function
quadratics
algebra
quadratic formula
complex numbers

[quote="saudaltamimi"]hello guys
determain the vertex and x intercept and y intercept of the given quadratic function ?

1-h(x) =(x+1)^2   + 4
2-f(x)  =2x^2   -4x   +2
3-h(x) = -(x+2)(x+6)[/quote]

Here is a hint: $ f(x) \equal{} a(x\minus{}h)^2 \plus{} k$ where $ (h,k)$ is the vertex.  Also, the y-intercept occurs when $ x\equal{}0$.

[hide="1."] Let $ y\equal{}h(x)$. The equation can be rewritten as $ y\minus{}4\equal{}(x\plus{}1)^2$. Thus, the vertex is $ (\minus{}1,4)$. The $ x$-intercepts are the ponts were $ y\equal{}0$. Expanding, we have $ 0\equal{}x^2\plus{}2x\plus{}5$. This has complex roots, so no $ x$-intercepts in the set of real numbers. Finally, the $ y$-intercept is where $ x\equal{}0$. Plugging this in, we have $ y\equal{}5$, so the $ y$-intercept is at the point $ (0,5)$.[/hide]

[hide="2."] Once again, let $ y\equal{}f(x)$. Then the expression can be factored as $ y\equal{}2(x\minus{}1)^2$. Thus, the vertex is a t $ (1,0)$. This point is also the $ x$-intercept, since its the point where $ y\equal{}0$. Plugging in $ x\equal{}0$, we find that $ y\equal{}2$, so the $ y$-intercept is at $ (0,2)$. [/hide]

I'll leave the last one up to you. :)

[hide="Solutions"]Note: $ a(x \minus{} h) \plus{} k \implies V(h,k)$. We can say:
1) $ V( \minus{} 1,4)$
y-intercept: $ x \equal{} 0 \implies h(x) \equal{} 5$
x-intercept: $ (x \plus{} 1)^2 \plus{} 4 \equal{} 0 \implies x^2 \plus{} 2x \plus{} 5 \equal{} 0 \iff \Delta \equal{} \minus{} 16$ so there is no x-intercept in this function.

2)$ V\left(\frac { \minus{} b}{2a},f\left(\frac { \minus{} b}{2a}\right)\right) \implies V(1,0)$
y-intercept: $ x \equal{} 0 \implies f(x) \equal{} 2$
x-intercept: By the quadratic formula we found that the only x-intercept is $ x \equal{} 1$

3)Expanding we get $ \minus{} x^2 \minus{} 8x \minus{} 12 \implies V\left(\frac { \minus{} b}{2a},f\left(\frac { \minus{} b}{2a}\right)\right) \implies V( \minus{} 4,4)$
y-intercept: $ x \equal{} 0 \implies h(x) \equal{} \minus{} 12$
x-intercept: By the quadratic formula we have $ x_{1,2} \equal{} \frac {8 \pm 4}{ \minus{} 2} \implies x_1 \equal{} \minus{} 6,x_2 \equal{} \minus{} 2$[/hide]

Let me edit your problem to make it easier to look at:

Determine the vertex and $ x$-intercept and $ y$-intercept of the given quadratic functions.

1-$ h(x)\equal{}(x\plus{}1)^2\plus{}4$
2-$ f(x)\equal{}2x^2\minus{}4x\plus{}2$
3-$ h(x)\equal{}\minus{}(x\plus{}2)(x\plus{}6)$

[hide="Solution"]The vertex of any given quadratic function is $ (h, k)$ if the function is written in the form $ y\equal{}a(x\minus{}h)^2\plus{}k$
Therefore, right off the bat we can state the vertex of function 1 is $ \fbox{(\minus{}1, 0)}$.

To determine the y-intercept, we set the $ x$ terms equal to zero.  We get $ \fbox{(0, 5)}$.

The x-intercept(s) of any function are known as their [b]roots.[/b]

Expanding 1, we get $ x^2\plus{}2x\plus{}5\equal{}0$.  Using discriminant, we see that $ b^2\minus{}4ac$ is less than 1, meaning there are [u][i][b]no real solutions.[/b][/i][/u]  You could, in theory, figure out the roots in complex numbers, but the graph does not intersect the x-axis.

For 2, we can pull 2 out, and get $ 2(x^2\minus{}2x\plus{}1)$, or $ 2(x\minus{}1)^2$.  The vertex of this equation is $ \fbox{(1, 0)}$, the y intercept is $ \fbox{(0, 2)}$, and the x-intercept is $ \fbox{(1, 0)}$ as well.

For 3, the roots, or the x-intercepts are clearly $ \fbox{(\minus{}2, 0)}$ and $ \fbox{(\minus{}6, 0)}$.  For the y-intercept, we substitute $ x\equal{}0$.  We get $ \minus{}12$, so the y-intercept is $ \fbox{(0, \minus{}12)}$.  For the vertex, we need to expand and complete the square.  We get: $ \minus{}(x^2\plus{}8x\plus{}12)$, or $ \minus{}(x\plus{}4)^2\minus{}12$, meaning the vertex is $ \fbox{(\minus{}4, \minus{}12)}$.[/hide]

that was fast wow i didnt think people will replay that quick thanks alot


limit
function
real analysis
real analysis unsolved

I think (though I'm not sure) that some trigonometric functions might work, but I can't find a "smart" substitution. I tried this: a_n=1/x_n. We get a_n <1 for any n>=1. If we put a_n=sinx then a_(n+1)=tan(x/2). It would have been nice to get sth with sin instead of tan but... Who knows? maybe this doesn't even help...

Where did you find this anyway? I'm getting pretty obsessed withit! :D  :D

Err, that wasn't a joke...

The problem was taken from Gazeta Matematica 10/2002. I also thought about trigonometric functions but got no result. Maybe if we use arcsin or arctan...

I found it there too (there was a more general case there, with x1>a^1/2 and x_(n+1)=x_n+(x_n^2-a)^1/2, but it makes no notable difference).

I remember I've struggled with it for a long time and found nothing. I showed it to my teacher and he couldn't solve it either.

I've been trying a different approach:
Let b_n=(x_n)-1. By doing all the calculations we get 2*b_n/b_(n+1) = b_(n+1)/(b_(n+1)+1). By multiplying all these relations from 1 to n we get (after multiplying with 2 and replacing b_i with (x_i)+1)

L*(x1-1)=2*Product {i>=2} (1-1/xi) (*)

Then, by using x_n=(x_(n+1)^2+1)/2x_(n+1) and doing sth similar to the above, we get

L*x1=2*Product {i>=2} (1+1/xi^2) (**)

Maybe (*) and (**) help? I really doubt it but who knows? Like I said, I'm becoming obsessed...

[quote="pcalin"]Let (x n) n \geq 1 be a sequence defined by xn+1=x n+  \sqrt (x n 2 -1), for all n  \geq 1 and x 1>1. Compute the limit  n  \lim   \infty  2 n/x n.[/quote]

hi grobber have you try x_n = ch(y_n) where ch(y)= (e^y+ e^(-y))/2

we get ch(y_(n+1)) = e^(y_n)

y_(n+1) = ln( e^y_n +  \sqrt( e^(2y_n) -1) )

y_(n+1) = y_n + ln( 1+  \sqrt (1- e^(-2y_n))

y_(n+1) - y_n =  ln( 1+  \sqrt (1- e^(-2y_n)) ---> ln2 when n-->+oo

( y_n -->+oo)

Cesaro theorem => lim ( (y_n- y_(n-1))+...+(y2-y_1))/(n-1) = ln 2

y_n ~ ln2^n

x_n = ch(y_n) = (e^(y_n)+e^( -y_n))/2

we have

2^n/x_n ~ 2^n.(2/(2^n + 1/2^n))

so lim 2^n/x_n = 2

do you think it is correct ?
What is the best romanian cake ?

Thanks Moubi! I'm over my obsession now! :D  :D  :D

There is a nice generalization of this problem:

Let a be a real number and x1>0 and they are fixed. The sequence (xn)n>=1 is given by the relation: x_(n+1)=xn+sqrt(a+xn^2), for any natural n. Compute 2^n/xn.

cheers! :D :D

[quote="Lagrangia"] nice generalization

Let a be a real number and x1>0 and they are fixed. The sequence (xn)n>=1 is given by the relation: x_(n+1)=xn+sqrt(a+xn^2), for any natural n. Compute 2^n/xn.
[/quote]

Hi Lagrangia,

If a<0, a=-b with b>0, ch(y_n)=x_n/b

x_(n+1)=x_n+ \sqrt (x_n-b)=x_n+b \sqrt ((x_n/b)-1)
consider x_n = b.ch(y_n)
x_(n+1)=x_n +bsh(y_n)= b.exp(y_n)
ch(y_(n+1))=exp(y_n)
as above y_n ~ ln2^n
2^n/x_n ~2^n/b.(2/(2^n + 1/2^n)) ~2/ \sqrt (-a)

If a>0 no idea i tried sh(y_n) = x_n/ \sqrta i found y_(n+1)-y_n -->0
in this case Cesaro don't give an equivalent of y_n
hum seems more difficult  :?
chers
Moubinool

I think that if a>0 the substitution I first tried works (either that or another direct trigonometric fction, not a hyperbolic one). Have you tried that?


algorithm
number theory unsolved
number theory

A positive integer $k$ is in the set $S_{n}$ iff

(1) $k$ has at least $n$ distinct prime factors
(2) There are $n$ distinct positive integers $1=x_{1}, x_{2}, ... x_{n}$ such that
$1+x_{2}+...+x_{n}= k$ and $x_{i}$ divides $k$ for all $i$

Find all $n$ such that $S_{n}$ is not empty.

Okay let first prove that if [b]$S_{n}$ [/b]is not empty then [b]$S_{n+1}$[/b] is not empty.Let k belongs to [b]$S_{n}$[/b] then we shall prove that [b]k(k+1[/b]) belongs to
[b]$S_{n+1}$.[/b]First note that [b]k(k+1)[/b] has at least n+1 prime factors,since k has at least n and[b] (k,k+1)=1[/b].Now if [b]$\sum{x_{i}}=k$[/b], let  [b]$y_{i}$=(k+1).$x_{i}$[/b]
for i>1.$y_{1}=1$, and[b] $y_{n+1}=k$.[/b]THen clearly [b]$\sum{y_{i}}=k(k+1)$[/b]
and y-s are n+1, and all of them are divisors of k(k+1).So [b]$S_{n+1}$[/b] is not empty.
It is not hard to see that [b]$S_{3}$[/b] and [b]$S_{2}$ [/b]are empty

For n=6 k=1806.1807 is in $S_{n}$.

It remains to look the cases n=4 and n=5, but i think it is not that hard.

Wow, great solution!

Anyway, for $n=6$, which six numbers add up to $(1806)(1807)$?

you use the above algorithm starting with numbers 1+2+3=6.The first step you get 1+14+21+6=42 but unfortunately 42 has only 3 prime factors.
The next step you get 1+14.43+21.43+6.43+42=43.42, but again since 43 is prime 42.43 has only 4 prime factors. Iterating one more time this argument you get what you want since 43.42+1=1807 has two prime divisors (13 and 139)

The actual numbers are:
[b]1+14.43.1807+21.43.1807+6.43.1807+42.1807+1806=1806.1807 [/b]


geometry
circumcircle
incenter
perpendicular bisector
geometry unsolved

Through the midpoints of the sides of a triangle $T$ the lines perpendicular to the corresponding angle bisectors of a triangle pass. These lines bound a triangle $T_{1}$. Prove that the circumcenter of $T_{1}$ is the midpoint of the line segment that joins the incenter and the orthocenter of $T$.

Let $\triangle ABC$ be the triangle $T$ with incenter I and orthocenter H. Let A', B', C' be the midpoints of BC, CA, AB. The internal angle bisectors of the medial triangle $\triangle A'B'C'$ are parallel to the internal angle bisectors of the original triangle $\triangle ABC.$ The external angle bisectors of the medial triangle pass through A', B', C' and they are perpendicular to the corresponding internal angle bisectors. Thus the triangle $T_{1}$ is the excentral triangle $\triangle E_{a}E_{b}E_{c}$ of the medial triangle $\triangle A'B'C'.$ The 9-point circle (N) of the original triangle $\triangle ABC$ is the circumcircle of its medial triangle $\triangle A'B'C'.$ It passes through the foot D of the A-altitude AH and through the midpoint D' of the segment AH. Since $D'D \equiv AD \perp BC \equiv A'D,$ A'D' is a diameter of the circle (N). But (N), being the circumcircle of $\triangle A'B'C',$ is also the 9-point circle of its excentral triangle $\triangle E_{a}E_{b}E_{c},$ cutting $E_{b}E_{c}$ at A' and again at the midpoint K of $E_{b}E_{c}.$ Since A'D' is a diameter of (N), the angle $\angle A'KD'$ is also right, so that KD' is the perpendicular bisector of the segment $E_{b}E_{c}$ and $KD' \parallel AI$ (both perpendicular to $E_{b}E_{c}$). Since D' is the midpoint of AH, KD' is a midline of $\triangle AIH$ cutting IH at its midpoint P. Similarly, the perpendicular bisectors of $E_{c}E_{a},\ E_{a}E_{b}$ also cut IH at its midpoint P, they concur at P, and P is the circumcenter of $\triangle E_{a}E_{b}E_{c}.$


Is that possible to find Cos(10) or Sin(10) ?

Yup.... triple angle formula.
$\sin{\theta}= \sin{10}$.
$\sin{3\theta}= \sin{30}$.

Apply the triple angle formula and solve from there. I'm not sure you'll get a nice number though. Too lazy to do it. :)

$\cos{3\theta}=4\cos^3{\theta}-3\cos{\theta}$, then let $\cos{10}=x$

$4x^3-3x-\frac{\sqrt{3}}{2}=0$

you should note that we get some extra roots when doing this (can you see why?)

[hide="hint"]
the argument is similiar to the way that you find the nth complex roots of a number
[/hide]

then to solve the polynomial, this does not solve that elegantly, but you can use the cubic formula (look for it on google)


geometry
3D geometry
sphere
linear algebra
matrix
vector
function

Suppose I have a 3D curve in two variables, i.e. $ x(s, t)$, $ y(s, t)$ and $ z(s, t)$. For instance, the curve could be a sphere with a fixed radius. Now, suppose I transform my curve with a matrix $ M$. How can I calculate how much the area "changes" for a given $ s$ and $ t$? I'm just talking about local changes.

