In Algebraic Inequality (page 334), there is the following generalization

$\sum \frac{1+pab}{a^{2}+rab+b^{2}}\ge \frac{1+p}{2+r}+2$

for any $r>-2$ and $r+\frac 1{4}\le p \le 2(r+2)^{2}+\frac r{2}$.

To NguyendungTN : I think your problem is Wrong

I think problem two is wrong I will find an example  :wink:
P/S : Who rate my post is spam ? Why?  :maybe:

[quote="NguyenDungTN"]Let $a,b,c \geq 0, a^{2}+b^{2}+c^{2}=2$.Prove that:
$\frac{1+ab}{a^{2}+b^{2}}+\frac{1+bc}{b^{2}+c^{2}}+\frac{1+ca}{c^{2}+a^{2}}\geq 3$[/quote]

Feel free to rate this post as a spam, but I would write as a Subject of this topic like "a^{2}+b^{2}+c^{2}=2" and its Subject description like "frac{1+ab}{a^{2}+b^{2}}+\frac{1+bc}{b^{2}+c^{2}}+\frac{1+ca}{c^{2}+a^{2}}\geq 3", rather than "No name!" and "help me!".

 :lol:

conceive


function
real analysis
real analysis unsolved

Can it be proven that for x >= 10, the number of primes up to 2x is less then twice the number of primes up to x?

Something to begin with: $\binom{2n}{n}=\frac{(2n)!}{(n!)^{2}}$ is an integer, and its size can be estimated.

Some examples:

$\binom{6}{3}=20=2^{2}\cdot 5.$ Note that $5$ is the only prime between $4$ and $6.$

$\binom{10}{5}=252=2^{2}\cdot3^{2}\cdot 7.$ Note that $7$ is the only prime betwen $6$ and $10.$

$\binom{40}{20}=137846528820= 2^{2}\cdot3^{2}\cdot5\cdot7\cdot11\cdot13\cdot 23\cdot29\cdot31\cdot37.$

Note that $23,\, 29,\, 31,$ and $37$ are all of the primes between $21$ and $40.$

Can you make a general argument out of this?

Oops...
It turns out that I miswrote the question. It should be

Can it be proven that for [b]SOME[/b] x >= 10, the number of primes up to 2x is less then twice the number of primes up to x?

Can you help with that?


probability geometry circumcircle calculus

This random problem that I came up with.

Square ABCD has vertices A(6,6) B(6,-6) C(-6,-6) D(-6,6)
Three points are selected at random in ABCD.
What is the probability that these three dots lie in a circle of radius 4?

I'm not sure we can do this, but why can't we draw a circle of radius 4 within the square?  Then the question becomes geometric probability, and the answer is pi/9.

I assume that the circle of radius 4 does not have to be contained entirely within the square. only the four points. For example, (4,3), (4,0), (4,-3) lie in a circle of radius 4, but it is not contained within the square.

A very similar problem has been discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=201012]in this thread[/url], and the consensus was that there's no pretty solution.

The circumradius of the triangle with the three points as vertices must be less than 4. After a torturous bash with Heron's, if a, b, and c are the sides of the triangle, then the points lie in a circle of radius 4 if
$ a^2b^2c^2\le16(2(a^2b^2\plus{}a^2c^2\plus{}b^2c^2)\minus{}a^4\minus{}b^4\minus{}c^4)$.
But the problem is probably harder if you look at it in terms of the lengths of the sides, because all side lengths are not uniformly probable.
I don't think there is an easy way to do this either.

Could this possibly be done with calculus?


They don't mean double each number and add one to get the next one, they mean this:

Start with 1.

Double it to get 2.

Add one to get 3.

Double it to get 6.

Add one to get 7.

Double it to get 14.

Add one to get 15.

etc.

So we multiply 1 by two, to get two, and then add 1 to get the third number in the sequence, 3. Then double 3 to get 6, and add one to get 7. Double 7 to get 14, and add one to get 15, etc.

EDIT: Oh, sorry! I was typing this, then left to go do something, and that thing that says "another post has been posted blahblah" didn't come up for some reason, so I never saw your post when I came back and submitted it.

I'm pretty sure that's what I just said.

Care to read the thread, izzy?  :P  :P  :P

[quote="joyofpi"]I need some help with this pattern
1, 2, 3, 6, 7, 14, 15... What are the next three numbers? What is the pattern?
Everyone says its double plus one, but I don't see how. Thanks.[/quote]

anyway, to answer the first question (what are the next three numbers)
the next three numbers would be 30, 31, and 62 based on what izzy and fishy said

I thought a new problem:
Find 100th term of this pattern.

2k th term = $ 2^k \plus{} (2k \minus{} 2)$ th term

so 100th term = $ 2^{50} \plus{} 2^{49} \plus{} ... \plus{} 2^2 \plus{} 2^1 \equal{} 2*\frac {2^{50} \minus{} 1}{2 \minus{} 1}$

with a little help of my calc, its 2251799813685246.

answer : 2251799813685246

nice problem, bachukas.


Represent the intersection(s) of a horizontal line with a y-intercept of A and $ y = Bsin^{-1}(sin(cos^{-1}(cos(x^{\circ})))) $ Assume that $B > 0$ and that both graphs intersect. Find in terms of the variables.

This is a problem that i thought of on my own, i am not sure i have the right answer but i am pretty sure about my solution.

Can you plz explain your problem. It's abit unclear!!!

there are 2 lines that intersect, a horizontal line and another line specified already, it does not give a constant answer, however, there is an answer in terms of variables
i suppose that the answer should be in a cooridinate pair form

but seeing that it is not really clear what i am going for

[hide]since there are sin and cos functions, the graph repeats every 180 degrees, since there are the inverses there also, the graph behaves normally so the first solution is (a/b,a), then if you look at the graph (on a calc,etc) you see that the next solution is the supplement

so the next solution would be like (180-a/b,a)

so the solutions would be something like (a/b + 180n , a)  and (180-a/b , a) where n is an integer
or like (+or- a/b + 180n, a)[/hide]

The general solution will involve intervals of length $ \frac{pi}{2} $. e.g. in $ [0. \frac{pi}{2} ] $ all the inverse-function pairs $ %Error. "asin" is a bad command.
\sin{x} $ etc. will give x so we het $y = Bx $ and can easily find point of intersection.....

Such cases can be continued and a general solution can be formed by analysing 4 different cases in $ [0,.2 \pi ] $ But i am telling you it will be quite complicated if you know the formulae for inverse-function pairs.

Graphing will definitely be the best way!!!!


function
analytic geometry
symmetry
graphing lines
slope

Hi! I have got a problem.
How to write a formula of function which plot (in coordinate system) will be regular hexagon (centre of symmetry in point (0;0) and side of hexagon has length $a$)?

Example of function which plot is hexagon:  $\mid y \mid+ \mid x+1 \mid + \mid x-1 \mid=4$.

I'm sorry for my poor English and I hope You have the sense of what I wrote :)

Nearly the same function you've written will work; you just need to make the slope match.  The slope of the side of a regular hexagon is $\sqrt3$, so each of your terms $|x\pm a|$ needs to be changed to $\frac{\sqrt3}2|x\pm a|$ and you need to choose the constant to make it work out right.

I suppose the left side is $|y| +\frac{\sqrt3}2|x + 1|+\frac{\sqrt3}2|x- 1|$. I did by trial and error right side but nothing match ($\sqrt3$, $2$, $4$ etc.). The hexagon is too '	expanded' or doesn't exist :?  So could you write the whole equation?

Try $2\sqrt3$.  (You have to multiply it by the same factor of $\frac{\sqrt3}2$.)


Se considera paralelogramul ABCD cu centrul in O . Se noteaza cu M si N mijloacele segmentelor BO si CD . Sa se arate ca daca triunghiurile ABC si AMN sunt asemenea , atunci ABCD este patrat.

Demonstatie :

trgAMN asemenea trgABC =>AM:AB=AN:AC ungNAC = ungCAB => AMBasemenea cu ANC => ungABD = ungACD =>ABCD dreptunghi .Fie {S}=MN :cap: BC , {T}=MN :cap: AB =>ungANS=ungACB . IN trgDBC aplicam Menelaos fata de N,M,S obtinem BS:BC=1:2 =>BS=ON , BS paralel cu ON => MN=MS in trg ANS dreptunghic AM inaltime si mediana =>ANS isoscel =>m(MAN)=45 =m(BAM) + m(MAC) = m(BAC) =>ABCD patrat

Observ ca ai postat multe probleme, numai ca daca ai folosi LaTeX am intelege mai bine textul, eventual si rezolvarea :P. keep posting  ;)

Aveti dreptate dar nu stiu de unde sa-l iau Latex-ul asta

$\LaTeX$:

http://prdownloads.sourceforge.net/miktex/small-miktex-2.4.1705.exe?download

$\LaTeX$ Help:

http://www.mathlinks.ro/Forum/files/latexhelp_140.pdf

 ;)


Try to simplify it.  Multiply through by something that will result in a factorable expression.

I think you may need logs to do this as well, but I'm not sure.

Unless there's some meaning I'm not familiar with, it does refer to calculus.

Notice in the announcements with a big, yellow exclamation point how it says "PLEASE DON'T POST CALCULUS PROBLEMS IN THIS FORUM!"

Then again, there might be some way to do it without calculus.  Maybe we can use the facts that:

[hide]$ 5\plus{}2\sqrt{6}\equal{}\frac{1}{5\minus{}2\sqrt{6}}$ or that $ 5\pm 2\sqrt{6} \equal{} (\sqrt{3}\pm \sqrt{2})^2$[/hide]

Taylor series are an awful idea.

I see no reason at all to suspect that this equation has "nice" solutions.  I can say that there are two roots in $ [0, 2\pi]$, symmetric around $ \frac {3 \pi}{4}$ (essentially due to the first symmetry krsattack notes: $ \sin\left(\frac {3\pi}{2} \minus{} x\right) \equal{} \minus{} \cos x$ and $ \cos\left(\frac {3\pi}{2} \minus{} x\right) \equal{} \minus{} \sin x$, so sending $ x \mapsto \frac {3\pi}{2} \minus{} x$ just interchanges the two summands, preserving the value of the sum), but there is absolutely nothing to suggest that they can be expressed in a nice form.


geometry
trapezoid
search
geometric transformation
reflection
rotation
rectangle

A related problem, with probably the same kind of answer :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1309814342&t=164125

Pierre.

I think I have solution, but it is very messy and involves far too much casework.

$ n\equal{}15$ does not work: Name the points $ 0,1,2,3 ... 14$ in cyclic order. Colour points $ 1,2,3,5,9$ red, points $ 6, 7, 8, 10, 14$ blue, and $ 4, 11, 12, 13, 0$ green. Then we can easily check that there is no iscoceles trapezium with all vertices one colour (since the pentagon formed by each of these $ 3$ set of points are congruent, it suffices to check that there is no red iscoceles trapezium). By fiddling around slightly with this colouring, we can easily prove that $ n \leq 15$ doesn't work either.

