modular arithmetic
induction
number theory proposed
number theory

Let $ a\in\mathbb{N}$ be a natural number of the form $ 8m \plus{} 3$,or $ 8m \minus{} 3$,and $ b$ be the number of the form $ 8l \plus{} 1$,where $ l,m\in\mathbb{N}$.Prove that $ \forall n\geq 4$,there exist $ k\in\mathbb{N}$,such that
$ a^k \minus{} b$ is divisible by $ 2^n$.
See here for a proof and particular case :
http://www.mathlinks.ro/viewtopic.php?t=187197

[quote="pbornsztein"]Come on Erken!
With $ 650$ posts, you could find a better title than that :wink:

Pierre.[/quote]
For example?
I have no idea,how to name it.

It's false. There's no integer $ k$ such that $ 7^k\equiv 9\pmod{16}$.

If $ a\equiv \pm 3\pmod{8}$ and $ b\equiv 1\pmod{8}$ then it's true, and my proof given in the link can be generalized to prove this (all that mattered was $ a^2\equiv b\pmod{8}$ and $ a^2\minus{}1$ containing exactly three factors of 2).

[quote="scorpius119"]It's false. There's no integer $ k$ such that $ 7^k\equiv 9\pmod{16}$.

If $ a\equiv \pm 3\pmod{8}$ and $ b\equiv 1\pmod{8}$ then it's true, and my proof given in the link can be generalized to prove this (all that mattered was $ a^2\equiv b\pmod{8}$ and $ a^2 \minus{} 1$ containing exactly three factors of 2).[/quote]
Yes,you are right.
I've already edit the statement.Thank you.

[quote="ZetaX"]And when editing it, why did you not chancge the titel¿
And don't tell me you have no idea how to name it, describe the problem in the title![/quote]
Before it was "Nice,and not hard".Now the title is "Generalization of the well known problem".
How should i describe it?
Should i write Multiple of $ 2^n$,or smth else?

[quote="quangpbc"][color=green]Dear Erken, you can use a sequence and induction to complete the solution. And the same idea for a problem from VMO. Goodluck  :) .[/color][/quote]
 Which problem from VMO?
 To the first problem,i've already known two solutions,my own,and scorpius119's one.So if you know another,which uses sequences and
induction,please post it. :wink:


Determine all integers n > 1 such that $ \frac{{2^n  \plus{} 1}}{{n^2 }}$ is an integer

[quote="MR.129"]Determine all integers n > 1 such that $ \frac {{2^n \plus{} 1}}{{n^2 }}$ is an integer[/quote]
This a a problem from IMO 1990
You can see solution at topic http://www.mathlinks.ro/viewtopic.php?p=849318#849318 :P


geometry
geometric transformation
reflection
incenter
inradius
ratio
quadratics

In trapeze $ ABCD$ the length of basis $ BC$ is 13, the angle $ BAD$ is sharp and twice more than angle $ ADC$. The circle with center on line $ BC$ touches lines $ AC, AD$ and side $ CD$. Find the area of trapeze $ ABCD$, if  radius of the circle is 5

Assuming trapeze = trapezoid/trapezium, basis = base, sharp = acute, twice more than = twice as large as, touches = is tangent to, this is the picture I think you mean, and I really have no idea how the center of that circle could possibly be on BC. :?:

The $ (K)$ be the circle tangent to the lines $ AC, CD, DA$ and centered on the line $ BC.$ (It is the excircle of the $ \triangle ACD$ in the angle $ \angle DAC.$) The excenter $ K$ lies on the external bisector of the $ \angle ACD$ and this is identical with the line $ BC \parallel DA.$ This means that the $ \triangle ACD$ is isosceles with $ AC \equal{} CD.$ Let $ M$ be the midpoint of $ DA,$ so that $ CM$ is the perpendicular bisector of $ DA.$ Reflect the line $ AB$ in $ CM$ into $ DB'.$ The lines $ AB, DB'$ intersect at a point $ P$ on the perpendicular bisector $ CM$ of $ DA$; the $ \triangle APD$ is also isosceles with $ AP \equal{} PD.$ Since $ AC, DC$ bisect the angles $ \angle DAP \equiv DAB$ and $ \angle ADP \equiv ADB',$ the point $ C$ is the incenter of the $ \triangle APD.$ Its inradius $ r \equal{} CM \equal{} 5$ is the same as that of the given excircle $ (K).$ Denote $ b \equal{} BC \equal{} 13,$ $ a \equal{} AP \equal{} PD$ and $ p \equal{} DA.$ By Pythagorean theorem, $ PM \equal{} \frac {\sqrt {4a^2 \minus{} p^2}}{2}.$ The incenter $ C$ divides $ PM$ in the ratio $ \frac {CM}{PM} \equal{} \frac {p}{2a \plus{} p},$ hence

$ r \equal{} CM \equal{} \frac {p}{2} \sqrt {\frac {2a \minus{} p}{2a \plus{} p}} \equal{} 5$

$ 2b \equal{} 2 BC \equal{} BB' \equal{} \frac {2ap}{2a \plus{} p} \equal{} 26$

We got 2 equations for 2 unknowns $ a, p$; we need only $ p \equal{} DA.$ Calculating $ a$ from the 2nd equation,

$ a \equal{} \frac {bp}{p \minus{} 2b}$

Squaring the 1st equation,

$ 2ap^2 \minus{} p^3 \equal{} 4r^2 (2a \plus{} p)$

Substituting for $ a,$

$ \frac {2bp^3}{p \minus{} 2b} \minus{} p^3 \equal{} 4r^2 \left(\frac {2bp}{p \minus{} 2b} \plus{} p\right)$

Reducing by $ p$ and multiplying by $ p \minus{} 2b,$

$ 2bp^2 \minus{} p^3 \plus{} 2bp^2 \equal{} 4r^2 (2b \plus{} p \minus{} 2b)$

Again, reducing by $ p,$

$ 4bp \minus{} p^2 \equal{} 4r^2$

Solving this quadratic equation for $ p,$

$ p \equal{} 2\left(b \pm \sqrt {b^2 \minus{} r^2}\right) \equal{} 2\left(13 \pm \sqrt {169 \minus{} 25}\right) \equal{} 50$ or $ 2.$

Obviously, the angle $ \angle BAD$ is (very) obtuse for $ p \equal{} 2,$ hence $ p \equal{} 50.$ Area of the trapezoid $ ABCD$ is then

$ S(ABCD) \equal{} r \cdot \frac {p \plus{} b}{2} \equal{} \frac {315}{2}.$

[color=darkblue]I"ll simplify little the your [u]nice proof[/u], [b]Yetti[/b] !

Using the same notations and ideas, denote $ PC=x$ , $ PB=y$ .

Observe that $ \{\begin{array}{ccc}
BC=BA=13 & \mathrm {and} & MC=5\\\\
\frac {PC}{CM}=\frac {PB}{BA} & \implies & \frac {x}{5}=\frac {y}{13}\\\\
PC^2+CB^2=PB^2 & \implies & x^2+13^2=y^2\end{array}\|$ $ \implies$

$ \frac {x^2}{5^2}=\frac {y^2}{13^2}=$ $ \frac {y^2-x^2}{13^2-5^2}=\frac {13^2}{12^2}\implies$ $ \{\begin{array}{c}
x=\frac {65}{12}\\\\
y=\frac {169}{12}\end{array}$ . Therefore, $ PM=x+5$ and

$ \frac {PC}{PM}=\frac {CB}{AM}$ $ \implies$ $ \frac {x}{x+5}=\frac {13}{AM}$ $ \implies$ $ AM=25$ . In conclusion, $ AD=50$ and

$ [ABCD]=\frac 12\cdot CM\cdot (BC+AD)=$ $ \frac 52\cdot (13+50)$ $ \implies$ $ [ABCD]=\frac {315}{2}$ .[/color]


function calculus derivative trigonometry logarithms integration limit

We know that if f'(x)=0 and f' does not change sign at x then f has a point of inflexion at x. I need to give an example to show that this is not always true.

Thank you

If you mean: "$ f^{\prime}(x_0) \equal{} 0$ and either $ f^{\prime}(x) < 0$ in a neighborhood of $ x_0$ either $ f^{\prime}(x) > 0$  in a neighborhood of $ x_0$, then has a point of inflection at $ x_0$" then this is [u]always true[/u] from the definition of the inflection point.

the function $ f(x)\equal{}\arctan(x)$ at $ x\equal{}0$ is as required.

hamidreza wrote:
[quote]the function $ f(x) \equal{} arc\tan(x)$  at  $ x \equal{} 0$  is as required.[/quote]
But $ f^{\prime}(x) \equal{} \frac {1}{1 \plus{} x^2}$ and $ f^{\prime}(0) \equal{} \frac {1}{1 \plus{} 0} \equal{} 1\neq0$.

So this it isn't an example!

The "example" must be a function such that $ f^{\prime}(x_0) \equal{} 0$ and either $ f^{\prime}(x) < 0$ in a neighborhood of  $ x_0$ either $ f^{\prime}(x) > 0$ in a neighborhood of  $ x_0$ and $ x_0$ it isn't a inflection point of $ f$. Such "example" simply doesn't exists!

$ P$^$ Q$ is false if and only if at least one of them is not true

hamidreza wrote:
[quote]$ PQ$  is false if and only if at least one of them is not true[/quote]

I can also reply with $ P\ \wedge\ Q \longrightarrow\ R$,  but...
in a problematic question such as that of nikolany, comments with  such generality  doesn't help at all! Sorry I can't follow!
I give an exact answer in an exact proposition which is " $ f^{\prime}(x_0)\equal{}0$ and either $ f^{\prime}(x)<0$ in a neighborhood of $ x_0$, either  $ f^{\prime}(x)<0$ in a neighborhood of $ x_0$, then has a point of inflection at $ x_0$" .

The standard notion of an inflection point is about the second deriavtive changing sign from $ \plus{}$ on the left to $ \minus{}$ on the right or vice versa at the point. Of course, the second derivative cannot [i]preserve[/i] its sign near the point under the conditions of the problem, but there is one more option that amounts to neither preserving the sign, nor changing it from one sign on the left to another on the right. ;)

Anyway, it is all about terminology: the example I have in mind can also be classified as an "inflection point" if you really wish to do so though, as far as I know, nobody is willing to go that far :).

fedjia wrote:
[quote]The standard notion of an inflection point is about the second deriavtive changing sign from  on the left to  on the right or vice versa at the point. [/quote]
Because it is not given that $ f$ is twice differentiable, the definition of inflection point I had in mind is that of changing from convex to concave, or from concave to convex.

fedjia wrote:
[quote]Anyway, it is all about terminology: the example I have in mind can also be classified as an "inflection point" if you really wish to do so though, as far as I know, nobody is willing to go that far.[/quote]
Can you describe, in short, what has this special function you mention, so you can say that has an inflection point instead of others?
Let me guess: It isn't twice differentiable.

P.S. The funny think is that I edited my first post to “correct” “either.., or..” to “either.., either..”.

Actually my example is $ C^2$ (or even $ C^\infty$, if you wish). The idea is to create a function whose second derivative changes sign arbitrarily close to the point on both sides and still satisfies your assumptions :).

fedja wrote:
[quote]Actually my example is  $ C^2$ (or even $ C^{\infty}$, if you wish). The idea is to create a function whose second derivative changes sign arbitrarily close to the point on both sides and still satisfies your assumptions[/quote]

Let $ f$ has a continuous second derivative $ f^{\prime\prime}$. Because $ f^{\prime}$ preserves sign in a neighborhood of $ x_0$, so has an extreme in $ x_0$ from $ Fermat$, we have that $ f^{\prime\prime}(x_0)\equal{}0$. Also it can't be $ f^{\prime\prime}\equiv0$ in a neighborhood of $ x_0$. Then the continuity of $ f^{\prime\prime}$ doesn't provide a neighborhood of $ x_0$ where $ f^{\prime\prime}$ has no roots?
Or you mean something else?

You are right that we must have $ f''\equal{}0$ at the point but the fact that $ f''$ is not identically $ 0$ near the point doesn't imply that it cannot change sign infinitely often near the point ;). (I'm still trying to make you to invent the example by yourself; when you get too tired of it, let me know and I'll present a full formal construction).

Let clear up some things:

[b]DEFINITION 1.1.[/b] Let  $ f: \mathbb{R}\longrightarrow\mathbb{R}$ a  function which is differentiable at  $ x_0\in{\cal{D}}(f)$. We say that $ f$ is [b]strictly convex[/b] at $ x_0$ iff  exists $ \delta>0$, such that for every   $ x\in\left({x_0\minus{}\delta,\,x_0}\right)\cup\left({x_0,\,x_0\plus{}\delta}\right)$, holds   $ f(x)\minus{}f(x_0)\minus{}f^{\prime}(x_0)(x\minus{}x_0)>0$.

[b]DEFINITION 1.2. [/b] Let  $ f: \mathbb{R}\longrightarrow\mathbb{R}$ a  function which is differentiable at  $ x_0\in{\cal{D}}(f)$. We say that $ f$ is  [b]strictly concave[/b] at $ x_0$ iff  exists $ \delta>0$, such that for every   $ x\in\left({x_0\minus{}\delta,\,x_0}\right)\cup\left({x_0,\,x_0\plus{}\delta}\right)$, holds  $ f(x)\minus{}f(x_0)\minus{}f^{\prime}(x_0)(x\minus{}x_0)<0$.

[b]DEFINITION 1.3. [/b] Let  $ f: \mathbb{R}\longrightarrow\mathbb{R}$ a  function which is differentiable on an interval $ \Delta$. We say that $ f$ is [b]strictly convex[/b]  ( respectively [b]strictly concave[/b] ) on $ \Delta$ iff  is strictly convex ( respectively strictly concave )  at every $ x\in\Delta$.

[b]DEFINITION 2.1.[/b] Let  $ f: \mathbb{R}\longrightarrow\mathbb{R}$ a differentiable function and  $ x_0\in{\cal{D}}(f)$. We say that $ x_0$ is an [b]inflection point[/b] of $ f$ iff the following holds:
(a) the tangent line of graph of $ f$ at $ x_0$ exists, and
(b) either $ f$ is strictly convex on $ \left({\alpha,\,x_0}\right)$ and strictly concave on $ \left({x_0,\,\beta}\right)$, or $ f$ is strictly concave on $ \left({\alpha,\,x_0}\right)$ and strictly convex on $ \left({x_0,\,\beta}\right)$.