If I were talking about a ball instead of a sphere (which requires three variables), I could take the determinant of the jacobian matrix. But now I only have two variables, and Im interested how much a area differential changes.

It is more complicated; it depends both on the direction of the surface and the transformation matrix.

Write the surface as a vector-valued function $ f(s,t)$, with tangents $ \frac{\partial f}{\partial s}(s,t)$ and $ \frac{\partial f}{\partial t}(s,t)$. The area differential is the length of the cross product $ \left|\frac{\partial f}{\partial s}(s,t)\times \frac{\partial f}{\partial t}(s,t)\right|$. We apply a transformation by $ M$ (a more general differentiable map would work, replacing $ M$ by the derivative at the relevant point). Now, $ \frac{\partial Mf}{\partial s}(s,t)=M\frac{\partial f}{\partial s}(s,t)$ and similarly for the other partial derivative. The new area differential is $ \left|M\frac{\partial f}{\partial s}(s,t)\times M\frac{\partial f}{\partial t}(s,t)\right|$.

That's still more complicated than we'd like; we'd rather base everything on the normal to the surface, which should be possible. Call the tangents $ u$ and $ v$, and their cross product $ w$. Furthermore, let's consider the case in which $ u$ and $ v$ are the first two standard basis vectors $ e_1$ and $ e_2$. Now $ M=\begin{bmatrix}Mu&Mv&Mw\end{bmatrix}$ in block form. Now, where can we bring in cross products nicely...
The trick is to look at the inverse in the classical adjoint form: $ M^{-1}=\frac1{\det M}\begin{bmatrix}(Mv\times Mw)^T\\(Mw\times Mu)^T\\(Mu\times Mv)^T\end{bmatrix}$ This is easily verified: $ M^{-1}M=\frac1{\det M}\begin{bmatrix}(Mv\times Mw)\cdot Mu&(Mv\times Mw)\cdot Mv&(Mv\times Mw)\cdot Mw\\(Mw\times Mu)\cdot Mu&(Mw\times Mu)\cdot Mv&(Mw\times Mu)\cdot Mw\\(Mu\times Mv)\cdot Mu&(Mu\times Mv)\cdot Mv&(Mu\times Mv)\cdot Mw\end{bmatrix}$, with the off-diagonal entries clearly zero and the diagonal entries all forms of the same determinant. So then $ (Mu\times Mv)^T=\det M\cdot w^TM^{-1}$, or taking transposes, $ Mu\times Mv=\det M(M^{-1})^Tw$. This formula applies equally to any orthonormal frame. It also scales properly if we scale the lengths, and so our area scaling factor is $ \frac{|\det M(M^{-1})^Tw|}{|w|}$ for any orthogonal frame of tangent/tangent/normal $ u,v,w$. This expression doesn't have any $ u$s or $ v$s in it, so we don't even need $ u$ and $ v$ to be orthogonal anymore. Thus, the final answer: the area scales by a factor of
$ \det M\frac{|M^{-1}w|}{|w|}$, where $ w$ is a normal to the surface at that point such as $ \frac{\partial f}{\partial s}(s,t)\times \frac{\partial f}{\partial t}(s,t)$.

Jmerry has pipped me to the post with a much more developed explanation. But here is mine anyway...which I suppose broadens his first paragraph a little more gently...

Let $ r(s,t) \equal{} (x(s,t),y(s,t),z(s,t))$, where $ r$ is a vector function. We will write partial derivatives using suffices.

We consider an area element on our surface at $ (s,t)$. For simplicity we suppose that it is a parallelogram with edges parallel to the tangents $ r_s$ and $ r_t$. Then the differential vector edge of this parallelogram corresponding to differential elements $ ds$ and $ dt$ are $ dr_s \equal{} r_s ds$ and $ dr_t \equal{} r_t dt$ respectively (this can be seen from noting that the vector line segment coresponding to an increase $ ds$ is $ r(s\plus{}ds,t) \minus{} r(s,t)$ which gives the previous result when one takes the first non-vanishing term in the Taylor series, and similarly for $ dt$). Then the area of the surface parallelogram is $ dA \equal{} |dr_s \times dr_t| \equal{} |r_s \times r_t|dsdt$.

Now we multiply the surface by a matrix. We need to look at what happens the edges of the parallelogram. This is quite simple, since $ M$ represents a linear map, and coordinate differentials transform in the same way to the coordinates themselves. So the edges of the parallelogram under the matrix multiplication become $ M dr_s \equal{} (Mr_s)ds$ and $ M dr_t \equal{} (Mr_t)dt$. So the area of the new parallelogram is $ dA' \equal{} |(Mr_s)ds \times  (Mr_t)dt| \equal{} |Mr_s\times Mr_t|dsdt$. And we see that the fractional change in area is $ \frac{|Mr_s\times Mr_t|}{|r_s \times r_t|}$

That's awesome, thanks a lot!

Actually, what I want to do is the following (it's for computer graphics). I have a sphere, from which I want to select some truly random points. So I generate two random numbers $ s, t \in [0,1]$ and use them as the two angles. They are completly random because of the symmetry.

But I'd like to do this with a sphere that has been transformed with a matrix. So I thought that I could build a distribution function from the "area differential" function, hence the areas that will be transformed to much larger areas will have a higher probability of containing my random point. Will this work, and are there any better ways? Perhaps the calculus forum is the wrong place for statistical questions :)

Careful with your "random" points- if you do that the naive (uniform) way in the usual spherical coordinates, you're going to bias your distribution toward the north and south poles, where there is much less area in a given latitude band.

The simplest way to get something uniform on the surface of a unit sphere is to choose the $ z$ coordinate uniformly on $ [\minus{}1,1]$ and the longitude $ \theta$ uniformly on $ [0,2\pi]$. Then the final point is $ (\sqrt{1\minus{}z^2}\cos\theta,\sqrt{1\minus{}z^2}\sin\theta,z)$.

The problem with constructing probability distributions is that what you really need are cumulative distribution functions, not just densities. Those cumulative distribution functions will involve surface area integrals for an arbitrary ellipsoid, which are elliptic integrals. That can't be done exactly with elementary functions, so you're basically stuck with numerical approximations.

Oh, so simply generating a random $ s,t \in [0,1]$ and plug it into the formula, doesn't work. I can see that now, silly me :) Thanks for the solution!

As for the original problem, I'll follow your advice :) Actually, I think it's sufficient for me to handle rotations and uniform scalings, which shouldn't be a problem. Thanks!


algebra
polynomial
number theory
relatively prime

Let p(x) be a polynomial of the form p(x) = ax^2+bx+c, where a,b and c are integers, with the property that 1< p(1)< p(p(1))< p(p(p(1))). Show that a ≥ 0.

[hide]Assume to the contrary.
a<0 so 1<a+b+c=k<$ k^{2}a+kb+c$.
a+b+c=k and $ k^{2}a+kb+c=k_{2}$ and $ {k_{2}}^{2}a+k_{2}b+c=k_{3}$
Now, three linear equations involving a and b now exist.
$ (k-1)[(k+1)a+b]=k_{2}-k$
$ (k_{2}-k)[(k_{2}+k)a+b]=k_{3}-k_{2}$
$ (k_{2}-1)[(k_{2}+1)a+b]=k_{3}-k$
If these equations have a solution, some rational number z can be multiplied to any of the equations so that when subtracted with another, one achieves 0=0.
Since the b term has a coefficient of 1, the term that multiplies the $ y_{coef.}a+b$ term when multiplied by z must equal another equation's similar term.  Since subtracting would yield 0=0, either:
$ k=k_{2}$
$ 1=k_{2}$
$ k=1$, which are all obviously false.  Thus, the three equations are relatively prime and there's no solution when a<0 so $ a\geq 0$[/hide]
Show that this property is always true for when the leading term is greater than 0.


Δινεται παρ/μο $ ABCD$ και οι παρεγεγραμμενοι κυκλοι του τριγωνου $ BAC$ εντος της γωνιας $ BAC$ και του τριγωνου $ DAC$   εντος της γωνιας $ DAC$. Aν $ M$ ειναι το σημειο τομης των καθετων που αγονται απ τα κεντρα $ O_1$ και $ O_2$ των παραπανω κυκλων αντιστοιχα προς τις $ AB$ και $ AD$ αντιστοιχα και $ U_1$, $ U_2$ τα ιχνη των καθετων απ τα $ O_1$ και$ O_2$ στην $ AM$ ν.δ.ο.
Το γινομενο των αποστασεων των σημειων επαφης των παραπανω κυκλων με την $ AC$  απ το $ A$ ειναι ισο με $ (O_1U_1)(O_2U_2)$

[b]ΚΑΛΟ ΚΑΛΟΚΑΙΡΙ[/b]

Μήπως υπάρχει κάποιο λάθος στην εκφώνηση: Τα σημεία επαφής με την AC ταυτίζονται, αλλά ακόμα και έτσι δεν μοιυ προκύπτει η ζητούμενη ισότητα
Ανδρέας

Οντως ταυτιζονται τα σημεια επαφης εστω $ K$ (αυτο δεν ηθελα να φαινεται αμεσως απ την εκφωνηση) . Εδω δεν εχει λαθος .
Απλα τωρα που την προσεχω ξανα εκανα typo δηλαδη ξεχασα να γραψω ενα $ C$ δηλ. ολοκληρωμενο '' το γινομενο των αποστασεων απ τα $ A$ και [b]$ C$[/b]''.Οποτε θελουμε να αποδειξουμε οτι $ (KA)(KC)\equal{}(O_{1}U_{1}))(O_{2}U_{2})$.
Aντρεα συγνωμη καθως και απ αυτους που ταλαιπωρησα.


help with this qn please(:
sorry first time using LaTeX  :blush:


Let $S=\{1,2,\cdots,21\}$. Find the minimum value of $n$ such that in any $n$ element subset $P$ of $S$ there exists $a,b,c,d\in P$ with $\{a,b\}\neq \{c,d\}$ and $a+b=c+d$.($a,b$ possibly equal.)

please don't keep me waiting like that, say something.


1. Let x, y, z > 0, with xyz = 1. Prove that :

x + y + z  <= x 2 + y 2  + z 2 .

2. Let x, y, z  >= 0 with xyz = 1. Find the minimum value of

x/(y+z) + y/(x+z) + z/(x+y).

3. Let a, b, c  >= 0. Prove that:

1/a + 1/b + 4/c + 16/d  >= 64/(a + b + c + d).

I am stuck on these three problems. Can anyone give some hints? Thanks.

[hide="1"]
try to take advantage of the fact that $xyz=1$
and if you know arrangement inequality, this should come to you easy[/hide]

[hide="2"]
this is famous inequality, known as the nesbitts.[/hide]

For 3: [hide]When is equality acheived?  What do you notice about the numerators?[/hide][/hide]

1)
We have
[hide]
$\sum x^2\geq \frac {1}{3}\left(\sum x)\right)^2\geq \sum x$ because $x+y+z\geq 3(xyz)^{\frac {1}{3}}=3$[/hide]

2)

use the rearrangement inequality - it's just Nesbitt

3)

We have

[hide]$\left(\frac {1}{a}+\frac {1}{b}+\frac {4}{c}+\frac {16}{d}\right)\left(a+b+c+d\right)\geq (1+1+2+4)^2=64$ by Cauchy[/hide]


algebra
polynomial
calculus
integration
function
domain
Ring Theory

Suppose $ f(t)$ and $ g(t)$ are nonzero polynomials over an integral domain $ D$ contained in $ C$, and that $ g(t)$ is irreducible over $ D$. Suppose also that there is a complex number $ w$ (not necessarily in $ D$) for which $ f(w)\equal{}g(w)\equal{}0$. Prove that, for some polynomial $ h(t)$ over $ D$, $ f(t)\equal{}g(t)h(t)$.

$ g$ is the minimal polynomial for $ w$ and hence must divide $ f$.

btw, it would be more appropriate to post problems in abstract algebra in the College Playground.

This is [b]not[/b] true!

$ g(x)\equal{}x^2\minus{}x\minus{}1$ is irreducible over $ D\equal{}\mathbb{Z}[\sqrt{5}]$
and has with $ f(x)\equal{}2x\minus{}1\plus{}\sqrt{5}\in\mathbb{Z}[\sqrt{5}][x]$ the common root $ w: \equal{}{1\minus{}\sqrt{5}\over 2}$.


Prove or disprove that any rational number in the interval $(0,1)$ with odd denominator can be written in form $\frac{xyz}{x^2+y^2+z^2}$ for some integers $x,y,z$

However the answer is no.
Notice the degree of the numerator and denominator. The degree of the numerator is bigger, which means beyond a certain point the numerator will exceed the denominator, when x,y,z get big enough. This means there is only a finate number of values this can take in the given interval. However there are infinate values in this interval with an odd denominator. so we cannot possible make all of them.

Are you sure about your argument? Say $x=1$ and the numerator degree is down to $2$. Then it can take infinite value.

But if $z=2$, don't you have $x^2+y^2+4>2xy$? Already there are an infinite number of values that $x$ and $y$ can take, that will give a smaller numerator than denominator.

Try $\frac{1}{2}$.
$2xyz=x^2+y^2+z^2$
Let $2^{n}$ be the largest power of two that divides $x$.
If $n$ is odd, then define $m=n+1$. If $n$ is even, let $m=n$.
Since $2^m$ divides the LHS, it must divide the RHS, so it turns out that $2^{m-1}$ divides $y,z$.
Then $2*2^{3m-3}$ divides the LHS. If $3m-3$ is odd, let $k=3m-2$. Else, let $k=3m-3$.
Since $2^k$ divides the LHS, it must divide the RHS again, so $2^{k-1}$ divides $x,y,z$.

Now let's check some things. The LHS is divisible by $2$, so one of $x,y,z$ is even. Then the LHS is divisible by $4$, so in fact, all of $x,y,z$ are even. LHS is now divisible by $16$, so all of $x,y,z$ are divisible by $4$. The LHS is divisible by $64$, so all of $x,y,z$ are divisible by $8$.