To prove $ n\equal{}16$ does work: by Pigeon Hole Principle, there are $ 6$ vertices that are colored one same colour. Thus it suffices to prove, that in any $ 6$ vertices of a $ 16\minus{}gon$, there must be an iscoceles trapezium in these vertices. Assume to the contrary. Name the points $ 0,1 ... 15$ in cyclic order. Consider the six points named $ a_{1}, a_{2} ... a_{6}$. Working in $ mod$ $ 16$, lets consider the set $ S$ of numbers, which may be expressed as $ a_{i} \minus{} a_{j}$ for some $ i \ne j$, with repeated elements allowed. $ S$ consists of $ 30$ elements. Suppose some element occurs $ 3$ times. Then $ a_{i} \minus{} a_{j} \equal{} a_{k} \minus{} a_{l} \equal{} a_{m} \minus{} a_{n} \neq 0$, but it's easy to see that in the $ 3$ pairs $ (i,j), (k,l), (m,n)$, some $ 2$ pairs must be pairwise distinct, so if say $ a_{i} \minus{} a_{j} \equal{} a_{k} \minus{} a_{l}$ with $ j \ne k$, then the points $ a_{i}, a_{j}, a_{k}, a_{l}$ form an iscoceles trapezium. Contradiction.
So, each element in $ S$ occurs at most twice. Since there are at most $ 16$ distinct elements in $ S$ (since we're working $ mod$ $ 16$), but $ 30$ elements in $ S$ together, at least $ 7$ of the elements in $ S$ occur twice. If the pairs that generates this difference $ (a_{i}, a_{j}), (a_{k}, a_{l})$ are pairwise distinct, then since $ a_{i} \minus{} a_{j} \equal{} a_{k} \minus{} a_{l}$, those four points will form an iscoceles trapezium, contradiction. So thus these pairs cannot be distinct - since $ a_{i} \minus{} a_{j} \equal{} a_{k} \minus{} a_{l}$, the only possibility is $ j\equal{}k$ (since $ i\equal{}k, j\equal{}l$ is unacceptable). Thus we must have $ 7$ different pairs $ a_{i}, a_{j}, a_{k}$ so that $ 2a_{j} \equal{} a_{i} \plus{} a_{k}$, within our $ 6$ points. Using  this, and the assumption that no iscoceles trapezium exists within our $ 6$ points, we will now establish a contradiction.
Let the point named $ a_{i}$, for $ i\equal{}1...6$ be $ A_{i}$. Call three points $ A_{i}, A_{j}, A_{k}$ 'good' if $ 2a_{j} \equal{} a_{i} \plus{} a_{k}$ (order of $ i,j,k$ doesn't matter). Note three points are good iff, the triangle formed is iscoceles.  First there must exist $ 4$ points, $ A_{i}, A_{j}, A_{k}, A_{l}$ so that $ A_{i}A_{j}A_{k}$ and $ A_{i}A_{j}A_{l}$ are both good triangles. [To prove this: Consider a 'good' triple, $ A_{i}, A_{j}, A_{k}$, and call the remaining $ 3$ points, $ B_{1}, B_{2}, B_{3}$.  If $ B_{1}B_{2}B_{3}$ are 'good' points, then some triangle $ B_{r}B_{s}A_{l}$, with $ l \equal{} i,j$ or $ k$, and $ r,s \equal{} 1,2,3$ must be 'good' (since there are at least $ 7$ good triangles together) - in that case, just pick the points $ B_{1}, B_{2}, B_{3}, A_{l}$ - they satisfy the required property. Otherwise, then $ A_{i}A_{j}B_{r}$ is not 'good', for $ r\equal{}1,2,3$, and the same holds when replacing $ (i,j)$ by $ (j,k)$ or $ (i,k)$ (since otherwise, the points $ A_{i}, A_{j}, A_{k}, B_{r}$ satisfy required property). Then the only possible good triangles, apart from $ A_{i}, A_{j}, A_{k}$  have $ 2$ of their vertices from $ (B_{1}, B_{2}, B_{3})$, and one from $ (A_{i}, A_{j}, A_{k})$. But some quick computations show that at most $ 2$ out of $ 3$ sets of points, $ (A_{i}, B_{1}, B_{2}), (A_{i}, B_{1}, B_{3}), (A_{i}, B_{2}, B_{3})$ can be 'good'. Thus, at most $ 6$ triangles, apart from $ (A_{i}, A_{j}, A_{k})$ are 'good' - but there are at least $ 6$ such triangles, so there are exactly $ 6$ triangles. Thus from the set of triangles, $ A_{i}B_{1}B_{2}, A_{i}B_{1}B_{3}, A_{i}B_{2}B_{3}$, exactly $ 2$ of them are 'good'. But then, if say $ A_{i}B_{1}B_{2}$ and $ A_{i}B_{1}B_{3}$ are both good, the $ 4$ points $ A_{i}B_{1}B_{2}B_{3}$ satisfy the required property].
Now consider our $ 4$ points $ A_{i}, A_{j}, A_{k}, A_{l}$ so that $ A_{i}A_{j}A_{k}$ and $ A_{i}A_{j}A_{l}$ are both good triangles. It's easy to see, by considering some simple cases, that either the quadrilateral $ A_{i}A_{j}A_{k}A_{l}$ either [1]: consists of $ 3$ vertices compromising a right-angled triangle, and the $ 4$-th vertice being the reflection of the remote vertice about the hypotenuse, OR [2] a quadrilateral where, after re-ordering vertices, $ A_{i}A_{j} \equal{} A_{j}A_{k}$ and $ A_{i}A_{k} \equal{} A_{k}A_{l}$. We need to prove now, that in each case, it is impossible to find $ 2$ other vertices, so that no iscoceles trapezium can be found in the $ 6$ vertices.
[1] Working in $ mod$ $ 16$, label one vertice which harbors the right angle, $ 0$. Then the other vertices will be $ a, 2a, a\plus{}8$ for some $ a,  1 \leq a \leq 3$ (well, this isn't always the case, but it's easy to see that any other case is just a rotation or reflection of this case). So we have $ 3$ possibilities - the vertices are labelled $ (0, 1, 2, 9)$, or $ (0, 2, 4, 10)$ or $ (0, 3, 6, 11)$. We need to check each case to see, that adding any other $ 2$ vertices makes an iscoceles trapezium somewhere. Here's a fairly quick way of doing that:
For example, let's do the case $ (0,1,2,9)$ - suppose there exist two other vertices $ (b,c)$ so that there is no iscoceles trapezium in the $ 6$ points. We know $ (b,c) \neq 3, 15, 8, 10, 7, 11$, so $ b,c \equal{} (4, 5, 6, 12, 13, 14)$. But, $ b\plus{}c \ne 1,2,9,3,10,11$ and $ b\minus{}c \ne 1,2,9,7,8$ (since $ b\plus{}c$ cannot equal the sum of two numbers from $ (0,1,2,9)$). Thus, we can check that no pair of numbers from $ (4, 5, 6, 12, 13, 14)$ can work.
[2] Using similar reasoning, when we name the points from $ 0$ to $ 15$, all such quadrilaterals in this case ($ A_{i}A_{j}A_{k}A_{l}$, where $ A_{i}A_{j} \equal{} A_{j}A_{k}$ and $ A_{i}A_{k} \equal{} A_{k}A_{l}$), can be described by the form $ (0, a, 2a, 4a)$ for $ a \ne 4, 1 \leq a \leq 7$.
For the two cases $ (0,2,4,8)$ and $ (0, 6, 12, 8)$, we can dismiss them using the method described above. For case $ (0, a, 2a, 4a)$ with $ a$ odd, we can use a slight shortcut, and nuke them all in one case:
The points $ 3a, 15a$ would complete an iscoceles trapezium with the $ 3$ points $ 0, a, 2a$ (working in $ mod$ $ 16$). Similarly, we can dismiss the points $ 3a, 15a, 5a, 13a, 6a, 14a$. Since the set $ (0, a, 2a ... 15a)$ covers all points (since $ a$ is odd), the only possible points for the remaining two points are $ 7a, 8a, 9a, 10a, 11a, 12a$. However, we note $ b\plus{}c \ne (a, 2a, 4a, 3a, 5a, 6a)$ and $ b\minus{}c \ne (a, 2a, 3a, 4a)$, so no pair of points from $ 7a, 8a ... 12a$ actually works.

[quote="Fang-jh"]Find the least possible positive integer $ n$ with the following properties: each vertex of  a regular $ n$-gon is colored by one of the colors red, yellow, or blue, arbitrarily. then there exists an isosceles trapezoid with its vertices having the same color.[/quote]
In the Chinese version, isoceles trapezoid cannot be rectangle.
And the answer should be 17.

So we're allowed to have rectangles then?

In that case, $ n\equal{}16$ cannot work; ordering vertices cyclically, a counterexample is: colour $ 0,1,5,8,9,11$ red, $ 12,13,14,2,6$ blue and $ 10,15,3,4,7$ green. Easy to check that there is no iscoceles trapezium (apart from the rectangle, which is allowed).
To prove that $ n\equal{}17$ works: Since $ 17$ is odd, we cannot have a rectangle, so we can ignore rectangles. Thus, the proof I used above to prove that $ n\equal{}16$ for question where rectangles are not allowed, that same method can be used to prove $ n\equal{}17$ works. The only difference, is that the cases we needed to check at the very end are slightly different - but those few cases are just computations.


Determine all integers $ n$ for which $ \frac{n^3\plus{}n\plus{}12}{n\minus{}2}$ is an integer.

So far I found $ \{\minus{}9,3,,4,13\}$ and am not sure if these are all of them.

$ \dfrac{n^{3}\plus{}n\plus{}12}{n\minus{}2} \in \mathbb{Z}$ if and only if $ \dfrac{n^{3}\minus{}4n\plus{}12\plus{}5n}{n\minus{}2} \in \mathbb{Z}$ if and only if
$ \dfrac{n(n^{2}\minus{}4)\plus{}12\plus{}5n}{n\minus{}2} \in \mathbb{Z}$ if and only if $ \dfrac{n(n\plus{}2)(n\minus{}2)\plus{}12\plus{}5n}{n\minus{}2} \in \mathbb{Z}$ if and only if
$ \dfrac{12\plus{}5n}{n\minus{}2} \in \mathbb{Z}$ if and only if $ \dfrac{12\plus{}5n}{n\minus{}2} \minus{} 5 \in \mathbb{Z}$ if and only if
$ \dfrac{12\plus{}5n\minus{}5n\plus{}10}{n\minus{}2} \in \mathbb{Z}$ if and only if $ \dfrac{22}{n\minus{}2} \in \mathbb{Z}$. So we have
$ n\minus{}2 \equal{} \pm 1, \pm 2, \pm 11, \pm 22$.
That give us the values $ n \equal{} 3, 1, 4, 0, 13, \minus{}9, 24, \minus{}20$.


[quote="makar"]Let $ ABC$ be an acute-angled triangle and let $ D,E,F$ be the feet of perpendiculars from $ A,B,C$ respectively to $ BC,CA,AB .$ Let the perpendiculars from $ F$ to $ CB,CA,AD,BE$ meet them in $ P,Q,M,N$ respectively. Prove that the points $ P,Q,M,N$ are collinear.[/quote]
Nice.
Let $ H$ be the orthocenter of the triangle.
Consider the complete quadrilateral $ HECDBA$.Then the Miquel point of the complete quadrilateral is the $ F$.So $ P,Q,M,N$ are collinear(Simson line :wink: ).

Note that $ F$ is the Miquel point of the complete quadrilateral because:
1) $ EHDC$ is inscribed,so the miquel point is on the line $ AB$ and
2) $ DBAE$ is inscribed so the miquel point is on the line $ CH$.

So the miquel point is the intersection of these two lines.

http://www.mathlinks.ro/viewtopic.php?t=236802

[Moderator edit: also http://www.mathlinks.ro/viewtopic.php?t=291532 ]

[quote="makar"][quote="jayme"]Dear Mathlinkers,
the line PQMN is also named "the Miquel's line of..."
Sincerely
Jean-Louis[/quote]

Dear friend,
Can you provide me some reference for the Miquel's line or some link which explain it in detail. It is totally new for me,though I already gone through Miquel's Point and Miquel's Theorem but never heard about Miquel's line.

thanks in advance,

regards

Makar[/quote]
Actually the ''miquel'' line is the simson line of the complete quadrialateral,which is parallel to Aubers line and perpendicular to Gaus's line :wink:

[quote=jayme]Dear Mathlinkers,

see for example
http://jl.ayme.pagesperso-orange.fr/Docs/Miquel.pdf  p. 15-17   and more...

Sincerely
Jean-Louis[/quote]
Is there an english translation of that book [b]jayme[/b]??

[quote=maheshwariaaryan2005]I had a question. Do we need to prove theorems like sine law and cosine law in RMO?[/quote]

No. We don't need to.Just write in brackets that you are using sine rule/cosine rule.

[quote=PhysicsMonster_01][hide=Solution]$BFEC$ is cyclic. For $\triangle{BEC}$, $F$ lies on $(BEC)$ $\implies PQ$ is the simson line of triangle $\implies$ $P, N, Q$ are collinear.
Similarly, $BFHD$ is cyclic. $\implies$ For $\triangle{BHD}$, $F$ lies on $(BHD)$ so again by simson line $\implies P, M, N$ are collinear.
By the two results we get $P, Q, M, N$ collinear.
[/hide]

Simson Line trivialises this problem, so will we need to prove it if it appears in a contest?  (Not that it takes time to prove it.)[/quote]

Just mention it  in brackets.

Dear Mathlinkers,
my oldest reference

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=603020

Sincerely
Jean-Louis

[quote=makar]Let $ ABC$ be an acute-angled triangle and let $ D,E,F$ be the feet of perpendiculars from $ A,B,C$ respectively to $ BC,CA,AB .$ Let the perpendiculars from $ F$ to $ CB,CA,AD,BE$ meet them in $ P,Q,M,N$ respectively. Prove that the points $ P,Q,M,N$ are collinear.[/quote]


logarithms
inequalities
inequalities unsolved

[quote="darkdevil"]$a,b,c \in (1,\infty)$ with $a+b+c=6$ Prove:
$\sum_{cyc}\log_{a}{\left(\sqrt[3]{bc}+a\right)}\geq \frac{11}{2}$[/quote]
darkdevil,$a,b,c<?$

solved it in 2 ways: with jensen or am-gm

Interesting problem!But I don't know when will the equality hold?


trigonometry
inequalities
ratio
geometry proposed
geometry

[b][size=100][color=DarkBlue]A triangle $ \bigtriangleup ABC$ is given and let $ D$ be, an arbitrary point on its side-segment $ BC$ $ ($ $ D,$ between $ B$ and $ C$ $ ).$ Through the vertex $ A,$ we draw two lines isogonal conjugates with respect to the angle $ \angle A$ and we denote as $ M,\ N,$ their intersection points, from the lines through $ B,\ C$ respectively, and parallel to $ AD.$ Prove that the line segment $ AP,$ where $ P\equiv BN\cap CM,$ is isogonal conjugate of $ AD,$ wrt $ \angle A.$[/color][/size][/b]

Kostas vittas.

It's straightforward with Trigonometric Ceva. Let $ \angle NBA \equal{} \beta_1$, $ \angle NBC \equal{} \beta_2$, $ \angle MCB \equal{} \gamma_1$, $ \angle MCA \equal{} \gamma_2$, $ \angle BAC \equal{} \alpha$, $ \angle BAM \equal{} \angle CAN \equal{} x$ and $ \angle CAD \equal{} y$.
Since $ BM, CN$ are parallel to $ AD$ we have $ \angle MBA \equal{} \alpha \minus{} y$, $ \angle MBC \equal{} 180^{\circ} \minus{} (\gamma_1 \plus{} \gamma_2 \plus{} y)$, $ \angle BCN \equal{} \gamma_1 \plus{} \gamma_2 \plus{} y$, $ \angle ACN \equal{} y$.
Trigonometric Ceva applied to the lines $ CM, AM, BM$ yields:
\[ \frac {\sin{\gamma_1}}{\sin{\gamma_2}} \cdot \frac {\sin{(\alpha \plus{} x)}}{\sin{x}} \cdot \frac {\sin{(\alpha \minus{} y)}}{\sin{(\gamma_1 \plus{} \gamma_2 \plus{} y)}} \equal{} 1.
\]
And to the lines $ BN, CN, AN$:
\[ \frac {\sin{\beta_1}}{\sin{\beta_2}} \cdot \frac {\sin{(\gamma_1 \plus{} \gamma_2 \plus{} y)}}{\sin{y}} \cdot \frac {\sin{x}}{\sin{(\alpha \plus{} x)}} \equal{} 1.
\]
So after multiplying it out we get
\[ \frac {\sin{\gamma_1}}{\sin{\gamma_2}} \cdot \frac {\sin{\beta_1}}{\sin{\beta_2}} \cdot \frac {\sin{(\alpha \minus{} y)}}{\sin{y}} \equal{} 1.
\]
Now, after using Trigonometric Ceva to lines $ AP, BP, CP$ we immediately obtain that:
\[ \frac {\sin{(\alpha \minus{} y)}}{\sin{y}} \equal{} \frac {\sin{(\alpha \minus{} \angle BAD')}}{\sin{\angle BAD'}}
\]
and it's easy to prove that it implies $ \angle BAD' \equal{} y$.

Thank you dear [b][size=100]Tomci0[/size][/b], for your interest and solution with a different approach, new to me.