[b]DEFINITION 2.2. [/b]Let  $ f: \mathbb{R}\longrightarrow\mathbb{R}$ a twice differentiable function and  $ x_0\in{\cal{D}}(f)$. We say that $ x_0$ is an [b]inflection point[/b] of $ f$ iff  the following holds:
(a) $ f^{\prime\prime}(x_0)\equal{}0$ and
(b) either $ f^{\prime\prime}$  is positive on $ \left({\alpha,\,x_0}\right)$ and negative on $ \left({x_0,\,\beta}\right)$, or $ f^{\prime\prime}$  is negative on $ \left({\alpha,\,x_0}\right)$ and positive on $ \left({x_0,\,\beta}\right)$.

[b]QUESTION 1. [/b] If $ f: \mathbb{R}\longrightarrow\mathbb{R}$ is a twice differentiable function then DEFINITION 2.1. and DEFINITION 2.2. are equivalent?

[b]PROPOSITION 1. [/b] Let  $ f: \mathbb{R}\longrightarrow\mathbb{R}$ a twice differentiable function and  $ x_0\in{\cal{D}}(f)$. If  $ f^{\prime}(x_0)\equal{}0$ and either $ f^{\prime}$  is positive on $ \left({\alpha,\,x_0}\right)\cup\left({x_0,\,\beta}\right)$, or $ f^{\prime}$  is negative on $ \left({x_0\minus{}\delta,\,x_0}\right)\cup\left({x_0,\,\beta}\right)$, then $ x_0$ is an inflection point of $ f$.

[b]QUESTION 2.[/b] Is PROPOSITION 1. true according the above definitions?

[b]PROPOSITION 2.[/b] Let  $ f: \mathbb{R}\longrightarrow\mathbb{R}$ a twice differentiable function and  $ x_0\in{\cal{D}}(f)$. If  $ f^{\prime}(x_0)\equal{}0$ and either $ f^{\prime}$  is positive on $ \left({\alpha,\,x_0}\right)\cup\left({x_0,\,\beta}\right)$, or $ f^{\prime}$  is negative on $ \left({x_0\minus{}\delta,\,x_0}\right)\cup\left({x_0,\,\beta}\right)$ and $ f^{\prime\prime}$ change sign infinitely often near the point $ x_0$, then $ x_0$ is an inflection point of $ f$.

[b]QUESTION 3.[/b] Is PROPOSITION 2. true according the above definitions?

[b]2.1[/b] and [b]2.2[/b] are [i]almost[/i] equivalent. What 2.1 is [i]exactly[/i] equivalent to for twice continuously differentiable functions is "$ f''\ge 0$ on $ (\alpha,x_0)$ and $ f''$ is not identically $ 0$ on any subinterval of $ (\alpha,x_0)$, etc.". What we really need to know for this problem is that either of these definitions implies

[b]If $ f\in C^2$ has an infelction point at $ x_0$, then $ f''$ preserves sign on $ (\alpha,x_0)$ and on $ (x_0,\beta)$[/b]

Proposition 1 is false according to the above definitions.

Proposition 2 should really end with "is [color=red]not[/color] an inflection point of $ f$". Then it becomes true and is exactly what my example is based upon. As written, it is false.

fedja wrote:
[quote] If  $ f\in{C}^{2}$ has an infelction point at $ x_0$, then  $ f^{\prime\prime}$ preserves sign on $ \left({\alpha,\,x_0}\right)$ and on  $ \left({x_0,\,\beta}\right)$. [/quote]
You mean  $ f^{\prime}$ preserves sign...
____________________________________________________________________________________________________________

fedja wrote:
[quote] Proposition 1 is false according to the above definitions. [/quote]
[b]QUESTION 1.[/b] Is Proposition 1 false according to DEFINITION 2.1. ?

[b]QUESTION 2.[/b] Is Proposition 1 false according to DEFINITION 2.2. ?

[b]ANSWERS:[/b] Yes & Yes

[quote="grigkost"]
You mean  $ f^{\prime}$ preserves sign...
[/quote]
Nope, $ f''$ (the second derivative). The preserved sign can (and must) be different on $ (\alpha,x_0)$ and $ (x_0,\beta)$ but on each interval it does not change.

Yes, Proposition 1 is false under both definitions.

You forgot about me   :ninja:

Is this example correct?

f(x)=x, x<0
f(x)=x^2, 0<=x<1
f(x)=x, x>=1

f'(x)>=0 for all x  but there is no inflexion change at 0

f'(0)=0

Thx :)

The function $ f(x)\equal{}\left\lbrace{\begin{array}{l}
\displaystyle\frac{1}{6}\left({{\rm{Ci}}\left({\textstyle\minus{}\frac{1}{x}}\right)\plus{}4x^3\plus{}x^2\,\cos{\textstyle\frac{1}{x}}\plus{}\left({2x^3\minus{}x}\right)\sin{\textstyle\frac{1}{x}}}\right)\hspace{0.1cm},\hspace{0.3cm}x<0\vspace{0.2cm} \\
0\hspace{0.1cm},\hspace{7.4cm}x\equal{}0\vspace{0.2cm} \\
\displaystyle\frac{1}{6}\left({{\rm{Ci}}\left({\textstyle\frac{1}{x}}\right)\plus{}4x^3\plus{}x^2\,\cos{\textstyle\frac{1}{x}}\plus{}\left({2x^3\minus{}x}\right)\sin{\textstyle\frac{1}{x}}}\right)\hspace{0.1cm},\hspace{0.6cm}x>0
\end{array}}\right.$,

where $ {\rm{Ci}}(x)\equal{}\gamma\plus{}\ln{x}\plus{}\displaystyle\int_{0}^{x}{\frac{\cos{t}\minus{}1}{t}\,dt}$,  $ x\in\left(0,\plus{}\infty\right)$,  $ \gamma$ is $ Euler's \ constant$, is twice differentiable in  $ \mathbb{R}$.

$ f^{\prime}(x)\equal{}\left\lbrace{\begin{array}{l}
x^2\left({2\plus{}\sin{\textstyle\frac{1}{x}}}\right)\hspace{0.1cm},\hspace{0.3cm}x\neq0\vspace{0.2cm} \\
0\hspace{0.1cm},\hspace{2.3cm}x\equal{}0
\end{array}}\right.$ is continuous in $ \mathbb{R}$ and positive in $ \mathbb{R}^{*}$.

$ f^{\prime\prime}(x)\equal{}\left\lbrace{\begin{array}{l}
4x\plus{}2x\,\sin{\textstyle\frac{1}{x}}\minus{}\cos{\textstyle\frac{1}{x}}\hspace{0.1cm},\hspace{0.3cm}x\neq0\vspace{0.2cm} \\
0\hspace{0.1cm},\hspace{3.55cm}x\equal{}0
\end{array}}\right.$ it isn't continuous at $ x\equal{}0$ and for the sequenses $ x_{\nu}\equal{}\displaystyle\frac{1}{2\nu\pi}$, $ \nu\in\mathbb{N}$ and  $ y_{\nu}\equal{}\displaystyle\frac{1}{2\nu\pi\minus{}\pi}$, $ \nu\in\mathbb{N}$, holds: $ \mathop{\lim}\limits_{\nu\rightarrow{\plus{}\infty}}x_{\nu}\equal{}\mathop{\lim}\limits_{\nu\rightarrow{\plus{}\infty}}y_{\nu}\equal{}0$ and  $ \mathop{\lim}\limits_{\nu\rightarrow{\plus{}\infty}}{f^{\prime\prime}(x_{\nu})}\equal{}\minus{}1$ and  $ \mathop{\lim}\limits_{\nu\rightarrow{\plus{}\infty}}{f^{\prime\prime}(y_{\nu})}\equal{}1$.
So in every neighborhood of $ 0$, $ f^{\prime\prime}$ has positives and negatives values. Obviously $ x\equal{}0$ is not an inflection point of  $ f$.


Here's another problem from [b]Valentin Vornicu's book[/b]. I found the solution on my own. It's not really that hard but it's a nice problem:

a,b,c>0 ab+bc+ca=1 Show that:

a/(a^2-bc+1) + b/(b^2-ca+1)+ c/(c^2-ab+1)>= 1/(a+b+c)

Strange. Is this wrong ? Should be ...

a/(a - bc + 1)
= a/(a - bc + ab + bc + ca)
= a/(a + ab + ac)
= 1/(a + b + c)

Hence a/(a - bc + 1) = b/(b - ca + 1) = c/(c - ab + 1).
So the left hand side equals 3/(a + b + c).
And now it's obvious that 3/(a + b + c) :ge: 1/(a + b + c).

I must have read something in the wrong way ... :?

Cauchy works ...

LHS x (a(a - bc + 1) + b(b - ca + 1) + c(c - ab + 1)) :ge: (a + b + c)

so it suffices to show that if ab + bc + ca = 1/3 then

(a + b + c) :ge: a(a - bc + 1) + b(b - ca + 1) + c(c - ab + 1).

Now the last inequality is equivalent to

3 Symsum(ab) + 9abc :ge: a + b + c.

Homogenizing we have to prove that

Symsum(ab) + 3abc :ge: (a + b + c)(ab + bc + ca)

but this is an EQUALITY. However, the inequality holds :D


induction
combinatorics unsolved
combinatorics

Let $a_{n}$ denote the number of nonempty subsets $S$ such that
(1) $S \subseteq \{1,2,\dots,n\}$;
(2) all elements of $S$ have the same parity;
(3) each element $k \in S$ satisfies $k \geq 2|S|$ (where $|S|$ denotes the number of elements in $S$).

Prove that
$a_{2m-1}=2(F_{m+1}-1)$ and $a_{2m}=F_{m+3}-2$
for all $m \geq 1$, where $F_{n}$ is the nth Fibonacci number.

[quote="nebula42"][quote="who"]its originally due to i kaplansky can be done using formula for fibonacci nos which involves binomial coeff.[/quote]
Could you please show the solution? I'm not sure how to solve the problem.  :oops:[/quote]

If you're unsure, you can do it using induction on m, and then you'll see...


If $ x,y$ are non-zero real numbers such that $ |x|\plus{}y\equal{}3$ and $ |x|y\plus{}x^3\equal{}0$, what is the integer nearest to $ x\minus{}y$.

We could sketch the graphs of $ y\equal{} 3 \minus{} \mid x \mid$ and $ y\equal{} \minus{}x^{2} (x>0); y \equal{} x^{2} (x<0)$.

They meet where $ x\plus{}3\equal{}x^{2}$.

$ x$ is the positive root of thios quadratic and $ y$ is $ x^{2}$.

Then we want the nearest integer to $ x\minus{}y$. I think this might be $ \minus{}3$.


Έστω η συνάρτηση $f: (0,+\infty)\rightarrow(0,+\infty)$ η οποία  για κάθε $x,y>0$ ικανοποιεί τη σχέση $f(xy)\leq f(x)f(y).$ Για κάθε $x>0$ και για κάθε $n\in\mathbb{N^*},$ να αποδείξετε ότι:
$f(x^n)\leq f(x)\cdot(f(x^2))^{\frac{1}{2}}\cdots(f(x^n))^{\frac{1}{n}}.$

Το φτιάχνω σε Latex.....

[quote="jj-the-king"]Θα δουλέψω επαγωγικά χρησιμοποιώντας τη δεύτερη μορφή της επαγωγής ( δηλ θεωρώ ότι ισχύει το p(1),...,p(n) και δείχνω ότι ισχύει το p(n+1) και όχι θεωρώ απλά ότι ισχύει το p(n)!).

  Για n=1 ok.

 Έστω ότι ισχύει μέχρι και για $n=k$ ,δηλαδή

  $f(x)\leq f(x)$
  $f(x^2)\leq f(x)(f(x^2)^{\frac{1}{2}})$
  .
  .
  .
  .
  $f(x^k)\leq f(x)( f(x^2) ^ {\frac{1}{2}}).....( f(x^k) ^{\frac{1}{k}})$

  Θα δείξω ότι ισχύει για  $n=k+1$.

  Πολλαπλασιάζω και έχω:
  $f(x)f(x^2)...f(x^k) \leq ( f(x)^k )( f(x^2) ^{\frac{k-1}{2}}) ... ( f(x^k) ^{\frac{1}{k}}) ( f(x)f(x^2)...f(x^k) )^2\leq ( f(x)^{k+1})( f(x^2) ^{\frac{k+1}{2}}) ... ( f(x^k) ^{\frac{k+1}{k}})$

  όμως  $f(x^m)f(x^{k+1-m})\geq f(x^{k+1})$ και άρα

 $( f(x^{k+1}))^k \leq ( f(x)f(x^2)...f(x^k) )^2 \leq ( f(x) ( f(x^2) ^{\frac{1}{2}})... ( f(x^k) ^{\frac{1}{k}} )^{ k+1} )$

  $( f(x^{k+1}) ) ^{k+1} \leq ( f(x) ( f(x^2) ^{\frac{1}{2}} ) ... ( f(x^k) ^\frac{1}{k}) ( f(x^{k+1}) ^ {\frac{1}{k+1}) ) ^ { k+1}}$

  και άρα
$f(x^{k+1}) \leq f(x) ( f(x^2) ^{\frac{1}{2}}) ... ( f(x^k) ^{\frac{1}{k}} ) ( f(x^{k+1}) ^{\frac{1}{k+1}}) )$
  και έτσι τελειώνει η επαγωγή![/quote]


geometry
3D geometry
tetrahedron
geometry proposed

Let $ABCD$ be a regular tetraedron with sides 1 and $P$ it is a point in its interior.
 Prove that:

                                       \[ PA+PB+PC+PD<3. \] ;)

See http://www.mathlinks.ro/Forum/viewtopic.php?t=17028 or http://www.mathlinks.ro/Forum/viewtopic.php?t=19485 .

  darij


Prove that if $x\geq y\geq 0$ then
\[x\geq\frac12(\sqrt{x^2+y^2}+\sqrt{x^2-y^2})\]

Actually this is AM-QM for the numbers $\sqrt{x^2+y^2}$ and $\sqrt{x^2-y^2}$, nothing more...

  Darij


linear algebra
matrix
superior algebra
superior algebra unsolved

So i know that there are only two groups of order four up to isomorphism but I have a question regarding their multiplication tables.

There are three ways to complete the table so that we have a group. One gives us Klein's group and the other two are shown here.

