geometry
incenter
power of a point
radical axis
geometry unsolved

Dear everyone.

please think of the following.

let the incircle of triangle ABC touch BC, AC, AB at D, E, F respectively. M and N are the midpoints of AE and AF respectively. MN meets DF at K. draw two tangents KP and KQ through K to incircle. PQ meets MN at L and EF meets PQ at R. (I=incenter) prove IR is parallel to AL.

thanks.

$ NA^2\equal{}NF^2, MA^2\equal{}ME^2$ so $ MN$ is the radical axis of $ (A,0)$ and $ (I)$.
$ K\in MN \Rightarrow KA^2\equal{}KP^2 \Rightarrow A\in (K,KP)$.
$ L\in MN\Rightarrow LA^2\equal{}LP.LQ$ therefore $ LA$ is a tangency of $ (K,KP)$
$ \Rightarrow KA\perp AL$
But $ KA$ is the polar of $ R$ wrt $ (I)$ then $ IR\perp KA$ and we are done.


vector
induction
linear algebra
linear algebra solved

let $E$ be an euclidian vector space of dimension $n$.
What is the maximum number of vectors $(v_1,..,v_k)$ we can find such that
$(v_i|v_j) < 0$ for every $i \neq j$

$k\leq n+1$

Hint

Let $x_1,..,x_k \in R, \; x=\sum_{i=1}^{k}x_iv_i, \; y=\sum_{i=1}^{k}|x_i|v_i$

Compare $||x||$ and $||y||$

Prove that if $x=0$ then all $x_i$ are 0 or the $x_i$ are not all 0

Let $f : R^{k}\rightarrow E$, $f(x_1,...,x_k)=\sum_{i=1}^{k}x_iv_i$
prove that $dim Ker(f) \leq 1$
Finally deduce that $k\leq n+1$

I think that now you can solve http://www.mathlinks.ro/Forum/viewtopic.php?t=31806 with ideas like this

How about this:

All Euclidean spaces of dimension $n$ are orthogonally isomorphic (i.e. among any two of them there's an isomorphism from one of them to the other one which also conserves the scalar products). This means that we may assume that we're dealing with $\mathbb R^n$ with the usual scalar product.

Suppose we have $n+2$ such vectors. We may assume WLOG that one of them is $(-1,0,0,\ldots,0)$, so we know that the other $n+1$ have a positive first coordinate. Let these $n+1$ vectors be $v_i,\ i\in\overline{1,n+1}$, and let $v'_i$ be the $n-1$-dimensional vector obtained from $v_i$ by disregarding its first coordinate. In $\mathbb R^{n-1}$ with the usual scalar product, we have $(v'_i|v'_j)<0,\ \forall i\ne j,\ i,j\in\overline{1,n+1}$, so we can find $n+1$ vectors with negative pairwise scalar products in an $n-1$-dimensional Euclidean space, meaning that we can solve the problem by induction.

Is it Ok? :?

I worked on this (funny, imho) problem, and came up with two different ideas to attack it.

At first, let me restate it :

[color=darkblue]Let $E$ be an euclidean vector space of dimension $n \geq 1.$
A family of $p$ vectors $(x_{1}, \ldots, x_{p})$ is said to be obtuse-angled if and only if for any $(i, j)$ such that $i \neq j, \ \ (x_{i}|x_{j}) < 0.$
Then we have $p \leq n+1.$[/color]

The first one is by induction on $n,$ namely we're going to show that, if $(x_{i})_{1 \leq i \leq p}$ is obtuse-angled, $p \leq n+1.$
For $n=1,$ an obtuse-angled family cannot contain at least three vectors since if it contains the vectors $x_{1}$ and $x_{2}$ satisfying $x_{1}\cdot x_{2}< 0,$ an arbitrary vector $x_{3}$ is either zero (and then $x_{3}\cdot x_{1}=0$), or has the same orientation as $x_{1}$ (and then $x_{1}\cdot x_{3}> 0$), or has the same orientation as $x_{2}$ (guess what?). So $p \leq 2.$
Let $n \geq 1.$ Let's assume that any obtuse-angled family of an $n-$dimensional space has a cardinal less than or equal to $n+1.$
Let $(x_{i})_{1 \leq i \leq p}$ be an obtuse-angled family of an $n+1-$dimensional space $E.$ If $p=1,$ there's nothing to say. Let us assume that $p \geq 2.$ $x_{p}$ is non-zero and $H = x_{p}^{\perp}$ is an hyperplane of $E$ and thus has dimension $n.$
Let, for $1 \leq i \leq p-1, \ y_{i}= x_{i}-\frac{(x_{i}|x_{p})}{\|x_{p}\|^{2}}x_{p}$ be the orthogonal projection of $x_{i}$ onto $H.$
Let us check that the family $(y_{i})_{1 \leq i \leq p-1}$ is an obtuse-angled family. Let $(i, j) \in \{1, \ldots, p-1\}$ with $i \neq j.$
$y_{i}\cdot y_{j}= x_{i}\cdot x_{j}-\frac{(x_{i}|x_{p})(x_{j}|x_{p})}{\|x_{p}\|^{2}}-\frac{(x_{j}|x_{p})(x_{i}|x_{p})}{\|x_{p}\|^{2}}+\frac{(x_{i}|x_{p})(x_{j}|x_{p})(x_{p}|x_{p})}{\|x_{p}\|^{4}}= x_{i}|x_{j}-\frac{(x_{i}|x_{p})(x_{j}|x_{p})}{\|x_{p}\|^{2}}< 0.$
But then, from the induction hypothesis, $p-1 \leq 1+\text{dim}H = n+1,$ and hence $p \leq n+2.$ We're done.

(Sorry for double posting, but I decided to write my second solution in another post not to make the previous one longer than needed. Of course, a mod can feel free to delete this comment and put the second solution in the previous post, if he feels like this is inappropriate.)

Let's show that if the family $(x_{i})_{1 \leq i \leq p}$ is obtuse-angled, the family $(x_{i})_{1 \leq i \leq p-1}$ is linearly independent. Let us assume, for the sake of contradiction, that there exists a family of scalars $(\lambda_{i})_{1 \leq i \leq p-1}$ which are not all zero and are such that $\sum_{i=1}^{p-1}\lambda_{i}x_{i}= 0 \ \ (*).$
Up to multiplying the two sides of $(*)$ by $-1,$ we may assume that there is at least one real $\lambda_{1}> 0.$ Let $I$ be the set of the indices $i$ such that $\lambda_{i}> 0$ and $J$ be the set of the indices $i$ such that $\lambda_{i}\leq 0.$ $I$ and $J$ are disjoint.
$(*)$ rewrites $\sum_{i \in I}\lambda_{i}x_{i}= \sum_{i \in J}\lambda_{i}x_{i}.$ We have :
$0 \leq \left \|\sum_{i \in I}\lambda_{i}x_{i}\right \|^{2}= \left(\sum_{i \in I}\lambda_{i}x_{i}\right) \cdot \left(-\sum_{i \in J}\lambda_{i}x_{i}\right) = \sum_{(i, j) \in I \times J}\lambda_{i}(-\lambda_{j})x_{i}\cdot x_{j}\leq 0.$
Thus, $\left \|\sum_{i \in I}\lambda_{i}x_{i}\right \|^{2}= 0$ then $\sum_{i \in I}\lambda_{i}x_{i}= 0.$
But, when doing the scalar product with $x_{p},$ we get $\left(\sum_{i \in I}\lambda_{i}x_{i}\right) \cdot x_{p}= \sum_{i \in I}\lambda_{i}(x_{i}\cdot x_{p}) < 0,$ which is a contradiction.
Since the family $(x_{i})_{1 \leq i \leq p-1}$ is linearly independent, its cardinal $p-1$ is less than or equal to the dimension $n$ and therefore $p \leq n+1.$


geometry
geometric transformation
rotation

How many possible complete Sudoku boards are there?

ie.
Given a 9x9 grid with 9 3x3 non-overlapping boxes, fill each 1x1 square with an integer $1 \leq x \leq 9$ such that each column, row, and 3x3 box contains every number from 1-9.  How many ways are there to do this?

I'm not sure of the difficulty of or answer to this problem, it was just something that popped into my mind.

The answer really depends on if rotations count...
think of it as a coloration problem.
I'm working on it.

I found the number of 4x4 Sudokus there are once.  I did it by hand, too.
[hide="The answer"]$288$[/hide]

[quote="sapphyre571"][quote="Rzeszut"]It is very diffcult. A result confirmed many times is $6\ 670\ 903\ 752\ 021\ 072\ 936\ 960$. It was obtained of course with computer, by Bertram Felgenhauer and Frazer Jarvis.[/quote]

wow. that is a huge number.
but is odd because as far as i'd gotten, the result should be divisible by 6^6, my calculator said that huge number only divided 6^5.
but i'm calculating with a ti83, so...[/quote]
See [url]http://www.afjarvis.staff.shef.ac.uk/sudoku/[/url]. How did you get that number of sudoku squares must be divisible by $6^{6}$?

I have part of the derivation. There's some way a gigantic prime number slips in that I don't quite remember.


Consider the regions $ R$ where $ \minus{}6\le x \le \minus{}4,\, |y|\le 1$ and $ Q$ where $ 8 \le x \le 12,\, \minus{}1 \le y \le 3.$ Find the mass center of $ R\cup Q.$

I really don't know how to find the mass center for $ R\cup Q,$ I was thinking on finding the mass center by separately, but I can't keep going.

[quote="0714446459923"]Can't you find the center of mass of each part and then take the weighted average?[/quote]
Yes, I can do that, but what do you mean by saying "take the weighted average?"

Suppose $ (a,b)$ is one mass center and $ (m,n)$ the other one, then the one I'm looking for is $ \left( \frac{a\plus{}m}{2},\frac{b\plus{}n}{2} \right),$ is that?

[quote="Null"]
Suppose $ (a,b)$ is one mass center and $ (m,n)$ the other one, then the one I'm looking for is $ \left( \frac {a \plus{} m}{2},\frac {b \plus{} n}{2} \right),$ is that?[/quote]
Nope, that'd be the case only if the regions have same mass (or equivalently same area).
Just consider the system as a combination of 2 point masses at say acc. to you, $ (a,b)$ and $ (m,n)$ (The masses of the points being proportional to the areas of the regions they're representing).

Could you please tell me what's the expression which gives the mass center for $ R\cup Q$ ? I just don't see it. :(

Following savy2020's suggestion:

let r and q be  R's and Q's area, then the mass center will be
\[ ({{r.a \plus{} q.m}\over{r \plus{} q}}\; ,\; {{r.b \plus{} q.n}\over{r \plus{} q}})\]

Perfect! It makes sense now! Thanks a lot for all your help!


Hi!

let $0<x,y,z<1$ be three real numbers such that $xy+yz+za=1$. Complete
the inequalities:
\[
??\leq x+y+z+xyz\leq ??
\]

[quote="Kantor"]Hi!

let $0<x,y,z<1$ be three real numbers such that $xy+yz+zx=1$. Complete
the inequalities:
\[
??\leq x+y+z+xyz\leq ??
\][/quote]

you can find without a lot of difficulties that  -1<(x-1)(y-1)(z-1)<0  (( use -1<x-1<0))

so -1<xyz+x+y+z-(xy+yz+xz+1)<0
  -1<xyz+x+y+z-2<0
 1=<xyz+x+y+z=<2

I think we can uses the method of mixing variables.
    Let f(x,y,z) = x+y+z+xyz.
    Let t=(yz +zx)/2.
    Let's prove f(a,b,c)<=f(a,t,t) (or >=). The min or max when x=y=z or x=y=1,z=0

This problem was proposed in the October 2004 issue of Crux.

Please do not post problems from journals that are less than one year old.


algebra
polynomial
superior algebra
superior algebra unsolved

Show that a finite field of size p^n is isomorphic to F_p [x] / (q(x)), where q(x) is an irreducible polynomial of degree n in F_p [x].

F_p[x] denotes the ring of polynomials with p elements.

I started off by letting |F|=p^n
Then we want to show that F_p[x] is contained in F, and F is contained in F_p[x].

You'll be better off by considering the action of the Frobenius map $ x \mapsto x^p$ on $ \mathbb{F}_{p^n}$.


calculus
function
geometry
inequalities
complex numbers
IMO

WHAT IS INCLUDED IN THE OFFICIAL BASIC THEORY NEEDED FOR IMO ?IS IT POSSIBLE FOR A PROBLEM OF IMO TO NEED ADVACED CALCULUS,FURRIER'S ANALYSIS... ,?

The IMO is a pre-college contest, and as such treats pre-college material.  You may find that some knowledge of the properties of convex functions is useful, but that's about it.

You forgot the difference between needed and usefull...

[quote="BanishedTraitor"]What about complex numbers and matrices? I have done some IMO questions which are quite standard if you use these techniques.[/quote]

Complex numbers and matrices are included in high school.

That's not strictly true (e.g. New Zealand courses (NCEA) don't have matrices neither does the HSC course in Australia).

I read somewhere that Calculus, Complex numbers, Inversion in geometry, Solid geometry are excluded topics  :maybe:

I AM A FIRST YEAR STUDENT IN THE UNIVERSITY OF ATHENS (DEPARTMENT OF MATHEMATICS) AMD MY SISTER HAS ENDED THE PRIMARY SCHOOL.SO,SHE WANTS TO PARTICIPATE AT IMO AT THE NEXT 2-3 YAERS AND SHE WANTS,FROM ME, TO GIVE HER LESSONS.PLEASE ,IF YOU WANT SUGGEST ME WHAT SHE MUST STUDY (WITH THE APROPRIATE SERIE).THANKS

At first, please stop posting in capitals...

Inequalities, Geometry, Number Theory, Algebra  :D

http://www.mathlinks.ro/index.php?f=217


limit
calculus
derivative
function
real analysis
real analysis unsolved

i'm a little confused by this problem. i went through it and think i have an answer, but i found a theorem that contradicts my answer.....i think. here it is:

let n be a natural number and let f:R->R be defined by f(x)=x^n for x>=0 and f(x)=0 for x<0. for which values of n is f' continuous at 0? for which values of n is f' differentiable at 0?

i got that when n=1 f' is not continuous but for all other n it is. also, i have when n=1 it is differentiable, but for all other n it is not. however, doesn't differentiability imply continuity? i'm puzzled.

Is $f$ differentiable at $0$?

$f'(0)=\lim_{h\to0}\frac{f(h)}h$ (since we have $f(0)=0$.)

