let G be a finite group with card(G)=n. and let H a subgroup of G such that card (H)=p. p is the greatest prime divisor for n. prove or disprove that p<=card (Z(G)). Z(G) is the center for G.

uhm..
i suppose you mean to ask for $H$ to be a normal subgroup of $G$ (it'd be an empty hypothesis, by cauchy), otherwise any group with trivial center is working: $S_{p}$ has a subgroup of order $p$ (which is the biggest prime divisor of $p! = |S_{p}|$), and $Z(S_{p})$ is trivial.

anyway, even if you $H$ is normal, for $p>2$, you can consider the dihedral group $G = D_{p}= \langle r,s | rsrs , r^{p}\rangle$, and $H=\langle r \rangle \lhd G$. $Z(G)$ is trivial, so $1<p$.

if $p=2$, then $G$ is a $2-group$, then it has non-trivial center (by the class formula), and so $|Z(G)|>1$, which implies $|Z(G)|\ge 2$.


trigonometry
function
algebra proposed
algebra

I have a idea for part ii
Let $f(x)=8.27^{x}-38.18^{x}+57.12^{x}-27$. Note that $f(x)>0\forall x\in\mathbb{R}$ so $f(x)$ is a increasing function on $\mathbb{R}$. Therefore the equation $f(x)=0$ has only a root. Obviously $x=0$ is an unique root of equation

i) 2^(sin^2 (x)) =.....<=0.  so x<= -1. if x<-1 then 3^(x+1) - 1 is negative. impossible because 2^(sin^2 (x)) is greater or equal then 0. x= -1 doesn`t chek the equation so there are no solutions in R


There is a rectangular box with a length of 6.212 units, a width of 5.201 units,
and a height of 4.179 units.  Just by looking at these three dimensions,
can you pick which following lettered response gives the closest answer for the number of cubic units for the box's volume?

A)  120

B)  130

C)  140

I can tell that it's too close to tell without multiplying out.  [hide]We have something that's about $ 6 \times 5 \times 4$, and then we increase each dimension by something in the neighborhood of $ 4 \%$, so the answer is in the neighborhood of $ 120 \cdot (1.12) \equal{} 134$.  That's too close too call.  In fact, the product is $ 135.017\ldots$.[/hide]  It's not at all clear to me what the appeal of such a question is.

jmerry,

hypocrite,

turn in ALL of your stars here, all of your alleged educational degrees,
AND STOP with the giving out namecalling on these boards with what you
"think" someone is.

You and your alleged subjective garbage are irrelevant.

Stop changing the subject; either just contribute to the thread, OR simply do NOT enter the thread.

Hurry up and understand.

"It's not at all clear to me..."

What, with your continual arrogant, stuck-up, know-it-all attitude?  Hmm?
You are your own undoing!
The humble intelligent learner (and simultaneous teacher) is OPEN.

You, on the other hand, are biased, a closed box.
Also, your comments are INSINCERE, and you are of sour grapes.

Again, try catching up and to understand.
You are all about namecalling.  I already know the source.

"Hurry up and understand." [i]-- previous post[/i]

Why don't [i]you[/i] hurry up and understand there's really no point in wasting others' time?

This is new... Posting in the same thread on different accounts? (Refer to some of the links jmerry provided.)

Fundamental principle of online communications: [i]don't feed the trolls[/i].  If bothering other people was problematic to it, it wouldn't behave the way it does.  Instead, it takes pleasure (for whatever bizarre reason) in being troublesome.  Telling it that it is being troublesome is positive reinforcement: it will enourage it to act in the same way.  The best way to make it stop being bothersome is to ignore it.  I suggest that you simply avoid responding to it on any thread, no matter what.

[quote="jmerry"]Well- this is another [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167086]name[/url] for the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164380]same[/url]  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=71819]old[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59286]troll[/url].

