Έστω a,b,c,d πραγματικοί αριθμοί που  ικανοποιούν τις σχέσεις 1/2=<a,b,c,d=<2   και abcd=1

Να βρεθεί η μέγιστη τιμή αυτής της παράστασης

[a+(1/b)][b+(1/c)][c+(1/d)][d+(1/a)]

Να βάλει καποιος την εκφώνηση σε Latex

Nice one  :)  που την βῤήκες ?

Λοιπόν θα είναι  $ (4a \minus{} c)(4c \minus{} a)\ge 0$, άρα θα έχουμε μετά τις πράξεις  $ a \plus{} c\le\frac {5}{2}\sqrt {ac}$

 και με ύψωση στο τετράγωνο  $ \frac {a}{c} \plus{} \frac {c}{a}\le\frac {17}{4}$.
Εντελώς όμοια  $ b \plus{} d\le\frac {5}{2}\sqrt {bd}$ και  $ \frac {b}{d} \plus{} \frac {d}{b}\le\frac {17}{4}$.
Πολλαπλασιάζοντας κατά μέλη θα είναι
  $ (a \plus{} c)(b \plus{} d)\le\frac {25}{4}\sqrt {abcd} \equal{} \frac {25}{4}$,
οπότε
$ (ab \plus{} 1)(bc \plus{} 1)(cd \plus{} 1)(da \plus{} 1) \equal{} 4 \plus{} \frac {a}{c} \plus{} \frac {c}{a} \plus{} \frac {b}{d} \plus{} \frac {d}{b} \plus{} 2(a \plus{} c)(b \plus{} d)\le$
$ 4 \plus{} \frac {17}{4} \plus{} \frac {17}{4} \plus{} 2\frac {25}{4} \equal{} 25$

 :D

Ωραία λύση .

Έστω θετικοί πραγματικοί a,b,c,d
Να βρεθεί η ελλάχιστη τιμή αυτής της παράστασης:

$ (a\plus{}\frac{1}{b})(b\plus{}\frac{1}{c})(c\plus{}\frac{1}{d})(d\plus{}\frac{1}{a})$

Δεν είναι δύσκολη, αλλά μοιάζει με την αρχική.

[quote="maniopas"]Έστω θετικοί πραγματικοί a,b,c,d
Να βρεθεί η ελλάχιστη τιμή αυτής της παράστασης:

$ (a \plus{} \frac {1}{b})(b \plus{} \frac {1}{c})(c \plus{} \frac {1}{d})(d \plus{} \frac {1}{a})$

Δεν είναι δύσκολη, αλλά μοιάζει με την αρχική.[/quote]

Από την $ (a \plus{} \frac {1}{b})\geq 2\sqrt{\frac{a}{b}}$ προκύπτει ότι η παράσταση είναι μεγαλύτερη ή ίση από 16 :wink:

Την έβαλα με το σκεπτικό ότι για τη λύση της θέλει αναδιάταξη και ότι $ x \plus{} \frac{1}{x} >\equal{} 2$.
Προφανώς όμως αυτή η λύση είναι πιο ωραία.


sorry! :(  the nos. are not divisible by 5 for topic rmo.(it was missed by me while writing)

[color=green]You should have posted that on the original topic.

Anyway, thanks. You made me laugh  :D
Really, this was the most unnecesary topic I have ever seen xD

Now you can feel like a good for nothing  :D

Best wishes (?)
CVaneG[/color]


vector
linear algebra
matrix
linear algebra unsolved

Let $ V$ be a vector space and let $ T_1,T_2: V\to V$ be linear transformations. Then, $ T_1,T_2$ commute if and only if the set of all common eigenvectors of $ T_1,T_2$ is a basis for $ V$.

($ \alpha$ is a common eigenvector for $ T_1$ and $ T_2$ if $ T_1\alpha\equal{}c_1\alpha$ and $ T_2\alpha\equal{}c_2\alpha$ for scalars $ c_1,c_2$)

I came up with this myself. Is it true? If not, do you know of anything related?

This is true if your transformations $ T_1$ and $ T_2$ are diagonalizable. You have then more or less found the (important) fact that two diagonalizable endomorphisms of a finite-dimensional vector space commute if and only if they are simultaneously diagonalizable (i. e. they become diagonal in one and the same base).

Does anyone know such criteria for non-diagonalizable endomorphisms?

  darij

@darij grinberg:Thanks

I have one more question: Is there a non-trivial example of two transformations that do [b]not[/b] commute but have a common eigenvector?

[quote="darij grinberg"]
Does anyone know such criteria for non-diagonalizable endomorphisms?
[/quote]
Well, you have the same result with trigonalizable endomorphisms :
if two endomorphisms are trigonalizable and commute, then they're simultaneously trigonalizable.

[quote="shreyas_patankar"]
I have one more question: Is there a non-trivial example of two transformations that do [b]not[/b] commute but have a common eigenvector?[/quote]
Sure.
Take, for example, $ A \equal{} \begin{pmatrix}0 & 1 \\
0 & 0\end{pmatrix}$ and $ B \equal{} \begin{pmatrix}1 & 0 \\
0 & 0\end{pmatrix}$

On a side note, you can try to prove the following statement :
let $ (A, B) \in (\mathcal{M}_n(\mathbb{C}))^2$ such that $ AB \equal{} 0.$ Then $ A$ and $ B$ have a common eigenvector.
(note that my above example falls under the conditions of that problem)


geometry
circumcircle
3D geometry
prism
inradius
geometric transformation
homothety

Given a triangle $\triangle ABC$, let $S$ and $S'$ be its circumcircle and its incircle respectively. Choose any other point $A'$ on $S$, and let $B'$ and $C'$ be the intersections of the tangents to $S'$ trhough $A$ with $S$. Show that $B'C'$ is tangent to $S'$. (I read somewhere that this can be generalized to any n-agon, so that if we have an n-agon that is both inscribible and circumscribible, and we have n points on it circumcircle such that n-1 sides are tangent to the incircle, then the last side must also be tangent)

Look at [url=http://mathworld.wolfram.com/PonceletsPorism.html]Poncelet's Porism[/url].

For a triangle, this can be proved by a simple contradiction. Let $(O, R), (I, r)$ be the circumcircle and incircle of a triangle $\triangle ABC$, $R, r$ the circumradius and inradius and $d = OI$. The following relation holds for $R, r, d$:

$\frac 1 r = \frac{1}{R + d} + \frac{1}{R - d}$

This can also be written as $R^2 - d^2 = 2Rr$. Let $D$ be an arbitrary point on the circumcircle $(O, R)$ different from the vertices $A, B, C$. Since the incircle $(I, r)$ is always inside of the circumcircle $(O, R)$, the 2 tangents from the point $D$ to the incircle $(I, r)$ intersect the the circumcircle $(O, R)$ at points $E, F$ different from the point $D$. Assume that the chord $EF$ is not a tangent to the incircle $(I, r)$, for example, it intersects it. Continuously compress the incircle $(I, r')$, keeping its center $I$ fixed and let the points $E', F'$ continuously move on the circumcircle $(O, R)$ in such a way that the chords $DE', DF'$ are tangents to the circle $(I, r')$ being compressed. Eventually, the chord $E'F'$ will be a tangent to the smaller circle $(I, r')$. But then the circles $(O, R), (I, r')$ are the circumcircle and incircle of the triangle $\triangle DE'F'$ and

$\frac{1}{r'} = \frac{1}{R + d} + \frac{1}{R - d}$

Since $r' < r$, $\frac{1}{r'} > \frac 1 r$ and

$\frac{1}{r} < \frac{1}{r'} = \frac{1}{R + d} + \frac{1}{R - d}$

which is a contradiction with the original relation between $R, r, d$. The possibility that the chord $EF$ does not have any common points with the incircle $(I, r)$ is disproved similarly, by expanding the incircle $(I, r')$ and reversing the direction of all inequalities.

The above relation between $R, r, d$ can be proved by an inversion. Let $D, E, F$ be the tangency points of the incircle $(I)$ with the triangle sides $BC, CA, AB$. Obviously, the circle $(I)$ is the circumcircle of the triangle $\triangle DEF$. Invert the triangle $\triangle ABC$ in the incircle $(I)$. The vertices $A, B, C$ are carried into the intersections of the angle bisectors $AI, BI, CI$ with the chords $EF, FD, DE$, i.e., into their midpoints $A', B', C'$, and the circumcircle $(O)$ into the circumcircle $(O')$ of the triangle $\triangle A'B'C'$. Since $A', B', C'$ are the midpoints of the chords $EF, FD, DE$, the triangle $\triangle A'B'C'$ is the medial triangle of the triangle $\triangle DEF$. These 2 triangles are (centrally) similar with homothety coefficient $-\frac 1 2$ and so are their circumcircles $(O'), (I)$. (The external homothety center of the circles $(O'), (I)$ is the orthocenter of the triangle $\triangle DEF$ and the internal homothety center is the common centroid $G$ of the triangles $\triangle A'B'C', \triangle DEF$.) Thus the radius of the circle $(O')$ is $r' = \frac r 2$. The circles $(O), (O')$ are also centrally similar and their (internal) homothety center is the inversion center $I$. This is an internal homothety center, because the circle $(O)$ is outside and the circle $(O')$ inside of the inversion circle $(I)$. The homothety coefficient of the circles $(O'), (O)$ is equal to the power of inversion (square of the radius of the inversion circle) divided by the power of the inversion center $I$ to the circle $(O)$. Let the line $OI$ intersect the circle $(O)$ at points $P, Q$. This homothety coefficient is then equal to

$-\frac{r^2}{IP \cdot IQ} = -\frac{r^2}{(R - d)(R + d)}$

Consequently,

$\frac{r'}{R} = \frac{r}{2R} = \frac{r^2}{(R + d)(R - d)}$

$\frac 1 r = \frac{2R}{(R + d)(R - d)} = \frac{R + d + R - d}{(R + d)(R - d)} = \frac{1}{R - d} + \frac{1}{R + d}$

For a bicentric quadrilateral, there is a similar relationship between $R, r, d$:

$\frac{1}{r^2} = \frac{1}{(R - d)^2} + \frac{1}{(R + d)^2}$

which can be proved by an inversion. The fact that an arbitrary point of the circumcircle of a bicentric quadrilateral can be selected as a vertex of another bicentric quadrilateral with the same circumcircle and incircle can then be proved by a contradiction, similarly as for the triangles.

For an n-gon with a circumcircle and an incircle, this is known as Poncelet's theorem (or Poncelet's porism). The proof is by an inversion, but it is a rather nasty one.

Thanks!, very nice way of provin Euler result by inversion  :)


Whenever I compile my Texnic Center File, it shows no errors, warnings, or anything. However, when I click view, Acrobat Reader says:

THERE WAS AN ERROR IN OPENING THIS DOCUMENT: ACCESS IS DENIED

I am 100 PERCENT SURE THAT THERE IS NOTHING WRONG WITH MY CODE (I even copied it directly from AoPS Wiki).

I followed all the instructions on the Wiki, it used to work before but I don't know what happened now!

Please help; I have to make some problems because of my *Challament* in the MATHCOUNTS sub-forum.

Does this happen with one, or with all your files?

[quote="akhil0422"]Whenever I compile my Texnic Center File, it shows no errors, warnings, or anything. [/quote]
That means the file is not being compiled. Look in the folder where your tex file is. Are there any other files there with the same name? If your file was called example.tex then there should be example.log, example.aux and example.pdf and maybe others.
If there is an example.log then please post its contents here which will tell us what happened.

When I said no errors or anything, I meant it said: 0 errors, 0 warnings, 0 boxe(s), 1 page.

Also, I can't find any ____.log file. They are all ____.aux or ____.bbl.

OK I see what you meant. It makes no sense that no log or no pdf is created as
0 errors, 0 warnings, 0 boxe(s), 1 page
is taken from the log and 1 page means 1 page of a pdf.
Scrolling up above the errors and warnings line TexnicCentre shows the log. Is anything there?
Start again with a brand new tex file with a distinctive name and save it to an empty folder. Are any files created there when you compile? Search for the distinctive name in case, for some reason, files have been saved elsewhere.

No other files with the same name exist. I named mine "test". I searched and came up with nothing.

??????????? Suddenly, it works again. I don't get that. :huh:


algebra
function
domain
ratio
calculus
integration
algebra unsolved

Find all f:R+->R+ such that
f(x)^2 + f(y)^2 = f(x+y) [ f(f(x)) + f(y) ]

Hint.[hide]The problem has only f(x)=c as its solutions.[/hide]

[u][b]Edited: The domain and range is postive reals only. (0 excluded)[/b][/u]

Maybe it will be a good idea if you could read my hint.
other than that your solution is incomplete.

Why its incomplete :?
I dont need a Hint to solve such easy problem :rotfl:

Dear Aymane!
I think that Ali want you to say in conclusion:

$ So,$ $for$ $all$ $x$ , $f(x)=f(0)$

Sorry, if i wasn't clear.
The first post is now edited.