Compute the right hand limit and the left hand limit. The left hand limit is always zero. The right hand limit is

$\lim_{h\to0^+}\frac{h^n}h.$

This is $1$ for $n=1$ and $0$ for $n>1.$ Thus the derivative does not exist at zero for $n=1$ but does for all larger $n.$ In this particular example, (a "spline"), the function is differentiable at zero in exactly the same cases for which the derivative is continuous.

[quote="patthew"]however, doesn't differentiability imply continuity?[/quote]
Let's be careful with that! Yes, there is the following theorem:

If $f$ is differentiable at $a$ (that is, $f'(a)$ exists) then $f$ is continuous at $a.$

But in all cases here, $f$ is continuous; this theorem is utterly irrelevant to the problem.

If $f'$ is continuous at $a$ then $f'(a)$ exists; that's just because a function has to be defined at a point to be continuous there; this is not deep.

But there are examples (which don't look at all like this example) for which $f'(a)$ exists but $f'$ is not continuous at $a.$ In other words, differentiability does not imply continuity of the derivative.


Initially an urn contains 100 black marbles and 100 white marbles. Repeatedly, three marbles are removed from the urn and replaced from a pile outside the urn as follows:

\[ \text{MARBLES REMOVED} \qquad \qquad \qquad \text{REPLACED WITH} \]
\[ \text{3 black} \qquad \qquad \qquad \text{1 black} \]
\[ \text{2 black, 1 white} \qquad \qquad \qquad \text{1 black, 1 white} \]
\[ \text{1 black, 2 white} \qquad \qquad \qquad \text{2 white} \]
\[ \text{3 white} \qquad \qquad \qquad \text{1 black, 1 white} \]

Which of the following sets of marbles could be the contents of the urn after repeated applications of this procedure?

$\text{(A)} \ \text{2 black marbles} \qquad \text{(B)} \ \text{2 white marbles} \qquad \text{1 black marble} \qquad \text{(D)} \ \text{1 black and 1 white marble} \qquad \text{(E)} \ \text{1 white marble}$

An operation leaves the number of white marbles modulo 2 invariant.  It's also clear that we can't arrive at 0 white marbles.  Answer B.


Find the solubility of silver chloride, at 25ºC, in:

(a) water
(b) a 0.01 M aqueous solution of NaCl

(Ks = 1.82x10^{-10})

Let s be the solubility of $ \ce{Ag \plus{} }$ and $ \ce{Cl \minus{} }$
$ \ce{AgCl} \leftrightharpoons \ce{Ag\plus{}} \plus{} \ce{Cl\minus{}} \rightarrow (1)$
There are no common ions for $ \ce{AgCl}$ in $ \ce{H2O}$
$ \implies$ the equation for $ K_{sp}$ is $ K_{sp} \equal{} s^2$
$ \implies s \equal{} sqrt{K_{sp}}$

nacl is a strong electrolyte, so molar conc of cl- is 0.01 due to nacl. i suppose u can neglect the cl- due to agcl. equate ionic product of [cl-][ag+] to ksp.

Let the solubility in 0.01 M aq $ \ce{NaCl}$ be $ s_{1}$
$ \ce{AgCl} \leftrightharpoons \ce{Ag \plus{} } \plus{} \ce{Cl \minus{} } \rightarrow (1)$
...................$ s_{1}$......$ s_{1} \plus{} 0.01$
$ \ce{NaCl} \longrightarrow \ce{Na \plus{} } \plus{} \ce{Cl \minus{} } \rightarrow (2)$
...................0.01......0.01
$ K_{sp} \equal{} s_{1} \cdot (s_{1} \plus{} 0.01) \approx 0.01 \cdot (s_{1})$
$ \implies s_{1} \equal{} 100 \cdot K_{sp}$
From this value, we can justify the approximation :)

PS: How do i write the conc values under the ions without putting dots like this? :maybe:

EDIT: sorry, didnt see varun posted.


geometry
incenter
calculus
trigonometry
geometry unsolved

Dear Mathlinkers,
let ABCD a square, M a point on segment CD, I, X, Y the incenters wrt MAB, MAD, MBC,
P, Q the meetpoints of the parallel to AD passing through X with BI, AC,
S, R the meetpoints of the parallel to AD passing through Y with AI, BD.
Prove or disprove that MQR is an isosceles triangle.
Sincerely
Jean-Louis

Well, your supposing is correct  :roll:

Denote $ QX \cap BC \equiv E, RY \cap BC \equiv F$.
We immediately have $ QE \equal{} EC, RF \equal{} FD$

So we have:
$ MQ^2 \equal{} MR^2$
$ \leftrightarrow EC^2 \plus{} EM^2 \equal{} FD^2 \plus{} FM^2$

It's easy to check that
$ ME \equal{} \frac{MD \plus{} MA \minus{} AD}{2}$ and $ DE \equal{} \frac{DA \plus{} DM \minus{} AM}{2}$
So by calculating (not too difficult  :roll: ), we complete the proof

I am really curious, what is jayme going to say to this.

Dear Yetti and Mathlinkers,
I made the calculus until the end... and the egality holds. Howewer, perhaps it is your question, dear Yetti, I ask me if a synthetic proof can be found? I have no idea in that way...
Sincerely
Jean-Louis

No, the question is: Can you do the calculation with Pythagoras theorem as above ? BTW, I posted a proof elsewhere, starting with similar right triangles, followed by a short calculation (not synthetic).

Dear Yetti and Mathlinkers,
after the Pythagoras theorem, I need to use the al-Kaschi 's theorem (a generalisation of the last one) and it works...
The proof will be put on my website when I will finished my article.
If you want some more information, tell me.
Sincerely
Jean-Louis

See EDIT at [url]http://www.mathlinks.ro/viewtopic.php?t=276743[/url]

What is al Kaschi theorem ? Cosine theorem ?

In French, the law of cosines is named Théorème d'Al-Kashi .

Dear Mathlinkers,
for a proof see:
http://perso.orange.fr/jl.ayme  vol.5,  le résultat de Larrosa Canestro, p.17
Sincerely
Jean-Louis


[quote="Phelpedo"]Binet's? Eww..[/quote]

not really...just use variables for the ratios

There is a cute combinatorial interpretation here, and also one involving the matrices corresponding to the Fibonacci sequence.


geometry
perimeter
inequalities
trapezoid
geometry proposed

Here is a problem that I think is quiet hard.

In triangle $ABC$ denote $A_1, B_1, C_1$ intersections of bisectors with opposite
sides and $A_2, B_2, C_2$ intersections of medians similarly.

Prove that $P_{A_1 B_1 C_1}\leq P_{A_2 B_2 C_2}$, where $P$ we denote perimeter of the triangle.

nice! i like it..
  and I think I solved it,  but my sln is toooooooooo long, so I don't have time to post it..
  the basic ideas are these:
 Clearly we have to show that $\frac{a+b+c}{2}\geq DE+EF+FD$ with $D,E,F$ the intersections of the angle bisectors with the opposite sides of the triangle.
 first show by geometric arguments that $\frac{2abc}{ab+2bc+ca}\geq EF$ and the same with $DE,FD$...
 Then the problem reduces to prove that $\sum{\frac 1{ab}}\geq 4\sum {\frac 1{ab+2bc+ca}}$
 I don't have a nice proof of this, but put $x=bc,y=ca,z=ab$, and after clearing all the denominators (it's not so much) the last inequality follows by Muirhead... (something like $(\sum{xy})(16xyz+\sum{(2x^3+7x^2y+7x^2z)})\geq 4xyz\sum{(11xy+5x^2)}$, where the sums are cyclic...)

Well done RaMLaF.
My solution is also long, so I think I will post it later too.
Maybe somebody else, who thought on it will post a short and beautiful one.

well now I realized that my last inequality $\sum{\frac 1{x}}\geq \sum {\frac 4{x+2y+z}}$ can be solved in a very easy way: by AM-HM we have \[\sum{\frac 1{x}}= \frac 14\sum {(\frac 1{x}+\frac 2{y}+\frac 1{z})} \geq \sum {\frac 4{x+2y+z}}\]  :lol:

About

[quote]$\displaystyle \frac{2abc}{ab+2bc+ca}\geq EF$ [/quote]

Considering that the bisector from A cut the opposite side a in two parts

$\frac{ab}{b+c}$ and $\frac{ac}{b+c}$

I have found applying the triangular inequality at $CDE$ that

$ED \leq \frac{ab}{a+c} +\frac{ab}{b+c}$

which is different of your relation.

Can you please explain to me how to deduce it?

Thank you.

Well the geometric aproach for proving that is diferent.. i mean, that won't work because of the case of equality, you will never have equality with the triangular inequality in that triangle.
  My idea is to consider the parallel lines to $EF$ through $B,C$ which intersect lines $AC,AB$ at $M,N$ respectively... Then try to show that in the trapezoid $BMCN$ we have $BC\geq \frac{BM+CN}2$, and it is very easy to calculate the segments $BM,CN$ in terms of $EF$; doing that you obtain the desired result.

A hint, please, for this one

$\displaystyle BC\geq \frac{BM+CN}2 $

Thank you.

I see, the $BC\geq \frac{BM+CN}{2}$ inequality has been posted by Manlio in http://www.mathlinks.ro/Forum/viewtopic.php?t=30992 now. As for the original problem, namely to prove that $P_{A_1B_1C_1}\leq P_{A_2B_2C_2}$, you can find different solutions in

Nikolaos Dergiades, [i]The Perimeter of a Cevian Triangle[/i], Forum Geometricorum 2 (2002) p. 131-134;

Toshio Seimiya, Michel Bataille, [i]Solution of problem #2502[/i], Crux Mathematicorum 27 (2000) p. 53-54.

I enclose copies of both of these papers to this post (note that the first paper should be availiable online from the [url=http://forumgeom.fau.edu]website of Forum Geometricorum[/url], but this website is off right now).

  darij

Sorry Darij I didn't know that this problem exists.
But I really invented and solved it alone.


function
calculus
derivative
real analysis
real analysis unsolved

Let $ D$ be an open subset of $ R^n n\ge 2$  and $ f: D\minus{}>R$, contiunous, diferentiable in $ x_0$ with $ f'(x_0)\ne 0$ Prove that in any $ V$ neighbourhood of $ x_0$ there exists $ x\ne x_0$ such that $ f(x)\equal{}f(x_0)$.

The nonzero derivative guarantees that in any neigborhood of $ x_0,$ there exist $ y$ and $ z$ such that $ f(y) < f(x_0) < f(z).$ (Just follow the tangent plane approximation to find the "high side" and the "low side.")

Now consider the punctured neighborhood $ B(x_0;\delta)\setminus\{x_0\}.$ This is a connected set. (That's where we use that $ n\ge 2.$) Hence the image of this set under $ f$ is a connected subset of $ \mathbb{R}$ that contains $ y$ and $ z,$ hence all of $ [y,z],$ hence $ f(x_0).$


limit
function
calculus
calculus computations

Use continuity to evaluate the limit.

lim 5 + sqrt{x}/sqrt{5} + x as x--->4

What are the steps?

Do I plug 4 for x and simplify?

Do I need to use the conjugate or do I rationalize the denominator?

You wrote $\lim_{x\to4}5+\frac{\sqrt{x}}{\sqrt{5}}+x.$

Is that what you meant? Or did you actually mean

$\lim_{x\to4}\frac{5+\sqrt{x}}{\sqrt{5}+x}$ ?

Parenthesis are not a luxury to be used only if you feel like it - use what you have to use to make the expression mean what you want it to.

I try to express my questions as precise as possible.

Yes, I meant your second choice for Latex.

I'm not exactly sure what you mean by using continuity to solve the limit but this is my approach. First of all, find out if the function is defined a 4. Yes it is. Then take the limit as x--> 4- and x--> 4+ of the function. If they are the same, then the function is continuous. You can straight plug the 4 in, no need to conjugate.


Αν οι $ x,y,z$ ειναι θετικοι και $ x \plus{} y \plus{} z \equal{} 1$ ν.δ.ο.

$ \frac {y \plus{} z}{x \plus{} \sqrt\frac {yz}{x}} \plus{} \frac {x \plus{} z}{y \plus{} \sqrt\frac {xz}{y}} \plus{} \frac {x \plus{} y}{z \plus{} \sqrt\frac {xy}{z}} > \frac {\sqrt\frac {yz}{x}}{x \plus{} \sqrt\frac {yz}{x}} \plus{} \frac {\sqrt\frac {xz}{y}}{y \plus{} \sqrt\frac {xz}{y}} \plus{} \frac {\sqrt\frac {xy}{z}}{z \plus{} \sqrt\frac {xy}{z}}$

Για να μην χαλάσω την προσπάθεια των άλλων η λύση σε hide
[hide]Για να απαλλαγούμε από τις ρίζες θέτουμε $ x \equal{} a^2$ κτλ  t;ote $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
Οπότε η ανισότητα είναι ισοδύναμη με

$ \sum\frac {a(b^2 \plus{} c^2)}{a^3 \plus{} bc} > \sum\frac {bc}{a^3 \plus{} bc}$  ή ισοδύναμα

$ \sum\frac {a(1 \minus{} a^2)}{a^3 \plus{} bc} > \sum\frac {bc}{a^3 \plus{} bc}$ ή ισοδύναμα

$ \sum\frac {a}{a^3 \plus{} bc} > 3$  (1)

Αν έχουμε $ a > a^3 \plus{} bc$ και $ b > b^3 \plus{} ca$ kai $ c > c^3 \plus{} ab$ τότε προφανώς  η (1) ισχύει .