Now look at the first part again. Since $x$ is divisible by at least $8$, $n\ge3$. Then $k-1=3m-4>n$. Thus, we have a (very unclearly stated) contradiction to $2^n$ being the largest power of two that divides $x$.

Seems your argument on $\frac{1}{2}$ is good, but the problem is:

Prove or disprove that any rational number in the interval $(0,1)$ with [b]odd denominator[/b] can be written in form $\frac{xyz}{x^2+y^2+z^2}$ for some integers $x,y,z$

Oh my, that certainly makes me feel useless! :?
Other thoughts?

So next time you should read the question more carefully  :D


Given $ a\star b \equal{} (a \plus{} b)^2 \minus{} ab$, express $ x\star 1$ in terms of $ x$.

We just have to plug in the numbers here:

$ (x\plus{}1)^2\minus{}x(1) \equal{} x^2 \plus{} 2x \plus{} 1 \minus{}x \equal{} \boxed{x^2\plus{}x\plus{}1}$


trigonometry
complex analysis
complex analysis unsolved

Find all limit points of the following complex sequence $ z_n\equal{}e^{in}$.

[quote="Ibrogimov"]Find all limit points of the following complex sequence $ z_n \equal{} e^{in}$.[/quote]
I hope I'm not writing stupidity
$ e^{in}\equal{}\cos n\plus{}i\sin n\equal{}\cos n\pm i\sqrt{1\minus{}\cos^2 n}$

Let $ x_n\equal{}2n\pi\plus{}\alpha,\: \alpha\in[0,2\pi]\: \mbox{then}\: x_n\stackrel{n\to\infty}{\to}\infty$
So $ \cos x_n\stackrel{n\to\infty}{\to}\cos\alpha\equal{}x\in[\minus{}1,1]\Rightarrow e^{ix_n}\stackrel{n\to\infty}{\to}x\pm i\sqrt{1\minus{}x^2},\: x\in[\minus{}1,1]$


geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations
difference of squares

This section (Simplifying Rational Expressions) was quite challenging to me. Below was a problem I ran into during an exercise and couldn't really grasp.

[url=http://img410.imageshack.us/my.php?image=simplifyuu5.png][img]http://img410.imageshack.us/img410/6758/simplifyuu5.png[/img][/url]
[url=http://g.imageshack.us/img410/simplifyuu5.png/1/][img]http://img410.imageshack.us/img410/simplifyuu5.png/1/w640.png[/img][/url]

why don't you success this simplification?
what was you doing?

hello, i have let $ t_1\equal{}\frac{z^2}{x^3y^2}\minus{}\frac{x^3}{y^2z}$ and $ t2\equal{}\frac{y^2}{z^2x}\minus{}\frac{z^2}{x^3y^4}$ then we have $ t_1\equal{}\frac{z^3\minus{}x^6}{x^3y^2z}$ and $ t_2\equal{}\frac{x^2y^6\minus{}z^4}{x^3y^4z^2}$ so we have
$ \frac{t_1}{t_2}\equal{}\frac{(z^3\minus{}x^6)y^2z}{x^2y^6\minus{}z^4}\equal{}\minus{}\frac{(x^2\minus{}z)(x^4\plus{}x^2z\plus{}z^2)y^2z}{(y^3x\minus{}z^2)(y^3x\plus{}z^2)}$.
I think this is the simplest form.
Sonnhard.

it seems big for a simplest form. But i think it's good

Just find common denominators, and divide.

[hide="in here"]$ \frac{\frac{z^2}{x^3y^2}\minus{}\frac{x^3}{y^2z}}{\frac{y^2}{z^2x}\minus{}\frac{z^2}{x^3y^4}}\equal{}\frac{\frac{z^3\minus{}x^6}{x^3y^2z}}{\frac{y^6x^2\minus{}z^4}{x^3y^4z^2}}\equal{}\frac{(z^3\minus{}x^6)x^3y^4z^2}{x^3y^2z(y^6x^2\minus{}z^4)}\equal{}\frac{y^2z(z^3\minus{}x^6)}{y^6x^2\minus{}z^4}$[/hide]

Then Dr. Graubner used the difference of Cubes and the difference of Squares factorizations, which IMO don't put it in a very [i]simple[/i] form, but it [i]is[/i] factored completely.


Euler
quadratics
modular arithmetic
algebra
quadratic formula

Number 2 is [hide]$69$[/hide]

EDIT, found my mistake :blush:

Question 1:
[hide="#1"]
$2n+1 = a^2$
$3n+1 = b^2$

Using mod $4$.
From the first equation $2 | n$.
Let $n = 2k$
From the second equation $2 | k$
Let $n = 4k$

Using mod $5$.
From the first equation $k \equiv 0,1,3 \mod{5}$
From the second equation $k \equiv 0,2,4 \mod{5}$
Therefore $k \equiv 0 \mod 5$
Let $n = 20k$

So far $20 | n$.

[/hide]

[hide="Conclusion to 1"]Use mod 8 on the second equation.

A square must be 0, 1, or 4.  This gives us $3n$ as 0, 3, or 7 in mod 8.  Clearly, it is 0, so eight divides $n$ and the proof is complete.[/hide]

thanks, now #$3$ needs to be solved.

Ok, my solution to both #$1$ and $2$ were poor, so here is a second attempt at 3$2$.

According to Euler's Generalization to Fermat's Little Theorem.

$a^\phi (n) \equiv 1 \mod {n}$ For relative primes $a$ and $n$.
We obviously want $n = 100$ and we're in luck since when $a=3$, $a$ and $n$ are relatively prime.

so our next step is to determine what $\phi (100) = \phi (2^2 5^2)$ is.

Thus $\phi(100) = 40$.

Then $3^{40} \equiv 1 \mod {100}$

$3^{1234} \equiv 3^{34} \mod {100}$

Note $100 = 4 \cdot 25$

$3^34 \equiv (-1)^{34} \equiv 1 \mod 4$

By Euler Phi $\rightarrow 3^{20} \equiv 1 \mod 25$

$3^{14} \equiv (27)^4 \cdot 9 \equiv 16 \cdot 9 \equiv 19 \mod {25}$

$69 \equiv 1 \mod {4}$
and
$69 \equiv 19 \mod {25}$

Thus,
$\boxed{69}$

I hope that is well written.

[hide="A start on No. 3"]1) $x$ and $y$ cannot both be even.

Let us consider mod 4.  If they were both even, then the left side would be 0.

2) $y$ is not even.

If $y$ were even, then on the left side, we would have $1+6x$, in mod 4.  Clearly, this cannot be 2.

3) If $x$ is even, then $x\equiv 0 \mod 4$.

If $x\equiv 2\mod 4$, then the left side in mod 4 is $2-2\equiv 0$.

4) If $x$ is odd, then $x\equiv y \mod 4$.

This time we re-write the variable side thus :  $xy-3y^{2}+(x+y)^{2}$.  In mod 4, this is $xy+1$, so $xy\equiv 1\mod 4$.[/hide]

Heres a hint if you'd like it

[hide]Use the quadratic formula to solve for x, then look at it modulo 17.
Can it ever be a perfect square?) [/hide]

[quote="G-UNIT"]1.Prove that if 2n + 1 and 3n + 1 are both perfect squares, then n is divisible by 40.

2.What are the last two digits of 3^1234?

3.Prove that there are no integers x and y for which

x^2 + 3xy − 2y^2 = 122

  All problems extracted from: MA 491 Problem set #4[/quote]

[hide="1"]
We need to prove that $n$ is divisible by 8 and 5.

The quadratic residues $\mod 5$ are $0^2 = 0$, $1^2=1$, $2^2=4$, $3^2 \equiv 4$, and $4^2 \equiv 1 \pmod {5}$. Thus they are 0, 1, and 4.

If $2n + 1 \equiv 0 \pmod {5}$, then $2n \equiv 4 \pmod {5}$ and $n \equiv 2 \pmod {5}$. If $2n+1 \equiv 1 \pmod {5}$, then $n \equiv 0 \pmod {5}$. If $2n+1 \equiv 4 \pmod {5}$, then $2n \equiv 3 \equiv 8 \pmod {5} \implies n \equiv 4 \pmod {5}$.

If $3n + 1 \equiv 0 \pmod {5}$, then $3n \equiv 4 \equiv 9 \pmod {5}$ and $n \equiv 3 \pmod {5}$. If $3n+1 \equiv 1 \pmod {5}$, then $n \equiv 0 \pmod {5}$. If $3n+1 \equiv 4 \pmod {5}$, then $3n \equiv 3 \pmod {5} \implies n \equiv 1 \pmod {5}$.

The only way that $n$ satisfies both of these is if $n \equiv 0 \pmod {5}$; thus $5|n$.

Next we need to prove that it is divisible by 8. Quadratic residues $\mod 8$ are $0, 1, 4$.

If $2n+1 \equiv 0 \pmod 8$, then $2n \equiv 7 \pmod 8$, so there are no solutions.
If $2n+1 \equiv 1 \pmod 8$, then $2n \equiv 0 \pmod 8$, so $4|n$.
If $2n+1 \equiv 4 \pmod 8$, then $2n \equiv 3 \pmod 8$, so there are no solutions.

If $3n+1 \equiv 0 \pmod 8$, then $3n \equiv 7 \equiv 15 \pmod 8$, so $n \equiv 5 \pmod 8$.
If $3n+1 \equiv 1 \pmod 8$, then $3n \equiv 0 \pmod 8$, so $8|n$.
If $3n+1 \equiv 4 \pmod 8$, then $3n \equiv 3 \pmod 8$, so $n \equiv 1 \pmod 8$.

The only time both of these is satisfied is when $4|n$. Let $n=4k$. $3(4k)+1=12k+1$ is a perfect square, so $\mod 8$, $4k+1 \equiv 0, 1, \text { or } 4 \pmod 8$, so $k$ is even, and $n=4k=8j$, so $8|n$ and $5|n$, hence $40|n$.
[/hide]

[hide="2"]
$3^{1234} = 9^{617} = (10-1)^{617}$
$= \sum_{k=0}^{617} \binom{617}{k} \cdot 10^{617-k} \cdot (-1)^k$
Mod 100, all terms cancel out except when $k = 617, 616, 615$. In these cases,

$\binom{617}{617} (10)^0 (-1)^{617}=-1$
$\binom{617}{616} (10)^1 (-1)^{616}=6170 \equiv 70 \pmod {100}$
And the last two digits are $-1+70 = 69$.
[/hide]

[hide="3"]
$x^2 + 3xy - 2y^2 = 122$
The discriminant (in x) is $\sqrt{b^2 - 4ac} = \sqrt{(3y)^2 - 4(-2y^2 - 122)} = \sqrt{17y^2 + 488}$.  Taking $\bmod 17$, we see that $12 \equiv k^2 \bmod 17$ in order for there to be a solution.

Looking at squares $\bmod 17$, we see $1, 4, 9, 16, 8, 2, 15, 13, ...$.  Hence $x$ cannot be an integer if $y$ is an integer.  Q.E.D.
[/hide]


function
limit
calculus
calculus computations

:lol:
if : $ f$ is continuous at $ l$  and $ \underset{x\rightarrow\infty}{\lim}g( x) = l) ) \rightarrow( \underset{x\rightarrow\infty}{\lim}( f\circ g) ( x) = f( l) = \underset{y\rightarrow\underset{x\rightarrow\infty}{\lim}g( x) }{\lim}f( y))$
PROOF:
( proved ) if : $ f$ is continuous at $ l$  and $ \underset{x\rightarrow p^{ + }}{\lim}g( x) = l) ) \rightarrow( \underset{x\rightarrow p^{ + }}{\lim}( f\circ g) ( x) = f( l) )$
Let : $ h( \frac {1}{x}) = g( x)$

( $ f$ is continuous at $ l$ ) and ( $ l = \underset{x\rightarrow\infty}{\lim}g( x) = \underset {x\rightarrow0^{ + }}{\lim}g( \frac {1}{x}) = \underset {x\rightarrow0^{ + }}{\lim}h( x) )$ )

$ \Rightarrow f( l) = \underset{x\rightarrow0^{ + }}{\lim }( f\circ h) ( x) = \underset{x\rightarrow\infty }{\lim}( f\circ h) ( \frac {1}{x}) = \underset {x\rightarrow\infty}{\lim}f( h( \frac {1}{x}) ) =$ $ \underset{x\rightarrow\infty}{\lim}f( g( x) )$
 :lol:

[color=red]Is it correct above ?[/color]

 :lol:

[color=orange]How to prove this question?[/color]

if: $ f$ is continuous  and  $ \underset{x\rightarrow\infty}{\lim}g( x) = \infty$ ,then: $ \underset{x\rightarrow\infty}{\lim}( f\circ g) ( x) = \underset{y\rightarrow\underset{x\rightarrow\infty}{\lim}g( x) }{\lim}f( y)$
 :lol:

Is it correct? No, because the reasoning is circular. The equivalence of $ \lim_{x\to\infty}g(x)$ and $ \lim_{x\to0^\plus{}}g\left(\frac1x\right)$ is itself an instance of the theorem you are trying to prove.

Essentially, you cannot prove anything about limits without defining what limit means. Here's one of the definitions we need:

$ \lim_{x\to\infty}g(x)\equal{}L$ iff $ \forall\,\epsilon>0,\exists\,M$ such that if $ x>M,$ then $ |g(x)\minus{}L|<\epsilon.$

Hey, I had doubt my proof long time.

now i try to prove it again from your guidance.

I had proved it by define of limit , ya. And this is a wrong statement :

if: $ f$ is continuous  and  $ \underset{x\rightarrow\infty}{\lim}g\left( x\right) \equal{} \infty$ ,then: $ \underset{x\rightarrow\infty}{\lim}\left( f\circ g\right) \left( x\right) \equal{} \underset{y\rightarrow\underset{x\rightarrow\infty}{\lim}g\left( x\right) }{\lim}f\left( y\right)$


For this, rote memorization is good:
I just memorize $ 2^5\equal{}64$ and $ 2^10 \equal{}1024$, and that suffices, because I can either divide or multiply by 2 to reduce or increase the power of 2:
$ 512 \equal{} \frac{1}{2}\times1024$, so $ 512\equal{}2^9\minus{}1\equal{}2^8.$

[quote="phr33k"]For this, rote memorization is good:
I just memorize $ 2^6 \equal{} 64$ and $ 2^{10} \equal{} 1024$, and that suffices, because I can either divide or multiply by 2 to reduce or increase the power of 2:
$ 512 \equal{} \frac {1}{2}\times1024$, so $ 512 \equal{} 2^{10 \minus{} 1} \equal{} 2^9.$[/quote]

I think this is what phr33k means.  The final answer should simply be $ x\equal{}\boxed{9}$


Let's define a permuted Fibonacci sequence u' (k) this way :

u(k) = u(k-1) + u(k-2); u(0) = a, u(1) = b>a, k=>2

u'(k) = u'(k-1) + u'(k-2); u'(0) = b, u'(1) = a

if gcd(a, b) = 1, does exist a couple of values (a, b) such that u(k) and u'(k) are composite for all k ?