So you can see that your soltuion does not apply anymore.
again it was my fault. ;)

Interchange $x$ and $y$. We get $f(f(x))+f(y)=f(f(y))+f(x)$. So, $f(f(x))-f(x)=c$ is a constant. Denote by $A$ the set of all values of a function $f$. Then we have $f(a)=a+c$ for arbitrary $a$ in $A$. If $c<0$, then a sequence $a,a+c,a+c+c$ etc has negative terms, a contradiction. If $c>0$, then $f(x+y)<\frac{f^2(x)+f^2(y)}{f(x)+f(y)}\le \max(f(x),f(y))$ But take $y=c,\,x\in A$. We get $x+2c=f(x+c)<\max(x+c,f(c))$. It is not so for large $x$ (and $a$ contains arbitrary large elements). So, $f^2(x)+f^2(y)=f(x+y)\cdot(f(x)+f(y))$. Put $x=y$, get $f(2x)=f(x)$. Put $2x$ instead $x$ and compare with pair $(x,y)$. We get $f(2x+y)=f(x+y)=\frac{f^2(x)+f^2(y)}{f(x)+f(y)}$. But $x+y$ and $2x+y$ are arbitrary positive reals, whose ratio is less then 2. So, f is constant on all segments $[1.9^k,1.9^{k+1}]$ for integral $k$, hence $f$ is constant on $\mathbb{R}_+$


factorial
probability
number theory open
number theory

Hi everybody,

Let N=77131267

a1= N mod(3)=1
a2= N mod(5)=2
a3= N mod(7)=3
a4= N mod(11)=4
a5= N mod(13)=5
a6= N mod(17)=6
a7= N mod(19)=7
a8= N mod(23)=8

8!=a1*a2*a3*....a*8

The next N has to be

a9=N mod(29)=9

N=2531152837 gives 9!

My questions are :

Is there any way to find any k! such as described above?

Is there some link between N and k! ?

I did not find an answer I'm still thinking

Thank you for any idea or trick.

Can we state that if k is big enough (example 1000) then N will be prime with high probability?
I used another way (other than chinese theorem) to find my sequence

I begun with N=7

7 mod(3)=1
7 mod(5)=2

Then I added an increment of 30=2*3*5
37
67
97
and so on
and each time I checked if N mod(7)=3

I found N=157

157mod(3)=1
157mod(5)=2
157mod(7)=3

I changed my increment to 210=2*3*5*7

267
367

Each iteration I checked if N mod(11)=4

367 was candidate

and so on until
N=106046244517 wich fit with k=10!

If someone can program it we can have all the list until the limits of the machine.
So please if someone can do it and publish (format txt or excel) the list with the couple (k,N) it will be useful for me and for anyone who is interested on such sequence.

Thank you for any help.

Ps : I'm just an amateur.

Hi!
I can only give you the coefficients
I calculated until 13!, after that I have overflow errors

What I mean "coefficients":

Let $ \{p_n\}$ be the sequence of the primes (starting with $ p_1\equal{}2$)

You created a sequence $ \{a_n\}\equal{}\{7,7,157,367,14227,...\}$ with the property $ a_n \equiv m \mod p_{m\plus{}1}, \ \forall m\leq n$

The recursive relation was $ a_{n} \equal{} c_n\prod_{k\equal{}1}^{n}{p_k}\plus{}a_{n\minus{}1}$

For example
$ a_1 \equal{} 0\cdot (2) \plus{} 7 \equal{} 7\\
a_2 \equal{} 0\cdot (2\cdot 3) \plus{} 7 \equal{} 7\\
a_3 \equal{} 5\cdot (2\cdot 3\cdot 5) \plus{} 7 \equal{} 157\\
a_4 \equal{} 1\cdot (2\cdots 7)\plus{} 157 \equal{} 367\\
a_5 \equal{} 6\cdot (2\cdots 11)\plus{} 367 \equal{} 14227$

So the coefficients are $ \{c_1, c_2, c_3, c_4, c_5,\cdots\} \equal{} \{0,0,5,1,6,\cdots\}$

I made the calculations in modulo of each $ p_n$. So, I'm not sure that they are all correct. Just test them

I attach a file with the first 1000 coefficients

[i][u]Comment[/u][/i]
We could create the sequence such that it gives $ N\equiv 1\mod 2,\ N\equiv 2\mod 3,\ N\equiv 3\mod 5$ etc

Thank you for your data.
I'm going to download some software able to handle big numbers (thousands of digits) and try to check the primality using protheus.


function
limit
real analysis
real analysis unsolved

Let $f: X\to Y$ be a continuous function such that its restriction to a dense subset of $X$ is uniform continuous. Prove that $f$ is uniform continuous.

[Assuming that part of the hypothesis is that $X$ and $Y$ are metric spaces - not necessarily complete.]

First prove this lemma: If $x_{n}\to x$ and $y_{n}\to y$ in $X,$ then $\lim_{n\to\infty}d(x_{n},y_{n})=d(x,y).$


AMC
AIME
USA(J)MO
USAMO
AMC 12
trigonometry
search

While you guys are aiming for USAMO, all i want is to qualify for the AIME. :blush:  I have the AMC12 next week. Currently, i'm reading through old algebra II content that i forgot (trig, logs, complex #'s), and doing AMC12 problems. Is there anything else i should do to qualify for the AIME?

Thanks alot. :)

Please use Search because threads like this one are constantly appearing so there is no reason to keep making new ones.

See [url=http://www.hiddendiscoveries.com/]Silverfalcons's blog[/url]-one part contains information on this.

Im in the same boat as you elite.. going for aime.

For the 12, I've heard that you should master all the geometry in AoPS 1 as well a above-average knowledge of the rest of the topics in it. One of Silver Falcon's guide's tells you about AoPS 2. If you do not have these books, then I suggest doing old tests  (which it seems like your doing) and cover geometry, algebra, p
basic pre-calc, number thry, ...ect


A circle with radius $\sqrt{2}$ is centered at $(0,0)$. What is the area of the smaller region cut from the circle by the chord from $(-1,1)$ to $(1,1)$?

[hide]
Use figure for reference.

We know that $BD=1$.  $AB=BC=\sqrt{2}$, since they are both radii.  By Pythagorean theorem, $AD=CD=1$.  Now, you can see that $\triangle ABD$ and $\triangle CBD$ are isosceles right triangles.  Therefore, $m\angle ABD=m\angle CBD=45$, so $m\angle ABC=90$.  The area of circle B is $2\pi$, so the area of sector $ABC$ is $\frac{90}{360}2\pi=\frac{\pi}{2}$.  $[ABC]=1$, so   the area we are looking for is $\frac{\pi}{2}-1$.
[/hide]


find all $x,y \in Z$ satisfy:
                   $2(x^5+y^5+1)=5xy(x^2+y^2+1)$
I have the solution but it isn't short and  have a lot of cases (3 cases) .So i want the elegant solution.Good luck :)  :)

i doubt there's an elegant solution; problems like this often do require case-by-case analysis.  and even if there is an elegant solution, it may just be a coincidence.


It is given that \[x^3-2=2y^3-2y=2y(y^2-1)=2(y-1)y(y+1).\] Because $12|2(y-1)y(y+1)$ for each $y\in\mathbb Z$, we know $4|x^3-2$. The residues of $x^3-2$ modulo $4$ are $\{2,3,2,1\}$. None of them is $0$, so there are no solutions.

Please excuse me if I am wrong.


geometry
incenter
circumcircle
geometry unsolved

In a triangle $ ABC$, $ I$ and $ I_a$ denote the incenter and the excenter corresponding to side $ BC$. Let $ A'$and $ M$
respectively  be the intersections of $ II_a$ with $ BC$  and the circumcircle of $ \triangle ABC$ let $ N$ be the midpoint of arc
$ MBA$, and let $ S, T$ be the intersection points of rays $ NI$ and $ NI_a$ with the circumcircle of $ \triangle ABC$.Prove that $ S,T$ and $ A$ are collinear

your problem is posted before(i'm sure)
but unfortunately i can't find it  :(
anyway,
we can easily prove that $ SITI_a$ is cyclic.
denote the intersection of $ ST$ and $ II_a$ by $ L$.
then we know that $ I_aL.LI\equal{}TL.LS$
but $ LHS$ is equal to the power of $ L$ wrt circle with diameter $ II_a$ and $ RHS$ is equal to the power of $ L$ wrt circle $ (O)$ which is the circumcircle of $ \Delta ABC$.
so $ L$ lies on the radical axe of the two circle $ (O)$ and $ (M)$.
but the radical axe of these two circle is $ BC$ so $ BC,II_a,ST$ are concurrent and we r done.

[quote="shoki"]..................we can easily prove that $ SITI_a$ is cyclic.
.........................[/quote]

Please, can someone explain this ?

Thanks in advance!

Babis


Katrina can bicycle 50 miles in four hours. At the same speed, how many miles does she bicycle in six hours?

Six hours is $ 3/2$ of four hours, so in six hours, she bicycles $ 3/2\times50$ miles, or $ 75$ miles.


analytic geometry
vector
algebra proposed
algebra

Take three points $A(0,1,2),\ B(2,3,0)$ and $P(5+t,\ 9+2t,\ 5+3t)$ on the coordinate space with the origin $O.$
Prove that there exist a real numbers $t$ such that segment $OP$ has an intersection point with segment $AB.$
Then find the coodinate of this intersection point.

The radius-vector of the general point $X$ of the segment $AB$ is given by $\overrightarrow{OX}=\overrightarrow{OA}+\lambda\overrightarrow{AB}$ where $0\leqslant\lambda\leqslant 1$. Substituting the numbers we get

$\overrightarrow{OX}=\langle 2\lambda, 2\lambda+1,-2\lambda+2\rangle\qquad(*)$

The radius-vector of the general point $Y$ of the segment $OP$ is given by $\overrightarrow{OY}=\mu\overrightarrow{OP}$ where $0\leqslant\mu\leqslant 1$. Substituting the given data we get

$\overrightarrow{OY}=\langle\mu(5+t),\mu(9+2t),\mu(5+3t)\rangle$

The condition $X\cong Y$ yields the system

\begin{eqnarray} 2\lambda &=& \mu(5+t)\\ 2\lambda+1 &=& \mu(9+2t) \\ -2\lambda+2 &=& \mu(5+3t)\end{eqnarray}

Substituting $(1)$ into $(2)$ and $(3)$ we get

$1=\mu(4+t)\qquad(4)$
$2=\mu(10+4t)\qquad(5)$

Multiplying $(4)$ by $-2$ and adding to $5$ we get

$\mu(2+2t)=0$

Therefore, either $\mu=0$ or $t=-1$. The first possibility is in contradiction with $(4)$ and $(5)$, hence $t=-1$. Now $(4)\implies\mu={1\over 3}$, and $(1)\implies\lambda={2\over 3}.$ Since both $\lambda$ and $\mu$ are less than one, the desired point exists, and its coordinates are given by $(*)$: $X({4\over 3},{7\over 3},{2\over 3})$


help i have know idea how to factor these types of these questions... :blush:

Hmm...do you know how to do FOIL? Usually you can do some guessing and get it pretty easily. But I'm sure there's a much better way...
By the way, the factored form of that is $(3w+2)(w-9)$.

no i dont know how to use foil, if you could describe me or give me a link

FOIL stands for First Outer Inner Last
when you multiply 2 binomials, $(w+x)(y+z)$, you:
Multiply the FIRST[[i]([b]w[/b]+x)([b]y[/b]+z)[/i]] terms $wy$
Multiply the OUTER[[i]([b]w[/b]+x)(y+[b]z[/b])[/i]] terms $wz$
Multiply the INNER[[i](w+[b]x[/b])([b]y[/b]+z)[/i]] terms $xy$
Multiply the LAST[[i](w+[b]x[/b])(y+[b]z[/b])[/i]] terms $xz$
Add them together $wy+wz+xy+xz$

The reason for that is that the distributive property of multiplication over addition expands it into $w(y+z)+x(y+z)$ if you look at $y+z$ as the $c$ term in $(a+b)c=ac+bc$
then you expand both terms into $wy+wz+xy+xz$

I googled and found a link for FOIL http://www.freemathhelp.com/using-foil.html

The reason this helps with factoring is becausethe OUTER and INNER terms combine in the case of $(x+a)(x+b)$. so expanded, it is $x^{2}+(a+b)x+ab$
you have to find 2 numbers that add up to the middle coefficient, and multiply together to the constant. Then you jus put it in $(x+a)(x+b)$ form.
when the coefficient of $x^{2}$ isn't 1, it is in the form $(cx+a)(dx+b)$
when expanded, it is $cdx^{2}+(ad+bc)x+ab$. The best way in this case is to guess and check integers that might be candidates by prime factorizing the first coefficient and the constant, and making factors and adding them.
Another way is to just put it in that form and solve for a b c and d

Hmm, ProtestanT has already explained FOIL.
Let's do an example, $3w^{2}-25w-18$.
$3w^{2}$ can only be $(3w)(w)$; so far our factored form looks like $(3w+x)(w+y)$. Our $x$ and $y$ must multiply to give us $-18$, and $x+3y$ (the inside and outside terms) must add up to give us $-25$. This gives us the only possible solution, $x=2$ and $y=-9$.

[quote="7h3.D3m0n.117"][quote="Glen Mann"]$\3w^{2}-25w-18$
find two numbers which adds to $-25$, and multiplies to $3\cdot-18 =-54$

those are 2, and -27

$(3w+2)(3w-27)$
$(3w+2)(w-9)$

roots are $9$ and $-\frac{2}{3}$[/quote]

hehe close but it was $3w^{2}-25w-18$ not $w^{2}-25w-18$[/quote]

i just forgot the three when displaying the equation.. the rest of the work however works with the $3w^{2}$

[quote="now a ranger"]why not just use quadratic equation? works for me.[/quote]
because its long and tedious to use, and there is no point in using it when given an equation that factors so easily?


Let $p$ be a prime number such that $p$ [i]mod[/i] $4$ is $1$. Show that there is an integer $n$, $1 \leq n \leq p-1$ such that $p$ divides $n^2+1$.

Take $\left(\frac{p-1}{2}\right)! \mod p$. Indeed, this was posted before.


Eighty percent of the students in a class (group A) share $ 40\%$ of the candy equally. The remaining $ 20\%$ of the students (group B) share the other $ 60\%$ of the candy equally. The ratio of the amount of candy a student in group A has to the amount of candy a student in group B has is equal to what common fraction?

Lets think of numbers.