He hasn't behaved too badly lately, but I've never liked his taste in problems.[/quote]
I don't get it. What's going on? Whos the troll?

wow this is kinda sad
look at all these "birch and swinnerton dyer" people talking away
and there have only been two pertinent responses to the question
what a troll?

okay: it looks like
[hide]
130 to me, because it goes up .2, up .2, and then down .2 from 120. 120 to 140 is up 16%, which seems kinda high.
120 is low, so it must be 130.
[/hide][/hide]

bowei, ProtestanT:

A [b]troll[/b] is defined as a person on an online interactive resource, such as a forum, that posts something argumentative or otherwise to purposefully stir up trouble or insult a person.

I think jmerry was saying the starter of this thread was in fact one of the several accounts of a troll.

This thread should be locked.

[quote="bowei"][quote="Temperal"]
This thread should be locked.
[/quote]
agreed. we are getting nowhere.[/quote]

this thread really isn't in need of [i]more[/i] backseat moderation, you know.


The line $ax+by+c=0$  is tangent to a circle with equation $x^{2}+y^{2}=\frac{1}{5}$. If $a,b,c,$ are the sides of a $\triangle ABC$ with sides  [usual notation] and centroid the origin,with the circle given as the circumcircle, find the equation of the median through  if the median through  is $x+3y=0$


trigonometry
inequalities
inequalities solved

Prove that for x in [0,pi/3) we have:
1/(sin(x+pi/3))+1/(sin(x-pi/3))>=4sqrt(3)/3

take x = 0. and your ineq doesn't hold...

but I think it can be handled like this:
1/(sin(x+\pi/3))+1/(sin(x-\pi/3))
 = (sin(x+\pi/3)+sin(x-\pi/3)) / (sin(x+\pi/3)sin(x-\pi/3))
 = 2 sin x cos \pi/3 / ((sin x)^2/4 - 3(cos x)^2/4)
 = sin x / ((sin x)^2 - 3/4)
 = 1/(sin x - 3/(4sinx))

u - 3/(4u) increases on [0,\infty) and sin x increases on ??

so 1/(sin x-3/(4sin x)) decreases on ??. thus, we get the minimum value of it.

the correct ineq is
1/(sin(x+pi/3))+1/(sin(pi/3-x))>=4/\sqrt 3

Yes, the corect inequality is:
1/(sin(x+pi/3))+1/(sin(-x+pi/3))>=4sqrt(3)/3

liyi you are right! sorry 4 my mistake! :(

cheers! :D:D

well, i give the proof here. just using the method above

1/(sin(x+\pi/3))+1/(sin(\pi/3-x))
 = (sin(x+\pi/3)+sin(\pi/3-x)) / (sin(x+\pi/3)sin(\pi/3-x))
 = 2 cos x sin \pi/3 / (3(cos x)^2/4 - (sin x)^2/4)
 = \sqrt 3 cos x / ((cos x)^2 - 1/4)
 = 4\sqrt 3/(4cos x - 1/cos x)

4u - 1/u increases on [0,\infty) and cos x increases on [0,\pi/3) so
so 1/(4cos x-1/cos x) increases on [0,\pi/3). thus,
4\sqrt 3/(4cos x - 1/cos x) \geq (take x=0) 4\sqrt 3/3


geometry
geometric transformation
vector
analytic geometry
geometry solved

A polygon in the xy-plane has area greater than n. Prove that it contain points (x_i,y_i), i \leq i \leq n + 1 such that x_i - x_j and y_i - y_j are integers for all i and j.

Let P be the given polygon.
Divides the plane into unit squares, with sides parallel to the axis of coordinates.
Let C_0 be the unit square with opposite vertices O(0,0) and A(1,1).
Let C be any of the unit squares, and let M(a,b) be the down-left vertex of C.
We use the translation t with vector u = MO, to superpose C onto C_0.
We perform this operation for each unit square (which is to consider the torus). Note that u has integer coordinates.

Let P' be the union of the sets t(C \cap P) where t ranges over easch of the translations above. The sets C \cap P have at most a segment in common, the sum of the area of the sets t(C \cap P) which is equal to the sum of the area of the sets C \cap P which in turn is equal to the area of P. Since C_0 has area 1, the pigeon-hole principle ensures us that there is some point, say X, in C_0 is covered at least n+1 times by the sets C \cap S when C ranges over all the unit squares.
It means that X is the image of at least n+1 distinct translations, and therefore is the image of at least n+1 distinct points from P by these translations, say X_1,...,X_(n+1).
Since each of the translations has vector with integer coordinates, it follows that each of the vectors X_iX_j (which are the difference of to of the vector of the translations) has integer coordinates, and we are done.