$ \begin{matrix}e & a & b & c \\
a & e & c & b \\
b & c & a & e \\
c & b & e & a \end{matrix}$

$ \begin{matrix}e & a & b & c \\
a & b & c & e \\
b & c & e & a \\
c & e & a & b \end{matrix}$

The two groups having these multiplication tables certainly are isomorphic since both are cyclic but I cannot find a "structural" likelihood between the two tables nor can I see an obvious bijection between the elements(e $ \rightarrow$ e, a$ \rightarrow$ b, b$ \rightarrow$a c$ \rightarrow$c works but it took me a while).

My main question is: to what extent can you manipulate one table in order to get an equivalent one and therefore say the two groups whose tables you are manipulating are isomorphic? For example I doubt that you can switch rows or columns but I am not so sure about taking the transpose or turning the table on its side.

Also enlightening facts about how to see isomorphism from the tables more clearly are also welcomed.

You know that the only other group of order $ 4$ is cyclic.  $ b, c$ are generators of the first table and $ a, c$ are generators of the second table.

There is no reason you aren't allowed to switch rows and columns or to transpose; this respects the multiplication table.  The more useful thing you can do, since you're writing your multiplication tables "symmetrically," is to do whatever you to do to the rows to the columns as well (in other words, to conjugate by permutations).

I can see that now. Also swapping the two middle rows and columns  for the first table in my previous post you get the second one. It is sort of like row equivalent matrices from this perspective.


geometry
perpendicular bisector
angle bisector
analytic geometry

[b]See also the figure ![/b]

   We have a triangle $ABC$ , $AE = BF$ and $D$ is the second intersection point of the the circles $(A,F,C)$ and $(B,C,E)$.Prove that $CD$ bisects angle $ACB$.

[u] Babis[/u]

[color=red] Edited: Jan thank you. I have not read carefully the original german text.[/color]

[quote="Jan"]Dear Babis,

I'm afraid that your problem is false. However, it seems logical to me that $E$ should be on $AC$.
I think the problem is true if you make this correction.

Here's my proof:
Because $BDEC$ and $ADFC$ are quadrilaterals with a common angle, we can easiley see that $\angle ADF = \pi-\angle ACB = \angle BDE$. This implies of course that $\angle ADE = \angle FDB$.
Also, $\angle DAC = \pi-\angle DFC = \angle DFB$.

These two equal angles imply that $\Delta ADE$ and $\Delta FDB$ are similar, and because $\left|AE\right|=\left|BF\right|$, they are congruent.

We deduce that $\left|BD\right|=\left|DE\right|$, so $D$ is on the perpendicular bisector of $[BE]$, so $CD$ is the angle bisector of $\angle BCA$.

[img]http://img512.imageshack.us/img512/9628/stergiucorrectionuu9.jpg[/img]

Best regards,

Jan[/quote]

 [color=blue]Dear Jan ![/color]

 Really sorry ! It seems ,it was too late for me :(  to read carefully the german text , so instead  to take the point $E$ on $AC$ , I took it on side $AB$.
 I apologize to all of you and I congatulate  you for the nice solution !

 [u]Babis[/u]

[quote="silouan"]Hi , your solution Jan is similar to mine , nice work .
Mr Babis do you remember this problem ?(I have solved it when you was at Trikala  :wink: )[/quote]

Oh yes !

 Was it a problem in woot, or somewhere in a paper ? A! Now I remember ! It was a problem from the german competition '' Bundeswettbewerb - Mathematik ''. You where asked from someone in Germany to solve all the problems!And you had solved all of them. Last night I saw the solutions to those problems (in german) .To this geometry problem there are also solutions with analytic geometry and other solutions. The best is like Jan's and yours.
  Very nice silouan! Keep on solving hard problems.IMO  2007 is very close :) .
 [u] Babis[/u]


geometry
3D geometry
modular arithmetic
AMC
AIME

Find the smallest positive integer whose cube ends in 888.

Posted and solved: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=75142[/url]

My solution's slightly different from the other two.

[hide="Solution"]Let the desired number be $10^{2}a+10b+c.$ By inspection, $c=2.$

To find the tens digit, we examine $(10^{2}a+10b+c)^{3}$ modulus $10^{2}.$ Note that \begin{eqnarray*}(10^{2}a+10b+c)^{3}&\equiv & (10b+2)^{2}\pmod{10^{2}}\\ &\equiv & 20b+8 \pmod{10^{2}}. \end{eqnarray*} Since we want this quantity to equal $88,$ $b=4,9.$

To find the hundreds digit, we use modulus $10^{3}.$ For $b=4,$ \begin{eqnarray*}(10^{2}a+42)^{3}&\equiv & 3\cdot 42^{2}\cdot 10^{2}a+42^{3}\pmod{10^{3}}\\ &\equiv& 200a+88 \equiv 888 \implies a=4. \end{eqnarray*} For $b=9,$ \begin{eqnarray*}(10^{2}a+92)^{3}&\equiv & 3\cdot 92^{2}\cdot 10^{2}a+92^{3}\pmod{10^{3}}\\ &\equiv& 200a+688 \equiv 888 \implies a=1. \end{eqnarray*} Hence, $a=1$ yields the smallest value of $10^{2}a+10b+c$ and our answer is $192.$
[/hide]

The ones digit must be 2.
The number must be greater than 888.

888^(1/3) = 9^3  + C

So we start:
12^3
22^3
32^3
...
192^3

Kind of brute force and stupid, but another solution if you blank out under time pressure. 19 multiplications would not take long if you use:

(x + 2)^3

Where in our case, x = 1 ,2 , 3, , 4 , 5, 6, ...19

Again, this is stupid i kno. don't make fun of me.

[hide="Well sorry to revive but ive managed never to have seen this aime problem before and I have a new solution"]
Notice $888\equiv 400-8^{3}\mod{1000}$. This nicely adapts to the form of $(100a-8)^{3}\equiv 200a^{2}-512$ choosing $a=2$ for minimality shows that $\boxed{192}$ is a rather small solution.

Now the other solutions are clearly $10n+2$ so assume smaller solution $0<n<19$; then $192^{3}-(10n+2)^{3}=1000k=10(19-n)(192^{2}+(10n+2)^{2}+192(10n+2))$ so $192^{2}+(10n+2)^{2}+192(10n+2)\equiv 12 \mod{10}$ means that $5^{2}|19-n$ which is impossible.
Thus $192$ is minimal... :)
How was that for a completely nice and new method (but there might be a mistake somewhere...)
[/hide]

My solution I think is similar to the first solution I think, but I what I did is to easy look for what the 1's digit must be (2 is the only option after cubing each even number 0-9)

so you can write the number out as 100A+10B+2
squaring once and removing all terms that are perfectly divisible by 1000 (0mod1000)
you get 400A + 100B^2 + 40B + 4

multiplying the above by 100A + 10B + 2 again and taking the equation mod 10000 to get
200A + 600B^2 + 120B + 8
Setting this equal to 888 we get

200A + 600B^2 + 120B = 880

since 200A and 600B^2 are divisble by 100, the 120B term is the only term that will not be 0 mod 100. So 120B = 80 mod 100 so it can be 4 or 9

Plugging in 4 gives A = -2 but plugging in 9 gives A = 1

Main part of the problem is just to keep taking each equation mod 1000 to simplify.

Let the number be $N$. So $N^3\equiv 8\pmod {10}$, which means $N\equiv 2\pmod {10}$. Let $N=10a+2$. We have $(10a+2)^3\equiv 88\pmod {100}$, or $a\equiv 4\pmod 5$. Let $a=5b+4$. We have $(50b+42)^3\equiv 888\pmod {1000}$, or $b\equiv 3\pmod 5$ ([hide="proof"]\begin{align*}
(25b+21)^3 &\equiv 111\pmod {125} \\
3\cdot 25b\cdot 21^2+21^3-111 &\equiv 0\pmod {125} \\
3\cdot 25b\cdot 21^2+(25-4)^3+14 &\equiv 0\pmod {125} \\
3\cdot 25b\cdot 21^2+3\cdot 25\cdot 4^2-50 &\equiv 0\pmod {125} \\
3b\cdot 21^2+3\cdot 4^2-2 &\equiv 0\pmod 5 \\
3b+1 &\equiv 0\pmod 5 \\
b &\equiv 3\pmod 5.
\end{align*}[/hide]). Let $b=5c+3$. So all numbers of the form $250c+192$ satisfy the condition, the smallest positive one being $250(0)+192=\boxed{192}$. $\blacksquare$


function
calculus
derivative
limit
integration
logarithms
real analysis

Here is a nice problem.
Given function $f:\mathbb{R}{\rightarrow}(0,+\infty)$ which is differentiable,increasing function with bounded derivative and with property $\lim_{x{\rightarrow}\infty}{f(x)}=\infty$.Further let us denote $F(x)=\int_{0}^{x}{f(t)dt}$.Define $a_{0}=1,a_{n+1}=a_{n}+\frac{1}{f(a_{n})} (n{\geq}0)$,also $b_{n}=F^{-1}(n)$.Prove that $\lim_{n{\rightarrow}\infty}{(a_{n}-b_{n})}=0$.
Please don't write down the official solution.I have seen it and it differs from mine and I want to see more different and interesting solutions.

Let me try to prove the statement under the weaker assumptions that $f$ is just continuous and increasing to $+\infty$ (the boundedness of the derivative condition makes the statement of the problem much less useful than it could be). First of all, we have $\int_{a_i}^{a_{i+1}}f(t)\,dt\geqslant (a_{i+1}-a_i)f(a_i)=1$, so $\int_{a_0}^{a_n} f(t)\,dt\geqslant n$, whence $a_n\geqslant b_n$ for all $n$. Now, define the auxiliary sequence $c_n$ by $c_0=0$, $c_{n+1}=c_n+\frac1{f(c_{n+1})}$ (sic!). It is not hard to see that $c_n$ is well defined, i.e., we can find a unique $c_{n+1}>c_n$ once we know $c_n$. Now we have $\int_{c_i}^{c_{i+1}}f(t)\,dt\leqslant (c_{i+1}-c_i)f(c_{i+1})=1$, so $\int_{c_0}^{c_n} f(t)\,dt\leqslant n$, whence $c_n\leqslant b_n$ for all $n$.

The hard part of the solution is to show that $a_n-c_n\to 0$ as $n\to\infty$.

The first observation is that it is enough to show that $\liminf_{n\to\infty}(a_n-c_n)=0$. Indeed, using the condition that $f$ tends to $+\infty$, it is not difficult to show that $a_{n+1}-a_{n}\to 0$ and $c_{n+1}-c_n\to 0$. Suppose now that $c_n>a_n-\delta$ and $n$ is very big. If $c_{n+1}>a_n$, then $c_{n+1}>a_{n+1}-\delta$ as well. Otherwise, $a_{n+1}-c_{n+1}=a_n-c_n+\frac 1{f(a_n)}-\frac 1{f(c_{n+1})}\leqslant a_n-c_n<\delta$ again. So, once the difference is less than $\delta$ for one sufficiently large $n$, it remains below $\delta$ forever.

Assume now that there exists $\delta>0$ such that $a_n-c_n>3\delta$ for all sufficiently large $n$. Then, $a_n-c_{n+1}>2\delta$ for large $n$ and, using again the formula   $a_{n+1}-c_{n+1}=a_n-c_n+\frac 1{f(a_n)}-\frac 1{f(c_{n+1})}<a_n-c_n-\left(\frac 1{f(a_n-2\delta)}-\frac 1{f(a_{n})}\right)$, we see that we must have $\sum_n\left(\frac 1{f(a_n-2\delta)}-\frac 1{f(a_{n})}\right)<+\infty$. Let us combine together the terms in this series with $a_n\in (m\delta, (m+1)\delta))$ for very large $m$. Note that there are at least $\frac 12\delta f(m\delta)$ such terms for big $m$ (because the step of the sequence $a_n$ on the interval $(m\delta, (m+1)\delta)$ does not exceed $\frac 1{f(m\delta)}$ and this step is much less than $\delta$ for large $m$). Also, each such term is at least $\frac 1{f((m-1)\delta)}-\frac1{f(m\delta)}$. Thus, we must have $\sum_m \left(\frac{f(m\delta)}{f((m-1)\delta)}-1\right)<+\infty$. But for any positive sequence $F_m$ increasing to $+\infty$, we have $\sum_m\left(\frac {F_m}{F_{m-1}}-1\right)\geqslant \sum_m \ln\frac{F_m}{F_{m-1}}=+\infty$, so we've got a contradiction.

The end.