There are 100 kids, and 100 pieces of candy.

80% of the kids (80 people) share 40% of the candy (40 pieces) equally. Each gets 40/80=1/2 piece.
20% of the kids (20 people) share 60% of the candy (60 pieces) equally. Each gets 60/20=3 pieces.

(1/2)/3=1/6


Let $A_{k}(k,k^{3}),k=1,2006$ be points in the plane.
Prove that the polygon $A_{1}A_{2}...A_{2006}$ is convex.

No ideas?
I think we should use that the function $f(x)=x^{3}$ is convex for $x>0$


logarithms inequalities geometry calculus derivative

[hide]
$\log{\frac{1}{2}} < 0$ ;)

It took me too long to figure that out... [/hide]

log of fractions, is a negative number...so when you cancell the logs, the inequality changes sign..
edit-paladin8, you beat me to it.

That was really cool and fast ,
A lot of studnets took a hell lot of time to solve it.
If anyone has one like this pls fwd them I can use them in class
Anything interesting in Squares and squareroot

[quote="probability1.01"]There is a fairly unobvious one that proves all triangles are isosceles.[/quote]

It's in Kedlaya's geometry notes, right? Quite interesting, actually.

I haver this one yet thanks
You wont beleive it many of my students actually got the isoceles triangle one

[quote="Devendra Patel"]That was really cool and fast ,
A lot of studnets took a - lot of time to solve it.
If anyone has one like this pls fwd them I can use them in class
Anything interesting in Squares and squareroot[/quote]
i believe there is a book written on these stuff....

Has anyone seen an illogical proof where it was very difficult to discover the error yet?

I like this one:

$x^2 = x+x+...+x$ ($x$ times)

Taking the derivative of both sides...

$2x = 1+1+...+1$ ($x$ times)
$2x = x$
$2 = 1$

Nice, there's no very obvious flaw to it :).

yeah the flaw is where you say:
$2x=x$ that is wrong... the $x$ on the left side a variable, while the right side is only a constant number

Yeah, well I just posted it because it's a fun "proof"... you guys don't need to get overly critical about it... :D

[quote="probability1.01"][quote="blahblahblah"]How about the hugely obvious flaw that you're holding one of the $x$'s constant?[/quote]
Well, I didn't consider that hugely obvious....[/quote]

Sorry if I came across as condescending.  I probably shouldn't be spoiling other people's fun anyways  :?

[quote="Magnara"]Or you lose all continuity, because x clearly must be an integer there.[/quote]

Not really, if we write

$x=1+1+1+\ldots +\{x\}$ and hence $x^2=x+x+x+\ldots +\{x\}x$ we can patch up the argument.  But the same flaw exists.

look at this proof for $1=2$:

let $a=b$
$a=b\quad |\cdot a\\
a^2=ab\quad |+a^2\\
2a^2=a^2+ab\quad |-2ab\\
2a^2-2ab=a^2-ab\quad |\div (a^2-ab)\\
2=1\quad\blacksquare$

h4ve fun

a=b so a2-ab=o.  We cant divide by zero so the proof is flawed.

What you can't do is go directly from
Here
$(4-\frac{9}{2})^2=(5-\frac{9}{2})^2$
to here
$4=5$

You would have to take +/- square root of each side, and then you know that there are situations which are false.

[quote="Michael"][quote="afaouzi"]\begin{eqnarray*}
(-1)&=&(-1)^{1}\\
(-1)&=&(-1)^{2/2}\\
(-1)&=&((-1)^2)^{1/2}\\
(-1)&=&(1)^{1/2}\\
(-1)&=&1
\end{eqnarray*}[/quote]

The relation $(a^n)^m=a^{n\times m}$ is valid only for integers $n$ and $m$ :P  :P[/quote]

Actually, no, it's not. That part of the "proof" is valid. The flawed step is in taking the square root at the end.

whats about this one?
$-1=i^2=i\cdot i=\sqrt {-1} \cdot \sqrt {-1} = \sqrt{(-1)\cdot (-1)} =1$

to peeta
simply the idea that $\sqrt[m]x \sqrt[m]y = \sqrt[m]{xy}$ is just valid for the ;positive numbers . you want to use this rule then prove it  ;) ?


Triangle ABC has vertices A=(0,0); B=(0,3); C=(5,0). Point P is inside the triangle and is $\sqrt{10}$ units from point A and $\sqrt{13}$ units from point B. How many units is it from point C? Simplest radical form, please.

[quote="biffanddoc"]nasty problem :bomb:
[hide]
Let the point be $(x,y)$
Using the distance formula, we set up the equation
$\sqrt{x^2+y^2}=\sqrt{10}$
$\sqrt{x^2+(y-3)^2}=\sqrt{13}$
Square both to get
$x^2+y^2=10$
$x^2+(y-3)^2=13$

Subtract the first from second.
$y^2-(y-3)^2=-3$
$y^2-(y^2-6y+9)=6y-9=-3$
So $y=1$
Substitute back into $x^2+y^2=10$, to get $x=3$
Now find the distance from $(3,1)$ to $(5,0)$
$\sqrt{2^2+1^2}=\sqrt{5}$
[/hide][/quote]

Ha. Ha. Ha. I laugh at biffanddoc. Everyone knows the best solution is to guess, and then improve your answer from there. Me, with my luck, got it in one guess. :lol:

Draw a basic graph, and judge around where it would be, then get the most nice answer thats around there.
Check to make sure its right.  :P

It was really easy... P is obviously 3 and 2 points away from B and 3 and 1 points away from A so it must have been (3,1)

It was really easy but at the time I didn't have time so I guessed on it...

guess what I guessed..

THE SQUARE ROOT OF FIVE HHAAHAH.

Yeah, when I was doing this, I didn't feel like using algebra, so I noticed that $10=3^2+1^2$ and $13=3^2+2^2$.

So I graphed it and found the position of point P, which was $(3,1)$, so then the distance was $\sqrt{2^2+1^2}$.

I was incredibly stupid. First I figured that $1^2+3^2=10$ and $3^2+2^2=13$, then I mixed up A and B so that point P was at $(3,2)$ instead of $(3,1)$. So my answer came up to be $2\sqrt{2}$ instead of $\sqrt{5}$

I simply ran out of time but biffanddoc over did it. MathCounts would probably not put something like that on a State Sprint and you should just fiddle arround with sums of squares.

I pulled a completely random number out of thin air: $4\sqrt{5}$. If I'd dropped the 4, I might have beaten the first place guy.

Point P is sqrt (3^2+1^2) from A, so it is either at (3,1) or (1,3).
(3,1) is sqrt(2^2+3^2) from B, or sqrt(13) from B, so P is (3,1).
This means P is sqrt(2^2+1^2) from C, or sqrt(5).

[hide]All you post complex solutions, ehehheehee is lazy, he just sees 3^2+2^2=13, and 3^2+1^1=10, yes I know that it wouldn't work in most cases, but I am lazy...

and 2+1 conviently add to three, so p is at 3,1 and radical 5 away from C[/hide]


symmetry
AMC
AMC 10
geometry
geometric transformation
rotation
linear algebra

Let $ a$, $ b$, and $ c$ be real numbers such that $ a \minus{} 7b \plus{} 8c \equal{} 4$ and $ 8a \plus{} 4b \minus{} c \equal{} 7$. Then $ a^2 \minus{} b^2 \plus{} c^2$ is

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$

[hide="Solution"]
Rearranging the equations to take advantage of the symmetry,
\begin{align*}
a + 8c &= 7b + 4 \\
8a - c &= -4b + 7
\end{align*}Squaring both equations gives
\begin{align*}
a^2 + 64c^2 + 16ac &= 49b^2 + 16 + 56b \\
64a^2 + c^2 - 16ac &= 16b^2 + 49 - 56b
\end{align*}Summing these two equations gives $ 65a^2 + 65c^2 = 65b^2 + 65 \Longrightarrow a^2 - b^2 + c^2 = 1 \Rightarrow \textbf{(B)}$.

[/hide]

I got it quickly by making up numbers, but I thought this was unusually hard for and AMC10 #20. It took me a long time to come up with a rigorous solution.

Altheman and I discuss this problem in depth [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170124&search_id=105981621]here[/url].  In some sense, the "correct" way to write this problem (that is, that makes the result obvious) is to write

$ \left[ \begin{array}{cc} 8 & 1 \\ -1 & 8 \end{array} \right] \left[ \begin{array}{c} c & a \end{array} \right] = \left[ \begin{array}{cc} 4 & 7 \\ -7 & 4 \end{array} \right] \left[ \begin{array}{c} 1 & b \end{array} \right]$

and then to recognize the rotation matrices, but the geometric interpretation in terms of ruled surfaces is also quite nice.

[hide="Cheap"]
Let $ a\equal{}0$, and we get
\[ \minus{}7b\plus{}8c\equal{}4\\
4b\minus{}c\equal{}7
\]

Multiply the second by 8 to get
\[ \minus{}7b\plus{}8c\equal{}4\\
32b\minus{}8c\equal{}56
\]

Sum:
\[ 25b\equal{}60\implies b\equal{}60/25\equal{}12/5
\]

And so $ 8c\equal{}4\plus{}7\cdot12/5\equal{}104/5\implies c\equal{}13/5$

Then $ c^2\minus{}b^2\equal{}169/25\minus{}144/25\equal{}25/25\equal{}\boxed{\textbf{(B)}\ 1}$.
[/hide]

Just a clarificaton of grn_trtle's solution:
Value of Expresion $ a^{2} \minus{} b^{2} \plus{} c^{2}$ is indipedent from the values of $ a$,$ b$ and $ c$ as t0rajir0u wrote.
So we take $ a \equal{} 0$ and the other part is like grn_trtle wrote ...

[b]ee"[/b]

Well, not independent of all the variables, but enndb0x is correct; because we're given that there are real number solutions to the equation, letting one variable equal 0 doesn't make a difference, though it simplifies computations.

The great thing about general questions like these on the AMC is that you can assume whatever you want and consider it a special case with the same result (though this only works [i]if the problem is general enough[/i]).

There are more unknown variables than the number of equations in this system, so there is a free variable, allowing a system with an infinite number of solutions. For instance, in the augmented matrix, we reduce it to row canonical form,

$ \left[\begin{array}{ccccc}
1&\minus{}7&8&|&4\\
8&4&\minus{}1&|&7\\
0&0&0&|&0
\end{array}\right]
\sim
\left[\begin{array}{ccccc}
1&0&\frac5{12}&|&\frac{13}{12}\\
0&1&\minus{}\frac{13}{12}&|&\minus{}\frac5{12}\\
0&0&0&|&0
\end{array}\right]$

So $ a\equal{}\minus{}\frac5{12}c\plus{}\frac{13}{12}$ and $ b\equal{}\frac{13}{12}c\minus{}\frac5{12}$. If you substitute them into $ a^2\minus{}b^2\plus{}c^2$, it yields a constant.

To further investigate why this works well, check out t0rajir0u's discussion on rotational matrices.


number theory
relatively prime
number theory unsolved

How to solve the equation $ a^b\equal{}b^a$ if $ a,b\in\mathbb{Z}_\plus{}$? I guess the only solutions are $ a\equal{}b$.

The only other integer solution is $ 4^2 \equal{} 2^4$.  (Think divisibility.)  The rational solutions are given by $ a \equal{} \left(1 \plus{} \frac{1}{k} \right)^k, b \equal{} \left( 1 \plus{} \frac{1}{k} \right)^{k\plus{}1}$.

I tried to think divisibility. Let $ b\equal{}ka$. Then $ a^{ka}\equal{}(ka)^a$ so $ (a^a)^{k\minus{}1}\equal{}k^a$. So $ k\mid a$ i.e. $ k\equal{}la$... I don't find an exact proof why this process halts.

The conclusion we can reach is stronger than $ b \equal{} ka$.

[b]Lemma:[/b]  If $ a^x \equal{} b^y$ then there exists an integer $ c$ such that $ a, b$ are both powers of $ c$.

[quote="t0rajir0u"]The rational solutions are given by $ a \equal{} \left(1 \plus{} \frac {1}{k} \right)^k, b \equal{} \left( 1 \plus{} \frac {1}{k} \right)^{k \plus{} 1}$.[/quote]
This looks a nice generalization. How to prove it?

[b]MeKnows[/b] —

$ a^{b} \equal{} \left(1 \plus{} \frac {1}{k} \right)^{k \left(1 \plus{} \frac {1}{k}\right)^{k \plus{} 1}} \equal{} \left(1 \plus{} \frac {1}{k} \right)^{(k \plus{} 1)\left(1 \plus{} \frac {1}{k}\right)^{k}} \equal{} b^{a}$

For any $ k \in \mathbb Q$.

Of course, that's not a proof that the rational solutions can only take that form.  Write $ a \equal{} \frac{a'}{d}, b \equal{} \frac{b'}{d}$ and think along the same lines as the integer solutions.

Of course, but he didn't really ask for that, did he? :P

$ x \equal{} y$ is a trivial solution. Otherwise assume without loss of generality that $ x < y$, and let $ z \equal{} y/x$, which is a rational number greater than 1. From $ x^y \equal{} y^x$ we get $ x^{zx} \equal{} (zx)^x$, or $ (x^x)^z \equal{} z^x \cdot x^x$, which gives $ x^{z \minus{} 1} \equal{} z$, or $ x \equal{} z^{1/(z \minus{} 1)}$. Since $ \frac {1}{z \minus{} 1}$ is rational and positive, we can write $ \frac {1}{z \minus{} 1} \equal{} \frac {u}{v}$ for some relatively prime positive integers $ u$, $ v$. This is equivalent to $ z \equal{} \frac {u \plus{} v}{u}$. Since this means $ x \equal{} z^{u/v}$, and $ x$ and $ z$ are rational, $ z$ must be the $ v$-th power of some rational number. But since $ z \equal{} \frac {u \plus{} v}{u}$, both $ u \plus{} v$ and $ u$ must be $ v$-th powers of some rational numbers, since they are relatively prime. It is well known that $ |a^c \minus{} b^c| > c$ for all positive integers $ a$, $ b$ and reals $ c > 1$. Therefore, this can only be true if $ v \equal{} 1$, and then $ z \equal{} \frac {y}{x} \equal{} u \plus{} 1$. Therefore, $ x \equal{} \left(\frac {u \plus{} 1}{u}\right)^u$ and $ y \equal{} \left(\frac {u \plus{} 1}{u}\right)^{u \plus{} 1}$ for all positive integers $ u$.