Pierre.


Let ABCD be a convex quadrilateral and O be the intersection of AC and BD. Let M, N be points on AB such that AM = MN = NB, and let P, Q be points on CD so that CP = PQ = QD. Show that triangles MOP and NOQ have same area.

I believe this problem was posted once before, but only one very complicated solution was posted (with 20 different extra points and some obscure (to me) theorems), and I'm hoping for a better solution.

S(EON)=S(FOM)=4/9S(ABD)
S(EOQ)=S(FOP)=4/9S(ACD)
S(FPSM)=S(EQRN)=4/9S(ABCD)
S(NOQ)=1/2S(EQRN)- S(EON)- S(EOQ)
S((MOP)=1/2 S(FPSM)- S(FOM)- S(FOP)
DONE!

Here is a solution with vectors:

Let $ O$ be the origin.  We see that $ \vec{M} \equal{} \frac{2 \vec{A} \plus{} \vec{B}}{3}$, $ \vec{N} \equal{} \frac{\vec{A} \plus{} 2 \vec{B}}{3}$, $ \vec{P} \equal{} \frac{2 \vec{C} \plus{} \vec{D}}{3}$, and $ \vec{Q} \equal{} \frac{\vec{C} \plus{} 2 \vec{D}}{3}$.  Then the vector corresponding to the area of triangle $ MOP$ is
\[ \frac{1}{2} \vec{M} \times \vec{P} \equal{} \frac{1}{18} (2 \vec{A} \plus{} \vec{B}) \times (2 \vec{C} \plus{} \vec{D}) \equal{} \frac{1}{9} (\vec{A} \times \vec{D} \plus{} \vec{B} \times \vec{C}).\]
Similarly, the vector corresponding to the area of triangle $ NOQ$ is
\[ \frac{1}{2} \vec{N} \times \vec{Q} \equal{} \frac{1}{18} (\vec{A} \plus{} 2 \vec{B}) \times (\vec{C} \plus{} 2 \vec{D}) \equal{} \frac{1}{9} (\vec{A} \times \vec{D} \plus{} \vec{B} \times \vec{C}).\]
Hence, $ [MOP] \equal{} [NOQ]$.

Wow, I didn't realize chosing a convenient origin would be so helpful =D. Thanks for both great solutions!

It seems the equality holds for M,N and P, Q just symmetrical about the midpoints of their respective sides, AB and CD and AM/AB = DQ/DC.

Best regards,
sunken rock


function
number theory unsolved
number theory

prove that for all integer x>2, $ 8 / [(\sqrt [3]{x}\plus{}\sqrt [3]{2\plus{}x})^{3}]\plus{}1$.
as [\cdot]:the integer part .
goog luck :)

i am really sorry, $ x$ is an integer :blush:


abstract algebra
LaTeX
group theory
superior algebra
superior algebra unsolved

Is the following a group? an Abelian group? Justify.

$ ({1,\xi,\xi^2,\xi^3,...\xi^{n \minus{} 1}},\cdot)$ where $ \cdot$ is ordinary multiplication and $ \xi$ is the complex $ n$th root of 1

Which group axiom are you having trouble with? You should be able to check them all by hand. Also, the group is generated by $ \xi$, so it is cyclic (and hence abelian).

I think I'm just not familiar with complex roots but could you show me existence of an inverse?

$ \xi^n \equal{} 1$ by definition.  Therefore ....  (You have to already know what the unit is at this point, but I think this shouldn't be too tricky.)

If you want curly brackets in LaTeX, you need to use \{ and \}.

write each root of unity in the form $ e^{i \theta}$ and then you see that the arguments add when you multiply roots.


hello to everyone,
i was lately thinking about this one: i find it obvious that one can place at most countably many pairwise disjoint open balls in the usual euclidian plane for one can choose a rational point (p, q) in every ball and thus the set of all balls is bijective to some subset of QxQ which is obviously countable. But what about placing 'T' letters, where
1) the horizontal and vertical lines are of equal size
2) the horizontal and vertical lines are of arbitrary size
What can be said about the largest cardinality in both cases ? I suppose in case one again aleph_0 because some argument floats in my mind like this: u can try to associate a circle with each letter of the same size such that two T letters intersect each other iff their circles have common points. But for case 2) i need some help.

i'm sorry, case 1) needs to be more precisely formulated: each letter T has the same size of its horizontal and vertical segment, and that size is constant for ALL letters.