Αν για για τουλαχιστον ένα ισχύει η αντίθετη φορά
(έστω χωρίς βλάβη της γεν ότι $ a\leq a^3 \plus{} bc$  άρα  $ a(1 \minus{} a^2)\leq bc$ άρα $ a(b^2 \plus{} c^2)\leq bc$ και ισοδύναμα

$ \frac {ab}{c} \plus{} \frac {ac}{b}\leq 1$  (2)

Είναι από Cauchy
$ \sum\frac {a}{a^3 \plus{} bc} \equal{} \frac {a}{a^3 \plus{} bc} \plus{} \frac {1}{b^2 \plus{} \frac {ca}{b}} \plus{} \frac {1}{c^2 \plus{} \frac {ab}{c}} \equal{} A$

και $ A\geq\frac {a}{a^3 \plus{} bc} \plus{} \frac {4}{b^2 \plus{} c^2 \plus{} a(\frac {b}{c} \plus{} \frac {c}{b})} > \frac {a}{a^3 \plus{} bc} \plus{} \frac {4}{1 \plus{} ax}$

όπου $ x \equal{} \frac {c}{b} \plus{} \frac {b}{c}$

Άρα μένει νδο $ \frac {a}{a^3 \plus{} bc} \plus{} \frac {4}{1 \plus{} ax}\geq 3$  ή ισοδύναμα αφαιρώντας το 1 από το πρώτο κλάσμα

$ \frac {a(b^2 \plus{} c^2) \minus{} bc}{a^3 \plus{} bc} \plus{} \frac {4}{1 \plus{} ax}\geq 2$ ή ισοδύναμα

$ \frac {ax \minus{} 1}{\frac {a^3}{bc} \plus{} 1} \plus{} \frac {4}{1 \plus{} ax}\geq 2$ ή ισοδύναμα με $ y \equal{} \frac {a^3}{bc}$

$ \frac {ax \minus{} 1}{y \plus{} 1} \plus{} \frac {4}{1 \plus{} ax}\geq 2$ ή με απαλοιφή παρονομαστών

$ a^2x^2 \minus{} 1 \plus{} 4y \plus{} 4\geq 2y \plus{} 2yax \plus{} 2 \plus{} 2ax$ ή ισοδύναμα $ (ax \minus{} 1)^2 \plus{} 2y \minus{} 2yax\geq 0$ ΄

που ισχύει αφού $ 2y\geq 2yax\ \leftrightarrow ax\leq 1$  που ισχύει λόγω της (2) and a big QED  :)[/hide]

Αρκει να δειξουμε οτι $ \sum\frac{x}{y}<3$  :wink:

Sil αλλη φορα θα βαλω απαγορευτικο για εσενα..........
mailo δεν νομιζω οτι ισχυει αυτο που λες .Δοκιμασε για 1/2,1/3,1/6.........

Ναι εχεις δικιο xp. Η ΑΜ-ΓΜ δεν δουλεψε εδω. Παντως την απλοποιησα λιγο οποτε αρκει νδο

$ \frac {1}{x \plus{} \sqrt\frac {yz}{x}} \plus{} \frac {1}{y \plus{} \sqrt\frac {xz}{y}} \plus{} \frac {1}{z \plus{} \sqrt\frac {xy}{z}}>3$

δεν ξερω αν βοηθαει αυτο...


Where do we save a file to make it a template?

More info required! Template for what? Which operating system, what program?

It's just possible you mean TexnicCenter where you can save templates in one of the directories in C:\Program Files\TeXnicCenter\Templates and create new such directories in Tools, Options, Directories.

Yeah, I meant TexNicCenter.
Thanks for the speedy response!


calculus integration function algebra polynomial inequalities derivative

Find the minimal value of the integral between -1 and 1 of the function $|f(x)|$ where $f$ is a monic polynomial of degree n.

with Tchebychev polynomials you can get a lower value that $\frac{1}{2^{n-1}}$. I think. And I think that the minimum is not achieved with Tchebychev polynomial but i'm not sure. Harazi, what is your sln ?

I think that the minimum is achieved in polynomial $ P_{n}(x) $, such that
$ P_{n}(cos(\theta)) = c_{n} \frac{sin((n+1) \theta) }{sin( \theta ) } $,
where $ c_{n} $ is the appropriate constant to make $ P_{n} $ to be monic.

That is the right value. I think this problem is extremely difficult. Here is the main idea: consider the polynomial $\frac{1}{(n+1)\cdot 2^n}\cdot (T_{n+1})'$ where $ T_n$ are the Chebyshev polynomials. Prove that for any polynomial $f$ of degree at most $n-1$ we have $ \int_{-1}^{1}{f(x)g(x)dx}=0$ where $ g(x)$ is the sign of the above considered polynomial. Then compute the integral of $ x^n g(x)$ between -1 and 1.

I think that this problem can be solved in the following way:
Consider a monic polynomial $ f(x) $ and consider the integral $ \int_{-1}^{1} | f(x) | dx $.
Make a change of variables $ x = cos( \theta ) $, $ 0 \leqslant \theta \leqslant \pi $.
Then we obtain the integral $ \int_{0}^{ \pi } | f( cos(\theta)) sin(\theta)| d \theta $.
Now it is easy to see that for every $ k \in \mathbb{N} $, the function
$ \frac{sin(k \theta) }{sin(\theta)} $ is a polynomial of degree $ k-1 $, therefore for every
monic polynomial $ f(x) $ we can write in unique way $ f(cos(\theta)) =
c_{n+1} \frac{sin((n+1) \theta) }{sin(\theta)} + c_{n} \frac{sin(n \theta) }{sin(\theta)} +
\ldots + c_{1} \frac{sin( \theta) }{sin(\theta)} $, where $ c_{n+1} $ is some appropriate
constant. Therefore we can write
$  \int_{0}^{ \pi } | f( cos(\theta)) sin(\theta)| d \theta =
\int_{0}^{ \pi } | c_{n+1}sin((n+1) \theta) + c_{n} sin(n \theta) + \ldots + c_{1}sin( \theta)
| d \theta $. Now, the function
$ F( \theta ) = c_{n+1}sin((n+1) \theta) + c_{n} sin(n \theta) + \ldots + c_{1}sin( \theta) $
is odd and $ 2 \pi $ -periodic, so $ \int_{0}^{ \pi } | F( \theta ) | d \theta =
\frac{1}{2} \int_{0}^{ 2 \pi } | F( \theta ) | d \theta $. Now we see that
$ \frac{1}{n+1} \sum_{k=0}^{n} F( \theta + \frac{2\pi k}{n+1} ) = c_{n+1} sin((n+1) \theta ) $,
therefore $  \int_{0}^{ 2 \pi } | c_{n+1}sin((n+1) \theta) | d \theta  = $
$ \int_{0}^{ 2 \pi } | \frac{1}{n+1} \sum_{k=0}^{n} F( \theta + \frac{2\pi k}{n+1} ) | d\theta
\leqslant \frac{1}{n+1}  \sum_{k=0}^{n} \int_{0}^{ 2 \pi } | F( \theta + \frac{2\pi k}{n+1} ) | d \theta
= \frac{1}{n+1}  \sum_{k=0}^{n} \int_{0}^{ 2 \pi } | F( \theta ) | d \theta =
\int_{0}^{ 2 \pi } | F( \theta ) | d \theta $, therefore we see that the minimum is attained
when $ c_{n} = c_{n-1} = \ldots = c_{1} = 0 $, so
$ F( \theta ) = c_{n+1}sin((n+1) \theta) $, so $ f( cos(\theta) ) =
c_{n+1} \frac{sin((n+1) \theta) }{sin(\theta)} $.

if $P$ is a polynomial such that $\int_{0}^{1}|P(x)|dx$ is minimal then for
any polynomial $f$ of degree $deg(f) \leq n-1$ we know that the derivative of
$H_{f}(t)=\int_{0}^{1}|P(x) + tf(x)|dx$ is null for $t=0$. But this derivative
is precisely $\int_0^{1} g(x)f(x)dx$ . But I think it should be very interesting
to explain how to find these $\frac{1}{(n+1).2^n}.(T_{n+1})'$. I searched for
a long time such a polynomial but I found nothing. How one can find this ?

[quote="harazi"]Really, no idea how they found this one.[/quote]
I try to post, in the next week, the  proof of Korkin and Zoloraev result  [see below:]

For $ k\in {\mathbb R}\; \; a\in {\mathbb C} , $ let $   (a)_k=a(a+1)\cdots (a+k-1)\; , \; (a)_0:=1 ,$ and consider the  [i]Chebychev polynomial of second kind[/i]
\[
U_n(x)=\sum\limits_{k=0}^{n}{n \choose
k}\frac{(n+2)_k}{\left(\frac{3}{2}\right)_k}
\left(\frac{x-1}{2}\right)^k= a_{0,n}x^n+\ldots \] where $
a_{0,n}=\displaystyle \frac{2^{n}}{n+1} .$ Note that for $ x\in
(-1,1) $  we have $ U_n(x)=\frac{\sin\left((n+1)\cdot\arccos{x}\right)}{(n+1)\sqrt{1-x^2}}\;  .  $
Further  $ U_n^{*}:=\displaystyle \frac{1}{a_{0,n}}U_n(x) $ is the  monic Chebychev polynomial of the second kind. Denote by $ \Pi_n^* $ the set of all monic   algebraic polynomials of degree $ n$ , with real coefficients . Also consider the $ L_1-$ norm of polynomials  $
 ||P||_1:=\int\limits_{-1}^{1}|P(t)|\; dt\; .
$
In [b]1873[/b], A.N.Korkin and E.I. Zolotarev (see [1]) proved that  if  $ P(x)=\displaystyle x^{n}+\sum\limits_{k=1}^{n}b_kx^k ,\; b_k\in
{\mathbb R},\;  $    then
\[
\min\limits_{P\in \Pi_{n}^*}{||P||_1}=||U_n^*||_1=2^{1-n}\; .
\]
Therefore,[b] if $ p(x)=b_0x^n+b_1x^{n-1}+\ldots+b_n \; ,\; b_0\ne 0\; ,\;  $ then
\[
 \int\limits_{-1}^{1}|P(t)|\; dt\; \ge \frac{|b_0|}{2^{n-1}}   \; ,
\]
the equality being attained for $ p(x)=\displaystyle \frac{b_0}{a_{0,n}}U_n(x)\; .  $[/b]

 [b]References:[/b]
[1] A.N. Korkin(e)  and E.I.Zolotarev, [i]Sur un  certain minimum[/i], Nouv.Ann.Math.S\'er., 2 , [b]12[/b] (1873) 337-355.
[2] P.K. Suetin, [i]Classical Orthogonal Polynomials[/i],(Russian), Nauka,Moskow,1976 , see pp.98-101.
[3] C. Visser, [i] A simple proof of certain inequalities  concerning polynomials[/i], Ned.Akad.Wetenschap., Proc., [b]48[/b] (1945) 276-281 ,
[=Indag.Math., [b]7[/b](1945) 81-86. ]

[quote] For $k\in {\mathbb R}\; \; a\in {\mathbb C} ,$ let $(a)_k=a(a+1)\cdots (a+k-1)\; , \; (a)_0: =1 ,$ and consider the  [i]Chebychev polynomial of second kind[/i]
\[ U_n(x)=\sum\limits_{k=0}^{n}{n \choose k}\frac{(n+2)_k}{\left(\frac{3}{2}\right)_k} \left(\frac{x-1}{2}\right)^k= a_{0,n}x^n+\ldots  \] where $a_{0,n}=\displaystyle \frac{2^{n}}{n+1} .$ Note that for $x\in (-1,1)$  we have $U_n(x)=\frac{\sin\left((n+1)\cdot\arccos{x}\right)}{(n+1)\sqrt{1-x^2}}\; .$
Further  $U_n^{*}: =\displaystyle \frac{1}{a_{0,n}}U_n(x)$ is the  monic Chebychev polynomial of the second kind. Denote by $\Pi_n^*$ the set of all monic   algebraic polynomials of degree $n$ , with real coefficients . Also consider the $L_1-$ norm of polynomials  $||P||_1: =\int\limits_{-1}^{1}|P(t)|\; dt\; .$
In [b]1873[/b], A.N.Korkin and E.I. Zolotarev (see [1]) proved that  if  $P(x)=\displaystyle x^{n}+\sum\limits_{k=1}^{n}b_kx^k ,\; b_k\in {\mathbb R},\;$    then
\[ \min\limits_{P\in \Pi_{n}^*}{||P||_1}=||U_n^*||_1=2^{1-n}\; .  \]
Therefore,[b] if $p(x)=b_0x^n+b_1x^{n-1}+\ldots+b_n \; ,\; b_0\ne 0\; ,\;$ then
\[ \int\limits_{-1}^{1}|P(t)|\; dt\; \ge \frac{|b_0|}{2^{n-1}} \; ,  \]
the equality being attained for $p(x)=\displaystyle \frac{b_0}{a_{0,n}}U_n(x)\; .$[/b] [/quote]