If so, can you produce one of these (a, b) pairs?


hehe i solved the easy problem :|
[hide]let $6\left(x-\frac{1}{y}\right)=3\left(y-\frac{1}{z}\right)=2\left(z-\frac{1}{x}\right)=xyz-\frac{1}{xyz}=k$ then $x-\frac{1}{y}=\frac{k}{6}$, $y-\frac{1}{z}=\frac{k}{3}$, and $z-\frac{1}{x}=\frac{k}{2}$
\begin{eqnarray*}\left(\frac{k}{6}\right)\left(\frac{k}{3}\right)\left(\frac{k}{2}\right)&=&\left(x-\frac{1}{y}\right)\left(y-\frac{1}{z}\right)\left(z-\frac{1}{x}\right)\\ &=&\left(xyz-x-y-z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{xyz}\right)\\ &=&\left(xyz-\frac{1}{xyz}-z+\frac{1}{x}-y+\frac{1}{z}-z+\frac{1}{x}\right)\\ &=&\left(k-\frac{k}{2}-\frac{k}{3}-\frac{k}{6}\right)\\ &=& 0 \end{eqnarray*} so $x=\frac{1}{y}$, $y=\frac{1}{z}$, and $z=\frac{1}{x}$ so $x=y=z$ and $(1,1,1)$ or $(-1,-1,-1)$
[/hide]


geometry
ratio
analytic geometry
graphing lines
slope
similar triangles

We have an equilateral triangle ABC, with side length 1. BC is extended 1 unit to point D. We draw a line segment from D to the midpoint of AB (which is E), intersecting AC at F. What is the area of quadrilateral EBCF? Express your answer in simplest radical form.

If anyone needs a diagram for this, I will provide one.

It's not from your friend  :roll:  It is a modified version of the 1991 national targets  :P

[hide="hint from how I did it"]Draw perpendiculars to BC from E and F. These lines are parallel to each other. I see similar triangles. [/hide]

Yes, and my friend got it from the 1991 national targets.

[hide="hint from how I did it"]
What is true about the triangles formed by the medians?
[/hide]

This is my solution.

[hide]

We have given an equilateral $\triangle ABC$ with each side equal to $1$.

$BC$ is extended $1$ unit to point $D$.

By drawing a line from $D$ to $A$ you will find that $AD$ is $\bot$ to $AB$.

Now by drawing a line from $C$ to $E$ another $\bot$ is formed with $AB$.

There are now two sets of similar triangles that are both at a $2: 1$ ratio.

One set is $\triangle ABD$ and $\triangle EBC$. The other set  is $\triangle AFD$ and $\triangle EFC$.

If you add the areas of $\triangle EBC$ and $\triangle EFC$ together, you will have the area of the quadrilateral $EBCF$.

Next, we find the area of $\triangle EBC$:

  It's area is $\frac{(\frac{1}{2})(\frac{1}{2} \sqrt 3)}{2}$ or $\frac{\sqrt 3}{8}$

Then we find the area of $\triangle EFC$:

   The first thing to do is to draw a line through $F$ that is $\bot$ to both $AD$ and  $CE$. Call this line $GH$, where $G$ intersects  $AD$ and $H$ intersects $EC$.

   Since $\triangle AFD$ and $\triangle EFC$ are in the ratio of $2: 1$, $GF: FH = 2: 1$

   We know that $GF + FH = AE = \frac{1}{2}$ so $GF = \frac{1}{3}$ and $FH = \frac{1}{6}$

   We can now find the area of $\triangle EFC$ which is $\frac{(\frac{1}{2} \sqrt 3)(\frac{1}{6})}{2}$ or $\frac{\sqrt 3}{24}$

Finally, we can determine the area of the quadrilateral $EBCF$:

   The total area is $\frac{\sqrt 3}{8} + \frac{\sqrt 3}{24} = \frac{1}{6} \sqrt 3$

[/hide]

Objection!!!

How do you know they're perpendicular? (I don't think they are) They're medians, not altitudes.

The answer I got was mentioned somewhere in your solution, but I'm not telling you where.

in an equilateral triangle the medians and the altitudes are one and the same.

But noooo. DF and AC are medians of the triangle ABD, which has a base of 2 and one side length 1. It isn't that easy.

can I please have a diagram? and ur signiture is funny

I'll try to explain this without a diagram if I can.

Yes it's true that $DE$ and $AC$ are medians and not $\bot$ to the opposite sides.

What I was saying is that if you draw a line from $D$ to $A$, then THAT line will be $\bot$ to $AB$.

How do I know that?

When you extend $BC$ out to $D$ you are creating a side of $\triangle ABD$ that is  twice as long as $BC$ which is the corresponding side of $\triangle ECB$. Now, we know that by drawing a line from $C$ to $E$ will be the $\bot$ bisector of $AB$. That will make $AB$ and $EB$ a $2: 1$ ratio. $BD$ and $BC$ are also in a $2: 1$ ratio. Also, the two triangles share a common angle, which is angle $B$, which would make the two triangles similar to each other.
Therefore, if the triangles are similar to each other, and $CE$ is at right angles to $AB$, then $AD$ would also be at right angles to $AB$.

I have a tedious anaylitic geometry solution -- that probably wouldn't be in MC. :lol:

[hide]
Put DB on the X axis, where D is the origin.  Point C is at point $(1, 0)$, and point A is at point $(1.5, \frac{\sqrt{3}}{2})$.  The slope of the line connecting A and C is $\sqrt{3}$, and the x-intercept is $(1, 0)$, so the equation is $y=\sqrt{3}x-\sqrt{3}$.  Construct segment $EQ$ where $EQ$ is perpendicular to $CB$.  The length of EQ is $\frac{\sqrt{3}}{4}$, because $\triangle EQB$ is a 30-60-90 triangle.  Thus, the equation of line DE is $y=\frac{\sqrt{3}}{7}x$.   Now, we can use these equations to solve for the point the 2 lines intersect, $F$.  Solving, we get the point $(\frac{7}{6}, \frac{\sqrt{3}}{6})$.  $[DEB]$ would equal $\frac{\frac{\sqrt{3}}{4}\cdot2}{2} = \frac{\sqrt{3}}{4}$.  $[DCF]$ would equal $\frac{1\cdot\frac{\sqrt{3}}{6}}{2}=\frac{\sqrt{3}}{12}$.  Subtracting $[DCF]$ from $[DEB]$, we get $\boxed{\frac{\sqrt{3}}{6}}$.

Any mistakes?
[/hide]

So, yeah, I agree w/ritchjp.

Ignore my hint. I approached the problem incorrectly.

ccy, I probably would think about your way  :D , but it's a time waster.

I finally drew up a diagram. Attached. Have a nice day.

499th post... almost there!


i am making a research proposal with latex. what command do i put my name on the bottom like so? i use \documentclass{article}

Anika Friedmann
Göteborgs universitet
18. Februar 2008

Is there also maybe an example of a research proposal like with how you put addresses and so?

Thanks!

If you mean by 'at the bottom' putting it in the footer then use [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Packages#fancyhdr]fancyhdr[/url].
Different institutions have different rules about formatting documents so best to ask your university for their rules and examples of proposals.


ratio
number theory proposed
number theory

Find all sets $X$ consisting of at least two positive integers such that for every two elements $m,n\in X$, where $n>m$, there exists an element $k\in X$ such that $n=mk^2$.

I will show that all such sets X are sets of the forms $\left\{a;\;a^3\right\}$, where a can be an arbitrary integer > 1.

In fact, let X be a set consisting of at least two positive integers such that for every two elements m and n of the set X, where n > m, there exists an element k of X such that $n=mk^2$. In other words, we demand that for every two elements m and n of the set X, where n > m, the ratio $\frac{n}{m}$ is the square of an element of X.

Write our set X in the form $\left\{a_1;\;a_2;\;...\right\}$, where $a_1 < a_2 < ...$. Hereby, the dots don't imply that the set X must be infinite; the set X can be finite as well. We don't know a priori how many elements X has, but we know that the set X has at least 2 elements, so that we can consider the smallest two elements $a_1$ and $a_2$ of X.

First we show that 1 is not an element of X. In fact, if 1 were an element of x, it would be the smallest element of X; thus, we would have $a_1=1$, and since $a_2 > 1$, it would follow that $\frac{a_2}{1}$ is the square of an element of X; in other words, $a_2$ is the square of an element of X; but this latter element of X would be > 1 and $< a_2$ (this is because $a_2 > 1$), what contradicts the assumption that there is no element of X between $1=a_1$ and $a_2$ (in fact, remember that $a_1 < a_2 < ...$). Hence, 1 is not an element of X. Thus, all elements of X are > 1, i. e. we have $1 < a_1 < a_2 < ...$.

Since $a_2 > a_1$, the ratio $\frac{a_2}{a_1}$ is the square of an element of X. But this latter element of X can only be $a_1$, since $\sqrt{\frac{a_2}{a_1}} \leq \frac{a_2}{a_1} < a_2$ (because of $1 < a_1$), and $a_1$ is the only element of the set X which is $< a_2$. Hence, $\frac{a_2}{a_1} = a_1^2$, and $a_2 = a_1^3$. If the set X has only 2 elements, then this already yields our family of solutions: $X = \left\{a;\;a^3\right\}$, where a is an arbitrary integer > 1.

Now consider the case when the set X has at least 3 elements, i. e. there is an element $a_3$. Since $a_3 > a_1$, the ratio $\frac{a_3}{a_1}$ is the square of an element of X. But this latter element of X must be either $a_1$ or $a_2$, since $\sqrt{\frac{a_3}{a_1}} \leq \frac{a_3}{a_1} < a_3$ (this is because $1 < a_1$), and $a_1$ and $a_2$ are the only elements of the set X which are $< a_3$. In the case $\frac{a_3}{a_1}=a_1^2$, we get $a_3=a_1^3$, so that $a_3=a_2$, contradicting with $a_2 < a_3$. Hence, only the case $\frac{a_3}{a_1}=a_2^2$ remains possible; since $a_2=a_1^3$, we thus have $a_3=a_1a_2^2=a_1\left(a_1^3\right)^2=a_1^7$. But on the other hand, since $a_3 > a_2$, the ratio $\frac{a_3}{a_2}$ is the square of an element of X. But $\frac{a_3}{a_2} = \frac{a_1^7}{a_1^3} = a_1^4 = \left(a_1^2\right)^2$, and $a_1^2$ is not an element of X (in fact, if it would be an element of X, it would lie between $a_1$ and $a_3 = a_1^7$, so it would be equal to $a_2$, but $a_2 = a_1^3 \neq a_1^2$). So we get a contradiction, and hence, the set X cannot have more than 2 elements. And thus, the family of solutions we found above, $X = \left\{a;\;a^3\right\}$ for $a \in \left\{2;\;3;\;4;\;...\right\}$, yields all solutions.

This is in fact Darij's solution, re-posted.


geometry
geometric transformation
homothety
reflection
angle bisector
geometry proposed

$ABC$ is a triangle, $DEF$ is its orthic triangle. The incircles of $AEF$, $BDF$ and $CDE$ touch the sides of $DEF$ at $C'$, $A'$ and $B'$.
Show that $AA'$, $BB'$ and $CC'$ concur.

As the small triangles are oppositely similar to the reference triangle, they concur at the isogonal conjugate of the Gergonne point (of the reference triangle). For example, if X, Y, and Z are the intersections of the incircle of ABC with its sides, we have that AEF is the image of ABC under a homothecy about A and a reflection across the angle bisector of A, so AA' is the image of the AX under a homothecy about A and a reflection across the angle bisector of A. The homothecy doesn't do anything, so AA' is simply the reflection of AX across the A-angle bisector. Similarly for the other two; we know the reflections of three concurrent lines are concurrent at the isogonal conjugate.


Let $ G$ be a simple graph. Suppose that size of largest independent set in $ G$ is $ \alpha$. Prove that:
a) Vertices of $ G$ can be partitioned to at most $ \alpha$ paths.
b) Suppose that a vertex and an edge are also cycles. Prove that vertices of $ G$ can be partitioned to at most $ \alpha$ cycles.

a) Consider a path partition of $G$ with minimal cardinality $p$. Let $P_i = v_{i,1}v_{i,2}\ldots v_{i,\ell_i}$ for $1\leq i \leq p$. Clearly $v_{i,1}v_{j,1}$ is not an edge for any $1\leq i \neq j \leq p$ (or else we can "glue" the two paths $P_i$ and $P_j$). Therefore the set $\{v_{i,1} \mid 1\leq i \leq p\}$ is an independent set, and so $p\leq \alpha$.
The result remains true for digraphs; it's the Gallai-Milgram theorem (with a more arduous proof).

b) This is a theorem of Lajos Pósa.


LaTeX
real analysis
real analysis unsolved

May we assume that's $\sum_{k=0}^na{n\choose k}(a-kx)^{k-1}(b+kx)^{n-k}$ ?

Because ${k\choose n}$ would make no sense.

you don't  know that  $C_n^k={n \choose k}$ :o

That's a rather strange and dangerously unclear notation- it helps to use $\LaTeX$ for the unambiguous standard.

In what I've seen of other notations, the $n$ is always listed first, and I would read being on top as being first.

ok ; but in my country we use  $C_n^k$ more then ${n \choose k}$  :)


AMC AIME geometry articles Alcumus HMMT ARML

how are you preparing before the 10/12 b?

by preparing for AIME (though that doesn't help much with speed...)