This problem has been inspired from a problem of [b][size=100]Virgil Nicula[/size][/b], in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=208213]Hard geo[/url], of [b][size=100]inequalities master[/size][/b] and let me to present another approach, based on the double ratio theory.

$ \bullet$ We denote the points $ B'\equiv AC\cap BM$ and $ C'\equiv AB\cap CN.$

The pencils $ A.C'NCD,\ A.B'MBP,$ are intersected from the line segments $ BN,\ CM$ respectively and so, we have that $ (A.C'NCD) \equal{} (B,N,E,Z)$ $ ,(1)$ and $ (A.B'MBP) \equal{} (C,M,F,P)$ $ ,(2)$ where $ E\equiv AC\cap BN,\ Z\equiv AD\cap BN$ and $ F\equiv AB\cap CM.$

But, $ (B,N,E,Z) \equal{} (C.BNEZ)$ $ ,(3)$ and $ (C,M,F,P) \equal{} (B.CMFP)$ $ ,(4)$

We see now, that the points $ \infty\equiv CN\cap BM,\ A\equiv CE\cap BF$ and $ Z\equiv CZ\cap BP$ are collinear and so, because of the pencils $ C.BNEZ,\ B.CMFP,$ have the line segment $ BC$ as their common ray, we conclude that they have congruent double ratios.

That is, we have that $ (C.BNEZ) \equal{} (B.CMFP)$ $ ,(5)$

From $ (1),$ $ (2),$ $ (3),$ $ (4),$ $ (5)$ $ \Longrightarrow$ $ (A.C'NCD) \equal{} (A.B'MBP)$ $ ,(6)$

From $ (6)$ and because of $ \angle C'AN \equal{} \angle B'AM$ and $ \angle NAC \equal{} \angle MAB,$ we conclude that $ \angle CAD \equal{} \angle BAP$ $ ,(7)$ and the proof is completed.

Best regards, Kostas vittas.

This problem is the same as this one here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1186774#1186774 since\[ \angle BAP \equal{} \angle CAD\iff \angle MAP \equal{} \angle NAD\iff \angle MAP \equal{} \angle ANC\]since $ \angle BAM \equal{} \angle NAC$ and $ DA\parallel NC$.


calculus
integration
trigonometry
function
algebra
partial fractions
calculus computations

$ \int{\frac{x\plus{}1}{(x^2\plus{}9)^2}}dx$

I'm having a little trouble with this one. Any help would be greatly appreciated :)

This is not partial fractions at all- it's already reduced.

The $ \frac{x}{(x^2\plus{}9)^2}$ term is easy by substituting $ x^2\plus{}9$. For the $ \frac1{(x^2\plus{}9)^2}$, you should probably just do the trig substitution.

First point: there's no role here for partial fractions. You use partial fractions when you have at least two different factors in the denominator. That's not the case here. (You may be using "partial fractions" as a shorthand way of saying "all of the techniques for integrating rational functions, including partial fractions.")

Second point: break it up into $ \int\frac {x}{(x^2 \plus{} 9)^2}\,dx \plus{} \int\frac {1}{(x^2 \plus{} 9)^2}\,dx.$

The first integral can be handled by the "natural" substitution $ u \equal{} x^2 \plus{} 9.$

That leaves us with the second integral. There are many methods; I'm sure several more of them will be posted soon. I'm going to assume that you are a beginning calculus book, and what you know is what you might get out of a typical calculus book the first time through. So let's try something you might know: a trig substitution.

Let $ x \equal{} 3\tan\theta.$ Then $ x^2 \plus{} 9 \equal{} 9\sec^2\theta$ and $ dx \equal{} 3\sec^2\,d\theta.$

$ \int\frac {1}{(x^2 \plus{} 9)^2}\,dx \equal{} \frac {1}{27}\int\frac {\sec^2\theta}{\sec^4\theta}\,d\theta \equal{} \frac1{27}\int\cos^2\theta\,d\theta$

$ \equal{} \frac1{27}\int\frac {1 \plus{} \cos2\theta}{2}\,d\theta$

$ \equal{} \frac1{54}\left(\theta \plus{} \frac12\sin2\theta\right) \plus{} C$

$ \equal{} \frac1{54}(\theta \plus{} \sin\theta\cos\theta) \plus{} C$

Use that $ \sin\theta \equal{} \frac {x}{\sqrt {x^2 \plus{} 9}}$ and $ \cos\theta \equal{} \frac {3}{\sqrt {x^2 \plus{} 9}}.$

$ \equal{} \frac1{54}\left(\arctan\left(\frac {x}{3}\right) \plus{} \frac {3x}{x^2 \plus{} 9}\right) \plus{} C.$

Thanks, that's what I needed to see :)  I had better try a few more now.

This question was in the partial fractions section of my book; so I assumed I needed them for this.

Thanks for the help!


[b]1st easier part:[/b]

1. If you had only one match and entered a dark room containing an oil lamp, a newspaper, and some kindling wook, what would you light first?

2. A plane full of English tourists flies from Holland to Spain. It crashes in France. Where should the survivors be buried?

3. If a redhouse is made of red wood and a - is made of white wood, what is a green house made of?

4. If Mr. Jones' peacock laid an egg in Mr. Brown's garden, who is the rightful owner of the egg?

5. Mrs. Taylor's bungalow is decorated entirely in pink. Her lamps, walls, carpets, and ceilings are all pink. What color are her stairs?

6. Write this down as one number: 15 thousand, 15 hundred, 15.

7. What do Kemit the Frog and Attila the Hun have in common?

8. What do you sit on, sleep on, and brush your teeth with?

9. How many times can you take 3 from 25?

[b]2nd harder part: [/b]

1. John is a popular boys' name in the United States, and many Spanish boys are called Jose. What is a boy commonly called in France?

2. To enter the water, why do scuba divers sit on the edge of the boat and fall out backwards?

3. If you started in Houston, Texas, to dig a hole through the center of the earth, where would you come out?

4. If folk is spelled F-O-L-K and joke is spelled J-O-K-E, how do you spell the word for the white of an egg?

5. What could you be sure to find right in the middle of Toronto?

6. A daddy bull drinks 10 gallons of water per hour and a baby bull drinks 4 gallons. How much would a mommy bull drink?

7. Who was the last man to box Joe Louis?

8. How many grooves are there on the average longplaying record?

9. Farmer Giles has three black pigs, two brown pigs, and one pink pig. How many of Farmer Giles' pigs could say that they are the same color as another of Farmer Giles' pigs?

10. A rope ladder over the side or a ship has rungs which are 1 foot (0.3 meters) apart. Exactly 11 rungs are showing above the water level. The tide rose 8 feet (2.4 meters); how many rungs would now be showing?

11. Is it likely or unlikely that the next U.S. men's tennis champion will have more than the average number of arms?
[b]PUT ALL ANSWERS IN SPOILERS[/b]

Careful...these are tricky!!

Posted by MrMcGeek

1) [hide]the match[/hide]

2) [hide]it would be best if you didn't bury live people[/hide]

3) [hide]glass and metal, usually, although plastics would work.[/hide]

4) [hide]The peacock?[/hide]

5) [hide]There aren't any stairs, although technically a bungalow can have an attic (dictionary.com[/hide]

6) [hide]16515? Is that what you want?[/hide]

7) [hide]"the"[/hide]

8) [hide]time?[/hide]

9)[hide]I don't get it. Is there a trick, or is the answer 24/3=8?[/hide]

10)[hide]Whatever the french word for "boy" is.[/hide]

11) [hide]So that their goggles, tank, fins, and such stay in the right spot, and because it's really the only way for them to do it without it being ackward[/hide]

12) [hide]You wouldn't make it through.[/hide]

13) [hide]W-H-I-T-E O-F A-N E-G-G[/hide]

14) [hide]Yourself?[/hide]

15) [hide]As much as she wanted?[/hide]

16) [hide]No idea, and unless "arms" is a tennis term of some sort, it's rather unlikely[/hide]

17) [hide]One, or two if there are two sides.[/hide]

18) [hide]Um, five?[/hide]

19) [hide]11 rungs, but 1 foot is not .3 meters.[/hide]

Those weren't tricky.

1. correct
2. correct
3. correct
4. [b]yes, i said peacock[/b]
5. correct
6. correct
7. correct
8. [b]wrong[/b]
9. [b]wrong[/b]

1. [b]what is it[/b]
2. [b]wrong[/b]
3. [b]you didn't answer the question[/b]
4. [b]wrong[/b]
5. [b]wrong[/b]
6. [b]wrong[/b]
7. [b]didn't answer (you answered the unnumbered one which is now is #11)[/b]
8. correct
9. [b]wrong[/b]
10. correct
11. [b]wrong[/b]

Posted by MrMcGeek

1) [hide]I don't speak french; Babel Fish says garon.[/hide]

1. [hide]garcon, with a sedilla under the c.[/hide]

4.[hide] W-H-I-T-E[/hide]

6. [hide]There is no such thing as a mommy bull.[/hide]

7. I don't know, but my name sounds just like Joe Louis when you say it quickly.

9. [hide]None-pigs can't talk[/hide]

10. [hide]The answer should still be 11 -- the boat should have gone up with the tide.[/hide]

11. [hide]I would be inclined to say unlikely, which makes me suspect that the answer must be likely, but who knows.[/hide]

[hide]1) match

2) survivors don't get buried

3) glass

4) the peacock

5) no stairs in bungalow

6) 16515

7) the

8) a chair, a bed, a toothbrush

9) once

1) Garcon

2) because falling forward would put them in the boat, not the water

3) you'd be burned to bits

4) T-H-E W-O-R-D F-O-R T-H-E W-H-I-T-E O-F A-N E-G-G

5) on

6) there's no such thing as a mommy bull

7) the funeral home

8) 2, one on front and one on back

9) pigs don't talk

10) 11, the ship rises with the tide

11) likely, because some people have less than 2 arms, so that affects the whole average[/hide]

[hide]um...it doesn't belong to the peacock. female "peacocks" are called peahens, i.e. peacocks are masculine. i sure hope he doesn't lay an egg[/hide]

mathfiend- all right except 4, 3, 4, 5, 7
mystic- good job on 4 on 1st set

So far none of you got 3, 4, 5, 7 second set...keep trying

Posted by MrMcGeek

Doesn't this one belong in the Etc. Forum?  Not that I don't find this sort of thing fun, it just isn't math...

3. [hide]Not sure if this one is supposed to be a trick question or not.  I'm inclined to say "Houston", because you probably wouldn't get too far.  The earth is pretty hard to dig through.  If you mean, "Where would a diameter of the earth with one end at Houston have its other end?", a glance at an atlas suggests that the answer is "In the middle of the Indian Ocean" (Although it is late and I may have made a mistake...).[/hide]

4. [hide]The word is "ALBUMEN".[/hide]

5. [hide]I'm pretty sure we're looking for "the letter 'O'".[/hide]

7. [hide]His undertaker.[/hide]

And a new one:

Before Austrailia was discovered, what was the largest island in the world?

(Answer in spoiler for this one too.)

etc. is now fun and games. in answer to your australia thing, i think [hide]australia[/hide] is the largest island in the world? ([hide]even before it was discovered...[/hide])

well unless you mean like [hide]Pangaea[/hide] or something

Technically Austrailia is considered a continent, not an island. But I think the answer should be [hide]Greenland[/hide]

[quote="Setofallsets"]Stupid me didn't realize that the forums had been restructured.  There is no longer an "Etc." forum.  *wham* *wham* *wham* <bangs head on computer>
I'm still not sure that this thread belongs in a math forum, tho.[/quote]
all you bases belongs to me...er, this thread belongs in Fun and Games :)

Can't believe I missed this...I got like half of part one and a third of part two :(

gauss202 is correct, and for precisely his reason ([hide]Greenland is the biggest island - Austrailia is a continent, not an island[/hide]).  For some reason, this one always seems to trip people up.

Anyone else have some more questions of a similar sort?

Try these.. a bit easier though I think.

1) If a ton of coal costs $30 and a ton of coke costs $25, what will a ton of firewood come to?

2) Removing an appendix is called an appendectomy, removing 3) tonsils is called a tonsillectomy. What is it called when they remove a growth from your head?

3) Why are U.S. soldiers forbidden to carry rifles any longer?

4) What three things that you can eat can never have for breakfast?

5) If a farmer raises wheat in dry weather, what does he raise in wet weather?

6) What would you call a person who did not have all his fingers on one hand?

7) Which is greater: six dozen dozen or half a dozen dozen?

8) What is the best way to get down from a camel?

9) How could a man be severely injured being hit by some tomatoes?

10) Why do Chinese men eat more rice than Japanese men?

And five more..

1) With which hand do Arabs stir their coffee?

2) Train A leaves New York travelling at 50 mph. Train B leaves Boston travelling at 45 mph. 90 minutes later, they meet. Which train is further from New York?

3) Two of every seven people in town have an unlisted telephone number. If there are 14,000 names in the phone book, how many of them have unlisted numbers?

4) How many successful parachute jumps does a US Army recruit have to make before graduating from jump school?

5) A plane from Houston is halfway to Miami cruising at 5,000 feet when it has engine trouble. The pilot jettisons a 100kg sack of mail and a 100kg steel rod. Which hits the ground first?

[hide]

1) A stove/fireplace/fire

2) Surgery

3) Because the ones they have are long enough

4) Brunch, Lunch, Supper

5) His umbrella

6) Normal

7) six dozen dozen, by a factor of a dozen

8) carefully

9) Getting hit by a truck carrying tomatoes

10) There are more of them

1) Hopefully they use a spoon or something

2) When they leave, the one in Boston is. When they meet, they're the same distance from NY (assuming that the track goes straight)

3) None

4) One, and if he doesn't...

5) The plane, because they'd be over the ocean. However, the terminal velocity of the steel rod is probably a bit higher, so it would hit the ocean before the mail. Hopefully the plane doesn't hit the ocean.

[/hide][/hide]

1)[hide] I don't actually know the answer, but I bet it's based on the use of the words "come to."  Something like ashes, or a fireplace.  I dunno.

