I_1}{I_1\cap I_2}\simeq\frac{I_1+I_2}{I_2}. \] $\frac{I_1+I_2}{I_2}$ is noetherian being a submodule of the noetherian $R$-module $R/I_2$. Hence $\frac{I_1}{I_1\cap I_2}$ is noetherian. We have a short exact sequence of $R$-modules
\[ 0\to\frac{I_1}{I_1\cap I_2}\to\frac{R}{I_1\cap I_2}\to\frac{\frac{R}{I_1\cap I_2}}{\frac{I_1}{I_1\cap I_2}}\to 0.  \]
$\frac{\frac{R}{I_1\cap I_2}}{\frac{I_1}{I_1\cap I_2}}\simeq\frac{R}{I_1}$.
We know that if we have a short exact sequence of $R$-modules $0\to E\to F\to G\to 0$ such that $E$ and $G$ are noetherian, then $F$ is noetherian. The conclusion follows.

Assume we have solved the problem for $n$ and we want to solve it for $n+1$.
By the induction hypothesis $\frac{R}{\cap_{i=1}^nI_i}$ is noetherian. By the case $n=2$, $\frac{R}{\cap_{i=1}^{n+1}I_i}$ is noetherian.


geometry
ratio
circumcircle
geometry proposed

Let be given a triangle $ABC$ with incircle $(I)$ and the altitude $AH$.$(I)$ touches $BC$ at $D$,.Let $M$ be the midpoint of $AH$,$L$ be the intersection of the line $DM$ and the circle $(I)$($L\neq D$).Prove that $\widehat{BLD}=\widehat{CLD}$

If the incircle touches the sides $AB$, $AC$ in the points $F$, $E$ respectively and $S\in BC\cap EF$, then must prove $LS\perp LD$, i.e. the point $L$ belongs to the Appolonius' circle for the side $BC$ and the ratio $\frac{p-b}{p-c}$ (the circle with the diameter $SD$).

Here is my solution (which is based on a problem in ISL2003):
 [i]Lemma 1[/i]: Let $(T)$ be a circle and $BC$ be a chord. Suppose $D$ is a point lying on the segment $BC$. Let $(I)$ be the circle that touches $(T)$ at $L$ and touches $BC$ at $D$. Then $LD$ is the bisector of $\measuredangle{BLC}$.
 [i]Lemma 2[/i]: $LD$ passes through the A-excenter $I_a$.
 Back to the problem, we will show that the circumcircle of $BLC$ touches $(I)$ at $L$. If we can, by applying the lemma 1, we will obtain $\measuredangle{BLD} = \measuredangle{DLC}$.
 Let $E$ be the intersection of the tangent of $T$ at $L$ and $BC$. Then $IE$ is perpendicular to $LD$ at $N$. Hence $EN.EI = ED^2 =EL^2$. [b](1)[/b]
 Since $LD$ passes through $I_a$ (lemma 1), we obtain that $\measuredangle{I_aNI} = 90^0 = \measuredangle{I_aBI}= \measuredangle{I_aCI}$. Therefore, $B, N, I, C, I_a$ lie on a circle. This implies that $EN.EI = EB.EC$. [b](2)[/b]
 From [b](1)[/b] and [b](2)[/b], we obtain that $EL^2 = EB.EC$. Hence $EL$ touches the circumcircle $BLC$ at $L$. Therefore, $(I)$ touches $(BLC)$ at $L$. And the problem is solved.


algebra
polynomial
Rational Root Theorem

Find $ x \in R$ of this equation: $ 16x^4\plus{}8x^3\minus{}12x^2\minus{}2x\plus{}3\equal{}0$

I'm using newton's method...

it appears one solution is $ x\equal{}\minus{}.9285308364$ or $ x\equal{}\minus{}.6326154648$.  The graph makes it clear that these are the only real solutions.

The equation was calculated by programs.
I found it.But,I couldn't solve it by hand because it was so difficult and complex.

By rational root theorem, if $ x$ is rational it's numerator is a factor of $ 3$ and its denominator is a factor of $ 16$.  None of these values work.

So now we will try to factor the quartic into two quadratics.  First, divide the polynomial through by $ 16$.  Then we have

$ (x^2\plus{}bx\plus{}c)(x^2\plus{}ex\plus{}f) \equal{} x^4\plus{}1/2x^3\minus{}3/4x^2\minus{}1/8x\plus{}3/16$
$ x^4 \plus{}(e\plus{}b)x^3 \plus{} (f\plus{}c\plus{}be)x^2 \plus{} (bf\plus{}ce)x \plus{} cf \equal{} x^4\plus{}1/2x^3\minus{}3/4x^2\minus{}1/8x\plus{}3/16$

Equating coefficients:
$ e\plus{}b \equal{} 1/2$
$ f\plus{}be\plus{}c \equal{} \minus{}3/4$
$ bf\plus{}ce\equal{}\minus{}1/8$
$ cf \equal{} 3/16$

Now lets use the substitutions $ e\equal{}1/2\minus{}b$ and $ f\equal{}\frac{3}{16c}$.  Then we have:
$ 3/(16c)\plus{}b(1/2\minus{}b)\plus{}c \equal{} \minus{}3/4$
$ 3b/(16c)\plus{}c(1/2\minus{}b)\equal{}\minus{}1/8$

Using the first equation:
$ 3\plus{}8bc\minus{}16b^2c \plus{} 16c^2 \equal{} \minus{}12c$
$ 16c^2 \plus{} (8b\plus{}12)c \plus{} 3\minus{}16b^2 \equal{}0$
$ c \equal{} \frac{\minus{}8b\minus{}12 \pm \sqrt{(8b\plus{}12)^2\minus{}4(16)(3\minus{}16b^2)}}{32}$
$ c\equal{} \frac{\minus{}8b\minus{}12 \pm \sqrt{1088b^2\plus{}192b\minus{}48}}{32}$
$ c\equal{}\frac{\minus{}2b\minus{}3 \pm \sqrt{68b^2\plus{}12b\minus{}3}}{8}$

Now from the 2nd equation:

$ 3b \plus{} (8\minus{}16b)c^2 \plus{}2c \equal{} 0$
...

I don't think this method is getting me anywhere.  But it works on some 4th degrees. Anyway, mathematica tells me this gives a 6th degree polynomial to solve.  sucks.

[quote="pythag011"][quote="123456789"]What is the point of this? Mathematica yields enormous answers...[/quote]

There should be no enormous answers. If they are enormous, the 16x^4 term will be much bigger than all the orhter terms, so it won't be a solution.[/quote]

I meant enormous as in lots of numbers and square root signs.


function
inequalities
algebra unsolved
algebra

Let f(x) be a continuos function defined on [0,1] such that:
i) f(0)=f(1)=0
ii) 2f(x)+f(y)=3f(${{2x+y\over 3}}$)
Prove that f(x)=0 for all $x,y \in \mathbb[0,1]$

Since ${2\over 3}f(x)+{1\over 3}f(y)=f\left({2\over 3}x+{1\over 3}y\right)$, then by Jensen the function is neither concave nor convex, meaning it's linear, giving $f(x)=ax+b$. From $0=f(0)=b$ and $0=f(1)=a+b$ we get $a=b=0\iff f(x)\equiv 0$.

You need continuity at the very least to get from $af(x)+(1-a)f(y)=f(ax+(1-a)y)\implies f$ linear for [i]only one [/i]$a$.

OK then, we can reduce it to Cauchy equation:

For $x=0$ we get ${1\over 3}f(y)=f\left({1\over 3}y\right)$.

For $y=0$ we get ${2\over 3}f(x)=f\left({2\over 3}x\right)$.

Hence the equation becomes $f\left({2\over 3}x\right)+f\left({1\over 3}y\right)=f\left({2\over 3}x+{1\over 3}y\right)$, which is Cauchy equation, whose only continuous solution is linear function. The conclusion follows.

we can easly check that  F(a/3^n)=0.then let A={a/3^n \n  natural number}.we can prove that A is dense in [0.1].(prove like prove densite Q in R).then f(x)=0 because  f is continuos.azbi :ninja:


inequalities
trigonometry
inequalities proposed

Let $a,b,c$ be length of the sides of triangle. Prove that:
$(t+1)(a^3b+b^3c+c^3a) \geq ab^3+bc^3+ca^3+tabc(a+b+c)$
for $\forall t \geq 1$