Indeed nice solution(you caught the wave [b]fedja[/b] :)).
My solution uses two main ideas.Here is my solution.
First of all some notes.$a_{n}$ monotnically tends to infty,besides $a_{n+1}-a_{n}$ decreases to zero since $f$ monotonically tends to infty.So does $b_{n}$.Next instead of $b_{n}$ I considered a sequence $c_{n}$ s.t. $\int_{a_{0}=1}^{c_{n}}{f(t)dt}=n$ and $c_{0}=1$.Again it is easy to show that $c_{n}$ monotonically goes to infty and $c_{n+1}-c_{n}$ decreases to zero:
$\frac{1}{f(c_{n+1})}=\frac{\int_{c_{n}}^{c_{n+1}}{f(t)dt}}{f(c_{n+1})} <c_{n+1}-c_{n}<\frac{\int_{c_{n}}^{c_{n+1}}{f(t)dt}}{f(c_{n})}=\frac{1}{f(c_{n})}$
Besides $c_{n}-b_{n}$ tends to zero:
$0<c_{n}-b_{n}<\frac{\int_{b_{n}}^{c_{n}}{f(t)dt}}{f(b_{n})}=\frac{\int_{0}^{1}{f(t)dt}}{f(b_{n})}$.
Now it stays to show that $\lim_{n{\rightarrow}\infty}{(a_{n}-c_{n})}=0$.
$\int_{a_{0}}^{a_{n}}{f(t)dt}=\sum_{i=0}^{n-1}{\int_{a_{i}}^{a_{i+1}}{f(t)dt}}=\sum_{i=0}^{n-1}{\int_{a_{i}}^{a_{i+1}}{(f(t)-f(a_{i}))dt}}+\sum_{i=0}^{n-1}{f(a_{i})(a_{i+1}-a_{i})}=\sum_{i=0}^{n-1}{\int_{a_{i}}^{a_{i+1}}{(f(t)-f(a_{i}))dt}}+n=\Sigma_{n}+\int_{c_{0}}^{c_{n}}{f(t)dt}$
so
$\int_{c_{n}}^{a_{n}}{f(t)dt}=\Sigma_{n}>0$(as a consequence $c_{n}<a_{n}$).
[b]Idea 1.[/b]
$\lim_{n{\rightarrow}\infty}{\frac{\Sigma_{n}}{f(a_{n})}}=0$.
[i]Proof.[/i]
$0<\frac{\Sigma_{n}}{f(a_{n})}<\frac{\Lambda_{n}}{f(a_{n})}$,where
$\Lambda_{n}=\sum_{i=0}^{n-1}{(f(a_{i+1})-f(a_{i}))(a_{i+1}-a_{i})}=\sum_{i=0}^{n-1}{(\frac{f(a_{i+1})}{f(a_{i})}-1)}$.
$0<\frac{f(a_{n+1})-f(a_{n})}{f(a_{n})}{\leq}K{\cdot}\frac{a_{n+1}-a_{n}}{f(a_{n})}=K{\cdot}\frac{1}{f^{2}(a_{n})}{\rightarrow}0$,so
$\frac{f(a_{n+1})}{f(a_{n})}-1{\sim}\ln{\frac{f(a_{n+1})}{f(a_{n})}}$,thus $\frac{f(a_{n+1})}{f(a_{n})}-1<2{\cdot}\ln{\frac{f(a_{n+1})}{f(a_{n})}}$ for sufficently large $n$,so $\Lambda_{n}<\Lambda_{N}+2{\cdot}\ln{\frac{f(a_{n})}{f(a_{N})}}$,$n>N$ and we get what we wanted to prove.
[b]Idea 2.[/b]
$0<\delta_{n}=a_{n}-c_{n}$ is bounded.
[i]Proof[/i].
I will prove that $\delta_{n}{\leq}a_{1}-a_{0}$ for all $n$.
For $n=1$ it is clear.Now suppose we have that
$\delta_{n}{\leq}a_{1}-a_{0}$.
If $c_{n+1}>a_{n}$ then $\delta_{n+1}<a_{n+1}-a_{n}{\leq}a_{1}-a_{0}$,if
$c_{n+1}{\leq}a_{n}$
$f(a_{n})(a_{n+1}-a_{n})=1=\int_{c_{n}}^{c_{n+1}}{f(t)dt}{\leq}f(c_{n+1})(c_{n+1}-c_{n}){\leq}f(a_{n})(c_{n+1}-c_{n}) $,thus $\delta_{n+1}{\leq}\delta_{n}{\leq}a_{1}-a_{0}$.
And the final step is the following:
$\Sigma_{n}=\int_{c_{n}}^{a_{n}}{f(t)dt}=f(a_{n})\delta_{n}-\frac{1}{2}f'(\zeta_{n})(\delta_{n})^{2}$,so
$\delta_{n}=\frac{\Sigma_{n}}{f(a_{n})}+\frac{f'(\zeta_{n})(\delta_{n})^{2}}{2f(a_{n})}$
and the conclusion folows from two main ideas.That's all folks.
 :)
How do you like it?


I think there was a typo. What do you mean by the probability being n+1 2(2n-1)

yes I saw this problem too. I don't understand the solution though.

[quote="The Quadratic Kid"]That is stupid. You are dumb eagle2. It is so easy I can answer it![/quote]
@The Quadratic Kid, please do not post useless insults at other users.  That is just spam.  We are here to learn, not to boast and make fun of others.  Everyone is at their own level.  No one should be ashamed of that.  Making fun of others only has negative affects on them, so please do not make fun of others.  Remember, we are here to learn, [b]not to brag[/b].

[quote="The Quadratic Kid"]That is stupid. You are dumb eagle2. It is so easy I can answer it![/quote]

Exactly, thank you for expressing all our thoughts JSGandora.  Now The Quadratic Kid, if you really think you know the solution then post it.

[quote="The Quadratic Kid"]WHO CARES? HE IS AS DUMB AS A DIME! NO WORSE!
The answer is so easy you can tell me it. You are stupid too then?[/quote]

Yes, exactly, I'm stu---.  Now show us that you are "smart", which I doubt since you haven't posted the solution yet.

JEEZ! The quadratic Kid! I really shouldn't have shown you to Aops.

[quote="The Quadratic Kid"]I'll tell you the answer if you tell me yours. All these Aopsers are retarded![/quote]
@The Quadratic Kid Again, do not flame others.  If you have nothing better to do than insult other users, then you shouldn't be here on AoPS.  I ask you to stop the insults.

how come you aren't answering the question, The Quadratic Kid?

[quote="The Quadratic Kid"]You did a good thing inviting me to Aops. Now I can insult everyone![/quote]
Doing so can very likely get you banned.  Personally, it is quite unfortunate to be banned from such a useful resource of mathematics, but the choice is yours.  :no:

Okay, guys, just ignore The Quadratic Kid and figure out the prob. This is a forum for discussing problems. [b]NOT[/b] for insulting people.
Btw, The Quadratic Kid, I would stop insulting people, if I were you. JUST a suggestion, if you don't want to get banned. (Cuz you will..., the admins will find out 'bout you.)

[quote="wovenfractal"]A regular polygon with 2n + 1 sides is inscribed in a circle. Three of the polygon’s vertices are
selected at random and the triangle formed by connecting these vertices is drawn. Prove that
the probability that the center of the circle lies inside the triangle is n+1
2(2n−1) .[/quote]
About the problem, are you sure this is a problem from Alcumus?  If I am not mistaken, Alcumus does not ask for proofs, because you can only type in an answer...  :what?:

If this is not an Alcumus problem, I think it should be moved to a different forum or subforum.

[quote="JSGandora"][quote="wovenfractal"]A regular polygon with 2n + 1 sides is inscribed in a circle. Three of the polygon’s vertices are
selected at random and the triangle formed by connecting these vertices is drawn. Prove that
the probability that the center of the circle lies inside the triangle is n+1
2(2n−1) .[/quote]
About the problem, are you sure this is a problem from Alcumus?  If I am not mistaken, Alcumus does not ask for proofs, because you can only type in an answer...  :what?:[/quote]

Pretty sure this is a problem from Alcumus.  Maybe wovenfractal modified the problem a bit.  Maybe the original problem was

[quote="Alcumus"]
A regular polygon with 2n + 1 sides is inscribed in a circle. Three of the polygon’s vertices are
selected at random and the triangle formed by connecting these vertices is drawn.  What is the probability that the center of the circle lies inside the triangle?
[/quote]

yeah probaly. Sorry I looked at the first few words and it looked the same so sorry!

I posted this question in another forum, and someone gave this answer:

[quote="Jason27603"]Using complementary counting, this is equivalent to counting the number of obtuse triangles, since the circumcenter of any obtuse triangle is outside of the triangle.
Let $\angle ABC$ be obtuse (so that $A$,$B$,$C$ are on the same half of the circle), and let there be $m$ vertices between $A$ and $C$, so that $B$ is one of those $m$ vertices. For each $m\leq n-1$, there are $2n+1$ choices for $A$ and $C$, and for each choice of $A$ and $C$ there are $m$ choices for $B$. Thus, the total number of obtuse triangles is
\[\sum_1^{n-1}m(2n+1)=(2n+1)\sum_1^{n-1}m=\frac{(2n+1)n(n-1)}{2}\]
The total number of possible triangles is $\binom{2n+1}{3}$, so the fraction of obtuse triangles is $\frac{(2n+1)\frac{n(n-1)}{2}}{\binom{2n+1}{3}}=\frac{\frac{(2n+1)n(n-1)}{2}}{\frac{(2n+1)(2n)(2n-1}{6}}=\frac{\frac{n-1}{2}}{\frac{(2n-1)}{3}}=\frac{3n-3}{4n-2}$, so the answer to the question is $1-\frac{3n-3}{4n-2}=\frac{n+1}{4n-2}$.[/quote]

Which is exactly what wovenfractal said.

[quote="joycezhou1"]Um, has anyone noticed??
[quote="joycezhou1"] You know, mice are actually sorta smart. Yeah, I think that wovenfractal changed the problem a bit. [/quote]
[quote="nsato"]User "The Quadratic Kid" has been banned from the website.[/quote]
[quote="eagle2"]Thank God. Sorry if that offended anyone[/quote]
These posts are on both page 1 [b]and[/b] 2 of this thread.[/quote]
Yes, we have noticed, and why did you bring this up?

Uh... because it was unusual. I was just curious about why it is like this.

Hey Joyce, did you notice your avatar is weird now, it cant be found or something.

The fact that he was banned on a topic I started makes me queasy. :(

Yeah, I altered an alcumus problem. I had an idea for a generic identity. Thank you, pd. The proof is quite elegant. Makes me happy. :)

[quote="eagle2"]i think he means $\dfrac{n + 1}{2(2n-1)}$[/quote]

Not that it matters, but I'm a girl. :)

@rahster: Yea, I know. I don't know what happened with my avatar. It just disappeared and I didn't even change it or anything. Anyway, it's fine now.

Really?? I can see it, though. It looks perfectly fine to me.

I can see your avatar, but I don't think this should be discussed here?

Sorry, it shouldn't be discussed here, but now I will close it since I started it.

@joyce: After I saw your post I could not see it, but now I can. Now we can stop discussion.


Take a look on Wilson's Theorem:
http://mathworld.wolfram.com/WilsonsTheorem.html

I don't think you need Wilson...

It is true for all non-prime numbers. When a number is prime, it is obviously not true.
When a number is composite, it can be written as the product of at least 2 prime factors. Those prime factors are smaller than $n-1$ (Proof is trivial), so $n$ divides $(n-1)!$. :D

Jan: Looking at it better, I think that there's a problem. See for example $n=4$...

[quote="José"]Then, the answer would be: For all composite $n$, $n\geq 6$... please if any agree or disagree please quote this post...[/quote]

You forgot 1. And use the Edit button.
Jan's proof is correct, except that 4 has only one prime factor, which occurs only once in 3!.

Really, the proof needs a separate case for prime powers.  Then, when you examine it, every $p^n$, $n \geq 3$ is also fine because you have a $p$ and $p^2$ factor, so you only need to deal with the squares.  And then it's fairly clear that $2^2$ is the only one that will be problematic.

[quote]Really, the proof needs a separate case for prime powers.[/quote]
Of course you are right, thanks


For $ a, b, c> 0$ prove
\[ \sum \frac {1}{4a} \plus{} \sum\frac {1}{a \plus{} b}\geq 3\sum \frac {1} {3a \plus{} b}.\]

I think the only known solution of this one uses "Harazi integration". See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83217]here[/url]  :lol:

Here is some more: http://www.mathlinks.ro/Forum/viewtopic.php?t=83869


trigonometry
geometry
inradius
Pythagorean Theorem
trig identities
Law of Sines

I was teaching myself some Trig, and there were a few problems that seemed pretty good.

8. In isosceles triangle $ ABC$, the base angles $ B$ and $ C$ have a measure of $ 50^\circ$. The sides of the triangle each touch a circle of radius 2. Calculate the length of $ BC$

9. A ladder of length $ 5 m$ is leaning against a vertical wall at an angle of $ 60^\circ$ to the horizontal. The foot of the ladder moves in by $ 50$ cm. By how much does the top of the ladder move up the wall?

[hide="My Answers"]
8. 8.58 cm
9. 25.24 cm
[/hide]

[hide]
i got 28.4 cm for the 2nd one

here is my equation.
$ x \equal{} 2.5$
$ y \equal{} \frac {5\sqrt {3}}{2}$

$ x^2 \plus{} y^2 \equal{} 5^2$
$ (x \minus{} .05)^2 \minus{} (y \plus{} k)^2 \equal{} 5^2$

and solve for $ k$[/hide]

I agree with happyme on the 2nd one.
[hide]Originally, the base of the ladder is 2.5m out on the horizontal axis, and $ 2.5 \sqrt {3}$ up the vertical (30-60-90 right triangle). When the base moves in 50 cm, or .5 m, it is now 2m out on the horizontal axis, and by the Pythagorean theorem, $ \sqrt {21}$ up the vertical axis. $ \sqrt {21} \minus{} 2.5\sqrt {3} \approx .252448676\text{...}$. It asks in centimeters, so the answer is $ 25.2448676 \text{...}$[/hide]

I too agree with happyme, this is because $ 50$ cm$ \equal{} .5$ m [b]NOT[/b] $ .05$ m

and then for the first prob I was able to get my work down to an equation with one variable by using the fact that the inradius of a triangle is $ \frac{2A}{a\plus{}b\plus{}c}$, the law of sines, and the Pythagorean theorem.  The Pythagorean theorem I used to write the area in terms of $ a$ and $ b$.  Then the law of sines was used to write $ b$ ($ BC$) in terms of $ a$ and then it was just algebra.  I would have shown all my algebra but I have a LOT of reading to catch up on and I've already spent the majority of today MATHing.  I never came to a numerical value for $ BC$ because at the point where I was, the rest was just tedious calculator work.

Solution for problem #1:

[hide]
We know that the intersection point of 2 bisectors of 2 angles is the center of the incircle, which is $ M$.
$ F$ is the midpoint of $ BC$.
Furthermore we know that $ FM = 2$, which in the right triangle $ BFM$ is a cathetus.

Then $ tan(25^{\circ}) = \frac 2{x}$ and thus $ x = \frac 2{tan(25^{\circ})}$

$ 2*x = 8.58$ is the solution.[/hide]

The answer is correct for the second problem, it is 25.24 cm


trigonometry
equation
algebra
Trigonometric Equations
IMO
IMO 1961

Solve the equation $\cos^n{x}-\sin^n{x}=1$ where $n$ is a natural number.

The question is pretty simple, felipesa, you're making things too complicated :D

45 years ago IMO problems were rather different from what they are now  !

If $n$ is even then $cos(x)^n = 1 + sin(x)^n$
RHS is $> 1$ except when $sin(x) = 0$

Solutions are $k \times \pi$

If $n$ is odd, same reasonning, we must have $cos(x) = 0$ and $sin(x) < 0$ or else $cos(x) > 0$ and $sin(x) = 0$

Solutions are $2 \times k \times \pi$ and $2 \times k \times \pi - \pi/2$

I solved an IMO #3 in 10 minutes :o

Suppose $n$ is even. Then $\cos^n x$ and $\sin^n x$ are positive, so the only time $\cos^nx-\sin^nx=1$ is when $\cos x=1,\,-1$ and $\sin x=0$. This happens at $x=k\pi$ for all $k\in \mathbb{Z}$.

Now suppose $n$ is odd. Then, $\cos^n x$ must be greater than or equal to $0$ (and thus $\cos x\ge0$) and $\sin^n x$ must be less than or equal to $0$ (and thus $\sin x\le 0$). This only happens in the interval $\left[\frac{3\pi}2+2k\pi,2\pi+2k\pi\right]$ for all $k\in\mathbb{Z}$. We claim that the equation is satisfied only at the endpoints. We can verify this works.