The following   is similar with the proof from [2].
[b]Lemma[/b] [i]If $\mbox{sign}[t] =\left\{\begin{array}{rcl} 1 & , & t>0 , \\ 0 &, & t=0 , \\ -1& , & t>1 , \end{array}\right. \; ,\;$ and  $S_n(t)=(n+1)U_n(t)= \frac{\sin\left((n+1)\cdot\arccos{t}\right)}{\sqrt{1-t^2}}\; ,$
then
\[ I_{n,k}: =\int\limits_{-1}^{1}t^k \mbox{sign}\left[S_n(t)\right]\; dt= \left\{\begin{array}{llc} 0 &,& 0\le k\le n-1,\\ &&\\ \displaystyle 2^{1-n} & , & k=n\; . \end{array}\right.  \]
[/i]
[b]Proof[/b] Observe that  $I_{n,k}=\int\limits_{0}^{\pi}(\cos{\phi})^k\sin{\phi}\mbox{sign} \left[\sin{(n+1)\phi}\right]\; d\phi \; .$ From the theory of Fourier series it is known that
\[ F(t): =\mbox{sign}[\sin{t}]=\frac{4}{\pi}\sum\limits_{j=0}^{\infty}\frac{\sin(2j+1)t}{2j+1}\; ,  \]
which implies ( $t=(n+1)\phi$) that
\[ F\left((n+1)\phi\right)\sin{\phi}=\frac{2}{\pi}\sum\limits_{j=0}^{\infty}\frac{1}{2j+1}\left(\cos(N_j-1) -\cos(N_j+1)\phi\right)\; \; , \]
\[ \mbox{with}\; \; \; N_j: =(2j+1)(n+1)\; .  \]
After multiplication  above equality by $\cos^k\phi \; ,\;$ and then   integrating, one finds that \[ (**)\; \; \; \; I_{n,k}=\frac{2}{\pi}\sum\limits_{j=0}^{\infty}\frac{1}{2j+1}\left(J_{k,N_j-1}-J_{k,N_j+1}\right)  \]
where for $k,N \in {\mathbb N}$ we have denoted         \[ J_{k,N}: =\displaystyle \int\limits_{0}^{\pi}\left(\cos{\phi}\right)^k\; \cos{(N\phi)} \; d\phi =\displaystyle \left\{\begin{array}{lcl}\displaystyle \pi 2^{-k}{k\choose s} &,& k\ge N, k+N=2s\\ 0 &,& \mbox{otherwise}. \end{array}\right.  \]
When $j\ge 0$ we have $N_j\ge n+1$;  because we are interested  only in the situation $0\le k\le n$ , all  integrals on the right-hand side in (**) vanish, except $J_{k,N_j-1}$ for $j=0$ and $k=n$. Thus  $I_{n,k}=0$ for   $0\le k\le n-1,$ and
\[ I_{n,n}=\frac{2}{\pi(n+1)}J_{n,n}=\frac{1}{2^{n-1}}\; .  \]
Further I use the same notation  as in my last message on this topic.
 [b]Theorem (A.N. Korkin and E.I. Zolotarev-1873).[/b] [i] If $P(x)$  is a monic polynomial of degree  $n$ , then
\[ ||P||_1\ge \frac{1}{2^{n-1}}  \]
the equality case being valid if and only if $P(x)=U_n^*(x)=\displaystyle \frac{n+1}{2^n}U_n(x) ,$ i.e. when $P(x)$ is the monic Chebychev polynomial of the second kind.  [/i]
[b]Proof.[/b] Let  $I[P]=||P||_1=\int\limits_{-1}^{1}|P(t)|\; dt.$  According to above Lemma , we have
\[ \int\limits_{-1}^{1}P(t)\mbox{sign} \left[S_n(t)\right]\; dt =\frac{1}{2^{n-1}}  \]
which implies $||P||_1=I[P]\ge \displaystyle \frac{1}{2^{n-1}}.$ At the same time,  $||U_n^*||_1=\frac{1}{2^n}\int\limits_{-1}^{1}S_n(t)\mbox{sign}\left[S_n(t)\right]\; dt=\frac{1}{2^{n-1}}.$
The unicity question in the equality case is easy to investigate.
[b]References:[/b]
 [1] A.N. Korkin(e)  and E.I.Zolotarev, [i]Sur un certain minimum[/i], Nouv.Ann.Math.S\'er., 2 , [b]12[/b] (1873) 337-355.
[2] G.V.Milovanovi\'c , D.S. Mitrinovi\'c and Th.M.Rassias , [i]Topics in Polynomials: Extremal Problems ,Inequalities,
Zeros,[/i] World Scientific ,1994.
[3] P.K. Suetin, [i]Classical Orthogonal Polynomials[/i],(Russian), Nauka,Moskow,1976 , see pp.98-101.
[4] C. Visser, [i] A simple proof of certain inequalities  concerning polynomials[/i], Ned.Akad.Wetenschap., Proc., [b]48[/b] (1945) 276-281 [=Indag.Math., [b]7[/b](1945) 81-86.


A point $E$ is chosen at random from within square $ABCD$.  Express as a decimal to the nearest hundredth the probability that $\Delta ABE$ is obtuse.

I got [hide]pi/8 which i think is about .39

E has to inside d semicircle with diameter AB cuz any point on d circle would make it a right triangle[/hide]

[hide]I got 0.39.[/hide]

I think I saw that problem in saab.org

[hide="Solution"]
$S$ must be inside semicircle of diameter $MN$ inside the square. If $r$ is the radius of the semicircle, then the area is $\frac{\pi r^{2}}{2}$. Also, the area of the square is $4r^{2}$. So, the answer is simply $\frac{\frac{\pi r^{2}}{2}}{4r^{2}}\approx\boxed{0.39}$.[/hide]

[quote="binonunquineist"]lol...i still want to know who this 'Kevin Chen Xi' is. btw aren't we supposed to hide our answers?[/quote]

I think his middle name is chen, unless xi is a nickname (or meaning he's 11 years old).

I don't this question works out when I think about it in my head (not really thinking, but w/e)

jorian

If you wanna know so bad,

my English name is Kevin Chen and my Chinese name is Chen Xi...

maybe I'll be lucky and make it to the nats... there'll be 2 Kevin Chens on the roster

[quote="jhredsox"]i'm really confused. could someone draw a picture?

thanks,
jorian[/quote]

[hide]

For triangle ABE to be obtuse, point E needs to be inside the gray region, which is a semicircle.

Area of semicircle/Total area of square

if square has sides of length 2:

$\frac{\frac{\pi*1^{2}}{2}}{2*2}= \frac{\pi}{8}\approx .39$

mad mspaint skills :D
[/hide]

Extention: What is the probability that point E forms an obtuse triangle with two of the sides of the square?

[hide]
To find the are of the shaded regions:

$\frac{\pi1^{2}* 2-2^{2}}{2^{2}}= \frac{2\pi-4}{4}\approx .57$
[/hide]

[quote="mwpl11"]I don't understand how you would get that figure from the extension...[/quote]

The shaded areas are areas of intersection of two semicircles.

EDIT (ehh i need to learn to be more specific): since a point E in one of those semicircles forms an obtuse triangle with the side that forms the diameter of the said semicircle.

[quote="mwpl11"]wouldn't it be four semicircles?[/quote]

Umm.. E is in two semicircles since it has to form an obtuse triangle with two of the sides.. so that would be any intersecting areas of two semicircles.

okay, i'm really confused right now.

your scores seem eerily familiar to someone i know(i doubt it's coincidence). i'll ask u to explain it to me at school.


Let $ z_j$ ($ j\equal{}1,\dots, k$) be $ k$ points on the complex plane none of which lies on the real line. Is it always true that the function $ F(x)\equal{}\sum_j \frac{1}{|x\minus{}z_j|^2}$ has at most $ k$ local maxima on the real line?

Do you know from where comes functions like  $F(x)$?  :)
I mean, dependence the   energy on the temperature in the ising model.
And have you solved the problem?


Somebody knows how to compute the sum of all the fractions $\frac m n$ where $gcd(m,n)=1$ and $m<n<1000$. If we don't ask for $gcd(m,n)=1$, then it is pretty easy, but if we do, what happens?

Best,

[quote="Myth"]It is a question for Intermediate  :P
Just note that $\gcd(m,n)=\gcd(n-m,n)$.[/quote]

Probably it is and I just haven't really spend the proper time thinking on it, but what about $\gcd(m,n)=\gcd(n-m,n)$ so what? I don't really see how it can help to the computation of the sum.

Best,

[quote="Myth"]Really?
Suppose $n>2$, then we have $\frac{\varphi(n)}{2}$ pairs $(n,m)$ and $(n,n-m)$ and these pairs give us 1 to the sum. So answer is $\frac{\varphi(n)}{2}$.[/quote]

Yep, you are right. Thank you very much!

Best,


calculus
inequalities
calculus computations

Hi there.

I'm taking Advanced Calculus I this semester and I'm having a hard time doing what is somehow meant to be an easy problem/proof.  The problem goes like so:

Let epsilon > 0.  Determine how large any natural number N must be to ensure that the given inequality is satisfied, and use the Achimedian Property to establish that such n exist.

The inequality is like so:

1/n < epsilon

To put the Archimedian Property into context, it says in my textbook:  If epsilon > 0 and if M >0, then there exists a natural number n such that n * epsilon > M.

I seem to be at a loss of where to begin here and I feel like a moron for not seeing anything here.

I appreciate any help ahead of time. Thanks.

Let's consider M = 1. Then, for a given $ \varepsilon > 0$, the archimedean property of the real numbers allow us to be sure of the existence of a natural $ n$ such that $ n \varepsilon > 1 <\equal{}> \varepsilon > \frac1n$.

Carcul already showed how to use the Archimedean property to establish the existence of such $ n$. Now you have to determine what $ n$ should be to satisfy the inequality $ \frac{1}{n}<\epsilon$.
You probably see that the larger you take $ n$, the smaller $ \frac{1}{n}$ becomes. So if you take large enough $ n$, the inequality will be satisfied. It's up to you give an explicit expression for this statement. Note: your $ n$ should depend on $ \epsilon$. Give it a try.


Simplify the following expression

\[\frac{\sin \theta+\sin 3\theta+\sin 5\theta+\cdots+\sin (2n-1)\theta}{\cos \theta+\cos 3\theta+\cos 5\theta+\cdots+\cos (2n-1)\theta}\]

3X.lich wrote:Simplify the following expression

Well, it's easy to find the pattern [hide]tan(nx), [/hide]but somewhat harder to prove it, I think.  Looks like some routine plug and chug with De Moivre's.

Even if a student never heard of DeMoivre's form, the problem is still doable within the rules of trigonometry. Not hard at all.


A single large banquet table is created by joining the sides of several small square tables, each of which seats one person for each open side.  What is the maximum number of people that could be seated at a table which uses 10 small tables?

This looks like a chemistry problem to me. How many hydrogen atoms can there be in a 10-carbon hydrocarbon molecule?

[quote="Treething"][hide="Answer"]
40. I don't believe that the tables have to be fully connected - connecting half of one side to half of another side should do :P .
[/hide][/quote]
[hide]It says a single large table.  Anyway, I get 22.[/hide]


calculus
integration
function
geometry
algebra
binomial theorem
real analysis

Here is a challenging problem to satiate some AOPS discussion:

Prove the "Binomial Theorem" for positive integral exponent $n$ using TAYLOR's Theorem:

${(a+x)^{n}= a^{n}+{n}{a^{n-1}}{x}+{\frac{n(n-1)}{2}}{a^{n-2}}{x^{2}}+\frac{(n)(n-1)(n-2)}{2\times3}}{a^{n-3}}{x^{3}}+\ldots+x^{n}$

Thanks.

This is indeed a nice problem but I doubt if it is a challenging one!

[quote="keta"]OK, I was wrong, it's not challenging for AOPSers, but anyway, could you show me how to do it?[/quote]
I am sorry if I misuderstood you earlier but I thought you knew how to do the problem; I assumed you just wanted to pose this problem to other forum members.

If you haven't solved it yet, then my question to you is, what is the Taylor series expansion for an infinitely differentiable function $f(x)$ centered at $x = 0$? (We would call this a Maclaurin series.)

[quote="now a ranger"]what course do you use this theorem in? geometry? alg 2?[/quote]
This is taught in either Calculus I or II (I don't remember exactly which) in college though some people might learn this in high school.

[quote="keta"]Here is a challenging problem to satiate some AOPS discussion:
[/quote]

More accurately, don't call a problem challenging just because you think it will give you more replies.

[quote="cyqwhu"][quote="keta"]
${(a+x)^{n}= a^{n}+{n}{a^{n-1}}{x}+{\frac{n(n-1)}{2}}{a^{n-2}}{x^{2}}+\frac{(n)(n-1)(n-2)}{2\times3}}{a^{n-3}}{x^{3}}+\ldots+x^{n}$
[/quote]

I'm sorry. Isn't it the binomial theorem?[/quote]
If you read the original post by keta, that is exactly what he mentioned.

[quote="keta"]I'm guessing to prove this theorem all we need to do is differentiate the function $f(x) = (a+x)^{n}$ infinitely many times and plug in $x = 0$ for $f(x), f'(x), f''(x), \ldots,{f^{n}}(x), \ldots$, you get the point.

Then we write the Maclaurin Series to get the binomial theorem.

Is my logic right? I'll post a full proof tomorrow.[/quote]
Yes, it is correct.

Let $f(x) = (a+x)^{n}$. So we have $f'(x) = n(a+x)^{n-1}, f''(x) = n(n-1)(a+x)^{n-2}, f'''(3) = n(n-1)(n-2)(a+x)^{n-3}, etc...$. Now, take $f(0)$ of each of these, to get $f(0) = a^{n}, f'(0) = n(a)^{n-1}, f''(0) = n(n-1)(a)^{n-2}, f'''(0) = n(n-1)(n-2)(a)^{n-3}$, etc... We can then write out the Taylor series for this, but my question is how can we show $R_{n}(b,x)$ tends to $0$ as $n$ gets real large?


[hide]The domain of x is limited only by the square root, so we have to find the intervals that work in the square root:
$\sqrt{x^2-3x+2} = 0$
$(x-1)(x-2) = 0$
$x = 1, x = 2$
Intervals:
$(-infinity, 1) (1,2) (2, infinity)$
We test and see that the first and third intervals work, and that:
$(-infinity, 1] [2, infinity)$ is the domain; now we can find the range by plugging those numbers into the original equation and we get R := {for all y elements R, there exists y <=1Vy>=2}
I hope that is correct
[/hide]

I fail to understand the notation and the logic at the end of Apprentice's solution, so here is
[hide="mine."]The range of a function is the domain of its inverse, so we solve for $y$ in terms of $x$ :

\begin{eqnarray*}y^2-2xy+x^2&=&x^2-3x+2\\ x&=&\frac{y^2-2}{2y-3}\\ \end{eqnarray*}

Clearly, then, $y\neq \frac{3}{2}$.  Having made that condition, we substitute :
\begin{eqnarray*}y&=&\frac{y^2-2}{2y-3}+ \sqrt{\left(\frac{y^2-2}{2y-3}-1\right)\left(\frac{y^2-2}{2y-3}-2\right)}\\ &=&\frac{y^2-2}{2y-3}+\sqrt{\frac{(y-1)^2(y-2)^2}{(2y-3)^2}}\\ &=&\frac{y^2-2}{2y-3}+\left|\frac{(y-1)(y-2)}{2y-3}\right|\\\end{eqnarray*}

If $y\in \left[1,\frac{3}{2}\right)\cup[2,\infty)$, then the expression inside the absolute value bars is non-negative, and clearly
\begin{eqnarray*}\frac{y^2-2}{2y-3}+\frac{y^2-3y+2}{2y-3}&=&\frac{2y^2-3y}{2y-3}\\ &=&y\end{eqnarray*}

If $y\in (-\infty,1)\cup\left(\frac{3}{2},2\right)$, however, then the expression inside the absolute value bars is negative, and we have
\begin{eqnarray*}y&=&\frac{y^2-2}{2y-3}-\frac{y^2-3y+2}{2y-3}\\ &=&\frac{3y-4}{2y-3}\\ y&\in&\{1,2\}\text{ ,}\\ \end{eqnarray*}
a contradiction.  Thus the range of the function is $\left[1,\frac{3}{2}\right)\cup[2,\infty)$.

Q.E.D.[/hide]

That is completely correct... sorry for my bad math... I also tried using the inverse but that got me nowhere...  I like your solution though   :D


[quote="long14893"]$ x,y \in R$ ,$ cos (n \pi x)$ is not equal to $ cos (n \pi y)$
if $ n \in N*$  has the limited values ,then prove that $ x,y \in Q$[/quote]

This is wrong.