Hint: Put $a=y+z,b=x+z, c=x+y$ then we get $x,y,z>0$ and ... :D

I have a non-brute force solution.
\[
\begin{gathered}
  (t + 1)\left( {\sum {a^3 b} } \right) \hfill \\
   = (t + 1)\left( {\sum {ab(b^2  + c^2  - 2bc\cos A)} } \right) \hfill \\
   = (t + 1)\left( {\sum {ab^3 }  + \sum {abc(a + b + c) - 2abc\left( {\sum {a\cos A} } \right)} } \right) \hfill \\
   = (t + 1)\left( {\sum {ab^3 }  + \sum {abc(a + b + c) - 2abc\left( {\sum {\frac{{a^2 b^2  + a^2 c^2  - a^4 }}
{{2abc}}} } \right)} } \right) \hfill \\
   = (t + 1)\left( {\sum {ab^3 }  + \sum {abc(a + b + c) - 2abc\left( {\sum {\frac{{a^2 b^2  + a^2 c^2  - a^4 }}
{{2abc}}} } \right)} } \right) \hfill \\
   = (t + 1)\left( {\sum {ab^3 }  + \sum {abc(a + b + c) - 2\left( {\sum {a^2 b^2 } } \right) + \sum {a^4 } } } \right) \hfill \\
\end{gathered}
\]
Now it remains to prove
\[
t\sum {ab^3 }  + abc(a + b + c) + (t + 1)\sum {a^4 }  \geqslant 2(t + 1)\left( {\sum {a^2 b^2 } } \right)
\]
which is easy, as it's only the sum of
\[
\begin{gathered}
  \frac{t}
{2}\sum {a^4 }  + t\sum {ab^3 }  \geqslant \frac{{3t}}
{2}\sum {a^2 b^2 }  \hfill \\
  \sum {a^4 }  + abc(a + b + c) \geqslant 2\sum {a^2 b^2 }  \hfill \\
  \frac{t}
{2}\sum {a^4 }  \geqslant \frac{t}
{2}\sum {a^2 b^2 }  \hfill \\
\end{gathered}
\]
which is amgm and schur.

[quote="Vasc"]$\sum{a \cos A}$
or
$\sum{b \cos A}$ ?[/quote]

Oh, I am wrong.... :(

[quote="keira_khtn"]My proof:
We need prove:
[tex]ab[(t+1)a^2-tc^2-b^2)+bc[(t+1)b^2-ta^2-c^2]+ca[(t+1)c^2-tb^2-a^2]\geq\ 0[/tex];
Assume [tex]a=max{a,b,c}[/tex],there fore:[tex]ca\geq\ bc [/tex]
[tex]L.H.S\geq\ ab[(t+1)a^2-tc^2-b^2)+bc[tc^2+b^2-(t+1)a^2]=(ab-bc)[(t+1)a^2-tc^2-b^2]=b(a-c))[(t+1)a^2-tc^2-b^2]\geq\ 0[/tex] (since [tex]a\geq\ b,c[/tex])[/quote]

I am sorry, but this step has some problem:
[tex]L.H.S\geq\ ab[(t+1)a^2-tc^2-b^2)+bc[tc^2+b^2-(t+1)a^2][/tex]

For this to be true, just [tex]ca\geq\ bc [/tex] is not sufficient, u also need to have $(t+1)c^2-tb^2-a^2 \geq 0$. Indeed, if $(t+1)c^2-tb^2-a^2 <0$, then $ca[(t+1)c^2-tb^2-a^2]\leq bc[(t+1)c^2-tb^2-a^2]$ and it would become
[tex]L.H.S \leq ab[(t+1)a^2-tc^2-b^2)+bc[tc^2+b^2-(t+1)a^2][/tex]

Sorry ,I'm wrong...
But i think we can prove it with condition:a,b,c are lengths of a triangle.

Okay, I have a proof without Ravi's substitution. But it's still a big algebraic mayhem. Anyway, for those who are interested, let's go:

[color=blue][b]Lemma 1.[/b] If x and y are two nonnegative numbers, then $x^3+3y^3-2x^2y-y^2x\geq 0$.[/color]

[i]Proof of Lemma 1.[/i] If $x\leq y$, then also $x\leq 2y$, so that $x-y\leq 0$ and $x-2y\leq 0$, and hence $\left(x-y\right)\left(x-2y\right)\geq 0$, so that

$x^3+3y^3-2x^2y-y^2x=\left(x-y\right)\left(x-2y\right)\left(x+y\right)+y^3\geq 0$.

If $2y\leq x$, then also $y\leq x$, so that $x-2y\geq 0$ and $x-y\geq 0$, and thus

$x^3+3y^3-2x^2y-y^2x=\left(x-y\right)\left(x-2y\right)\left(x+y\right)+y^3\geq 0$.

Hence, the only case remaining to consider is the case $y\leq x\leq 2y$. But this case is not too difficult either: We have $\left(x-y\right)\left(x-2y\right)\geq -\frac14y^2$ (since $\left(x-y\right)\left(x-2y\right)-\left(-\frac14y^2\right)=\frac14\left(2x-3y\right)^2\geq 0$), and, of course, $\left(x-y\right)\left(x-2y\right)$ is $\leq 0$ (since $y\leq x$ yields $x-y\geq 0$, while $x\leq 2y$ yields $x-2y\leq 0$). On the other hand, $x\leq 2y$ yields $x+y\leq 2y+y=3y$, and of course, we have $x+y\geq 0$. Thus, $\left(x-y\right)\left(x-2y\right)\left(x+y\right)\geq -\frac14y^2\cdot 3y=-\frac34y^3$. Hence,

$x^3+3y^3-2x^2y-y^2x=\left(x-y\right)\left(x-2y\right)\left(x+y\right)+y^3\geq -\frac34y^3+y^3=\frac14y^3\geq 0$.

Thus, we have $x^3+3y^3-2x^2y-y^2x\geq 0$ in all cases, and Lemma 1 is proven.

[color=blue][b]Lemma 2.[/b] If a, b, c are the sidelengths of a triangle, then

$2a^3b+2b^3c+2c^3a-b^3a-c^3b-a^3c-a^2bc-b^2ca-c^2ab$
$=b\left(a+b\right)\left(a-b\right)\left(a-c\right)+c\left(b+c\right)\left(b-c\right)\left(b-a\right)+a\left(c+a\right)\left(c-a\right)\left(c-b\right)$
$\geq 0$.[/color]

[i]Proof of Lemma 2.[/i] The identity

$2a^3b+2b^3c+2c^3a-b^3a-c^3b-a^3c-a^2bc-b^2ca-c^2ab$
$=b\left(a+b\right)\left(a-b\right)\left(a-c\right)+c\left(b+c\right)\left(b-c\right)\left(b-a\right)+a\left(c+a\right)\left(c-a\right)\left(c-b\right)$

can be checked by straightforward computation; thus, it will be enough to prove the inequality

$2a^3b+2b^3c+2c^3a-b^3a-c^3b-a^3c-a^2bc-b^2ca-c^2ab\geq 0$.

Since this inequality is cyclic, we can WLOG assume that a is the greatest among the numbers a, b, c. Then, $a\geq b$ and $a\geq c$, so that the numbers x = a - b and y = a - c are nonnegative. Hence, Lemma 1, applied to the numbers x and y, yields $x^3+3y^3-2x^2y-y^2x\geq 0$. Also, b + c - a > 0 (since b + c > a, because a, b, c are the sidelengths of a triangle), and $a^2+b^2+c^2-bc-ca-ab\geq 0$  (since

$a^2+b^2+c^2-bc-ca-ab=\frac12\left(\left(b-c\right)^2+\left(c-a\right)^2+\left(a-b\right)^2\right)$).

Now, after some algebra,

$2a^3b+2b^3c+2c^3a-b^3a-c^3b-a^3c-a^2bc-b^2ca-c^2ab$
$=2\left(b+c-a\right)^2\left(a^2+b^2+c^2-bc-ca-ab\right)+\left(b+c-a\right)\left(3y^3+x\left(y^2+xy+3\left(x-y\right)^2\right)\right)$
$+y^4+x\left(y^3+\left(x^3+3y^3-2x^2y-y^2x\right)\right)\geq 0$.

Thus, Lemma 2 is proven.

Finally, the problem asks us to prove the following theorem:

[color=blue][b]Theorem 3.[/b] If a, b, c are the sidelengths of a triangle, and $t\geq 1$ is a real number, then

$\left(t+1\right)\left(a^3b+b^3c+c^3a\right)\geq b^3a+c^3b+a^3c+tabc\left(a+b+c\right)$.[/color]

[i]Proof of Theorem 3.[/i] By Lemma 2, we have

$2a^3b+2b^3c+2c^3a-b^3a-c^3b-a^3c-a^2bc-b^2ca-c^2ab\geq 0$.

By the IMO 1983 problem 6 (see http://www.kalva.demon.co.uk/imo/isoln/isoln836.html or http://www.mathlinks.ro/Forum/viewtopic.php?t=34606 or http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 posts #8-#10), we have

$a^2b\left(a-b\right)+b^2c\left(b-c\right)+c^2a\left(c-a\right)\geq 0$.

Finally, $t\geq 1$ implies $t-1\geq 0$. Thus,

$\left(t+1\right)\left(a^3b+b^3c+c^3a\right)-\left(b^3a+c^3b+a^3c+tabc\left(a+b+c\right)\right)$
$=\left(t-1\right)\left(\left(a^2b\left(a-b\right)+b^2c\left(b-c\right)+c^2a\left(c-a\right)\right)+\frac12\left(\left(ca-ab\right)^2+\left(ab-bc\right)^2+\left(bc-ca\right)^2\right)\right)$
$+\left(2a^3b+2b^3c+2c^3a-b^3a-c^3b-a^3c-a^2bc-b^2ca-c^2ab\right)\geq 0$,

so that $\left(t+1\right)\left(a^3b+b^3c+c^3a\right)\geq b^3a+c^3b+a^3c+tabc\left(a+b+c\right)$, and Theorem 3 is proven.