[/hide]

2) [hide]A haircut.[/hide]

3) [hide]Because the ones they have are long enough?[/hide]

6) [hide]Normal -- I don't have all my fingers on one hand, I have them on two hands.  Or, possibly, that would depend on what his name was.[/hide]

2) [hide]Oh, this one is a classic[/hide].

3) [hide]None, of course, if they all are in the phonebook.[/hide]

TripleM wrote:2, 8, and 9 are wrong (well, you can't say wrong because I'm sure theres lots of crazy answers you can give, but they're not the ones I have ) (oh and the second number 3 would actually be [hide]all of them[/hide] but yeah.)

Um, you might want to read your wording on the second #3 again. Unlisted numbers aren't in the phone book.

Sorry, I meant the second number 4. Jumping out of a plane thing.

1. fire/charcoal/burnt/fireplace
2. haircut
3. they're long enough
4. lunch, dinner, snack
5. umbrella
6. normal
7. six dozen dozen
8. you don't get down, you get off
9. if he got hit in the mouth, the tomato got stuck and he suffocated
10. there are more of them

1. do they even drink coffee??
2. neither
3. none
4. after 1 unsuccessful jump, he's dead
5. they don't hit the ground, they hit the water

Posted by MrMcGeek

Ok we've pretty much got all of them apart from number 8 which is just funny nobody would ever come up with it..
You don't get down from a camel, you get down from a duck.
And anyway..
they were tinned tomatoes, and they use a spoon not their hand, as someone said.

Yeah, don't worry, it took me a while to get to. Think of 'down' as a noun :P

isnt the word for the white of an egg albumin(en)

most of you have probably heard this but...
You should be warned that the following joke can be infuriating. It is said that it was used at Cambridge University as an intelligence test for undergraduates - those that got it were intelligent.
If you get it - well done. If you don't get it - well neither do most people when they hear it. It is a regular (though rather subtle) joke which requires no special knowledge for its understanding. Here it is:

Two climbers were climbing roped together in the Scottish Highlands. They saw some eagles soaring above them. Later the climbers slipped over the edge of a precipice and unfortunately plunged to their deaths. Their souls left their mortal bodies and ascended to heaven. As they rose they saw the same eagles and one soul cried out to them, 'Ah - Eagles' But the eagles, being polite, said nothing.


function calculus derivative limit algebra polynomial LaTeX

The usual definition of the Schwartz class of functions is the set of functions $f \in C^\infty$ (that is, $f$ has continuous derivatives of every order) such that

$\lim_{x \to \infty}p(x) f^{n}(x) = 0$

for any polynomial $p(x)$ and any degree derivative $n$.

Prove that the definition is equivalent if we drop the $n$ and only make that requirement about the function itself, or exhibit a function which shows that the two definitions are not equivalent.

[hide="Extremely dangerous and almost unparseable LaTeX code"]${f(x)=e^{-x^{2}}\sin(e^{e^{e^{e^{e^{x}}}}}})$[/hide] :D


AMC
AIME
AMC 10
AMC 8
USA(J)MO
USAJMO
AMC 12

Nope

You qualify for AIME through AMC10/12

None of the AMC exams require qualification.  However, the qualification for the AIME requires exceptional performance on the AMC exams.

There are three "branches" of the AMC series, where two of them sort of overlap:

AMC8 --> nothing, other than a certificate or plaque if you do really good

AMC10 --> AIME --> USAJMO --> Red (?) MOP

AMC12 --> AIME --> USAMO --> MOP --> IMO

Anyone can take the AMC8/10/12, but from there on (AIME, USA(J)MO, MOP, IMO), you need to qualify. For information about the competitions themselves or qualifying scores, see [url=http://www.unl.edu/amc/whatswhat.shtml]here[/url].

I think you can make USAMO if you score 11 on the AIME and take the AMC 10.

[quote="batteredbutnotdefeated"]ok, are you allowed to register for the AMC10 beforehand even if you do get invited later through the AMC8? Can someone please post a link?[/quote]

There are no invitations to the AMC10 through the AMC8.
From what I've heard, top-scoring participants (I think 20+) on the AMC8 get a letter suggesting that they might enjoy taking the AMC10, [i]not[/i] an invitation to the AMC10. As PoP said, anyone can register and take the AMC10 or 12.

yes, I realize that andersonw, but [b]how[/b] would you register? The website says you can only register in like bundles... How do you register for a single person? This is not for me, btw.

Qualification for USAMO through AMC/AIME index: (10AIME + AMC)
Qualification for AIME through AMC 10/12
Qualification for AMC 10/12 from... not really

[quote="AMCDirector"]The answers to all these questions are in the Frequently Asked Questions Section of the AMC Website:
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

[quote]wait... You have to qualify for the AMC 10 through the AMC 8, correct?[/quote]

FAQ AMC 8 #1  and FAQ AMC 10/12 #3

[quote]ok, are you allowed to register for the AMC10 beforehand even if you do get invited later through the AMC8? Can someone please post a link?[/quote]

FAQ AMC 8 #1  and FAQ AMC 10/12 #3

[quote]but how would you register? The website says you can only register in like bundles... [/quote]

FAQ #4.  Also read the information brochure at:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2009-10pub/AMC1012/2009-10-AMC1012-bro.pdf

Register with the forms at:
http://www.unl.edu/amc/b-registration/b1-archive/2009-2010/10AMC1012AB-register.pdf

Steve Dunbar
Director, MAA American Mathematics Competitions[/quote]

yes, but that registration is for bundles... How can you register for the AMC10 such that you can just go take it at a regular, local high school or something?

How it works is that a teacher/coach/school buys the bundles for either or both dates, and then it is up to them. The website doesn't have a specific school registration.

Most schools (mostly high schools), just advertise it within the school, and anyone who wants to take it can contact the teacher who is coordinating it. If you are homeschooled or go to a middle school, you will probably have to have your parents talk to a counselor or someone to see if it is possible for you to come during the time of the testing.


Which is a better Lewis acid and why: $ \ce{AlCl3}$ or $ \ce{GaCl3}$?

I think $ \ce{AlCl_3}$ > $ \ce{GaCl_3}$. Because generally the thirst for electrons decreases as we go down the group and metallic character starts dominating.

My guess is that GaCl3 > AlCl3. Because Ga and Al have similar size but Ga has more positive charge and thereby increasing thirst for electrons. Morover Al is more metallic so by using reverse logic, I'd say Its ability to accept e- is less.

P.S Carcul, please give more of this reasoning kind of questions.

Thanks

Ga is more metallic I think. I first thought that size determines the lewis acid strength then Ga>Al. :(

i think we might take into account the ineffective shielding when it comes to Ga thus making $ \ce{GaCl3}$ a better lewis acid  :maybe:

Chemrock's answer is right: because the Galium atom is bigger than Al, the corresponding attraction between the nuclear charge and an incoming electron pair would be weaker. Therefore, $ \ce{AlCl3}$ is a better Lewis acid.

can it be bcoz the back bonding in case of gacl3 is greater.

well, since ga and al are electron deficient, there is a transfer of electrons from the halogen atoms to the metal atom which has vacant p orbitals(similar to charge transfer in case of MnO4- i suppse)

And so? Is it difficult to provide a detailed answer?

Since the size of ga is greater than al, the p orbitals of ga can overlap better and accept electrons from cl better than al, which makes it less electron def and hence a weaker acid.

Isn't $ Cl$ too large an atom for back bonding to occur??

then how do you explain the foll order of acidity:
BI3>BBr3>BCl3>BF3?


Not quite. The first property applied with $x=1$ gives $f(1+h)=2+ah+2h^{2}$, that is, the leading coefficient of $f$ is $2$.

to answer to this question you simply need to use f'
f'(x)=ax
then
f(x)=1/2ax^2+c
with f(1)=2=1/2a+c
f(2)=1=2a+c

therefore

a=-2/3
c=7/3

[quote="goldbach"]f'(x)=ax[/quote]
I don't see why. :maybe:

Fair enough. But the condition $f(x+h)-f(x)=axh+2h^{2}$ also implies $f''(x)=4$, which is not true for your function. Apparently, there is no solution.

[quote="Mr.Mathematics"]\[\begin{array}{l}f(x+h)-f(x) = axh+2h^{2}\\ f(1) = 2 \\ f(2) = 1 \end{array}\][/quote]

A quick demonstration that these conditions are incompatible: taking $x = h = 1$ gives $-1 = a+2$, but taking $x = 2$ and $h =-1$ gives $-2a+2 = 1$, and these two results are inconsistent.  There are many ways to turn this into a question which actually makes sense.  For instance, find all functions $f$ such that $f(x+h)-f(x) = axh+b h^{2}$, $f(1) = 2$ and $f(2) = 1$.


The sides of $ AD$, $ AB$, $ CB$, $ CD$ of the quadrilateral $ ABCD$ are divided by the points $ E$, $ F$, $ G$, $ H$ so that $ AE: ED \equal{} AF: FB \equal{} CG : GB \equal{} CH: HD$. Prove that $ EFGH$ is a parallelogram.

It would also be nice if somebody could prove the converse:
If a parellelogram $ EFGH$ is drawn on quadrilateral $ ABCD$ such that each vertex of $ EFGH$ lies on a side of the quadrilateral, then $ AE: ED \equal{} AF: FB
 \equal{} CG: GB \equal{} CH: HD$

OK - I don't know how to use latex to make mathematical symbols, but here goes...

Let *AB* denote Vector AB.

Let n*AB*=*AE*

Then n*BC*=*BF*

*EF*=*EB*+*BF*=n*AB*+(1-n)*BC*

Similarly, *GH*=n*DC*+(1-n)*AD*

Since *BC*=*AD*, *AB*=*DC*

We have *EF*=*GH*. This solves the problem.[/quote]


Solve for positive  integer solutions $ x,y$ for  which  $ 2^{x} \plus{} 25^{y}$ is a square of an integer.

[hide="Solution"]

Solve for natural numbers x,y,k
$ 2^{x} \plus{} 25^{y} \equal{} k^{2}$

Now $ k^{2}  \equiv 1,0\ (mod 3)$
If $ k^{2}  \equiv 1\ (mod 3)$
then $ 2^{x}  \equiv 0\ (mod 3)$ which is impossible

Therefor $ k^{2} \equiv 0\ (mod 3)$
which implies $ 2^{x}  \equiv \minus{}1\ (mod 3)$

Therefore $ x$ is odd. ( 1 )

Similarly applying mod 5,  we get x= 4n+2 or x = 4n for any natural n.
Both of these are even and thus contradicts ( 1 ).

Therefore there are no solutions.

Forgive me if my solution is not neat.
[/hide]

er....sorry i am getting a contradiction but the same final answer
$ 2^{x} \plus{} 5^{2y} \equal{} n^{2}$
$ 2^{x} \equal{} n^{2} \minus{} 5^{2y}$
$ 2^{x} \equal{} (n \plus{} 5^{y})(n \minus{} 5^{y})$
Now it is obvious that .......
$ (n \plus{} 5^{y} \equal{} 2^{m}$........$ (n \minus{} 5^{y} \equal{} 2^{l})$.......$ m \plus{} l \equal{} x$
By manipulation ......
We get
$ 5^{y} \equal{} 2^{l \minus{} 1}(2^{m \minus{} l} \minus{} 1 )$
$ \implies l \equal{} 1$ and $ m \equal{} x \minus{} 1$
Plugging this we get  $ n \minus{} 5^{y} \equal{} 2$
$ 5^{y} \equal{} n \minus{} 2$
Adding the two equations at the beginning we get $ n \equal{} 2^{m \minus{} 1 } \plus{} 1 .... \implies 5^{y} \equal{} 2^{m \minus{} 1 } \minus{} 1$
since $ 5^{y}$ ends with $ 5$
$ \implies 2^{m \minus{} 1}$ ends with 6 so $ m \minus{} 1 \equiv 0\mod 4$
$ \implies m \plus{} 1 \equiv 2\mod 4$
But $ x \equal{} m \plus{} 1$.....so   $ x \equiv 2\mod 4$
$ \implies x$ is even  :(
But I suppose harekrishna got $ x$ is odd.
Am I being dumb......?  :oops:

I believe harekrishna has made a mistake, as one solution is $ (3, 0)$.

[hide="solution"] Firstly, $ x$ and $ y$ must be nonnegative, otherwise $ 2^x+25^y$ can clearly not be a square of an integer.

Suppose $ 2^x+5^{2y}=z^2$. We write this as $ 2^x=(z+5^y)(z-5^y)$. Clearly, both $ z+5^y$ and $ z-5^y$ must be powers of $ 2$. This means that a power of $ 2$ must also divide their difference, $ 2\cdot{5^y}$. Let $ z-5^y=2^a$. Then $ 2^a$ must divide $ 2\cdot{5^y}$, so $ a=1$ or $ a=0$.

[hide="case 1, a=1"]Thus, $ z=2+5^y$ and $ 2+2\cdot{5^y}$ must be a power of $ 2$. Letting $ 2+2\cdot{5^y}=2^{x-1}$, we have $ 5^y=2^{x-2}-1$. If $ x\ge{4}$, $ 2^{x-2}\equiv{0}\pmod{4}$, so $ 5^{y}\equiv{1}\equiv{3}\pmod{4}$, a contradiction. Checking $ x=1, 2, 3$, we find $ (3, 0)$ is the only solution.[/hide]

[hide="case 2, a=0"] Then $ z=1+5^y$ so ${ z+5^y}=1+2\cdot{5^y}=2^{x}$. Suppose $ x\ge{2}$; then $ 2^x\equiv{0}\pmod{4}$ while $ 1+2\cdot{5^y}\equiv{3}\pmod{4}$, so no solutions. Checking $ x=0, 1$, there are no solutions here. [/hide]

Thus, our only solution is $ (3, 0)$.[/hide]

Now, I was not sure if it was necessary to do a lot of those steps. For example, did I even need to consider when $ a=0$?

ooooooooops sorry  :oops:
I forgot to mention to solve for [b]positive[/b] integers $ x,y$
oh I am terribly sorry :wallbash:

hello .....can any one point what is the mistake in my  solution ....I checked mine  1000 times and still not able to find any mistake .... :?:

I actually meant $ 5^0$ does not.
If you accept $ (3,0)$ as a solution (the question now says positive integers), then this is why it was missed.

[quote="chrischris"] then this is why it was missed.[/quote]
May I know what this means my friend.....