This solution is by chess64.

When I want to find positive integer solutions to $ a^{(b^k)} = b^{(a^j)}$ ($ a,b,k,j \in \mathbb{N}$ Assume $ k \ge j$)
I usally write $ a = \prod_{i = 1}^r p_i^{x_i}$ and $ b = \prod_{i = 1}^r p_i^{y_i}$ Remember that $ p \mid a \iff p \mid b$

(Note: We just wanna find solutions other than $ a = 1$ or $ b = 1$)

Then we have that $ a^{(b^k)} = b^{(a^j)} \iff x_i = \frac {y_i \cdot a^j}{b^k} \forall i \in \{0,1,..,r\}$

Set $ \frac {x_i}{y_i} = \frac {a^j}{b^k} = \frac {p}{q}, (p,q) = 1, p,q \in \mathbb{N}$

But $ \frac {a^j}{b^k} = \frac {\prod p_i^{x_i \cdot j}}{\prod p_i^{y_i \cdot k}} = \prod p_i^{x_i \cdot j - y_i \cdot k} = \prod p_i^{j \cdot y_i \cdot \frac {a^j}{b^k} - y_i \cdot k} =$ $ \left ( \prod p_i^{y_i} \right ) ^ {j \cdot \frac {b^k}{a^j} - k} = b^{j \cdot \frac {b^k}{a^j} - k} = \frac {p}{q}$

We clearly have $ p = 1$ or $ q = 1$. But we also have $ a = b^{\frac {b^k}{a^j}} = b^{\frac {q}{p}}$ and $ b = a^{\frac {a^j}{b^k}} = a^{\frac {p}{q}}$ or $ b^q = a^p$

So now we only have to cases
Case $ q = 1$: Then we have $ b = a^p$. Substitution gives $ a^{a^{pk}} = a^{p \cdot a^j} \iff a^{pk} = p \cdot a^j \iff a^{pk - j} = p$. Substitution again gives $ a^{pk - j} = p$ Assume that a prime divides $ p$, $ m$ times. But any prime which divides a will divide $ a$ at least $ 2^mk - j$ and $ 2^mk - j \ge 2^m - 1 > m$ unless $ m$ = 1 and $ k = j$ and thus $ p = 2$. So if $ k = j$ we have $ a^k = 2$ implying $ k = j = 1$. So if $ k = j = 1$ we have the solution $ a = 2, b = 4$. One thing i almost forgot!! If $ p = 1$ it implies $ a = b$ and so $ k = j$. So if $ k = j$ we have $ (a,b) = (m,m) \forall m \in \mathbb{N}$

Case $ p = 1, q > 1$: We have $ b^q = a$. With substitution as before we get $ q = b^{qj - k}$ If $ k = j = 1$ we have as before $ a = 4, b = 2$ and if $ k = j > 1$, no solutions. So now you just have to find solutions for $ j < k$ There should be relatively few for each $ q,j$..


trigonometry
inequalities unsolved
inequalities

Let triangle $ ABC$. Find Max of
$ \cos A\plus{}\cos B\plus{}\cos C$

[quote="sylvect"]Let triangle $ ABC$. Find Min of
$ P \equal{} \frac {\cos^2 A}{1 \plus{} \cos A} \plus{} \frac {\cos^2 B}{1 \plus{} \cos B} \plus{} \frac {\cos^2 C}{1 \plus{} \cos C}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59649


[quote="ProtestanT"][quote="7h3.D3m0n.117"]Two different prime numbers are selected from the first eight prime numbers. What is the probability that the sum of the two chosen prime numbers is greater than or equal to 16? Express your answer as a common fraction.

Hmm I might have counted wrong but I want to see what you guys get.[/quote]
[hide]
the first 8 primes are 2,3,5,7,11,13,17,19.
the probability of choosing either a 17 or a 19 is $\frac{1}{4}+\frac{3}{4}\frac{2}{7}$
out of the rest of them, the probability of choosing a 13 and something higher than a 2 (not including 17 and 19) are $\frac{1}{8}\frac{4}{7}+\frac{4}{8}\frac{1}{7}$
the probability of choosing 11 and something higher than a 3 (not including 17, 19, or 13) is $\frac{1}{8}\frac{2}{7}+\frac{2}{8}\frac{1}{7}$
the rest of them cannot be combined to 16+ without the help of 11, 13, 17, and 19
adding them together, it is $\frac{19}{28}$
[/hide][/quote]
When you put two fractions together with nothing between them, are you multiplying them?


Are there infinitely many primes p such that 2 is a primitive root mod p?

(by guas in another thread:)
Is this solved? I know that this is Artin conjecture:
every positive integer which is not a square is a primitive root for
an infinitely many p. But I know only how it can be deduced from
extended Riemann conjecture, which is very far from being proved.

Funnily enough, it is known that either 2, 3, or 5 is a primitive root modulo infinitely many primes.
[color=#fff]reap[/color]


$\frac{1}{2-\sqrt[4]{2}}$

Rationalize the denominator.

[hide="hint"]
start off by rationalizing it just like you would a normal square root and you should then end up with a square root in the denominator..

go from there.[/hide]
[hide="answer"]
multiply by the conjugate of the denominator
$\frac{1}{2-\sqrt[4]{2}}\cdot \frac{2+\sqrt[4]{2}}{2+\sqrt[4]{2}}$
this gives us
$\frac{2+\sqrt[4]{2}}{4-\sqrt{2}}$
multiplying then we multiply again by our new denominators conjugate:
$\frac{4+\sqrt{2}}{4+\sqrt{2}}$
which gives us our answer of
$\boxed{\frac{8+2\sqrt{2}+4\sqrt[4]{2}+\sqrt[4]{8}}{14}}$
[/hide]


Harvard
college
MIT
Princeton
Stanford
University of Chicago

Hi!

Harvard, MIT, and Princeton and probably the best schools to study maths in the world. However there must probably be some differences? Where do you people think is the best place to study Mathematics as an undergraduate?

Greets,
psykyk

I once saw a personal Web site by a young man who said that the best way to study math is to do as he did: study undergraduate at Harvard and graduate school at MIT. However, he has removed that statement from his personal Web site since becoming a lecturer at Stanford  :P  .

In other words, don't worry about it too much. Do ask questions to see what program fits you best, but since you can't be sure of getting into every school that you apply to, be prepared to adapt to a different environment if you get in somewhere else.

What does everyone else think?

beside the top math schools(MIT, CA Tech...)

try Harvey Mudd college, it "seeks to educate engineers, scientists and mathematicians"

I know rankings are somehow stupid... but do you guys know any rankings concerning undegraduate concentration on maths? Furthermore... what could be the differences between one of these top universities and another?

greets,
psykyk

One difference between math undergraduate programs is how coherent their overall undergraduate curriculum is. The University of Chicago has a very well thought-out undergraduate curriculum, for example, while some other universities of otherwise equal prestige mostly have a collection of courses, not necessarily well connected to one another.

[quote="juicybooty911"]better money[/quote]

Thats a very conceited way of looking at learning.


Triangle ABC with B=90.CD=8*2^.5 and AE=9 both angles bissectors.Calculate AC.

Let $ ABC$ be an $ A$-right triangle. Denote the length $ l_a$ of the $ A$-bisector of $ \angle BAC$ a.s.o. Then $ \left\|\begin{array}{c} 2ab^2 \equal{} (a \plus{} b)l_c^2 \\
 \\
2ac^2 \equal{} (a \plus{} c)l_b^2 \\
 \\
b^2 \plus{} c^2 \equal{} a^2\end{array}\right\|$

and $ 8\underline a^4 \equal{} 4\underline a^2\left(l_b^2 \plus{} l_c^2\right) \plus{} \left(l_b^4 \plus{} l_c^4\right) \plus{} l_b^3\sqrt {8\underline a^2 \plus{} l_b^2} \plus{} l_c^3\sqrt {8\underline a^2 \plus{} l_c^2}\ \implies$ the value of $ \boxed {a}$ when we know $ l_b$ and $ l_c$ .

This solution is very difficult to solve the equation.Did you have another way?


This question is very easy. I knew the solution a few weeks ago but now it is entirely eluding me. Maybe I should drink some coffee?

Show that $ 2^{2^{n}}+1| 2^{2^{n}*2k+1}+1$

I will assume you meant $ 2^{2^{n}}+1|2^{2^{n}(2k+1)}+1$, because as it's stated now, it's false (take $ n=k=1$).

We prove the more general statement.

$ x+1|x^{2k+1}+1$.  This is true because $ x^{2k+1}\equiv (-1)^{2k+1}\mod{x+1}\equiv-1\mod{x+1}$. (or $ x^{2k+1}+1=(x+1)(x^{2k+1}-x^{2k-2}+x^{2k-2}-...-x+1)$).

[quote="Altheman"]This is really just a direct application of the following [trivial] factorization: $ a^{k}+b^{k}=(a+b)(a^{k-1}-a^{k-2}b+...-...+b^{k-1})$ where $ k$ is odd.[/quote]

Well yeah of course...that was what I was encouraging.  (ie, in the argument given above, $ -1$ works as well as any number)


search
inequalities
LaTeX
inequalities solved

For any two natural numbers $m,n$ and real $x\in[0,1],$ prove that
\[\left(1-x^n\right)^m+\left(1-(1-x)^m\right)^n \geq 1.\]

I wrote it in POP, and it had been written already. :)

More :

Prove that for all $ m,n\in\mathbb{N}$ and two real sequences $ x_1,x_2,..,x_n,y_1,y_2,...,y_n\in\;[0,1]$ we have

$ (1\minus{}\prod_{i\equal{}1}^n x_i)^m\plus{}\prod_{i\equal{}1}^n(1\minus{}y_i)^m\ge 1$

the generalization as it is stated is not correct, e.g., set all variables equal to 1, then the LHS is equal to zero

Sorry, i misstake  :blush: $ x_i\plus{}y_i\equal{}1,\forall i\equal{}\overline{1,n}$

but this one is wrong as well for $ m>1$, e.g. set $ \prod_{i\equal{}1}^nx_i\equal{}\frac{1}{2}$, then the LHS is equal to $ 2^{1\minus{}m}$

[quote="hjbrasch"]but this one is wrong as well for $ m > 1$, e.g. set $ \prod_{i \equal{} 1}^nx_i \equal{} \frac {1}{2}$, then the LHS is equal to $ 2^{1 \minus{} m}$[/quote]

Please tell me what are two sequences to disprove this problem, after edited, it is a true inequality  :mad:

I think I have already stated the counterexample, set $ x_i\equal{}\left(\frac{1}{2}\right)^{\frac{1}{n}}$ this implies $ y_i\equal{}1\minus{}\left(\frac{1}{2}\right)^{\frac{1}{n}}$ and therefore

$ \prod_{i\equal{}1}^nx_i\equal{}\prod_{i\equal{}1}^n(1\minus{}y_i)\equal{}\frac{1}{2}\Longrightarrow LHS\equal{}2\cdot\left(\frac{1}{2}\right)^m\equal{}2^{1\minus{}m}$ which is stricly smaller than 1 for $ m>1$

I think my $ \text{\LaTeX}$ has some misstake  :wacko:

Prove that for all $ m,n\in\mathbb{N}$ and two real sequences $ x_1,x_2,..,x_n,y_1,y_2,...,y_n\in\;[0,1]$ and $ x_i+y_i=1$ we have

$ (1 - \prod_{i = 1}^n x_i)^m + \prod_{i = 1}^n(1 - y_i^m)\ge 1$

I am very so sorry  :(

I have a nice probabilistic proof of your corrected statement:

Define for $ i\equal{}1,\ldots,n$ and $ j\equal{}1,\ldots, m$ independent Bernoulli-variables $ D_{i,j}\simeq Ber(x_i)$. Define the row and column sums via $ R_i\equal{}\sum_{j\equal{}1}^mD_{ij}$ resp. $ C_j\equal{}\sum_{i\equal{}1}^nD_{ij}$. Then we have the identifications:

$ \left(1\minus{}\prod_{i\equal{}1}^nx_i\right)^m\equal{}\mathbb{P}(C_1<n,\ldots,C_m<n)$ and $ \prod_{i\equal{}1}^n(1\minus{}y_i^m)\equal{}\mathbb{P}(R_1>0,\ldots,R_n>0)$.

Now observe, that the complement of the event $ \{C_1<n,\ldots,C_m<n\}$ is given by $ \{\exists j\in\{1,\ldots,m\}: C_j\equal{}n\}$. The event $ C_j\equal{}n$ for some $ j\equal{}1,\ldots,m$ implies $ \forall i\equal{}1,\ldots,n: \,D_{ij}\equal{}1\Longrightarrow R_1>0,\ldots,R_n>0$, i.e. we have the set inclusion $ \{\exists j\in\{1,\ldots,m\}: C_j\equal{}n\}\subseteq\{R_1>0,\ldots,R_n>0\}$ from which $ 1\minus{}\mathbb{P}(C_1<n,\ldots,C_m<n)\leq \mathbb{P}(R_1>0,\ldots,R_n>0)$ follows verifying the claim.