  Darij

We can prove this in following way..

$WLOG$ $Let$ $a\geq b\geq c$

$t(\sum_{cy} a^3 b - abc \sum_{cy} a )+ (\sum_{cy} a^3 b - \sum_{cy} ab^3 ) \geq 0 $

$\Leftrightarrow t(b(a-b)(a^2 - bc)+a(b-c)(ab-c^2 )) + a(a-b)(ab-c^2)+c(b-c)(b^2 -ac) \geq 0$

Here, We know

$b(a-b)(a^2 - bc) \geq 0$ and $ta(b-c)(ab-c^2 ) \geq a(b-c)(ab-c^2 )$

so, It is enough to prove that

$ a(b-c)(ab-c^2 ) +a(a-b)(ab-c^2) + c(b-c)(b^2 - ac) \geq 0$

$\Leftrightarrow a(a-c)(ab-c^2 ) + c(b-c)(b^2 -ac) \geq 0$

But this is true cuz

$ |a| \geq |c| $ , $|a-c| \geq |b-c| $, $|ab-c^2 | \geq |b^2 - ac|$

and, $a, a-c, ab-c^2 \geq 0$

$Q.E.D$

It can be prove similar way.

if $a\geq c\geq b$

ineq can be arranged as $t(c(c-b)(ac-b^2 )+ b(a-c)(a^2 - bc) ) + a (a-c)(ab-c^2 ) + b(c-b)(a^2 -bc) \geq 0$

and from same reason as above

$c(c-b)(ac-b^2 ) \geq 0$ and $ tb(a-c)(a^2 - bc)\geq b(a-c)(a^2 -bc)$

it is enought to prove that

$b(a-c)(a^2 -bc)+ b(c-b)(a^2 - bc) + a (a-c)(ab-c^2 ) \geq 0$

$\Leftrightarrow b(a-b)(a^2 - bc) + a(a-c)(ab-c^2 ) \geq 0$

We can let $b=b, c=b+x, a=b+x+y $ , $x, y\geq 0$ then

$\Leftrightarrow b^2x^2 + 2xb^2 y + bx^3 + bx^2 y + 3bxy^2 +3b^2y^2 +2by^3 - yx^3 -x^2y^2 \geq 0$

Here, from $b+c \geq a$ we get $b\geq y$

so, $bx^3 \geq yx^3$, $b^2y^2 \geq x^2 y^2 $

$Q.E.D.$

Sorry for Omit this case  :D  :D


find all prime $p$ such that $p$ can be represented as $p=m^{2}+n^{2}$ and $m^{2}+n^{2}|m^{3}+n^{3}-4$

see that:

$0\equiv-2(m^{3}+n^{3}-4) \equiv 2mn(m+n)+8 \equiv (m+n)^{3}+8$
$\equiv (m+n+2)((m+n)^{2}-2(m+n)+4) \ mod p$.

Now as $(m+n,8)=1$ we conclude that $(m+n+2,(m+n)^{2}-2(m+n)+4)=3$ or $((m+n+2,(m+n)^{2}-2(m+n)+4)=1$, hence $p|m+n+2$ or $p|mn-m-n+2$ (as $3\not |p$) and it should be easy to finish from here (some estimations)

Thank you,Megus.Nice solution.
(I have edited the topic tittle because I don't think it is hard any long after seeing your solution,though it still hard for me :blush: )


How many positive factors of 36 are not multiples of 4?

a. 2
b. 3
c. 4
d. 5
e. 6

The factors of 36 that ARE multiples of 4 are just 4, 12, and 36. Since 36 has 9 factors, the answer is 6.

Find the number of factors in 18, since 18 is $ \frac{36}{2}$, which factorized is $ 2 \cdot 3^2$. There are 6 factors, so the answer is E, 6.


Show that in any pitagoric primitive tern exists at least one number multiple of $4$.

[i]Hint: primitive terns are those of the type $(a,b,c)$ with $a > b$, $a > c$, $a^{2}= b^{2}+c^{2}$ and $a$, $b$ and $c$ are positive integers.[/i]

you can get that [url=http://www.math.rutgers.edu/~erowland/pythagoreantriples.html]here[/url]

[hide="Full Solution"]Assume that a Pythagorean triple exists such that it does not contain a multiple of 4.  Since $n^{2}\equiv 0,1\pmod4$, all terms are odd.  But this leads to $\text{odd}+\text{odd}=\text{odd}$, which is false.[/hide]

Dude... the squares of all even numbers are $\equiv 0 \pmod{4}$... like $3^{2}+6^{2}\equiv 7^{2}$

[hide]
A Pythagorean triple can be expressed in the form $(m^{2}+n^{2},m^{2}-n^{2},2mn)$

If at least one of $m$ or $n$ is even, then $4|2mn$

Otherwise, both $m$ and $n$ are odd, and $m^{2}-n^{2}=(m+n)(m-n)=\text{even}\cdot \text{even}$ is divisible by 4.[/hide]

I know.  But I assumed that none of them are multiples of 4, so they must be $\equiv 1\pmod 4$, hence they are odd.


Let R be a rectangle. How many circles in the plane of R have a diameter both of whose endpoints are vertices of R?

$ \textbf{(A)}\ 1\qquad
\textbf{(B)}\ 2\qquad
\textbf{(C)}\ 4\qquad
\textbf{(D)}\ 5\qquad
\textbf{(E)}\ 6$

[hide]There can be 5 circles:

One for each side of the recangle and one for the diagonals.[/hide]

Beautiful, magnificent, and colorful solution by drawing without words  ;)

Exactly.

Explanation:

Four circles on each corner + two circles by diagonal.

BUT!

Two circles by diagonal are actually same circle, so it's 4+2-1 = 5.

you call that easy?
it's from
AHSME

or something which is for 12th graders!!

well it gave me a brain freeze for a moment

he he he

Not all AHSME are hard. I say that first few questions of AHSME can be handled by any MC students.

After #10 or some before is where it gets harder. :)


integration
calculus
derivative
calculus computations

Let $g$ be continuous on $[a, b]$, and let $f(x) = \int_{a}^{x}(x-t)g(t)dt$.
Prove that $f$ is a solution of the differential equation $f''=g$ and the initial conditions $f(a)=f'(a)=0$.

$f(x) = \int_{a}^{x}(x-t)g(t)dt=x\int_{a}^{x}g(t)dt-\int_{a}^{x}tg(t)dt$.
Use the differentiation of the product and F.T.C, we have

$f'(x)=\int_{a}^{x}g(t)dt+x\cdot g(x)-xg(x)=\int_{a}^{x}g(t)dt$,yielding $f''(x)=g(x).$

Then plugging $x=a$ into $f(x),\ f'(x)$ gives $f(a)=f'(a)=0.$


Let $ a,b$ are real numbers such that:$ a,b$ not root of equation:$ x^2 \plus{} mx \plus{} n \equal{}$ where $ m,n\in Z$.Let:$ S_n \equal{} a^n \plus{} b^n$.
Prove or disprove that:sequence $ S_n$ not contains infinitely integer number

I think it's locked for again naming a couple of problems very badly.


[hide="Incomplete solution - one large hole"]
Observe first that $ f(x) = ax+1$ for all nonnegative $ a$ works.

First, suppose that $ f(a) = f(b) = c$. Then $ cf(y) = f(a+yc) = f(b+yc)$.

Suppose $ a \neq b$, then WLOG $ a > b$. Let $ d = \frac{a-b}{c}$. Then we get that $ cf(y) = f(a+yc) = f(b+(y+d))c = cf(y+d)$. Thus $ f$ is periodic with period $ d$.