[color=darkred]Let $ ABC$ be an equilateral triangle. $ P$ is a variable point internal to the triangle and its perpendicular distancies to the sides are noted by $ a^2$ , $ b^2$ and $ c^2$ for positive real numbers $ a$ , $ b$ and $ c$ . Find the locus of points $ P$ so that $ a$ , $ b$ and $ c$ can be the sides of a nondegenerate triangle.[/color]


function
number theory
number theory open

prove or disprove :for all integer a,there exist infinitly many n such that $\phi (n) | n+a.$
do u know any approch or paper which its subject is dependent?

This problem is discussed in UPINT3 at B37.

Could you translate that for people who do not know what "UPINT3 at B37" means?

UPINT just stands for unsolved problems in Number Theory :)
So "UPINT3" refers to the book "Unsolved problems in number theory (3rd edition)" by Richard Guy; Springer Verlag, 2004, and B37 just means the above problem IS an open problem. I remembered asking [i]rgep[/i] the same question [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41151]before[/url]...


trigonometry
calculus
derivative
function
limit
integration
calculus computations

[quote="pureblack2003"]$f : \mathbb{R}\to \mathbb{R}$ , $f(0)=0$ and $f(x)=x^{\alpha}\sin \frac{1}{x}$ if $x \neq 0$ . find $\alpha$ for which f is derivable and f' is continous[/quote]
Note: I will use $n$ instead of $\alpha$.

First lets find the derivative for $(0,\infty) \cup (-\infty,0)$ which we can do by the chain rule:
$f'(x) = (x^{n})'\sin \frac{1}{x}+(x^{n})\sin ' \frac{1}{x}= nx^{n-1}\sin \frac{1}{x}+(x^{n})(-x^{-2})\cos \frac{1}{x}=\\ nx^{n-1}\sin \frac{1}{x}-x^{n-2}\cos \frac{1}{x}$

Note, the function $f(x)$ is in fact continous at $x=0$ because $\left|x^{n}\sin \frac{1}{x}\right| \leq x^{n}$ for $n\geq 1$ and so $\lim_{x\to 0}x^{n}=0$.

And the derivative by definition is:
$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}= \lim_{x\to 0}\frac{x^{n}\sin \frac{1}{x}}{x}= \lim_{x\to 0}x^{n-1}\sin \frac{1}{x}$
If $n=1$ then the limit doesn't exist, so it no even differentiable.
If $n=2$ then the limit equals $0$ (by squeeze theorem).
But at $n=2$ the derivative is not continous because,
$\lim_{x\to 0}nx^{n-1}\sin \frac{1}{x}-x^{n-2}\cos \frac{1}{x}$ doesn't exist.
Now for $n\geq 3$ it indeed works.

i have a question . could $n>2$ ? why not ? $n$ is a real number ... why can't $n<1$ ?

It could be. sylow_theory only stated the answer for positive integer values of $n$, but the information is there for the general case.

[hide="Answer"]$f$ is differentiable at zero if $\alpha>1$. It is $C^{1}$ (deriv. is continuous) if $\alpha>2$.[/hide]

The reason why $n$ has to be an integer is that in order to talk about limits at $0$ (which this entire discussion is about) we need that the function be defined on an open interval containing $0$ except possibly at $0$ itself (that is by defintion). So I am afraid to talk about rational (but not integer) and irrational exponents. I wonder why jmerry doesn't address this  :roll:

OK, just imagine that that's an $|x|^{\alpha}$.

[quote="pureblack2003"]i don't know why, but i think all real numbers except 1 and 2 are ok ...[/quote]

But that is clearly wrong. It cant be postive non-integral numbers (unless you think as jmerry).

How do you define the function for negative $x$?

sylow: even if you aren't willing to follow the jmerry route, you've completely neglected rational powers with odd denominators (for which that function is well-defined on the whole reals).

[quote="pureblack2003"]it is not clear for me . if you know more, please feel free to explain why those numbers aren't ok[/quote]

Look at happens at $n=2$
We require that,
$\lim_{x\to 0}f'(x)=f'(0)$
We already shown that $f'(0)=0$ above.
So we want,
$\lim_{x\to 0}nx^{n-1}\sin \frac{1}{x}-x^{n-1}\cos \frac{1}{x}$
At $n=2$ we have,
$\lim_{x\to 0}2x\sin \frac{1}{x}-\cos \frac{1}{x}$

Now the first summand, i.e. $\lim_{x\to 0}x\sin \frac{1}{x}= 0$ (explained above).

The second summand,
$\lim_{x\to 0}\cos \frac{1}{x}$ does not even exist. Why?
Because it it did then any way we approach zero on $\mathbb{R}-\{0\}$ we need that the function at these values to zero.
So, if we approach it as,
$x_{1}=\frac{1}{\pi}, x_{2}=\frac{1}{2\pi}, x_{3}=\frac{1}{3\pi},...$
You can see we are approaching 0 be yet,
$\cos \frac{1}{x_{1}}=-1 , \cos \frac{1}{x_{2}}= 1, \cos \frac{1}{x_{3}}=-1,...$
So we get,
$1,-1,1,-1,..$
Which doesnt converge.

what i meant was: why aren't all real numbers EXCEPT 1 and 2 ok ? 1 and 2 are not good, 1 for the it begin derivative, and 2 for not being continous , but what about the others ? for example, why isn't -2 good? or 3/2?

(Let's use the || form to avoid definition issues)

$f$ is differentiable at zero (with derivative zero) iff $\lim_{x\to 0}\frac{|x|^\alpha\sin\frac1x}{x}= \lim_{x\to0}\text{sgn}(x)|x|^{\alpha-1}\sin\frac1x=0$. $\text{sgn}(x)\sin\frac1x$ oscillates between $1$ and $-1$, so this is true if and only if $|x|^{\alpha-1}$ goes to zero- which happens exactly when $\alpha>1$.
Continuity of the derivative is similar, except that the threshold value of $\alpha$ is $2$ instead.


On the AMC 8 #21 from 2005, the problem says "How many distinct triangles can be drawn using three of the dots below as vertices?"  It then shows 6 dots arranged in a 2x3 grid.  I don't think this problem was worded very well because it never tells you that any 3 of the points are collinear, but according to the answer key, you do have to subtract out triangles that contain 3 collinear points.  I think that the answer that assumes that no 3 of the points are collinear (the only thing you can assume is a line is straight) should be counted as correct because the problem doesn't specify.

The answer they give to be correct is 18.  I think 20 should also be correct.

There are a total of ${6\choose3}=20$ points if no 3 of the points are collinear.

I very highly doubt they will change the answer key. In addition, I bet there is a clause somewhere in the test that states 'points that appear to be collinear are collinear'.

I checked the test and there aren't any clauses like that.  But it does say that not all figures are drawn to scale.

I think it would have been a safe assumption that the figure's accurate. :|

Anyway here's the exact wording:

$\boxed{21}\rule{0.1in}{0in}\text{How many distinct triangles can be drawn using three of the dots below as vertices?}$
$\cdot\rule{0.25in}{0in}\cdot\rule{0.25in}{0in}\cdot$
$\cdot\rule{0.25in}{0in}\cdot\rule{0.25in}{0in}\cdot$

My point is you can't assume that the points are collinear, despite what they look like.  Just like you can't say to triangles are congruent because they look like it.

If they had not intended for those points to be collinear, they would have drawn them obviously not collinear. They wouldn't make them look collinear but not really be just to trick you.

Hmmm...he has a point. They should have said 2x3 grid or something.

Yes, if they said it was an array or if they were lattice points, that would remove confusion.  However, you are generally allowed to assume straight lines, even though you cannot assume other things.  So, you can assume that these points are collinear.

Whoever made this thread has a point.

It's unfortunate, but I don't think they pay too much attention to things like this for AMC 8.

I was wondering, do any of you people know when we will get our AMC 8 contests back?

[quote="shinwoo"]and

on the test it said 3 points. and I assumed that you shouldn't pass the a dot or otherwise it would mean 4 points, so I put down the wrong answrer...

it was kinda weird[/quote]

It CAN pass through because it just states that the VERTICES have to be three of those points

Also, I think since the AMC 8 is for younger kids (<high schoolers) I don't think they are going to be as formal as for the AMC 12. Maybe if this was the 12 then they would have paid more attention to that.
Almost all middle schoolers are going to look at that and say that they are collinear.

I did 6 choose 3 without even thinking about it.  :oops:  :roll:


calculus
integration
real analysis
real analysis unsolved

Compute:
$ I \equal{} \int \frac {tan{(2x \plus{} \frac {\pi}{3})}}{cot{(2x \minus{} \frac {\pi}{4})}}dx$
$ J \equal{} \int \frac {tan{(2x \plus{} \frac {\pi}{4})}}{tan{2x \minus{} \frac {\pi}{4}}}dx$
$ H \equal{} \int \frac {2x \plus{} 1}{\sqrt {x^2 \plus{} 1} \plus{} 3x}dx$

Edited.
Sorry Carcul!

[hide="Number 1"]$ \frac{\tan(2x \plus{} \pi/4)}{\cot(2x \minus{} \pi/4)} \equal{} \tan(2x \plus{} \pi/4) \cdot \tan(2x \minus{} \pi/4) \equal{} \frac{\sin(2x \plus{} \pi/4)}{\cos(2x \plus{} \pi/4)} \cdot \frac{\sin(2x \minus{} \pi/4)}{\cos(2x \minus{} \pi/4)}$

$ \equal{} \frac{\cos(2x) \plus{} \sin(2x)}{\cos(2x) \minus{} \sin(2x)} \cdot \frac{\sin(2x) \minus{} \cos(2x)}{\cos(2x) \plus{} \sin(2x)} \equal{} \minus{}1$[/hide]

Number 3: substitution $ x\equal{}\cosh{t}$.

[quote="The Wicker Man"]Number 3: substitution $ x \equal{} \cosh{t}$.[/quote]
Can you explain for me. I substitute $ \sqrt{x^2\plus{}1}\equal{}t\minus{}x$ but it's long.


2-amino benzoic acid on treatment with NaNO2/HCl gives N. N was analyzed using a spectrometer. Of the fragments obtained, some 3 had masses 28, 44 and 76. Identify these and also identify N.

(worth 4 marks)

It does seems that the fragments are

76--Benzene minus 2H
28--N_2 (Cl ion dissociated)
44--CONH_2 (COOH in HNO2)

shouldn't it be N2 ,CO2 , and benzyne ??

it will be CO2 ..dunno wat i thought :P  :rotfl:


[quote="tuanct"]Let k be a positive integer and P be a Polynomio with integer coeficients.Prove that exists a n be positive integer such that:
P(1)+P(2)+...+P(2) is divisible by k[/quote]
you want say,
$ k|\sum_{i \equal{} 1}^{n}p(i)$

we take $ g(n) \equal{}\sum_{i \equal{} 1}^{kn}p(i)$
$ \exists i > j:$ such that $ g(i)\equiv g(j)(mod\ k)$
then $ \sum_{h \equal{} 1}^{k(i\minus{}j)}p(h)\equiv\sum_{h \equal{} kj\plus{}1}^{ki}p(h)\ (mod\ k)\equiv g(i)\minus{}g(j)\ (mod\ k)\equiv 0\ (mod\ k)$
so $ k|p(1)\plus{}p(2)\plus{}...\plus{}p(k(i\minus{}j))$


Suppose I have a penny, a dime, and a dollar coin, and I say: “If you make a
true statement, I will you one of the coins. If you make a false statement, I will give you
nothing.” What should you say to obtain the best coin?

[hide]
 you basically want to force to get the dollar...you could say like; you will give me the dollar; but you wont give me anything else.
but this wont work, b/c I can make this statement false by giving you a penny.[/hide]

[hide=Answer]
You: "You will give me nothing"
Him: Freaks out because it's a paradox and runs away.
You:  Take the money he dropped on the floor while running away.

The "best coin?"

That's an ambiguous phrase.

The "best coin" FOR ME at that moment may be any one of the three, depending on my strategy and who is asking what of me,
or what device needs me to feed one of the coins, etc.

Make your problem question include "the highest-valued coin"* as this is not equivalent to "the best coin" in meaning.

*or equivalent

[quote="tjhance"][hide]
but if he were to give you a penny then it would be false and he would have to give you nothing, so he couldn't give you a penny.
[/hide][/quote]
but then he would give you nothing; and that would make the statement false as well.

the correct answer is:
[hide]you wont give me a penny or a dime[/hide]


Σας παραθέτω μία σχετικά καλή άσκηση που είχε το διαγώνισμα στα μαθηματικά κατεύθυνσης...

Εκφώνηση:

Έστω $\vec a, \vec b$ δύο διανύσματα για τα οποία ισχύει: $\left|\vec a\right| = \left|\vec b\right| = 1$ και $\left(\vec a, \vec b\right)=\frac{\pi}{3}$.

i) Να βρείτε το διάνυσμα $\vec w$, παράλληλο στο $\vec a+\vec b$, ώστε το διάνυσμα $\vec w-\vec b$ να είναι κάθετο στο $\vec a$.

ii) Αν $\vec u = \vec a+x\vec b, x \in \mathbb{R}$ και $\vec v = 2\vec a+\vec b$, τότε:

a) Να βρείτε για ποιες τιμές του $x$ ισχύει $\vec u \perp \vec v$.
b) Να βρείτε ως συνάρτηση του $x$ το εμβαδόν $E$ του τετραγώνου που σχηματίζεται με μήκος πλευράς ίσο με $\left|\vec u+\vec v \right|$ και να αποδείξετε ότι για κάθε πραγματικό αριθμό $x$ ισχύει $E\geq\frac{27}{4}$

Can anyone translate this context into English?