It has been shown here on the forum that at most countably many non-intersecting [b]topological[/b] $T$'s can be placed on the plane. Clearly, this implies the fact that at most countably many $T$'s as you describe them can be placed on the plane.

For your question, simply choose, for each $T$, two rational points, one on each side of the leg of the $T$, and close to the point where the two segments of the $T$ are joined. I think you understand what I mean.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=topological&t=6468]Here[/url]'s a link that might interest you.

thanks for the comments :) are there any alphabet memers uncountably many of which could be non-intersectingly mapped to the plane? Does someone know how the situation described here is about Z's for instance?


Find all positive integer solutions to
a^2 + b^2 + c^2 = 2d^2
where d < 50

similarly, enumerate. However, one small note, we only need to check odd ds. Even ds, require all even a,b,c's, so we can divide throughout by 4. Thus, the only primitive ds are the odd ones.


geometry
circumcircle
incenter
trigonometry
parallelogram
trapezoid
geometry proposed

Let $ {O,I }$ be the circumcenter and the incenter of $ {\triangle ABC}$. Let $ {\Gamma}$ be the circumcircle of $ {\triangle ABC}$ , $ {M,N }$ is the midpoint of arc $ {AC, AB}$ (Doesn't contain point B and C) $ {AP // MN}$ .$ {P}$ is on the circumcircle .let $ {PI \cap \odot O \equal{} T}.$ Then

(1) Prove that : $ {PM*MT\equal{}PN*NT}$;
(2) Take point $ {Q \in arc BTC}$  let $ {I_1,I_2}$ be the incenters of $ {\triangle ABQ }$ and $ {\triangle ACQ}$ , prove that :points $ {I_1,I_2,Q,T}$ are concyclic.

a, It's easy to prove that $ P$ is the midpoint of arc $ BC$ which contains $ A$.
$ \Rightarrow T$ is the tangency of A-Mixtilinear incircle of triangle $ ABC$ with $ (O)$.
Denote $ K\equal{}TN\cap AB, L\equal{}TM\cap AC \Rightarrow I$ is the midpoint of $ KL$.
But $ KL\perp AI, MN\perp AI \Rightarrow KL//MN \Rightarrow TI$ is the median of triangle $ MNT$.
By the law of sine we get $ \frac{\sin \angle NMP}{\sin \angle MNP}\equal{}\frac{\sin \angle NTP}{\sin \angle MTP}\equal{}\frac{\sin \angle MNT}{\sin \angle NMT}$
$ \Rightarrow \frac{NP}{MP}\equal{}\frac{MT}{NT} \Rightarrow NT.NP\equal{}MT.MP$
b, We have $ N,I_1, Q$ are collinear, $ M,I_2,Q$ are collinear.
$ \frac{NT}{MT}\equal{}\frac{PM}{PN}\equal{}\frac{AN}{AM}\equal{}\frac{NI_1}{MI_2}$
$ \Rightarrow \Delta TNI_1\sim \Delta TMI_2 \Rightarrow \angle I_1TI_2\equal{}\angle NTM\equal{}\angle I_1QI_2 \Rightarrow QED$

Let $ L$ be the midpoint of the arc $ BAC.$ It is well-known that $ I$ is the orthocenter of $ \triangle MNL,$ hence antipode $ P$ of $ L$ WRT the circumcircle forms the parallelogram $ PMIN$ $ \Longrightarrow$ $TP$ is the T-median of $ \triangle TMN.$ From the isosceles trapezoid $ PAMN,$ we get $ \angle MTA \equal{} \angle NTP,$ which implies that $ TA$ is the T-symmedian of $ \triangle TMN.$ Then its T-Apollonius circle cuts its circumcircle $ (O)$ at $ A$ $ \Longrightarrow$ $ \frac {_{TN}}{^{TM}} \equal{} \frac {_{AN}}{^{AM}} \equal{} \frac {_{PM}}{^{PN}}.$