I'm just doing past amc exams.  Doing them just helps me not make careless mistakes.

cramming for the AIME way too early. :wink:

Doing a lot of combinatorics and geometry seeing as there was barely any on the 12A.

[quote="mathemonster"]i just read rrusckys article on not making stupid mistakes. How do you know how many boxes to put on each page when you premaking your scrap paper?[/quote]

I just draw a line after I do each problem.

I've been sick for the past couple of days, so I've been looking over this forum, past tests on the wiki, alcumus, etc. In other words, I have a lot of free time and I'm spending a lot of it doing math.

I'm doing all the Olympiad Geometry challenge set problems in one night.

 :|

Though I will start doing past AMC 12s... soon...

[quote="Zmastr"]Went to HMMT.  That probably helped the competition/confidence aspect of it all![/quote]

compared to this last hmmt, amc is baby stuff.

Combining the 3 facts that I still haven't even started typing up my Olympiad Geometry homework, and that I have ARML training homework to type up, and the USAMTS Round 4 stuff to do, I haven't had much time to think about the AMC.  But hooray, it's only in 2 more days!!!  :D

that HMMT was so hard it was unbelievable
amc can't be even compared to it.

I'm hoping there aren't too many counting problems. I should have done well on the AMC 12A because of the lack of counting problems, but I still managed to let stupid mistakes kill me. I hope I think clearly on Wednesday.

No, I like counting problems, they're easy and typically really quick.
I want less geometry problems. It's not that they're hard, they just suck up time, and time is basically the single factor that killed me on the 12A.
And I've had to do #2 on USAMTS for the last month... so I should probably get around to that.

And what was with all the complex numbers on the 12A??
I'm terrible with complex numbers...

I'm guessing it's really soon? I don't know the date but anyway...

GET GOOD NIGHT SLEEP

You'll be amazed how this deceptively simple suggestion can make a huge difference... Speaking of sleep, I should get some...since I'm running on 7 hours for past 48 hours.   :huh:


inequalities
function
integration
logarithms
real analysis
real analysis unsolved

Suppose $ f,g: [0,1] \to \mathbb R$ are continuous functions and $ f \geq 0,\ g > 0,\ \int_0^1f(x)dx \equal{} 1$. Prove that:
$ \int_0^1f(x)\ln g(x)dx \leq \ln\int_0^1f(x)g(x)dx.$

Isn't it just Jensen's inequality? Maybe I'm missing something :blush:


For what value of $ x$ will a line which passes through the points $ (5,3)$ and $ (x,7)$ have a slope of $ \minus{}\frac{1}{3}$?

The slope is $ \minus{}\frac13\equal{}\frac{7\minus{}3}{x\minus{}5}\implies12\equal{}5\minus{}x\implies x\equal{}\boxed{\minus{}7}$.

the game says the answer is 7: this is an error, right?

Yes, that is clearly an error. The correct answer is $ \boxed{\minus{}7}$.

If it was seven, then the slope would be positive.  The answer is -7 and not 7.


[quote="rem"]Where is the residence that we are supposed to go to for Winter Camp??? :maybe:[/quote]

Norman Bethune College, York University, Toronto. You can find the campus map at http://www.yorku.ca (it's the Keele campus).

Have fun in winter camp, and see you there! (I'll come to mark questions and terrorize you guys with my favourite IMO problem.)

Can you guys post some of the problems, you will be working at, here too?

[quote="malezgerd"]you have got your invitation to winter camp?

If yes, Congratulations Alex, First step toward IMO.[/quote]
I got it a while ago, I don't think they started sending out the invitations from COMC scores.

Thanks Victoria.

Has Professor Sands said anything about transportation yet? Has he told any of you guys?

When he emailed me a while back about winter camp, he said that CMS would take care of it all, but since time is almost there (im so excited!!!) im just wondering if he's said anything to you guys yet

Thanks!

Professor Sands told me that the olcal organizer will be sending another E-mail later in the month.
btw I will be coming late (about 7-8 pm). sorry for any inconvenience

Yeah, I'm rather excited about going to Winter Camp - this is going to be so cool :D

They already booked my transportation, I guess because I'm flying in.

Looking forward to a great time!

So Bobby is coming too.

Looking forward to seeing all you guys there!

Bobby is "nneonneo"? Hey!!!! Watsup Bobby!

hey bobby the Saskatchwanian!!! Lol. Are u the first guy from Sask ever entered winter camp? congratz.  :D

frank
btw, anyone noticed that Boris's not on the list again?

Whoa! Boris...not into math? how did that happen?.....gosh i remember him still, that really smart little russian kid

Well I think there are 12 E-mails there to ppl who did good on olympiads, and 15 ppl are invited,so they still have to send out 3 more invitations for ppl who pwned COMC, maybe Boris will be one of them.

By the way who is Kalimdor? :maybe:

Kalimdor is Frank Meng da pimp, cuz he's such a pimp man.

Dude, I thought Boris turned down his Winter Camp invitation last year.  Deja vu?  I'm not sure about the situation this year.

I'm going to get pwned next week dude.  Like, majorly pwn't dude.

Oh yeah btw dudes, there's lots of snow here and like two minutes out of town there's like no snow.  WTF.

Hope u get in, Lowest_Reynolds, otherwises I won't be able to enjoy whatever the game you've got on hand right now...  :wink:
And for the comments you give me, lol.  :rotfl:

my friend got 73 in comc which is same as me and she got invitation of the winter camp, but i didn't

could someone tell me why? ( cuz they need more girls??)

lolz. My conclusion: Prolly your friend's joking with you.
Last year the top three's all like 79 or 80. 73 may not be her real score, or she's just joking around

Frank.

impossible .. we are in the same skool, the 2 ppl got invited scored 75 and 73 in comc and i've seen the invitional letter written by Prof Sands

btw, the ppl got 75 wins ontario metro

I somehow question vincent_five's claims, though I do allow for the benefit of the doubt.

Danny Shi got a 77, and did not get an invitation to the Winter Camp yet.  And he's a sophomore.  You would think he's the ideal candidate...

[quote="startogalaxy"]you must be kidding?! 75 winning in Toronto...that hardly ever happen before....lol, the west is on the rise...btw, how did u know? winner for region are not listed on the school result page. Is the offical result out already?[/quote]
No, the official results are not up yet.

k w8 that makes no sense...how can someone with 73 get an invitation to CMO and someone who gets 77 is still waiting  :huh:

Yes..sorry i meant Winter Camp not CMO... :oops:

I got 68 on the COMC  :(
Though last year I got 74, so I was able to write CMO and APMO (CMO was hard though....I hated the first question :P)

So, does anyone know what we do exactly at winter camp?

bobby, congrats on...68, nono, dont feel bad, i only got 69  :( , i failed miserably too, but hmm

no idont know wat we gonna do at winter camp, but im guessing something similar to national camp and waterloo, only harder  :P

hmmm...any insight from lightrhee, or other experienced veterans?


...
[hide]84%.  Some guess and check, but I believe it is linear, so I said if 100 people said yes, and 100 said no (for every 200 people tested).  I could tell more people said yes, so I tested 110 yes, 90 no, and in a algorithmic sequence I figured that 120 people said yes and 80 people said no for every 200 tested.  Sure enough it works.  120 people saying yes times 70%=84.[/hide]

[quote="Silverfalcon"]This is simple problem but I just can't do it. :(

An equal number of juniors and seniors responded to the question "Do you like math," each responding with Yes or No. If 70% of those who said Yes were seniors and 80% of those who said No were juniors, then what percentage of the seniors polled said Yes?[/quote]
No problem is "simple", until you see how to solve it.

[hide="Formal Solution"]Investigate Bayes Theorem and the Total Probability Theorem.  But read the informal solution, first.[/hide]
[hide="Informal Solution"]For a smallish case, like this, I think it is instructive to make a table and fill in the values.  Each row adds to the total shown.  Each column adds to the total shown.

$\displaystyle \begin{array}{cccc}  & Yes & No & Total \\ Juniors & & 0.8n & 0.5 \\ Seniors & 0.7y & & 0.5 \\ Total & y & n &  \end{array}$

Let $y$ be the total portion of students who responded "Yes".  Let $n$ be the total portion of students who responded "No".  The columns of this table correspond directly to the statements in the problem.

Fill in the missing elements:

$\displaystyle \begin{array}{cccc}  & Yes & No & Total \\ Juniors & 0.3y & 0.8n & 0.5 \\ Seniors & 0.7y & 0.2n & 0.5 \\ Total & y & n & 1  \end{array}$

Then solve the simultaneous linear equations given by the horizontal rows:

$0.3y+0.8n=0.5 \\
0.7y+0.2n=0.5 \\
y = 0.6,\ n = 0.4 $

Plug the numbers into the table:

$\displaystyle \begin{array}{cccc}  & Yes & No & Total \\ Juniors & 0.18 & 0.32 & 0.50 \\ Seniors & 0.42 & 0.08 & 0.50 \\ Total & 0.60 & 0.40 & 1.00  \end{array}$

My final answer is $\frac{0.42}{0.5}=84\%$.[/hide]


i want to take the SAT2 math, but there are 2 levels, IC IIC
which one should i take?or both? i'm in Alg2 right now

Anything in your math background besides your coursework?  I have taken both tests and would advise you to take Math level II if you are well versed in competitions (able to qualify for AIME maybe?).  However, if what you learn in your regular math classes is the extent of your experience, go ahead and take the IC.  Or you could just take both like me (and ironically score higher on the II!)

Taking both is a horrible waste of time and money in my opinion. (Save up for the Aston Martin Vanquish instead. :-D )

If you have a 100 + on the AMC, go for the II. Easy as that. However, wait until you finish pre-calc.

The IIC has a much generous curve when it comes to scoring..
The average score is something like 80 points higher than IC.
So supposedly the IIC is easier to get higher scores on.

That's because the IIC is taken by a lot (a LOT) of people who are good enough to get a perfect score.  It's not quite as extreme as, say, the Chinese or Korean language tests where a perfect score is 75 percentile or below, but it's not because scores are scaled up, it's because the student pool for the IIC is very top-heavy.

I don't think there's any reason to take the SAT 2 (either of them) until your senior year. You should wait until then to decide; and then decide based on what your future plans are in college, and which one best suits your level.

I disagree. You should be taking SAT 2s in your junior year because you'll be applying to colleges in your senior year. If you feel comfortable in an ALgrebra 2 class, then definitely take the IIC test. It looks much better for colleges.

It's only more impressive if you can [i]do well[/i].

i'm only in the 9th grade right now, i already took the SAT and got 760. so i think i can score high on level 1, and i read in the princereview books that it's better to take both.
Is this gonna waste time and money or can taking the level 1 help me on level2?

Well, of course a company whose business is test prep will tell you to take lots of tests. No, you are young enough that you should wait: there is a good chance of both tests being revised before you use either one for college entrance. You will keep your math skills fresh by being involved in other math activities meanwhile, if you are like most AoPS participants.

If you can get a 760 as a freshman in IC

Then you should definitely wait a year or so and take the IIC and ace it.
It just covers a little more heavier into algebra 2, I believe. It is worth it.

If you are looking to apply to top schools, you should strive to challenge yourself...

my experience: took algII/trig/pre-cal in tenth grade. took math IIC in june. it took me about 20 minutes to do. played games on calculator rest of time. got 800 on it, 90th percentile. does this help you at all?

For me, I took this test (IIC) before I took precalculus (aka freshman year) and I did well on the test: 800
I do not think you should wait until you are older, you should just take when you feel you are ready, and take practice tests to gauge ability.

Do not take IC because = waste of time


Solve in $ \mathbb{Z^\plus{}}$ the equation $ 7^x\plus{}x^4\plus{}47\equal{}y^2$.

Here's a start.  Consider the equation (mod 4).  The right side is congruent to either 0 or 1.  If we subtract $ 47\equiv 3\text{(mod 4)}$, we get $ 7^x\plus{}x^4\equiv \text{1 or 2 (mod 4)}$.  A fourth power can only be congruent to 0 or 1 (mod 4) since it is also a square.  Either way, $ 7^x$ must be congruent to 1.  This only happens for even powers meaning $ x^4\equiv 0\text{(mod 16)}$.  And now I'm getting lazy, so...

First we work modulo $ 8$ and note that $ x$ odd would yield $ y^2\equiv (\minus{}1)^x\plus{}x^4\plus{}(\minus{}1)\equiv \minus{}1\bmod{8}$, which has no solution. Therefore, we can write $ x\equal{}2z$ with $ z\in\mathbb{Z}^\plus{}$ and get $ 49^z\plus{}16z^4\plus{}47\equal{}y^2$. Making the substitution $ y\equal{}7^z\plus{}t$ with $ t>0$ gives (*) $ 16z^4\plus{}47\equal{}2t\cdot7^z\plus{}t^2$. For $ t\geq 4$ we compute $ 16z^4\plus{}47\geq 8\cdot 7^z\plus{}16\Longrightarrow 2z^4\plus{}4>7^z$, a contradiction and we must have $ t\in\{1,2,3\}$. Analysing (*) modulo $ 7$ we get $ 2z^4\minus{}2\equiv t^2\bmod{7}\Longrightarrow z^4\equiv 1\plus{}4t^2\bmod{7}$, which has no solution for $ t\equal{}1,2$. Therefore, we must have $ t\equal{}3$ and (*) rewrites as $ 8z^4\plus{}19\equal{}3\cdot 7^z$. Clearly, the function $ f(z)\equal{}3\cdot 7^z\minus{}8z^4\minus{}19$ is stricly positive for $ z>2$. Since $ f(1)<0$ and $ f(2)\equal{}0$ we get only one solution determined by $ z\equal{}2$ and represented by $ x\equal{}2z\equal{}4$ and $ y\equal{}7^z\plus{}t\equal{}52$.


heres a question from the intermediate counting and probability question that im having trouble with...

we are given a table with j rows and k columns. a box in the m'th row and the n'th column is chosen. how many rectangles formed by boxes in the table contain the chosen box? (Your answer should be written in terms of j,k,m and n)

after this problem im done with the whole book  :D

[hide="Hint"] A rectangle can be represented in exactly two ways by considering two extremal boxes that it contains. [/hide]

another idea...