[quote="mostel"]

I BRING to publish,one good exercise that my professor gave me to solve....
let $\vec a, \vec b$
two vectors such that  $\left|\vec a\right| = \left|\vec b\right| = 1$ and$\left(\vec a, \vec b\right)=\frac{\pi}{3}$.

i) find the vector$\vec w$, parallel to  $\vec a+\vec b$, such that the vector $\vec w-\vec b$ is perpendicular to $\vec a$.

ii) if $\vec u = \vec a+x\vec b, x \in \mathbb{R}$ and $\vec v = 2\vec a+\vec b$, then

a) find the values of  $x$ so that $\vec u \perp \vec v$.
b) find the function $E(x)$ of the area of the square,who has side length equal to  $\left|\vec u+\vec v \right|$ and prove that for every real number  $x$ is valid $E\geq\frac{27}{4}$[/quote]

i hope helped u my friend kunny.... :lol:

$Ev\chi\alpha\rho\iota\sigma\tau\omega$ RAKAR :)

[hide="Answer"]

1) $\overrightarrow w=\frac{1}{3}(\overrightarrow a+\overrightarrow b)$

2)

a) $x=-\frac{5}{4}$

b) $E=\left(x+1+\frac{3}{2}\right)^{2}+\frac{27}{4}\geq \frac{27}{4}$

[/hide]

Nice kunny!

The whole point in the exercise it was to result in the following parabolic function: $x^{2}+5x+13$, and then make use of the type $-\frac{b}{2a}$. Then we would have $E = x^{2}+5x+13\implies \ldots \implies E\geq\frac{27}{4}$


Let $f(x) = ax^7+bx^3+cx-5$, where $a,b$ and $c$ are constants. If $f(-7) = 7$, the $f(7)$ equals

$\text {(A)} -17 \qquad \text {(B)} -7 \qquad \text {(C)} 14 \qquad \text {(D)} 21\qquad \text {(E)} \text{not uniquely determined}$

[quote="SnowStorm"]Hey, I've just noticed the Olympiad resources link you've provided with this post.
Can only moderators or other administrators add to this supply of problems?[/quote]

Yes. I think only "some" moderators can add to this list. I know that other than me, orl, Mithsapprentice, Arne, Peter VDD, and some other people are doing it.


Dear Mathlinkers,
let ABC be a triangle,
CB'B"A, AC'C"B, BA'A"C three squares erected externally on the sides of ABC,
and L, M, N respectively the points of intersection of A'B" and A"C', B'C" and B"A', C'A" and C"B'.
Prove that the outer (inner) Vecten point of ABC is also the inner (outer) Vecten point of LMN.
Sincerely
Jean-Louis

Sorry ，my english is very poor！ There are many Conclusion aboat the picture.

（1）：$ AO1$ is Perpendicular to $ O2O3$.

（2）：$ D$ is the point of intersection of $ AO1,CC',BB'',B'C''$ .
$ AD$ is  the radicalaxis of $ circle O2$ and $ circle O3$ ，empathy delimit $ E,F$ .
$ T$ is the point of intersection of $ AD,BE,CF$ ，$ TE/TF\equal{}sin EAT/sin FAT$.

（3）：Let $ triangle GMN$ be a sosceles right triangle，$ GK,GJ$ are Perpendicular to $ ML,NL$ .
Then $ GJ/GK\equal{}sin GNJ/sin GMK\equal{}sin DC'E/sin DB''F\equal{}sin EAT/sin FAT$

So $ TE/TF\equal{}GJ/GK$，So $ L,G,T$ are Collinear.
So：outer (inner) Vecten point of ABC is also the inner (outer) Vecten point of LMN

Attention：$ LG$ is Perpendicular to $ BC$，$ AG$ is the bisector of $ angel MGN$ .

Some Conclusions：

(1):$ PP'\parallel{}BC,QQ'\parallel{}AC,RR'\parallel{}AB$ .

(2): Δ$ PQR$ = Δ$ P'Q'R'$，Just need attention the importent point $ "T"$.

Another Conclusion：

$ G$ is the centroid of $ triangle ABC$，$ K$ is the isogonal conjugates of $ G$ with respect to  $ triangle ABC$ .

Then：$ O1X,O2Y,O3Z$ Intersect at $ S$， and $ G,K,S$ are Collinear .

Just need Isosceles $ triangle O1BC\sim triangle O2AC\sim triangle O3AB$

Let $ D$ be, the point of intersection, of the circles $ (O_{2}),\ (O_{3}),$ of the squares $ ACB'B'',\ ABC''C'$ respectively, the other than $ A$ and it is easy to show that the points $ C'',\ D,\ B'$ are collinear, from $ \angle ADC'' \equal{} 90^{o} \equal{} \angle ADB'.$

Also, because of $ \angle ADC' \equal{} 45^{o} \equal{} \angle AB''C,$ we conclude that the points $ C,\ D,\ C'$ are collinear and similarly, the points $ B,\ D,\ B'',$ are collinear too.

Based on the [b][size=100]Isogonic (Jacobi) theorem[/size][/b], we have that $ T\equiv AO_{1}\cap BO_{2}\cap CO_{3}$ and $ D\equiv AO_{1}\cap BB''\cap CC',$ where $ O_{1},\ O_{2},\ O_{3}$ are the centers of the squares $ BA'A''C,\ ACB'B'',\ AC'C''B,$ respectively.

So, we conclude that the points $ A,\ D,\ T,\ O_{1},$ are collinear.

$ \bullet$ Through the vertex $ A$ of the given triangle $ \bigtriangleup ABC,$ we draw the line parallel to $ C''B',$ which intersects the line segments $ A''C',\ A'B'',$ at points $ P,\ Q,$ respectively and we will prove that $ \bigtriangleup TPQ,$ is a right isosceles triangle with $ \angle PTQ \equal{} 90^{o}.$

We use the argument of a well known ( at least to me ) locus, as follows :

We consider as fixed, the line segment $ A'B''$ and the point $ A$ and then, the locus of the vertex $ C$ of the right isosceles triangle $ ACB'',$ with $ \angle CAB'' \equal{} 90^{o},$ where its vertex $ B''$ is a mobile point on the line segment $ A'B'',$ is the line $ CO_{3}\perp A'B'',$ as well.

Because of now, the right triangle $ \bigtriangleup ATQ$ with $ \angle TAQ \equal{} 90^{o},$ have its vertex $ Q,$ on $ A'B''$ and its vertex $ T,$ on $ CO_{3}\perp A'B'',$ we conclude that it is a right isosceles one and so, we have $ AT \equal{} AQ$ $ ,(1)$

By the same way we can prove that $ \bigtriangleup ATP,$ is a right isosceles triangle and then, we have $ AT \equal{} AP$ $ ,(2)$

From $ (1),$ $ (2)$ and from $ TA\perp PQ,$ we conclude that $ TPQ,$ is also a right isosceles triangle, with $ \angle PTQ \equal{} 90^{o}$ $ ,(3)$

$ \bullet$ It is easy to show now, that the right isosceles triangles $ \bigtriangleup TPQ,\ \bigtriangleup XNM,$ where $ X$ is the center of the square erected on $ MN,$ inwardly to $ \bigtriangleup LMN,$ are perspective, because of $ PQ\parallel NM.$

So, based on the [b][size=100]Desarques theorem,[/size][/b] we conclude that the line segment $ TX,$ passes through the point $ L\equiv PN\cap QM.$

That is, the line connecting the vertex $ L$ of $ \bigtriangleup LMN,$ with the point $ X$ as the center of the square erected on $ MN,$ inwardly to it, passes through the point $ T,$ as the outer [b][size=100]Vecten point[/size][/b] of $ \bigtriangleup ABC.$

By the same way, we can prove that also the lines $ MY,\ NZ,$ where $ Y,\ Z$ are the centers of the squares erected on $ LN,\ LM$ respectively, inwardly to $ \bigtriangleup LMN,$ pass through the point $ T.$

Hence, we conclude that the point $ T$ is the inner [b][size=100]Vecten point,[/size][/b] of the triangle $ \bigtriangleup LMN$ and the proof is completed.

Kostas Vittas.

the 2nd floor have given a way，let me given another way：

Δ$ O1O2O3$ $ \sim $ Δ$ LMN$，$ H$ is the midpoint of $ BC$，and we can easy get Δ$ HO2O3$ $ \sim $ Δ$ GMN$.
So，$ LG\parallel{}O1H$，$ LG$⊥$ BC$. Also，because of  $ TE*TB\equal{}TF*TC$，So we have：
∠$ BTL\plus{}$∠$ TBC\equal{}$∠$ BTL\plus{}$∠$ TFE\equal{}$∠$ BTL\plus{}$∠$ ELT\equal{}90$，it means $ LT$⊥$ BC$.
Hence，$ L,G,T$ are Collinear，......，Finally，we can see  the point $ T$ is the inner Vecten point of Δ$ LMN$.
That is say the outer Vecten point of ABC is also the inner Vecten point of LMN .

[quote="jayme"]Dear Lym and Mathlinkers,
how do you prove that
"Δ PQR= ΔP'Q'R'  "Sincerely
Jean-Louis[/quote]

I just say the rough essential：
(1):$ AD,BE,CF$ intersect at $ T$，$ TE,TF$ separately are the bisector of ∠$ AEC$ and ∠$ AFB$，∠$ TEP'\equal{}$∠$ TFP\equal{}45$.
(2):$ T,E,P,P',F$ have a circumcircle，Δ$ TPP'$ is a sosceles right triangle，empathy，Δ$ TRR'$ is a sosceles right triangle.

So，Δ$ TRP\equal{}$Δ$ TR'P'$，hence $ RP\equal{}R'P'$，empathy can prove $ PQ\equal{}P'Q',QR\equal{}Q'R'$，So Δ$ PQR\equal{}$Δ$ P'Q'R'$.

$ "T"$ is the similar whirling point of Δ$ PQR$ and Δ$ P'Q'R'$.

[quote="jayme"]Dear Lym and Mathlinkers,
thanks for your answer.
What is remarkable is that the triangle TRQ' repeat the beginning of the Vecten's figure with ABC and answer to a more complicated question concerning the triangle LMN which area is doble of PQR.
Sincerely
Jean-Louis[/quote]

My idea maybe is not brilliant，we hope to see your opinion，is it that you use $ sondat's theorem$ to found Vecten conclusion？
here just is some little conclusions，We can see the point “$ T$”is very work.

$ (1)$：Δ$ PQR\sim$Δ$ P'Q'R'\sim$Δ$ LMN$，area similar scale is $ 1: 1: 2$ .
$ (2)$：the point “$ T$”also is the inner Vecten point of Δ$ PQR$ and Δ$ P'Q'R'$.  Rotating angle of $ T$ in Δ is $ 45$ .
$ (3)$：$ AA' \equal{} CC'' \equal{} a,BB' \equal{} b,CC' \equal{} c,A''C' \equal{} sqrt2*a,O1O3 \equal{} BO2 \equal{} a/sqrt2$，empathy.......
$ (4)$：$ GH/AG \equal{} tan$∠$ A,empathy...,AG/CH \equal{} tan$∠$ C/tan$∠$ A,empathy...$，
so，$ SG\parallel{}KH,KJ\parallel{}SI,IH\parallel{}JG,AS*BJ*CH \equal{} AG*CI*BK$ .
$ (5)$：∠$ A'AA'' \equal{} m,$∠$ B'BB'' \equal{} n,$∠$ C'CC'' \equal{} p$，then $ cosn/cosp \equal{} a/b \equal{} PR/PQ,empathy...,PR: PQ: QR \equal{} a: b: c$.
Of course, there are else some of conclusions ，and we can calculate all segment .

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 33
Sincerely
Jean-Louis


Lol, so taking the AMC 10B today made me think about the state math team (I forgot. >.>).
So a couple questions?

How do I go about joining? And what are the requirements?
My school doesn't really participate in math contests anymore as the math department head left. So all we've got are AMCs and some calculus thing at Clemson. That being said, can I still get in?

Paavan Gami
Southside '11

We will select students for the ARML/All-State Math Team based on performance on local contests and the AMCs. With what you think you made on the AMC-10B, you would definitely make the team. Then we will have monthly or bi-monthly practices in Columbia starting in April to determine who's on the A Team and who's on the B Team.

See [url=http://scall-statemathteam.com/default.aspx]here[/url] for some of the practice problems from last year.

Marcus Neal
assistant coach of the SC All-State Team

haha, all right. thanks. This board makes me feel insignificant. T.T. My scores though. Turn em into a 120/126.  :blush:
Lol. Stupid mistakes. =/.


[quote="Nukular"]Find all solutions $ (A,B) \in \mathbb{Z}^2$ of the equation $ A^2 \plus{} 1 \equal{} 2B^2$.[/quote]

This is the most classical Pell equation.

All solutions $ \in\mathbb N^2$ may be obtained from $ (A,B)\equal{}(1,1)$ by using the transformation $ (u,v)\to(3u\plus{}4v,2u\plus{}3v)$

All solutions $ \in\mathbb Z^2$ may be obtained from solutions $ \in\mathbb N^2$ by using the transformations $ (u,v)\to(u,\minus{}v),(\minus{}u,v),(\minus{}u,\minus{}v)$

So the solutions $ \in \mathbb N^2$ are $ (1,1)\to(7,5)\to(41,29)\to(239,169)...$

And the solutions $ in \mathbb Z^2$ : $ (1,1),(1,\minus{}1),(\minus{}1,1),(\minus{}1,\minus{}1),$ $ (7,5),(7,\minus{}5),(\minus{}7,5),(\minus{}7,\minus{}5),...$

Indeed. This was a simpler case of my other post http://www.mathlinks.ro/viewtopic.php?t=308050, but I wanted to sanity check it with this simpler case (which ends up being easier it appears).


function
trigonometry
induction
algebra
functional equation
real analysis
real analysis unsolved

[quote="kiemkhach"]consider $f:R->R$ satisfies $f(x+y)+f(x-y)=2f(x)f(y) \forall x;y \in R$

Prove that:
 if $\exists a$ such that $f(a)>1$ then $f(x) \ge 1 \forall x \in R$[/quote]
Solution??? solution???? :(  pleaseee!

[quote="Moubinool"]$A\cosh x$[/quote]
I don't think so;I think $f(x)=Acosh x$ if f(x) is a continuous function! :(
But my problem doesn't have "[b]f(x) is a continuous function[/b]"
but if you can prove $f(x)=Acosh x$ ....can you post your solution? :D pleaseee!

[quote="kiemkhach"]consider $f:R->R$ satisfies $f(x+y)+f(x-y)=2f(x)f(y) \forall x;y \in R$

Prove that:
 if $\exists a$ such that $f(a)>1$ then $f(x) \ge 1 \forall x \in R$[/quote]
This problem is hard :( I haven't proved! And you?