$ cos (n \pi x) \neq cos (n \pi y)$
$ \Leftrightarrow \begin{cases} 2 \neq n(x\plus{}y) \\
2 \neq n(x\minus{}y) \end{cases}$

So $ n$ has the limited values just means either $ (x\plus{}y)$ or $ (x\minus{}y)$ or both are rational.

I don't know what is meant by a limiting value, but $ x\plus{}y$ and $ x\minus{}y$ being rational definitely implies their average and their difference over 2 are rational, which are $ x,y$ respectively.

[quote="rofler"]I don't know what is meant by a limiting value, but $ x \plus{} y$ and $ x \minus{} y$ being rational definitely implies their average and their difference over 2 are rational, which are $ x,y$ respectively.[/quote]

I meant the sufficient and necessary condition is that at least one of them ($ (x \plus{} y)$ and $ (x \minus{} y)$) is rational, for example, $ (x \plus{} y)$ is rational and $ (x \minus{} y)$ is irrational satisfies the condition also.

Actually I guess (and it's just a guess) that $ n$ has limited values means there exists natural number $ n$ such that the condition $ cos(n \pi x) \neq cos(n \pi x)$ is not satisfied.

The author of this topic , long14893 , has once asked me this problem (in Vietnamese)
the problem actually is :
Given $ x,y \in R$ , $ n \in N^{*}$ and denote
 $ S\equal{}cos(n \pi x) \plus{} cos(n \pi y)$
If the set of all values that S can take has finite different elements , prove that $ x,y \in Q$

[quote="long14893"]$ x,y \in R$ ,$ cos (n \pi x)$ is not equal to $ cos (n \pi y)$
if $ n \in N*$  has the limited values ,then prove that $ x,y \in Q$[/quote]
sr ....
 :P  last night I and HTA did'nt find the solution of this problem , so we decided to ask here ... but I'm too sleepy so I wrote a wrong problem ...>"< anyway,our teacher solved this problem for us today ,so thanks everyone for noticing this problem>"< and sr for spamming such a stupid post >"< mod plz delete this topic >"<


College Interviews are undoubtedly the 'scariest' part of the admission process while it is well known that its importance is NOT ' that great!'. Anyways, asking questions HAVE to be asked after an interview with an alumnus.

[size=150][color=blue][b]Please give suggestions on what sort of questions one can ask an alumni interviewer of a top college say MIT or an Ivy League????[/b][/color][/size]

Thanks!!! Any other suggestions on college interviews are welcome!!!  :lol:

Hey guys, I'll really appreicate if you can give your suggestions in the next 2-3 hours. I have my interview in the morning!!!  :lol:

I've interviewed a lot of students for college.  If you ask questions you are 'supposed to ask', i.e. those that people on this board will tell you to ask, your interview will know it and it will be more of a minus than a plus.

Ask questions about whatever you're really curious about.  For me as an interviewer, this was always among the most interesting and important parts of the interview - it showed what the applicant really cares about, and if they really are curious.

Well, that is definitely true!!! Thanks!!!

But you dont expect to ask an alumnus about the courses, whether this dept./etc... is good or not...

I was considering more personal questions like about life, games, etc...

Is that correct?? I'd appreciate if you can give one example of a question that would be considered 'excellent' by the interviwer .... thanks!! (just one!!!)

Mr. Rusczyk is correct. Just ask the human being in front of you questions about what YOU are curious about, knowing that he has a connection to a university you are interested in attending.

College Interviews are undoubtedly the 'scariest' part of the admission ..

Rushil - These intervies are (or shouldbe)  any  thing but  "scariest part"  ... You could ask something which you want to know, and think that the inverviewer is  ther person who could answer. (These interviews are generally conducted by alumini , so they have first hand knowledge of the institutions).  They may also ask you some quetions (They do have the background information about you which you sent to the college) to know about you too.

Just be yourself. Nothing to it, really.

[quote="Gyan"]
Just be yourself. Nothing to it, really.[/quote]

Exactly.  I don't 'expect' anything.  I've seen such a wide range of students that I know not to expect anything specific.  Some ask nothing, some ask all sorts of interesting questions.  Ask what you want to know, not what you think the interviewer wants you to want to know.

Well, I had my interview. Contrary to most of what I read even on the college's site, the interview was NOT Informal. That happens!!! The person who interviewed was president of a reputed private university in India so I can imagine he was very busy and just spect30 minutes with me. I was a bit disappointed that I could not convey fully my passion for that college and could not ask all MY questions. He asked me the standard questions given on the college's interview report(he made me fill half of it -he'll probably change it later!). But a silver lining is that he said I can email him any other info I'd like to be included in my interview report. He also gave me some 'secret' suggestions'. So anyways, although this wasn't the sort of interview I would have hoped for from that college, it was good enough! (I'm sure you must have guessed which college I'm talking about!)

 Anyways, thanks for your suggestion!


function
complex analysis
algebra
polynomial
complex analysis unsolved

Let $ f$ be a meromorphic function on $ \bf{C}$ and assume that $ f$ is one-to-one. Show that $ f$ is a linear fractional transformation.

At $ \infty$ the function $ f$ must either have a pole or a removable singularity: it cannot be an essential singularity, sinc eotherwise due to Weierstrass-Casorati it wouldn't have been injective in a neighborhood of $ \infty$.
Hence $ f(\infty)$ is either a finite number or $ \infty$. Then note that our function $ f$ cannot have more than one pole in the extended complex plane, since otherwise due to the open mapping theorem it wouldn't be injective in $ \mathbb{C}$.
Hence suppose that $ f$ has a pole at $ z_0$, which can be either finite or $ \infty$. This pole must be of the 1'st order, since if it were of the second order at least $ f$ wouldn't be injective in its neighborhood.
If $ z_0$ is finite, denote by $ A\equal{}res_{z\equal{}z_0}f(z)$ and $ f(z)\minus{}\frac{A}{z\minus{}z_0}$ is holomorphic in the whole extended plane, hence constant by Liouville and $ f(z)\equal{}\frac{A}{z\minus{}z_0}\plus{}B$.
If $ z_0\equal{}\infty$, then $ f$ must be entire and polynomial and it is easy to checkthat the only injective olynomial is of the form $ f(z)\equal{}\alpha z\plus{}B$.

[quote="eugene"]
Then note that our function $ f$ cannot have more than one pole in the extended complex plane, since otherwise due to the open mapping theorem it wouldn't be injective in $ \mathbb{C}$.
[/quote]
[quote="eugene"]
This pole must be of the 1'st order, since if it were of the second order at least $ f$ wouldn't be injective in its neighborhood.
[/quote]
I'm not with you on these claims... :blush:

[quote="eugene"]
Then note that our function $ f$ cannot have more than one pole in the extended complex plane, since otherwise due to the open mapping theorem it wouldn't be injective in $ \mathbb{C}$.
[/quote]

Suppose the function $ f$ has two poles $ z_1,z_2$ in $ \bar{\mathbb{C}}$. Then the open mapping theorem says that it maps neighborhoods of $ z_1$ and $ z_2$ [b]onto[/b] some neighborhoods of $ \infty$, hence it should take lots of values in an appropriate neighborhood of $ \infty$ at least twice, which contradicts to the injectivity of $ f$.

[quote="eugene"]
This pole must be of the 1'st order, since if it were of the second order at least $ f$ wouldn't be injective in its neighborhood.
[/quote]

Suppose $ f(z)$ has a pole of order $ n$ at $ z_0$. Then the function $ g(z) \equal{} \frac {1}{f(z)}$ is analytic in a neighborhood of $ z_0$ and must also be injective there, but $ g'(z_0) \equal{} 0$ if $ n\geq 2$,  which contradicts to the injectivity of $ f$.

Is it clearer now ?.


A moderator please. It has been there all day on For The Win!

Thanks!!

He/she's allowed to do that.  Mods can't delete games, anyway.

Anyway, wrong forum.


Does there exist a non-constant function $f$ from ordered pair of natural numbers to natural numbers such that
$f(x,y)+f(y,x)=f(x^2,y^2)+1$ ?

[quote="rogue"][hide="hint"] Let g(x,y)=f(x,y)-1. Then all we need is non-constant function $g$ from ordered pair of natural numbers to nonnegative integers such that $g(x,y)+g(y,x)=g(x^2,y^2)$.  We may look for $g$ of the form $g(x,y)=h(xy).$
[/hide][/quote]

But f goes from NxN into N (not into R).


$ a,b,c$be real numbers and $ a\not\equal{} b\not\equal{} c$
prove that $ \sum_{cycl}(\frac{2a\minus{}b}{a\minus{}b})^{2}\geq 5$

[hide]First consider the case where one variable is 0. WLOG let $ c\equal{}0$ since expression is cyclic. The inequality then becomes $ \left(\frac{2a\minus{}b}{a\minus{}b}\right)^{2}\geq 0$, which is true.

Otherwise, let $ x\equal{}\frac{2a\minus{}b}{a\minus{}b}$ etc. so $ x\minus{}1\equal{}\frac{a}{a\minus{}b}\equal{}\frac{1}{1\minus{}\frac{b}{a}}$, $ \frac{b}{a}\equal{}1\minus{}\frac{1}{x\minus{}1}\equal{}\frac{x\minus{}2}{x\minus{}1}$. So we get
\[ 1\equal{}\frac{x\minus{}2}{x\minus{}1}\cdot\frac{y\minus{}2}{y\minus{}1}\cdot\frac{z\minus{}2}{z\minus{}1}\Rightarrow (x\minus{}1)(y\minus{}1)(z\minus{}1)\equal{}(x\minus{}2)(y\minus{}2)(z\minus{}2)\]
\[ \Rightarrow 7\plus{}xy\plus{}yz\plus{}zx\equal{}3(x\plus{}y\plus{}z)\]
Multiply by 2 to get
\[ 14\plus{}s^{2}\minus{}(x^{2}\plus{}y^{2}\plus{}z^{2})\equal{}6s\]
where $ s\equal{}x\plus{}y\plus{}z$. Therefore
\[ x^{2}\plus{}y^{2}\plus{}z^{2}\minus{}5\equal{}s^{2}\minus{}6s\plus{}9\equal{}(s\minus{}3)^{2}\geq 0\]
as desired.[/hide]


An integer is called "decreasing" if each digit is less than he one to its left. For example, 5320 is decreasing. How many decreasing integers are there between 2000 and 7000?

I got the correct answer by writing them all out. Is there a way to do it with combinations?

There's really not that many. I think counting is perfectly appropriate.


Steel has a density of 7.8 g/cm$ ^3$. How many kilograms does a cube of steel weigh if its edges each measure 100 cm?

There are $ 100^3\equal{}1000000$ cm^3, so multiply by $ 7.8$ to get $ 7800000$.  Now to find kilograms, we divide by $ 1000$ to get $ 7800$.


inequalities
search
inequalities proposed

Let $ a,b,c \geq 0$. Prove that :

$ \frac{1}{\sqrt{2a^2\plus{}ab\plus{}bc}}\plus{}\frac{1}{\sqrt{2b^2\plus{}ac\plus{}bc}}\plus{}\frac{1}{\sqrt{2c^2\plus{}ab\plus{}ac}} \geq \frac{9}{2(a\plus{}b\plus{}c)}$

I hope that you can find out some interesting things in this inequality  :P  :roll:

[quote="forevercontinue"][quote="can_hang2007"][quote="forevercontinue"]Let $ a,b,c \geq 0$. Prove that :

$ \frac {1}{\sqrt {2a^2 \plus{} ab \plus{} bc}} \plus{} \frac {1}{\sqrt {2b^2 \plus{} ac \plus{} bc}} \plus{} \frac {1}{\sqrt {2c^2 \plus{} ab \plus{} ac}} \geq \frac {9}{2(a \plus{} b \plus{} c)}$

I hope that you can find out some interesting things in this inequality  :P  :roll:[/quote]
This inequality is very old ;), see here: http://www.mathlinks.ro/viewtopic.php?t=126274[/quote]

I know this problem is very old . But where can I find a beautiful solution for this prolem . :roll:

Can_hang2007 , can you show me your proof for this problem . Thanks , my brother !!! :lol:[/quote]
I have posted my proof once time on maths.vn forum 2 years ago, I dont remember where I posted it. You can search at that forum with my past ID: "toanhocmuonmau" ;)

P/s: why do you call me "brother"? Am I know you, my friend?  :maybe:

Thanks Can_hang2007 so much !! I think you 've never know me before but  I 've heard about you !!
 :P

You are older than me so ....   :oops:


Let $ x,y,z\geq 0$, with $ xyz\equal{}1$. Find the minimum value of

\[ \dfrac{x}{y\plus{}z}\plus{}\dfrac{y}{x\plus{}z}\plus{}\dfrac{z}{x\plus{}y}.\]

$ \frac{3}{2}$,it is easy has show with C.S and chebychev without using $ xyz\equal{}1$

[hide]By rearrangement, we have:

$ \dfrac{x}{y \plus{} z} \plus{} \dfrac{y}{x \plus{} z} \plus{} \dfrac{z}{x \plus{} y}\geq\dfrac{y}{y \plus{} z} \plus{} \dfrac{z}{x \plus{} z} \plus{} \dfrac{x}{x \plus{} y}$

$ \dfrac{x}{y \plus{} z} \plus{} \dfrac{y}{x \plus{} z} \plus{} \dfrac{z}{x \plus{} y}\geq\dfrac{z}{y \plus{} z} \plus{} \dfrac{x}{x \plus{} z} \plus{} \dfrac{y}{x \plus{} y}$

Adding the inequalities gives:

$ 2(\dfrac{x}{y \plus{} z} \plus{} \dfrac{y}{x \plus{} z} \plus{} \dfrac{z}{x \plus{} y})\geq3$

$ \dfrac{x}{y \plus{} z} \plus{} \dfrac{y}{x \plus{} z} \plus{} \dfrac{z}{x \plus{} y}\geq\frac{3}{2}$

with equality for $ x \equal{} y \equal{} z$[/hide]

The constraint can be ignored (or respecified) since the inequality is homogeneous.