[quote="hjbrasch"]I have a nice probabilistic proof of your corrected statement:

Define for $ i \equal{} 1,\ldots,n$ and $ j \equal{} 1,\ldots, m$ independent Bernoulli-variables $ D_{i,j}\simeq Ber(x_i)$. Define the row and column sums via $ R_i \equal{} \sum_{j \equal{} 1}^mD_{ij}$ resp. $ C_j \equal{} \sum_{i \equal{} 1}^nD_{ij}$. Then we have the identifications:

$ \left(1 \minus{} \prod_{i \equal{} 1}^nx_i\right)^m \equal{} \mathbb{P}(C_1 < n,\ldots,C_m < n)$ and $ \prod_{i \equal{} 1}^n(1 \minus{} y_i^m) \equal{} \mathbb{P}(R_1 > 0,\ldots,R_n > 0)$.

Now observe, that the complement of the event $ \{C_1 < n,\ldots,C_m < n\}$ is given by $ \{\exists j\in\{1,\ldots,m\}: C_j \equal{} n\}$. The event $ C_j \equal{} n$ for some $ j \equal{} 1,\ldots,m$ implies $ \forall i \equal{} 1,\ldots,n: \,D_{ij} \equal{} 1\Longrightarrow R_1 > 0,\ldots,R_n > 0$, i.e. we have the set inclusion $ \{\exists j\in\{1,\ldots,m\}: C_j \equal{} n\}\subseteq\{R_1 > 0,\ldots,R_n > 0\}$ from which $ 1 \minus{} \mathbb{P}(C_1 < n,\ldots,C_m < n)\leq \mathbb{P}(R_1 > 0,\ldots,R_n > 0)$ follows verifying the claim.[/quote]

Sorry but i don't understand your proof . It is quite easy, we can use Inductive method  :)

Pls note, the probabilistic proof is just $ \mathbb{P}(A)\plus{}\mathbb{P}(B)\geq 1$ ,since $ A^c\subseteq B$, i.e. a one-liner, and you only have to identify the right sets.

P.S which part do you not understand ?

[quote="quangpbc"]I think my $ \text{\LaTeX}$ has some misstake  :wacko:

Prove that for all $ m,n\in\mathbb{N}$ and two real sequences $ x_1,x_2,..,x_n,y_1,y_2,...,y_n\in\;[0,1]$ and $ x_i + y_i = 1$ we have

$ (1 - \prod_{i = 1}^n x_i)^m + \prod_{i = 1}^n(1 - y_i^m)\ge 1$

I am very so sorry  :([/quote]

I will use the same method I used here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=75884[/url]

Denote $ f(y_{1},....,y_{n})=1-\prod_{i = 1}^n (1-y_{i})$
The inequality is equivalent to $ f(y_{1},....,y_{n})^{m} \geq f(y_{1}^{m},....,y_{n}^{m})$
if $ x_{i}=1$ or $ y_{i}=1$ for some $ i$, we get equality. Let's assume that all x's and y's are different from 0 and 1.
for m=1: trivial
for m=2:
$ \frac{\prod_{i=1}^n (1+y_{i}) + \prod_{i=1}^n (1-y_{i})}{2} \geq 1$ it's true, because:
$ \frac{\prod_{i=1}^n (1+y_{i}) + \prod_{i=1}^n (1-y_{i})}{2} = 1+\sum y_{i}y_{j}+\sum y_{i}y_{j}y_{s}y_{t}+.... \geq 1$
If we assume that the claim is correct for some n=k>1, then the following holds:
$ f(y_{1},....,y_{n})^{k+1}=f(y_{1},....,y_{n})^{k}f(y_{1},....,y_{n}) \geq f(y_{1}^{k},....,y_{n}^{k})f(y_{1},....,y_{n})$
it's sufficient to show that:
$ f(y_{1}^{k},....,y_{n}^{k})f(y_{1},....,y_{n}) \geq f(y_{1}^{k+1},....,y_{n}^{k+1})$
or-
$ \prod_{i=1}^n (1-y_{i}^{m+1}) + \prod_{i=1}^n (1-y_{i}^{m}-y_{i}+y_{i}^{m+1}) \geq \prod_{i=1}^n (1-y_{i}^{m}) + \prod_{i=1}^n (1-y_{i})$
which is immediate from the following lemma:

If $ a_{i}, b_{i}, c_{i}, d_{i}$ are 4 sequences of length n, of positive numbers, such that for every i:
$ a_{i}+b_{i}=c_{i}+d_{i}, b_{i} \geq c_{i} \geq d_{i} \geq a_{i}$
then-
$ \prod a_{i} + \prod b_{i} \geq \prod c_{i} + \prod d_{i}$
(reminds me of majorization... and rearrangement)

I don't think it's hard... I will try tomorrow.


A circle of radius 5 is circumscribed about an isosceles triangle. If the length of the altitude to the base of this triangle is 9, what is the area of this triangle?

I'm not going to do the problem (too lazy :P ), but sanity-check the problem. If your answer is under or equal to 45, you know you got it wrong.

[hide]The radius goes through the goes through the height or altitude of the triangle leaving 4 at the other end. The radius also goes to the vertices at the base of the triangle, forming a 3-4-5 right triangle. This means the base is 6 so the area of the triangle must be 27. To further prove my answer, if the base is 6, the other two sides of the isoscoles triangle must be the sqrt of 30. Using the formula abc/(4A) where A=the area of the triangle and a, b, and c are the three sides, to find the radius of its circumscribed circle, we get 5.[/hide]

[hide]
27.  Draw the radii to the vertices of the circle and the altitude to the base.  Notice if you subtract radius (5) from the altitude (9), that gives you the altitude of one of the pedal triangles (4). Half of this pedal triangle is a 3,4, 5 triangle telling you the length of the base of the inscribed triangle is 2*3=6.  Therefore, the area of the inscribed triangle is $\frac{1}{2}6*9=27$ [/hide]
 :D

About. I can't read T_T .

*redoes the problem*


Compute: $ \frac{7!}{6!}\bullet \frac{3!2!}{(8\minus{}1)!}$Express your answer as a common fraction.

Note that since $ 7\equal{}8\minus{}1$, you can cancel $ 7!$ from the numerator and denominator ($ 7!$ is disguised as $ (8\minus{}1)!$). Afterward, you have $ \frac{3!2!}{6!}\equal{}\frac{3*2*2!}{6*5*4*3*2}\equal{}\frac{2!}{6*5*4}\equal{}\frac{2}{120}\equal{}\boxed{\frac{1}{60}}$.


This is a fairly simple way to multiply numbers quickly in your head.  It really only works for relatively close numbers of the same parity, but it's still good for impressing people, and is useful on the AIME.  As far as I can tell, most people don't seem to realize that they can do this.

Remember from algebra class that (x-y)(x+y)=x^2-y^2?  Well that's actually useful, believe it or not.  Suppose you want to find 43*37 quickly.  Well, you know that that's (40+3)(40-3)=1600-9=1591.

What if it was 56*58, and you don't know 57^2 off of the top of your head quickly?  Then you can find it just by doing (60-3)^2=3600-360+9=3249, and then it's just (57+1)(57-1)=3249-1=3248.

Or suppose you wanted to find 16*19, where they aren't of the same parity.  Then you can find 16*18=(17+1)(17-1)+16=289-1+16=304.

It takes a little practice, but it's pretty simple once you get into it.

It looks very good. Maybe 2-digit numbers aren't so hard, but what about 3-digit numbers? Are there any tricks (I mean in general case)? Or it just require long, long training? :D

I've read about one man, who has same ilness as the one in movie Rainman (I don't know the name of that ilness in english) and he was able to multiply 100-digit numbers in 8 hours... Do you know about him? His name or what else could he do?

Heh, there's a trick for multiplying two numbers barely over 100.

1ab x 1cd = 1 ab+cd ab*cd

so 105 x 106 = 1 11 30 or 11,130

Of course, if you have numbers where ab*cd > 99, you have to start adding to the ab+cd section, so...

[quote]..... don't know 57^2 off of the top of your head quickly? Then you can find it just by doing ...( ,[/quote]

I generally use the trick as multiply  60 and 54  and add 9.
3249.

Fore example if you want square of 48, use  46 (=48-2) and 50 (=48+2)   to  get 2300 and add 4 (square of 2) = 2304 , (Just reverse of your method.)

(Of course in my days, as I tell my sons,  we routinely remembered times table  up to  100  .. (also walked  uphill both ways in 5 feet of snow and 100 degree heat  going to school..) :)

Yeah, I should really get around to those times tables....

But thanks for the tip!

Lol I used to use that trick all the time, and when I finally found out hat x^2-y^2 is (x+y)(x-y), I thought, oh THAT's how that works.  I love that trick, it's so handy.  Oh btw all you future mathcounters out there learn this formula (if you don't already know it):

n(n+1)/2, where you want to find the sum of every counting number up to n.  so the first 10 counting numbers added 1+2+3+...+10 is 10(11)/2, or how I look at it, either 10 or 11 is gonna be even, so half that and times the other.  i.e. 5*11=[b]55[/b]

^--any successful MathCounts person know this formula is a nececity for target round and some of the latter sprint rounds.

If you're multiplying any two digit numbers:
ex. 34*72
(30+4)(70+2)=2100+60+280+8=2438
It gets faster once you do it a lot, this one has helped me a lot at number sense.

Really thanx.
I've got some informations about that man who I was writing about.
His name is Johann Martin Zacharias Dase (1824-1861) and he propably didn't have that ilness. He was able to multiply in his head two
8-digits numbers in 54 seconds
20-digits numbers in 6 minutes
40-digits numbers in 40 minutes
100-digits numbers in 8 hours and 45 minutes
He wasn't very good at mathematics, but he find first 200 digits of pi in 2 months and made lots of mathematical tables.
It's pretty amazing, isn't it?

[quote="ondrob"]It's pretty amazing, isn't it?[/quote]It's almost not belivable :)


[quote="Mathias_DK"][quote="Ligouras"]Let  $ ABC$ be an triangle . Prove or disprove that
$ 4\cos(\frac {A}{2})\cos(\frac {B}{2})\cos(\frac {C}{2})\ge\sin(A \plus{} B \plus{} C)$ .

______________
 :([/quote]
$ A\plus{}B\plus{}C \equal{} 180$ so $ sin(A\plus{}B\plus{}C) \equal{} 180$. On the other hand $ a,b,c \le 180$ so $ LHS$ is non-negative....[/quote]

$ 4\cos\frac {A}{2}\cos\frac {B}{2}\cos\frac {C}{2}>0=\sin(A \plus{} B \plus{} C).$


inequalities
logarithms
function
integration
real analysis
real analysis solved

$k^{\frac1k}=e^{\frac1k\ln k}>1+\frac{\ln k}k.$

Hence $\sum_{k=3}^nk^{\frac1k}> \sum_{k=3}^n(1+\frac{\ln k}k) = n-2+\sum_{k=3}^n\frac{\ln k}k.$

Since $\frac{\ln x}x$ is a decreasing function on $[3,\infty)$, we can estimate that $\sum_{k=3}^n\frac{\ln k}k> \int_3^{n+1}\frac{\ln x}xdx = \frac12\ln^2x|_3^{n+1}=\frac12(\ln^2(n+1)-\ln^23).$

Thus, $\sum_{k=3}^nk^{\frac1k}> n-2+\frac12(\ln(n+1)+\ln3)(\ln(n+1)-\ln3).$

That's right.I have no explanation more.

[quote="kunny"]Prove the following inequality.

\[\sqrt[3]{3}+\sqrt[4]{4}+\cdots\cdots\cdots+\sqrt[2004]{2004}>2002+\frac{1}{2}\left(\ln 3\cdot2005\right)\left(\ln \frac{2005}{3}\right)\][/quote]
:( I would expect more nice estimate.


number theory
least common multiple
MATHCOUNTS

how do you do this question ?it was from aops but i din
 see answer n tried to solve but coulndt solve. plz post the answer. :D .

One light flashes every 5 minutes. Another flashes every 12 minutes. The third light flashes every 8 minutes. If they all flash together at the same time at 10:00 AM, when is the next time they will flash together?

[quote="limac"]This problem requires the use of the Least Common Multiple. So we know that one of the lights flash every $ 5$ minutes and the other two flash in $ 12$ and $ 8$ minutes, respectively. Thus, we need to find $ LCM[5,12,8] \equal{} 120$

If I have done it correctly, the three lights flash together every $ 120$ minutes, which is equivalent to $ 2$ hours. Since they flashed together at $ 10: 00 AM$, the next time will be $ 10: 00 AM \plus{} 2: 00 \equal{} \fbox{12: 00 AM}$

EDIT: I am not sure if $ 12: 00$ noon, is supposed to be $ AM$ or $ PM$, but it is basically at noon.[/quote]

I think 12:00 Noon is 12:00 PM.

let's not get distracted and spam...

Anyways: You want to find the LCM of 12, 5, and 8. You can easily find that to be 120. So to figure out the time you simply add the minutes to the 10:00. 120 min=2 hours, so you get [b]12:00[/b].

I believe this an MOEM elementary problem.

[quote="PowerOfPi"]Noon is neither AM or PM. The following explains:
[url]http://wwp.greenwichmeantime.com/info/noon.htm[/url][/quote]
Wow, I never knew that. I've always used 12:00 am for midnight and 12:00 pm for noon, which seems to be pretty standard convention. If this were a MATHCOUNTS test, I would hope they would accept [b]Noon[/b] as an answer.

thnx all of you my doubts are now vanished completely. :lol:  :)


modular arithmetic
quadratics
abstract algebra
number theory proposed
number theory

Prove that if $ x$ and $ y$ have the common factor then every odd factor of
\[ x^{2^n} \plus{} y^{2^n},
\]
where $ n$ is positive integer, is of the form $ 2^{n \plus{} 1}m \plus{} 1$.