You can use the same argument above to show that $ f(x+d/f(x)) = f(x)$ for all $ x$. Assume there is an $ x$ with $ f(x) > 1$ and apply this identity infinitely many times to get a contradiction. So now $ f(x) \leq 1$. Here's the part of the solution I haven't finished: I am pretty sure there is a way to proceed from here to get $ f(x) = 1$ for all $ x$, but I haven't figured out exactly how. (though it's entirely possible I missed an answer for when $ f(x) < 1$ for some $ x$)

Now we look for solutions other than those above, so we must remove our assumption that $ a \neq b$ above. Therefore $ a = b$ and $ f$ is injective. Observe $ f(x+yf(x)) = f(x)f(y) = f(y)f(x) = f(y+xf(y))$. By $ f$ injective, $ x+yf(x) = y+xf(y)$. Then plug in $ y = 1$ to get $ f(x) = 1+xf(1)-x$. Let $ f(1)$ be any value greater than 1 and we get the remaining solutions: $ f(x) = ax+1$ for $ a$ positive.
[/hide]

[quote="delegat"]Find all functions $ f: \mathbb{R+}\to\mathbb{R+}$   which satisfy the following identity:
$ f(x)f(y)=f(x+yf(x))$ for all positive reals $ x$ and $ y$[/quote]

We have the property $ P(x,y)$ : $ f(x)f(y)=f(x+yf(x))$ $ \forall x,y>0$

$ 1).$ $ f(x)\geq 1$ $ \forall x>0$
Let $ a>0$ such that $ 0<f(a)<1$. Then $ \frac{a}{1-f(a)}>0$ and $ P(a,\frac{a}{1-f(a)})$ gives $ f(a)f(\frac{a}{1-f(a)})=f(\frac{a}{1-f(a)})$ which is impossible since $ f(a)\neq 1$ and $ f(\frac{a}{1-f(a)})\neq 0$. So, such $ a$ don't exist and $ f(x)\geq 1$ $ \forall x>0$.
Q.E.D.

$ 2).$ $ f(x)$ is a non decreasing function.
Since $ f(x)\geq 1$ $ \forall x>0$, $ P(x,\frac{h}{f(x)})$ gives $ f(x)\leq f(x)f(\frac{h}{f(x)})=f(x+h)$.
Q.E.D.

$ 3).$ Either $ f(x)=1$ $ \forall x>0$, either $ f(x)$ is a strictly increasing function, and so is injective.
If it exists $ a>0$ such that $ f(a)=1$, then $ P(a,x)$ gives $ f(x)=f(x+a)$ and, since $ f(x)$ is non decreasing, $ f(x)=1$ $ \forall x>0$
If $ f(x)>1$ $ \forall x>0$, $ P(x,\frac{h}{f(x)})$ gives $ f(x)< f(x)f(\frac{h}{f(x)})=f(x+h)$.
Q.E.D.

$ 4).$ If $ f(x)\neq 1$, then $ f(x)=ax+1$ for some $ a>0$
If $ f(x)\neq 1$, $ f(x)$ is injective. Compare then $ P(x,y)$ and $ P(y,x)$ :
$ P(x,y)$ : $ f(x)f(y)=f(x+yf(x))$
$ P(y,x)$ : $ f(y)f(x)=f(y+xf(y))$
So $ f(x+yf(x))=f(y+xf(y))$ and so ($ f(x)$ is injective) $ x+yf(x)=y+xf(y)$
So $ \frac{f(x)}{x}-\frac{1}{x}=\frac{f(y)}{y}-\frac{1}{y}$
So $ \frac{f(x)}{x}-\frac{1}{x}$ is the constant function and $ \frac{f(x)}{x}-\frac{1}{x}=a$ and so $ f(x)=ax+1$
Putting back this expression in $ P(x,y)$, we get that any value $ a$ fit but we need to have $ a>0$ in order to have $ f(x)>0$ and $ f(x)\neq 1$
Q.E.D.

The general solution of the equation is $ f(x)=ax+1$ for any value $ a\geq 0$ (the value 0 gives the constant solution $ f(x)\equiv 1$)

Solution :
Suppose there exists $ x_{0}$ so that $ f(x_{0})<1$ .Let $ y=\frac{x_{0}}{1-f(x_{0})}$ ,we have $ f(x_{0}).f(\frac{x_{0}}{1-f(x_{0})})=f(\frac{x_{0}}{1-f(x_{0})})\to f(x_{0})=1$ contracdition.
Then $ f(x)\ge 1$ for all $ x>0$ . Now we have two cases :

Case 1 . There exists $ x_{1}$ so that $ f(x_{1})=1$ .
For any $ u( 0<u<x_{1})$ there exists $ v( 0<v<x_{1})$ so that $ u+vf(u)=x_{1}$
Then $ f(u).f(v)=f(x_{1})=1$ but $ f(u),f(v)\ge 1$ so $ f(u)=f(v)=1$ .
Hence $ f(x)=1$ for all $ 0<x\le x_{1}$.
In the equation , replace $ x$ by $ x_{0}$ we have $ f(x_{1}+y)=f(y)$ so $ f(x)$ is a circulation fucntion with cycle less than $ x_{1}$ . Then we have $ f(x)=1$ for all $ x\in \mathbb{R}^{+}$ .

Case 2 : $ f(x)>1$ for all $ x\in \mathbb{R}^{+}$ . For any pairs $ (x,y)$ with $ x>y$ there exists $ t$ so that $ x=y+tf(y)$ .Then $ f(x)=f(y).f(t)\to f(x)>f(y)$ so $ f(x)$ is increasing fucntion
In the equation replace $ x$ by $ y$ we have :$ f(x).f(y)=f(x+yf(x))=f(y+xf(y))$.
Then $ x+yf(x)=y+xf(y)$.
Hence $ \frac{f(x)-1}{x}=constant$ .We get $ f(x)=ax+1$ for any value $ a\ge 0$  .

Very nice one! (Lebesgue overkill moment)
[b][color=#f00]Claim 1:[/color][/b] $f(x) \ge 1$ holds always.
[b][color=#f00]Proof:[/color][/b] Assume that there exists $x$ such that $f(x)<1$ then by $P \left(x, \frac{x}{1-f(x)} \right)$
$$f(x)=1 \; \text{contradiction!!} \; \implies f(x) \ge 1$$
[b][color=#f00]Claim 2:[/color][/b] $f$ is linear.
[b][color=#f00]Proof:[/color][/b] On the F.E. set $y$ to be enough smaller and use [b]Claim 1[/b] to get
$$f(x+yf(x)) \ge f(x) \implies f \; \text{increasing}$$
Now we use Lebesgue monotone diferentiable theorem, so let $x_0$ such that $f$ is diferentiable at $x_0$, using $P(x_0,y)$ and taking limit of $y$ going to $0$ and re-writing the F.E. we get
$$\lim_{y \to 0^+} \frac{f(y)-1}{y}=c \; \text{exists}$$
And since this one doesnt depend of $x$ we use $P(x,y)$ with $y \to 0^+$ to get $f'(x)=c$ thus we are done.
[b][color=#f00]Final step:[/color][/b] Note that we can use the existence of $c$.
$$\lim_{y \to 0^+} \frac{ay+b-1}{y}=\lim_{y \to 0^+} a+\frac{b-1}{y} \; \text{exists} \; \implies b=1$$
Hence all $f(x)=ax+1$ work thus we are done  :blush:

@below, oh yeah, i completely did it wrong xD, dont worry, its easy to prove that $b=1$ when $f(x)=ax+b$. Delete ur post pls, i'm correcting it right now

[quote=MathLuis][b][color=#f00]Final step:[/color][/b] Assume that $c \ne 0$ then take derivate of the F.E. w.r.t. $y$ to get
$$cf(x)=c(x+f(x)) \implies x=0 \; \text{contradiction!!} \implies f'(x)=0$$
Hence $f$ is constant and by checking we can easly get $f(x)=1$.
Hence $f(x)=1$ is the unique solution and we are done  :blush:[/quote]

Look at the above solutions. It is $f(x)=cx+1$.