[hide="Solution 1"]Add $ 1$ to each fraction. AM-HM to get that the minimum of the new expression is $ 9/2$. Conclude the minimum of the original is $ 3/2$.[/hide]

[hide="Solution 2"]Set the constraint to $ x\plus{}y\plus{}z \equal{} 3$ and rewrite the denominators. Show that $ \frac{x}{3\minus{}x}$ is convex, then use Jensen's.[/hide]

dnkywin, you made an error in the last line
$ \dfrac{x}{y\plus{}z}\plus{}\dfrac{y}{x\plus{}z}\plus{}\dfrac{z}{x\plus{}y}\geq \frac{3}{2}$

not $ \frac{x}{y\plus{}z}\plus{}\dfrac{y}{x\plus{}z}\plus{}\dfrac{z}{x\plus{}y}\geq 3$

This goes by the name of [url=http://en.wikipedia.org/wiki/Nesbitt%27s_inequality]Nesbitt's inequality[/url], and the most interesting thing about it is probably [url=http://en.wikipedia.org/wiki/Shapiro_inequality]the care that must be taken in generalizing it[/url].


AMC
USA(J)MO
USAMO
summer program
PROMYS
AIME
ARML

I saw a thread elsewhere on the board with some thoughts about how old most participants here are. So I thought I'd ask the question directly:

1) as of 1 September 2003 (Labor Day in the United States in this school year), how old were you?

2) what school grade are you in in this current school year?

My answers are

45 years old

and

graduated from law school some years ago  :shock: I never participated in math contests in my youth  :-x

Feel free to brag up any unusual accomplishments you've had for people your age (e.g., qualifying for USAMO or whatever)--that could be useful information for younger people planning goals for the next few years.

Have a happy new year!

1) 13 years old
2) 8th
No accomplishments.

1.) 14
2.) 8th for now, 9th in a month

IMLEM Math Team school captain but that's not much...

1)13
2)8th

i dont like to brag unless provoked by a lesser being, although, i am proud to be able to make easy mac. and what a lifes lesson that is

[quote]As of twelve votes cast, the modal age is a little younger than I expected. It's impossible to say, of course, how representative the early voters are of the whole AoPS Web site membership.[/quote]

Yeah..people have only had about 2 hours to vote so far. I believe that quite a few people here are in college, but they just haven't had a chance to vote yet.

[quote="TripleM"]18.
And I have no idea about american grades etc, but I'm starting university next year if that means anything.[/quote]

Hi, I see you list your location as New Zealand--a place I've always wanted to visit. The school year in your country runs mostly in correspondence with the calendar year, right? Here in the United States, and in [url=http://www.taipeitimes.com/News/edit/archives/2004/05/22/2003156522]Taiwan[/url], the only other country I've lived in for a long time, the school year runs generally from August (or September) through May (or June). This week and next week most schools in the United States have a short (Northern Hemisphere) winter break for Christmas and New Year. But New Zealand is enjoying summer right now, of course, and when you say "next year," don't you mean that you begin a new school year soon, as soon as you get back to school in early 2004? Thanks for any clarification you can provide to someone who has never been south of 13 degrees north latitude (Guam).

1) 11
2) 6

First place in local math league division (not much  :cry: ).

1) 16
2) Junior (11th Grade)

Qual. for USAMO freshman year... went to PROMYS 2003

13, 9th grade. i take pleasure in being part of the majority. i don't really have accomplisments.

17 (18 in a week), college freshman.

No really spectacular accomplishments. I made the AIME three times in high school but never got beyond that. Captain of Upstate NY "A" team for ARML senior year.

[quote="tokenadult"][quote="TripleM"]18.
And I have no idea about american grades etc, but I'm starting university next year if that means anything.[/quote]

Hi, I see you list your location as New Zealand--a place I've always wanted to visit. The school year in your country runs mostly in correspondence with the calendar year, right? Here in the United States, and in Taiwan, the only other country I've lived in for a long time, the school year runs generally from August (or September) through May (or June). This week and next week most schools in the United States have a short (Northern Hemisphere) winter break for Christmas and New Year. But New Zealand is enjoying summer right now, of course, and when you say "next year," don't you mean that you begin a new school year soon, as soon as you get back to school in early 2004? Thanks for any clarification you can provide to someone who has never been south of 13 degrees north latitude (Guam).[/quote]

Yep, yep, yep, yep. :) School is from around february to december here, though of course university here doesn't start til march.. so I still have a couple of months holidays left!

Well Now I'm 15 and after 2 months i'm gonna be in grade 10.
Achievements :blush:
This year in grade 9 I got first place in the district olympiad ,first place and first prize in the National Olympiad and finally got bronze medal at the JBMO.
while in grade 7 and 8 i've always got first place in the district and third prize in the
National Olympiad.
I hope to be in the albanian IMO team next year. :)

[quote="RobinHood3000"]Was 14 at time of poll, now 15 (my birthday's in October).  Currently entering 10th Grade.

As of now, what meager achievements I have are as follows:

1) 1530 on the SAT I (730 Verbal, 800 Math)
2) Went to MathCounts Nationals (I was on the same team with that "Arnav guy" from NC)
3) In 7th Grade, I placed first in the region and 3rd in the state in the Algebra I Division of a North Carolina High School Mathematics Competition (not sure of the official name); placed first in region and first in state in the Geometry Division the next year
4) Did that Duke TIP thing; I've got the baseball cap to prove it
5) Freshman year in high school, I represented the state of Virginia in the National High School Mathematics Championship.  I placed fifth overall, which was okay, considering I was a freshman.
6) Led my JV Scholastic Bowl team to 1st in the county and 2nd in the region (again, freshman year)
7) Part of a six-way tie in my grade for highest GPA[/quote]

...so that's who you are. Why, specifically, are you on the same team as "Arnav guy" as opposed to "Misha guy" or "Vivek guy" ? hrmph. stupid system anyway who the heck uses buzzers in the real world

[quote="RobinHood3000"]As of now, what meager achievements I have are as follows:

4) Did that Duke TIP thing; I've got the baseball cap to prove it[/quote]

Yeah, real meager.  :maybe:  Some of my friends are at the throat of my computer because of that quote.

I lost my precious red sox cap at mathcounts last year. :cry: :cry:

Hmm... let's see.
1)13 years
2) 8th grade
3) nothing at all. 16/25 on amc 8  :blush:  :blush:  :(  :(

Well, since Arnav was the only one on our team to get into the Countdown Round, I figured he got the most coverage.

I apologize if I underplayed my achievements too much, but compared to most people on these boards, some humility on my part seemed to be in order...

I just became 15 years old today. I'm in grade 10. I don't have many accomplishments :maybe: except first place in some contests just within the students of my grade in the school these past few years.

I'm 16
I don't know american system so...

I predict, come December, at least some of those on these forums will have a PSAT score added to their list of their achievements.

uh...
eighth grade
13 as of 10/17/04
no accomplishments
except survivng multiple attacks from my 3-year old brother  :cool:

[quote="cardwacko9999"]Previous Profile:

7th grade begin
11
No Contests (Except for Olympiads :blush: )

Now

8th grade begin
12
AMC 8 21/25, AMC 10 ? Others

Shouldnt we be starting a new poll-
everyone has gotten [b]one[/b] year older at the minimum[/quote]

No. That'd make it more complicated. Just go by the old system. you should be able to figure it out.


Find the Funtions $ R \to R$ such that $ f(x \plus{} y) \ge f(x).f(y) \ge 1997^{x \plus{} y}$.

Let $ g(x)\equal{}1997^x, h(x)\equal{}\frac{f(x)}{g(x)}$. Notice that $ h(x)$ is well defined on R, g satisfies $ g(x\plus{}y)\equal{}g(x)g(y)$ and f is always positive.

We have - $ f(x\plus{}y) \ge f(x)f(y) \ge g(x\plus{}y)\equal{}g(x)g(y) > 0$. Letting $ y\equal{}0$ gives $ f(x) \ge g(x)$, or:$ h(x) \ge 1$

The conditions are now equivalent to: $ h(x\plus{}y) \ge h(x)h(y), h(x) \ge 1$. Letting x=0 gives $ h(0) \le 1$, or: $ h(0)\equal{}1$.
Letting x=-y gives: $ 1 \ge h(x)h(\minus{}x) \ge 1$ ,or: $ h(x)\equal{}1$ for all $ x \in R$. Conclusion: $ f \equiv g \equiv {1997}^{x}$.

Hi,my friend
You can find $ f(0) \equal{} 1$,$ f(x)\ge 1997^x$,$ f(x)(f \minus{} x) \equal{} 1$
So $ f( \minus{} x)\ge 1997^{ \minus{} x}$ thus you can obtain $ f(x)\le 1997^x$
The result is $ f(x) \equal{} 1997^x$ :wink:
It is not hard


induction
graph theory
number theory unsolved
number theory

A set $P$ consists of $2005$ distinct prime numbers. Let $A$ be the set of all possible products of $1002$ elements of $P$ , and $B$ be the set of all products of $1003$ elements of $P$ . Find a one-to-one correspondance $f$ from $A$ to $B$ with the property that $a$ divides $f (a)$ for all $a \in A.$

Note that for all integers $n$, we have $\binom{2n+1}{n}= \binom{2n+1}{n+1}$.
Thus you only have to use induction on $n$, by considering apart two of the given prime, say $p$ and $q$.
In both sets, there are exactly the same number products with $p$ and $q$, with ony $p$, with only $q$, with none of them. Then you can use the induction for each of the four cases.
It is easy to find such a correspondance for a set three primes (and products of one or two of them).

Pierre.

Hello Pierre
[quote="pbornsztein"]...you only have to use induction on $n$, by considering apart two of the given prime, say $p$ and $q$.[/quote]

I'm not sure this induction is so simple:
For products containing just p or just q, it's OK :
We have 1001 (from 2003) numbers + p and we obtain 1002 (from 2003) numbers + p

But for products containing neither p nor q ? :
We have 1002 (from 2003) numbers and we want to have 1003 (from 2003). And the induction here seems not to work.

--
Patrick

With this problem i think we must FIND not must EXIST!

:oops:  A false computation...
But, did I still miss something or the desired result is a direct consequence of Hall's theorem about perfect matching?
I mean :
Let $S$ be the given set of primes.
Now, consider the bipartite graph $X-Y$ where $X$ is the set of all products of $1002$ from $S$ and $Y$ is the set of all products of $1003$ from $S$.
For each $x \in X$ an$y \in Y$, draw edge $xy$ if and only if $x$ divides $y$, that is $y=px$ for some $p \in S$ which does not divide $x$.
Thus, this graph is a $1003$-regular bipartite graph. Using Hall's theorem, this forces the graph to have a perfect matching, as desired.

Pierre.

That's right! This is same for me. But we need FIND not must need EXIST.

Ok! I will check your solution!
[quote="Rust"] $f(p_{i_{1}}...p_{i_{1002}})\to p_{i_{1}}...p_{i_{1002}}p_{j_{g(x)}}.$
[/quote]
Is it well-defined ?

Here is a way to motivate the construction.

We want to construct a function $f: S \rightarrow \{1, 2, 3, \cdots, 1003\}$, where $S$ is the set of size $1002$ subsets of $[2005]$, satisfying the following property. If we consider the $f(a_1, a_2, \cdots, a_{1002})$th smallest number which is not in $\{a_1, a_2, \cdots, a_{1002}\}$, then the map sending $\{a_1, a_2, \cdots, a_{1002}\} \rightarrow \{a_1, \cdots, a_{1002}\} \cup \{b\}$ is a bijection from the set of subsets of size $1002$ to the subsets of size $1003$ of $\{1, 2, 3, \cdots, 2005\}.$

Since $\binom{2005}{1002} = \binom{2005}{1003}$, we just need the function to be surjective. Indeed, we actually want every subset $\{a_1, a_2, \cdots, a_{1003}\} \in $ is mapped to exactly once. This is equivalent to exactly one of:

$$f(a_1, a_2, \cdots, a_{i-1}, a_{i+1}, \cdots, a_{1002}) = a_i - i +1,$$

for $1 \le i \le 1003$, being true.

From here it's not hard to find what to do. Let $f(x_1, x_2, \cdots, x_{1002}) = -(x_1 + x_2 + \cdots + x_{1002}$ modulo $1003.$

$\square$


ARML
probability
geometry
LaTeX
analytic geometry
function
cyclic quadrilateral

Whats the answer according to the answer key?

[hide="Team 7"] Let $3x-2y=a$; we have:
$2a-z=\sqrt{3}$
$a=r\sqrt{3}$
$3a-2z=r^{2}\sqrt{3}$

Taking twice the first equation and subtracting the second and third, we get:
$0=2\sqrt{3}-r\sqrt{3}-r^{2}\sqrt{3}\Longrightarrow r^{2}+r-2=(r-1)(r+2)=0$

Because $r\neq 1$, this tells us that $r=-2$ and therefore $a=-2\sqrt{3}$.

Putting this back into the first equation, we get $2a-z=-4\sqrt{3}-z=\sqrt{3}\Longrightarrow z=-5\sqrt{3}$.

Also, from the second equation, we get
$3x-2y=-2\sqrt{3}\Longrightarrow y=\frac{3x}{2}+\sqrt{3}$.

Then $(x,y,z)=(x,\frac{3x}{2}+\sqrt{3},-5\sqrt{3})$.[/hide]

[hide="Individual 1"]

It says in the problem that $z$ is in the first quadrant, so if $n>2$ then $z>90$.[/hide]

[quote="Art of Owna"]well is E2 individual (2x^2)/567?[/quote]
Yes, but please post your solution. And use $\LaTeX$.

[quote="13375P34K43V312"][quote="Art of Owna"]Is the ARML Team #10 11/36?[/quote]

That's what I got, but that's wrong according to the answer key.[/quote]

um.. you have to take into account that each of them stay 10 minutes..

well. I got $\frac{31}{72}$

[quote="cincodemayo5590"][quote="Allan Z"][quote="cincodemayo5590"][hide="Team 10"]
Well if $(x,y) \in [0,60]^{2}$ (ugh, bad notation, you know what I mean  :P) are the entrance times for the two ppl, we want
\[|x-y| \leq 10 \]
in the square determined by $(0,0)$ and $(60,60)$. We get a polygon bounded by the coordinates $(0,0), \ (0,10) \ (50,60), \ (60,60), \ (60,50), \ (10,0), \ (0,0)$, which has an area of $1100$.

Thus, the probability is that area over the area of the square, which is equal to:
\[\frac{1100}{60^{2}}= \boxed{ \frac{11}{36}}\]
[/hide]
I don't know why the solution only bounded it to those coordinates.  :)[/quote]

nope.. you left out something-the first guy stays 10 minutes as well.
 draw a line passes through (0,0) and (60,60). and you set one guy to be the "template".. the time he will be there is bounded by $(0,0)$, $(60,60)$, $(50,60)$, $(0,10)$. Therefore.. the second guy should be bounded by $(0,0)$, $(10,0)$,$(60,50)$, $(60,60)$, $(40,60)$, $(0,20)$..

