It seems obvious, but I'm not sure how to prove it.

If BCD is isosceles triangle, then prove that ABC is also an isosceles triangle.

As the picture is unclear, D is the point inside triangle ABC

NOTE: Let M lie on the line $ AC$ and touches the base of triangle $ ABC$.
$ \alpha \equal{} < CBM \equal{} < DCM$

You can prove it through congruency. Prove that $ ABD$ congruent to $ ACD$

Since $ BCD$ is isosceles then $ \boxed{BD \equal{} CD}$.
$ \boxed{AD}$ (common)

< sum of triangle $ BCM$ = $ 180^o$

< sum of triangle $ CMD$ = $ 180^o$

Since $ < MBD \equal{} < MCD$ and $ < CMD \equal{} < MBD \equal{} 90^o$

then $ < CDM \equal{} < MDB \equal{} 180 \minus{} (90 \plus{} \alpha)$

Therefore $ \boxed{ < CDA \equal{} < BDA}$ $ \equal{} 180 \minus{} (90 \minus{} \alpha)$

Hence
ABD congruent to ACD ([b]SIDE ANGLE SIDE[/b])

Since the two triangles are congruent then it can be said that
$ < ABD \equal{} < ACD \equal{} \beta$

Therefore triangle ABC is isosceles where the base angles = $ \alpha \plus{} \beta$

[hide="simpler solution?"]
since BDC is isosceles and the perpendicular is drawn, call the point of intersection between the perpendicular and BC E. so angles BDE and CDE are congruent. by SAS, triangle BAD is congruent to CAD and because corresponding parts of congruent triangles are congruent, BA=CA which implies ABC is isosceles.[/hide]


invariant
geometry
analytic geometry
modular arithmetic
geometric transformation
vector
polynomial

Thought this was pretty cool when I figured it out...

Six frogs are sitting at the vertices of regular hexagon, one frog on each vertex. Frogs can jump over other frogs in a straight line. If the frog jumping is at point F and the frog being jumped over is at P, then the new position of the frog jumping F' is collinear with F and P such that the order of the points is F, then P, then F'. The length of the jump depends on which of two available jumps the frog decides to perform:
Jump 1: F'P=2FP
Jump 2: F'P=(FP)/2

a) Is it possible for a frog to end up in the center of the hexagon (after a finite number of jumps!) when the frogs only jump using Jump 1?
b) Is it possible for a frog to end up in the center of the hexagon (after a finite number of jumps) if the frogs can jump using Jump 1 and Jump 2? (Bay Area Math Olympiad 2009, #6)

Hint: the answer is in the subject, and is pretty obvious by "Problem Construction Theory". Use it.

~CzarLog

Does the solution go something like this?

[hide="solution"]
$ 1)$  Using the solutions to the equation $ x^6 \equal{} 1$ we can (w.l.o.g) assign coordinate points to the vertecies of the hexagon. they are;$ (2,0),(1,\sqrt {3}),( \minus{} 1,\sqrt {3}),( \minus{} 2,0),( \minus{} 1, \minus{} \sqrt {3}),(1, \minus{} \sqrt {3})$ and there is also a center at $ (0,0)$

Lets say we have the points $ (a,b)$ and we want to "jump" them over $ (x,y)$ then the points $ (a,b)$ are mapped to $ (3x \minus{} 2a, 3y \minus{} 2b)$
Now notice that $ 3x \minus{} 2a \equiv a \mod 3$. So we have found our invariant. (That is $ (a,b)$ remains the same $ (\mod 3)$ under transformation)

So if we consider the $ x \minus{}$coordinates: $ \minus{} 2, \minus{} 1,1,2$. These are all equivalent to $ 1,2 \mod 3$. Since our goal is $ (0,0)$ and $ 0 \equiv (0\mod 3)$ it follow that our goal is unattainable. :)

[/hide]

On what basis is $ 3x\minus{}2a \equiv a \pmod{3}$?

You're basically assuming that $ 3(x\minus{}a)$ is a multiple of $ 3$, or equivalently, that $ x\minus{}a$ is an integer. Is this necessarily true?

[quote="TZF"]On what basis is $ 3x \minus{} 2a \equiv a \pmod{3}$?

You're basically assuming that $ 3(x \minus{} a)$ is a multiple of $ 3$, or equivalently, that $ x \minus{} a$ is an integer. Is this necessarily true?[/quote]

Initially $ x,a\in \{\minus{}2,\minus{}1,1,2\}$ therefore $ x\minus{}a$ is an integer. This integer then becomes a part of the set.
Therefore all $ x$ and $ a$ are integers, yes :)

Yes, this solution certainly works, but I believe if you look at the lattice from a different [i]angle[/i], it might work better (that is to say, without the use of square roots at all)

Now how can we use what we know to solve part b)?

~CzarLog

[quote="CzarLog"]Yes, this solution certainly works, but I believe if you look at the lattice from a different [i]angle[/i], it might work better (that is to say, without the use of square roots at all)

[/quote]

I dont think we can have a regular hexagon on the cartestian plane such that vertecies have rational co-ordinates...

Right, but you're using the wrong grid. Forget Cartesian coordinates.

//if you look at the lattice from a different angle, it might work better//

a
[hide]Consider the Cartesian Plane, the center is at (0,0), and the frogs are at (-1,1),(0,1),(1,0),(1,-1),(0,-1),(-1,0). Jumps can be in the N-S, E-W, or NW-SE directions. Thus JUMP1 is a horizontal translation by some multiple of 3 units, or a vertical translation, or both a horizontal and vertical translation. Thus, the coordinate of a frog is invariant mod 3, but none starts at a point with coordinates (0,0) mod 3, that of the center. [/hide]

b?
[hide]Given a path by a frog consisting of JUMP1 and JUMP2, the coordinates (x,y) of the frog at any point in it's path is a multiple of $ 1/2^i$ for some i, with largest i = n. Assume there is a path to the center. We traverse this backwards, scaling the Cartesian plane by $ 2^n$, so, say, if the forward path was from (0,1) to (0,0), we now go from (0,0) to (0,$ 2^n$). In this new plane, every point in the frog's path has unit coordinates. The same argument from part a applies here, every move translates the frog horizontally, vertically, or both, a multiple of 3 units. Since we start at (0,0), every point in the frog's path must have coordinates a multiple of 3. But since $ \pm 2^n \equiv (\minus{}1)^n \pmod{3}$ is not 0, this is a contradiction. and thus, there is no such path.[/hide]

Why does it work if we look at the hexagon from different angle?

Okay, I was being deliberately vague. Look at the first link if you'd like more info, and the second one for an explicit explanation for pianoforte's coordinate system.

[hide="coordinates info"] As ocha said, you can't have an orthonormal set of coordinates (that is to say, a set of coordinates where the x and y axes are perpendicular) where all the coordinates of the vertices of the hexagon are rational. But you can modify the coordinates so that it works. Think of the hint I gave before clicking on the next link.[/hide]

[hide="explicit explanation of coordinates"] Draw two diagonals, each from a vertex to a vertex directly opposite. These two lines (which are at a 60 degree angle) can act as axes for a coordinate system with the origin at the center of the hexagon. If you have experience with vectors, you can make a unit vector parallel to one of the diagonals, and another unit vector parallel to the other diagonal, and then write the coordinates of the frogs as linear combinations of those two vectors, that is to say, linear combinations of the form $ \alpha_1 v_1\plus{}\alpha_2 v_2$, where $ v_1$ and $ v_2$ are the vectors and $ \alpha_1$ and $ \alpha_2$ are integers.

If you don't have experience with vectors, try drawing the two diagonals, and then keep adding lines parallel to the axes like in the image below. The image lends itself to be thought of as a skewed set of coordinates, with the origin at the center and the diagonals as a skewed pair of axes.[geogebra]d3ce4533b26b1fcc69e031f2d6a86447441123bb[/geogebra]

When I first thought of this problem, I kept trying to work with the triangular lattice (e.g. Eisenstein integers). But it didn't really work. Somehow I managed to see rhombi instead of triangles and the solution worked.
[/hide]


Euler
function
logarithms
calculus
integration
modular arithmetic
group theory

Prove that if $ n\equal{}p_{1}^{\alpha_{1}}\cdots p_{r}^{\alpha_{r}}$, then $ \sum_{d\mid n}\frac{\phi(d)}{d}\equal{}\prod_{i\equal{}1}^{r}\left[1\plus{}\alpha_{i}\left(1\minus{}\frac{1}{p_{i}}\right)\right]$.
I tried to do a decomposition but it didn't work.

Prove that if $ f(n)\equal{}\prod_{d\mid n}g(d)$, then $ g(n)\equal{}\prod_{d\mid n}f(d)^{\mu\left(\frac{n}{d}\right)}$, where $ \mu$ is the Mobius inversion.
The book I am using doesn't do a very good job explaining this part.

For the first one, show that it is a multiplicative function, so you need to check the result for primes only.

For the second one, use $ \log$ to reduce it to the additive version.

I know they're both multiplicative but where is the right side coming from?
Would it be $ \log f(n)\equal{}\log\left(\prod g(d)\right)\equal{}\log g(d_{1})\plus{}\cdots\plus{}\log g(d_{i})$?  But now where does this get me?  Is there some relation that I am missing?

I think ZetaX suggests you to see that $ logf(n)=\sum_{d\mid n}logg(d)$ implying $ logg(n)=\sum_{d\mid n}\mu\left(\frac{n}{d}\right)logf(d)$

Or , proving directly, lets start from $ \prod_{d\mid n}f(d)^{\mu\left(\frac{n}{d}\right)}=\prod_{d\mid n}\prod_{c\mid d}f(c)^{\mu\left(\frac{n}{d}\right)}$
Now lets fix c dividing n and using the fact that $ \sum_{d\mid n}\mu(d)=0$ for n >1, we get the result

Okay, I've solved the first one so ignore it.  I'm still stuck on the second one, though.

Edit: lajanugan posted while I was typing.  With regards to your first method (log version), I think I'm missing something because I still don't see the implication.  I'm probably overlooking something really simple. :oops:

Okay, got it. :)   The only question I have left is, why is it $ \mu\left(\frac{n}{d}\right)$ and $ f(d)$ instead of the other way around?  There is a theorem that says that if arithmetic functions satisfy $ f(n)\equal{}\sum_{d\mid n}g(d)$, then it satisfies $ g(n)\equal{}\sum_{d\mid n}\mu(d)f\left(\frac{n}{d}\right)$ and the other way around.

That theorem is what I was speaking about ("additive version").
And it makes no difference to exchange $ d \leftrightarrow \frac nd$ as it reorders the divisors, only.

Thanks.  One more question (sorry if this is annoying):

If $ g$ is a primitive root mod $ p$ ($ p$ is prime) and $ m\mid p\minus{}1$ ($ 0<m<p\minus{}1$), how many integral solutions are there to $ x^{m}\minus{}g\equiv 0\pmod{p}$?

I'm looking at the theorems at my disposal but I can't see how to apply them to the above congruence.

$ g$ is a generator, and $ x$ is not 0, so $ x \equal{} g^t$ for some $ 1 \le t \le p \minus{} 1$ (this $ t$ is unique). The congruence becomes $ mt \equiv 1 \pmod {p \minus{} 1}$. But $ m | p \minus{} 1$ implies $ m|mt \minus{} 1$ so $ m \equal{} 1$. So for $ m \neq 1$ there are no solutions and for $ m \equal{} 1$ there's one.

The reason: take the subgroup of $ m$th powers of the elements of $ (\mathbb{Z}/p\mathbb{Z})^*$. There's a solution to $ g \equal{} x^m$ iff this subgroup is the entire group (because $ g$ is a generator). But it can be shown that its order is $ \frac {p \minus{} 1}{(m,p \minus{} 1)}$. When $ (m,p \minus{} 1) > 1$ (in our case it obviously holds for $ m > 1$) this subgroup is of order smaller that the original group'.
To calculate the size of the subgroup we can use again generators (it is a general statement about cyclic groups)

I think it is quite elementary (If you're familiar with generators, you're supposed to be familiar with cyclic groups. Generators only exist in cyclic groups). But I can provide a more "specific" solution:
I say that if $ m>1, m|p\minus{}1$ then there are no solutions to $ x^m \equiv g \pmod p$, where g is a generator. Assume otherwise.
I will show that $ g$ [u]doesn't "generate"[/u] $ x^{m\minus{}1}$, obviously a contradiction.
$ g$ is a generator, thus it can be raised to a certain power such that it will equal $ x^{m\minus{}1}$: $ g^t\equal{}x^{m\minus{}1}$, or: $ g^{tm}\equal{}x^{(m\minus{}1)m}$.
$ x^m\equal{}g$ implies $ x^{m(m\minus{}1)}\equal{}g^{m\minus{}1}$. So $ g^{tm}\equal{}g^{m\minus{}1}$, or: $ g^{tm\minus{}m\plus{}1}\equal{}1$ (*)
$ g$s order is exactly $ p\minus{}1$ (the order of an element is the number of elements it generates)
we get: $ p\minus{}1|tm\minus{}m\plus{}1$ (if this divisibility doesn't hold, let $ tm\minus{}m\plus{}1\equal{}\lambda (p\minus{}1)\plus{}r$, where $ r$ is the remainder. You can verify that $ g^r\equal{}1$, yet $ r<p\minus{}1$, a contradiction)
But $ m|p\minus{}1 \implies m|tm\minus{}m\plus{}1 \implies m|1$, a contradiction.

Are "generator" and "primitive root" supposed to be synonyms for each other?

oh, yes. Sorry if that caused you confusion.

[quote="JRav"]I appreciate the solution but I am not familiar with algebra.  Do you have a more "elementary number theory" solution?[/quote]
It's not like he used any theorem from algebra  :wink:
You better get some algebraic background, purely elementary number theory gets very anoying after some time. Better use algebra (and others) to get a deeper understanding; as for here, a little bit of knowledge about groups and rings may give greater insight.


group theory
abstract algebra
induction
Galois Theory
number theory
relatively prime
superior algebra

1) If $ G$ is a finite $ p$- group , $ H \triangleleft G$ and $ H\not=\langle e \rangle$, then $ H \cap C(G) \not= \langle e \rangle.$

2) If $ N \triangleleft G$ and $ N \cap G^'=\langle e \rangle$ ,then $ N<C(G)$.