If $x$ is not in the endpoints, then $\cos^nx-\sin^nx=(\cos^{n-2}x)(\cos^2x)-(\sin^{n-2}x)(1-\cos^2x)$. Since $0<\cos x<1,\,0< \cos^{n-2}x<1$, and similarly $-1< \sin^{n-2}x<0$. Thus, $(\cos^{n-2}x)(\cos^2x)-(\sin^{n-2}x)(1-\cos^2x)< (\cos^2x)-(-(1-\cos^2x))=1$, and the equation is not satisfied.


function
limit
calculus
calculus computations

f(x) is a continuous function . We have 2 case :
1 ) $ f: [a, \plus{} \infty ) \minus{} > R$ and  $ \lim_{x \to \plus{} \infty}{f(x)} \equal{} a$
2 ) $ f: ( \minus{} \infty , \plus{} \infty ) \minus{} > R$ and  $ \lim_{x \to \plus{} \infty}{f(x)} \equal{} a$ , $ \lim_{x \to \minus{} \infty}{f(x)} \equal{} b$

Prove that in all 2 cases , we have a Real number $ M$ such that : $ |f(x)| \leq M$

There are many analogous cases in this proof, so I won't go into every detail - those analogies and the motivation for this solution seem obvious enough.

2) clearly follows from 1): divide $ f$ into two parts: $ f_1: \langle \minus{} \infty,a]\to\mathbb{R}$ and $ f_2: [a, \plus{} \infty\rangle\to\mathbb{R}$ such that $ f\mid_{\langle \minus{} \infty,a]} \equal{} f_1$ and $ f\mid_{[a, \plus{} \infty\rangle} \equal{} f_2$. Both functions are continuous. From 1), we'd get that $ f_2$ is bounded and also, by analogy which will be obvious in the proof of 1), we get that $ f_1$ is bounded too. Take the maximum of these two bounds and thus $ f$ is bounded.

Now, about 1). Take $ S \equal{} \{f(x): x\in[a,\ \plus{} \infty\rangle\}$. Since $ S$ is not empty, it has a supremum $ t$ in $ \overline{\mathbb{R}}$. We want to prove that $ t\in\mathbb{R}$. In that case, $ f$ has an upper bound. An analogous proof will work for $ \inf S$, so $ f$ will have a lower bound too, and thus it will be bounded.

Right, so let's prove that $ t\in\mathbb{R}$. Let's suppose otherwise: $ t \equal{} \plus{} \infty$. Thus, there exists a sequence $ (x_n)_{n\in\mathbb{N}}$ such that $ f(x_n) > n$ for all $ n\in\mathbb{N}$. Since $ x_n$ must have a limit point in $ \overline{\mathbb{R}} \equal{} \mathbb{R}\cup\{\minus{}\infty, \plus{} \infty\}$, take its subsequence $ x_{p_n}$ such that $ x_{p_n}\rightarrow L$ for some $ L\in\overline{\mathbb{R}}$. Obviously, $ L\geq a$ so we don't have any problems with $ f(L)$. Now, if $ L\in\mathbb{R}$ , by continuity we have $ f(x_{p_n})\rightarrow f(L)$. That is a contradiction since $ f(x_{p_n}) > p_n > n$ for all $ n\in\mathbb{N}$. If $ L \equal{} \plus{} \infty$, we have $ f(x_{p_n})\rightarrow A$. However, that is again a contradiction (same reason: $ f(x_{p_n}) > p_n > n$). [size=75][i](I don't know if you've made a typo or not, but you use $ a$ in the sense of both $ [a, \plus{} \infty\rangle$ and $ f(x)\rightarrow a$ for $ x\rightarrow \plus{} \infty$. Of course, a stronger claim is still correct, we can have $ \lim_{x\rightarrow \plus{} \infty}f(x) \equal{} A$ for any $ A\in\mathbb{R}$, not necessarily $ A \equal{} a$.)[/i]
[/size]

Yes it is a foundation problem in analytics  :)


How do you do a square root?

$ sqrt(2)$

[code]\sqrt{insert number here}[/code]

So

$ \sqrt{81}$ is simply \sqrt{81}.

[quote="istos"]Thanks!

$ \sqrt(2)$[/quote]

Actually, use the { } bracket keys. They will make your square roots look better.

Also, to add on to that, you can use the following code to make cube, quadruple, quintuple etc. roots:
[code]$\sqrt[x]{y}$[/code]
produces
$ \sqrt[x]{y}$
:)


A wheel has a radius of 5 feet. What is the minimum number
of complete revolutions that the wheel must make to roll at least
1,000 feet?

[hide]
find the circumfrence, which is 10(pi), then divide this into 1000 to get appox. 31.83[/hide]


can anyone give me some websites with prook techniques or books that cover this area.

I don't know if this helps:
[url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php[/url]

basically a site other then the art of problem solving site.  I've looked through most of the links.

Google the word "proof" in combination with words like "introduction," "beginner's guide," "how to write," and so on. I've gradually landed on quite a few interesting articles on the Web about proof, but I haven't organized my bookmarks yet.


abstract algebra
group theory
superior algebra
superior algebra solved

Let $G$ be a group of order $n$ with the property (*): exist $a,b\in G-\{e\},a\ne b$  s.t $a^2=b^2=e$. Prove that:
a)$n$ is even
b)if $n=4k+2,k\ge1$ then $G$ is not abelian
c)if $n=4k,k\ge2$ then there exist two groups $G_1,G_2$ with the property (*) s.t. $G_1$ is abelian and $G_2$ is not abelian.

[i]Marian Ionescu and Rica Zamfir[/i]

First two points are easy.
[b]a)[/b]If $a^{2}=e,a\neq{e}$ then the degree od the element $a$ is $2$ so by the Lagrange theorem $2$ must divide the order of $G$ so $n$ is even.
$b)$Now let $n=4k+2,k\geq{1}$.Now assume that $G$ is abelian.The subgroup of  group $G <a>$ is a normal divisor so we can consider the factorgroup $G/<a>$ and the  order of last one is $n/2=2k+1$.Now let's look at ${b<a>}\in{G/<a>}$,we have that
${b<a>}^{2}={{b}^{2}}<a>=e<a>=<a>$ and $b<a>\neq{<a>}$ because $b\notin<a>={e,a}$ so the degree of the element $b<a>$ in group $G/<a>$ is $2$ which is controdiction with Lagrange theorem because $2$ does not divide $2k+1$.So $G$ is not abelian.
Now I have a question about point $c)$.You mean that in the case $n=4k$ we cannot say that G is abelian or does not and we have to demonstrstse two examples $G_{1}$ and $G_{2}$.

for a) I have the same solution
for b)
Suppose that $G$ is abelian.
 $|G-\{e,a,b\}|=4k-1$ results that there are another odd number of elements in $G-\{e,a,b\}$ with the property that $x^2=e$.
Let's consider $H=\{x\in G|x^2=e\}$. From the previous we conclude that ord$H$ is even.
Obviously $H\le G$ ($H$ is a subgroup of $G$).
I think it's a well known result that if a group $G'$ with the property that $x^2=e,\forall x\in G'$ then ord$G'=2^t$ for some $t\in\mathbb N^*$.
Using this and the fac that $H$ has at least 4 elements results that 4|ord$H$|ord$G$ $\Rightarrow4|4k+2$ contradiction. So $G$ is not abelian.

c)we have to prove that if $n=4k$ then there exist two groups $G_1$ and $G_2$ with the property (*) s.t. $G_1$ is abelian and $G_2$ is not abelian.
So if $n=4k$ and a group $G$ has the property (*) we cannot say precisely if it is abelian or not.

Now for [b]c)[/b] $G_{1}=Z/2Z\times{Z/2Z}$ which is abelian and $G_{2}=S_{4}$(symmetric group) which is not abelian.
Yes,one more question.The statement that if all elements of the finite group have degree $2$ than the order group is the power of $2$,is the corollary of the Cachy theorem
which says that if the prime number $p$ divides the order of the group than the group has an element of the degree $p$,or you have more simple prove?

hmmm... I don't really get it with that corollary of Cauchy theorem... :( would you please be more explicit?

the proof I know for that statement is as follows:
Let $H\le G$ a subgroup of $G$ with the property that $|H|=2^t$ and there does not exist another one with the order $2^q,q>t$.(we have such a $H$ because if we take $H=\{e,x\},x\in G$ then $t=1$).
Assume that $G\ne H$. Then $xH\cap H=\emptyset$ with $x\in G-H$ and $xH\cup H$ is a subgroup of $G$ with the order $2^{t+1}$ which is a contradiction.
So $G=H$ and ord $G=2^t,t\in\mathbb N$

For [b]c[/b], we need a general construction for two groups of the [i]same[/i] order $4k$. The abelian group can be $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2k\mathbb{Z}$. The nonabelian group, for $k\ge 2$, is the dihedral group (symmetries of a regular polygon with $2k$ sides).

Also, clearing up part [b]b[/b]: If the group is abelian, consider the subgroup generated by two distinct elements of order two. This subgroup has four elements. Since the order of a subgroup divides the order of the group, we have a contradiction and the group is not abelian.

doesn't $\mathbb Z/2\mathbb Z$ x $\mathbb Z/2k\mathbb Z$ have an infinite order?  :?

Yes,you're right but here $Z/nZ=Z_{n}$(actually they can be identyfied with each other)


For problem 111 (chapter 6) in AoPS Vol 1, the answers are -9 and 26. But the solution manual says that 26 is the only answer. It says that -9 is an extraneos root. I don't understand why -9 is an extraneos root. Can anyone answer why -9 is NOT an answer?

I have encountered several other questions in AoPS where I got 2 answers, where the solution
manual gives only one. When I substitute my both answers in the equation, both of them prove to be correct.

Problem 111 (Chap 6) in AoPS Vol 1 is as follows:

$\sqrt{x+10} - \frac{6}{\sqrt{x+10}} = 5$

[quote="JesusFreak197"]Squaring the equation results in extraneous roots at times, i.e. roots that are answers to the squared equation but not the original one. Try plugging $-9$ back into the equation; it won't work. ;)[/quote]

But in this case if I plug in $-9$ into $\sqrt{x+10}$, it is $\sqrt{1} = \pm 1$. Using $+1$ doesn't give the right answer. But using $-1$, we get,
$\sqrt{x+10} - \frac{6}{\sqrt{x+10}} = -1 - \frac{6}{-1} = 5$, which is the correct answer. So $-9$ is a root of this equation as well!

The square root of 1 is just 1, not negative 1.  If x^2 = 1, then x can be either, but if you are strictly looking for the square root of 1, then 1 is the only answer.  I believe the technical term is principle root.

If you are going to use negative square roots, it must have a negative sign in front of it, or say $x^2$.  $-\sqrt{1+3}=-2$, the solution to $\sqrt{1+3}=2$, and $x^2=1+3$ gives $-2$ and $2$.

$\sqrt{x}$ is defined in such a way that it only gives one value (otherwise it is not a valid function) - that is, it will only return the positive square root of $x$.  Saying that " $-1$ is a square root of $1$ " means that $x^2 = 1$ has $x = -1$ as one of its solutions, but it does NOT mean that $\sqrt{1} = -1$.  (the distinction might become clearer once you learn more about inverse functions.)

Similarly, you would not expect me to say $\sqrt[3]{8} = -1 + \sqrt{3} i$, would you?  But it is a valid solution to the equation $x^3 = 8$  :D


algebra
polynomial
modular arithmetic
superior algebra
superior algebra solved

If $a$ belongs to a field $F$ of characteristic $p \neq 0$, such that $a$ is not a power of $p$, prove that $X^{p}-a$ is irreducible in $F[X]$.

I got a proof but I feel I am missing some very easy argument.

Well, the equation $x^p=a$ has no root by definition. Take derivation to see that there is only one root $\zeta$, and every factor of $x^p-a$ would be of the form $(x-\zeta)^k$ with small $k$. Well, look at the second highest coefficient $a_{n-1}$ of a factor $x^n +a_{n-1} x^{n-1} + ... + a_0$: it lies in the field and equals $-k \zeta$, which is just possible for $k=0,p$.

$\zeta$ is a root of which polynomial?? Is it a root of  $(X^p-a)'$ ?It has only 0 as root. Can you give more detailes?  :?

[quote="steven128"]every root is of the form $\zeta \sqrt[p]{a}$, where $\zeta$ denotes a primitive $p$th root of unity, so the constant term of any factor is $\zeta^n \sqrt[p]{a}^k$, where $k$ denotes the degree of the factor. If $\sqrt[p]{a}^k \in K$ for $1 \leq k < p$, also $\sqrt[p]{a}$ because $k$ is invertible modulo $p$ ($p$ prime): $kk' \equiv 1 \pmod{p}$, so $(\sqrt[p]{a}{}^k){}^k' = a^m\sqrt[p]{a} \in K$[/quote]

I still have some questions.. You have that $\zeta^n \sqrt[p]{a}^k\in K$ not $\sqrt[p]{a}^k\in K$. How the conclusion follows?

$\zeta$ is a root of $f(x)=x^p-a=0$ in an extension of $K$. $f'(x)=px^{p-1}=0$,because $K$ have caracteristic $p$. Right? I think I understood now.. :)

[quote="ZetaX"]$\zeta$ is the root of $x^p-a=0$ !
There is no other root and no primitive root of unity to consider, just see that $(x-\zeta)^p = x^p - \zeta^p = x^p-a$ ($p=2$ is easy and neglected here).[/quote]ugh, I thought that the characteristic should be [i]different[/i] from $p$


If $a,b,c$ are positive numbers then
$\frac{\sqrt{a^2+bc}}{b+c}+\frac{\sqrt{b^2+ca}}{c+a}+\frac{\sqrt{c^2+ab}}{a+b} \geq \frac{3\sqrt{2}}{2}$.