Circles centered at the midpoints $ N,M$ of the arcs $ AB,AC$ with radii $ NB$ and $ MC$ contain the incenters $ I_1,I_2$ of $ \triangle ABQ$ and $ \triangle AQC$ $ \Longrightarrow$ $NI_1 \equal{} AN$ and $ MI_2 \equal{} AM.$ Thus, in the latter proportion we have $ \frac {_{TN}}{^{TM}} \equal{} \frac {_{NI_1}}{^{MI_2}}.$ Since $ \angle I_2MT \equal{} \angle I_1NT,$ we deduce that $ \triangle I_1NT \sim \triangle I_2MT$ $ \Longrightarrow$ $ \angle QI_1T \equal{} \angle QI_2T$ $ \Longrightarrow$ $TQI_1I_2$ is cyclic.

See here: http://forum.mathscope.org/showthread.php?t=8757
Or here for more solutions: http://forum.mathscope.org/showthread.php?t=9366.

I think that this is a problem which belongs to Iran MO 1997,so check it out here also:
http://www.mathlinks.ro/viewtopic.php?p=349634#349634

Here is my solution for this problem:[Sorry for my bad English :blush: ]
1.Easy to get that $ {B,I,M}$ and $ {C,I,N}$ are collinear , With $ {I}$ the incentre of $ {\triangle ABC}$we easily get $ {IM \equal{} MA}$ and $ {IN \equal{} NA}$ With equal arcs we get that $ {APNM}$ is an isosceles trapzoid , then $ {PM \equal{} AN \equal{} IN,PN \equal{} AM \equal{} MI}$ $ {\rightarrow}$  $ {MINP}$ is a parallelogram , let $ {NM \cap IP \equal{} K}$ we get $ {NK \equal{} KM}$ with $ {\frac {PM}{NT} \equal{} \frac {PK}{NK} \equal{} \frac {PK}{KM} \equal{} \frac {PN}{MT}}$ $ {\Rightarrow}$ $ {\boxed {PM*MT \equal{} PN*NT}}$.
2.Only need to prove that :$ {\angle I1I2T \equal{} \angle TMN}$ We only need to prove that $ {\triangle I1I2T}$ ∽$ {\triangle NMT}$ $ {\Leftrightarrow}$ $ {\triangle NI1T}$∽$ {\triangle MI2T}$ we have $ {\angle QNT \equal{} \angle QMT}$ , We only need to prove :$ {\frac {NI1}{NT} \equal{} \frac {MI2}{MT}}$ ........@. With $ {NPMB }$ is an isosceles trapzoid  (Easy to get) and $ {NI1 \equal{} NB}$(the property of incentre) we finally get that $ {NI1 \equal{} PM }$ , similary , We get $ {MI2 \equal{} PN}$ so with the conclusion of (1) we get @ , So we can get $ {\angle I1QT \plus{} \angle TI2I1 \equal{} \pi }$ $ {\rightarrow}$ $ {I[1],I[2],Q,T}$ are concyclic. QED.
[hide]This problem is actually from national maths competition for seniors in China in 2009 (held in October) :) [/hide]

Hello!

The point (2) of the proposed problem can be generalized:
Let $ I$ be the incenter of $ \Delta ABC$, $ \Omega$ is a circle through the points $ B$, $ C$ and $ \left\{X,B\right\} \equal{} AB\cap\Omega$, $ \left\{Y,C\right\} \equal{} AC\cap\Omega$.
$ A'$, $ M$, $ N$ are the midpoints of the arcs $ \widehat{XY}$, $ \widehat{CY}$, $ \widehat{BX}$ ($ A'\notin\widehat{XBY}$, $ M\notin\widehat{CBY}$, $ N\notin\widehat{BCX}$), $ P\in\Omega$ such that
$ A'P\parallel{}MN$ and $ \left\{T,P\right\} \equal{} PI\cap\Omega$. If $ Q\in\widehat{BTC}$ and $ I_{1}$, $ I_{2}$ are the incenters of $ \Delta QXB$, $ \Delta QYC$ then prove that
the points $ I_{1}$, $ I_{2}$, $ Q$, $ T$ are concyclic.