3) If $ G$ is a finite nilpotent group, then every minimal normal subgroup of $ G$ is contained in $ C(G)$ and has prime order.

4) If $ H$ is a maximal proper subgroup of a finite solvable group $ G$, then $ [G : H]$ is a prime power.

1) Consider action $ G$ on $ H$ by conjugation and apply class equation.

2) Suppose $ x\in G, y\in N$. Then exists $ y_{0} \in N$ st $ xyx^{ \minus{} 1} \equal{} y_{0}$, so $ e \equal{} xyx^{ \minus{} 1}y^{ \minus{} 1} \equal{} y_{0}y^{ \minus{} 1}$. Thus $ y_{0} \equal{} y$, and $ N < C(G)$.

3) I assume that 'minimal normal subgroup' mean 'minimal nontrivial normal subgroup'.
Since $ G$ is nilpotent then it is direct product of its Sylow subgroups. If $ N\triangleleft G$, and $ P$ is Sylow subgroup of $ G$ then $ N\cap P\triangleleft G$. So minimal normal subgroup of $ G$ is subgroup of one of its Sylow subgroups. By 1) we obtain desired result.

4) Follows immediately from Hall's Theorem.

edit: typo..

4) I don't know Hall' theorem, but the proof is easy.

Take $ K\equal{}\bigcap_{g\in G} gHg^{\minus{}1}\unlhd G$ and some minimal normal subgroup $ L/K$ of $ G/K$.
(this always means nontrivial and we only need $ G/K\not\equal{}1$)
Since $ G/K$ is solvable we must have $ L/K\not\equal{}1$ some elementary abelian $ p$-group for some prime $ p$.
If $ L\le H$, then $ L\equal{}gLg^{\minus{}1}\le gHg^{\minus{}1}$ for all $ g\in G$, which gives $ L\le K$ and $ L/K\equal{}1$. Contradiction!
So $ L\not\le H$ and $ H\lneq LH\le G$ and $ H$ maximal gives $ G\equal{}LH$.
And now since $ K\le L\cap H$ we get $ |G|\equal{}|LH|\equal{}{|L|\cdot |H|\over |L\cap H|}$
and $ |G: H|\equal{}{|L: L\cap H|}$ divides $ |L/K|$ a power of $ p$.

I thought about the following result due to Philip Hall:
[i]If $ G$ is a solvable group of order $ mn$, where $ m, n$ are relatively prime natural numbers, then $ G$ has a subgroup of order $ m$.[/i]
But I must admit that I don't remember what I meant by "follows immediately by Hall's Theorem" (probably I misunderstood the question  :oops:)
Anyway, thanks for nice proof [b]olorin[/b] :)

Can you prove point (1) using induction?

[quote="MegaRockman"][quote="ehsan2004"]
3) If $ G$ is a [b]finite[/b] nilpotent group, then every minimal normal subgroup of $ G$ is contained in $ C(G)$ and has prime order.
[/quote]
Can we omit the hypothesis "G is finite"[/quote]

Yes.  If $N$ is a minimal normal subgroup of $G$, then $N\cap Z(G)$ is also a normal subgroup of $G$, and is non-trivial.  Thus, by the minimality of $N$, $N\le Z(G)$.  Since then any subgroup of $N$ would be normal in $G$, $N$ has no non-trivial subgroups; so $N$ is cyclic of prime order.

Steve

[quote="ehsan2004"]
4) If $ H$ is a maximal proper subgroup of a finite solvable group $ G$, then $ [G : H]$ is a prime power.[/quote]

Any minimal normal subgroup $N$ of $G$ has prime power order.  If $N\le H$, we can look at $G/N$ and proceed by induction on $|G|$; otherwise, $HN=G$, and $H\cap N=\{1\}$ (why?).  Thus $[G:H]=|N|$.

Steve


calculus
derivative
function
real analysis
real analysis solved

Is it possible that the derivative of non-constant everywhere derivable function on interval vanishes
a) at all irrational point of this interval
b) at subset, which dense in this interval   ?

I think I have seen somewhere that the set of zeroes of a derivative is a $G_\sigma$ set.  With this the answer should follow. The answer to the second question is true: there are such functions and it seems that Pompeiu was the first to construct a strcitly increasing function whose derivative vanishes on dense set.

I have an idea for b), but it is incomplete.
Let $q_n$ be some enumeration of rationals, then consider \[f(x)=\sum_{n=1}^{\infty} c_n(x-q_1)^2...(x-q_n)^2.\] We choose $c_n$ in such a way that $f'(q_{n+1})=0$. The only problem is proper convergence...

2harazi, haven't you seen the prove or maybe you could tell me where you've seen this one, because i haven't known the fact directly

b)  has already been discussed in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2822[/url]. BTW, Eugene, what is the title of your "problem book" (if it is one book, of course...)

Yes, fedja, it is not exactly one book, but a lot of problems, which source was named "problem book" are from book, you have already mentioned "Selected Problems in Real Analysis" 2004y-Russian edition. And this problem is also from this book.
i'm sorry for posting old problem, but i'm just interesting in a).
Is it true that the set of zeroes of a derivative is a $G_\delta$ set ?

it seems to me that it is not $G_\delta$, because the set of irrationals point is itself $G_\delta$ and in this case the answer for should be positive, but it's negative. I think that as follows from $Darboux$ theorem the set of our points must be union of closed subsets and it is $F_\delta$

[quote="Myth"]The only difference that you borrowed it from the book and I try to solve it myself...[/quote]
I am aware that one's own solutions are much preferred here but my excuse is that I was answering a post unanswered for a year and it just happened that I knew the solution. Anyway, if you like such stuff, try to solve the problem about the differentiable function whose derivative changes sign on each open interval.  ;) If I don't mistake, the first example of this kind was due to Denjoy and it hasn't been discussed on the forum yet.
[quote="eugene"]Is it true that the set of zeroes of a derivative is a $G_\delta$ set ?[/quote] Yes, this is true but useless in your problem. Note that Harazi never said that every $G_\delta$ set is a set of zeroes of some derivative, so the argumentation in post #8 is not valid. To show Harazi's claim, define $G_k$ to be the union of all open intervals $I$ such that $|I|<\frac 1k$ and $\max_I f-\min_I f<\frac1k|I|$. Then the set of points where $f'=0$ is exactly $\bigcap_{k\geq 1}G_k$ (provided that $f'$ exists everywhere).
[quote="eugene"]I think that as follows from Darboux theorem the set of our points must be union of closed subsets and it is $F_\sigma$ [/quote]
That would solve $a)$, but, unfortunately, it is false.

If (with the setting in our problem) $f'\equiv 0$ on the complement of a countable set, then $f$ is constant.

Assume there is an $x\in[a,b]$ s.t. $f'(x)>0$, say. Then $(0,f'(x))\subset f'([a,b])$, but this is a countable set, so we have a contradiction.

I think this takes care of (a).


algebra
polynomial
calculus
integration
factorial
combinatorics unsolved
combinatorics

Find a closed form formula for $1^5+2^5+...+n^5$

Maple says
$\frac{1}{6}\, \left( n+1 \right) ^{6}-\frac{1}{2}\, \left( n+1 \right) ^{5}+{\frac {5}
{12}}\, \left( n+1 \right) ^{4}-\frac{1}{12}\, \left( n+1 \right) ^{2}$.
I do know that $\sum n^k$ can be attained using $\sum n^{k-1}$(I forgot how to do this).

Is there any way you could tell me what you did in maple to get this answer?

here is a general trick to solve these kinds of questions.
since you wish to evaluate $\sum_{k=1}^n k^5$ start with $(k+1)^6-k^6= \sum_{i=0}^5 \binom{6}{i} k^i$. sum both sides over $k$.the LHS gives a telescopic sum. the RHS involves powers of $k$ starting with $5$ and going lower;so inductively you can sum those as well.
to give an example, $(k+1)^2-k^2=2k+1$ so that $\sum_{k=1}^n(k+1)^2-k^2 = \sum_{k=1}^n 2k + n$.so
$\sum_{k=1}^{n} k = ((n+1)^2-1-n)/2$.

There is a general formula to express $\sum_{k=1}^nk^m$ as a polynomial in $n$; the coefficients are Bernoulli numbers times binomial coefficients.

For something like this, I would use undetermined coefficients: assume the sum is $an^6+bn^5+cn^4+dn^3+en^2+fn$ and take a difference of that.
It's a lot of algebra, but it will work.

[quote="jmerry"]There is a general formula to express $\sum_{k=1}^nk^m$ as a polynomial in $n$; the coefficients are Bernoulli numbers times binomial coefficients.
[/quote]

oh yes! that is because if we consider the 'falling factorials',i.e. $f_k(x)=x(x-1)\cdots(x-k+1)$ then
$\sum_{x=0}^{n-1} f_k(x) = \frac{f_{k+1}(x)}{k+1}\large{\vert^{n}_{0}}$. this is similar to the idea of integration!
and the ordinary powers when written in terms of the falling factorials have the Bernoulli numbers occuring as coeffs.


logarithms
trigonometry
calculus
calculus computations

For:

$\sum_{k=1}^{\infty}\ln cos\frac{1}{k}$

Show me what convergence test do you use to  quickly see if this converges?

$\cos\frac1k$ resembles $1-\frac1{2k^{2}},$ so that since $\ln(1+u)$ resembles $u,$, the whole thing resembles $-\frac1{2k^{2}}.$

Hence the test I would use would be the limit comparison test, comparing this series to $\sum\frac1{k^{2}}.$ It converges.


A-Team
8/10 - Woo Young Park
7/10 - Erez Bosch, Ravi Fernando
6/10 - Paul Goodwin, Sam Lite (I think)
...etc.

B-Team
5/10 - Mike Erickson
...etc.

MO B had a correct super relay answer.

Ravi, Sam, Stephen got 8 our our 14 relay points.

We lost T-shirt design to IA.

Neh, I think everything else is in the results.

How the heck is Missouri ARML semi-legit?

Oh dang I pwned the MO team...

the MO team really fails

[quote="andersonw"][quote="Mewto55555"]Random fact of the day: Junu Bae used to go to my Kumon center.[/quote]

Oh hey, I remember him telling me about how he used to live in Missouri at nats last year. :P
Unfortunately, I only saw him twice for short periods of time at ARML.

[/gets out of MO forum][/quote]

Yeah at nats he was behind me and he's like "whoa are you max?" and im like "yeah who the heck are you" and then blahblahblah.


let(C)be a circle inwich its center is inaccessible.and let P be a point wich is in the exterior of circle (C).

draw aline from P,wich is tangent to circle(C).

draw two lines wich go through P.one of them intersects the circle in points A,B.and the otherone intersects the circle

in points C,D.let AC and BD intersect each other in point Q.

let AD and BC intersect each other in point Z.let QZ intersect the circle in points X and Y.

then PX and PY are tangents to the circle.

see:http://mathlinks.ro/forum/viewtopic.php?t=51865


sin 4x = -2 sin 2x

Solve for all values for x in the domain [0, 2pi)

$ \sin4x\equal{}2\sin2x\cos2x\equal{}\minus{}2\sin2x\implies\sin2x(\cos2x\plus{}1)\equal{}0$, so either

$ \sin2x\equal{}0\implies x\equal{}\left\{0,\frac\pi2,\pi,\frac{3\pi}2\right\}$ or

$ \cos2x\equal{}\minus{}1\implies x\equal{}\left\{\frac\pi2,\frac{3\pi}2\right\}$

$ \implies x\equal{}\boxed{\left\{0,\frac\pi2,\pi,\frac{3\pi}2\right\}}$.


[quote="laracroftheoriginal"]ΔΕΝ ΕΧΩ ΚΑΤΑΛΑΒΕΙ ΑΚΡΙΒΩΣ ΤΙ ΓΙΝΕΤΑΙ ΜΕ ΤΗΝ ΑΚΤΙΝΑ ΣΥΓΛΙΣΗΣ ΣΤΟ ΑΝΑΠΤΥΓΜΑ TAYLOR. ΑΦΟΥ Η ΣΕΙΡΑ TAYLOR ΔΕ ΣΥΓΛΙΝΕΙ ΓΙΑ ΟΛΑ ΤΑ Χ ΤΟΤΕ ΠΩΣ ΜΠΟΡΟΥΜΕ ΝΑ ΑΝΑΠΤΥΣΣΟΥΜΕ ΜΙΑ ΣΥΝΑΡΤΗΣΗ ΓΙΑ ΕΝΑ ΤΥΧΑΙΟ Χ ΚΑΙ ΝΑ ΒΡΙΣΚΟΥΜΕ ΜΕ ΤΗ ΒΟΗΘΕΙΑ ΤΟΥ ΑΝΑΠΤΥΓΜΑΤΟΣ ΟΡΙΑ ΚΑΙ ΟΛΟΚΛΗΡΩΜΑΤΑ ΤΗΣ? ΔΙΑΦΩΤΙΣΤΕ ΜΕ ΛΙΓΟ ΔΕΝ ΜΠΟΡΩ ΝΑ ΤΟ ΚΑΤΑΛΑΒΩ ΑΥΤΟ...ΕΠΙΣΗΣ ΓΙΑΤΙ ΥΠΑΡΧΕΙ ΣΦΑΛΜΑ? ΑΦΟΥ ΕΧΟΥΜΕ ΑΠΟ ΤΗΝ ΑΡΧΗ ΠΑΡΑΔΕΧΤΕΙ ΟΤΙ Η ΝΙΟΣΤΗ ΠΑΡΑΓΩΓΟΣ ΤΗΣ F(Χ) ΙΣΟΥΤΑΙ ΜΕ ΤΗ ΝΙΟΣΤΗ ΠΑΡΑΓΩΓΟ ΤΟΥ ΠΟΛΥΩΝΥΜΟΥ?  :blush:[/quote]

na valoume ta pragmata sti thesi tous...alla oxi polu analutika.(ta xontra mono)

proapaitoumenes gnoseis : simeia susoreusis, akolouthies, seires

proxorame...proseggistiko poluonumo. opos les i niostes paragogoi einai ises, alla to poluonumo auto dinei mono mia proseggisi stin f kai gi auto parxei to "sfalma" i upoloipo opos to sunantame sti vivliografia.
to upoloipo einai iso me ti diafora tou poluonumou apo ti sinartisi. $ R_n(x)\equal{}f(x)\minus{}P_n(x)$.
 an $ R_n(x)\equal{}r_n(x)(x\minus{}a)^{n\plus{}1}$ tote $ f(x)\equal{}P_n(x)\plus{}r_n(x)(x\minus{}a)^{n\plus{}1}$ kai an theseis opou x to a+h lamvaneis mia isodunami morfi pou kaleitai aanaptugma taylor.
tora exoume to theorima taylor pou leei oti an i f einai n fores continuously differentiable sto [a,x] kai n+1 fores differentiable sto (a,x) tote