[quote=delegat]Find all functions $ f: \mathbb{R^+}\to\mathbb{R^+}$   which satisfy the identity:
\[f(x)f(y)=f(x+yf(x))\] for all positive reals $ x$ and $ y$[/quote]
I think I have a slightly different solution.
$\color{blue}\boxed{\textbf{Answer: }f\equiv ax+1}$
$\color{blue}\boxed{\textbf{Proof:}}$
$\color{blue}\rule{24cm}{0.3pt}$
$$f(x)f(y)=f(x+yf(x))...(\alpha)$$
$\color{red}\boxed{\textbf{If }f\textbf{ is inyective}}$
$\color{red}\rule{24cm}{0.3pt}$
In $(\alpha) x=y, y=x:$
$$\Rightarrow f(x)f(y)=f(y+xf(y))$$
By $(\alpha):$
$$\Rightarrow f(x+yf(x))=f(y+xf(y))$$
Since $f$ is inyective:
$$\Rightarrow x+yf(x)=y+xf(y)...(\beta)$$
In $(\beta) y=1:$
$$\Rightarrow x+f(x)=1+xf(1)$$
$$\Rightarrow f(x)=x(f(1)-1)+1$$
$$\Rightarrow \boxed{f\equiv ax+1\text{ is a solution}}$$
$\color{red}\rule{24cm}{0.3pt}$
$\color{red}\boxed{\textbf{If }f\textbf{ isn't inyective}}$
$\color{red}\rule{24cm}{0.3pt}$
$$\Rightarrow \exists a>b/ f(a)=f(b))=c$$
In $(\alpha) x=a:$
$$\Rightarrow cf(y)=f(a+yc)...(I)$$
In $(\alpha) x=b:$
$$\Rightarrow cf(y)=f(b+yc)...(II)$$
By $(I)$ and $(II):$
$$\Rightarrow f(x)=f(x+a-b), \forall x\in\mathbb{R}^+_{>a}$$
$$\Rightarrow f \text{ is periodic}, \forall x\in\mathbb{R}^+_{>a}$$
$$\Rightarrow f\text{ is constant or periodic with period }t>0, \forall x\in\mathbb{R}^+_{>a}$$
$\color{green}\boxed{\textbf{If }f\textbf{ is constant}, \forall x\in\mathbb{R}^+_{>a}}$
$\color{green}\rule{24cm}{0.3pt}$
$$f(x)=m, \forall x\in\mathbb{R}^+_{>a}$$
In $(\alpha) x>a, y>a:$
$$\Rightarrow m^2=m$$
$$\Rightarrow m=1$$
In $(\alpha) x>a:$
$$\Rightarrow f(y)=1, \forall y\in\mathbb{R}^+$$
$$\Rightarrow \boxed{f\equiv 1 \text{ is solution}}$$
$\color{green}\rule{24cm}{0.3pt}$
$\color{green}\boxed{\textbf{If }f\textbf{ is periodic with period }t>0, \forall x\in\mathbb{R}^+_{>a}}$
$\color{green}\rule{24cm}{0.3pt}$
In $(\alpha) y=y+t:$
$$\Rightarrow f(x)f(y+t)=f(x+yf(x)+tf(x))$$
$$\Rightarrow f(x)f(y)=f(x+yf(x)+tf(x))$$
By $(\alpha):$
$$\Rightarrow f(x+yf(x))=f(x+yf(x)+tf(x))$$
$$\Rightarrow f(y)=f(y+tf(x)), \forall y>x$$
$$\Rightarrow tf(x)\ge t$$
$$\Rightarrow f(x)\ge 1, \forall x\in\mathbb{R}^+...(\blacktriangle)$$
In $(\alpha) y=\frac{t}{f(x)}$
$$\Rightarrow f(x)f(\frac{t}{f(x)})=f(x)$$
$$\Rightarrow f(\frac{t}{f(x)})=1$$
In $(\alpha) x=\frac{t}{f(x)}$
$$\Rightarrow f(y)=f(y+\frac{t}{f(x)})$$
$$\Rightarrow \frac{t}{f(x)}\ge t$$
$$\Rightarrow f(x)\le 1$$
By $(\blacktriangle):$
$$\Rightarrow f\equiv 1$$
$$\Rightarrow t=0(\Rightarrow \Leftarrow)$$
$\color{green}\rule{24cm}{0.3pt}$
$$\Rightarrow \boxed{f\equiv ax+1}_\blacksquare$$
$\color{red}\rule{24cm}{0.3pt}$
$\color{blue}\rule{24cm}{0.3pt}$


Consider that $\alpha$ is a real number satisfy:$\alpha\ge 1$ and $\forall\ m,n\in \mathbb{N}, m,n\neq 0$,if $n$ divide  into $m$ then $[n\alpha]$ divides into$[m\alpha]$.
Prove that:$\alpha$ is an integer.
[$x$] is the greatest integer $\le x$

We must first assert that $m \neq n$ because otherwise the hypothesis is trivially false.

Let $\alpha = a + b$ where $a$ is an integer and $0\leq b < 1$, b real.

$na + [nb] = kma + [kmb]$ (where $k \geq 2$)

Have to go now, be back to finish later, sorry.

It is equavalent for x=alpha $\frac{[mnx]}{[nx]}$ are integer when m,n natural. We have
$\frac{[mnx]}{[nx]}=m+\frac{[m\{nx\}]}{[nx]}\Longrightarrow {nx}=0$, when m<n. Therefore x is integer.


Prove that there is some integer power of 2 that begins 2005 · · · .

[...]

[EdiT]
thx [b]mozza314[/b], good to know :P my mistake...

I don't have the time to solve this for myself at the moment (actually I'm not sure if I can), but mdk meant some number $ 2^{n}=2005$..., not $ n^{2}=2005...$


[quote="dduclam"]Let $ a,b,c$ are non-negative real numbers.Prove that
$ a^2(a \minus{} b)^2 \plus{} b^2(b \minus{} c)^2 \plus{} c^2(c \minus{} a)^2 \plus{} 2\sqrt {(a^4 \plus{} b^4 \plus{} c^4)(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}\ge2(ab^3 \plus{} bc^3 \plus{} ca^3)$[/quote]

From Cauchy, $ (a^4 \plus{} b^4 \plus{} c^4)(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)\ge \left(\sqrt {(a^4)(c^2a^2)} \plus{} \sqrt {(b^4)(a^2b^2)} \plus{} \sqrt {(c^4)(b^2c^2)}\right)^2$
$ \equal{} (ab^3 \plus{} bc^3 \plus{} ca^3)^2$

$ \implies 2\sqrt {(a^4 \plus{} b^4 \plus{} c^4)(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)}\ge2(ab^3 \plus{} bc^3 \plus{} ca^3)$

On the other hand we have $ a^2(a \minus{} b)^2 \plus{} b^2(b \minus{} c)^2 \plus{} c^2(c \minus{} a)^2\ge 0$.

Adding the two inequalities, we get the desired one.

Is this the problem you wanted to post? It's correct, but far too simple I suppose. :huh:

Yes,nayel.This problem is very easy. The origin problem is
$ \sqrt {a^4 \plus{} b^4 \plus{} c^4}\plus{}\sqrt{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}\ge \sqrt{2ab(a^2\plus{}b^2)\plus{}2bc(b^2\plus{}c^2)\plus{}2ca(c^2\plus{}a^2) }$
I prove it by AM-GM.

[quote="dduclam"]Yes,nayel.This problem is very easy. The origin problem is
$ \sqrt {a^4 \plus{} b^4 \plus{} c^4} \plus{} \sqrt {a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}\ge \sqrt {2ab(a^2 \plus{} b^2) \plus{} 2bc(b^2 \plus{} c^2) \plus{} 2ca(c^2 \plus{} a^2) }$
I prove it by AM-GM.[/quote]
Of cause AM-GM: $ a^4\plus{}a^2b^2\geq2a^3b.$ :wink:


trigonometry
combinatorics unsolved
combinatorics

Show that among any four points contained in a unit circle, there exist two whose distance is at most $\sqrt 2$

Let the four points be $A,B,C,D$; let the centre of the circle be $O$. Because the four points are contained in the circle, we have $AO, BO, CO, DO\le 1$. Wolog we have $ABCD$ a simple quadrilateral( maybe concave, but has sides not intersecting each other). Then we have $\angle{AOB}+\angle{BOC}+\angle{COD}+\angle{DOA}\le 360^{o}$, therefore one of the four angles on left side, wolog $\angle{AOB}\le 90^{o}$. By Cosine Law in triangle $AOB$ we have:
$AB=\sqrt{AO^{2}+BO^{2}-2AO(BO)cos(AOB)}$
$\le \sqrt{1^{2}+1^{2}-2*1*1*0}=\sqrt{2}$
The result follows.


Two numbers x & y are chosen at random (without replacement) from the numbers 1,2,3,............,2004.
The probability that $ x^{3} \plus{} y^{3}$ is divisible by 3 is ?

Thanks for any help.

[hide="Hint"]Use Modular Arithmetic.  Find any relationship between the remainder of $ x$ and $ x^3$?  Remember that you must consider several cases, i.e.:

$ x^3 \equiv 0, y^3 \equiv 0 \pmod{3}$

$ x^3 \equiv 1, y^3 \equiv 2 \pmod{3}$

$ x^3 \equiv 2, y^3 \equiv 1 \pmod{3}$.[/hide]

[quote="nittanylion"][hide="Hint"]$ x^{3} \plus{} y^{3} \equal{} (x \plus{} y)(x^{2} \minus{} xy \plus{} y^{2})$.[/hide][/quote][hide="Better Hint"]$ x^{3} \plus{} y^{3} \equal{} (x \plus{} y)((x \plus{} y)^{2} \minus{} 3xy)$.[/hide]


What is the smallest positive number which is equal to the cube of one positive integer and also is equal to the fourth power of a different positive integer?

Show your work

The number is the twelfth power of a positive integer. If it is $ 1\equal{}1^3\equal{}1^4$, then the numbers are the same. So the smallest number is $ 2^{12}\equal{}4096\equal{}16^3\equal{}8^4$.


function
algebra
polynomial
algebra proposed

It's a geometric sequence... Hence:

$f(x_{n}) = k x_{n}, \forall k \in \mathbb{R}_{+}^{*}$

[quote="afaouzi"]Find all continuous functions $f : \mathbb{R}\to \mathbb{R}$ s.t.:

 $\forall x \in \mathbb{R}$ the sequence defined by:

  * $x_{0}= x$
  * $x_{n+1}= f(x_{n}), \forall n \in \mathbb{N}$
 verifies the relation : $x_{n+1}x_{n-1}= x^{2}_{n}, \forall n \in \mathbb{N}$[/quote] \[\frac{x_{n+1}}{x_{n}}=\frac{x_{n}}{x_{n-1}}=\frac{x_{2}}{x_{1}}\] then \[\frac{x_{n+1}}{x_{n}}=k=\frac{f(x_{n})}{x_{n}}\] i.e. \[f(x)=kx\] for all \[x=x_{n}\] But we cannot deduce $f(x)=kx$for all $x\in R$
if $f(x)$is a polynomial and $k\not=0,1$ then we can.


geometry
calculus
integration
derivative
trigonometry
function
trig identities

Consider a curve [tex]y=f(x)[/tex] with the property that the area beneath the curve between [tex]x=a[/tex] and [tex]x=b[/tex] is equal to the length of the curve between [tex](a,f(a))[/tex] and [tex](b,f(b))[/tex] for all real [tex]a[/tex] and [tex]b[/tex]. The line [tex]y=1[/tex] is one such curve. Are there any others?