Actually, the (continuous) solutions are $\cos tx$ where $t$ is either pure real or pure imaginary.

I can't rule out horrible discontinous solutions yet.

We have of course $f(0)=1$ and $f(2x)=2f^2(x)-1$ Thus $f(x)>-1$ for all $x$ and even more the sequence $f(2^na)$ is clearly unbounded (it has first term greater than 1 and you can find the reccurence relation from the first relation I wrote). Suppose that for a certain x we have $f(x)<1$. THEN since $ f(x+(n+1)a)+f(x+(n-1)a)=2f(x)f(x+na)$ we find that the sequence $f(x+na)$ and also $ f(x-na)$ are bounded. But then the sequence $ f(na)$ must be bounded, false.

Help me!
Who can translate in french the harazi's post because I want understand it but I don't understand the english  :(

Ok, sorry for the mistakes I am going to make:
   Il est clair que $ f(0)=1$ et que $ f(2x)=2f^2(x)-1$, donc $ f(x)>-1$ pour chaque x. De plus, la condition $f(a)>1$ nous assure que la suite $ f(2^na)$ n'est pas bornee, donc $ f(na)$ aussi. Maintenant, supposons qu'on a un x pour lequel $ f(x)<1$. Alors, comme $ f(x+(n+1)a)+f(x+(n-1)a)=2f(x)f(x+na)$ pour chaque entier n , on en deduit que la suite $ f(x+na)$ est bornee. Mais comme $ f(x+na)+f(x-na)=2f(x)f(na)$, on trouve que la suite $f(na)$ est bornee, faux.

Thanks harazi  :)
I promise that I learn english soon  :huh:

[quote="harazi"] since $ f(x+(n+1)a)+f(x+(n-1)a)=2f(x)f(x+na)$ [/quote]

is incompatible with the statement $f(x+y)+f(x-y)=2f(x)f(y)$

[quote="kiemkhach"]consider $f:R->R$ satisfies $f(x+y)+f(x-y)=2f(x)f(y) \forall x;y \in R$

Prove that:
 if $\exists a$ such that $f(a)>1$ then $f(x) \ge 1 \forall x \in R$[/quote]
I think the problem itself missing some conditions. As a counter example,

Let $L_3 = \{a+b\sqrt{3}, a, b \in \mathbb{Q}\}$ and consider

$f(x) = \cosh{x}$ if $x\in L_3$ and $f(x) = 0$ elsewhere.

We have $f(2) > 1$ and  $f(x+y)+f(x-y)=2f(x)f(y)$ holds for all $x,y \in R$ but $f(x) < 1$ for any $ x\not\in L_3$

[quote="darij grinberg"][quote="vuhung"]I think the problem itself missing some conditions. As a counter example,

Let $L_3 = \{a+b\sqrt{3}, a, b \in \mathbb{Q}\}$ and consider

$f(x) = \cosh{x}$ if $x\in L_3$ and $f(x) = 0$ elsewhere.

We have $f(2) > 1$ and  $f(x+y)+f(x-y)=2f(x)f(y)$ holds for all $x,y \in R$ but $f(x) < 1$ for any $ x\not\in L_3$[/quote]

Are you sure about f (x + y) + f (x - y) = 2 f (x) f (y) ? What happens, e. g. if $x+y\in L_3$, but $x\notin L_3$ and $y\notin L_3$?

  darij[/quote]
Yes, you are right.

Here is another approach:

Suppose that f takes only positive values.

Put x = y = 0, we have $f(0)(f(0) - 1 ) = 0$.

1. If f(0) = 0, f(x) = 0 for all x
2. If f(0) = 1

Let $A= f(a ) = \cosh{\alpha}$. Put x = y = a we have $f(2a) = \cosh{2\alpha}$. Put x = na, y = a and prove by induction, we have

\[f(na) = \cosh{n\alpha}\, \text{ for all } n \in \mathbb{N}\]

Let consider a natural number n, we will try to compute $f(a/n)$.

If $f(a/n) \leq 1$ we can assume that $f(a/n) = \cos \beta$ for some beta. By compute $f\left(\frac{1}{n}a+\frac{1}{n}a\right), f\left(\frac{1}{n}a+\frac{2}{n}a\right),\cdots, f\left(\frac{1}{n}a + \frac{n-1}{n}a\right)$ in that order, we have

\[ f(a) = f(n\frac1na) = \cos n\beta \leq 1\]

whichs is a contradiction.

If $f(a/n) > 1$, put $f(n) = \cosh\beta$ and proccess in a similar fashion:

\[ f(a) = f(n\frac1na) = \cosh \beta = \cosh \alpha \]

and therefore

\[f(a/n) = \cosh\frac1n\alpha\]

We can now prove that

\[ f(ra) = \cosh{ra}\, \text{ for all } r \in \mathbb{Q}\]

IM-very-very-HO, the problem missed some conditions :), that's all sorry if it dissapointed you :)

[quote="kunny"]If $f''(0)=-1$ then $f(x)=\cos x$[/quote]
In fact, if f is continuous, we can have immeidately:

1. $f(x) = \cosh{kx}$ if $f(a) > 1$ for some a
2. $f(x) = \cos{kx}$ if $f(a) < 1$ for some a

fyi 4 any1 who interested: http://aoclife.ddo.jp/tmp/Chebyshev-Polynomials.zip

Ok! my problem is true! I have sloved this problem
My proof:
It's easy to show $f(0)=1$ and $f(2x)=2f^2(x)-1$
we have $f(x+y)+f(x-y)=2f(x)f(y) \forall x;y\in R$

And $f(2x)+f(2y)=2f(x+y)f(x-y) \forall x;y \in R$

$\leftrightarrow f^2(x) +f^2(y)-1=f(x+y)f(x-y) \forall x;y \in R$

we have $f(x+y)$ and $f(x-y)$ are root of this equation:

$t^2-2f(x)f(y)t+[f^2(x)+f^2(y)-1]=0$

we have $\Delta \geq 0$

$\leftrightarrow [f^2(x)-1][f^2(y)-1] \geq 0 \forall x;y \in R$

Done!

[quote="vuhung"][quote="kiemkhach"]
$\leftrightarrow [f^2(x)-1][f^2(y)-1] \geq 0 \forall x;y \in R$
Done![/quote]
Very nice, we have $\left| f(x) \right| \geq 1$ for all $x \in R$ but how come you conclude that $f(x)$ is positive-valued?[/quote]
okie!
we have $|f(x)| \geq 1$

and it's easy to show $f(x)=2f^2(\frac{x}{2})-1 \rightarrow f(x) \geq -1 \forall x \in R$

so we have $f(x) \geq 1$ or $f(x)=-1$

if $\exists x$ such that $f(x)=-1$ then $f(\frac{x}{2})=0$(contradition! because $|f(\frac{x}{2})| \geq 1$)

[quote="kiemkhach"]
@:[b]vuhung[/b]câu h?i v? hàm cộng tính thế nào rồi?[/quote]

Hoa chư?ng minh lại phần f(x) > 1 đi, như ở trên là chưa hoàn hảo.

Nhưng bài này không được cho vào box college rồi.

btw, Ho?a ra người quen, dạo này vh bận lă?m, mâ?y bài câ?p III thê? không co? thời gian ngo? tơ?i. ?ang bị gia?o mă?ng vì lười kia kìa hehe

[quote="kiemkhach"]
okie!
we have $|f(x)| \geq 1$

and it's easy to show $f(x) \geq -1$

so :if $\exists x$ such that $f(x)=-1$ then $f(\frac{x}{2})=0$(contradition! because $|f(\frac{x}{2})| \geq 1$)

@:[b]vuhung[/b]câu h?i v? hàm cộng tính thế nào rồi?[/quote]
Edit lại 3 lần mà vẫn sai kìa :D

Seeing your solutions & trying the problem myself I think the problem should be continuous or some condition like that.

I still didnt understand why the function should be positive-valued. :(

i think my proof is true;i haven't seen my mistake :(

[quote="the game"]Ok guys heres the proof:
You will all accept that either f(0)=1 or 0.
Consider the case where f(0)=0.
in the above functional equation put x=0 so we have

f(y)=-f(-y) .Thus f is an odd function. Automatically the claim that all x>1 is not tenable at all.

You cant reject my claim...

So the fun begins...........[/quote]
Hmm...I don't understand your idea :(
if $f(0)=0$ put $y=0$ we have $f(x)=0 \forall x\in R$
but $\exists a$ such that $f(a)>1$ so $f(0)=1$ Then....

I know what happens when u put y=0 but u saw what happens when u put x=0 Thats something i pointed out to you.

I hope anybody here has any questions about my proof that f is an odd function.heres the last funny part.
In the functional eqn put y=-y & see that
f(x+y)+f(x-y)=f(x)f(-y)
considering that f(x)is not 0 we have f(y)=f(-y) for all y.well guys what do you think about this....

The story has just begun........

[quote="the game"]I hope anybody here has any questions about my proof that f is an odd function.heres the last funny part.
In the functional eqn put y=-y & see that
f(x+y)+f(x-y)=f(x)f(-y)
considering that f(x)is not 0 we have f(y)=f(-y) for all y[/quote]

Hey man, what you showed is that f(x) is an [i]even[/i] function, not an [i]odd[/i] one! And thus there is no problem with $f\left(x\right)\geq 1$.

  darij

If f(0)=0, then 2f(x)=f(x+0)+f(x-0)=2f(x)f(0)=0, of course f is even and odd.
[quote="the game"]Ok guys heres the proof:
You will all accept that either f(0)=1 or 0.
Consider the case where f(0)=0.
in the above functional equation put x=0 so we have

f(y)=-f(-y) .Thus f is an odd function. Automatically the claim that all x>1 is not tenable at all.

You cant reject my claim...

So the fun begins...........[/quote]

As Grinberg has pointed out I have shown that f is constant or generally equal to 0 when f(0)=0,but my result is inconclusive when f(0)=1. of course f will still be an even function.But that I dont think is gonna help much.
Sorry folks I have made a ghastly mistake.


calculus
integration
geometry
calculus computations

Find the area bounded by the graph of $ y\equal{}\sqrt[3]{x\plus{}\sqrt{x^2\plus{}1}}\plus{}\sqrt[3]{x\minus{}\sqrt{x^2\plus{}1}}$, the line $ x\minus{}y\minus{}1\equal{}0$ and the $ x$ axis.

[hide]
Let $ a^3 \equal{} x \plus{} \sqrt{x^2\plus{}1}$, $ b^3 \equal{} x \minus{} \sqrt{x^2\plus{}1}$.  Then

$ (ab)^3 \equal{} x^2 \minus{} (x^2\plus{}1) \equal{} \minus{}1$, and

$ y^3 \equal{} (a\plus{}b)^3 \equal{} a^3 \plus{} b^3 \plus{} 3ab(a\plus{}b) \equal{} 2x \plus{} 3(\minus{}1)y$.

Therefore,

$ x \equal{} (y^3 \plus{} 3y)/2$.

We wish to find the intersection of the line $ x \equal{} y\plus{}1$ with the above cubic.  Solving, we find

$ 0 \equal{} y^3 \plus{} y \minus{} 2 \equal{} (y\minus{}1)(y^2 \plus{} y \plus{} 2)$,

from which we see that $ (x,y) \equal{} (2,1)$ is the only real solution.  Thus the area of the region of interest is

$ \int_{y\equal{}0}^1 (y\plus{}1) \minus{} \frac{y^3\plus{}3y}{2} \, dy \equal{} 5/8$.
[/hide]


geometry
quadratics
linear algebra
matrix
trigonometry
calculus
integration

I was hoping if somebody could direct me to a good physics book. About the level from an honors high school class to the Ap level. My backround mathematical knowledge is Algebra 2. If there is some other math book to complement it it would be nice if I don't know the math. Also where I could find this book.

I advise you "Feynman Lectures on physics",  Very clear and gives all mathematical tools you need

For now you can ask the question, but in the future, please look up in the stickies for the resource, as it will be loaded. ;)

by the way, you can get the Feynman lecture notes at Amazon.com or any other bookstore.

Thanks a bunch, I am sorry I didn't look at the sticky thread.

About how long does it take to get through the three volumes?

Thanks again, and sorry for the late reply, I haven't been able to get on aops in a while.

The Universe in a nutshell--Stephen Hawking;
The elegant Universe--W. Heisenburg
Two New Sciences--Galileo Galelei
Principia--I.Newton

Those are my favourite books though I don't understand them so well...

 :D

I wouldn't exactly call the Feynman lectures at the high school level...

I think Young and Freedman does a decent job with the material without going too much into more advanced math. It's a tough job though, because almost all physics calls for math above the high school level.

i would like to know what maths do you all do in highschool
if you can each one can say what topics on math have been covered to compare
what is beeing done in our own countries.
im in terminal year so i have all maths covered.
ill try to tell you what i have covered in my math classes.

synthethic geometry
quadratic equation
polinomials and viete equations
complex set
real set
irrational set
rational set
trigonometric form of complex numbers
solving of system of equations using determinants of matrix assigned to the system
analytic geometry
trigonometry
matrices and determinants
newton's binome
calculus
from derivates to integrals including
fermat,rolle,lagrange,cauchy,mean-value theorem for derivates and integrals,centers of gravity for
space under functions
i will try to add some more things if i forgot them

little probability stuff
combinatorics
inequalities
groups
rings
morphisms
vectorial spaces i think you call these fields


There are $8$ different symbols designed on $n$ different
$T-shirts$, where $n > =2$. It is known that each shirt contains
 at least one symbol, and for any two shirts, the symbols on them
are not all the same. Suppose that for any $k$ symbols, $1< = k < = 7$,
 the number of shirts containing at least
one of the $k$ symbols is even. Find the value of $n$.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=t-shirt&t=5242

Pierre.


Prove or disprove: $x_{10+a} = (10+a)_x + a \ \ \forall a,x \in \mathbb{Z}^+, x \ge 10$

Please ammend to the problem, since for some values $a,x$ the expressions aren't defined. For example, if $x=2$ and $a=3$, then we get $2_{13}=13_2+3$ where $13_2$ doesn't have sense coz $3$ isn't a digit in binary system.