[hide="..."]how many rectangles can be formed that do [b]not[/b] contain the chosen box?[/hide]

Is the answer as simple as
[hide]Jk (total) - (m-1)(n-1)... (m-1)(n-1) being the rectangles non-inclusive of m,n.  or is it more complicated?  Sorry, just putting out a guess.[/hide]


$ x,y,z$ are positive reals such that $ x\plus{}y\plus{}z\equal{}1$. Prove that
$ 2(\frac{1}{1\plus{}x}\plus{}\frac{1}{1\plus{}y}\plus{}\frac{1}{1\plus{}z})\leq\frac{6}{1\plus{}xy\plus{}yz\plus{}zx}\leq\frac{1}{1\minus{}x}\plus{}\frac{1}{1\minus{}y}\plus{}\frac{1}{1\minus{}z}$

Since we have  $ 1 \plus{} xy \plus{} yz \plus{} xz \equal{} (x \plus{} y \plus{} z)^2 \plus{} xy \plus{} xz \plus{} yz \equal{} \sum_{cyc}(x \plus{} y)(y \plus{} z)$

and $ \sum_{cyc}\left(\frac {1}{1 \minus{} x}\right) \equal{} \frac {\sum(x \plus{} y)(y \plus{} z)}{\prod(x \plus{} y)}$

we only need to prove that

$ (\sum(x \plus{} y)(y \plus{} z))^2\geq 6(x \plus{} y)(y \plus{} z)(z \plus{} x)$

and using $ (ab \plus{} bc \plus{} ca)^2\geq 3abc(a \plus{} b \plus{} c)$

we get  $ (\sum(x \plus{} y)(y \plus{} z))^2\geq 3(x \plus{} y)(y \plus{} z)$$ (x \plus{} z)(x \plus{} y \plus{} y \plus{} z \plus{} x \plus{} z) \equal{} 6(x \plus{} y)(y \plus{} z)(z \plus{} x)$

Done !!

The other Side is equivalent to

$ \frac {(x \plus{} y) \plus{} (y \plus{} z) \plus{} (z \plus{} x)}{(x \plus{} y)(y \plus{} z) \plus{} (y \plus{} z)(x \plus{} z) \plus{} (x \plus{} z)(x \plus{} y)}\ge \frac {2}{3}\sum_{cyc}\left(\frac {1}{(x \plus{} y) \plus{} (y \plus{} z)}\right)$

Let $ x \plus{} y \equal{} a$ and $ y \plus{} z \equal{} b$ and $ z \plus{} x \equal{} c$

Then rewrite the inequality as

$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}\ge \frac {2}{3}\sum_{cyc}\frac {1}{a \plus{} b}$

$ \Leftrightarrow 3(a \plus{} b \plus{} c)\ge 2.\sum_{cyc}\left(\frac {c(a \plus{} b) \plus{} ab}{a \plus{} b}\right)$

$ \Leftrightarrow a \plus{} b \plus{} c \ge \sum_{cyc}\frac {2ab}{a \plus{} b}$

Which is Obvious since we have by Am_GM

$ \frac {a \plus{} b}{2} \equal{} \frac {(a \plus{} b)^2}{2(a \plus{} b)}\ge \frac {2ab}{a \plus{} b}$  :)

Nice solution, Evariste-Galois :)
My solution use Chebyshev Inequality.

$ 2\equal{}(1\minus{}x)\plus{}(1\minus{}y)\plus{}(1\minus{}z)\equal{}\frac{1\minus{}x^2}{1\plus{}x}\plus{}\frac{1\minus{}y^2}{1\plus{}y}\plus{}\frac{1\minus{}z^2}{1\plus{}z}$
$ \geq\frac{1}{3}(3\minus{}x^2\minus{}y^2\minus{}z^2)(\frac{1}{1\plus{}x}\plus{}\frac{1}{1\plus{}y}\plus{}\frac{1}{1\plus{}z})$
$ \equal{}2(1\plus{}xy\plus{}yz\plus{}zx)(\frac{1}{1\plus{}x}\plus{}\frac{1}{1\plus{}y}\plus{}\frac{1}{1\plus{}z})$

Then, we get left side inequality. We can prove other side by similar method

[quote="jedaihan"]Nice solution, Evariste-Galois :)
My solution use Chebyshev Inequality.

$ 2 \equal{} (1 \minus{} x) \plus{} (1 \minus{} y) \plus{} (1 \minus{} z) \equal{} \frac {1 \minus{} x^2}{1 \plus{} x} \plus{} \frac {1 \minus{} y^2}{1 \plus{} y} \plus{} \frac {1 \minus{} z^2}{1 \plus{} z}$
$ \geq\frac {1}{3}(3 \minus{} x^2 \minus{} y^2 \minus{} z^2)(\frac {1}{1 \plus{} x} \plus{} \frac {1}{1 \plus{} y} \plus{} \frac {1}{1 \plus{} z})$
$ \equal{} \boxed{\frac {2}{3}}(1 \plus{} xy \plus{} yz \plus{} zx)(\frac {1}{1 \plus{} x} \plus{} \frac {1}{1 \plus{} y} \plus{} \frac {1}{1 \plus{} z})$

Then, we get left side inequality. We can prove other side by similar method[/quote]

Very nice solution it's simpler than mine  :lol:


search
number theory proposed
number theory

Find all $ x,n\in\mathbb{Z}$,such that
                       $ x^2\plus{}7\equal{}2^n$.

$ 2^n\minus{}8\plus{}1$ is square if and only if $ n\equal{}3$ or $ n\equal{}4$.  (These proplem considered in these forum many times).

[quote="Rust"]$ 2^n \minus{} 8 \plus{} 1$ is square if and only if $ n \equal{} 3$ or $ n \equal{} 4$.  (These proplem considered in these forum many times).[/quote]

I think the cases n=5 or n=7 match this problem ($ 2^5\minus{}7\equal{}5^2,\, 2^7\minus{}7\equal{}11^2$).

I am sorry.
If n=2m is even, then $ (2^m\minus{}x)(2^m\plus{}x)\equal{}7\to 2^m\plus{}x\equal{}7,2^m\minus{}x\equal{}1\to n\equal{}4,x\equal{}3$.
If $ n\equal{}2m\plus{}1$ is odd, then $ x^2\minus{}2(2^m)^2\equal{}\minus{}7\to x\plus{}2^m\sqrt 2 \equal{}(1\plus{}2\sqrt 2 )(1\pm \sqrt 2)^{2k}$ or
$ 2^m\equal{}\sum_{i\equal{}0}^k2^{i\plus{}1}C_{2k}^{2i}\frac{k\plus{}i\plus{}1}{2i\plus{}1}$ or $ 2^m\equal{}\sum_{i\equal{}0}^k2^{i\plus{}1}C_{2k}^{2i}\frac{3i\plus{}1\minus{}k}{2i\plus{}1}$.
$ k\equal{}0\to m\equal{}1,n\equal{}3$,
$ k\equal{}1\to m\equal{}3,2\to n\equal{}7 \ or \ n\equal{}5.$
If $ k>1$ consider by mod 8 give $ k\equal{}1\mod 4$ for first case and $ k\equal{}3\mod 4$ for second.

It's still wrong since $ x\equal{}181, n\equal{}15$ is also a solution. This problem is rather very hard and it was discussed already, I think.

It's Ramanujan-Nagell equation
See here: http://en.wikipedia.org/wiki/Ramanujan-Nagell_equation
 http://mathworld.wolfram.com/RamanujansSquareEquation.html

Actually this problem was at one of preparation olympiad,and i still don't know a solution.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=865863058&t=66245]here[/url] (where I give an incorrect solution - forgive me, I was young!  :oops: ) and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=504745393&t=17135]here[/url].

Anyone has an elementary proof?
I don't understand in algebraic number theory solution.

I doubt it.  Even an elementary proof would need to explain the large last solution.  The fact that Ramanujan couldn't resolve this problem suggests it's fairly difficult.


If $ a,b,c$ are [b]real[/b] numbers such that $ abc(a \plus{} b \plus{} c) > 0,$ then

$ \frac {a^2}{2b^2 \minus{} bc \plus{} 2c^2} \plus{} \frac {b^2}{2c^2 \minus{} ca \plus{} 2a^2} \plus{} \frac {c^2}{2a^2 \minus{} ab \plus{} 2b^2} \ge 1.$

Did you meant:
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} abc(a \plus{} b \plus{} c) \geq\ 2\sum\ a^2b^2$
with $ abc(a\plus{}b\plus{}c) \geq\ 0$
 :wink:

[quote="nguoivn"]Did you meant:
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} abc(a \plus{} b \plus{} c) \geq\ 2\sum\ a^2b^2$
with $ abc(a \plus{} b \plus{} c) \geq\ 0$
 :wink:[/quote]

+)If $ a;b;c <0$ we can prove that a easy way similar $ a;b;c >0$

+) If not $ a;b;c <0.$

Let $ a\plus{}b\plus{}c\equal{}p;ab\plus{}bc\plus{}ca\equal{}q;abc\equal{}r$

we have $ pr \ge\ 0$

Case1: $ p\ge\ 0 \rightarrow r \ge\ 0$

Assum $ p\equal{}1$

$ Ineq \leftrightarrow 9r \ge\ 4q \minus{} 1(1)$

.If $ q \le\ 0 \rightarrow LHS \ge\ RHS$

. If $ q \ge\ 0 \rightarrow a;b;c \ge\ 0 \rightarrow (1)$ true follow schur

Case2: $ p\le\ 0 \rightarrow r \le\ 0$. Assum $ p\equal{}\minus{}1$

And we have in$ (a;b;c)$ have one number $ \le\ 0$.

$ Assum a \le\ 0;b,c \ge\ 0$ so $ |a| \ge\ b\plus{}c \rightarrow q \le\  0$

$ Ineq \leftrightarrow 9r \le\ 1\minus{}4q$ which obvious with $ q \le\ 0$ and $ r \ge\ 0$

We have done! :)

[quote="onlylove_math"][quote="nguoivn"]Did you meant:
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} abc(a \plus{} b \plus{} c) \geq\ 2\sum\ a^2b^2$
with $ abc(a \plus{} b \plus{} c) \geq\ 0$
 :wink:[/quote]

+)If $ a;b;c < 0$ we can prove that a easy way similar $ a;b;c > 0$

+) If not $ a;b;c < 0.$

Let $ a \plus{} b \plus{} c \equal{} p;ab \plus{} bc \plus{} ca \equal{} q;abc \equal{} r$

we have $ pr \ge\ 0$

Case1: $ p\ge\ 0 \rightarrow r \ge\ 0$

Assum $ p \equal{} 1$

$ Ineq \leftrightarrow 9r \ge\ 4q \minus{} 1(1)$

.If $ q \le\ 0 \rightarrow LHS \ge\ RHS$

. If $ q \ge\ 0 \rightarrow a;b;c \ge\ 0 \rightarrow (1)$ true follow schur

Case2: $ p\le\ 0 \rightarrow r \le\ 0$. Assum $ p \equal{} \minus{} 1$

And we have in$ (a;b;c)$ have one number $ \le\ 0$.

$ Assum a \le\ 0;b,c \ge\ 0$ so $ |a| \ge\ b \plus{} c \rightarrow q \le\ 0$

$ Ineq \leftrightarrow 9r \le\ 1 \minus{} 4q$ which obvious with $ q \le\ 0$ and $ r \ge\ 0$

We have done! :)[/quote]
Nice ,my friend , :lol: And the ine here is also true
With $ a,b,c$ are [b]real[/b] number we have $ a^4 \plus{} b^4 \plus{} c^4 \plus{} abc(a \plus{} b \plus{} c)\ge \sum ab(a^2\plus{}b^2)$

[quote="nguoivn"]Did you meant $ a^4 \plus{} b^4 \plus{} c^4 \plus{} abc(a \plus{} b \plus{} c) \geq\ 2\sum\ b^2c^2$ with $ abc(a \plus{} b \plus{} c) \geq\ 0$? :wink:[/quote]Yes, see : http://www.mathlinks.ro/viewtopic.php?t=229307
[quote="Allnames"]With $ a,b,c$ are [b]real[/b] number we have $ a^4 \plus{} b^4 \plus{} c^4 \plus{} abc(a \plus{} b \plus{} c)\ge \sum bc(b^2 \plus{} c^2).$[/quote]See : http://www.mathlinks.ro/viewtopic.php?t=1166


Let $ M$ denote the point of intersection of the diagonals $ AC$ and $ BD$ of a quadrilateral $ ABCD$, and let $ N$ be any point on the line $ AB$. Let the line $ MN$ intersect the line $ CD$ at a point $ L$. Prove that

$ \frac{DL}{LC}\cdot\frac{MC}{MD}\cdot\frac{MB}{MA}\cdot\frac{AN}{NB}\equal{}1$.

[url=http://imageshack.us][img]http://img263.imageshack.us/img263/1134/ayanalm5.jpg[/img][/url]
Shot at 2007-07-15

This is a trivial consequence of the following fact (applied to n = 4, the 4-gon ABDC, and the collinear points N, M, L, M on its sidelines AB, BD, DC, CA):

[color=blue][b]Menelaos theorem for polygons.[/b] Let $ A_{1}A_{2}...A_{n}$ be a $ n$-gon, and let $ P_{1}$, $ P_{2}$, ..., $ P_{n}$ be points on its sidelines $ A_{1}A_{2}$, $ A_{2}A_{3}$, ..., $ A_{n}A_{1}$, respectively. If the points $ P_{1}$, $ P_{2}$, ..., $ P_{n}$ are collinear, then $ \prod_{i=1}^{n}\frac{A_{i}P_{i}}{P_{i}A_{i+1}}=\left(-1\right)^{n}$, where the segments are directed and the indices are cyclic modulo $ n$ (so that $ A_{n+1}=A_{1}$).[/color]

The converse holds only for $ n=3$.

Proving the above fact is an exercise.