(Any moderator who wants to Tex what I'm writing, feel free )

What's the integral for arclength -- [tex]\int \sqrt{f'^2 + 1}\; dx[/tex]?

Beginning of solution, assuming that that formula is correct:

[hide][tex]\int_a^b f(x)\; dx = \int_a^b \sqrt{f'(x)^2 + 1}\; dx[/tex]; taking a derivative with respect to [tex]b[/tex], we get

[tex]f(b) = \sqrt{f'(b)^2 + 1}[/tex] for all [tex]b[/tex]; so we need to solve the differential equation

[tex]y^2 = (y')^2 + 1[/tex].

[tex]y^2 - (y')^2 = 1[/tex].  Hmm.  This reminds me strongly of trig identities -- in fact, it reminds me more strongly of hyperbolic trig identities.  In fact, checking quickly, we see that [tex]y = \cosh x[/tex] and [tex]y = \sinh x[/tex] are also both solutions.  Yee-Haw!  I'll think about doing something more general, but for now, that'll be all.[/hide]

Oh, right you are -- (a - b) does not in fact equal (b - a).  So I have one other solution.  That's not a bad start.

I believe that this problem is dependent on our choice of a,b

In fact, if we want to pick a function for which the integral vanishes identically, we proceed as JBL did.  And cosh is the only solution to that differential equation.  To solve it, we note that:

dy/dx = sqrt(y^2-1)

dy/sqrt(y^2-1) = dx

integration yields: x = arccosh(y)+C

whence y = cosh(x-C)

EDIT: In fact, I am guessing that for just about any function which is continuous, we can satisfy this problem over fixed limits of integration just by multiplying the function by some constant.

[quote]
EDIT: In fact, I am guessing that for just about any function which is continuous, we can satisfy this problem over fixed limits of integration just by multiplying the function by some constant.
[/quote]

actually you can just ADD some constant to the function, that would do it as well

[quote="ComplexZeta"]Well, yes, but this problem asks for any [tex]a[/tex] and [tex]b[/tex].[/quote]

That makes it almost trivial, then, doesn't it?

No, I don't see why it would be trivial.


Given a 7 x 9 rectangle (with gridlines drawn) in which a 2 x 1 rectangle has been cut off from top left corner and a 1 x 1 square has been cut off from the lower right corner leaving a 60 sq unit piece.

(It looks like the figure as shown in the attachment.)

Bisect the figure into two congruent pieces by cutting along the gridlines.

~ Shamik

At first, I thought it should be easy, but it ISN'T.
[hide="HINT"]Start drawing at the sixth point from the left in the lowest line, and imagine how they can be congruent.[/hide]
The answer is below.


floor function
number theory
prime factorization

How many factors of 8000 are perfect squares?

[hide="Solution."]$8000=2^{6}\cdot5^{3}$
For a number to be a perfect square, all exponents of its prime factorization must be even.  For this problem, we have
$2^{0}$ can be multiplied by $5^{0}$ or $5^{2}$ to make a perfect square.
$2^{2}$ can be multiplied by $5^{0}$ or $5^{2}$.
$2^{4}$ can be multiplied by the same
$2^{6}$ also.
Altogether, we have $8$ perfect square factors.

To be done fast, we just see $2^{0}$,$2^{2}$,$2^{4}$,$2^{6}$ and $5^{0}$,$5^{2}$  giving us $4\cdot2=8$.[/hide]

[hide="Generalization"]Suppose the problem were:
$ \textrm{How many factors of }n\textrm{ are perfect powers of }m\textrm{?}$

First, find the prime factorization of $ n$:
$ n = p_{1}^{e_{1}}\cdot p_{2}^{e_{2}}\cdots p_{r}^{e_{r}}$

Then, using a method analogous to buzzer11, there must be
$ \left( \lfloor \frac{e_{1}}{m}\rfloor+1 \right)\cdot \left( \lfloor \frac{e_{2}}{m}\rfloor+1 \right)\cdots \left( \lfloor \frac{e_{r}}{m}\rfloor+1 \right)$[/hide]


I have Firefox 3.0 and my OS is Fedora 9. I also have the latest version of Java, but I'm unable to view GeoGebra applets nor can I create one.

EDIT: Okay, thanks.

It is a local Java issue on your end. I just created (and viewed) a geogebra applet on my Firefox 3.0 on Windows.


modular arithmetic
floor function
combinatorics proposed
combinatorics

It was nice, [b]Johan Gunardi[/b] :)

2. It is well-known that for every prime number $ p$ such that $ p\equiv 3 \pmod{4}$, $ p | (a^2 \plus{} b^2) \Leftrightarrow p|a \text{ and } p|b$

By this, since $ 7 \equiv 3 \pmod{4}$, we have that $ (x^2 \plus{} y^2) \vdots 7 \Leftrightarrow x\equiv y\equiv 0 \pmod{7}$ and so $ (x^2 \plus{} y^2) \vdots 49$. Therefore, the problem became to count the number of unordered pairs $ (x,y) \in ((0;10^6)\cap \mathbb{Z})^2$, which is $ \lfloor \frac {10^6}{7}\rfloor ^2 \equal{} 20408122449$


floor function
induction
modular arithmetic
algorithm
search
continued fraction
Diophantine equation

For a positive integer $ n,$ $ a_{n}\equal{}n\sqrt{5}\minus{} \lfloor n\sqrt{5}\rfloor$. Compute the maximum value and the minimum value of $ a_{1},a_{2},\ldots ,a_{2009}.$

Is there actually an easy (by easy i mean non-brute force lol) way to solve this without using a machine?
I heard that the team leaders of the participating countries attempted to solve this by hand, then resorted to excel and still got different answers.

[quote="CatalystOfNostalgia"]Alright if there's nothing better than this then the problem is completely disgusting...it's not too hard to see that we want solutions to 5n^2-m^2=+/- (small), which we can solve in the usual way. the positive solutions correspond to the minimal values of a_n while the negative correspond to maximal. for (small)=1, the smallest positive solution is (305,682) but then after some horrible bounding with the aid of a calculator i can get that we have to check (small)=2, 3, 4, 5, 6 for better solutions. the smallest negative solution is (1292, 2889) which is a little better, but still the computation is just terrible.[/quote]

why did I get maximum at $ 1292$ and minimum at $ 1597$? and yes, I took [b]csl824[/b]'s "advice" and used MS Excel  :D

Well, is it just me or does it seem that Fibonacci numbers have to do with this problem as well.  1597 is the largest fibonacci number less than 2009 so it might have to do with the problem

I don't see 1292 though.

1292 is half a Fibonacci number.

Here's a few things I noticed (via computer simulations, not proofs).  The values of $ n$ that have solutions to $ m^2 \minus{} 5n^2 \equal{} 4$ are $ F_0$, $ F_2$, $ F_4$, $ F_6$, and so on.  Hence the values of $ n$ that have solutions to $ m^2 \minus{} 5n^2 \equal{} 1$ are half of $ F_0$, $ F_6$, $ F_{12}$, $ F_{18}$, and so on.  The values of $ n$ that have solutions to $ m^2 \minus{} 5n^2 \equal{} \minus{}4$ are $ F_1$, $ F_3$, $ F_5$, and so on.  Hence the values of $ n$ that have solutions to $ m^2 \minus{} 5n^2 \equal{} \minus{}1$ are half of $ F_3$, $ F_9$, $ F_{15}$, and so on.  By modular arguments, $ m^2 \minus{} 5n^2$ can never be $ \pm 2$ or $ \pm 3$.  If we could prove some of the patterns above, we could probably solve the given problem.

Well, we're looking at $ n\sqrt{5}$ modulo 1. This is the same as $ 2n\left(\frac{1\plus{}\sqrt{5}}{2}\right)$ modulo 1, so we're looking at the sequence $ 2\phi,\, 4\phi,\, 6\phi,\ldots, 4018\phi$ modulo 1, where $ \phi \equal{} \frac{1\plus{}\sqrt{5}}{2}$.