$\frac{\sqrt{a^2+bc}}{b+c}+\frac{\sqrt{b^2+ca}}{c+a}+\frac{\sqrt{c^2+ab}}{a+b} \geq \frac{3\sqrt{2}}{2}\Leftrightarrow\sum_{cyc}\left(\frac{\sqrt{a^2+bc}}{b+c}-\frac{1}{\sqrt2}\right)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{2a^2-b^2-c^2}{(b+c)(\sqrt{2(a^2+bc)}+b+c)}\geq0\Leftrightarrow\sum_{cyc}(2a^2-b^2-c^2)P_a\geq0,$  where
$P_a=\frac{1}{(b+c)\sqrt{2(a^2+bc)}+(b+c)^2},P_b=\frac{1}{(a+c)\sqrt{2(b^2+ac)}+(a+c)^2}$
and $P_c=\frac{1}{(a+b)\sqrt{2(c^2+ab)}+(a+b)^2}.$
Let $a\geq b\geq c.$ Hence, $P_a\geq P_b\geq P_c.$
$P_a\geq P_b$ since, $(b+c)^2\leq(a+c)^2$ and $(b+c)\sqrt{2(a^2+bc)}\leq(a+c)\sqrt{2(b^2+ac)}\Leftrightarrow$
$\Leftrightarrow c(a-b)(a^2+b^2+c^2-ab+ac+bc)\geq0.$
Similarly $P_b\geq P_c.$
But $\sum_{cyc}(2a^2-b^2-c^2)P_a\geq0\Leftrightarrow$
$\Leftrightarrow(a^2-b^2)(2P_a-P_b-P_c)+(b^2-c^2)(P_a+P_b-2P_c)\geq0.$ :)

Nice solution, arqady.
A more difficult inequality is the following:
If $a,b,c$ are positive numbers then
$\frac{\sqrt{a^2+2bc}}{b+c}+\frac{\sqrt{b^2+2ca}}{c+a}+\frac{\sqrt{c^2+2ab}}{a+b} \geq \frac{3\sqrt{3}}{2}$.

[quote="arqady"]$\Leftrightarrow\sum_{cyc}\frac{2a^{2}-b^{2}-c^{2}}{(b+c)(\sqrt{2(a^{2}+bc)}+b+c)}\geq0\Leftrightarrow\sum_{cyc}(2a^{2}-b^{2}-c^{2})P_{a}\geq0,$  where ...

Let $a\geq b\geq c.$ Hence, $P_{a}\geq P_{b}\geq P_{c}.$
[/quote]

From here, we can apply Chebishev ineq. (note that rearrange or Chebishev ineq do not need positive real number condition.)

As $a \geq b \geq c$, $2a^{2}-b^{2}-c^{2}\geq 2b^{2}-c^{2}-a^{2}\geq 2c^{2}-a^{2}-b^{2}$. So by $P_{a}\geq P_{b}\geq P_{c}$, $\sum_{cyc}(2a^{2}-b^{2}-c^{2})P_{a}\geq \frac{1}{3}\sum_{cyc}(2a^{2}-b^{2}-c^{2}) \sum_{cyc}P_{a}=0$.

From the above inequalities, we can pose the following nicer one (though weaker):

Let $x,y,z$ be positive real numbers such that \[x^{2}+y^{2}+z^{2}=1.\] Prove that \[\frac{\sqrt{1+xy}}{x+y}+\frac{\sqrt{1+yz}}{y+z}+\frac{\sqrt{1+xz}}{x+z}\geq3.\]

[quote="pvthuan"]
Let $x,y,z$ be positive real numbers such that \[x^{2}+y^{2}+z^{2}=1.\] Prove that \[\frac{\sqrt{1+xy}}{x+y}+\frac{\sqrt{1+yz}}{y+z}+\frac{\sqrt{1+xz}}{x+z}\geq3.\] [/quote]
$z^{2}-xy$ again. :wink:

[quote="Vasc"]If $ x,y,z$ are positive numbers, then

$ \frac {\sqrt {x^2 + yz}}{y + z} + \frac {\sqrt {y^2 + zx}}{z + x} + \frac {\sqrt {z^2 + xy}}{x + y} \geq \frac {3}{\sqrt {2}},$

with equality if and only if $ x=y=z.$[/quote]By Hölder inequality, we obtain:

$ \left(\sum{\frac{\sqrt {x^2 + yz}}{y + z}}\right)^2\sum{(x^2 + yz)^2(y + z)^2}\geq\left(\sum{x^2 + yz}\right)^3.$

Thus, it remains to prove that

$ \left(\sum{x^2 + yz}\right)^3\geq\frac {9}{2}\sum{(x^2 + yz)^2(y + z)^2},$

with equality if $ x=y-z=0.$ But

$ \left(\sum{x^2 + yz}\right)^3 - \frac {9}{2}\sum{(x^2 + yz)^2(y + z)^2}$

$ = \sum{x^6} + 3\sum{x^5(y + z)} - 3\sum{x^4(y^2 + z^2)} - 2\sum{y^3z^3} + 6\sum{x^3(y^2z + yz^2)}$

$ - 33x^2y^2z^2$

$ = \left[\sum{x^6} - \sum{x^4(y^2 + z^2)} + 3x^2y^2z^2\right] + 2\left[\sum{x^5(y + z)} - \sum{x^4(y^2 + z^2)}\right]$

$ + \left[\sum{x^5(y + z)} - 2\sum{y^3z^3}\right] + 6\left[\sum{x^3(y^2z + yz^2) - 6x^2y^2z^2\right]\geq0,}$

which is true by Schur inequality and Muirhead theorem.

[quote="Vasc"]If $ a,b,c$ are positive numbers then
$ \frac {\sqrt {a^2 \plus{} 2bc}}{b \plus{} c} \plus{} \frac {\sqrt {b^2 \plus{} 2ca}}{c \plus{} a} \plus{} \frac {\sqrt {c^2 \plus{} 2ab}}{a \plus{} b} \geq \frac {3\sqrt {3}}{2}$.[/quote]
By Hölder inequality, we obtain:

$ \left(\sum\frac{\sqrt{a^2\plus{}2bc}}{b\plus{}c}\right)^2\sum\frac{(a^2\plus{}2bc)^2}{b\plus{}c}\ge\left(\sum\frac{a^2\plus{}2bc}{b\plus{}c}\right)^3.$

Thus, it remains to prove that

$ \left(\sum\frac{a^2\plus{}2bc}{b\plus{}c}\right)^3\ge\frac{27}4\sum\frac{(a^2\plus{}2bc)^2}{b\plus{}c}.$

But

$ 4\left(\sum\frac{a^2\plus{}2bc}{b\plus{}c}\right)^3\minus{}27\sum\frac{(a^2\plus{}2bc)^2}{b\plus{}c}$

$ \equal{}4\sum a^{12}\plus{}12\sum a^{11}(b\plus{}c)\plus{}9\sum a^{10}(b^2\plus{}c^2) \plus{} 27\sum a^8(b^4\plus{}c^4) \plus{} 54abc\sum a^9 \plus{} 69abc\sum a^8(b\plus{}c) \plus{} 159abc\sum a^7(b^2\plus{}c^2) \plus{} 99 abc\sum a^6(b^3\plus{}c^3) \plus{} 396\sum a^2b^2c^2\sum a^6 \plus{} 756 a^2b^2c^2\sum a^5(b\plus{}c) \plus{} 388a^3b^3c^3\sum a^3$

$ \minus{}17\sum a^9(b^3\plus{}c^3) \minus{} 3\sum a^7(b^5\plus{}c^5) \minus{} 64 \sum a^6b^6 \minus{} 177 abc\sum a^5(b^4\plus{}c^4) \minus{} 45a^2b^2c^2\sum a^4(b^2\plus{}c^2) \minus{} 786 a^2b^2c^2\sum a^3b^3 \minus{} 726 a^3b^3c^3\sum a^2(b\plus{}c)\minus{}954a^4b^4c^4 \ge0,$

which is true by Muirhead theorem. :rotfl:

My old similar result:
$ \frac {\sqrt {a^2 \plus{} 256bc} }{b \plus{} c} \plus{} \frac {\sqrt {b^2 \plus{} 256ca} }{c \plus{} a} \plus{} \frac {\sqrt {c^2 \plus{} 256ab} }{a \plus{} b} \geq\ 10$
And my solution is here: :)
http://www.mathlinks.ro/viewtopic.php?t=216366

[quote=xzlbq]$a,b,c$ are sides of a triagle,prove that:

$\sqrt {{\frac {{a}^{2}+3\,bc}{{b}^{2}+{c}^{2}}}}+\sqrt {{\frac {{b}^{2
}+3\,ac}{{c}^{2}+{a}^{2}}}}+\sqrt {{\frac {{c}^{2}+3\,ab}{{a}^{2}+{b}^
{2}}}}\leq 3\,\sqrt {2}
$[/quote]
By C-S $\left(\sum_{cyc}\sqrt {{\frac {{a}^{2}+3\,bc}{{b}^{2}+{c}^{2}}}}\right)^2\leq\sum_{cyc}\frac{a^2+3bc}{b+c}\sum_{cyc}\frac{b+c}{b^2+c^2}$.
Id est, it remains to prove that $\sum_{cyc}\frac{a^2+3bc}{b+c}\sum_{cyc}\frac{a+b}{a^2+b^2}\leq18$, which is true, but my proof of the last inequality is very ugly.

The following inequality is also true.
Let $a$, $b$ and $c$ be sides-lengths of triangle, $k\geq\frac{39+15\sqrt{13}}{32}$. Prove that:
$$\sqrt{\frac{a^2+kbc}{b^2+c^2}}+\sqrt{\frac{b^2+kac}{a^2+c^2}}+\sqrt{\frac{c^2+kab}{a^2+b^2}}\leq3\sqrt{\frac{1+k}{2}}$$


articles
quadratics
geometry
3D geometry
algebra

Hi everyone,
I couldn't solve a few problems...can u guys please help me out?
And please show your steps so I can check..

1)A retailer buys an article at a discount of 15% on the printed price from a wholesaler. He marks up the price by 10%. Due to competition in the market, he allows a discount of 5% to a buyer. If the buyer pays 451.44 for the article INCLUSIVE OF 8% sales tax, find
i)The printed price of the article
ii)The profit percentage of the retailer.

2)Solve the equation and obtain values for x-
$ 10ax^{2}\minus{}6x\plus{}15ax\minus{}9 \equal{} 0$ a is not equal to 0.

3)A, P and B are collinear points. P being between A and B such that AB=6cm and $ AB.AP \equal{} BP^{2}.$ Obtain a quadratic equation and find the length of BP.

4)A straight line makes positive intercepts on the co-ordinate axes. The sum of the intercepts is 7. If the line passes through (-3,8), find the equation of the line.

5)A solid consisting of a right circular cone surmounted on a hemisphere, is placed upright in a right circular cylinder,full of water, and touches the bottom. Find the volume of water left in the cylinder, having been told that the radius of the cylinder is 3cm and its height is 6cm, the radius of the hemisphere is 2cm and the height of the cone is 4cm. Give your answer to the nearest cubic centimetre.

I have a math test tomorrow and these are the only sums I have a doubt about so please help me out...quick!!
Thank u!!!

Oops, Sorry!
it's $ 10ax^{2}\minus{}6x\plus{}15ax\minus{}9\equal{}0$ not 15a
Sorry...

Any solutions to the rest? Help anyone..please... :(

Usually, it's better to show what you tried to do, and people help you from there, rather than having them do all the work for you.

[quote="moogra"][quote="LevatorNasolabialis"]The awnser to 4 is:
[hide]X=3,0  Y=0,4[/hide]
you should really try by yourself[/quote]

we can double post in 10 min now?
[hide]
#1:
.85*.10*.95x=451.44*1.08=total price (solve x)
451.44*1.08/x - 1 is percentage profit[/hide][/quote]

Ya exactly 10 min


limit
trigonometry
calculus
calculus computations

$ \lim_{x\rightarrow \pi} (\frac{1}{x\minus{} \pi}( \sqrt{\frac{4cos^2x}{2\plus{}cos x}}\minus{}2))$

$ \lim_{x\rightarrow a^2} \frac{\sqrt{x}\plus{}x\minus{}a\minus{}a^2}{x\sqrt{x}\plus{}x\sqrt{a}\minus{}a^2\sqrt{x}\minus{}a^2\sqrt{a}}$ with a>0

$ \lim_{x\rightarrow \plus{}\infty} x^{2007} (\frac{x}{x\minus{}1}\minus{}(1\plus{}\frac{1}{x}\plus{}\frac{1}{x^{2007}}))$

$ \lim_{x\rightarrow 3}  \frac{\sqrt{x\plus{}6}\minus{}sin(x\minus{}3) \minus{}3}{(x\minus{}3)cos(x\minus{}3)}$

[quote="greatestmaths"]$ \lim_{x\rightarrow \pi} (\frac {1}{x \minus{} \pi}( \sqrt {\frac {4cos^2x}{2 \plus{} cos x}} \minus{} 2))$[/quote]

[url=http://www.mathlinks.ro/viewtopic.php?t=187879]See[/url]

[quote="greatestmaths"]$ \lim_{x\rightarrow 3} \frac {\sqrt {x \plus{} 6} \minus{} \sin (x \minus{} 3) \minus{} 3}{(x \minus{} 3)\cos (x \minus{} 3)}$[/quote]

$ \equal{} \lim_{x\rightarrow 3} \frac{\frac {\sqrt {x \plus{} 6} \minus{} 3}{x \minus{} 3} \minus{} \frac {\sin (x \minus{} 3)}{x \minus{} 3}}{\cos (x \minus{} 3)}$

$ \equal{} \frac {\left[\frac {d}{dx} \sqrt {x \plus{} 6}\right]_{x \equal{} 3} \minus{} 1}{\cos 0} \equal{} \frac {\frac {1}{6} \minus{} 1}{1} \equal{} \minus{} \frac {5}{6}$


Let $ f(x)\equal{}36x^{2}\minus{}1$ and define a sequence $ \{a_{n}\}$ as $  a_{1}\equal{}\minus{}1,\ a_{n\plus{}1}\equal{}a_{n}\minus{}\frac{f(a_{n})}{f'(a_{n})}\ (n\equal{}1,\ 2,\ 3,\ \cdots)$.

(1) Show that $  a_{n}<\minus{}\frac{1}{6}$ for all postive integers.

(2) Show that  $  a_{n}\plus{}\frac{1}{6}<2\left(a_{n\plus{}1}\plus{}\frac{1}{6}\right)$ for  all postive integers.

(3) Find $ \lim_{n\to\infty}a_{n}$.


Euler
geometry
geometric transformation
reflection
incenter
geometry proposed

Let ABC be triangle. with intouch triangle DEF.
Let H and H' be the orthocenters of triangle ABC and DEF.
Let A' be the reflection of A wrt. euler line OH.
Prove that simson line wrt A' is parallel to HH' iff G, I and H' are collinear.
where G and I are centroids and incentre of triangle ABC

Pls check your problem! I think it has a mistake.