  Darij

Or alternativelly, it is enough to prove that $ \frac{LD}{LC}\cdot \frac{MC}{MD}= \frac{NB}{NA}\cdot \frac{MA}{MB}$ $ ,(1)$

We denote the point $ P\equiv AD\cap LN$ and apply the [b][size=100]Menelaus theorem[/size][/b] in triangles $ \bigtriangleup ACD,$ $ \bigtriangleup ABD,$ taken as their transversal the segment line $ LN,$ we have the equalities:

$ \frac{LD}{LC}\cdot \frac{MC}{MA}\cdot \frac{PA}{PD}= 1$ $ \Longrightarrow$ $ \frac{(LD)\cdot (MC)}{LC}= \frac{(PD)\cdot (MA)}{PA}$ $ ,(2)$

$ \frac{NB}{NA}\cdot \frac{PA}{PD}\cdot \frac{MD}{MB}= 1$ $ \Longrightarrow$ $ \frac{(NB)}{(NA)\cdot (MB)}= \frac{(PD)}{(PA)\cdot (MD)}$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac{PD}{PA}\cdot \frac{MA}{MD}= \frac{PD}{PA}\cdot \frac{MA}{MD}$ $ \Longrightarrow$ $ 1 = 1$ and the proof is completed.

Kostas Vittas.


algebra
polynomial
function
calculus
advanced fields
advanced fields unsolved

While solving another problem, I bumped into this recurrence.

$ s_{n}\equal{}\frac{1}{r_{1}\minus{}r_{1}^{n}}\sum_{k\equal{}1}^{n\minus{}1}s_{k}B_{n,k}\left(
r_{1},r_{2},...,r_{n\minus{}k\plus{}1}\right)$

Assuming that $ \left\{ r_{k}\right\}$ is known and $ \left\{ s_{k}\right\}$ is to be determined.  Also s_1 = 1 and $ B_{n,k}$ is the partial Bell polynomial given by a formula on this page:
http://en.wikipedia.org/wiki/Bell_polynomials

I'm unaware of any recurrences known of the partial Bell polynomials that might help.  I could very well be doing something wrong but I am also not seeing how the generating function for the partial Bell polynomials helps.

Any clues, thoughts, insights?

[quote="t0rajir0u"]This looks very similar to the recurrence mentioned under "representation of sequences of binomial type."  Can you apply umbral methods to $ s_n$?  If there is a sequence of polynomials $ s_n(x)$ of binomial type such that $ s_n(1) \equal{} s_n$, then all you have to do is figure out what the linear functional $ L$ that sends $ x^n$ to $ s_n(x)$ is.  Or, well, you could if it weren't for that first term.[/quote]
Thanks for replying.  I'll look into those techniques.  I've heard of binomial type sequences and ubmral calculus but never looked into them really.  Thanks.

I have attached a pdf which explains more about the problem.


function
vector
algebra
domain
real analysis
real analysis theorems

Please supply definitions if you know because I need them in order to solve some practice problems.
I know the definition for a uniform norm of a bounded function, but not the following:
Consider the vector space $R^{n}$. Choose $v \in R^{n}$. What is the definition of $\parallel v\parallel _{u}$?

Calling it the "uniform" norm is appropriate when you're thinking of the elements of the vectors space of functions. Now, we can think of the elements of $\mathbb{R}^{n}$ as functions if we want to - they're functions whose domain is the set $\{1,2,\dots,n\}.$ But we usually don't. The other name for the same norm is to call it the "infinity" norm. It's also the "sup" norm.

In $\mathbb{R}^{n}$ the infinity norm of a vector $\mathbf{x}$ is $\|\mathbf{x}\|_{\infty}=\max_{1\le j\le n}|x_{j}|.$

In $\mathbb{R}^{2},$ the unit ball under this norm would be the square with corners $(\pm1,\pm1).$

Thank you so much Dr.Merryfield!
I will go ahead and use that definition.


inequalities
function
trigonometry
linear algebra
matrix
inequalities unsolved

No equality cases!!!!!     :D
It seems like the solution above is a wrong solution

The farthest I can go is $ z^{2}=3+xy$...I don't know if that goes anywhere...

If you put $ z\equal{}\minus{}(x\plus{}y)$ you have that $ x^{2}\plus{}xy\plus{}y^{2}\equal{}3$ (from $ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}6$). Hence, you only have one variable, and if you find the maximum of the function of one variable, you get that the maximum of $ x^{2}y\plus{}y^{2}z\plus{}z^{2}x$ is $ 6$, and equality in many cases, for example $ x\equal{}2\cos\left(\frac{\pi}{9}\right)$.

Yes, xavi, you are right!
$ x^{2}y\plus{}y^{2}z\plus{}z^{2}x\leq 6\Leftrightarrow x^{3}\plus{}3x^{2}y\minus{}y^{3}\leq6.$
If $ x^{3}\plus{}3x^{2}y\minus{}y^{3}\geq0$ we obtain that $ x^{3}\plus{}3x^{2}y\minus{}y^{3}\leq6\Leftrightarrow$
$ \Leftrightarrow3(x^{3}\plus{}3x^{2}y\minus{}y^{3})^{2}\leq4(x^{2}\plus{}xy\plus{}y^{2})^{3}\Leftrightarrow(x^{3}\minus{}3x^{2}y\minus{}6xy^{2}\minus{}y^{3})^{2}\geq0.$

thank you Xavi and Arqady . it was a very nice solition .

[quote="zaya_yc"]x,y,z  real number , x+y+z=0 and $ x^{2} \plus{} y^{2} \plus{} z^{2} \equal{} 6$  find max --->
$ x^{2}y \plus{} y^{2}z \plus{} z^{2}x$[/quote]
The generalization is
If $ \left\{\begin{matrix} x\plus{}y\plus{}z &\equal{}m \hfill\\ x^2\plus{}y^2\plus{}z^2 &\equal{}n^2 \hfill\\ \end{matrix} \right.$ where $ m,n$ $ (3n^2>m^2)$ are the given constant. Prove that
\[ \minus{}\frac{2}{9}\left( \frac{3n^2\minus{}m^2}{2}\right)^{3/2}\plus{}\frac{m^3}{9} \le x^2y\plus{}y^2z\plus{}z^2x \le \frac{2}{9}\left( \frac{3n^2\minus{}m^2}{2}\right)^{3/2}\plus{}\frac{m^3}{9}\]


What are the first  100 decimal places of (7 + sqrt(50))^100?

They are all 9. Consider what happens when you add the very small positive number $ (7\minus{}\sqrt{50})^{100}<\frac{1}{10^{100}}$. To put it frankly, you get an integer. Pwned.

$ (7\plus{}\sqrt{50})^{100}\plus{}(7\minus{}\sqrt{50})^{100}$ is an integer because it satisfies a simple recurrence relation that I will not delve into. Hence, since $ (7\minus{}\sqrt{50})^{100}$ is so small, $ (7\plus{}\sqrt{50})^{100}$ needs 9's in it's first 100 decimal places so that their sum will be an integer.

$ (7\plus{}\sqrt{50})^{100}\plus{}(7\minus{}\sqrt{50})^{100} \equal{} \sum_{n\equal{}0}^{100}\binom{100}{n}\cdot\left(7^{100\minus{}n} \cdot \left(\left(\sqrt{50}\right)^n\plus{}\left(\minus{}\sqrt{50}\right)^n\right)\right)$. Simplifying, we have $ \sum_{n\geq 0, even}\binom{100}{n}\cdot 7^n\cdot 2\cdot 50^{n/2} \in \mathbb{Z}$ and the inequality follows as $ \sqrt{50}\minus{}7<\frac{1}{10}$.

[quote="rofler"]$ (7 \plus{} \sqrt {50})^{100} \plus{} (7 \minus{} \sqrt {50})^{100}$ is an integer because it satisfies a simple recurrence relation that I will not delve into. Hence, since $ (7 \minus{} \sqrt {50})^{100}$ is so small, $ (7 \plus{} \sqrt {50})^{100}$ needs 9's in it's first 100 decimal places so that their sum will be an integer.[/quote]
Hmm, I understand it as an integer by doing the binomial expansions, but what is the recurrence that you hinted?
(By the way, what does "Pwned" mean?)

Well, consider the recurrence $ f(0)\equal{}2$ $ f(1)\equal{}14$ and $ f(n\plus{}2)\equal{}14f(n\plus{}1)\plus{}f(n)$.

Pwned :P

Repeating portions of answers you've learned before that are only comprehensible to people who already know the answer isn't "pwning" anything, it's just obnoxious.  Much like appending "Pwned" to the end of your posts.

Aww, come on, just having fun. Like that thread where people were trying to find the meaning of 1337.

Pwned is an internet meme.

Edit: $ f(n)$ defined above can be rewritten as $ (7\plus{}\sqrt{50})^n\plus{}(7\minus{}\sqrt{50})^n$ because they are the roots of the characteristic equation $ x^2\equal{}14x\plus{}1$. You can look up more on this on wikipedia.

I am sure that JBL already knew and recognized this, and I know that I did, but you ought to explain your work clearly (as you just did) without jumping to conclusions, to help others who are curious.  :wink:


Given a trapezoid $ABCD \hspace{0,2cm}(BC \parallel AD)$ for which $2AB=CD$, $\angle BAD+\angle CDA = 120^{\circ}$.
Find the angles of $ABCD$.

[color=darkblue][b]Very simple problem for Minsk Olympiad ![/b] ! Denote the point $E\in (AD)$ for which $CE\parallel AB$ and the middlepoint $M$ of the side $[CD]$.Then $AB=CE=CM=MD$, $A=120-D$, $m(\widehat{AEC})=60+D$, $m(\widehat{ECD})=60$, $m(\widehat{MED})=60-D$. Therefore, $CE=CM$ and $m(\widehat{ECM})=60$ $\Longrightarrow$ $EM=MC=MD$ $\Longrightarrow$ $\widehat{MED}\equiv\widehat{MDE}$ $\Longrightarrow$ $60-D=D$ $\Longrightarrow$ $D=30$ $\Longrightarrow$ $A=B=90$, $C=150$, $D=30\ .$
[/color]

My solution was slightly different

$A+D=120^{o}$, so the lines $AB,DC$ intersect at $F$, to the direction of $B$ and $C$. Also, $\angle AFD = 60^{o}$

$\frac{BF}{CF}= \frac{AB}{DC}= \frac{1}{2}\Rightarrow \angle CBF=90^{o}\Rightarrow \boxed{A=B=90^{o}, \ D=30^{o}, \ C=150^{o}}$

I just drew perpendiculars from $B$ and $C$, took the sines of $A$ and $D$, and because one is double the other, and the two angles sum to 120 degrees, I got one was 90 degrees and the other 30 degrees.  The rest followed from the right triangles formed.


I think one of those equals signs was supposed to be a plus, otherwise the problem's trivial.

[quote="Seo-Yeon, Sohn"]fnd all (x,y,z)

x^2 = xy = y^2 + xz

x,y,z are all integers[/quote]
$ x(x\minus{}y)\equal{}0$ ........... easy

Solution? (long)
[hide]$ x^2\plus{}xy\minus{}xz\equal{}y^2$
$ x(x\plus{}y\minus{}z)\equal{}y^2$
Case 1:, $ x\equal{}y$
Then $ x\plus{}y\minus{}z\equal{}y$, $ z\equal{}y$...$ (n,n,n)$
Case 2: $ y^2\equal{}x^2a^2$
Then $ x\plus{}y\minus{}z\equal{}xa^2$
$ 1\plus{}a\minus{}\dfrac{z}{x}\equal{}a^2$
$ a^2\minus{}a\minus{}1\plus{}\dfrac{z}{x}\equal{}0$
$ a\equal{}\dfrac{1\pm \sqrt{1\minus{}4(\minus{}1\plus{}\dfrac{z}{x})}}{2}$
$ \sqrt{5\minus{}\dfrac{4z}{x}}$ should be $ 0, 1, 4$.
When it's zero, $ z\equal{}\dfrac{5x}{4}$ and $ y\equal{}\dfrac{x^2}{4}$, or $ (4n, 4n^2, 5n)$, but checking shows this works only when $ n\equal{}\dfrac{1}{2}$, so it doesn't work.
When it's one, $ z\equal{}x$, $ y\equal{}x^2$, $ (n, n^2, n)$, but checking shows this works only for $ n\equal{}0,1$, so no new solutions.
When it's four, $ z\equal{}\dfrac{x}{4}$, $ y\equal{}\dfrac{9x^2}{4}, \dfrac{x^2}{4}$...$ (4n, 36n^2, n)$ or $ (4n, 4n^2, n)$.  Checking the first yields $ n\equal{}\dfrac{1}{6}, \dfrac{\minus{}1}{8}$, so no new solutions.  Checking the second yields radicals.
Thus, $ x\equal{}y\equal{}z$.
(I will be very mad if I messed up somewhere.)[/hide]

Err...

$ z\equal{}x\plus{}y\minus{}\frac{y^2}{x}$

Thus you need all pairs $ (x,y)$ so $ x|y^2$ and you are done. There are clearly more solutions... ex. $ x\equal{}4$, $ y\equal{}2$.

Let $ (x,y)\equal{}d$, $ x\equal{}ad$, $ y\equal{}bd$. Then $ a|b^2d$ so $ a|d$. $ d\equal{}ma$. Then $ x\equal{}ma^2$, $ y\equal{}bam$ where $ (a,b)\equal{}1$ and $ m$ is an integer. $ z$ is whatever is left. This works since $ x|y^2$.


geometry
trigonometry
search
circumcircle
power of a point
radical axis
geometry proposed

sorry if it posted before.
$ T_a$ is the tangency point of incircle and a circle wich passes through points $ B,C$ of triangle $ ABC$.name the line $ AT_a$ with $ \ell_a$.
$ \ell_b,\ell_c$ are defined in a similar way.Prove that $ \ell_a,\ell_b,\ell_c$ are concurrent.

I'll be glad to see a nice proof.

Denote by $ A', B', C'$ the points of tangency of the incircle with the sides $ BC, BC, AB$ respectively. Then by Steinbart`s theorem, the lines $ l_{a}, l_{b},l_{c}$ are concurent if $ A'T_{a}, B't_{b}, C'T_{c}$ are concurrent(you can prove this via trig ceva.)