$ f(x)\equal{}f(a)\plus{}f'(a)(x\minus{}a)\plus{}...\plus{}\frac{f^{(n)}(x)}{n!}(x\minus{}a)^{n}\plus{}\frac{f^{(n\plus{}1)}(\xi)}{(n\plus{}1)!}$ for some $ \xi\in (a,x)$

Yparxoun kai isodunames morfes (gia paradeigma vale opou a to x opou x to x+h kai opou $ \xi$ to $ x\plus{}\theta h$ me $ 0<\theta<1$ kai pisteuo oti einai eukolo na vreis tis isodunames morfes-den uparxei logos na tis grapso)

gia a=0 anti na leme talor tha leme mac laurin..den allazei tpt.

tora to parapano theorima tha prepei na to apodeikseis..allios den prokeitai na katalaveis tpt. i am just gonna say Cauchy's Mean Value Theorem and nothing more. vsk auto pou tha apodeikseis einai i lagrange formula gia to remainder (to "upoloipo"), ta alla einai to proseggistiko poluonumo taksis n.

tora proxorame stis seires taylor...
edo exoume oti an h f einai apeires fores paragogisimi kai to orio sto apeiro tou upoloipu einai iso me 0, tote h f einai isi me ti seira taylor (kai mono pou leo ti leksi seira, u should be able to find the formula given the taylor expansion) dedomenou oti i seira suglinei. simeioste oti an oi paragogoi einai fragmenes (opos sumvainei stis triognometrikes), tote to orio tou upoloipou einai 0 (giati?).
episis kathe sunartisi pou topica dinetai apo suglinousa dunamoseira leme oti einai analytiki.
similarly maclauurin series.

to simantiko tora!

[b] The convergence of the Taylor series does not imply that the series converges to f [/b]

counterexapme i sunartisi tou Cauchy...

$ f(x)\equal{}e^{\frac{\minus{}1}{x^2}} for x\neq0$ kai $ f(0)\equal{}0$.

*efarmoges: vres tis seires ma laurin tou imitonou, tou sunimitonou, tis ekthetikis, tis arctan, tis arcsin, tou ln(1/(1-x))

hope this helps.den eimai polu analutikos, exo paraleipsei kai merika pragmatakia...k ena theorimataki gia to remainder alla gia arxi as to afisoume edo kai epanerxomaste.

[quote="laracroftheoriginal"]ευχαριστώ πολύ για την απάντηση σου. με βοήθησε αρκετά αλλά τι γίνεται με την ακτίνα σύγλισης? αν για παράδειγμα μου ζητάνε να βρώ ένα όριο μια συνάρτησης και αυτό βγαίνει μόνο με ανάπτυγμα taylor τότε πώς εγώ ξέρω οτι η συνάρτηση προσεγγίζεται από τη σειρά taylor? βασικά αυτό το σημείο με δυσκολεύει περισσότερο. επίσης δεν κατάλαβα γιατί το πολυώνυμο δίνει απλά μια προσέγγιση της f ?ελπίζω να μη σου βάζω δύσκολα. :oops:  σε παρακαλώ διαφώτισε με λίγο.[/quote]

kataavaino oti kapou exeis berdeutei...kai einai logiko-otan ta vlepeis gia proti fora kapos etsi einai...na sou po tin alitheia den katalavaino ola ta erotimata sou 100% (einai ki arga vevaia kai den vlepo brosta opote isos na m ftais ki esu...) alla tha proteina to eksis:
1) arxika diavase apo to vivlio tous louis brand gia paradeigma tin enotita pou milaei gia tis taylor
2)tora pou to thimithika ..to pio sosto einai na milame gia anaptugma taylor about a point a dld guro apo to simeio a.
3) an exeis sugekrimena paradeigmata..auta pou les me ta oria kai tis aktines suglisis, vale tis askiseis edo kai prosekse tous tropous luseis
4) an eksakoloutheis na exeis apories kai den grapsei edo kanenas apo tous "giteutes" tis analusis (exoume arketous edo trust me! kai eimai eutuxismenos gi auto) tote tha grapso ena akoma post.

ps. auto me to approximation polynomial...to poluonumo dinei apla mia proseggisi..dld esu ti theleis na dinei?ti sunartisi?....skepsou se poia monadiki periptosi sumvainei auto.
tha apantiso mono se mia erotisi sou..gia na se valo sto paixnidi...i apantisi vsk einai sto theorima taylor...apodeiknueis ti suglisi tis seiras taylor arxika ki episis deixneis oti to "upoloipo" suglinei sto 0 kathos to n teinei sto apeiro...ara tote kai mono tote i seira taylor isoutai me ti sunartisi. episis hdh sto proigoumeno post mou anefera oti an i niostes paragogoi gia sufficiently large n einai fragmenes tote to orio tou upoloipou einai pragmati 0 (giati?)

*xrisimopoihse kai ti wikipedia!

όταν θέλουμε να βρούμε την ακτίνα σύγκλισης π.χ. στο ανάπτυγμα Maclaurin τότε παίρνουμε το κριτήριο ρίζας για το πηλίκο της νιοστής παραγώγου της συνάρτησης προς το ν παραγοντικό. αυτό εννοείς όταν λες ότι κοιτάμε αν το υπόλοιπο τείνει στο 0? αλλά αφού το υπόλοιπο δεν ισούται με την νιοστή παράγωγο της συνάρτησης προς ν παραγοντικό...

ασχετο...πως μπορώ να γράψω εδώ μαθηματικά σύμβολα όταν θα θέλω να δημοσιεύσω μια άσκηση?[/hide]

Ας πάρουμε ένα παράδειγμα: $ f(x) \equal{} (1\minus{}x)^{\minus{}1}$

Θέλουμε να υπολογίσουμε το ανάπτυγμα Taylor γύρω από το $ 0$. Το πρώτο πράγμα που πρέπει να κάνουμε είναι να υπολογίσουμε τις παραγώγους. Από επαγωγή έχουμε $ f^{(n)}(x) \equal{} n! (1\minus{}x)^{\minus{}(n\plus{}1)}$, άρα $ f^{(n)}(0) \equal{} n!$.

Το ανάπτυγμα Taylor λοιπόν είναι $ f(x) \equal{} 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} \ldots$.

Προς το παρών αυτό δεν μας λέει τίποτε για τις τιμές του $ f(x)$. Το επόμενο βήμα είναι να υπολογίζουμε την ακτίνα σύγκλισης. Τι είναι η ακτίνα σύγκλισης. Υπάρχει ένα θεώρημα της ανάλυσης που μας λέει το εξής:

Για κάθε δυναμοσειρά $ f(x) \equal{} \sum\limits_{n\equal{}0}^{\infty} a_n x^n$, υπάρχει ένας μη αρνητικός αριθμός $ R$ (επιτρέπουμε $ R \equal{} \infty$) έτσι ώστε

1) Για κάθε $ x$ με $ |x| < R$ η δυναμοσειρά συγκλίνει.
2) Για κάθε $ x$ με $ |x| > R$ η δυναμοσειρά αποκλίνει.
3) Αν $ |x| \equal{} r$ τότε οτιδήποτε μπορεί να συμβεί.

Πώς μπορούμε να υπολογίσουμε την ακτίνα σύγκλισης; Υπάρχουν διάφορα θεωρήματα που αν δουλέψουν μας δίνουν την ακτίνα. Το πιο απλό είναι το

Ratio test: $ R \equal{} \lim_{n \to \infty}|a_n/a_{n\plus{}1}|$ αν αυτό το όριο υπάρχει.

Στην περίπτωσή μας το ratio test δίνει. $ R \equal{} 1$. Υπάρχουν όμως πολλές δυναμοσειρές για τις οποίες αυτό το όριο δεν υπάρχει και το ratio test δεν μας λέει τίποτα. Ένα θεώρημα που πάντα δουλεύει (αλλά συνήθως είναι πιο δύσκολο να εφαρμόσεις από το ratio test) δίνει  $ R \equal{} \frac{1}{\limsup \sqrt[n]{|a_n|}}$.

Ας πάμε πίσω στο παράδειγμά μας. Θα θέλαμε να πούμε ότι για $ |x| < 1$, $ \frac{1}{1\minus{}x} \equal{} \sum_{n\equal{}0}^{\infty}x^n$. Όπως όμως έχει εξηγήσει ο Νίκος, αυτό δεν μπορούμε να το πούμε προς το παρόν! (Το μόνο που μπορούμε να πούμε είναι ότι η σειρά συγκλείνει).  Για να δείξουμε ότι πράγματι $ \frac{1}{1\minus{}x} \equal{} \sum_{n\equal{}0}^{\infty}x^n$ χρειαζόμαστε το "θεώρημα Taylor με υπόλοιπο" για το οποίο έχει γράψει ο Νίκος. Περισσότερα όμως αργότερα.

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

Για να γράψεις μαθηματικά: Γράφουμε μαθηματικά μέταξύ δύο δολλαρίων. Π.χ. για να γράψω $ x^2$, γράφω $ \$$ x^2 $ \$$. Αν κάνεις click πάνω στους μαθηματικούς τύπυς που έγραψα, θα δεις την formula με την οποία τους έγραψα.

σκέφτομαι να μελετήσω λίγο ακόμα το όλο θέμα μήπως και το καταλάβω καλύτερα και να επανέλθω με τυχόν απορίες. Nick και Demetres σας ευχαριστώ πολύ για την πολύτιμη βοήθεια σας.

[quote="r_boris"][quote="Nick Rapanos"]polu oraia...gia to latex uparxei kai to antistoixo stickie sto forum pou leei pos to xrisimopoioume...just search for it.
episis prospathise na vreis ti seira mac laurin tis arctan...auto pou kanei o dimitris den tha se voithisei. prepei na skefteis kati allo! (tora pou to thimithika...mallon den anaferthika sti monadikotita tou anaptugmatos taylor, alla fantazomai tha diavaseis gi auta..)
ki alli mia efarmogi ton seiron pou eixame se ena question sheet sto panepistimio

na upologisthei to olokliroma $ I \equal{} \displaystyle \frac {2}{\pi}\int _0^{\frac {\pi}{2}}{cos(xsint)dt}$[/quote]
Bessel  $ J_0(x)?$[/quote]

Ακριβώς κ. Μπόρη!  :blush:


[quote="7h3.D3m0n.117"]As a temporary solution, you [i]could[/i] take an image of the screen (using Alt + Print Screen) and then crop the particular section out.[/quote]

Thanks, I'll try that.

[quote="GoldenFrog1618"]:help:

I'm taking an online class.  I am wondering if there is (or if you could make) a feature to print just the wording of a particular post or topic without having to print everything on the page (or using the "print what you select" feature).

Thanks![/quote]
you could highlight the stuff u want and copy&paste it into a document.


vector
algebra
polynomial
algebra unsolved

Let $ k$ be a nonnegative real, and let $ \left(u_{n}\right)_{n\in\mathbb{N}}$ be a sequence satisfying the recurrent equation $ u_{n \plus{} 1} \equal{} \frac {k \plus{} u_{n}}{1 \minus{} u_{n}}$ for all $ n\in\left\{1,2,3,...\right\}$.

Assume that $ u_{13} \equal{} u_{1}$. Find the value of $ k$.

[quote="Rust"]
\[ u_{n\plus{}1}\equal{}\frac{\frac{1\plus{}u_{1}/\sqrt{\minus{}k}}{2}(1\minus{}\sqrt{\minus{}k})^{n}\plus{}\frac{1\minus{}u_{1}/\sqrt{\minus{}k}}{2}(1\plus{}\sqrt{\minus{}k})^{n}}{(1\minus{}\sqrt{\minus{}k})^{n}\plus{}(1\plus{}\sqrt{\minus{}k})^{n}}\]
And for any k>0 exist $ u_{1}$, suth that $ u_{13}\equal{} u_{1}$.[/quote]

Hello Rust!

Your formula is likely wrong (it gives, for example, $ u_{1}\equal{} 1$).

I found $ u_{n}\equal{}\sqrt{\minus{}k}\frac{(u_{1}\minus{}k\minus{}\sqrt{\minus{}k}(u_{1}\plus{}1))(1\plus{}\sqrt{\minus{}k})^{n}\plus{}(u_{1}\minus{}k\plus{}\sqrt{\minus{}k}(u_{1}\plus{}1))(1\minus{}\sqrt{\minus{}k})^{n}}{\minus{}(u_{1}\minus{}k\minus{}\sqrt{\minus{}k}(u_{1}\plus{}1))(1\plus{}\sqrt{\minus{}k})^{n}\plus{}(u_{1}\minus{}k\plus{}\sqrt{\minus{}k}(u_{1}\plus{}1))(1\minus{}\sqrt{\minus{}k})^{n}}$

And this changes the conclusion since we no longer have a 1_degree equation to find $ u_{1}$ (with your formula) but a 2_degree equation which may have no solution. (example, take $ n \equal{} 2$ and not $ n \equal{} 13$ and the equation $ u_{2}\equal{} u_{1}$ becomes $ u_{1}^{2}\equal{}\minus{}k$ without real solutions).