Very cute problem!

I think you meant to say in the first line that $ x,y$ are coprime (which makes sense).
It is enough to show this for prime factors of $ x^{2^n} \plus{} y^{2^n}$, because the product of the numbers of the form $ 2^{n \plus{} 1}m \plus{} 1$ is also of this form (these are the numbers that equal to 1 modulo $ 2^{n \plus{} 1}$).

Consider $ p|x^{2^n} \plus{} y^{2^n}$, $ p$ a prime. Then $ (xy^{ \minus{} 1})^{2^n} \equiv \minus{} 1 \pmod p$.
It means that the order of $ xy^{ \minus{} 1}$ mod $ p$ is $ 2^{n \plus{} 1}$.
By Lagrange's Theorem (order of an element divides the order of a group), $ 2^{n \plus{} 1} | p \minus{} 1$, and the conclusion follows.

I believe it also implies the existence of infinitely many primes of the form $ 2^{n \plus{} 1}m \plus{} 1$  :)

[EDIT: I used the fact that $ p$ is odd when I said that the order of $ xy^{ \minus{} 1}$ mod $ p$ is $ 2^{n \plus{} 1}$ - if $ p\equal{}2$, the order is always 1]

Well,Here is some results
1) If $ p$ is an odd prime and $ p|2^{2^n} \plus{} 1$ then
\[ p \equal{} k.2^{n \plus{} 2} \plus{} 1
\]
2) My generalization : Let $ b;d$ be evens ,$ a;b$ be natural numbers satisfying $ gcd(a,c) \equal{} 1$
Prove that if $ p$ is an odd prime and
\[ p|a^{b^n} \plus{} c^{d^n}
\]
Then
\[ 2^{n \plus{} 1}|p \minus{} 1
\]

That generalisation isn't really one  :wink:

That problem is known, see e.g. http://www.mathlinks.ro/viewtopic.php?t=150543 . See http://www.mathlinks.ro/viewtopic.php?t=126566 for a real generalisation.

Sorry,ZetaX,But in your first link you only shows $ p \equal{} k.2^{n \plus{} 1} \plus{} 1$  :lol: .The problem above is stronger and harder (with me  :)  ).Indeed My proof is very long within some non-elementary knowledge,So  can  you give a nice one,ZetaX ?Thank you.
For the second link,Thank you again.

[quote="Allnames"]
2) My generalization : Let $ b;d$ be evens ,$ a;b$ be natural numbers satisfying $ gcd(a,c) \equal{} 1$
Prove that if $ p$ is an odd prime and
\[ p|a^{b^n} \plus{} c^{d^n}
\]
Then
\[ 2^{n \plus{} 1}|p \minus{} 1
\]
[/quote]

Let $ b \equal{} 2^{b_1}b_2, d \equal{} 2^{d_1}d_2, d_2 \equiv b_2 \equiv 1 \pmod 2$. Now take $ x \equal{} a^{b_2^n}, y \equal{} c^{d_2^n}$ and apply the original problem.

I would be very interested in seeing the proof to your first proposition.

[quote="Allnames"]1) If $ p$ is an odd prime and $ p|2^{2^n} \plus{} 1$ then
\[ p \equal{} k.2^{n \plus{} 2} \plus{} 1
\]
[/quote]
You need $ n \geq 2$ here (it indeed ir wrong for $ n\equal{}0,1$).
For $ n \geq 2$, we get that $ p \equiv 1 \mod 2^{n\plus{}1}$ by the previous, thus $ p \equiv 1 \mod 8$. So $ 2$ is a quadratic residue, say $ 2 \equiv a^2 \mod p$. Now use the same proof again for $ a$:
It has an order dividing $ 2^{n\plus{}2}$, but not dividing $ 2^{n\plus{}1}$, thus the order is $ 2^{n\plus{}2}$. This gives $ 2^{n\plus{}2}|p\minus{}1$.


geometry
3D geometry
induction
MATHCOUNTS

Find all numbers which are equal to:

1. The sum of it's digits squares
2. The sum of it's digits cubes

Edit: I got that the only one solution to #1 is 1. I checked some numbers less than 99, because 99 is the biggest number which is not greater than the sum of it's digits squares.

One thing to notice for number $ 2$ is that the number must be $ 4$ digits or less.

For $ 1,2$, examining $ \bmod8,9$ might help.

Another answer to #1 is 0, considering you didn't say that the numbers had to be positive.

Is there a way to prove or definitively find those bounds on the digits?

If we are looking at the sum of the $ n^{th}$ powers of the digits and there are $ k$ digits, then we must have $ k\cdot9^n < 10^{k \minus{} 1}$ to have no solutions, whose solution can probably be proven by induction (after you find the bounds and stuff).

For #1, 1 isn't the only answer..
81 works too

It's the sum of its [b]digits' squares[/b], not the sum of its [b]digits, squar[u]ed[/u][/b].

That is, we want $ \overline{a_0a_1\cdots a_n}\equal{}\sum_{k\equal{}0}^{n}a_k^2$.

not $ \left(\sum_{k\equal{}0}^{n}a_k\right)^2$.

Uh, $ 1^2 \plus{} 5^2 \plus{} 3^2\neq 153$

Edit: Ah, right, $ 1^3\plus{}5^3\plus{}3^3\equal{}153$. lol, wasn't thinking.
Was there some method you used to find that, or was it just brute force?

a mathcounts team round our coach made up stated that there are 4 three digit numbers that are the sum of the cubes of their digits, 3 of which were 153, 370, and 407, then asked for the fourth

and mewto's referring to the second problem

According to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1341292#1341292]this[/url], those three-digit numbers are $ 153,370,371,407$. As I said before, for problem $ 1$, investigating $ \bmod8$ eliminates many choices, as does $ \bmod9$ for problem $ 2$. Also, we have concluded that $ 3$ digits (less than $ 243$) is the maximum for problem $ 1$ and $ 4$ digits (less than $ 2916$) is for problem $ 2$.

Edit: According to my calculator, $ \boxed{\{0,1\}}$ are the only solutions to $ 1$ and $ \boxed{\{0,1,153,370,371,407\}}$ are the only ones to the second one.


Congragulations to the Pakistan team for improving considerably on last year's performance! We have to remember that this is only the 2nd time Pakistan has participated. Btw, why aren't the results of PAK 4 available?

yes,
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101847
Valentin posted the results. :)

The individual results are posted here:
[url]http://imo2006.dmfa.si/results.html[/url]

thanks rem and inspired nature... :wink:

u really helped me.


Let $x,y,z$ are positive real number such that $x+y+z=1$. Prove that:
$\sqrt{x+y^{2}}+\sqrt{y+z^{2}}+\sqrt{z+x^{2}}\ge 2$

:blush: She is so pretty. May I know her name? Ah by the way, may I have her phone number, too? :blush:

[quote="Naphthalin"]11833 - wir sind die auskunft[/quote]
What do you mean? Please use English :blush:  ,Thanks.

[quote="Naphthalin"]11833 - wir sind die auskunft[/quote]
I think it is spam  :P

of course this is SPAM. but no spam ;) there is a little difference^^. do you understand it? :)
ask google to translate...
the "auskunft" is an institution where you can ask to get phone numbers or something else^^ and there was a spot in germany a while ago that used this sentence^^ because langtupanda asked all the time about your name and number^^ (@zaizai-hoang) ;)

greetings from germany^^

Naphthalin


calculus
integration
function
limit
real analysis
real analysis unsolved

A bounded function $f$ is said to be Darboux-integrable if $lim S_{\Pi}$ and $lim s_{\Pi}$ exist and are equal as the mesh of the partitions go to $0$, and where $S$ and $s$ are the usual Darboux sums.

It seems to me that those limits *always* exist, by the bounded assumption. However, the author of my book keeps proving the existence all the time in order to show Darboux (or equivalently, Riemann)-integrability. Is there anything I overlooked?

I don't think you have overlooked anything. Anyway, I know the definitions of upper/lower Darboux integrals with $\sup$/$\inf$ insteam of $\lim$, in which case they always exist.


If community members want websites where they can buy purchase (yes, with real dollars and cents) cheats for World of Warcraft, I'm sure they would direct their attention to Google. Please refrain from advertising on this board.

Locked.


There are a lot of pretty good posts in this forum and stickying all of them will crowd the forum.  So this thread is going to fix the problem by having links to all of these threads.

[b]General Interest[/b]
$\bullet$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151467]Mock ARML 2007 (I) by Altheman[/url]
$\bullet$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30264]ARML Practice Series[/url]
$\bullet$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32273]ARML Practice Series Scoreboard/Solutions[/url]
$\bullet$[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86168]ARML Practice Series 2[/url]
$\bullet$[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86171]Arml practice series 2 scoreboard/Solutions[/url]

[b]Iowa[/b]
$\bullet$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147370]General Discussion Thread for 2007[/url]
$\bullet$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151503]Prof. David C. Kelly's Talk - The Euler Spoilers[/url]

[b]Las Vegas[/b]
$\bullet$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148086]Friday Afternoon in Las Vegas[/url]

[b]Penn  State[/b]
$\bullet$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150340]Join Sides Fellow Warriors[/url]
$\bullet$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151503]Dr. Sarah-Marie Belcastro's Talk - Surprising Binary Sequences and Mysterious Expanding Tilings[/url]
$\bullet$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142888]AoPS Meeting[/url]

If you have additions to this list, please post them in this topic.  I'll probably delete them afterwards to keep this post as neat as possible...so don't worry about spamming or anything.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30264]ARML Practice Series[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32273]ARML Practice Series Scoreboard/Solutions[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86168]ARML Practice Series 2[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86171]Arml practice series 2 scoreboard/solutions[/url]

the ARML Monthly challenge problem link seems dead

ARML Practice Tests from 08:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=206892]Altheman's Practice ARML test[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=207581]solutions[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=206547]e^pi*i=-1's mock test[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=206880]Twiz' mock test[/url]

I would like to add the following link: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=348&t=151670

It is the Mock ARML 2007 #2.


Support
geometry
geometric transformation
reflection
set theory

I'm Mennonite, which is a Christian religion for those of you who don't know. I am involved in JMYF (Junior Mennonite Youth Fellowship), which is like youth group. We meet every other week and I go whenever possible. I go to church with my family every Sunday unless, for some reason, I can't make it.

Lutheran and I go to church every Sunday as well as Catechism, youth group sorta bible study too.

I don't want to offend anyone, but I am atheist. I don't believe the bible is what really happened, but I've gone to church and they praise the bible like its the only true piece of scripture. They can't stand to see the amounts of contradiction there are. The answer to every question : "Because God made it that way" and there is no way to disprove that or support it. that is why I am a nonbeliever.

P.S. This should be at the Round Table - move it there. You will get lots of better feedback.

Only mods can do it, so I did it. Done and done. (I also deleted an "i" and added a "?" too :))

You bar mitzbahed and then quit? Mmm...

Wow, I thought mathy people was athiest (I was kind of thinking along the line of math=logic=everything making sense=not religious but guess not)...interesting to know.

Mmm I suggest "Inherit the Wind" for you white_horse_king88.

I am part athiest, mostly atheist I guess. I have my own set of thoughts, more of Hinduism/Buddhism then the Western religions I guess. I don't believe in actual "worshipping" and stuff but I do believe in doing good for others and helping out and stuff, just being good in spirit/soul/etc.

P.S. I believe The Round Table now covers all controversial topics (and be peaceful with it too), congratulations!

Isn't it "Adonai"? At least that's what I heard during the bar mitzbah/bat mitzbahs here...(this here's a Jewish community, Democratic too)

I am religios but I take the mathmatical and scientific aproch to things.  I don't know if I am that religios though we don't go to church that often.

Keep in mind that being religious and believing what your religion has to say are two completely different things. I am reasonably religious, but I certainly take the religious creed with more than a grain of salt. Then again, we Jews are known for being rather rebellious and argumentative (the joke says that three Jews will have five opinions), and it seems that questioning fundamental religious beliefs is not just accepted but encouraged.

i have a question,
complexzeta are all people that have rubistein as their last name jews because there is like this famous piano guy who is one (arthur rubistein came to the u.s from poland due to the nazis.

I'm Catholic, and yes, I go to church every week.  I also go to CCD on wednesday evenings and youth group after church....

[quote="Scrambled"]two things you should never talk about:
Religion and politics.

The round table now talks about both...

im just curious.  Is the proper term for someone who doesnt have a religion an atheist or a nihilist?[/quote]
I'd say atheist since nihilism is more of a thought rather than a religious thing, it's really hard to differentiate but the first means "does not believe in the existence of God" and the latter means more of a "belief that nothing (including, but not limited to, the existence of God, spirit, and other religious stuff) exists" so the latter have a broader (and many different) meanings.

I already said that Scrambled...but oh well.

I doubt that all people named Rubinstein are Jews (it's a fairly common name, after all), but for sure an extremely large proportion of them are. On a somewhat unrelated note, people who spell the name Rubenstein are wrong. People who are named Rubenstein got fed up of correcting others and changed their name (probably largely due to Ellis Island). But I imagine that I can count the number of people named Rubenstein who live in Europe on my fingers and toes, if not just my fingers.

Mmm aside from your religious affiliation I guessed you were Italian because of "Salzedo" but that's about it.

I think it's quite hard to figure out which country/religion a person has come from/belongs to, unless they're like Korean or Japanese, then it's mostly obvious. Still, even then there are some exceptions...

I'm Catholic. I go to church almost every week and CCD afterwards. I'm not actually that religious though.

Salzedo is a Spanish name, but my father's family hasn't lived in Spain for about 500 years.