We can prove by induction that $ \phi^{k\plus{}1} \equal{} F_k\phi \plus{} F_{k\minus{}1}$, where $ F_k$ are the Fibonacci numbers with $ F_0 \equal{} F_1 \equal{} 1$. So in modulo 1, $ F_k\phi \equal{} \phi^{k\plus{}1}$. This is nice for us because $ \phi^{k}$ gets closer and closer to an integer as $ k$ increases. We can see this by noting that $ \left( \frac{1\plus{}\sqrt{5}}{2} \right)^k \plus{} \left( \frac{1\minus{}\sqrt{5}}{2} \right)^k$ is an integer for all positive integer $ k$, and $ \left( \frac{1\minus{}\sqrt{5}}{2} \right)^k$ tends to 0 as $ k \to \infty$. In particular, when looking at fractional parts, $ \phi^2,\, \phi^4,\, \phi^6,\ldots$ tends to 1 and $ \phi^3,\, \phi^5,\, \phi^7,\ldots$ tends to 0.

Finally, every positive integer can be written as the sum of distinct, nonconsecutive Fibonacci numbers. So letting $ 2n \equal{} F_{k_1} \plus{} F_{k_2} \plus{} \cdots \plus{} F_{k_r}$, with $ k_1 < k_2 < \cdots < k_r$ and no two consecutive, then $ 2n\phi \equal{} \phi^{k_1\plus{}1} \plus{} \phi^{k_2\plus{}1} \plus{} \cdots \plus{} \phi^{k_r\plus{}1} \pmod 1$.

I'll outline the rest of this: basically, if we want to maximize the fractional part if $ 2n\phi$, we first want $ k_1$ to be the largest odd number possible (so $ \phi^{k_1\plus{}1}$ is close to 1), then $ k_2,\, k_3,\ldots$ to be even numbers (which have to be at least $ k_1\plus{}3$ by the non-consecutive stipulation) so that $ \phi^{k_2\plus{}1},\, \phi^{k_3\plus{}1}, \ldots$ "push" the number even closer to 1. Given the properties of $ \phi$, it's not hard to prove that this "greedy" algorithm always works, and that we would always prefer a larger $ k_1$ over a smaller one. To minimize $ 2n\phi$, just switch "even" and "odd" in the above. For the maximum, the best we can do is $ 2n \equal{} F_{17} \equal{} 2584$, so $ n\equal{}1292$. For the minimum, we can't use $ 2n \equal{} F_{16} \equal{} 1597$ since 1597 is odd, so we have to use $ 2n \equal{} F_{14} \plus{} F_{17} \equal{} 610\plus{}2584 \equal{} 3194$, so $ n \equal{} 1597$.

I will post two solutions.
The first is based on continued fraction approximation
The second is based on the theorem of Pell equation

Very sorry for written in Chinese,I will explain if not clear :)

how to use the pell equation to solve it?

[hide="Here is a Python program that solved the problem pretty quickly:"][code]def f(x):
    k = 5**0.5
    f = k*x - int(k*x)
    return f

def main():
    maxx = 5**0.5 - 2
    maxi = 5**0.5 - 2
    minx = 1
    mini = 1

    for i in range (1,2010):
        if f(i) < minx:
            minx = f(i)
            mini = i
        elif f(i) > maxx:
            maxx = f(i)
            maxi = i

    print maxx, maxi, minx, mini

main()[/code]

It should print 0.999826929728 1292 0.000560067164315 1597. This means that the maximum is $ a_{1292} \equal{} 0.999826929728$ and the minimum is $ a_{1597} \equal{} 0.000560067164315.$[/hide]

Happy new year,I will give some more explanations on the first solution based on some simple facts of continued fraction approximation

As $ \sqrt{5}$ can be written as $ <2,4,4,4,...>$

Then we get the approximation-fraction of $ \sqrt{5}$:$ 2,\frac{9}{4},\frac{38}{17},\frac{161}{72},\frac{682}{305},\frac{2889}{1292},\frac{12238}
{5473}$,they are known as $ \frac{h_n}{k_n}$ for $ n=1,2,...$

An important property of continued fraction approximation :$ |\sqrt{5}-\frac{h_n}{k_n}|<\frac{1}{k_nk_{n+1}}$

Suppose $ \sqrt{5}=\frac{2889}{1292}-c$where $ 0<c<\frac{1}{1292*5473}$

Thus $ \{1292\sqrt{5}\}>1-1292c>\frac{5472}{5473}$

But for any $ k$,$ k\neq 1292$ and $ 0<k\le 2009$,then $ 2889k\neq0,mod1292$

Hence,$ \{k\sqrt{5}\}=\{\frac{2889}{1292}k-kc\}<\frac{1291}{1292}-kc<\frac{5472}{5473}$

which implies $ a_{1292}$ is the maximal

Let's search for the minimal:
$ 305+1292=1597$

$ \{1597\sqrt{5}\}=\frac{1}{1292}-1597c$

$ \{305\sqrt{5}\}=\frac{1}{1292}-305c>\{1597\sqrt{5}\}$

For any $ n$ which is not equal to $ 305$ or $ 1597$ we have $ 1292n\neq1,mod 1292$

Hence $ \{n\sqrt{5}\}>\frac{2}{1292}-2009c>a_{1597}$

which implies $ a_{1597}$ is the minimal :lol:

QED

[quote="Georg-A."][quote="1=2"][quote="hxy09"]$ <2,4,4,4,...>$ [/quote]
Could you please explain this notation? I don't quite understand it.[/quote]

There should be a semicolon $[2;4,4,...]$ and it means

\[ \sqrt{5}=2+\frac{1}{4+\frac1{4+\frac 1{4+...}}} \][/quote]

Both notations are correct.


[quote="Fuaran"]Sure it's right??
RHS is much larger. It has n+8 digits, while LHS is 3*(log n) digits or so.[/quote]
Read carefully problem, please.


let$ (a_n)$ defined by $ a_1 \equal{} 2009; a_{n \plus{} 1} \equal{} a_n \minus{}\left[ {\frac{{{a_n}}}{2}} \right]\plus{}1$. $ \forall n \ge 1$

prove that $ a_n$ has limt and compute $ \lim a_n$  :lol:

If $ a_n>3$, then $ 3\le a_{n\plus{}1}<a_n$. Therefore $ lim\equal{}3$.


modular arithmetic
number theory
Diophantine equation
relatively prime
prime numbers

The AoPS Wiki says the following about ax + by = c diophantine equations.

This Diophantine equation is known as a linear combination. If two relatively prime integers a and b are written in this form with c=1, the equation will have an infinite number of solutions. More generally, there will always be an infinite number of solutions when gcd(a,b)=c. If gcd(a,b)=c, then there are no solutions to the equation.

So, if gcd(a,b) = c, there are infinite solutions, and also no solutions?

The former condition should be $\gcd(a,b)\mid c$, and the latter $\gcd(a,b)\nmid c$

Yes, the first part should be $gcd(a,b)\mid c$.  If this is true, then the equation can be simplified to 3 relatively prime numbers, $a$, $b$, & $c$.

If $gcd(a,b)\nmid c$, the left hand side of the equation will be $0\pmod{gcd(a,b)}$, but the right hand side is not, so there is not equality, and the equation has no solutions.


Is it rated or unrated

But if it is rated some ppl won't come but if it is unrated then i think it would be a waste of time.

[quote="pacman2812"]OK, my REALLY weird idea is to create a normal FTW game with the biggest possible number of players.

It's just an idea, I will tell the signups limit, the setup and the date of the game when I look the biggest possible number of players.

If someone thinks that this is prohibited or spam, please tell me and I will try to delete this thread. Or, tell it to a mod, or delete it yourselves if you are mods :D

If you are interested just post that you are interested, and I will squeeze you in.[/quote]
huh.
try to fit me in.

[quote="FantasyLover"]biggest number of people that can join in a normal game - 20

ive seen 2 20-ppl game...

one with THECHAMP and 19 guests
and another filled with tippers[/quote]

eww... :(

Then that wouldn't be so cool... I thought that it was 30+... :(

If rated or not: I will think that later. Maybe we would make it popular voting

I will make it two split 20 players game (I will need another mod; if interested, please tell me), and the top 10 of each game would participate in the GREAT FINAL.
I will make it a weekend, possibly a sunday, because saturday morning I can't :(. 50 signups, 10 replacements. This is officialy a tournament.

SIGNUPS

Game A: 1)pacman2812 (mod)
Game B: 2)myyellowducky82

hopefully noone will try to stop this...

well if the most number of player is 20...whats the point?

i thought the highest # of ppl in a game wuz 30.

its 20.. maybe thats just the new version... but once i saw a 20-ppl game and when i tried to join it said game was full.

OK, signups list:

Game A: 1)pacman2812 (mod)
3) AIME15
Game B: 2)myyellowducky82
4) isabella2296

AIME15, YOU HAD SIGNED UP YET; DO NOT SPAM

Game selection criteria: odd=A, even=B.

Is someone interested on modding Game B? Mods also can participate. They are just the creators of the game in FTW.

yes, well, whats the fun of that?

I HAVE 500 POSTS FINALLY!!!