So we can not conclude immediatly that  for any k>0 it exists $ u_{1}$ real, suth that $ u_{13}\equal{} u_{1}$.

$ u_{n+1}={1\ k\choose-1\ 1}^{n}{u_{1}\choose 1}$, were $ {a\ b\choose c\ d }{x\choose y}=\frac{ax+by}{cx+dy}$.
I calculate, that own values are $ 1\pm\sqrt{-k}$ and own vectors are $ {\mp\sqrt{-k}\choose 1}$.
May be I have mistake in calculate.

[quote="toan hoc"]Let $ (u_{n})$  be a sequence given by $ u_{1}$  anh the formula $ u_{n\plus{}1}\equal{}\frac{k\plus{}u_{n}}{1\minus{}u_{n}}$,where $ k\geq 0;n \equal{} 1;2;...$
Gicen that $ u_{13}\equal{} u_{1}$.Find value of $ k$[/quote]
It is a problem from Vietnam mathematics'magazine,and this exercise haven't been expired

Let $ {a\ b\ \choose c\ a}\equal{}{1\ k\choose\minus{}1\ 1}^{12}$/ $ a \equal{} a(k),b \equal{} b(k),c \equal{} c(k)$ polinoms degre 6.
Equation $ u_{13}\equal{} u_{1}$ equavalent to $ b(k) \equal{} c(k)u_{1}^{2}$, solution is $ \pm\sqrt{\frac{b(k)}{c(k)}}$ may be complex.

Here is an observation:
$ f(x): \equal{}\frac{k\plus{}x}{1\minus{}x}$.

Then:
$ f^{2}(x) \equal{}\frac{(\minus{}2)(k\plus{}x^{2})}{k\plus{}2x\minus{}1}$
$ f^{3}(x)\minus{}x \equal{}\frac{(3\minus{}k)(k\plus{}x^{2})}{kx\minus{}3k\minus{}3x\plus{}1}$
$ f^{4}(x)\minus{}x \equal{}\frac{4(1\minus{}k)(k\plus{}x^{2})}{k^{2}\plus{}4kx\minus{}6k\minus{}4x\plus{}1}$
$ f^{6}(x)\minus{}x \equal{}\frac{(2)(3\minus{}k)(3k\minus{}1)(k\plus{}x^{2})}{k^{3}\plus{}6k^{2}x\minus{}15k^{2}\minus{}20kx\plus{}15k\plus{}6x\minus{}1}$

Of course, if $ g(x): \equal{} f^{d}(x)$, then $ g^{\frac{12}{d}}(x) \equal{} f^{12}(x)$ so if $ f^{d}(x)\equal{}x$, then $ f^{12}(x)\equal{}x$ as well where $ d|12$.

I may not have found all solutions.

My calculate $ u_{n\plus{}1}\equal{}\frac{a_{n}u_{n}\plus{}c_{n}k}{\minus{}c_{n}u_{1}\plus{}a_{n}}$, were
$ a_{n}\equal{} Re(1\plus{}i\sqrt k )^{n}\equal{} 1\minus{}C_{n}^{2}k\plus{}C_{n}^{4}k^{2}\plus{}.., c_{n}\equal{}\frac{Im(1\plus{}i\sqrt k)^{n}}{\sqrt k}\equal{} C_{n}^{1}\minus{}C_{n}^{3}k\plus{}...$.
From $ u_{n\plus{}1}\equal{} u_{1}, n>0$ get, that $ u_{1}^{2}\equal{}\minus{}k$ or $ u_{1}\equal{}\pm i\sqrt k$.

[quote="Rust"]My calculate $ u_{n\plus{}1}\equal{}\frac{a_{n}u_{n}\plus{}c_{n}k}{\minus{}c_{n}u_{1}\plus{}a_{n}}$, [/quote]
Why?
Can you do more clearly?

I think
If we pose $ u_n \equal{} \sqrt {k}.tanx_n$, and note $ \sqrt {k} \equal{} tana$ we have
$ tanx_{n \plus{} 1} \equal{} \frac {1}{\sqrt {k}}.tan(a \plus{} x_{n \plus{} 1}$

From the definition and with a lot of work:
$ u_{13}\equal{}
\frac{k(12 \minus{} 66 u1)\plus{} k^5(220 \minus{} 66 u1)\plus{}k^6(\minus{}12 \plus{} u1)\plus{} u1 \plus{} 99 k^4(\minus{}8 \plus{} 5 u1) \minus{} 132 k^3(\minus{}6 \plus{} 7 u1) \plus{} 55 k^2(\minus{}4 \plus{} 9 u1)}
{1 \plus{} k^6 \minus{} 12 u1 \plus{} 6 k^5(\minus{}11\plus{} 2 u1) \minus{} 55 k^4(\minus{}9 \plus{} 4 u1) \plus{} 132 k^3(\minus{}7 \plus{} 6u1) \minus{} 99k^2(\minus{}5\plus{} 8u1) \plus{} 22k (\minus{}3\plus{} 10 u1)}$

The equation $ u_1\equal{}u_{13}$ is equivalent of finding roots of a six degree polynomial on $ k$
with its trivial solutions: $ k\equal{}1,3,\frac{1}{3}$ or $ \minus{}u_1^2$ and the other two $ k\equal{} 7\plus{}\minus{} 4 \sqrt{3}$

Deadline over, problem can be discussed again.

  darij


geometry
circumcircle
incenter
angle bisector
radical axis
geometry proposed

Let a triangle $  ABC $ and a point $ D $ on side $ BC$ . Two circles are externally tangent on the same point $ M $  of  $ AD  $  and each of them is tangent to the circumcircle of triangle $\triangle ABC $ and  to side $ BC $ , the first on the  segment $ BD $  and the other on the  segment  $ DC $  .
Prove that  $AD $  is bisector of angle $\angle A $.

[u]Babis[/u]

 [i]( proposer and autor of the problem in that test is Dan Brinzei)[/i]

I remember finding a very ugly and long proof of this using inversion a couple of years ago, but I think the following one (also using inversion) is pretty cute :):

Let $T$ be the point where the line of centers of the two circles cuts $BC$. Consider the inversion of pole $A$ and power $AM^2$. It turns $B,C$ into $B',C'$ respectively s.t. $B'C'$ is the second common tangent of the two circles. Now, $B'BCC'$ is cyclic, and a simple angle chase will show that the angle bisectors of $\angle BTB',BAC$ are perpendicular. On the other hand, $TM$ is the angle bisector of $\angle BTB'$, and $AM\perp TM$. This means that $AM$ must be the angle bisector of $\angle BAC$, as desired.

I will offer a equivalent enunciation:

Let $w_1,w_2$ be two circles which are interior tangent to the circle $w$ in the points $X_1,X_2$ respectively and which are exterior tangent in the point $M$. Let $t=Y_1Y_2$ be a common exterior tangent to the circles $w_1,w_2$, where $Y_1\in w_1,Y_2\in w_2$. I note $\{B,C\} \subset Y_1Y_2\cap w,\ A\in t\cap w,\ D\in t\cap BC$. Then the line $t$ is the bisector of the angle $\widehat {BAC}$.

[b]A solution.[/b] I note $Z\in X_1Y_1\cap X_2Y_2,\ T\in X_1X_1\cap Y_2Y_2$ (I note the tangent $XX$ in the point $X$ to the circle w).
Prove easily that $T\in t\equiv \overline {AMD}$ (the point $T$ is radical center for the circles $w,w_1,w_2$) and the quadrilateral $X_1Y_1Y_2X_2$ is ciclic. Therefore, $ZY_1\cdot ZX_1=ZY_2\cdot ZX_2$, i.e. $Z\in t,\ ZB=ZC$ (a wellknown property: see the my reply in  http://www.mathlinks.ro/Forum/viewtopic.php?t=53084 ). Thus, $t\equiv \overline {TAMDZ},\ \widehat {BAD}\equiv \widehat {CAD}$.

Both are very  interesting and nice aproaches !

     An other solution( I saw it in GM) uses the theorem of Casey twice. But first we produce AD to meet circumcircle at N. Finally we take NB = NC ect.

   If necessary , I ' ll post the whole solution.

 Babis

[quote="stergiu"]Let a triangle $  ABC $ and a point $ D $ on side $ BC$ . Two circles are externally tangent on the same point $ M $  of  $ AD  $  and each of them is tangent to the circumcircle of triangle $\triangle ABC $ and  to side $ BC $ , the first on the  segment $ BD $  and the other on the  segment  $ DC $  .
Prove that  $AD $  is bisector of angle $\angle A $.[/quote]

More can be said: The point D is the incenter of triangle ABC. For solutions, see

http://www.kalva.demon.co.uk/short/soln/sh927.html
http://www.mathlinks.ro/Forum/viewtopic.php?t=1370 post #4
http://www.mathlinks.ro/Forum/viewtopic.php?t=18743
http://www.mathlinks.ro/Forum/viewtopic.php?t=24869

  darij

To Ph - An

 For a proof using CASEY , please see , [b]GM - 1997 , page152[/b]. If you can not find this volume , then I' ll write the solution. It is only three lines and simple.If you have it , please write down this beautiful solution for the other friends in this forum.They ' ll like it.Thanks -

  [u]Babis[/u]


MIT college calculus Support trigonometry LaTeX geometry

I'm sure all you guys and gals out there[color=brown] (no offence meant)[/color] know about the world famous IIT's . They are regarded almost at par with MIT and the like (forget the rankings). The admission procedure is the toughest in the world with a Joint entrance exam (JEE) in [b]Physics, Chemistry and Maths,(almost Olympiad level problems).[/b]If any of you out there are preparing for the JEE , this is the post for you.I'm vieing for both the IIT's and MIT (Ivy league of US).If you have any good [b]problems for the JEE [/b]in any of the three subjects , post them here! If you want any info or want to share some, please do so.If you know of any good online links for the JEE , please share them here.

[b]Come on indians , let's crack the JEE together.[/b]

[color=red]NOTE :-[/color] Since this post involves all three physical science subjects and Indian Colleges , I have posted it here.If the administrator thinks otherwise , i'll be happy to adjust.[b]Thanks!![/b] :)

I'm always curious about the admissions procedures for colleges outside the United States, and I'd be happy to see replies to this thread, which I think is very on-topic for this forum.

I haevn't seen many good sites for IIT JEE exams, but India has plenty of books and magazines featuring lots of problems from them that you can probably find online somewhere (or at any Indian book store).

As for JEE's difficulty, the math section is not that difficult for skilled AOPSers, but in conjunction with the physics and chemistry this exam is as a whole VERY difficult.  I dont think that people in USA (I am born and I live in USA despite my Indian flag!!!) can really cope with this exam to tell the truth, and I dont really want to sound disparaging or anything. The physics is far far beyond AP Physics, the chemistry far far beyond AP Chemistry, and the math is far far far beyond anything you do in ordinary high school math.  Of course there are great students in any country that excel at ONE of the subjects, but you have to excel at all 3 to get in and given that 200,000+ people are competing for such few seats in IIT, it is extremely doubtful that one who hasn't been training for years can just get into IIT with a few months of training!

That being said, good luck to all who try!   :D

Has the application process changed a lot, because I remember my dad saying that since he got first place in a math competition in India, or something like that, he got automatic admission into IIT, without taking the JEE.  But I think it was a combined graduate and undergraduate thing, I don't really remeber.  And my mom also told me that if your really awesome at math, you'll get in.

[quote="Barnacle"]I haevn't seen many good sites for IIT JEE exams, but India has plenty of books and magazines featuring lots of problems from them that you can probably find online somewhere (or at any Indian book store).

As for JEE's difficulty, the math section is not that difficult for skilled AOPSers, but in conjunction with the physics and chemistry this exam is as a whole VERY difficult.  I dont think that people in USA (I am born and I live in USA despite my Indian flag!!!) can really cope with this exam to tell the truth, and I dont really want to sound disparaging or anything. The physics is far far beyond AP Physics, the chemistry far far beyond AP Chemistry, and the math is far far far beyond anything you do in ordinary high school math.  Of course there are great students in any country that [/quote]

Barnacle, please post some 20+ of the harder questions for all the math, physics and chemistry sections! Other than given [url=http://www.iitd.ac.in/jee/]here.[/url] :)

[quote="h_s_potter2002"]he got first place in a math competition in India, or something like that, he got automatic admission into IIT, without taking the JEE.  But I think it was a combined graduate and undergraduate thing, I don't really remeber. [/quote]

I cannot imagine that. As far as I know almost all good people leave IIT for their postgraduate level degrees and the level really declines. These IIT guys are also on top of Indian universities because they get all the financial support from the government they might need. And those guys of the rejected 200,000+ applicants who go to other Indian universities suck because the quality really drops compared to IIT. Is it true that subjects are not by choice but by ranking on the JEE?

[quote="blahblahblah"]I've seen the math portion, and it is laughable compared to olympiad math.  It is no more than slightly glorified cookbook calculus and linear algebra.[/quote]

Hold on. Those guys who knows the problems please do post and then everybody can judge for oneself. It would be also nice to have a full exam to see its total difficulty. But I think the combination of all three subjects makes it hard. I rather would like to have exams in math, physics and computer science.  ;)

thats very nice, i just saw the chemistry section and the problems are comparable to college level. i can do some of them too!
ty orl for the link, this is going to keep me busy for a while :)

[quote="orl"]
Hold on. Those guys who knows the problems please do post and then everybody can judge for oneself. It would be also nice to have a full exam to see its total difficulty. But I think the combination of all three subjects makes it hard. I rather would like to have exams in math, physics and computer science.  ;)[/quote]

We have examined the problems from the JEE before on this forum (maybe it was before the merge?).  They are not of olympiad quality, and a student with a decent background in calculus and linear algebra should find both simple.

yes blahlablah, but it's hard to be of olympiad quality in all 3 of math chemistry and physics.  the math is definitely not that hard for the skilled, but then again you have to raelize that the top scorers probably do get ~~100% on the math exam so you can't even make a slight mistake.. also you don't get nearly as much time as you do on olympiads.  I think that if I trained hard I could perform of qualification level on math/physics... but the chemistry.. eh.. I dont really like that much :)

orl if I can find my old magazines I will put some problems up... some are pretty interesting.

by the way orl, yeah it is true that if you get in, your major is based on ranking and not choice.. they only have a limited number of seats for a given major and the people up top get to choose majors first... it kind of sucks, and yet its interesting that people go to majors they are totally disinterested in just for the IIT brand name hahah.