Solution:
It's easy to prove that $ P$ is the midpoint of the arc $ \widehat{BA'C}$.
Let $ \Gamma$ be the circle that tangent to the lines $ AB$, $ AC$ and to the circle $ \Omega$ (the circles $ \Gamma$ and $ \Omega$ are internally tangent)
and $ \left\{T'\right\} \equal{} \Gamma\cap\Omega$ (the points $ T'$ and $ A$ are on different sides of $ BC$).
From here [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=284806[/url] $ \Rightarrow T' \equal{} T$.
The message #4 from here [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56605[/url] $ \Rightarrow$ the points $ I_{1}$, $ I_{2}$, $ Q$, $ T$ are concyclic.

Best regards, Petrisor Neagoe :)


a,b,c positive reals      and a+b+c=1
determine
$  infimum ( \frac{a^2}{a+bc}+\frac{b^2}{b+ac}+\frac{c^2}{c+ab})$

I think the solution is 3/2, but i need some prove....

thanx

[quote="pbornsztein"]Am I wrong, or for $a \rightarrow 0$ so that $b+c \rightarrow 1$ the desired expression tends to $1$?

Pierre.[/quote]
i posted again!
how did you conclude that teh desired expression tends to 1?

for $a=b=c=\frac1 3$
 $\frac{a^2}{a+bc}+\frac{b^2}{b+ac}+\frac{c^2}{c+ab}=\frac{3}{4}$
so I think the solution will be $\frac{3}{4}$
here is the proof:
$(a+b+c+ab+bc+ca)( \frac{a^2}{a+bc}+\frac{b^2}{b+ac}+\frac{c^2}{c+ab}) \geq(a+b+c)^2$
then $\frac{a^2}{a+bc}+\frac{b^2}{b+ac}+\frac{c^2}{c+ab} \geq \frac{(a+b+c)^2}{((a+b+c)^2+ab+bc+ca)}\geq \frac{3}{4}$

[quote="darkmaster"]I don't know why Various type "infimum" when this expression has a minimum.[/quote]
i know the different between minimum and infimum.
proving that this expression has an infimum is little more difficult than finding the minimum..
i


[hide="Solution"] Let $a = \frac{x+3}{x-1}, b = \frac{x-1}{3x+2}$.  Rewrite as

$4a^{2}+7ab-15b^{2}= 0 \Leftrightarrow (a+3b)(4a-5b) = 0$

And our solutions are

$a =-3b \Leftrightarrow (x+3)(3x+2) =-3(x-1)^{2}$.  After simplification, this gives $6x^{2}+5x+9 = 0 \Leftrightarrow x = \boxed{ \frac{-5 \pm \sqrt{191}\imath}{12}}$

And

$4a =5b \Leftrightarrow 4(x+3)(3x+2) = 5(x-1)^{2}$.  After simplification, this gives $7x^{2}+54x+19 = 0 \Leftrightarrow x = \boxed{ \frac{-27 \pm 2 \sqrt{149}}{7}}$ [/hide]

I was wondering why I wasn't getting the fourth real solution  :P


A pair of contact lenses costs 300.00. The warranty with the lenses states that if the lenses are damaged or lost within 12 months, they will be replaced at a cost proportional to the part of the 12 months they were used. If the lenses are scratched after 8 months, how much should it cost to replace them?

A. 67.00

B. 75.00

C. 100.00

D. 200.00

E. 225.00

8 months/12 months is 2/3.

The cost is 2/3 of $ \$$300.00 = $ \$$200.00
The answer is $ \boxed{D}$


What product will be obtained when 1-methylcyclopentene is treated with selenium dioxide?

It is allylic oxidation.
We will get 2-methylcyclopent-3-en-1-ol

Then it might be other allylic position
3-methyl cyclopentan-2-en-1-ol

Still wrong. It seems that your problem is on nomenclature.