The area is $60*60-\frac12*40*40-\frac12*50*50=1550$

$\frac{1550}{3600}=\frac{31}{72}$[/quote]

Umm... what are your coordinates representing?[/quote]

oh snap.. I am wrong..!!! my bad  :oops:

Nah, Im pretty sure Im correct.. you should reread the question, I know Bc/CD is 10/9, but it says to express in terms of x.. if you dont believe me set ur equation 10/9 equal to mine 2x^2/567 and solve for x

Then plug x back in( as you can find the four sides) Use Ptolomeys-do the summ of the product of opposite sides equal the product of the diagonals-yes.. so my answer and ur answer are both correct, except we are not looking for ur answer.. that would be toooo easy for arml..[/quote]

10/9 is a function of x...i don't see why you could just stop in the middle of a problem like that...you have to simplify as much as possible

[quote="13375P34K43V312"]In doing these problems, I found the solutions quite vague. I couldn't get them either.

[hide="Team 7"]There are infinite solutions to the simultaneous equations:

$6x-4y-z=\sqrt{3}$
$3x-2y=r\sqrt{3}$
$9x-6y-2z=r^{2}\sqrt{3}$

Given that the value of $r\neq1$, write the solution for $(x,y,z)$ in terms of $x$.[/hide]

[hide="Team 10"]Two people went in a store somewhere between 3:00 and 4:00 PM and each one stayed in the store for 10 minutes only. If it is equally probable that each person could have been in the store at any time during this hour, what is the probability they were in the store together at some time during this hour?

My problem here is that I set up the graph to do geometric probability and stuff, but for some reason in the solution the time frame is only bounded by $(5,5),(5,55),(55,55),(55,5)$. [/hide]

[hide="Individual C1"] Given $z$ is a complex number whose graph is in the first quadrant of the complex plane and satisfies the equation $z^{12}=2(1+i\sqrt{3})^{7}$. Write all solutions for $z$ in polar form, $r\text{cis },r>0,0^{\circ}<\theta<360^{\circ}$ and $r$ is expressed in simplest radical form.

My problem here is that it goes from writing $z=(2^{8})^{1/12}\text{cis }(\frac{60^{\circ}}{12}+\frac{360^{\circ}}{12}n)$, where $n=0,1,2,\ldots,11$ to $z=2^{2/3}\text{cis }(5^{\circ}+30^{\circ}n)$, where $n=0,1,2$. I don't understand why the set of values for $n$ changes.[/hide]

[hide="Individual E2"] Given a circle with arc AB congruent to arc AD, $AD=x$, $AE=15$, $EC=6$, and $BE=10$, find $BC: CD$ in terms of $x$. In the diagram given, $ABCD$ is a cyclic quadrilateral, and the intersection of its diagonals is $E$.[/hide]

[hide="Relay #3 2"] Given a circle O with radius R and $m\angle{A}=m\angle{B}-40^{\circ}$, find the area of segment II less than the area of segment I. In the given diagram, $A,O,B$ appear to be collinear, $A,B,C$ are on the circle, and segment II appears to refer to the area of the region bounded by segment AC and arc AC, and segment I appears to refer to the area to the region bounded by arc BC and segment BC.[/hide]

Thanks[/quote]
[hide="7"]
Clearly if there are infinite solutions, then the first and second equations imply the third.
Thus, \[9x-6y-2z=r^{2}\sqrt{3}\iff F(\text{ eq. 1})+G(\text{ eq. 2}) \]
In fact, we can solve for $r$ if we can achieve a left hand side from equations 1 and 2 which is equivalent to the left hand side of equation 3 (and nothing stupid which would result in introducing a non-$r$ variable on the r.h.s).
Since $(2)$ has no $z$ coefficient, this means that we must have twice the first equation. To match $x,y$ coeffs would involve subtracting the second equation. Thus we find: \[2(1)-(2)\equiv 9x-6y-2z=2\sqrt{3}-r\sqrt{3}\]
which, to yield infinite solutions, must be virtually equivalent to equation 3, i.e. $2-r=r^{2}\iff r^{2}+r-2=0 \iff (r-1)(r+2)=0\implies r=2$ since the other solution was disqualified.
It remains to solve the system: \[6x-4y-z=\sqrt{3}\\ 3x-2y=2\sqrt{3}\]
But now we perform: $2(2)-(1)$ to yield $z=3\sqrt{3}$, thus we have by $(1)$(originally) that $y=\frac32 x-\sqrt{3}$.

So, $(x,y,z)=(x,\frac32 x-\sqrt{3},3\sqrt{3}$.[/hide]

[quote="Altheman"][quote="Art of Owna"]I keep on gettin 11/36, I have no clue why this isnt the right answer..

Can someone tell me the right answer?

Ok now for E2, I dont have latex so bare with me..
[hide]
Triangle AED and BEC are similar ( same inscribed arcs, so by AA they are similar), ok so the using poportions, ED is 9, also by poportions, you can find that if AB is x, then BC is (2/3)x
THen by the congruent arcs, and BCA and ACD are congruent, so you can use the ANgle Bisector Theorm..
Denote DC as y, so we have y(2/3)x=10*9+6^2, so y then is equivalent to 189/x, now BC/DC is (2/3)(x)/(189/x), which is (2x^2)/567

[/hide]
If you would continue on this problem and use Ptolomys Theorm, you can actually solve for x..[/quote]

i am pretty sure that is incorrect then...especially when we can be like:

$\text{arc}\, AD=\text{arc}\, AB \implies \angle ACB=\angle ACD$, by power of a point, $ED=9$, so by the angle bisector theorem, $BC\, : \,CD=\frac{10}{9}$[/quote]

[hide="Yea, I get:"]
\[\frac{10}{9}\] also. Just look at all of the fun similar triangles! [/hide]

[quote="me@home"]
[hide]$2-r=r^{2}\iff r^{2}+r-2=0 \iff (r-1)(r+2)=0\implies r=2$ since the other solution was disqualified.[/hide][/quote]

Are you sure you have the right value of r?

[quote="Art of Owna"]I kinda get what you are talking about...but I dont think you were suppose to do taht..  Anyways the guy said my answer was correct.. 10/9 is not solving in terms of x because you canceled the x's out, the only way not to cancel the x out is by using the angle bisector theorm.  Plus, x is sqr315, try plugging that in for the sides.  I think that is correct.[/quote]

What do you mean, it's not solving in terms of x? Like Alex said, $f(x)=\frac{10}{9}$ is a function of x.

[quote="Art of Owna"][quote]What do you mean, it's not solving in terms of x?[/quote]

Ya it is, reread the question, it says to solve in TERMS of x[/quote]

....

[hide="Ok:"]
$\frac{10}{9}\cdot x^{0}$
[/hide]

I'm just saying that the solution given is badly stated. Or incomplete, since the situation is uniquely determined.

[quote="Art of Owna"]What if the question said to solve for x?  Then what would x be?[/quote]
[hide]You can use Stewart's Theorem to find x in $\triangle ABD$. I have that $x=\sqrt{315}$. [/hide]

Ya, that would have been a better question..

if the probability space is (x,y) | 0<x,y<6, we want the area in that space such that |x-y|<1

so whoever wants to draw that....i am getting

1-1/36(2*1/2*5^2)=11/36

huh?

does it mean that each stay 10 mins within the hour period?

I think there's a consensus that the answer key might need revising.


limit
search
inequalities
calculus
calculus computations

If $ {\{ a_n \} }$ is a positive sequence show that $ \mathop {\overline {\lim } }\limits_{n \to \infty } \left( {\frac {{1 \plus{} a_{n \plus{} 1}}}{{a_n }}} \right)^n \ge e$ and that this estimate cannot be improved upon.

(The limit with the bar overhead means superior limit)

$ a_n = n^2$ and the limit converge to $ 1$ and it's maximum value is app. the square root of 5. What is less than $ \e$

Sorry EulerIntegral. I have fixed the typo now.

It's been posted, not too long ago. I haven't found it yet because I'm not sure what to search for.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=178707
No proof, but a good suggestion made by [b]zhaobin[/b].

mlok it is not clear how to choose the subsequence

When it's not clear how to choose something, you can assume that something does not exist, and try to obtain a contradiction.

By contradiction assume for some sequence $ \{ a_n \}$, $ \mathop {\overline {\lim } }\limits_{n \to \infty } \left( {\frac{{1 \plus{} a_{n \plus{} 1} }}{{a_n }}} \right)^n  < e$

This means $ \exists k\forall n > k\left( {\frac{{1 \plus{} a_{n \plus{} 1} }}{{a_n }}} \right)^n  < \left( {1 \plus{} \frac{1}{{n \plus{} 1}}} \right)^n$. Or what is the same: $ \frac{{1 \plus{} a_{n \plus{} 1} }}{{a_n }} < 1 \plus{} \frac{1}{{n \plus{} 1}}$

If $ \{ a_n \}$has a convergent subsequence (to be denoted $ \{ a_n \}$ for simplicity) tending to a, then we have 1 + a <= a

If $ \{ a_n \}$is unbounded it is clear that we can choose a subsequence so as to fail the inequality (choose consecutive points to be 'far apart').

The only thing I don't get is how to show the estimate cannot be improved. I don't see how the example in the link was arrived at. :huh:

The above proof is completely wrong but I can't delete it. Here is a correct and shorter one:

If we assume a positive sequence {a(n)} exists for which the following holds then it must be decreasing and divergent:
$ \exists k\forall n > k\left( {\frac{{1 \plus{} a_{n \plus{} 1} }}{{a_n }} < 1 \plus{} \frac{1}{n}} \right) \Leftrightarrow \exists k\forall n > k(1 \le a_n  \minus{} a_{n \plus{} 1} )$

But decreasing positive sequences must converge.


let x,y,z be real numbers such that:
x²+z²=1
y²+2y(x+z)=6
prove that y(z-x) inf 4,dtermine when equality holds


geometry
circumcircle
geometric transformation
reflection
incenter
geometry solved

Part (a) follows from the angle version of Ceva's theorem.  They concur at the isogonal conjugate of the Nagel point.

Part (c) is problem 2 of the 1982 IMO:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1307]www.artofproblemsolving.com/Forum/viewtopic.php?t=1307[/url]

[quote="nsato"]Part (c) is problem 2 of the 1982 IMO:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1307]www.artofproblemsolving.com/Forum/viewtopic.php?t=1307[/url][/quote]
But we must also show that $A_{1}K_{a},B_{1}K_{b},C_{1}K_{c}$ concur on the $w_{2}$   :)
And what about [b](b)[/b] ? :roll:

[quote="cefer"]Let $O$ and $I$ be the centre of the circumcircle $w_{1}$ and the incircle $w_{2}$ of a triangle $ABC$. Denote by $D$, $E$, $F$ the points of tangency of the incircle $w_{1}$ with the sides $BC$, $CA$, $AB$, and denote by $K_{a}$, $K_{b}$, $K_{c}$ the reflections of these points $D$, $E$, $F$ in the lines $AI$, $BI$, $CI$, respectively. Denote by $A_{1}$, $B_{1}$, $C_{1}$ the midpoints of the segments $BC$, $CA$, $AB$.
[b](a)[/b] Prove that the lines $AK_{a}$, $BK_{b}$, $CK_{c}$ are concurrent.
[b](b)[/b] Prove that the points $S$, $I$, $O$ are collinear, where $S$ is the point of intersection of the lines $AK_{a}$, $BK_{b}$, $CK_{c}$.
[b](c)[/b] Prove that the lines $A_{1}K_{a}$, $B_{1}K_{b}$, $C_{1}K_{c}$ are concurrent at a point on the circle $w_{2}$.[/quote]

More or less the complete solution of this problem stands on ML. The only difficulty is to combine:

Since I is the incenter of triangle ABC, the lines AI, BI, CI are the angle bisectors of the angles CAB, ABC, BCA. Thus, instead of saying that the points $K_{a}$, $K_{b}$, $K_{c}$ are the reflections of the points D, E, F in the lines AI, BI, CI, we can say that the points $K_{a}$, $K_{b}$, $K_{c}$ are the reflections of the points D, E, F in the angle bisectors of the angles CAB, ABC, BCA.

Hence, the lines $AK_{a}$, $BK_{b}$, $CK_{c}$ are the reflections of the lines AD, BE, CF in the angle bisectors of the angles CAB, ABC, BCA. Thus, since the lines AD, BE, CF concur at the Gergonne point of triangle ABC, the lines $AK_{a}$, $BK_{b}$, $CK_{c}$ must concur at the isogonal conjugate of the Gergonne point of triangle ABC. Hence, problem [b](a)[/b] is solved, and we obtain that the point of intersection S of the lines $AK_{a}$, $BK_{b}$, $CK_{c}$ is the isogonal conjugate of the Gergonne point of triangle ABC. Hence, by [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=156692#156692]http://www.mathlinks.ro/Forum/viewtopic.php?t=23668 post #3[/url] Theorem 2, the point S is the internal center of similitude of the circumcircle and the incircle of triangle ABC. As a consequence, the point S lies on the line joining the centers of the circumcircle and the incircle of triangle ABC, i. e. on the line joining the points O and I. Thus, the points S, I, O are collinear, what proves problem [b](b)[/b].

According to the Additional Theorem in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=118923#118923]http://www.mathlinks.ro/Forum/viewtopic.php?t=14927 post #2[/url], the lines $A_{1}K_{a}$, $B_{1}K_{b}$, $C_{1}K_{c}$ pass through the Feuerbach point of triangle ABC. Of course, this Feuerbach point must lie on the incircle $w_{2}$ of triangle ABC. Thus, problem [b](c)[/b] is also proven.

Problem completely solved.

[b]PS.[/b]

[quote="nsato"]Part (a) follows from the angle version of Ceva's theorem.  They concur at the isogonal conjugate of the Nagel point.[/quote]

The "Nagel" should be replaced by "Gergonne" here.

  Darij


Given a circle (O,R). A point P lies inside the circle, OP=d, d<R. We consider quadrilaterals ABCD, inscribed in (O), such that AC is perp to BD at point P. Evaluate the maximum and minimum values of the perimeter of ABCD in terms of R and d.