So now we are left to prove that the $ A'T_{a}$-tipe lines are concurrent,which is equivalent to proving that their poles wrt the incircle are collinear.
Let $ M_{a}$ be the intersection between the tangent at $ T_{a}$ to the incircle and the side $ BC$.
Then the power of the point $ M_{a}$ wrt the incircle is $ M_{a}A'^{2}$ and the power of $ M_{a}$ wrt the circumscribed circle of triangle $ ABC$ is $ MB\cdot MC\equal{}M_{a}T_{a}^{2}\equal{}M_{a}A'^{2}$, therefore the point $ M_{a}$ belongs to the radical axis of the incircle and the circumscribed circle of triangle $ ABC$,  which is a line.
Thus $ M_{b}$ and $ M_{c}$ lie on the same line(by the same argument), therefore $ M_{a},M_{b}, M_{c}$ are collinear, and by the relations we have allready proven, we have that the $ l_{a}$-tipe lines are concurent, thus the problem is solved.

Andyciup, the Steinbart theorem states that $ l_{a}, l_{b},l_{c}$ are concurent if and only if $ A'T_{a}$, $ B't_{b}$, $ C'T_{c}$ are concurrent [b]or[/b] the intersections $ A'T_{a} \cap B'C'$, $ B'T_{b} \cap C'A'$, $ C'T_{c} \cap A'B'$ are collinear. Anyway in this case it works since one uses the converse.

For a more exact result, the lines $ A'T_{a}$, $ B'T_{b}$, $ C'T_{c}$ are concurrent on the line $ OI$ at $ X(57)$, the homothetic center of intouch and excentral triangles.

I remember also an other proof for the initial problem was posted in the Preolympiad section by Virgil Nicula, using only the trigonometric version of Ceva. (but I can't find the link right now)

Of course that`s the whole steinbart theorem. But I didn`t need the [b]or[/b] case in my solution, thus I didn`t state it.


Where can I find past Physics Olympiad problems (and solutions)? I'm looking for some more difficult problems that nonetheless basic material (I'm just starting physics this year), like the AMC is analogous to math.


could it be that either k has to be natural, or the question has to say k is the arithmetic average?

Since for example for a1=1,a2=2, the optimal k=1.5 ;)

WLOG $a_{1}\leq a_{2}\leq \cdots \leq a_{n}$

It's is clear that $|a_{i}-k| + |a_{n-i}-k|$  for $1\leq i \leq [\frac{n}{2}]$ is minimal iff $a_{i} \leq k \leq a_{n-i}$,

So summing it up we see that if $n$ is odd minimum occurs only when $k=a_{(n+1)/2}$
or if $n$ is even when $k \in [a_{(n)/2},a_{(n+2)/2}]$

Thus the sum is minimal when $k$ is the median of all the $a_{j}$ no matter if they are real or natural.

If $n$ is even you can choose $k=a_{n/2}$ if you want a natural $k$. :)

[quote="manuel"]Let $a_1,...,a_n$ be natural numbers.
Prove that $\sum_{i=1}^n |a_i-k|$ is minimum if $k$ is the median of $a_1,...,a_n$[/quote]

the conclusion seems wrong...
When a1=1, a2=2, a3=100, k=2 when the sum is minimum...


In triangle $ CAT$, we have $ \angle{ACT}\equal{}\angle{ATC}$ and
$ \angle{CAT}\equal{}36^\circ$. If $ \overline{TR}$ bisects $ \angle{ATC}$,
then how many degrees is $ \angle{CRT}$?
[asy]/* AMC8 2000 #13 Problem */
draw((0,0)--(.5,1.75)--(1,0)--cycle);
draw((1,0)--(.15,.5));
label("$R$", (.15,.5), W);
label("$C$", (0,0), SW);
label("$T$", (1,0), SE);
label("$A$", (.5,1.75), N);[/asy]

CAT is iscosceles, so angles C and ATC are (180-36)/2=72. RT is an angle bisector, so angle CTR=72/2=36. so anglr CRT=180-36-72=72


Given $ 2n\minus{}1$ integers,prove that one can choose exactly $ n$ of them whose sum is divisible by $ n$.

This is a very well known result:
 http://www.mathlinks.ro/viewtopic.php?t=14797
http://www.mathlinks.ro/viewtopic.php?t=28929&search_id=1825945745


induction
combinatorics unsolved
combinatorics

We are given n ( forgot how in English, but in romanian MUCHII) things that join vertices. They are labelled from one to n. How many trees can we form out of these things, such that for any "thing", its subtree does not containg a "thing" with a smaller number (we count everything from the root).

Anyway, I can't understand what did Dust mean?

What things are labelled? Vertices or edges?

I think I understand this correctly (correct me if not): the edges are labelled, and every path starting with the root (which must be labelled "1") and ending in a leaf must have the labels on its edges in increasing order. DusT, could you perhaps give an example please?

Ok, I am sorry, it was not too clear.
Given the edges labeled $1,2,\ldots ,n$ , how many rooted trees can we form using this edges, if for any edge, it's subtree must not contain any edges with number smaller than it.
For $n=2$ we have

      O               O                O
      |               / \              /   \
      1             1   2           2     1
      |            /       \        /         \
      0          O        O     O           O
      |
      2
      O
only 3 such trees. I wanted to show that order of edges matters!!!
I hope it's clear now!!! :)

Did anybody calculate required number for $n=3$? I have obtained 15, but I am not sure :blush:

Yes, I also got 15. I think $1.3....(2n-1)$ would be a good guess for the general case; each tree seems to spawn $2n-1$ new ones on the next level.

Edit: Actually, I can prove this. Suppose it is true for $n$. Then to create trees with $n+1$ edges which satisfy the condition, we just add edge $n+1$ to a valid tree with $n$ edges. There are $2n-1$ ways to do this for each tree: for leaves, attach the extra edge to the endpoint of the leaf or one side of the other end; for other edges, to either side of the bottom of the edge. This gives us $2n$; -1 since we can't add one below the root.

A closed expression is $\frac{(2n-1)!}{2^{n-1}(n-1)!}$.

Great!!!!!!
Why am I so blind! I have obtained 105 for $n=4$, but didn't see formula. :blush:

You didn't write what you exactly mean, Fiachra, so I will write my vision.
Consider arbitrary tree obtained for $n$. Then new edge $n+1$ has exactly $2n+1$ possibilities to be placed to the given tree. Indeed, $n+1$ must be leaf-edge and it can be attached to the arbitrary vertex in our tree. And if vertex has $k$ descendant then we have $k+1$ choices to attach $n+1$. Note that $k+1$ is a degree of this vertex if it is not a root; for root its degree is exactly $k$. Since tree has $n$ edges we conclude that overall sum of degrees is $2n$, add we must add 1 to obtain desired number of appropriate places for edge $n+1$.
Thus by induction we get $1\cdot 3\cdot 5\cdot...\cdot(2n+1)$ for $n+1$.

Hmm... sorry, yes. Thank you for your (far more accurate) explanation.

Thus we have just solved problem, which was unsolved during 1.5 year!!!
Congratulations! :)

What next?  ;)

Yep, this is the answer. A funny solution is by making an Eulerian parcourse of the tree. An Eulerian parcourse is done like this:
Take the root at first. For all it's neighbours, write the edge containing them, then EulerParcourse(neighbour), write again the edge.
Let me give an example:
For the picture, we hace 1221,1122,2211. Now see how the condition tranforms!!!

I guess order must be 1122, 2211,1221 ;)
In fact, we come to the same solution. Am i right?

Yes, it is like this. (My mistake). Also, this trees are in bijection with many other combinatorial structures.......


Whitney had to write numbers on the tickets for the school play.
The ticket numbers ranged from 1 to 325, inclusive. How many
digits did she have to write?

9(1)=9 digits form 1-digit numbers
90(2)=180 digits form 2-digit numbers
226(3)=678 digits form 3-digit numbers

total of 9+180+678=867 digits


Today we had a test in class, and we had one extra credit question. At the last minute, I took a guess, plugged it in, and saw that it worked. My teacher told me that I had the right answer, but Ifeel guilty because I really didn't know how to do it:

(8^3x-3)-(8^3x)=448

I know for sure that the answer is 2/3.
Thanks in advanced.

hello, do you mean the equation
$ 8^{3x\minus{}3}\minus{}8^{3x}\equal{}448$ ?
Sonnhard.

hello, in a few minutes i will post the solution of your equation.
Sonnhard.

hello, your equation has no solution, since
$ 8^{3x}(3^{\minus{}3}\minus{}1)\equal{}448$
$ 8^{3x}(\frac{1}{27}\minus{}1)\equal{}448$
$ 8^{3x}\equal{}\minus{}\frac{448\cdot27}{26}$
and this is impossible because $ 8^{3x}$ is positive for all real $ x$.
Sonnhard.

I'm very sorry!
What I meant to say was

8^3x+3 rather than minus 3

Sorry!

hello, also $ 2/3$ can not be the answer of your equation, because
$ 8^{3\cdot\frac{2}{3}\plus{}3}\minus{}8^{3\cdot\frac{2}{3}}\equal{}8^5\minus{}8^2\equal{}58985$
Sonnhard.

AGAIN! I"M VERY SORRY!!!

The true equation...

(8^3x+1)-(8^3x)=448

Factoring out $ 8^{3x}$, we have

$ 8^{3x} (8 \minus{} 1) \equal{} 448$

$ 8^{3x} \equal{} 64$

Clearly, $ 3x \equal{} 2$, since $ 8^2 \equal{} 64$. So,

$ \boxed{x \equal{} \frac{2}{3}}$


A pile of 2001 pebbles is placed in a vertical column. This configuration is modified according to the following rules. A pebble can be moved if it is at the top of a column which contains at least two more pebbles than the column immediately to its right. (If there are no pebbles to the right, think of this as a column with 0 pebbles.) At each stage, choose a pebble from among those that can be moved (if there are any) and place it at the top of the column to its right. If no pebbles can be moved, the configuration is called a [i]final configuration[/i]. Show that, no matter what choices are made at each stage, the final configuration obtained is unique. Describe that configuration as follows: Determine the number, $c$, of nonempty columns, and for each $i=1,2,\ldots, c$, determine the number of pebbles $p_i$ in column $i$, where column~1 is the leftmost column, column~2 the next to the right, and so on.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

According to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17459]this[/url] (even though I'm not that sure that's complete :)), the columns should be (from left to right) $62,61,\ldots,50,49,48,48,47,46,\ldots,3,2,1$.


I need a little help on this problem. I can't get the answer.  :huh:

[i]100 people put their coats in a coatroom. On their way home, each person randomly picks up a coat. What is the probability that 99 of them pick up their own coat? [/i]

0. Think about it.

If someone leaves with the wrong coat, (say person A with person B's coat), then clearly person B went home with the wrong coat too. Its impossible for just one to get it wrong.

Not true shank. Easiest way to disprove is counter-example. Suppose 97 people get the right coat. The other 3 can get it wrong in a few ways. One is $ A$ gets $ B$'s coat, $ B$ gets $ C$'s coat, $ C$ gets $ A$'s coat. Then 3 people have got it wrong and 3 is odd. Work similarly for all numbers and it is easy to see that it is possible for odd numbers.


limit
integration
trigonometry
real analysis
function
calculus
real analysis unsolved

How do you prove that neither of the limits below exist, how do we show this with a proof rather than using written out calculations? What does it look like?

\[\begin{array}{l}\mathop{\lim }_{\lambda \to \infty }\int_{-\infty }^\infty{\sin (x^{2})\cos (\lambda x)dx}\\ \mathop{\lim }_{\lambda \to \infty }\int_{-\infty }^\infty{\cos (x^{2})\cos (\lambda x)dx}\\ \end{array}\]

The Riemann-Lebesgue Lemma says that for $f\in L^{1}(\mathbb{R}),$

$\lim_{\lambda\to\infty}\int_{\mathbb{R}}f(x)\cos(\lambda x)\,dx=0.$

The classic proofs of this involve approximating $f$ in $L^{1}$ norm by some kind of "nice" function: $C^{1}$ compactly supported functions, maybe, or step functions - something like that.

Now, you can't approximate $\sin(x^{2})$ or $\cos(x^{2})$ by any of those things - in fact, you can't approximate them in $L^{1}$ norm (or in $L^{\infty}$ norm) by anything in $L^{1}.$ So that proof doesn't work. (In the back of my mind is the fact that $L^{\infty}$ is not a separable Banach space.)

That doesn't answer your question, of course - the failure of a proof is not itself proof (or disproof) of anything.

My inclination is to go ahead and make specific calculations - the insight will be there may be a patch of values of $x$ for $|x|$ near $\lambda$ at which $\cos(x^{2})$ and $\cos(\lambda x)$ will resemble each other closely enough to have "constructive interference" and to contribute at least a fixed positive amount to the sum, whereas elsewhere we have "destructive interference" and a small contribution to the integral. But that's going to have to be made concrete.

Ignore this: [color=yellow]On the other hand, if the integral exists at all (as an improper integral), then $\int_{-\infty}^{\infty}\sin(x^{2})\cos(\lambda x)\,dx \ne 0.$ It is an odd function, after all.[/color]

That's not an odd function- $\sin(x^{2})$ is even.
The integrals do exist for all $\lambda$; integrate by parts, cutting out a chunk near zero to deal with separately. Of course, that integration introduces factors of $\lambda$, which doesn't say anything nice about the limit near $\infty$.

The first integral:
$\int_{-\infty}^\infty \sin(x^{2})\cos \lambda x\,dx= \frac12\int_{-\infty}^{\infty}\sin(x^{2}+\lambda x)+\sin(x^{2}-\lambda x)\,dx$
$\int_{-\infty}^{\infty}\sin(x^{2}+\lambda x)\,dx= \int_{-\infty}^{\infty}\sin(u^{2}-\frac{\lambda^{2}}4)\,du$ (substituting $u=x+\frac{\lambda}2$)
$\int_{-\infty}^{\infty}\sin(x^{2}+\lambda x)\,dx= \int_{-\infty}^\infty \sin(u^{2})\cos\frac{\lambda^{2}}4-\cos(u^{2})\sin\frac{\lambda^{2}}4\, du$
The other term is identical, so we get $\int_{-\infty}^\infty \sin(x^{2})\cos\lambda x=\cos\frac{\lambda^{2}}4\int_{\infty}^{\infty}\sin(u^{2})\,du-\sin\frac{\lambda^{2}}4\int_{\infty}^{\infty}\cos(u^{2})\,du=\sqrt{\frac{\pi}{2}}\left(\cos\frac{\lambda^{2}}4-\sin\frac{\lambda^{2}}4\right)$

I haven't worked out the other integral, but the principles are the same.