My friend explained to me about this reaction and its mechanism long ago so that's why remember this. :)  :)

Homework:

1) What is the mechanism of this reaction?

2) What are the empirical rules that allow us to predict the products from it?

I think I remember the rules:
[hide="rules"]
1.The order of reactivity of different allylic positions is:
-CH2- > -CH3 > -CH
2.It occurs more easily at the substituted end.(Your example)
[/hide]


function
geometry
3D geometry
sphere
search
geometric transformation
reflection

I'm interested in the original solution (or any other solution) of:

Let S be the set of inner points of a sphere abd C be the set of the inner points of a circle. Find (with proof) if there exists or not a function $f:S->C$ (and some fixed real $ r>0$ ) such that $ |AB| <= r|f(A)f(B)| $ for any points $ A,B $ in $S$, where |XY| denotes here the Euclidian distance between the points X and Y.

Note. I tried many times to use Search button to see if this topic or some other topics already exists, but it seems that it doesn't work properly!

This is my solution.
Aswer: no.
"Divide" inner space of sphere into cubes with side $2x$ (put as many as possible). For $x$ very small their number will be close to $\frac k{x^3}$ with $k$ - constant. Consider the centers of these cubes. Distance between any 2 of them is at least $2x$. Thus distance between any 2 of their reflections is at least $2rx$. Consider circles with centers in these points (reflections) whith radius $x$. They don't intersect, their total area is $\pi nx^2$ ($n$ - number of points), which for $x\rightarrow0$ is close to $\frac{\pi k}x\rightarrow\infty$. It follows that area of $C$ is infinite. Contradiction.

My proof is "quite similar".
Does anyone know another solution?


Is there a way to find how many lattice points are enclosed in a given circle?

I would think that you would find the area of the circle, and then divide by something for the number of actual lattice points, and find the floor of that value.

i think you could cut the circle up in four quarter circles, find hte number of LP in one quarter, then multiply by 4.
----------------
maybe you could express the circle w/ center (a,b) and radius r as
(x-a)^2 + (y-b)^2 = r^2 ??
idk how thats suppposed to work.... just an idea....

Sadly, that does not work.  There are some on the borders also.

As for the original quesiton, you just have to count.

Well, in most cases you should be able to translate it (the center) to the origin, unless it's weird.
Then, I would probably find the number of latice points in the first quadrant, which should be easy by solving random inequalities, except for the ones on the x and y axes. Then you could multiply by four and add the ones on the axes. Yay.

I have had this type of question before in mathcounts team. First, you actually divide it into quarters... Then you start counting!!!  :blush:  I'm not really sure about the really big circles, but if it is a 10*10 lattice point or something, I'm sure you'll come around... (If there was an equation for this, it'd be VERY COMPLICATED)


Who do you think will win?

Before the start of the tournament, I would have never expected Nadal to make it to the final. But they have both been playing very well, from what I've seen.

Top-ranked Roger Federer has won the last three Wimbledons, and Number-2 Rafael Nadal was defeated early in the tournament last year in the 2nd Round. Nadal's best performance at Wimbledon prior to this year was reaching the 3rd Round.  The two have met four times this year (once on hard court and three times on clay), Nadal winning all.

Well, actually, Nadal didn't reach the final. He has to defeat Bagdathis, not an easy oponent. However, in my opinion, Federer will win by far. He's the Kind of Wimbledon

Nadal beat Baghdatis easily. but i still think federer will win.

ya he made it to the final like an hour or two before i made this topic. i watched the match on the TV live. sorry, i forgot that some people may not have been able to watch it live.

Federer because I don't follow tennis and I've never heard of Nadal.

I've seen Nadal play and all, I really like him, but I have to say Federer here.  He's a veteran, and if you look at the scoreboard, he "dominated" his opponents.  Nadal/Federer = 1/3

Yay, Federer wins...but drops a set. (I don't like Nadal much)

Yay, Federer. I forgot to watch the match this morning.  :( Federer is the better overall player, though he doesn't have a very good record against Nadal. 2-6 now, I believe.

I was hoping Nadal would be beaten by that guy in the second or third round, Hendrick, who had a 2-set lead.