Look, you don't have to be of olympiad quality to solve those problems.  You don't even have to be close to Olympiad quality.  There is nothing non-trivial about any of the problems - they are cookbook calculus problems - maybe difficult cookbook problems, but cookbook nonetheless.  It's absurd to compare getting 1 hour to solve an IMO problem with getting 10 minutes to solve a calculus problem.  If I ask you to fill out a times table in 5 minutes, are you going to complain that that's 169 'problems'?

Here is the 2003 exam for math.  You only get 2 hours for all 20 of the problems problems, thats only 6 minutes per problem; it's quite hard to get 100% in that much time, even though the problems are [i]relatively[/i] easy.  Please try it yourself without generalizing them all as cookbook calculus problems.  I assure you it's no easy feat.   :)

btw, you'll have to zoom up to see the problems.

ill put physics up soon.

You must have skipped over the post where I mentioned that we have discussed an entire paper of these problems on the forum, right?

physics 2003 exam, 3 pages.  first 3 or 4 are quite easy... but you have to see the solution to them [and the rest of the 16 problems] almost instantly if you want to be able to write solutions in time!  and the later problems are definitely nontrivial.  for physics it becomes less of a difficulty comparison than math though.. even physics olympiad isn't that hard.. its just beating the clock :)

By the way, even though this conversation should be split, I should note that problems $3$ and $20$ are very cute problems.

The solution to $3$ makes use of the very important (IMHO) identity $\sin x+\cos x=\sqrt{2}\sin(x+\pi/4)$, the solution to $20$ uses

$\binom{n}{k}=(2-1)^k\binom{n}{k}$ and then the identity

$\binom {n}{k}\binom {k}{i}=\binom {n}{i}\binom{n-i}{k-i}$.  I am still trying to come up with a combinatorial interpretation.

Why don't you post some of the noncalculus problems in Pre-Olympiad - they'd make nice exercises there.

[quote="Gyan"]Xposted from other thread:

Just a side remark in US congress (Resolution HR 227 on April 26, 2005), acknoledges the impact by
Indian Institute of Technology (IIT's) to US.  Note that accordiing to news papers, NO other  foreign  college has been so honored.  [/quote]

That's good news, but it is more likely a bone thrown to some Indian supporters by a congressman, perhaps from a district with many Indians (e.g. Delaware), than a unique attribute of IIT.   Moscow State, Beijing U, not to mention the much longer time horizon and therefore greater cumulative impact of contributions from Cambridge, Oxford (maybe ENS, Ecole Polytechnique, and others) are comparable to or larger than IIT.

I don't think the "rankings"  generally  t mean much, but  FWIW  UK based Times Higher  Education  Ranked "Technology universities" in the world.  IIT's  came up pretty high.  Top rankings (as  expeted ) were dominated by US  Institutions ..  For Technology MIT   got the top spot. IIT's were in the 3rd position.

Here is a news item about:

[url=http://us.rediff.com/money/2005/oct/10iit.htm]IIT Rankings[/url]

[quote]ITs, world's 3rd best tech universities

October 10, 2005 19:01 IST

There's another feather in the much-adorned IIT cap.

The United Kingdom-based Times Higher Education Supplement has ranked the Indian Institutes of Technology as the third best technology universities in the world for 2005.

In a statement, THES said: "Our peer review of the world's top technology universities shows that in 2004, the high praise for the Indian Institutes of Technology was no fluke. Up to third position in 2005 from fourth place last year, the IITs are a source of Indian national pride as well as innovation and wealth."

The technologists among our peer review panel regard Imperial College London as the UK's hottest university, ahead of Cambridge University and fifth in the world, the statement adds.

The rankings, THES said, is dominated by the United States, no surprise for the world's top high-technology nation. It appears 26 times in the top 101 institutions in our list. For once, this area is one in which Harvard University ranks outside the elite, appearing in 21st place.

"Our definition of technology covers the main engineering disciplines including information and communications technology. Viewed alongside the tables for science subjects, our rankings suggest it is possible for a university to be strong in science but not technology, and vice versa," THES said.

The top science institutions, Cambridge and Oxford universities, cut less of a dash in engineering and IT, placed 6 and 13, respectively, while the IITs come 36th in science despite being third in technology. Overall, 10 of the top 20 institutions in science are rated by our peer review panel as being in the top 20 for technology and IT.

In contrast to science, many technology academics appear rarely in the world's top research publications. This applies especially to institutions outside the English-speaking world but also to some well-rated universities in Australia and elsewhere. It suggests that big US institutions such as the Massachusetts Institute of Technology, the California Institute of Technology, Stanford University and the University of California, Berkeley generate not only highly regarded papers but also many commercial innovations and spin-offs, THES said in a media statement.
[/quote]

The congressman who introduced the IIT laudation is Tom Davis from Fairfax County, VA.
Fairfax County includes Thomas Jefferson HS and other fancy high schools, and the area is quite heavy on educated DC-area professionals, including a much higher proportion of engineers, scientists, and professors than most areas of the United States.  Anecdotally it has high concentration of Indians, though I have no statistical verification of this.   Davis himself is a former technology executive.

re: rankings of ITT, are the different campuses considered comparable within India?  Are the levels of selectivity similar, or are there flagship campuses and lesser, sattelite ones?

[quote="vaibhava10"]The JEE is the toughest engineering entrance exam in the world. Though the JEE syllabus is quite close to the olympiad, it is not exactly like the olympiad. The topics included in the syllabus are meant to train you specifically for engineering mathematics, physics and chemistry and not for an overall treatment of the subject as demanded by the olympiad. For example the pigeon-hole principle, which is required for the  mathematics olympiad, is not to be studied for the JEE. People bunk school like crazy(as a matter of fact students leave their hometowns and go to other towns to prepare for an entrance test and attend school only 9 days in the entire year.) Some bored loosers start preparing from Grade 8 itself. The people who compete for this exam are the real hungry, motivated students who are willing to do anything to get a global platform like the IITs. For example, a few years ago, the JEE all India Ranker 1 had not seen a train until he came first in the IIT for the simple region  that he lived in a village in rural India.The pressure of this exam is so high, that I am quite sure that even the brightest and most brilliant U.S. high school students will feel the heat of competition in India for this exam.[/quote]

How can you say the toughest engineering entrance exam in the world;you did not even went through it and dont' know what are the entrance exams in others part of the world! That's understandable to boast of one's elite schools but your arguments make no sense, same things are seen in every place where exams are contests.

well jee is REAL tough no doubt
i cleared it with rank 3374 mainly because i screwed up maths two years  ago.supposedly my strongest point
anyways the grudge i have is that there are too many coaching institutions oriented only for JEE preparations making students work for 10 hours together.
is all the effort worth getting in to IIT. and mind u i have seen better students in universities other than iit's.
u might call this a case of sour grapes but it isn't.
i dont see the point in taking a year off simply because u want to get into an IIT.
:)
no offense intended to anyone.it's more abt brand iit that the institute itself.sad that a student with GRE score of 1400 lands up at stanford with full scholarship from an iit and another with score of 1530 gets no scholarship at the same place because he does not have a recommendation from a proffesor of iit's stature

[quote="thealchemist"]Maxbl, I too feel that the JEE is "one of the toughest" :P  entrance exams. Just see the question paper and post your comments.[/quote]

WEll with the new format and aftre visting this math community,the math section of IIT doesn't seem as hard as before anymore(Physics is HARD).I could do 4 questions of screening easily at least :lol:

I did not say it was not hard, what I meant is you just cannot say an exam is the toughest in the world because its the toughest of your country.

Probably it's not the toughest exam in the world. But it's better to say that the rate of qualifying in that exam is the lowest in the world compared to any other university admission procedure. But then again, perhaps we are reviving some old debates in this forum...lol

I am a newone but i search on Internet and found that IIT has 7 location, such as Bombay, Delhi, Kharagpur... They are same or distinct ?
Can you explain about the difference between them ?
Thank you.

well those are districts (bombay,delhi...) and all of them have IIt(the college).Nothings different in any of them...(i think :? )

angentalcom:   These are separate IIT's (All called Indian Institute of Technology) in different locations. There is  one   entrance test, and how well you do in that test is  used  by all institutes as a criteria for admission in  the undergraduate program.

There are, of course, differences between the institutes and also a little rivalry. Some schools are good in one field or the other   but *all* follow a minimum standard and  all have a very good reputation, and generally all rank fairly high in various rankings.

Does anyone know where I can get a copy of JEE 2006 in order to take a look at the new format?

I just used Google to search for exactly the phrase "JEE 2006," and the first link was the official site for the exam, which includes a syllabus anyone may download and other information.


Nvm sorry, stupid question.   :oops:  :blush: Can someone delete this thread?

Well why not add a new question instead???

Like :

     Mary and Barry are running a race.  Mary runs at a rate of
     5.5 yds per second, while Barry runs at a rate of 3.75 yd/s.
     Barry gets an 18 yard head start and Mary wins by 3 yards.
     How many yards long is the race?

Im not sure if this is right but is it 66 yds (12 secs)

Well if someone would give a solution, maybe I'd comment on the answer :D

Mary: 5.5t=d where d is the length of the race
Barry:3.75t+18=d-3, since she finished 3 meters behind Mary

so 5.5t=3.75t+21, we get t equal to 12 seconds, so d=5.5*12=66 yards

[hide]Every second, that one dude gets ahead by $1.75$ yards. Therefore, that dude must win by $21$ yards. $21/1.75$ which is
[hide="..."]
12!
[hide="so"]
the answer would be

$\boxed{66}$

[/hide]
[/hide][/hide]

Edit: the genius beat me to it.

[hide]
I set up two equations. We know distance = rate x time, so  Mary's equation would be
$d=5.5t$.
Barry had an 18 yard headstart, and in addition traveled three yards less than Mary.  So, Barry's equation is
$d-18-3=3.75t$.

solving for $d$ and $t$ gives us
$d=66$
$t=12$
So the distance is 66... right?
I appreciate anybody telling me where I made my mistake in logic.  Thanks.[/hide]

EDIT never  mind.. two people posted before me.


In triangle $ \Delta ABC$ , $ D$ is a point inside it such that :

$ m\angle DAB \equal{} 10^0$

$ m\angle DAC \equal{} 40^0$

$ m\angle DCA \equal{} 30^0$

$ m\angle DBA \equal{} 20^0$

without using a calculator, prove that :

$ m\angle DCB \equal{} 20^0$

It is a problem from USA Math Olympiad - 1996 you may find a solution.


This is a generalization of http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204988

anyway,
[hide="solution"]
Each row is in the form
\[ \frac{x_1}{u\plus{}b_1}\plus{}\cdots\plus{}\frac{x_n}{u\plus{}b_n}\equal{}1\]
By clearing the denominators and bringing everything to the RHS, you get an $ n$th degree polynomial in $ u$, set equal to zero, with a leading coefficient of 1 and roots $ \{\minus{}a_1,\minus{}a_2,\ldots,\minus{}a_n\}$, which is equivalent to
\[ (u\plus{}a_1)(u\plus{}a_2)\cdots(u\plus{}a_n)\equal{}0\]
For simplicity, I'm only going to run through finding $ x_1$--the others are all done pretty much the same way. OK, so as of this point, we've pretty much ignored the $ b_n$'s. But notice that if you plug-in $ \minus{}b_1$ for $ u$ in the original (unsimplified) polynomial (the one we got after clearing the denominators), all of the $ x_k$'s disappear [i]except for $ x_1$[/i], which is left with a coefficient of $ \minus{}(b_2\minus{}b_1)(b_3\minus{}b_1)\cdots(b_n\minus{}b_1)$. Since this was also equal to plugging in $ b_1$ for $ u$ in our simplified polynomial, we have
\[ x_1\equal{}\minus{}\frac{(b_1\plus{}a_1)(b_1\plus{}a_2)\cdots(b_1\plus{}a_n)}{(b_2\minus{}b_1)(b_3\minus{}b_1)\cdots(b_n\minus{}b_1)}\]
[/hide]


Do you know those weird grid problems where there is a grid thing and it says "Nathan is at the point N and his school is at S. how many possible ways are there for Nathan to get to S?" There is a formula; where the starting point is (0,0), and the ending point is (n,r), the number is (n+r) choose r which is the same as n+r choose n. how do you prove that?   :D

In order to get to $(n,r)$ from $(0,0)$, you need to go $r$ times "right" and $n$ times "up", in no matter what order (they all end up at the same point). Hence you need to order $n$ "rights" and $r$ "ups", which gives ${n+r\choose n}={n+r\choose r}$

no i didn't mean to prove ${n+r\choose n}={n+r\choose r}$... i meant prove how you got ${n+r\choose n}$ in the first place :ninja:

Repetitive permutations? Doesn't ring a bell?

When you have $n$ objects of one kind and $r$ objects of the other kind, you can order them in ${(n+r)!\over n!r!}={n+r\choose r}$ ways.

Or - If you mark a path by 'R' (say to one unit right) and 'U' (say to go one unit up ) a typical path may look like:

(RURRRRUU...)

(A sequence of 'R' and 'U' - to be precise  n 'R's, and r 'U's (total (n+r) symbols...

totlal number of ways you can arrange (n+r) objects =$(n+r)!$
but there are n identcal (R's) and r identical (U's) so you divide by $n! r!$ and you are done..

you should the section in aops volume 2 that covers this section. The previous methods shown here work but when these problems have some places you can't pass there is a much easier way to do them by counting backwards.