Oh, sorry, problem edited (forgot $x \ge 10$). Oh, and for the number x, break up into single digits (for example there will be no 11's in base 12 even though it's possible, it would instead be 1, and another 1).

Ok then, let's disprove it:

Put $x=12, a=5.$ Then $12_{15}\neq 15_{12}+5$ (LHS is $17_{10}$, RHS is $22_{10}$)

Generally, choose $x=10+a$ and you immediately get $x_x\neq x_x+a$

For what values (x,a) is it true, and are there an infinite number of pairs? Oh, and new condition, $x \ge 11$. Having x=10 doesn't work for any case (pretty sure...).


how do you do this prob?

Find the centroid of the triangle with vertices (4,7,9) (5,-1,-17), and (-15,3,-1)

i don't know if this applies to triangles in 3-dimensions, but in two dimensions, you just find the average of the corresponding coordinates

[hide]The centroid of a triangle is the point of intersection of the three medians of the triangle.  It is like the "center of gravity" of the triangle, so its coordinates are the averages of all 3 coordinates of the vertices of the triangle.

$ (\frac{x_1\plus{}x_2\plus{}x_3}{3}, \frac{y_1\plus{}y_2\plus{}y_3}{3}, \frac{z_1\plus{}z_2\plus{}z_3}{3})$
$ (\frac{4\plus{}5\minus{}15}{3}, \frac{7\minus{}1\plus{}3}{3}, \frac{9\minus{}17\minus{}1}{3})$
$ \boxed{(\minus{}2, 3, \minus{}3)}$[/hide]


i get the following error message
[quote="error"]./inequalities.tex:17:Environment theorem undefined. \begin{theorem}[/quote]
what does that mean

The theorem environment can be used for theorems, propositions, lemmas etc etc so before you use it you must define it so it knows what you want. The definition uses \newtheorem and could be as simple as using \newtheorem{theorem}{Theorem}. But there's a lot more you can do, expecially if you use the amsthm package which you should do.
You can read all about using theorems in section 3.8 pages 58-61 of [url=http://tug.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction to LaTeX2e[/url]


1. Alice attends a college in which classes meet only once per week. There are 30 non overlapping classes to choose from, and 6 classes per day for Mon-Fri. Alice decides to register at random for 7 classes. What is the probability she will have class every day of the week? (M-F)

2. A fair coin is flipped 3 times, and the result is written on a slip of paper and thrown into a hat.
(a) Find the probabilility all 3 tosses were H, given at least two of them were H.
(b) Two of the slips of paper are drawn at random and both say H. Given this, find the probabililty all 3 heads are H.
(c) Explain the difference or equivalence of (a) and (b)

3. You are going to play 2 games of chess against an opponent who's equally like to be a master, an intermediate, or a beginner. Your P(winning) any single game is 90%, 50%, 30%, against beg, int, master, respectively.
(a) What is P(winning first game)
(b) You won game 1! Now, what's P(winning game 2)? Assume outcomes are independent
(c)You won game 2! What are the conditional probabilities that that ur opponent is a beg? int? master?

4. Two cards are randomly drawn w/o replacemtn from a standard deck. Like B = both cards are aces. Let A_s = ace of spades is chose. Let A = at least one card is an ace. Find
(a) P(B|A_s)
(b) P(B|A)

5. Suppose we have 10 coins such that if the i-th coin is flipped, heads will appear with probability i/10. When one of the coins is randomly selected and flipped, it shows heads. What is the conditional probability that it was the fifth coin?

6. A stock market investor owns shares in a stock whose present value is 25. She has decided taht she must sell her stock if it either goes down to 10 or up to 40. If each change of price is either up 1 point with Probability .55 or down 1 point with probability .45, and the successive changes are independet, what is the probability the ivnestor retires a winner? (i.e. sells at 40)

7. Let A, B, E, F be events, with A,B independent and P(A^B)>0.
(a) Are A and B necessarily conditionally independent given E?
(b)Suppose that P(A|E)>P(A), and P(A|F)>P(A), i.e. E and F separately provide evidence in favor of A. Is it necessarily true that P(A|E^F)>P(A)?

I'm just going to say it since some mod will tell you eventually: do not post more than one problem in any given post. It makes the forum cluttered and confusing.


I m getting atleast a 4 on aime. i wish to improve my score to 8 next in year. i have Acops and mathematical olympiad treasures, can you suggest some method and books that might help me achieve my goal. :?:

Thanks in advance.

[b]Practice.  A lot.[/b]  A lot means as much as you can, multiple hours per day.  That's the most important thing.  Just reading won't help much.

but to help you, there's a page of books on the AOPS wiki: [url]http://www.artofproblemsolving.com/Wiki/index.php/Math_books[/url]

Also, this has been posted before, so you should use the search button up there $\uparrow$

Also, this topic comes up more in the AMC forum rather than this one.  You should look there.  There's also mock AIMEs you can take. In addition if you press the contests button, the one next to the search button, you can browse through all the old tests.


i think this place is cool, especially the video:

[url]http://www.ultimateshowdown.org/[/url]

[quote="Lazarus"][quote="hunter34"]That was [b]uploaded onto[/b] newgrounds.[/quote][/quote]

 :rotfl:  nice.

[quote="Lazarus"][quote="hunter34"]That was [b]uploaded onto[/b] newgrounds.[/quote][/quote]

Owned. :gleam:

[quote="Lazarus"][quote="hunter34"]That was [b]downloaded onto[/b] newgrounds.[/quote][/quote]

No I mean it was first uploaded for Newgrounds, but it was stolen from them like "Charlie the Unicorn"

[quote="7h3.D3m0n.117"]You can't download onto a site, you can only upload it...[/quote]

Eh, it was just a pun.

WOW what a pun
downloaded and uploaded
the level of innovation is to be noted here  :rotfl:


factorial
number theory unsolved
number theory

p  is prime >2

p-1=a+b

a,b integers >0

A=n! is the [u][b]first factorial [/b][/u]bigger than p^4

[i]Example :

p=7 then p^4=7^4=2401 and A=7!=5040 which is > 2401

p=17 then p^4=17^4=83521 and A=9!=362880 which is > 83521[/i]

a and b (both) divide A

Prove (or disprove) that for any p there is at least one solution (a,b)

The smallest counterexample is $ p\equal{}31583$.


A, B and C are three towns forming a triangle. A man has to walk from one to the next, ride thence to the next, and drive thence to his starting point. He can walk, ride, and drive a mile in a, b, and c minutes, respectively. If he starts from B he takes a+c-b hours, if he starts from C he takes b+a-c hours, and if he starts from A he takes c+b-a hours. Find the length of the circuit. ( Assume that the circuit from A to B to C is counterclockwise)

p.s. the length is an actual number.

Just off the top of my head..

(spoiler)

[hide]

Add them all up, you get a+b+c hours, ie 60(a+b+c) minutes to walk, ride AND drive each leg of the circuit, and since he would do that in (a+b+c) times the number of miles, it must be 60 miles. I probably made a mistake somewhere. Does counterclockwise have any relevance? Or was it there to put someone off..[/hide]

About TripleM's solution:

[hide]It seems to me you have to divide by 3 since when you add them up, that's three circuits.[/hide]

Yeah thats right. I knew I was forgetting something somewhere.

No, on second thoughts..

[hide]

The answer is still 60 isn't it? Adding all three is the same as walking a whole circuit, riding a whole circuit and driving a whole circuit. So the total time in minutes should be ax+bx+cx where x is the length of circuit, and this equals 60(a+b+c), so x=60.

[/hide]


function
induction
combinatorics proposed
combinatorics

Let $ n$ be positive integer, $ A,B\subseteq[0,n]$ are sets of integers satisfying $ \mid A\mid \plus{} \mid B\mid\ge n \plus{} 2.$ Prove that there exist $ a\in A, b\in B$ such that $ a \plus{} b$ is  a power of $ 2.$

[hide]First, we prove that there exists a bijective function $ f: [0,n]\to[0,n]$ such that $ f(f(t))\equal{}t$ and such that $ t\plus{}f(t)\equal{}2^k$ for nonnegative integer $ k$.  Call such a function good.  We prove there exists a good function by induction on $ n$.  This is obviously true for $ n\equal{}1$.  Then $ f(0)\equal{}1,f(1)\equal{}0$.

We are given the set $ [0,n]$ with $ 2^{a\minus{}1}\le n<2^a$.  Assume there exists such a function for all $ m\le n\minus{}1$.  Then we construct $ f$ for $ [0,n]$ as follows: For $ 2^{a}\minus{}n\le t\le n$, let $ f(t)\equal{}2^a\minus{}t$.  By the inductive hypothesis, we are garunteed a good function $ g(t)$ for $ t< 2^a\minus{}n$ and so we let $ f(t)\equal{}g(t)$ for $ t<2^a\minus{}n$.

Now on to the main proof.  We prove a more general statement that any set of integers $ S$ ($ |S|\ge1$) subject to a good function with $ A,B\subseteq S$ and with $ |A|\plus{}|B|\ge |S|\plus{}1$ there must exist $ a\in A,b\in B$ with $ a\plus{}b$ a power or $ 2$.  We prove this by induction on $ |S|$.  For $ |S|\equal{}1$ both $ A$ and $ B$ must equal $ S$ but since the unique element $ s\in S$ has the property that $ f(s)\equal{}s$, we have $ s\plus{}s$ is a power of 2.

Assume the claim holds for $ |S|\le n$.  Suppose there exists a set $ S$ with $ |S|\equal{}n\plus{}1$ and $ A,B\subseteq S$ such that there do not exist $ a\in A,b\in B$ with $ a\plus{}b$ a power of $ 2$.  We are given that $ |A|\plus{}|B|\ge n\plus{}2$.  Since there are only $ n\plus{}1$ elements in $ S$, there exists an element in $ A\cap B$.  Let this element be $ e_1$.  Then $ f(e_1)$ cannot be in either $ A$ or $ B$.  Thus, $ A,B\subseteq S/f(e_1)$.  Consider $ A_1\equal{}A/e_1$ and $ B_1\equal{}B/e_1$.  We have $ A_1,B_1\subseteq S/\{e_1,f(e_1)\}$.  Obviously, $ e_1\ne f(e_1)$ and so $ |A_1|\plus{}|B_1|\ge n$ and $ |S/\{e_1,f(e_1)\}|\equal{}n\minus{}1$.  By the inductive hypothesis on the set $ S/\{e_1,f(e_1)\}$ and the function $ f$, this is impossible.  Contradiction!  [/hide]


analytic geometry
trigonometry
geometry
geometric transformation
rotation
circumcircle

Is there a triangle all of whose vertices are points with rational coordinates on the circle $x^2+y^2=1$ and whose vertex angles are 45,60, and 75?

[hide="A start"]First keep in mind $\sin 45$, $\cos 45$, $\sin 60$, $\sin 75$, $\cos 75$ are all irrationals.

Suppose that $(x,y)$ is a rational point on the unit circle. If we apply a rotation $\theta$, where $\theta=45,60,75$, we get new coordinates $(x',y')$. Now, is the point $(x',y')$ rational as well? I'll let you try to finish from here :)[/hide]

This could've been much shorter, but I felt like writing an informative and explanatory post.

If we have triangle ABC with angle A = 45, angle B = 60, and angle C = 75, we know the circumcenter lies inside the triangle. Drawing in point P, the circumcenter, we know AP = BP = CP by definition of a cirumcenter. Triangle APB has sides AP = BP, so it is isosceles, thus we have angle PAB and angle PBA congruent. The same holds on triangles BPC and CPA to give us:
angle PBC = angle PCB = b
and, angle PCA = angle PAC = c
along with the fact, angle PAB = angle PBA = a.
Now, we can solve for a, b, and c.
a + b = angle PBA + angle PBC = angle ABC = angle B = 60,
b + c = angle PCB + angle PCA = angle BCA = angle C = 75,
c + a = angle PAC + angle PAB = angle CAB = angle A = 45.

Now, for our problem, we want to know the angle a point rotates by to get to the next point. Since A, B, and C all lie on the circle and point P is the origin, if we can find the angles APB, BPC, and CPA, we know how much it rotates. From triangles APB, BPC and CPA, we have:
angle APB = 180 - angle PAB - angle PBA = 180 - 2a,
angle BPC = 180 - angle PBC - angle PCB = 180 - 2b,
angle CPA = 180 - angle PCA - angle PAC = 180 - 2c.
We can solve for these very easily:
180 - 2a = 180 + (b + c) - (a + b) - (c + a) = 180 + 75 - 60 - 45,
180 - 2b = 180 + (c + a) - (b + c) - (a + b) = 180 + 45 - 75 - 60,
180 - 2c = 180 + (a + b) - (c + a) - (b + c) = 180 + 60 - 45 - 75.
This gives us the proper angles so we know that
angle APB = 180 + 75 - 60 - 45 = 150
angle BPC = 180 + 45 - 75 - 60 = 90
angle CPA = 180 + 60 - 45 - 75 = 120.
From this we have point A rotates 150 degrees to point B which then rotates 90 degrees to point C which then rotates the 120 degrees back to point A. So if we say point B is x degrees past 0, we know A is (x - 150) degrees past 0 and C is (x + 90) degrees.
Thus we have our points:
Point B = (cos x, sin x) = (m, n)
Point C = (cos(x + 90), sin(x + 90)) = (m cos 90 - n sin 90, n cos 90 + m sin 90) = (-n, m)
Point A = (cos(x - 150), sin(x - 150)) = (m cos 150 + n sin 150, n cos 150 - m sin 150) = $\left(-m\frac{\sqrt{3}}{2} + n\frac{1}{2}, -n\frac{\sqrt{3}}{2} - m\frac{1}{2}\right)$.

Since we want all of our points to be rational, we say B is rational which makes m and n both rational. This immediately makes C a rational point. Since (1/2) is rational, we know n(1/2) and -m(1/2) are also rational, so in order for point A to be rational we must have both $-m\frac{\sqrt{3}}{2}$ and $-n\frac{\sqrt{3}}{2}$ be rational. Since m and n are already rational, only m = n = 0 will make both expressions rational. So, only the point (m, n) = (0, 0) will make all points rational, but (0, 0) does not lie on the circle x^2 + y^2 = 1. Thus, there is no such triangle.