I'm not that religious, but I do believe in God and being good to people and stuff like that. Some of the stuff in the Bible doesn't seem very likely that it happened that way, but I think the Bible isn't meant to be taken literally, more like a model to live your life the right way and not do anything really bad. I used to go to church-a Lutheran one, even though I was baptised Catholic. Kind of wierd.

I'm trying very hard not to preach right now, so bear with me.  The reason things don't seem likely in the Bible is because they weren't, that's the point.  God did it, it doesn't have to make sense.  If you think about it, the Big Bang is so unlikely that it was the right exploding force for planets to form that something had to help.  Just think about it.  I hope that wasn't too preaching.

Capitalize it!  Its Roman Catholic Christian!  don't say it like your ashamed!

She also didn't capitalize baptist or methodist.

True but as I am also Catholic I pay attention to it.  Understandable?

[quote="aalindsey100"]I say that Im catholic but really I dont practice it that often. I was first baptised in the catholic church when I was a baby. Then when I was 7 we converted to baptist and I was baptised in the baptist church. Then we went to the methodist church and I think I liked that one the most. Anyhow, I landed back on catholisism because I just like the structure it provides and it is still loose strange. I dont believe totally what any of the churches believe I believe in my theory on things which is a combination of the three actually. I love to study different religions and I firmly believe that there is a God but then again I believe there are more than one because you might find this weird but I believe in extra terrestrial life. I am different yet not so different. There is logic in science and math but that doesnt answer everything.[/quote]
Oh the irnoy! :lol:

raccoon:
I'm sorry if I offended you. I never said that I didn't think the things in the Bible were true, I just said there were some things that didn't seem likely from a logical approach. I really do believe in the Bible. Maybe my post was unclear (the result of not sleeping). There are other things that don't follow logic either and I believe in them, like ghosts and extraterrestrials (anybody: laugh if you'd like; I don't mind)^_^

Here's a question: What distinguishes a religion from another set of beliefs?

[quote="MithsApprentice"]religion
n.

   1.
         1. Belief in and reverence for a supernatural power or powers regarded as creator and governor of the universe.
         2. A personal or institutionalized system grounded in such belief and worship.
   2. The life or condition of a person in a religious order.
   3. A set of beliefs, values, and practices based on the teachings of a spiritual leader.
   4. A cause, principle, or activity pursued with zeal or conscientious devotion.

So, noting definition #3, a religion is a set of beliefs. And I guess each set of beliefs is it's own religion. Personally, if asked what religion i was, i'd say christian, but we don't believe in religion...our church calls it a lifestyle.[/quote]

As a mathematician, I am dissatisfied with these definitions.  They leave a lot open to interpretation.

I prefer the following:

A set of axioms that are believed to be true.

When I say "true", I don't mean "in general" -- I mean "absolutely and unwaveringly [i]true[/i]".

This seems to me to be a good way to differentiate religion from other belief systems.

[quote="gauss202"]But religion is supposed to be open to interpretation. The former definitions are better, because they reflect the sociological and humanic nature of religions. It's more than just a "set of beliefs". It has history and tradition and social structure all tied in with it.

Here's a cool list I found of the world's major religions by sizes - http://holtz.org/Thoughts/Philosophy/Metaphysics/Theology/Religion.html

The whole book that it comes from is rather of interesting actually - even the parts that I disagree with.[/quote]

Everything has history and tradition and social structure tied to it.  That's implicit in [i]everything[/i] humanic.

I suppose I could have said that religion is "a set of axioms that some human believes are true".

Talking about tradition and history all together with religion makes those things sound definitionally related.  Even athiests and agnostics have history and tradition.

I'm not saying that history and tradition don't [i]shape[/i] religion, but they couldn't logically be part of the meta-structure of the definition.

*sigh* It all comes dwon to this: Religion must make money, if it don't then it will cease to be since you need the money for making pamphlets or publishing books or wahtever. A set of beliefs you don't need to make money since you can believe in whatever you want and still go on with your life. This leads to the fact that religion requires more than one person in it, often a whole organization.

So in two words (plus a math symbol) religion=business

I dunno ... the Hindus, Daoists, and Jews all seem to have gotten along fairly well  have gotten along fairly well without central authority or conversion of other people or terribly much wide-scale money-making for the furtherance of the religion.  (That is, there isn't anything in any of those religions that parallels even remotely the Catholic Church or the evangelical-converter-types of many Protestant relgions

*ahem ahem* I agree for the first two...

I believe what you say is true religion does equal money precisely why I dont go to church half of the time. I think the reason catholics (notice Im not using capatials) dont get along with others well is because that religion is more set forth I guess thats how you put it. It has limits, yet its free. You dont have to hide behind things. Im not saying that other religions arent free or strict Im just saying this from haveing seen protistant as well as nonprotistant. I just would rather believe what I believe and not have to be bothered for why I dont follow a certain religion. I dont like our priest though hes awful. I dont like him at all. Another reason why I dont attend church but about once a month.

I'm a christian. It is a good thing to. because I need Jesus not only to get through life, but also these math questions. Its a good thing to have something to believe in.

i'm jewish but not religious in the least. I don't have the most traditional belief in god or a belief in all the old traditions and happenings. perhaps i'm agnostic, whatever that means.
fighting over religion is ridiculous, people ought let others think what they want.

I'm agnostic.  I couldn't really ever say that I was an atheist because i have never been to church or read the bible, so I haven't researched the topic well enough to form an opinion.  I can't tell.  From observing my religious friends, I enjoy being nonreligious because, I don't have to go to activities for young people, I can sleep in Sunday, and I can make my own standards and limits for myself based on my own experience.  The last one is the one I treasure most.  If I want to fill myself up to the brim with Diet Coke and coffee, or swear-then I will!  I don't like being told what to do if I see nothing wrong.  I'm a freelancer.

Proud to be Mormon and have traditional chinese beliefs. I am one conservative freak!!!!!!!!!!!!!!!

Yeah? Some interesting responses here. . .

I'm a hindu, pretty die-hard, but hinduism is a broad term for a lot of stuff. I have my own beliefs pretty much based off hinduism. Of course there are a lot of misconceptions people have, some of which are QUITE aggravating.

[quote="MCrawford"]
As a mathematician, I am dissatisfied with these definitions.  They leave a lot open to interpretation.

I prefer the following:

A set of axioms that are believed to be true.

When I say "true", I don't mean "in general" -- I mean "absolutely and unwaveringly [i]true[/i]".

This seems to me to be a good way to differentiate religion from other belief systems.[/quote]

In that case, I'm very religious. I unwaveringly, unfalteringly believe in the principles of Riemannian geometry and non-Cantorian set theory (with extra axioms). Oh, and anyone who says Paul Cohen was right is a nonbeliever and must be converted.

:P

And Kurt Godel was the ultimate atheist/agnostic/nonreligious man.


calculus
integration
logarithms
limit
derivative
function
real analysis

$f$:$[0,+\infty[\rightarrow C$ continous
derivable at $0$ such that
$\int_{1}^{+\infty}\frac{f(x)}{x}dx$ converge

$n>1$  integer
$a_1,...,a_n \in ]0;+\infty[$
$q_1,...,q_n  \in R , q_1+...+q_n=0$

Compute $\int_{0}^{+\infty}\frac{q_1f(a_1x)+...+q_nf(a_nx)}{x}dx$
------------------------------------------------------
I was thinking this classical one yesterday while waiting
Valentin and Pierre on the Champs-Elysées

uhm... are the two integrals taken on the right intervals?
not on the same one $[1,+\infty]$?

I read it as you :(
but i think there something wrong in the problem

The integrals are correctely written, and the problem is correct zhaobin

actually nice...
we've managed to solve it, after quite an interesting discussion.
now, let's come to the solution:

first of all, let us re-order $(a_i, q_i)$ as $(A_i, Q_i)$, such that $A_{i+1} < A_i$.
now, notice that \[\displaystyle \int_{\frac{C}{A_k}}^{\frac{D}{A_k}}\frac{f(A_kx)}{x}dx \; \; \; \; \; \; (1)\] is indipendent on $A_k$  (it's just a variable substitution).
now, let me write \[\displaystyle \int_{0}^{+\infty}\sum_{k=1}^{n}\frac{Q_kf(A_kx)}{x}dx = \int_{0}^{\frac{\varepsilon}{A_1}} \sum_{k=1}^{n}\frac{Q_kf(A_kx)}{x}dx + \sum_{j=1}^{n-1}\int_{\frac{\varepsilon}{A_j}}^{\frac{\varepsilon}{A_{j+1}}}\sum_{k=j}^{n}\frac{Q_kf(A_kx)}{x}dx + \sum_{k=1}^{n} \int_{\frac{\varepsilon}{A_k}}^{+\infty} \frac{Q_kf(A_kx)}{x}dx\] for $(1)$ we have: \[ \displaystyle \sum_{k=1}^{n} \int_{\frac{\varepsilon}{A_k}}^{+\infty} \frac{f(A_kx)}{x}dx = 0\] now, since $f$ is derivable in $0$, we can set, for an appropriate $\varepsilon$, $f(A_kx) = f(0) + f'(0)A_kx + o(x)$.
so \[ \displaystyle \int_{0}^{\frac{\varepsilon}{A_1}} \sum_{k=1}^{n}\frac{Q_kf(A_kx)}{x}dx = \int_{0}^{\frac{\varepsilon}{A_1}} \sum_{k=1}^{n}\frac{Q_kf(0)}{x}dx + \int_{0}^{\frac{\varepsilon}{A_1}} \sum_{k=1}^{n}Q_kf'(0)A_kdx + O(\varepsilon) = O(\varepsilon)\] and \[\displaystyle \sum_{j=1}^{n-1}\int_{\frac{\varepsilon}{A_j}}^{\frac{\varepsilon}{A_{j+1}}}\sum_{k=j}^{n}\frac{Q_kf(A_kx)}{x}dx = \sum_{j=1}^{n-1}\left(\int_{\frac{\varepsilon}{A_j}}^{\frac{\varepsilon}{A_{j+1}}}\sum_{k=j}^{n}\frac{Q_kf(0)}{x}dx +\int_{\frac{\varepsilon}{A_j}}^{\frac{\varepsilon}{A_{j+1}}}\sum_{k=j}^{n} Q_kf'(0)A_kdx \right)\]
now \[\displaystyle \lim_{\varepsilon \rightarrow 0^+}{\int_{\frac{\varepsilon}{A_j}}^{\frac{\varepsilon}{A_{j+1}}}\frac{Q_kf(0)}{x}dx = -f(0)\left(\ln {A_{j+1}} - \ln {A_j}\right)}\] and \[\displaystyle \lim_{\varepsilon \rightarrow 0^+}{\int_{\frac{\varepsilon}{A_j}}^{\frac{\varepsilon}{A_{j+1}}}Q_kf'(0)A_kdx = 0}\]
finally, we can write \[\displaystyle \int_{0}^{+\infty}\sum_{k=1}^{n}\frac{Q_kf(A_kx)}{x}dx = -f(0)\sum_{j=1}^{n-1} \sum_{k=j}^{n} Q_k\left(\ln {A_{j+1}} - \ln {A_j}\right) = -f(0)\sum_{i=1}^{n} q_i\ln{a_i}\] which is what we wanted to show... you can get it with some easy telescopic sum and computation.
moubinool, could you please check this solution? do you have any shorter one? any more elegant one?
ps. i hope that some moderator will set someway better this post..
pps. i don't know how long it took to be written!

uhm...
moubinool, are you sure you're going to find anything, looking to that way?
i've discussed i while about this way of solving the problem with another student here, but we didn't find it useful...
i mean, he told me that you can't take the derivative into the integral in such an easy way (not even if $f \in C^1$!)... anyway, you can't even define $f(+\infty)$, not even as $\lim_{x\rightarrow +\infty} f(x)$, because no one says this limit exists!
please, complete your solution with some more details... thanks.
ps. it could really be a nice solution, if just it worked :D

I have a solution only use a $lemma$,maybe similar with $ma_go's$(sorry $ma_go's$ I don't read your solution carefully)
the $lemma$ is: $f$ is the function with the condition give by $Moubinool's$ problem:we have
$\int_0^{+\infty} \frac{f(ax)-f(bx)}{x}dx=lnb-lna$.
then use Abel Vary  :)


modular arithmetic
number theory proposed
number theory

3^(5^2^n  -1 over 2^n+2) =(-5)^(3^2^n  - 1 over 2^n+2) mod 2^n+4
Prove this at n is positive integer


geometry
trapezoid
vector
geometric transformation
homothety

For a trapezoid ABCD,  DC//AB and DC:AB=1:3
AD meets BC at P and  EF passes through P, which PE=PF.
M, N are midpoints of AF, BE respectively.
Prove that MN passes through a fixed point.

When I write AB I 'll mean the vector AB. Let O be the midpoint of MN. We have: 2AO=AM+AN=AF/2+AB/2+AE/2=AB/2+AP(=constant). Thus, O is constant.

 Remark: The relation DC:AB=1:3 is not neccessary.

[i]Let $ABC$ be triangle and let $E$ and $F$ be such
that $C$ is midpoint of variable segment $EF$. If $M$ and $N$
halves $AE$ and $BF$ then line $MN$ passes trough fixed point.[/i]

[u]Proof[/u]: We have following map chain:
\[ M\stackrel{\mathcal{H}_{A,2}}{\longmapsto}E\stackrel{\mathcal{H}_{C,-1}}{\longmapsto}F\stackrel{\mathcal{H}_{B,{1\over 2}}}{\longmapsto}N\]
so composition of homothetys $\mathcal{H}_{B,{1\over 2}} \circ
\mathcal{H}_{C,-1}\circ \mathcal{H}_{A,2}$ which is again some
homothety $\mathcal{H}_{Q,-1}$ maps $M$ to $N$. So $M,N$ and $Q$
are always collinear and since $Q$ is fixed point we are done.