A sequence $ (a_n)$ is defined by $ a_1\equal{}1,a_n\equal{}\frac{n\plus{}1}{n\minus{}1}(a_1\plus{}a_2\plus{}...\plus{}a_{n\minus{}1})$. Determine $ a_{1998}$.


rotation
algebra
function
domain
combinatorics proposed
combinatorics

Here's something cool, although rather easy :):

We cover a $2m\times 2n$ chessboard with dominoes. Prove that the number of horizontal dominoes with a white left square equals the number of horizontal dominoes with a black left square.

I found it (actually, I found the particular case when $m=n=4$, i.e. an actual chessboard) in one of the papers (forgot which one) here: [url]http://www.cs.utexas.edu/users/EWD/indexEWDnums.html[/url].

unless i am much mistaken, if one starts from the tiling with all tiles horizontal, one can reach any other domino tiling by a series of steps, where in each step you simply change a $2\times 2$ square tiled by 2 dominoes to the 'other type',.i.e. if there are 2 adjacent tiles both horizontal or both vertical, then you simply rotate that square by $90^o$ keeping all the other tiles same of course.
since this operation changes the number of horizontal dominoes with a white left square by the same amount as it does the number of horizontal dominoes with a black left square and there are equally many in the 'all-horizontal' tiling,these numbers are same for any tiling.

Are you sure we can reach every tiling by performing that operation? :? It would be a very interesting fact, but it requires proof, of course :).

well, i guess i made it look like it was some trivial thing!probably its not.but it is a fact.in fact you can check this article by Thurston in the AMM:
W. Thurston. Conway’s tiling groups. Amer. Math. Monthly 97 (1990), 757-773.

in fact he proves it for any simply connected domain that is tileable by dominoes.

but of course this present problem is surely much simpler.its just that i thought of this fact as soon as i saw your post ... :)


Source: LBJ 2005

Homer is looking at the graph of a function $f(x)$ on his TI-89 with the screen set so that the point $(-24, -30)$ is in the lower-left corner of the screen and the point $(0,6)$ is in the upper-right corner. The graph appears exactly as the graph of the function $g(x)=x^3$ would appear in a window using $(-10,-10)$ and $(10,10)$ as its corner points. Write a formula for $f(x)$ in the form $f(x)=ax^3+bx^2+cx+d$.

Yes, sorry, -30 it is.  :blush:

I guess it would be a little hard to do otherwise. -.-


linear algebra
matrix
combinatorics proposed
combinatorics

A is a matrix with k row and n columns satisfy the condition:
1)a_{ij}=0 or 1
2)k<=n-2
3)in each row exists at least one 0
4)in each row exist at least two 1
Prove that we can rearrange the column of A such that for each row there is no block of 1s.(There is some 0 between some of the 1s)
 :D

I don't understand the question. Do you mean that there exist some 0's between two 1's?


March 11 is the probable date for GAME (the Garfield Annual Mathematics Extravaganza). If many of you have conflicts, we may be able to reschedule, although there aren't many dates left.

-Adam Hesterberg

11th is a semi-bad day for my team...although i will be there even if my team won't be there....
(you can put it after MAO, or right before it...if i recall correctly MAO is on the 25th)


Does there exist a positive integer $ d$ such that $ 2d \minus{} 1$, $ 5d \minus{} 1$, $ 13d \minus{} 1$ are perfect squares?

Try mathpages.com/. Is your third ques. a IMO one?...

Bomb

If 2d-1 is a perfect square, then it is of the form 4x+1, so d must be odd.
Hence 5d-1 and 13d-1 are even. if they are both square, write 5d-1=4b^2 and 13d-1=4c^2. susbtraction yields 8d = 4(c^2-b^2), or 2d = c^2 - b^2. so b and c have the same parity. but this implies that 2d = c^2 - b^2 is a multiple of 4, but this is impossible since d is odd.


calculus
integration
logarithms
calculus computations


A few difficult trigonometric identities:

(Assume O stands for theta is theta)

1. 8cos^4(O) = cos 4(O) + 4 cos 2(O) + 3

2. 2cos6(O)=64cos^6(O) - 96cos^4(O) + 36cos^2(O) - 2

3.
sin(O) + sin2(O) + sin3(O)
_______________________ = tan2(O)
cos(O) + cos2(O) + cos3(O)

4. sin5(O) + 2sin3(O) + sin(O) = 4sin3(O)cos&2(O)

5. 2sin^2(O/6) - sin&2(O/7) = cos^2(O/7) - cos (O/3)

6. sin2(O) + sin4(O) - sin6(O) = 4sin(O)sin2(O)sin3(O)

7.
sin2(O)cos(O) - 2cos2(O)sin(O)
________________________ = 2cos^2(O/2)
2sin(O)-sin2(O)

8. csc(O)tan(O/2) - {cos2(O)/[1+cos(O)]} = 4sin^2(O/2)

9. sin(O) + sin2(O)
_________________= tan(O/2)
2 + 3cos(O) + cos2(O)

10. If (alpha) + (beta) + (gamma) = pi, show that tan(alpha) + tan(beta) + tan(gamma) = tan(alpha)tan(beta)tan(gamma)

Hopefully the formatting comes out right  :lol:
EDIT: Theta wasn't showing up, replaced with O.

I've been having a lot of trouble figuring these out... can anyone help?

First, welcome to this forum. Second, please try to use latex.

[hide="Number 1"]$ 8 \cos^4x \equal{} \cos(4x) \plus{} 4 \cos(2x) \plus{} 3$.

We have $ 8 \cos^4x \equal{} 8(\cos^2x)^2 \equal{} 8\left[\frac{1}{2}(1\plus{} \cos(2x))\right]^2 \equal{} 2[1\plus{} \cos(2x)]^2$

$ \equal{} 2[1 \plus{} 2 \cos(2x) \plus{} \cos^2(2x)] \equal{} 2 \plus{} 4 \cos(2x) \plus{} 2 \cdot \frac{1}{2}[1 \plus{} \cos(4x)]$

$ \equal{} 2 \plus{} 4 \cos(2x) \plus{} 1 \plus{} \cos(4x) \equal{} 3 \plus{} 4 \cos(2x) \plus{} \cos(4x)$

as we wanted.[/hide]

[hide="Number 2"]Let $ z \equal{} \cos \theta \plus{} i \sin \theta$. Then $ 1/z \equal{} \cos \theta \minus{} i \sin \theta$ and so $ (z \plus{} 1/z) \equal{} 2 \cos \theta$. Also, $ z^n \equal{} \cos(n \theta) \plus{} i \sin(n \theta)$ and $ z^{\minus{}n} \equal{} \cos(n \theta) \minus{} i \sin(n \theta)$, and so $ z^n \plus{} z^{\minus{}n} \equal{} 2 \cos(n \theta)$. We now have:

$ (z \plus{} 1/z)^2 \equal{} z^2 \plus{} 2 \plus{} z^{\minus{}2} \equal{}> z^2 \plus{} z^{\minus{}2} \equal{} (z \plus{} 1/z)^2 \minus{} 2$.

$ (z \plus{} 1/z)^4 \equal{} z^4 \plus{} 4z^3z^{\minus{}1} \plus{} 6z^2z^{\minus{}2} \plus{} 4zz^{\minus{}3} \plus{} z^{\minus{}4} \equal{} (z^4 \plus{} z^{\minus{}4}) \plus{} 4(z^2 \plus{} z^{\minus{}2}) \plus{} 6$
$ \equal{}> z^4 \plus{} z^{\minus{}4} \equal{} (z \plus{} 1/z)^4 \minus{} 4(z^2 \plus{} z^{\minus{}2}) \minus{} 6 \equal{} (z \plus{} 1/z)^4 \minus{} 4(z \plus{} 1/z)^2 \plus{} 2$.

$ (z \plus{} 1/z)^6 \equal{} z^6 \plus{} 6z^5z^{\minus{}1} \plus{} 15z^4z^{\minus{}2} \plus{} 20z^3z^{\minus{}3} \plus{} 15z^2z^{\minus{}4} \plus{} 6zz^{\minus{}5} \plus{} z^{\minus{}6}$
$ \equal{} (z^6 \plus{} z^{\minus{}6}) \plus{} 6(z^4 \plus{} z^{\minus{}4}) \plus{} 15(z^2 \plus{} z^{\minus{}2}) \plus{} 20$
$ \equal{}> z^6 \plus{} z^{\minus{}6} \equal{} (z \plus{} 1/z)^6 \minus{} 6(z^4 \plus{} z^{\minus{}4}) \minus{} 15(z^2 \plus{} z^{\minus{}2}) \minus{} 20$
$ \equal{}> z^6 \plus{} z^{\minus{}6} \equal{} (z \plus{} 1/z)^6 \minus{} 6(z \plus{} 1/z)^4 \plus{} 24(z \plus{} 1/z)^2 \minus{} 12 \minus{} 15(z \plus{} 1/z)^2 \plus{} 30 \minus{} 20$
$ \equal{}> z^6 \plus{} z^{\minus{}6} \equal{} (z \plus{} 1/z)^6 \minus{} 6(z \plus{} 1/z)^4 \plus{} 9(z \plus{} 1/z)^2 \minus{} 2$.*

But $ z^6 \plus{} z^{\minus{}6} \equal{} 2 \cos(6 \theta)$, $ (z \plus{} 1/z)^6 \equal{} 2^6 \cos^6 \theta$, $ (z \plus{} 1/z)^4 \equal{} 2^4 \cos^4 \theta$, and $ (z \plus{} 1/z)^2 \equal{} 2^2 \cos^2 \theta$. Substituting this is the expression * we get as the final result

$ 2 \cos(6 \theta) \equal{} 64 \cos^6 \theta \minus{} 96 \cos^4 \theta \plus{} 36 \cos^2 \theta \minus{} 2$,

as we wanted.[/hide]

[hide="Number 3"]$ \frac{\sin \theta \plus{} \sin 2 \theta \plus{} \sin 3 \theta}{\cos \theta \plus{} \cos 2 \theta \plus{} \cos 3 \theta} \equal{} \frac{\sin \theta \plus{} \sin 2 \theta \plus{} \sin(\theta \plus{} 2 \theta)}{\cos \theta \plus{} \cos 2 \theta \plus{} \cos(\theta \plus{} 2 \theta)} \equal{} \frac{\sin \theta \plus{} \sin 2 \theta \plus{} \sin \theta \cos 2 \theta \plus{} \cos \theta \sin 2 \theta}{\cos \theta \plus{} \cos 2 \theta \plus{} \cos \theta \cos 2 \theta \minus{} \sin \theta \sin 2 \theta}$

$ \equal{} \frac{\sin 2 \theta \plus{} 2 \sin \theta \cos^2\theta \plus{} \sin \theta (\cos^2\theta \plus{} \sin^2\theta \plus{} \cos^2 \theta \minus{} \sin^2\theta)}{\cos 2 \theta \plus{} \cos \theta(\cos^2\theta \plus{} \sin^2\theta \plus{} \cos^2 \theta \minus{} \sin^2\theta) \minus{} 2 \sin^2 \theta \cos \theta}$

$ \equal{} \frac{\sin 2 \theta \plus{} 4 \sin \theta \cos^2 \theta}{\cos 2 \theta \plus{} 2 \cos \theta \cos 2 \theta} \equal{} \frac{\sin 2 \theta \plus{} 2 \cos \theta \sin 2 \theta}{\cos 2 \theta \plus{} 2 \cos \theta \cos 2 \theta} \equal{} \frac{\sin 2 \theta (1 \plus{} 2 \cos \theta)}{\cos 2 \theta (1 \plus{} 2 \cos \theta)}$

$ \equal{} \frac{\sin 2 \theta}{\cos 2 \theta} \equal{} \tan 2 \theta$.[/hide]

Wow, those were more difficult than I was told they would be!  Thanks for those 3  :D

As far as I can see using complex numbers with Euler's relation e^it= cos t + i sin t  ,(proven with the convergent power series)and its corollary,
de Moivre's identity (cos t + i sin t)^n = cos nt + i sint nt ,
make the first few at least, elementary.

The others require deliberate use of the sum and product formula etc.
With regard to the last,
say $ A \plus{} B \plus{} C \equal{} {\pi}$
then,
$ \tan {(A\plus{}B)} \equal{} \tan {({\pi}\minus{}C)}$  and by the projections,
$ \tan {(A\plus{}B)} \equal{} \minus{}\tan {C}$
plug in the addition formula and,
$ \tan {A} \plus{} \tan {B} \equal{} \tan {A} \tan {B} \tan {C} \minus{} \tan {C}$
and the result follows!


Prove
\[\sqrt[n]{n!}\leq \prod_{p|n!}p^{\frac{1}{p-1}}\]
[hide="solution"]
Lemma 1: $\lfloor \frac{n}{p}\rfloor+\lfloor \frac{n}{p^{2}}\rfloor+... \leq \frac{n}{p-1}$

Proof: remove the floor functions to get
\[\frac{n}{p}+\frac{n}{p^{2}}+... = \frac{n}{p-1}\]
Now, we note that
\[\prod_{p|n}p^{\frac{n}{p-1}}\geq \prod_{p|n}p^{\lfloor \frac{n}{p}\rfloor+\lfloor \frac{n}{p^{2}}\rfloor+...}= n! \]
Simply take the nth root and we are done. [/hide]

It looks nice but it is not true. Take $n=4$ to affirmate.
To improve just replace $p|n$ by $p\le n$ under product in RHS.

[quote="Phelpedo"]$\prod_{p|n}p^{\lfloor \frac{n}{p}\rfloor+\lfloor \frac{n}{p^{2}}\rfloor+...}= n!$[/quote]

Of course, here is the fatal flaw.  Replace $p | n$ with $p | n!$.

[quote="t0rajir0u"]Replace $p | n$ with $p | n!$.[/quote]
If $p$ is a prime then
$p|n! \iff p\le n$

Ah yes, excuse the typo. I think the $p|n!$ goes along more with the "spirit" of the problem, so I"ll put that.