One can say that mechanization in maths is bad. But if you think about it, when you decide how to attack a problem, you do this way because either you have seen something like that or you think this might work or something or an idea comes, then your brain makes some predetermined thinkings in order to analyse problem or hit it with some theorem, all that stuff.

But when you mechanize this way of attacking problems, the ideas become mechanized and then you have better and better ideas that keep in mechanizing. Is that it? Is a great idea a product of mechanizations, when it is not a spark that came from nothingness?

[quote]One can say that mechanization in maths is bad.[/quote]
If "mechanization" means thinking that $ (\ln 2)'\equal{}\frac12$, I'll agree that's bad. But if "mechanization" is understood as development of new tricks into more widely applicable methods (as you seem to use the word), then I don't know anyone who thinks it's a bad thing. A trick that solves one problem but does nothing else is likely to be forgotten. A prominent exception is the calculation of $ \int_{\minus{}\infty}^{\infty}e^{\minus{}x^2}dx$ by squaring it and using polar coordinates.

[quote="Pul de Algodoncito"] when you decide how to attack a problem, you do this way because either you have seen something like that or you think this might work or something or an idea comes, then your brain makes some predetermined thinkings in order to analyse problem or hit it with some theorem[/quote]

That describes quite well what you do in 99% of cases. Indeed, when facing a problem, the first reaction is to use all the standard methods you know on it and the formulation of the problem itself triggers certain moves you try (like if it is a problem about a compact set, you try to find some open covers, if it is about a singular integral operator, you try Calderon-Zygmund (or atomic) decomposition, etc., etc.). I usually call it "reflexive thinking" and tell my students that they should develop those "reflexes" in them (like if the problem is to prove some property of limits, the first five words: [i]take[/i] $ \varepsilon>0$ should come out of their mouth before they even finish reading the condition of the problem or if they see a compact set and an open cover of it, they should immediately choose a finite subcover regardless of whether they know what to do with it later or not). It is interesting that 99% of time these reflexes do a good job and basically nothing else is needed. They are also quite handy when you analize a new proof written by someone else. You just disregard all parts of it that fit your reflexes and immediately see what the key new ideas ("new" meaning "new for you") are.

[quote]
Is a great idea a product of mechanizations, when it is not a spark that came from nothingness?[/quote]
Yes, it is. I'm not sure whether it ever "comes from nothingness" but in 99% cases it is definitely not going beyond a slight modification of something that is a slight modification of something that is ... of something you know well enough.

As to remaining 1%, you just call it "strike of luck" or "genius". It is a great mistery for me where it really comes from but, if it is not the result of some direct telepatic communication between your living mind and the great shadows of the past, it should ultimately boil down to bringing to the conscious level something you noticed subconsciously somehow somewhere some time in your life, because mathematicians tend to convert everything they see or use (not only coffee) into problems and theorems :).

I think the implied "bad" here is that rote doesn't work in a creative field. If [i]all[/i] you know are these standard tricks and the knowledge to cover and choose a finite subcover when dealing with compact sets, then you aren't going to create something new.

Hutchinson's How To Think Creatively talks about creativity as something you must practice. If the only thing teachers assign are the standard rote procedures (been in those classes), then you probably won't actually be able to produce mathematics in that subject (or remember it for very long). This is bad. Along with those techniques people need to be bumping against new challenges without predescribed techniques. They must force their way through these in order to grow and be able to create new things. Blah...I could go on forever on this, so I'll stop.

I think fedja has it right. The mechanization is basically there in order to not waste brain power with clutter that could just be second nature. It is hard enough to think of how to prove something without automatically knowing how to start, and basically how the proof will most likely be structured.

These posts are great, and I agree with this stuff. I've taken two extremely useful classes at college on psychology: Learning and Memory; and Human Motivation. We spend a lot of time talking about these issues. In essence as a freshman, you don't have many techniques and haven't seen very much material. It makes sense to only give "straightforward" problems that require no additional insight. In fact, you are learning to internalize certain techniques. As you progress in your career, the best classes continue to force you to internalize the most common and useful techniques as we've seen above. This can help you get started on problems that require something more. But these classes will slowly shift towards more and more challenging problems that force you to come up with something you haven't ever done before. This helps develop creativity, as you can see in Hutchinson's work. Later on these "creative techniques" will carry over to your research where you may need to do things you've never done before. People who have not had years practicing creativity are at a huge disadvantage when that finally have to do it.

How do we start to start mastering skills?

I think every post touched on the answer: you need to master the basics. Work on problems that are just beyond what you're used to and ask others for help. What may seem like creativity is probably the result of experience and having gone through lots of ideas that didn't work.


Put the numbers 1 to 9 in a $3\times3$ grid such that each row, column and diagonal adds 15.

NO WAY

darn it 19 year bump

pre-tty good


In an $ h$-meter race, Sunny is exactly $ d$ meters ahead of Windy when Sunny finishes the race. The next time they race, Sunny sportingly starts $ d$ meters behind Windy, who is at the starting line. Both runners run at the same constant speed as they did in the first race. How many meters ahead is Sunny when Sunny finishes the second race?

$ \textbf{(A)}\ \frac {d}{h} \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ \frac {d^2}{h} \qquad \textbf{(D)}\ \frac {h^2}{d} \qquad \textbf{(E)}\ \frac {d^2}{h \minus{} d}$

[hide]
While Sunny runs h meters, Windy runs h-d meters.
While Sunny runs h+d meters, Windy runs x meters.
$\frac{h}{h-d}=\frac{h+d}{x}$
$x=h-\frac{d^2}{h}$

So the difference between Sunny's distance and Windy's distance is:
$(h+d)-(h-\frac{d^2}{h})=d+\frac{d^2}{h}$
But we have to subtract d, since we are calculating the distance between Sunny and Windy at the end of the race, not the difference in distances run. Sunny ran d meters more than h:

$\boxed{\frac{d^2}{h}}$

[/hide]

[hide="Answer"]For simplicity, we say that $h=100$ and $d=10$. Therefore, Sunny can run $\frac{10}{9}$ times as fast as Windy. Therefore, when Sunny has run $110$, then Windy has run $99$, so Sunny wins by $1$. The only answer choice that gives this distance with the chosen values of $h$ and $d$ is $\boxed{C}$.[/hide]

 :D

This is old, but I thought I had found a clever solution. [hide]Notice that Sunny outpaces Windy by $ \frac{d}{h} $. In the second race, Sunny will catch up to to Windy when there is a distance of $ d $ left. Treating that leftover distance as a proportionate version of the original track because their paces remain constant, we get $ \frac{d}{h} \times d $. Therefore the answer is $ \boxed{\frac{d^2}{h}} $[/hide]


inequalities Euler algebra polynomial Vieta

Its true for all triangle that its sides are the roots of the cubic equation:

$ t^3-2st^2+(s^2+r^2+4Rr)t-4sRr=0$

And is true too that $ a^2+b^2+c^2\le9R^2$
$ \implies (a+b+c)^2 \le 3^3R^2 \hspace{2cm} (1)$
and $ R\geq 2r \hspace{2cm} (2)$ (Euler inequality).

First part:
[hide]Then $ (r^2s)^{\frac{1}{3}} \le \sqrt{ \frac{r^2+ 4Rr}{3}}$

$ \Leftrightarrow 27rs^2\le (r+4R)^3$

$ \Leftrightarrow 27.2.r.(a+b+c)^2\le (2r+8R)^3$

By $ MA%Error. "geqMG" is a bad command.
$
$ 8R+2r\geq 9(2R^8.r)^{\frac{1}{9}}$

Then we have to prove that:
$ 2.r.(a+b+c)^2\le 3^3(2R^8.r)^{\frac{1}{3}}$

$ \Leftrightarrow (2r)^2(a+b+c)^6\le 3^9R^8$

And it is true by $ (1)$ and $ (2)$.[/hide]

For the second part:
Use that $ r^2+4.R.r=s^2-\frac{a^2+b^2+c^2}{2}$
and the inequality becomes so easy.

how did u get \[ \ ab\plus{} bc \plus{}ca\equal{} s^2\plus{}r^2\plus{}4Rr\]

In any triangle:
$ sin(\frac{A}{2})\equal{}\frac {r}{s\minus{}a}cos\frac{A}{2}$
$ \implies sin^2 A\equal{}\frac{r.a}{4R(s\minus{}a)}$
On the same way $ cos^2 A\equal{}\frac{a(s\minus{}a)}{4r.R}$
Like we know that $ sin^2 x\plus{}cos^2 x\equal{}1$, by the sum we obtain:

$ a^3\minus{}2sa^2\plus{}(s^2\plus{}r^2\plus{}4Rr)a\minus{}4sRr\equal{}0$
The result that you want is obtained from Vieta  Formulas.


[quote="Makaveli"]One of 2 circles of radii R passes through points A and B and the other one passes through B and D.Let M be their second common point. Prove that the radius of the circle circumscribed around triangle AMD is R.[/quote]

Can you please correct this problem? It cannot be true as stated.

  darij

[quote="Makaveli"]One of 2 circles of radii R passes through points A and B and the other one passes through B and D.Let M be their second common point. Prove that the radius of the circle circumscribed around triangle AMD is R.[/quote]
I've 2 questions:
1. thecircles have same radius?
2. If A,M,D conlinear then ...?


Show how you would carry out the following conversion:

Aniline $\longrightarrow$ 2,3-dimethylindole.

(Source: adapted from an Heterocyclic Chemistry examination, September 2001)

Carcul

[hide="Hint"]Use Fischer indole synthesis.[/hide]

React 2 butanone and Phenylhydrazine in the presence of acid and heat (Fischer Indole).
But how do you prepare phenyl Hydrazine from aniline?  :D

Good question. However, the procedure is very simple (two steps).

1.NaNO2/HCl,0c, followed by reduction with SnCl2
2. Fischer indole synthesis with butanone

its not a high yield rn  :D  atleast less yield than this  :P
Phenyl diazonium chloride reacts in solution with sodium sulfite (Na2SO3) to give phenyl diazonium sodium sulfonate, which when treated with sulphurous acid undergoes reduction to phenylhydrazine sodium sulphonate. The latter readily hydrolyses in the presence of concentrated hydrochloric acid to give phenylhydrazine hydrochloride, from which the free base can be isolated by the action of caustic alkalis.


email
search
pigeonhole principle
symmetry
graph theory
Ramsey Theory

It's probably a bit pretentious for me to post this in the "Advanced Problems" section, because it doesn't seem like it should be that hard, but it has vexed me for a long time and I consider myself fairly good at math...

Prove that in a group of ten people there is a group of four people who all know each other or all don't know each other.

I decided to word it like that in case anyone is not familiar with graph theory terminology. This is question 4.1.14 from "The Art and Craft of Problem Solving" for anyone with the solutions book (note to such people: it is necessary that you also tell me where to get one).

The solution isn't in the solution book. If you want the solution book, the best way to get one is to ask him for one.

Well the problem is that I don't know Zeitz personally. Is there a way I can contact him?

He's a professor at USF (University of San Francisco). His email is probably somewhere on the website for USF.

About your problem, this is a case of the more general type of "prove there are x who know each other or y who don't". Have a search for Ramsey numbers. I don't think its solved in general but specific cases for small numbers have been, at least. Usually involves pigeonhole principle somewhere. I may try solving it properly soon.

Hmm, the Ramsey number R(4, 4) = 18, which means the problem you quoted can't be right (there is a counterexample with as many as 17 people).  On the other hand, R(3, 4) = 9, so perhaps the problem asked about either 4 people who do know each other or 3 people who don't?

Yeah Ravi I was playing with it and got that you had to prove that N(3,4) was 5... which made absolutely <i>no</i> sense whatsoever.

Here's a quick counterexample: a graph on ten vertices with neither four vertices
that pairwise have no edges nor four vertices that pairwise have all six edges.

The vertices are the integers 0, 1, ..., 8, 9.
Two vertices are adjacent if (mod 10) they differ by 1 or 2.
So for example 9 and 1 have an edge.

Say there are four vertices that pairwise have no edges.

We have a 5-cycle among the odds, and similarly a 5-cycle
among the evens.  So if four such vertices exist, 2 are odd and
2 are even.  By symmetry, say we have the vertices 1 and 5.
Then 0, 2, 3, 4, 6, 7, and 9 can't be in the set.  So we get only
a set of 3 vertices.

Say there are four vertices that pairwise have all 6 edges.

We can take at most two evens.  Say 4 and 6.  Then the only
odd adjacent to both 4 and 6 is 5, so again we only get 3 vertices.

Hope that's clear,
--Dan


Parametric Function and its local extrema

What do you have so far? This is an ordinary textbook exercise- where are you stuck?


How many four-letter words are there? (ignoring spelling, so one of the words could be zzxd.)

Hmm, 26^4?
[hide]26 letters for the first letter, then 26 again and 26 and.. 26 so : 26*26*26*26 = 26^4[/hide]

yes, i think 26^4 is the answer as well.

So you basically want all 4 letter combinations, since the letters can be repeated, I agree it would be $26^4$

[hide]
For each letter slot, there are 26 choices, so it's $26^4$[/hide]


Theres a typo in the google chrome page for the classroom:
When the classroomw window is blocked, you should see the following at the bottom of your Chrome browser window.

Chrome is not really a well-known internet browser yet, whereas safari is, and the linux one is also because its a slightly popular OS, but if someone went and made a internet browser, and had like 3 ppl download it, someone can't really complain about it here can they?
Newer or less known internet browsers will have problems, and AoPS staff won't have much reason to pay attention, and also if its a typo its probably not because of your internet browser.

...

Its on the aops site that they made the typo.

[color=green]EDIT: Sorry for double posting this... I didn't realize how quickly it got sent to solved reports.

Moderator please delete this.[/color]


If I deal cards one by one from a well-shuffled pack, find the expectation of the number of cards I shall have to deal to secure a spade.

I know how to calculate this the long way; just wanted to know if there was a simpler way.


For how many integer values of $ n$ does $ n^2 \equal{} |n|$?

$ n^2\equal{}|n| \Rightarrow n^4\equal{}n^2 \iff n^2(n\minus{}1)(n\plus{}1)\equal{}0 \iff n\equal{}\minus{}1, 0, 1$.
Thus, $ \boxed{3}$.


The letters of the word ANNIVERSARY can be arranged in $\frac{11!}{(2!)^3}=4989600$ different ways. When these "words" are arrenged in alphabetical order, what is the tenth letter of the 50th word?