Note: As already pointed out in previous solution, fixed point $Q$ is midpoint of segment $MN$.

The problem diagram is equivalent to a segment $AB$ and a segment $EF$ with midpoint $P$ (points $C$ and $D$ are not relevant whatsoever), with $M$ and $N$ as the midpoints of $AF$ and $BE$, respectively. If we denote the midpoint of $AB$ as $J$, then clearly the midpoint of $MN$ is also the midpoint of $PJ$. However, $P$ and $AB$ (and thus $J$) are both fixed in the problem, and so the midpoint of $MN$ must be fixed as well.


number theory
greatest common divisor
number theory proposed

If $ n$ is odd and $ a_{1} ,a_{2},a_{3}.........a_{n}$ are odd integers prove that the greatest common divisor of $ (a_{1} .a_{2},a_{3}.........a_{n})$ is same as the greatest common divisor of $ \frac {a_{1} \plus{} a_{2}}{2},\frac {a_{2} \plus{} a_{3}}{2}....................\frac {a_{n} \plus{} a_{1}}{2}$

It is enough to prove that if $ 2 \mid n \plus{} 1$ and $ \text{gcd}(a_1,a_2,\ldots,a_n) \equal{} 1$ then $ \text{gcd}(b_1,b_2,\ldots,b_m) \equal{} 1$, where $ b_i: \equal{} 2^{ \minus{} 1}(a_i \plus{} a_{i \plus{} 1})$ for all $ i \equal{} 1,2,\ldots,n$ and $ a_{n \plus{} 1}: \equal{} a_1$. But it is obvios since if $ \text{gcd}(b_1,b_2,\ldots,b_m): \equal{} k \ge 2$ then $ k \mid b_i \minus{} b_{i \plus{} 1} \plus{} b_{i \plus{} 2} \minus{} \ldots \minus{} b_{i \plus{} n \minus{} 1} \plus{} b_{i \plus{} n} \equal{} a_i$, for all $ i\equal{}1,2,\ldots,n$, where indices are taken modulo $ n$, contraddicting the assumption. []


Find $ x\in R$ of this equation: $ \boxed{\left(\frac {1 \plus{} a^2}{2a}\right)^x \minus{} \left(\frac {1 \minus{} a^2}{2a}\right)^x \equal{} 1}$, with $ 0 < a < 1$.

Let $ a\equal{}exp(\minus{}b)$, then $ f(x)\equal{}chb^x\minus{}shb^x\equal{}1$. Obviosly $ x\equal{}2$ is solution. Let $ c\equal{}shb^2>0,y\equal{}x/2$, then $ g(y)\equal{}(1\plus{}c)^y\minus{}c^y\equal{}1$.
Then $ g'(y)\equal{}(1\plus{}c)^yln(1\plus{}c)\minus{}c^ylnc>0$. Therefore $ x\equal{}2$ is unique solution.

[quote="Rust"]Let $ a \equal{} exp( \minus{} b)$, then $ f(x) \equal{} chb^x \minus{} shb^x \equal{} 1$. Obviosly $ x \equal{} 2$ is solution. Let $ c \equal{} shb^2 > 0,y \equal{} x/2$, then $ g(y) \equal{} (1 \plus{} c)^y \minus{} c^y \equal{} 1$.
Then $ g'(y) \equal{} (1 \plus{} c)^yln(1 \plus{} c) \minus{} c^ylnc > 0$. Therefore $ x \equal{} 2$ is unique solution.[/quote]

Yes! $ x\equal{}2$ is root of equation.

We can done it by this way:

Equation $ \Longleftrightarrow \left(\frac {1 \plus{} a^2}{2a}\right)^x \equal{} \left(\frac {1 \minus{} a^2}{2a}\right)^x \plus{} 1$

$ \Longleftrightarrow 1 \equal{} \left(\frac {1 \minus{} a^2}{1\plus{}a^2}\right)^x \plus{} \left(\frac {2a}{1 \plus{} a^2}\right)^x$

Let $ a \equal{} \tan{t}$ we get equation $ 1\equal{}(sin2t)^x\plus{}(cos2t)^x$. Easy $ \Longrightarrow x\equal{}2$ is only root of equation.


Hi, I was wondering if anyone knew of any simple justifications for the following two polynomials rules:

1) Descartes' rule of signs
2) the method of finding the upper and lower bounds of the roots of a polynomials as described on pages 79-80 in AoPS v2

Thanks.

Fierytycoon

Look up the Sturm Theorem and Sturm sequences. 1) can also be proven via induction and the FTA.


Okay, I am starting a Hall of Fame post if it hasn't already been made. Here is how it works. Users pm me the names of people that they think deserve to be in the hall of fame, and I continue tallying up all the votes, and when a person gets 10 votes then they make it to the hall of fame. I already made my own votes for people that I think should make it to the hall of fame. You can vote as many times as you want and you can't vote for yourself. You might want to vote for administrators, people who got really high scores on FTW, people who have been on AoPS for a long time, people who have made a lot of posts, or just people who you really like or think deserve to be in the hall of fame.

I will be keeping the votes secret, until somebody makes the hall of fame. PMing starts now.

This is more sensible than the Most in AoPS topic...

People vote for the MathCounts national competitors! stevenmeow!

Given the general demographic that posts in FF frequently, which I do not believe adequately represents AoPS as a whole (seriously AoPS is not a bunch of middle schoolers making the same topic over and over again...), I don't think this should be called the AoPS Hall of Fame. Try Fun Factory Hall of Fame.

So far there is a person with 2 votes and about 30 other people that each have one vote.


Determine all triples $ (a,b,c)$ in reals for which the inequality

$ |x^3\plus{}ax^2\plus{}bx\plus{}c|\le |x^3|$

holds for all $ x$ in reals.

[hide]
Let's split the problem into cases where (1)$ x\geq0$ and (2)$ x<0$.
The inequality can be expressed as
(1) $ \minus{}x^3 \leq x^3\plus{}ax^2\plus{}bx\plus{}c \leq x^3$ and
(2) $ x^3 \leq x^3\plus{}ax^2\plus{}bx\plus{}c \leq \minus{}x^3$

In both cases, if $ x\equal{}0$, then $ 0\leq c \leq0$.  Thus $ c\equal{}0$.

Substituting $ c\equal{}0$ yields
(1) $ \minus{}x^3 \leq x^3\plus{}ax^2\plus{}bx \leq x^3$
(2) $ x^3 \leq x^3\plus{}ax^2\plus{}bx \leq \minus{}x^3$

Dividing by $ x$ yields the same inequality because of the sign reverse when dividing by a number less than 0.
(1) and (2) $ \minus{}x^2 \leq x^2\plus{}ax\plus{}b \leq x^3$

Now if $ x\equal{}0$, then $ 0\leq b \leq 0$.  Thus $ b\equal{}0$.

Substituting $ c\equal{}0$ yields
(1) and (2) $ \minus{}x^2 \leq x^2\plus{}ax \leq x^2$

Dividing by $ x$ again yields
(1) $ \minus{}x \leq x\plus{}a\leq x$
(2) $ x \leq x\plus{}a\leq \minus{}x$

If $ x\equal{}0$, then $ 0\leq a \leq 0$.  Thus $ a\equal{}0$.

The only triple in reals for $ (a,b,c)$ is $ (0,0,0)$.
[/hide]

[quote="pac-man"][hide]
Now if $ x \equal{} 0$, then $ 0\leq b \leq 0$.  Thus $ b \equal{} 0$.
[/hide][/quote]

Disclaimer: written at 2 A.M.

[hide]This step seems inaccurate. If you assume $ X\equal{}0$ Then you get $ |0\plus{}(0)A\plus{}(0)B\plus{}C|\leq0$ or $ |C|\leq0$ so it doesn't matter what B is equal to when $ X\equal{}0$

|X^3+AX^2+BX|<|X^3| X does not equal 0

if X^3+AX^2+BX<X^3 then X>0
AX^2+BX<0
X(AX+B)<0
X<0 or AX+B<0
AX<-B as X approaches 0 AX approaches 0, and therefore -b is positive or zero and b is negative or zero.

as X approaches infinity AX will approach infinity if A is positive

A is therefore negative or 0

X^3+AX^2+BX>X^3 then X<0
AX^2+BX>0
X(AX+B)>0
X>0 and AX+B>0 or X<0 and AX+B<0
AX+B<0

see above

I am too tired to finish this but if you test for when X^3+AX^2+X<0 and X>0 I think it ends up showing that 0,0,0 is the only correct answer. Essentially, I agree with your answer but not your methodology.
[/hide]

Yes, the matter is how to treat the case $ x \equal{} 0$ ,but the limitation theory that

''If $ f(x),g(x)$ is defined in the neighborhood of $ \alpha$,and constantly $ f(x)\le g(x)$,
we have $ \[ \lim_{x \to \alpha}f(x) \le \[ \lim_{x \to \alpha}g(x)$ (Suppose they have limitations)''

helps us.

[hide]

Now we easily get $ c \equal{} 0$ ,and $ |x^2 \plus{} ax \plus{} b|\le |x^2|$ (for all $ x\neq 0$).

By the theory,we have $ \[ \lim_{x \to 0}|x^2 \plus{} ax \plus{} b| \le \[ \lim_{x \to 0}|x^2| \Longrightarrow |b| \le 0 \Longrightarrow b \equal{} 0$

and $ |x \plus{} a|\le |x|$ (for all $ x\neq 0$)

By the same way we have $ a \equal{} 0$ and it turned out that $ a \equal{} b \equal{} c \equal{} 0$ is the only solution.

[/hide]

[b]Fact:[/b] $ |ab| \equal{} |a| \cdot |b|$.

[hide="Solution"]Testing at the origin, we see that $ c\equal{}0$.

We want to find $ (a,b)$ such that $ |x^3\plus{}ax^2\plus{}bx| \leq |x^3|$

Factor as $ |x| \cdot |x^2\plus{}ax\plus{}b| \leq |x| \cdot |x^2|$. This requires that $ |x^2\plus{}ax\plus{}b| \leq |x^2|$ for all $ x \neq 0$.

It is clear that for $ x$ approaching $ 0$, the LHS approaches $ |b|$ and the RHS approaches $ 0$. So unless $ b\equal{}0$, for $ x$ sufficiently close to $ 0$, the inequality will fail. So, set $ b\equal{}0$.

Now, we want to find $ a$ such that $ |x^3\plus{}ax^2| \leq |x^3|$

Factor as $ |x^2| \cdot |x\plus{}a| \leq |x^2| \cdot |x|$. This requires that $ |x\plus{}a| \leq |x|$ for all $ x \neq 0$.

Again, it is clear that for $ x$ approaching $ 0$, the LHS approaches $ |a|$ and the RHS approaches $ 0$, so we must have $ a\equal{}0$.

Therefore, $ (a,b,c) \equal{} (0,0,0)$.[/hide]


although i'm taking a number theory course, but i'm not sure what does the following question do with number theory. Please take a look,

[color=olive]A piggy bank contains 24 coins, all of which are nickels, dimes, or quarters. If the total value of the coins is two dollars, what combinations of coins are possible[/color]?

Sounds like it's a stat question, thanks for helping.

Why do you post all these problems in the algebra forum?!

@mods: please move them

[mod.: moved]


function
linear algebra
matrix
algebra
polynomial
rational function
superior algebra

Let $ k$ be a field, and let$ K\equal{}k(x)$ be the rational function field in one variable over $ k$. If $ u\in K$, show that $ K\equal{}k(u)$ iff $ u\equal{}(ax\plus{}b)/(cx\plus{}d)$ for some $ a,b,c,d\in k$ with $ \det\left(
\begin{matrix}
a&b\\
c&d
\end{matrix}\right)\not \equal{} 0.$

if $ u \in K \backslash k$, say $ u \equal{} f/g$ with coprime $ f,g \in k[x]$, then $ K/k(u)$ is finite of degree $ max(deg(f),deg(g))$. this can be checked with the minimal polynomial $ g(t)*u \minus{} f(t) \in k(u)[t]$. from this, the claim easily follows; also $ Aut_k(k(x)) \cong PGL_2(k)$.


Suppose there are $n$ points in a plane no three of which are collinear with the property that if we label these points as $A_1,A_2,\ldots,A_n$ in any way whatsoever, the broken line $A_1A_2\ldots A_n$ does not intersect itself. Find the maximum value of $n$.

[i]Dinu Serbanescu, Romania[/i]

The first four points must form a concave quadrilateral (otherwise the diagonals intersect), and considering the possible places to put the fifth point, the broken line can always intersect itself. So $n=4$

Consider the figeure fromed by the n points as a graph and the broken line as a tree in that graph. I will prove that the shortest tree in that graph has the requested property:

If two edges of the tree: $AB$ intersects $CD$ then the line obtained by eliminating $AB$ and $CD$ from the tree and inserting edges $AC$ and $BD$ will be shorter. (by triangle ineq) so we are done.

It is easy to see that if $4$ of the points are the vertices of a convex quadrilateral then the desired property is not satisfied.
Moreover, it is well-known that among any set of $5$ points in the plane, no three being colinear, we may find $4$ which are the vertices of a convex quadrilateral (this is the famous Esther Klein-Szekeres' result).
Thus the maximum number of points is at most $4$.
In another hand, considering the three vertices of an equilateral triangle and its center, wa have a 'good' configuration with $4$ points.

Pierre.

First we can say that,

if we choose 5 points in plane with no three are collinear,
there is a such 4 points this points are vertices of a convex quadrilateral.

and in quadrangle diognals are intersect

so that n=4 is maximal. If we choose 4 points such that this points are vertices of a concave quadrilateral, this is ok.