Oh, I have nothing against spirit.  :clap2:


[quote="ayush_2008"]a man has infinite amount of money. he distribute his money in n people in a multiple of 4,such that each person does not get more than 4n-1 rupees..in how many ways he can do that.[/quote]

Once is enough don't post twice!!
See here

http://www.mathlinks.ro/viewtopic.php?p=1643637#1643637

[quote="-Elixir-"]What do you mean: the total amount of money he gives out is a multiple of 4, or that each person gets a multiple of 4?[/quote]
total money to be given is  multiple of 4


I was discussing this with the user boxedexe by PM.
Instead I post it here on forums for other people to see.

[b]Problem:[/b] Show that if a rectangle is tiled with smaller rectangles that have the property that at least one side is of integer length then show that the big rectangle has at least one side of integer length.

[hide="As explained by Zeitz"]
I am going to be a little sloppy but I hope it makes sense.

1)Draw a coordinate plane so that the lower left corner is on a lattice point. (You will see why soon).

2)Now if this rectangle failed to have a side of integer length then that would mean only 1 corner is a lattice points. So we need to show that either 2 or 4 corners are lattice points. Hence this big rectangle has the above property iff the number of lattice corner points is a positive even integer.

3)Now the smaller rectangles have the same property. Either one has 0,2,4 corner lattice points. Always an even number.

4)Let S be the sum of all the total number of lattice points. It is even.

5)Note that is can be possible that we over counted some of the lattice corner points on the smalller rectangles. The next step says the general rule.

6a)The corner points of the big rectangle were counted exactly once and those are the only ones. Call then $ c_{1}$.

6b)The inside corner points are counted twice or four times. Never one of three times. Call them $ c_{2}$ and $ c_{3}$ respectively.

7)Then $ S=c_{1}+2c_{2}+4c_{2}\implies c_{1}= S-2c_{2}-4c_{2}$.

8)So $ c_{1}$ is even cuz $ S$ and $ -2c_{2}-4c_{2}$ are even.

9)That proves the claim.
[/hide]

Here's another proof; colour the smaller rectangles which have a vertical integer side red. Colour the rectangles with horizontal integer sides blue. If there is a red "path" from one side of the larger rectangle then we're done. Suppose there is not such a path; this implies that the makings of a completed path have been blocked, but for this to happen there must be a completed (horizontal) path of blue tiles whence again there is an integer side.

@EBz89:your proof is completely correct  :) ,and the beautiful idea that you used is a very useful one,for example there is another problem which can be proved by a (somehow) similar idea:
[b][hide]we have colored the cells of a $ m*n$ table arbitary into black and white.prove that either a rook can go along from the first row to the last row s.t it only passes thruogh black cells,or a king can go from the first column to the last column s.t it passes only thruogh white cells.[/hide][/b]

You can also place the rectangle in the plane such that its bottom left corner is at the origin and one of its sides on the positive x-axis. Colour the larger rectangle in a chessboard fashion such that each lattice square is coloured using $ 4$ squares.  Since at least one side of each smaller rectangle is a rectangle, each smaller rectangle covers an equal area of black and white.  This implies that the larger rectangles also covers an equal area of black and white.  It can be easily shown that this can happen if and only if at least one side of the larger rectangle is an integer.


inequalities
geometry
circumcircle
perimeter
trigonometry
inequalities proposed

Consider a triangle $ ABC$ and its circumcircle $ \Gamma(O,R)$. Let $ P$ be a point inside the triangle $ ABC$, and let $ A_1,B_1,C_1$ be the second intersection points of the lines $ AP,BP$ and $ CP$, respectively, with the cicumcircle $ \Gamma(O,R)$.Prove that:
$ S[ABC_1] \plus{} S[BCA_1] \plus{} S[CAB_1]\leq p(R \minus{} r)$,

where $ p,R,r$ are the well-known commonly used notations for the triangle $ ABC$.

I assume, by $ p$ you denote the semi-perimeter which I shall denote by $ s$. Then our inequality is equivalent to
\[ [ABC_1]+[BCA_1]+[CAB_1]+sr\le sR\iff [AC_1BA_1CB_1]\le sR\]
since $ sr=[ABC]$. Now verify that
\begin{align*}[AC_1BA_1CB_1]=\frac 12R^2(\sin\angle AOC_1+\sin\angle C_1OB+\angle BOA_1+\sin\angle A_1OC+\sin\angle COB_1+\sin\angle B_1OA)\end{align*}
First assume that $ \triangle ABC$ is acute i.e. $ O$ lies inside the perimeter of triangle $ ABC$. Then each of the angles on the right hand side of the above equality are in $ [0,\pi]$ and hence by Jensen's inequality we get
\[ \&\sin\angle AOC_1+\sin\angle C_1OB\le2\sin\frac{\angle AOB}{2}=2\sin C\]
Similarly,
\begin{align*}
&\sin\angle BOA_1+\sin\angle A_1OC\le 2\sin A\\
&\sin\angle COB_1+\sin\angle B_1OA\le 2\sin B\end{align*}
Therefore
\[ [AC_1BA_1CB_1]\le R^2(\sin A+\sin B+\sin C)=Rs\]
which was what we wanted.

I haven't thought much about the case when $ \triangle ABC$ is obtuse. I have an exam tomorrow so I'll think about it later.


summer program
Mathcamp
function
PROMYS
Ross Mathematics Program

er.......can anyone tell me For what annual income of a family does the mathcamp give full financial aid (including travel) for a student .
One more question ,for entering the mathcamp program held during july-august 2010 should we try and post the solutions for the qualifying quiz 2010 or will there be a separate qualifying quiz 2010 for the 2010 applicants?

precedently has there been people from my country india in these camps?? i am really interested and hope someone will reply.

There was one person from India this year, Aniruddha (Ani) Bapat. There were probably others in other years.

er........ya i know brutus (anirudhha) has been attending the camp from india
But I dont think he is active in the website now .
anyway.........what about the financial aid ?

Mathcamp used to offer automatic fee deductions based on a family's annual income and additional scholarships based on demonstrated merit/need. The automatic deductions (based on a copy of a family's tax return) were replaced with the aid application to allow more individualized assistance. I believe it is up to the administrators' discretion when to give a full scholarship for tuition and/or travel. That being said, Mathcamp gives very generous financial aid (over 100,000 US dollars annually, as the official site says), so is highly encouraged that you at least apply. If you are accepted, they'll work with you personally to enable you to attend.

There will be a new Qualifying Quiz released sometime around January 2010. We can only discuss it after the end of ALL applications (usually around May or June, including waitlists).

I hope that helps. If you need more clarification, please feel free to ask. :)

So if I apply and after some time if my parents are not able to pay I can withdraw right ?
Do you know anyone who have got a full aid ?
Thanks  for the info friend

Yes; you're not at all obligated to go if you are accepted. So why not apply? :)
  I think I know someone who got full aid once; however, I'm not absolutely sure about this and it's probably not very common to get it, if people have gotten it.
  Last year, I wasn't sure if I could go to camp because of the cost, but in the end received a good scholarship and was able to go :) . It wasn't full aid, but it was what we needed--MC doesn't want people who got in to not go just because of the cost, so they try to find a way to make it work. Also, MC is really, [i]really[/i] worth it.

Wow! I am happy at this info .Thanks friend .
Obviously if I get an aid which will make me attend ,I will be very happy.
Just one more question , In US you all have your vacation at that time when the camp functions . (july-august)
What about other internationals whose schools might be functioning that time  .
In my country we have summer vacation at the time april-june .

Anyway at any cost I want to attend the camp ,would be happy to be off with the horrilble uninteresting school  for a month :rotfl:  :rotfl:  :D

Hi Shrinath!

These good programms do give near complete fin aid and travel aid provided your family income is not very high. In any case your family income converted to dollers will reduce to a great extent..so do not worry abt fin aid and as of now concentrate on getting selected.

All Indian students who make it to US summer programmes have to miss school/college for 5 to 6 weeks and your family and your school should be ok with it. Mathcamp, SSP, RSI, PROMYS and ROSS ...all these programmes are really worth attending even at the cost of missing school/college and paying for travel.

All the best to you!

Which class are you in? Where are you studying?

@Srinath: I attended MC this year, and loved the experience. If you enjoy maths, this is the place for you.
I had similar doubts regarding financial aid last year, but as was pointed out: MC tries to ensure that anyone who qualifies should not find camp fees a problem.
All the best! Hope to see you at Mathcamp next year.

thnx for the info i am gonna try my best to qualify and btw if i attend this year am i automatically registered for the next yr or something?

[quote="swanand"]@geniusbliss
Yes you can attend provided you have not joined a college and they invite you again! :maybe:[/quote]

To further clarify, if you want to attend mathcamp for a second (or third, or fourth, etc.) time you do not need to reapply, you just need to tell them that you are coming. (The one exception to this is that if too many [i.e. >55] alums want to come back then the people who would be attending for their third consecutive year would have to skip that year. To the best of my knowledge this rule has not been applied since it was added [after '07]).

[quote="srinath.r"]
oh thanks friend ,I will try my level best to clear the quiz
Can you tell me what all I should prepare for so that I can do my best at the quiz in jan
[/quote]

Remember that the quiz is open-book open-internet open-anything and you have like 3-4 months to work on it, so I wouldn't worry much about studying for it. If you think you need to learn something in order to solve a problem, you're more than welcome to study that after you've looked at the quiz. (Also remember that these problems are designed not to need too much advanced knowledge.)


inequalities
geometry
inequalities proposed

[quote="vu thanh tung"]if x,y,z such that: xy + yz +zx >=0,x >0, y >0 ;a, b, c are 3 sides of a triangle; S is its area then:
(y + z)bc + (z + x)ca + (x + y)ab - xa^2 - yb^2 - zc^2  >= 4sqrt(yz + zx + xy)S[/quote]
If $x,y,z$ such that: $xy + yz +zx \geq0,x >0, y >0$ ;$a, b, c$ are 3 sides of a triangle; $S$ is its area then:
$(y + z)bc + (z + x)ca + (x + y)ab - xa^2 - yb^2 - zc^2 \geq  4S\sqrt{(yz + zx + xy)}$
Looks better now. :huh:

[quote="hungkhtn"]I had seen It some times. As I know, It has three nice solutions. But....  It's not new.[/quote]
I found it when I was in high school, inspired from a similar $xa^2+yb^2+zc^2\geq 4\sqrt{yz + zx + xy}S$


linear algebra
matrix
linear algebra unsolved

$ A$ in $ S^{\plus{}\plus{}}(n,R)$  real symetric definite positive matrix

$ B$ in $ S(n,R)$ real symetric matrix

prove that $ AB$ is diagonalisable

$ AB$ is similar to $ A^\dfrac{1}{2}BA^\dfrac{1}{2}$.

yes loup blanc
-----------------------------------------

Now suppose
$ A \in S^{\plus{}}(n,R)$  real symetric positiv (not definite) matrix

$ B \in S(n,R)$  real symetric matrix

Does $ AB$ still diagonalisable ?

The answer is NO:
Take $ A\equal{}\begin{pmatrix}1&1\\1&1\end{pmatrix}$ and $ B\equal{}\begin{pmatrix}0&\minus{}1\\\minus{}1&2\end{pmatrix}$.
Then $ AB\equal{}\begin{pmatrix}\minus{}1&1\\\minus{}1&1\end{pmatrix}$ is nilpotent.


Hi!
I very like [b]inversion[/b] of geometry. But I don't have a nice book about inversion. So, I want help from you. Thank you very muck.

[url=http://www.maths.gla.ac.uk/~wws/cabripages/inversive/inversive0.html]This[/url] might be helpful. :)

Thank you very much. but this page have a little of exercise and home work. Can you help me more.

Oh! Hi!!!!! Can you help me????
 :(   :maybe:   :o

[quote="thanhnam2902"]OH! Why no body help me? :oops:  :mad:[/quote]
dont be anxious,take care of ure self,look at the pm,
there are many links about inversion posted in this theorems and formulas,
http://www.mathlinks.ro/viewtopic.php?t=14958


algebra
polynomial
function
algebra unsolved

Find all polynomial funtion $ P(x)$ that
$ P(xy)\equal{}P(x)P(y)$ for all $ x,y$ real.