A) N
B) R
C) S
D) V
E) Y

[hide]The first word in the sequence is
AAEINNRRSVY.

The first 60 words will start with the letters AAEINN (there are 60 ways to arrange the last 5 remaining letters, 5!/2!), so the 50th word is within these 60 words.

Since we know the first 6 letters of the 50th word, ignore the letters AAEINN.

We turn to the 7th letter. The first 24 words, with the same reasoning as above, must start with AAEINNR, the next twelve with AAEINNS, and the next 12, or up to 48 words, with AAEINNV. The 49th word thus must start with AAEINNY, and the last 5 letters of this word is YRRSV.

Thus the 50th word is AAEINNYRRVS, 10th letter V.

[/hide]

This was confusing.
[hide]
The first word in the list would be AAEINNRRSVY. Looking at the last 4 letters (RSVY) shows that the first 4!=24 words are given by rearranging those. Adding one more letter (RRSVY) shows that 5!/2!=60 of the first words are given by rearranging those, so our 50th word must be among these.

I'll ignore all but the last 5 letters for now.

I'll start with the 24th word, which is given by arranging the last 4 letters in descending order: RYVRS

The 25th word is gotten by switching the first letter with the next largest letter: SRRVY

There are 4!/2!=12 arrangements of S _ _ _ _.

There are also 4!/2!=12 arrangements of the next largest set V _ _ _ _. The last word in this set is the 25+12+12=48th word and is VYSRR.

Following previous logic, the 49th word is YRRSV, and switching the last two letters gives the next largest and 50th word YRRVS.

The 10th letter in AAEINNYRRVS is V.

On a side note, I found this problem less confusing when assigning numbers to the letters. R=1, S=2, V=3, Y=4.
[/hide]


What is the probability that four different points chosen at random from the fifteen equally-spaced points shown are the vertices of a square? Express your answer as a common fraction.

I guess it's safe to assume that the "fifteen equally-spaced points shown here" are like this:
[asy]unitsize(1cm);
for(int i=1; i<6; ++i) for(int j=1; j<4; ++j) dot((i,j));[/asy]
cause otherwise it'd be pretty pointless..

counting the squares isn't that hard. 8 squares with side length 1 unit, and 3 squares with side length 2 units for a total of [b]11[/b] squares.

Total number of ways you can choose four points $ \equal{} \binom{15}{4} \equal{} 1365$

So the fraction would be $ \cfrac{11}{1365}$, I hope?

well...
thats not the answer

im confused and dont know how to solve this problem(pleaze excuse mygrammar)

Don't forget the squares with diagonal sides. There are 3 of those, giving 2/195.


In the reaction below is difference. Because when the base (LDA) concentration isn't high enough it attacks the carbon of highest degree.
p.s LDA is lithium diisopropyl amide

Question 7 (10 minutes)
When 2-methylcyclohexanone is treated with 1.2 equivalents of lithium diisopropylamide
followed by bromomethane, 2,6-dimethylcyclohexanone is formed. However when 2-
methylcyclohexanone is treated with 0.5 equivalents of lithium diisopropylamide followed by
bromomethane, 2,2-dimethylcyclohexanone is formed.

This interesting problem deserves some discussion.

I seem to remeber that $ LDA$ is a non-nucleophilic strong base used to deprotonate compounds. So I would say that at high concentration (of $ LDA$) there will be an equilibria between $ \text{2\minus{}methylcyclohexanone}$ and an enolate ion (double bond between carbon $ 1$ and carbon $ 6$). While at low concentrations we get the same equilibrium but the double bond is between carbon $ 1$ and carbon $ 2$.

Then the reaction between $ CH_{3}Br$ and the enolate ions follow a $ S_{N}2$ mechanism (because bromomethane is a primary halogenoalkane).

Yes, that's right, but the question here is why does a low concentration of LDA favours one enolate anion, while a higher concentration favours another enolate.

I wonder if it has anything to do with Kinetic and Thermodynamic Enolates. :maybe:

Yes, the explanation lies on kinetic versus thermodynamic control.

No, I will not simply provide the solution. Let's think a little bit, shall we? To start, try to answer the following two questions:

1) What kind of reaction is the enolate forming process?

2) What is the most acidic hydrogen of 2-methylcyclohexanone?

I want to come back to this question.

Shouldn't the hydrogen/proton on the $ \alpha$ carbon be the acidic one?

But there are two types of alpha hydrogens: the one on carbon 2 and the two on carbon 6.

Is that hinting at why the two different tautomerisms occur?

Yes. And as I told you before, answer the following two questions for a start:

1) What kind of reaction is the enolate forming process?

2) What is the most acidic hydrogen of 2-methylcyclohexanone?

Is the one on the $ alpha$ carbon without the methyl group the more acidic?

This is probably not an explanation but it is how I thought about it.

If the negative charge resides on the carbon with the methyl group then the alkyl groups (electron donating) won't allow the charge to be spread over a larger volume so the carbanion ion is less stable.

Ok, that's right. Now the first question: what kind of reaction is the enolate forming process?

What do you mean? (endergonic, exergonic, endothermic, exothermic or otherwise)

I wanted to mean that it is an acid-base reaction. Now some more questions:

3) Is LDA a weak, strong or very strong base?

4) What is the most stable enolate: the one from deprotonation of carbon 2, or the one from carbon 6?

3) LDA (Lithium DiisopropylAmine) is a very strong base, since it is the conjugate base of an amine (a very weak acid).

4) No.

Actually is it because the structure of the enolate formed when carbon 2 is deprotonated has a more highly substitued alkene double bond?

Yes, that's right. The carbon 2 enolate is more stable. So let's put together the information collected thus far:

- the hydrogens of carbon 6 are more acidic;

- the enolate formed from deprotonation of carbon 2 is the most stable;

- LDA is a very strong base.

What are now your thoughts about all this?

I think I might know what is being hinted at earlier with the thermodynamic/kinetic controls.

In Kinetic Control the major product is the one that has a lower activation energy (hence proceeds faster).

In Thermodynamic Control the major product is the more stable species.

Now do we spin these two ideas together with the data and the theory?

[quote="BanishedTraitor"]Now do we spin these two ideas together with the data and the theory?[/quote]

Yes. Now, tell me what happens if we use:

1) more than 1 mole of LDA for each mole of ketone?

2) less than 1 mole of LDA for each mole of ketone?

I am taking a guess here:

1) Would we have more collisions between molecules making it more Kinetically oriented?\

Or does the explanation lies in the acidity of the hydrogens (because if it does I can't see why).

Here's an hint:

[quote="BanishedTraitor"]In Kinetic Control the major product is the one that has a lower activation energy (hence proceeds faster). In Thermodynamic Control the major product is the more stable species.[/quote]

Another hint:

1) With a greater concentration of LDA relative to ketone is the reaction under kinetic or thermodynamic control?

2) With a greater concentration of ketone relative to LDA is the reaction under kinetic or thermodynamic control?

Well looking at the original reaction I see it as:

1) Forms a less stable enolate ion so it is kinetic control

2) Forms a more stable enolate ion so it is thermodynamic control

I just can't explain why.

A kinetic controled reaction is essentialy irreversible, while a reaction under thermodynamic control is an equilibrium. Does this help?

I know that. But how does the concentration of LDA affect whether the reaction is going to be reversible or not? Is it because the higher the $ [LDA]$ the more collisions take place and hence the more the reaction proceeds to completion while for low $ [LDA]$ the forward and reverse reaction comes to an equilibrium more easily?

Until now it was explored the influence of the concentration of base in an enolate forming reaction. However, there are other important (and perhaps unexpected) factors that are also important in determining the selectivity of the enolate formation process. They are:

1) The solvent.
2) The nature of the base.
3) The counterion of the base.
4) The temperature.
5) The reaction time.

Can anyone provide explanations on how these factors might influence which enolate ion is formed?


I have a problem explaining a question in my physics book about absorption spectrum and emission spectrum. We are taking a look at the spectrums for lithium. In the absorption spectrum we see that three lines are removed as those wave lenghts are absorpted by the lithium atoms. Taking a look at the emission spectrum from the lithium atom we see the same three line (320.00nm [UV], 323.26nm [UV] and 670.79nm [visuble red]) plus a forth visuble orange line (610.35nm). So my question is why we in the absorption spectrum see only the three lines removed while we see 4 lines in the emission spectrum? PS. The light comes from a light source with a continious spectrum.

I have read the chapter through once again and think this could be the reason: The lithium atoms don't absorp that much of the 610.35 wave lenght and therefore we will still see the orange colour in the absorption spectrum as the intensity of the orange colour in the continious spectrum is only slightly reduced.


Find all $m,n\in\ N$such that : $ n\mid\ m^3+1$  and  $m\mid\ n^3+1$.

I'm not sure whether we can give a precise answer to this problem...

For instance, let's define an increasing sequence $a_n$ of positive integers such that $a_1=a_2=1$
and $a_n^3+1=a_{n-1}a_{n+1}$ for $n \geq 2$. Two consecutive elements of the sequence
satisfie problem's conditions.
However, it's easy to see that we can construct several examples based on this method.

well, and why not use descent using information of your sequence? we can easily see, thah if $m$ and $n$ work, then $\frac{m^3+1}{n}$ also works, from here we construct your sequence but "going back" and we analise minimal solutions....

ok, but that doesn't have to denote all possible "classes" of solutions. Why should we even construct sequences ?
This example was to show that this problem is not interesting.

[quote="Rafal"]This example was to show that this problem is not interesting.[/quote]
You speak like Lord Vader  :D

Because maybe all such sequences are contained in some finite number of sequences we can describe!!!!!!! It is interesting!


algebra
function
domain
calculus
calculus computations

Let $\Omega = \{(x,y,z):\ \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\leq 1\}$.
Find the volume of the part of $\Omega$ which is cut by the plane $lx+my+nz=p$.

Just a small hint: Everything will become a lot easier if you first shrink the domain $a$ times in $x$-direction, $b$ times in $y$-direction and so on. Then the problem will become one of finding the volume of a spherical cap cut off the unit ball by a given plane. That volume depends only on the distance of the plane from the origin in a fairly simple way.


[hide]
$0.\overline{12}=\frac{12}{99}=\frac{4}{33}$
The sum of its numerator and denominator is $4+33$ or $37$.
[/hide]

You guys need to show you work, not just say some " special porperty" , beggining problem solvers can only become proficient in solving problems when they study [b]full solutions[/b].

  [hide="Full solution"]

Detonate the repeating decimal $0.12121212\cdots$ as $S$.

Thus, $100S=12.121212121212\cdots$

Now, if we subtract $S$ from $100S$, we will get an integer value as follows:

             $100S=12.1212121212\cdots$
          -     $S  =  0.1212121212\cdots$

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

$ 99S=12$

dviding each side by $99$, we get:

$ S= \frac{12}{99} = 0.\overline{12} $

Thus, $0.\overline{12}$ can be expressed in the fractional form $\frac{12}{99}$

  However, the problem wants the fraction in lowest terms so $\frac{12}{99}=\frac{4}{33}$

  Now to answer the question, $ 4+33=\boxed{37}$ [/hide]

[hide]$0.\overline{12}\times 100 = 12.\overline{12}$
$12.\overline{12} - 0.\overline{12} = 12$
Since it is a hundred $0.\overline{12}$s minus one $0.\overline{12}$, 12 must be 99 $0.\overline{12}$s. So $0.\overline{12} = 12/99 = 4/33$

The answer is 37[/hide]

Thanks for the solution G-unit and Lady Knight. I just learned something today.

Short cut method to convert a non terminating recurring decimal to a fraction is

Take as many nines in the denominator as there are the numbers repeated in the decimal and in the numerator, write the digits that are repeated.

 So we write $0.121212.... = \frac{{12}}{{99}} = \frac{4}{{33}}$

Therefore the sum of the numerator and denominator is

4 + 33 = 37

i know you are a new user but use the hide thing...like this
[hide]hi[/hide]

[quote="joejia"]When the repeating decimal $0.\overline{12}$ is expressed as a common fraction in lowest terms, what is the sum of its numerator and denominator?[/quote]

[hide]I remember this from the AoPS book.

$x=0.\overline{12}$

$100x=12.\overline{12}$

subtracting the equations from each other..

$99x=12$

$x=\frac{12}{99}=\frac{4}{33}$

$4+33=37$[/hide]

[quote="nikhil"]Short cut method to convert a non terminating recurring decimal to a fraction is

Take as many nines in the denominator as there are the numbers repeated in the decimal and in the numerator, write the digits that are repeated.
[/quote]

How about 0.34123123123123...?  ;)

[hide]

I got 37. I used algebra for this problem. $  x = 0.\overline{12} $

$  100x = 12 + 0.\overline{12} $

$  100x = 12 + x  $ ( the first equation tells you that) $  99x = 12  $

$  x = \frac{12}{99}  $

$  \frac{12}{99} = \frac{4}{33}  $; $  4 + 33 = 37  $ [/hide]


calculus
derivative
calculus computations

Find the x- and y- intercepts of the straight line that is tangent to the curve y=  at the point (1,3)??

sorry, will post the question again..

Find the x- and y- intercepts of the straight line that is tangent to the curve y= x^3+ x^2+x at the point (1,3)

thanks in advance

Well, what have you done so far? Do you know what "tangent to the curve" means?
(Hint: it has to do with a derivative)


Let $n_{1}<n_{2}<\cdots$ be a sequence of natural numbers such that for $i<j$, the decimal representation of $n_{i}$ does not occur as the leftmost digits of the decimal representation of $n_{j}$ (for example 134 and 13498 can not both occur in the sequence). show that:

$\sum_{i=1}^{\infty}\frac{1}{n_{i}}\leq 1+\frac{1}{2}+\cdots+\frac{1}{9}$

Suppose a finited sequence is given, and let $M=10N+d$ be the largest element of the sequence, with $0\le d\le 9$.  Then $N$ does not belong the sequence.  Furthermore, removing $10N, 10N+1,...,10N+9$ from the sequence if they occur and adding $N$ gives another sequence whose sum of reciprocals is:

$\sum_{i}\frac{1}{n_{i}}+\frac{1}{N}-\sum_{i=0}^{9}\frac{1}{10N+i}\geq \sum_{i}\frac{1}{n_{i}}$.

Thus we can repeatedly make such subsitution and never decrease the sum of reciprocals.  This process must terminate (since the sequence is finite), and nly does so when the sequence is $\{1,2,...,9\}$, so this sequence has the largest sum of reciprocals.
so we are done. :)