Take a look at this: http://www.mathlinks.ro/viewtopic.php?t=138970


induction
combinatorics unsolved
combinatorics

Dear all reader, I have this problem that try to solve form one day and a half...
I m not able to find a way to encode this problem as math induction problem...

Let a_1 = 5 and lei a_(n+1) = (a_n) Prove that the last n digits of a_n are
the same as the last n digits of a_(n+1)

Some hint??

Thanks

All the best.

[quote="joksy"]a_(n+1) = (a_n)[/quote]

[img]http://img362.imageshack.us/img362/1563/orlyowl9jk.jpg[/img]

Let a_1 = 5 and lei a_(n+1) = (a_n) Prove that the last n digits of a_n are
the same as the last n digits of a_(n+1)

Sorry guys i miss the most important part ....

Let a_1 = 5 and lei a_(n+1) = (a_n)^2 Prove that the last n digits of a_n are
the same as the last n digits of a_(n+1)

Ah, this explains much. Some hints:

$ a_n\equal{}5^{2^{n\minus{}1}}$ (note that $ 5^{2^{n\minus{}1}}$ means $ 5^{\left(2^{n\minus{}1}\right)}$, not $ \left(5^2\right)^{n\minus{}1}$).
You have to prove that $ 10^n\mid a_{n\plus{}1}\minus{}a_n$, i. e. that $ 10^n\mid 5^{2^{n}}\minus{}5^{2^{n\minus{}1}}$.
Do you see why it is obviously enough to show that $ 2^n\mid 5^{2^{n\minus{}1}}\minus{}1$.
Do you see how to prove this by induction over $ n$ ? Think about how $ 5^{2^n}\minus{}1$ inherits from $ 5^{2^{n\minus{}1}}\minus{}1$ the divisibility by $ 2^n$ and how does it get an "additional" divisibility by $ 2$.

Isn't $ 2^n\mid 5^{2^{n\minus{}1}}\minus{}1$ a nice fact? Can you generalize it? (Hard question.)

  darij


AMC
AIME
geometry
area of a triangle
Heron\u0027s formula

This was an Aime problem in 92 i think. Ive tried everything and cant seem to find the answer. Can someone please help?

In Triangle ABC........AB=9 and   BC:CA = 40:41.

What is the largest possible area of this triangle?

Thanks

Assuming that AC and BC are integer values?

no you can t assume that.  I looked the problem up and it is number 13 on the AIME

A nice problem indeed.

Let $A$ and $B$ be two fixed points. The locus of points $P$ such that $\frac{AP}{BP}$ is fixed is a circle centered on the line $AB$. Let $AB = 9$.  The two possible $P$ on the line $AB$ are the $P$ on the line $AB$ but not on $\overline{AB}$ such that $AP = 360, BP = 369$ and the point $P \in \overline{AB}$ such that $AP = 9 \cdot \frac{40}{40 + 41}$. Hence, the diameter of the circle of Apollonius (as the locus of $P$ is sometimes called) is $360 + \frac{40}{9}$; its radius is $\frac{1640}{9}$. Since $P$ must be selected at most a distance of $\frac{1640}{9}$ away from line $AB$, the maximum area is $\frac{1}{2}\cdot\frac{1640}{9}\cdot 9 = 820$.


Finally, after nearly 2 weeks of forgetfulness and slacking on my part, here they are!

#1. ToBe (graphic)

[url]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=309[/url]

#2. ccy (graphic)

[url]http://www.artofproblemsolving.com/Forum//album_pic.php?pic_id=369[/url]

You have one week to vote. Next Wednesday we will find out the winner! I will vote in case of a tie.

NOTE: The competitors CAN vote for themselves.

Wow, hi guys. This is my first post in the graphical design section... I'm not a very good artist but I really like your drawing contest and perhaps I would like to partcipate. Is this thing going to continue some day?

Yes, ccy has just won Round Eight and we are waiting for him to give us the topic for Round Nine.

How do you create these things?  Do you need Photoshop?  Sorry, I am a complete newbie, and would appreciate some help.  Thanks!

If i'm not mistaken, ccy's graphic was created using some type of modeling software (such as Blender). If not, it could have been created by some type of modeling software (such as Blender).

Btw, Blender is completely free. :)


Why the number of the posts of Strzelec and Aziz are both '0'?

oops ~ I can't understand, maybe because they are reposting, I mean thiss posts were posted before - don't know, have no idea  :huh:

davron latipov

No..there is a problem with the post count..everybody has it.

[quote="modeler"]So will this post count?

(This is just a test)

(It really doesn't count! :D )[/quote]
 :?:

[quote="Jimbombinio"][quote="Strzelec"]$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq3$

$abc=1 \Rightarrow c=\frac{1}{ab}$

$\frac{1+ab}{1+a}+\frac{1+b\cdot\frac{1}{ab}}{1+b}+\frac{1+\frac{1}{ab}\cdot a}{1+\frac{1}{ab}}\geq3$

$\frac{1+ab}{1+a}+\frac{a+1}{a\cdot(1+b)}+\frac{(1+b)\cdot ab}{(ab+1)\cdot b}\geq3$
$\Rightarrow \frac{\frac{1+ab}{1+a}+\frac{a+1}{a\cdot(1+b)}+\frac{(1+b)\cdot ab}{(ab+1)\cdot b}}{3}\geq1$

By AM$\geq$GM:

$\frac{\frac{1+ab}{1+a}+\frac{a+1}{a\cdot(1+b)}+\frac{(1+b)\cdot ab}{(ab+1)\cdot b}}{3}\geq\sqrt[3]{\frac{1+ab}{1+a}\cdot\frac{a+1}{a\cdot(1+b)}\cdot\frac{(1+b)\cdot ab}{(ab+1)\cdot b}}= 1$.[/quote]

There is an easy way to solve it!!!

put instead of 1 $abc$

$\frac{abc+ab}{1+a}+\frac{abc+bc}{1+b}+\frac{abc+ac}{1+c}= \frac{ab(c+1)}{1+a}+\frac{bc(a+1)}{1+b}+\frac{ac(b+1}{1+c}\geq3\sqrt[3]{a^{2}b^{2}c^{2}}$ By AM-GM

and $3\sqrt[3]{a^{2}b^{2}c^{2}}=3$[/quote]
Maybe your solution is totally same as Strzelec's :D


Answer with explanations ( i do not have the answers)

1) Solid, Liquid and gaseous water are all present at very low pressure near 0 deg celsius. How do distances between molecules in the 3 states compare under these conditions?

a) Distances are equal in all states
b) Distances are similar in solid and liquid, which are smaller than that in the gas
c) Distances are smallest in the solid, and similar in the liquid and the gas.
d) Distances are smalllest in the liquid, and similar in solid and the gas.

2) In which gas sample do molecules have the greatest average kinetic energy?

a) $ H_2$ at 100K
b) $ CH_4$ at 273K
c) $ H_2O$ at 373K
d) $ CH_3OH$ at 353K

3) Which of the following factors affect the equilibrium vapour pressure of a liquid in a container?

a) Temperature
b) Surface of the liquid
c) Volume of the container

4) Which one the following species can act as both a bronsted - lowry acid and base in aqueous solution?

a) $ CH_3COOH$
b) $ NO_3^{ \minus{} }$
c) $ H_2PO_4^{ \minus{} }$
d) $ OH^{ \minus{} }$

5) The $ K_a$ value for an acid is $ 1.0*10^{ \minus{} 2}$. What is the $ K_b$ value for its conjugate base?

a) $ 1.0*10^{ \minus{} 2}$
b) $ 1.0*10^{ \minus{} 4}$
c) $ 1.0*10^{ \minus{} 10}$
d) $ 1.0*10^{ \minus{} 12}$

1) Do we really need to answer that question?

2) Isn't the average kinetic energy directly proportional to temperature?

4) The answer should be (c), although other species can also act as acids and bases in the appropriate conditions.

5) Perhaps you want to mean $ K_b$? And at what temperature?

yea i meant $ K_b$

I am not testing anyone with these questions, i know they are really easy, i just want to check my answers and see how different people interpret them so that i can learn a little more and perhaps something new

for the second one, even though K.E is proportional to the temperature, i thought that the type of interrmolecular bonding would play a role but i guess i'm wrong.

just guessing
r these answers rite??

1.(c)
2.(c)
3.(a) {most appropriate}
4.(c)
5.(d) at 25C


geometry
geometric transformation
circumcircle
incenter
national olympiad

[b]Translations are due to the courtesy of [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=2616]zhaoli[/url] and [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=235]psykyk.[/url][/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42562]1[/url] For all primes $p \geq 3,$ define $F(p) = \sum^{\frac{p-1}{2}}_{k=1}k^{120}$ and $f(p) = \frac{1}{2} - \left\{ \frac{F(p)}{p} \right\}$, where $\{x\} = x - [x],$ find the value of $f(p).$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42565]2[/url] Let $n \geq 2, n \in \mathbb{N}$, $a,b,c,d \in \mathbb{N}$, $\frac{a}{b} + \frac{c}{d} < 1$ and $a + c \leq n,$ find the maximum value of $\frac{a}{b} + \frac{c}{d}$ for fixed $n.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42563]3[/url] A graph $G=(V,E)$ is given. If at least $n$ colors are required to paints its vertices so that between any two same colored vertices no edge is connected, then call this graph ''$n-$colored''. Prove that for any $n \in \mathbb{N}$, there is a $n-$colored graph without triangles.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42566]4[/url] Find all integer solutions to $2 \cdot x^4 + 1 = y^2.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42567]5[/url] Let $S = \{(x,y) | x = 1, 2, \ldots, 1993, y = 1, 2, 3, 4\}$. If $T \subset S$ and there aren't any squares in $T.$ Find the maximum possible value of $|T|.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42568]6[/url] Let $ABC$ be a triangle and its bisector at $A$ cuts its circumcircle at $D.$ Let $I$ be the incenter of triangle $ABC,$ $M$ be the midpoint of $BC,$ $P$ is the symmetric to $I$ with respect to $M$ (Assuming $P$ is in the circumcircle). Extend $DP$ until it cuts the circumcircle again at $N.$ Prove that among segments $AN, BN, CN$, there is a segment that is the sum of the other two.

Problem six is an extremely hard one, probably the hardest ever appeared in the China TST history. Only 3 of the 6 team members solved this problem.


[quote="marquez1234"]$ x,y,z > \equal{} 0, n > \equal{} k > 0 \\
(\frac {x^{n \plus{} 2} \plus{} y^{n \plus{} 2} \plus{} z^{n \plus{} 2}}{3})^{n \minus{} k}\cdot(\frac {x^{n} \plus{} y^{n} \plus{} z^{n}}{3})^{k \minus{} 1} > \equal{} (\frac {x^{n \plus{} 1} \plus{} y^{n \plus{} 1} \plus{} z^{n \plus{} 1}}{3})^{n \minus{} k}\cdot(\frac {x^{n \minus{} 1} \plus{} y^{n \minus{} 1} \plus{} z^{n \minus{} 1}}{3})^{k}$[/quote]

This follows trivially from the power means inequality. Let $ p_k \equal{} \sqrt [k]{\frac{x^k \plus{} y^k \plus{} z^k}3}$ be the $ k$th power mean. We have

$ LHS \equal{} p_{ n \plus{} 2}^{(n \plus{} 2)(n \minus{} k)}\cdot p_{n }^{n(k \minus{} 1)} \equal{} p_{ n \plus{} 2}^{(n \plus{} 1)(n \minus{} k)}p_{ n \plus{} 2}^{n \minus{} k}\cdot p_{n }^{nk \minus{} n}$ $ \ge p_{ n \plus{} 1}^{(n \plus{} 1)(n \minus{} k)}p_{n \minus{} 1 }^{n \minus{} k}p_{n \minus{} 1 }^{nk \minus{} n} \equal{} RHS$. :lol:


geometry
3D geometry
tetrahedron
trigonometry
distance
IMO
IMO 1971

All faces of the tetrahedron $ABCD$ are acute-angled. Take a point $X$ in the interior of the segment $AB$, and similarly $Y$ in $BC, Z$ in $CD$ and $T$ in $AD$.

[b]a.)[/b] If $\angle DAB+\angle BCD\ne\angle CDA+\angle ABC$, then prove none of the closed paths $XYZTX$ has minimal length;

[b]b.)[/b] If $\angle DAB+\angle BCD=\angle CDA+\angle ABC$, then there are infinitely many shortest paths $XYZTX$, each with length $2AC\sin k$, where $2k=\angle BAC+\angle CAD+\angle DAB$.

[hide]Rotate the triangle BCD around the edge BC until ABCD are in one plane. It is clear that in a shortest path, the point Y lies on the line connecting X and Z. Therefore, XYB=ZYC.
Summing the four equations like this, we get exactly <ABC+<ADC=<BCD+<BAD.

Now, draw all four faces in the plane, so that BCD is constructed on the exterior of the edge BC of ABC and so on with edges CD and AD.

The final new edge AB (or rather A'B') is parallel to the original one (because of the angle equation). Call the direction on AB towards B "right" and towards A "left". If we choose a vertex X on AB and connect it to the corresponding vertex X' on A'B'. This works for a whole interval of vertices X if C lies to the left of B and D and D lies to the right of A. It is not hard to see that these conditions correspond to the fact that various angles are acute by assumption.

Finally, regard the sine in half the isosceles triangle ACA' which gives the result with the angles around C instead of A, but the role of the vertices is symmetric.
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Prove that for every $0\leq a,b,c<\frac{1}{\sqrt{3}}$ we have $\frac{a+b}{1-ab}+\frac{b+c}{1-bc}+\frac{c+a}{1-ca}\leq 2\frac{a+b+c-abc}{1-ab-bc-ca}$.

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let $a=\tan(x),b=\tan(y),c=\tan(z)$

We then have

$\tan(x+y)+\tan(y+z)+\tan(z+x)\le 2\tan(x+y+z)$

This is clearly true if $x=y=z=0$

I will show that if the inequality holds for some values $x,y,z\in [0,\frac{\pi}{2}]$ (meaning a, b, and c $<\frac{1}{\sqrt{3}}$) then I can increase of the variables by a value and if it is still within the required interval, the inequality will still hold.

Say I increase y by k.

$\tan(x+y+k)+\tan(y+z+k)+\tan(z+x)\le \tan(x+y+z+k)+\tan(x+y+z+k)$

Since the tangent increases faster as the angle gets higher (within the given interval), and the tangents on the right side have larger angles, they will increase faster than the the two tangents increased on the left side. Thus the inequality continues to hold true.

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