Answer: $ 0,1,x^n$.
[hide="Hint"]If $ a_nx^n\plus{}\cdots\plus{}a_0\equal{}b_mx^m\plus{}\cdots\plus{}b_0$ then $ m\equal{}n$ and $ b_i\equal{}a_i$.[/hide]


geometry
circumcircle
geometric transformation
trigonometry
power of a point
radical axis
perpendicular bisector

Let $ABC$ be a triangle and denote by $D,E,F$ the foot of the internal angle bisectors such that $D \in (BC)$, $E \in (CA)$, $F \in (AB)$.  $(I_a,r_a),(I_b,r_b),(I_c,r_c)$ are their three excircles. Prove that there exists a point $M$ on the incircle of the triangle $ABC$, such that:
\[MD \cdot \frac{I_{a}I}{I_{a}D}\cdot \sqrt{R+2r_{a}}\pm ME \cdot \frac{I_{b}I}{I_{b}E}\cdot \sqrt{R+2r_{b}} \pm MF \cdot \frac{I_{c}I}{I_{c}F}\cdot \sqrt{R+2r_{c}}=0\]

Let $(I,r),(O,R)$ be the incircle and circumcircle of $\triangle ABC.$ Since $IAI_bC$ and $IAI_cB$ are both cyclic, we have $\overline{EI} \cdot \overline{EI_b}=\overline{EA} \cdot \overline{EC}$ and $\overline{FI} \cdot \overline{FI_c}=\overline{FA} \cdot \overline{FB},$ i.e. $E,F$ have equal powers WRT the circumcircles $(U),(O)$ of $\triangle II_bI_c$ and $\triangle ABC$ $\Longrightarrow$ $EF \perp OU$ is the radical axis of $(O),(U).$ Since $I_a$ is the exsimilicenter of $(O) \sim (U),$ then $OI_a \perp EF.$ Since $(O)$ cuts $I_aI_c$ again at the midpoint of its arc $ABC,$ it follows that $OQ \perp AC$ is the perpendicular bisector of $\overline{AC},$ thus $\angle QOI_a=\angle AEF.$ By sine law for $\triangle BOI_a$ and $\triangle QOI_a,$ we obtain

$\frac{\sin \widehat{OI_aB}}{\sin \widehat{OBI_a}}=\frac{\sin \widehat{OI_aB}}{\sin \widehat{BEA}}=\frac{R}{OI_a} \ , \ \frac{\sin \widehat{QOI_a}}{\sin \widehat{OI_aB}}=\frac{QI_a}{R}=\frac{I_bI_c}{2R} \Longrightarrow$

$\frac{\sin \widehat{QOI_a}}{\sin \widehat{BEA}}=\frac{\sin \widehat{AEF}}{\sin \widehat{BEA}}=\frac{I_aI_c}{2OI_a} \Longrightarrow EF= \frac{2 \sin A}{ \sin \widehat{BEA}} \cdot \frac{AF \cdot OI_a}{I_aI_c} \Longrightarrow$

$EF=\frac{2BE}{bc} \cdot \frac{bc}{a+b} \cdot OI_a \cdot \sin \frac{B}{2}=\frac{4ac}{(a+b)(a+c)} \cdot OI_a \cdot  \cos \frac{B}{2} \sin \frac{B}{2}=$

$=\frac{abc}{(a+b)(a+c)} \cdot \frac{OI_a}{R}=\frac{abc}{(a+b)(a+c)} \sqrt{ \frac{R+2r_a}{R}} \ \ (1)$

By cyclic exchange of elements, we get the expressions for $FD,DE$

$FD=\frac{abc}{(b+c)(b+a)} \sqrt{ \frac{R+2r_b}{R}} \ \ (2) \ , \ DE=\frac{abc}{(c+a)(c+b)} \sqrt{ \frac{R+2r_c}{R}} \ \ (3)$

Let $M$ be an arbitrary point on the circumcircle of $\triangle DEF,$ different from $D,E,F.$ By Ptolemy's theorem for $DEFM$ we get $MD \cdot EF \pm ME \cdot FD \pm MF \cdot DE=0.$ Substituting $(1),(2)$ and $(3)$ into the latter equation and simplifying terms gives

$MD \cdot (b+c) \sqrt{ R+2r_a} \pm ME \cdot (c+a) \sqrt{ R+2r_b} \pm MF \cdot (a+b) \sqrt{R+2r_c}=0$

But $\frac{I_aI}{I_aD}=\frac{r+r_a}{r_a}=\frac{b+c}{s} \ , \ \frac{I_bI}{I_bE}=\frac{r+r_b}{r_b}=\frac{c+a}{s} \ , \ \frac{I_cI}{I_cF}=\frac{r+r_c}{r_c}=\frac{a+b}{s}$

$MD \cdot  \frac{I_aI}{I_aD}\sqrt{ R+2r_a} \pm ME \cdot  \frac{I_bI}{I_bE}\sqrt{ R+2r_b} \pm MF \cdot  \frac{I_cI}{I_cF}\sqrt{R+2r_c}=0$

Thus, $M \in (I)$ is either the Feuerbach point $X_{11}$ or the 8th Stevanovic point $X_{3024}.$


Here is the disproof of Godel's theorem:

Gödel accepted the false but widely held tradition that Richard argued that truth in number theory cannot be defined in number theory.  It turns out that what is undefined in Richard’s argument (as he himself pointed out) is the number crucial to making the argument.  However, Gödel added to Richard’s argument the idea that provability in number theory can be defined in number theory, and came up with mistaken result that if the provable formulae are all true, then there must be some true but unprovable formulae.  An enticing notion, except that in his argument there are no formulae.

Well, the fact that the continuum hypothesis has been shown to satisfy Godel's theorem, Godel's theorem is back to being true.

[quote="tokenadult"]Check [url=http://64.233.161.104/search?q=cache:eVFqwZOCvUgJ:www.physicsforums.com/showthread.php%3Fgoto%3Dlastpost%26t%3D59016+John+Ryskamp]this[/url], moderators.

[i]Moderator responds: Thanks for the tip.  I'm inclined just to let this post float its way down the forum, since it does not, of itself, comprise objectionable behavior.[/i][/quote]

The link doesn't work anymore.

I believe physicsforums got really mad at him for spamming and deleted him and all his posts.

Google revealed an ENORMOUS spam volume at the time, some of which has since been deleted from other forums.


polynomial
algebra
algorithm
number theory
greatest common divisor
superior algebra
superior algebra unsolved

While writing tests to assure that my implementation of Eisenstein works correctly
I accidentally stumped over this polynomial $ (X\plus{}1)(X\plus{}2)(X\plus{}3)\equal{}x^3 \plus{} 6x^2 \plus{} 11x \plus{} 6$
The algorithm I'm implementing is trying to apply Eisenstein to every polynomial(which  seems to be wrong to start with).
So the algorithm is , if gcd(all coefficients of a polynomial except for the one of the dominant term) is different from 1 then apply eisenstein because there will be a prime satisfying the conditions needed to apply it.
If not, then succesively shift with 1,2,3,... until the above mentioned gcd becomes greater than 1 so there will be a p that divides all coefficients(except for the dominant term's coefficient) so we can apply Eisenstein.
But the above mentione polynomial is extremely weird in the sense that no matter how much
I shift, it's coefficients are relatively prime.
(I don't know how to prove this but I've tested until about $ f(x\plus{}200)$ or so )

For example here's some pairs of the coefficients(except for the first dominant term which is of less interest):

shift $ f(x\plus{}1)$ gives coefficients: 24,26,9
shift $ f(x\plus{}2)$ gives coefficients: 60,47,12
shift $ f(x\plus{}3)$ gives coefficients: 120,74,15
shift $ f(x\plus{}4)$ gives coefficients: 210,107,18
shift $ f(x\plus{}5)$ gives coefficients: 336,146,21
shift $ f(x\plus{}6)$ gives coefficients: 504,191,24
shift $ f(x\plus{}8)$ gives coefficients: 720,242,27

So my question is , to which polynomials in Z[X] can I apply Eisenstein.
This polynomial, is there a way to apply Eisenstein to it ?

What methods other than Eisenstein are there to decide if a polynomial is  reducible/irreducible ?

P.S.: It's obviously reducible but I'm faking I don't know it is and I want to prove it by using Eisenstein.

Per the criterion I mentioned in your other Eisenstein thread, since $ \rm{disc}(f) \equal{} \minus{} 4$ we need only check if $ f(x)$ or $ f(x \plus{} 1)$ is Eisenstein for $ p \equal{} 2$. It's not:  $ f(x)$ has odd coef of $ x$ and $ f(x\plus{}1)$ has constant coef divisible by 4.


Given $ f(x) \equal{} x^2 \plus{} 1/5$. Prove that:

$ f_{10}(x) \geq x^4$

I think this question was posted before in the form:

$ f_{5}(x) \geq x^2$

I hope this is true: $ f(x) \geq x^c$ with $ c^5 \equal{} 2$

moderators, plz help me. I remember that question was post before.

Anyway, thx, I did solve it.  :roll:

[Moderator edit: This follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=61954 .]


limit
number theory proposed
number theory

$n$ is powerful if for any prime divisor $p$ of $n$, $p^{2}$ also divides $n$.
Prove that there exist $k$ consecutive numbers that are not powerful.

If I'm not drunk now, then the sum of inverses of all powerful numbers is convergent, so the conclusion follows.

Let $g(N)$ is number of powerful numbers, less or equal N. It is not hard to show, that \[\lim_{N\to \infty }\frac{g(N)}{N}=0.\] Therefore for any $k\le \frac{N}{g(N)+1}$ we have consequitive k numbers less than N and not powerfull numbers.

Let $p_{1},p_{2},\ldots,p_{k}$ be distinct prime numbers. Due to the Chinese Remainder Theorem there exists $x$ such that \[\left\{ \begin{array}{ll}x\equiv p_{1}-1 (mod\ p_{1}^{2}) \\ x\equiv p_{2}-2 (mod\ p_{2}^{2}) \\ x\equiv p_{2}-3 (mod\ p_{3}^{2}) \\ \ldots \\ x\equiv p_{k}-k (mod\ p_{k}^{2}) \\ \end{array}\right.\] It's easy to notice, that all the numbers $x+1, x+2,\ldots,x+k$ are not powerful.

can we ask for existence of k con. squarefree nos. :welcomeani:  for any natural k?

It is not true. If k>3, then exist x,suth that $4|x$.

[quote="who"]sorry,neglect powers of 2.[/quote]
Then for $k\geq9$ one of this numbers is divisible by 9...

I think that more interesting is, for example, to show that there are infinitely many pairs of two consecutive powerful integers.

Start with 8 and 9 and if $(x,x+1)$ is a pair, so is $(4x(x+1), (2x+1)^{2})$.

It is simple question. For example we can consider Pel's equation $x^{2}-Ay^{2}=\pm 1$. It had infinetely many solution with $rad(A)|y$. It give solution $(x^{2}-1,x^{2})$ or $(x^{2},x^{2}+1)$.
It is interesting question about 3 consequitive powerful numbers.


A few girls ask Lindsay for her weight (for some reason). Lindsay would not give it away directly, but she will give a riddle. She says, "Turn 0.025555... into a common fraction, and the numerator would be my weight in pounds."
What is Lindsay's weight in kilograms (1 kg is 2.2lbs)? Express your answer as a decimal to the nearest hundredth.

I think there's a sensible way to do this problem but I never knew how to do all this decimal conversion stuff so Im just going to do it the most savage way I know:

[hide]
infinite series -_-;;

.0255555 = .02 + .00555555.. = .02 + (5/10^3 + 5/10^4 + ...) = .02 + (5/10^3)/(1-1/10) = 23/900.

23 lb?  If I did it right, she's pretty clever for someone that young.  23 lb / 2.2(lb/kg) = 10.45 kg...

Something seems fishy.
[/hide][/hide]

[hide]
x = 0.02555[u]5[/u]

100x = 2.555[u]5[/u]
1000x = 25.555[u]5[/u]
900x = 1000x - 100 x = 25.555[u]5[/u] - 2.555[u]5[/u] = 23
[tex]x = \frac{23}{900}[/tex]

Lindsay's weight is 10.45 kg...er....yea....there must be a problem  :blush:
[/hide]

EDIT: OK, I'm not the only one who has found this  :D
Barnacle, I understand nothing to what you did but it seems you have found the right weight  :clap: how did you do ?

[hide="clik dis!"]i put the decimal into a fraction:
23/90
23lb is how much Lindsay weighs(whoa, that's messed up)
convert 23 to kilograms and you get$\boxed{10.45 \text{kg}}$[/hide]

Hi Tanynep,
Here's what I did:

[hide]
I used infinite series.  There is a simple formula to sum a geometric infinite series.  An infinite geometric series is a series of the form:

a + ar + ar^2 + .... [going on forever]

For example if a = 1 and r = 2, we get the infinite geometric series 1 + 2 + 4 + 8 + ....

It turns out that if -1<r<1, then the sum of the infinite geometric series doesn't go off to infinity (as it does in the example series 1 + 2 + 4 + 8 + ... where r = 2) but instead converges to a sum S = a/(1-r)   [easy to show, but I won't bother].

In this case, notice that .005555... is a series of that form where a = 5/1000 and r=1/10:

.005555... = 5/1000 + 5/10000 + 5/100000 + .... = 5/10^3 + 5/10^4 + ....

Thus we can use that formula to get the sum in a fractional form:

S = a/(1-r) = 5/10^3 / (1-1/10) = (5/1000) / (9/10) = 1/180.

Then we add this to the .02 that we took out initially so that our final sum is:

.02 + 1/180 = 2/100 + 1/180 = [b]23/900.[/b]

Clearly your way is much better.  Thanks for showing it to me.  ;) [/hide]

[hide]Barnacle, a simple way to express an infinate expression without summations is:

$a + ab + ab^2 + ab^3 + ...$

These were just random variables.[/hide]

Thank you for showing me your way too, Barnacle, maybe it's more complicated but this formula is very interesting  :)

Can you just tell me if I have understood ?
If I have [tex]a = 10[/tex] and [tex]r = \frac{1}{2}[/tex], the sum is :

[tex]S = 10 + \frac{10}{2} + \frac{10}{4} + \frac{10}{8} + \frac{10}{16}[/tex]....
And it can be calculated like this : [tex]S = \frac{a}{1 - r} = \frac{10}{\frac{1}{2}} = 20[/tex]
Am I right ?

Thanks again,

yup thats correct!

a fun series problem to test your ability to see when to use that formula is to evaluate S where S is:

S = 1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32 + ... etc.

you made a small error in the second part for getting S2.  r was 1/4 so the denominator should be 3/4.  Otherwise its right.

alternatively you could've said:

S = (1 - 1/2) + (1/4 - 1/8) + (1/16 - 1/32) + ...
   =  1/2 + 1/8 + 1/32 + ...

a = 1/2, r = 1/4

S = 1/2/(1-1/4) = 1/2 / 3/4 = 2/3

I tried to give a weight that would throw everyone off. Ever thought about very short people?

Here's how I do repeating decimals into fractions.

Let 0.025555 be n.
Then 1000n = 25.55555
And 100n = 2.555555
Subtract.
.555555 cancels

900n = 23

n = $\frac{23}{900}$

Her weight is 23 pounds, or 10.45 kilograms.

[quote="Barnacle"]Yeah factorial thats how your supposed to do it.  I'm just dumb.  :ninja:[/quote]
Well, you can also see that .02555555 is equal to
0.02 + 0.0055555, and since you know that .02 is 2/100 and .005555 is 5/900, the sum of those is 18/900 + 5/900 or 23/900.
I'm not sure if that is at basics level however. So you are not dumb.  ;)