[quote="ernie"][quote="vallon22"]that sorta defeats the purpose of this...
this was only created to try to reach the limit for a normal game[/quote]

Here's an idea.  Make 19 guests, and play a game with those guests and you!  Sounds kinda lonely though, but it will work.[/quote]
That's a strange idea, but a good one.

Hmm that really does work.
That kinda defeats the purpose as stated before but if you really want to, you could.

duh...

THE FUN IS HAVE A BIG GAME TO PLAY NOT SPAM!

Okay, so we should probably have a zillion guests,

[quote="ernie"]Here's an idea. Make 19 guests, and play a game with those guests and you! Sounds kinda lonely though, but it will work.[/quote]

[quote="vallon22"]y don't u just start a normal game, get whoever's on at that time to join, and fill in the rest of the spots with guests?[/quote]

My idea.  :P

And I don't see any other way.  So you can either do this or drop the whole idea.

Getting 20 USERS on at the same time is virtually impossible.
Ernie's right =]

umm...
this was from a few weeks ago
the creator (i forget who) has already cancelled this


AMC
USA(J)MO
USAMO
inequalities
trigonometry
geometry
3D geometry

I ask that, please, no discussion of the USAMO problems take place until
Thursday, April 29, 8:00 am EDT.

Thank you!

Steve Dunbar
AMC Director

good luck to everyone tomorrow...and wednesday...

Yes, the intended meaning is absolutely, positively no discussion until the designated date. No discussion of "how hard did you think it was?" or "What was the easiest problem?" or ANYTHING at all about USAMO, through any channel, until after the date designated by the contest director.

[quote="probability1.01"]Just wondering... can we discuss the problems of the first round before the specified date?

Also, can we discuss them privately with people who we know have taken the test already (via private messages)?[/quote]

Do not discuss it until the time the AMC Director has set.

We will delete any mention of the test on this site before that time, including even seemingly harmless comments like 'Wow, that was hard' or 'I think I got 2, how did you do?'

See everyone for discussion on Thursday. Good luck!

I laughed out loud when I heard that there is a rule saying something like unofficial students should not be allowed to write the usamo...

It seems to me that this thread is borderline discussing the USAMO before the approved date....better close it.

nothing's been said about the content on this year's USAMO...

This is really killing me.  I just sat locked in a room twice for a grand total of nine hours, and I can't even discuss this...  I need to know what the answers are!  Oh well, just 10 more hours...

[quote="probability1.01"]We can start discussing USAMO now, right?[/quote]

You posted at 8:29am EDT, so I think you're good to go. The request was to wait until 8:00am today, and it's past 8:00am.

I felt confident with solutions to two and four.

Will hopefully be able to check in at lunch and see how others did (and hopefully some solutions - especially problem five!)

Here is my attempt at #1

Lemma 1     AD+BC=AB+DC

Proof:  By equality of tangents

Lemma  2  Each angle of the quadrilateral lies between 60 and 120 degrees.

Proof:   Definition of exterior angle and supplements.

Lemma  3 (AB)(AD) :le: (AB)^2+(AD)^2-(AB)(AD) :le: (BC)^2+(BC)(CD)+(CD)^2

Proof:   The left inequality follows from the expansion of (a-b)^2 and the right inequality from the law of cosines applied with  :ang: A to  :tri: ABD and again with  :ang: C to  :tri: CBD, using Lemma 2.

Proof of the theorem:    Now from Lemma 3, used twice,

          (AB)^2+(AD)^2+(AB)(AD) :le: 3((BC)^2+(BC)(CD)+(CD)^2)

Lemma I  gives  AB-AD=BC-CD, from which difference of cubes yields

          |AB^3-AD^3| :le: 3|BC^3-CD^3|

Reversing the roles of the sides gives the other desired inequality.

Equality may arise from AB=AD, forcing BC=CD, or else equality throughout the inequality of Lemma 3, which also necessitates AB=AD, etc.  It seems then that equality occurs only when quadrilateral ABCD is a kite.

The solution of #5 given above is very elegant.

I appreciate any comments/corrections of this attempt at #1.

Thanks,

afanofpolya

Nukular wrote:Soln to #5

[hide]
Lemma: a^5 - a^2 + 1 >= a^3, for positive a.

Proof:
     This is equivalent to (a^3-1)(a^2-1) >= 0, which has equality at a = 1.

Now, to apply this to the inequality:

(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) >= (a+b+c)^3    (1) .
By Holder's inequality,

((a^5-a^2+1)+1+1)^(1/3) (1+(b^5-b^2+1)+1)^(1/3) (1+1+(c^5-c^2+1))^(1/3) >= (a^5-a^2+1)^(1/3) + (b^5-b^2+1)^(1/3) + (c^5-c^2+1)^(1/3)
>= (a^3)^(1/3) + (b^3)^(1/3) + (c^3)^(1/3) = a + b + c.

Cubing both sides, and simplifying the LHS gives the result (1).

[/hide]

On mathworld, Holder's inequality only involves the product of two terms. Can this be generalized for more than 2? For example (I can't tex), does Cauchy Schwarz also imply that ( :Sigma: a^2)( :Sigma: b^2)( :Sigma: c^2) :ge: ( :Sigma: abc )^2?

[edit]-Wait, of course that's true. :Sigma: a^2 * :Sigma: b^2 = :Sigma: (ab)^2, so now we've reduced it again.

[edit 2]- Wait, that product isn't true. What am i THINKING? This is why I sucked on the USAMO

Awesome proof Nukular . I used some stupid calculus method to try and grab a few points, but I think they'll still give me 0.

To MathMan Jay,

The generalization that Nukular used of the Holder inequality is this

 [tex]$\sum  a_kb_kc_k \le  |a|_p|b|_q|c|_r $[/tex]

where 1/p+1/q+1/r=1, and [tex]|a|^p[/tex] denotes the little L p norm
of the sequence.  Further generalizations occur in the same way, the main
thing being that the sum of the reciprocals be one. In his case
it was 1/3+1/3+1/3=1.

afanofpolya

according to  Kiran Kedlaya's inequalities packets:

Provided that  [tex]\displaystyle \sum_{i=1}^n w_i = 1[/tex] with [tex]w_i[/tex] positive reals,

then for positive reals [tex]a_{ij}[/tex]

[tex]\displaystyle \prod_{i=1}^n(\sum_{j=1}^m a_{ij})^{w_i} \ge \sum_{j=1}^m (\prod_{i=1}^n a_{ij}^{w_i})
[/tex]


Source: 360 PMC
Define the sequence $(a_{n})_{n\geq 1}$ by $a_{1}=1$ and
$a_{n+1}=2a_{n}+\sqrt{3a_{n}^{2}-2}$ for all integers $n\geq 1$. Prove that $a_{n}$ is an integer for all $n$.

[quote="mffmartinelli"]We can easily prove by induction that $a_{n+2}=4a_{n+1}-a_{n}$, for all natural $n$. But we have $a_{1}=1$ and $a_{2}=3$, and so all $a_{n}$ is integer by an easily induction again.[/quote]
yes, induction works here.
but u can also derive the recurrence forumula by doing some algebra...

do we derive the closed form of $a_{n}$ from $a_{n+2}=4a_{n+1}-a_{n}$. however how is it that we notice that  $a_{n+2}=4a_{n+1}-a_{n}$?

We can calculate the first terms and notice that formula. Then it´s just prove it by induction...


Let $ \Delta ABC$. Let $ P$ a point such that $ \measuredangle APC=\frac{\pi}{2}$ and let $ E$ a point on $ AC$ such that $ B-E-P$ and $ AP^{2}=AE\cdot AC$. Let Q a point such that $ \measuredangle AQB=\frac{\pi}{2}$ and let $ F$ be a point on $ AB$ such that $ C-F-Q$ and $ AQ^{2}=AF\cdot AB$. Let $ T$ the intersection point of $ BP$ and $ CQ$ and let $ M$ the intersection point of $ AT$ and $ BC$. Let $ AB=2\cdot BM$. Determine the measure of $ \angle ABC$

If we see, the propierties: $ AP^2 = AE*AC $ and $ AQ^2=AF*FB $ tell us that $QF$ and $PE$ are the heights of $ \triangle {QBA}$ and $ \triangle {CPA}$ respectively. That implies that $\angle {BFC} $ and $\angle {BEC} $ are equal to 90°. So, T is the orthocenter of $ \triangle {ABC} $, and $AM$ it's a height. To calculate $\angle {ABC} $ we have:
$cos \angle {ABC} = \frac {BM} {BA}$ (but $ BA= 2MB $).
That is equal to say: $cos \angle {ABC} = \frac {BM} {2BM} = \frac {1}{2} $
And we know that $cos\angle {60} = \frac {1}{2} $. So $ \angle {ABC} = 60 $


What product is obtained when tert-butyl cyanide is treated with benzylmagnesium bromide, followed by $ Li/NH_3$?

I think Imine might be the product. :maybe:  :maybe:  :maybe:

Thanks  :D  It took the pains outta me to draw that  :) And i hope there is soon going to be a drawing helper on aops  :)