[quote="shadysaysurspammed"]Yeah thnx a lot roger,tennis is so unpredicktable now.[/quote]

Well, true, but he's the best at grass, nobody can defeat him, that's the true thing

[quote="shadysaysurspammed"]Yeah thnx a lot roger,tennis is so unpredicktable now.[/quote]Nadal has gotten a lot better very quickly, and he's still young.  Nadal also owns Federer on clay, and Federer only has a 1-6 record vs. Nadal this year.  (I think)

The only thing you can really predict is that one of the two will win.

Nadal has gotten better VERY fast.  He's only 20 and can beat the best in the world (but not on grass ^_^)  I think that the rivalry between the two will grow now that federer actually won a match against him.

But we can't deny that Federer is far better. I mean, he has almost 7000 points in the ATP Ranking, while Nadal is near 4500. I mean, Nalbandian is nearer to Nadal that Nadal from Federer.

Did someone close the poll?  I can't vote, and I'm SURE I have not voted yet...BTW, I was voting for Nadal just because he has no votes.  :D

Wimbledon is over, so it would make sense that you can't vote on a prediction anymore.

[quote="lizcheang"]Nadal should just go back to soccer. Haha jk[/quote]

Or maybe he may just continue winning tournaments

[quote="junggi"][quote="blahblahblah"][quote="shadysaysurspammed"]Yeah thnx a lot roger,tennis is so unpredicktable now.[/quote]

No.  Nadal wins on clay, Federer everywhere else.[/quote]

I thought Nadal beat Federer on Hard court too  :maybe:[/quote]

Well, I'll have to disagree. Federer is much stonger than Nadal in hard courts. And, maybe, Lubicic or Nalbandian bet Nadal in hard court too, and Nalbandian too in grass


There is a 25x25 empty grid. Every day, everyone alive must make a valid knight move (post), depositing a mine in their previous position. You die if you land on a mine.

When you die, all mines you have laid down are safely diffused.

Winner is the person who kills the most number of people; signups begin now. I take 8. Please specify starting position.

[b]Signups:[/b]
(start position)
M3 -
E5 -
U5 -
C13 -
W13 -
E21 -
U21 -
M23 -

Wait, I don't get this.

Can you show me the grid?

Sign up W13

[hide="here"]

Signups:
(start position)
M3 - King1
E5 -
U5 - chompy
C13 -ernie
W13 - AOD
E21 -
U21 -
M23 - phirecalc
[/hide]

What is there to show? You have a 25x25 grid. Label one axis A...Y, and the other axis 1...25.

Signups:
(start position)
M3 - KING1
E5 -
U5 - Chompy
C13 - ernie
W13 - AceOfDiamonds
E21 -
U21 -
M23 - PhireKaLk6781

3 spots left

I wanna spot E21

Signups:
(start position)
M3 - KING1
E5 -
U5 - Chompy
C13 - ernie
W13 - AceOfDiamonds
E21 - Mr. Happy
U21 -
M23 - PhireKaLk6781

2 spots left

Due to unexpected popularity I have added 4 more spots. (J10,P10,J16,P16)

[b]Signups: [/b]
(start position)
M3 - KING1
E5 - EggyLv.999
U5 - Chompy
C13 - ernie
W13 - AceOfDiamonds
E21 - Mr. Happy
U21 -
M23 - PhireKaLk6781
J10 -
P10 -
J16 -
P16 -

5 spots left.

um dragon stop trying to act pro and pick a fixed location.

Yes someone went boom.
[hide="..."]...ernie. dragon96's mine killed ernie![/hide]

[hide="Scoreboard:"]
dragon96 - 3 pts
cf249 - 2 pts
Mewto55555 - 1 pt
[/hide]

[quote="ernie"]Oh, did someone start there?  YAY.[/quote]
haha YAY? :)

Yeah, I was gonna drop out.  But why go out like a wuss? :D

i guess this is abandoned then... i win? :)

no do not let AoD win...ITS HIS FAULT I GOT KILLED :mad:
ah well this is dead anyway yay AoD for winning...wait why does AoD win?

dragon should win but i dont want him to...

[quote="AceOfDiamonds"]I win cause i said so and i'm awesome.[/quote]

 :|  :|  :| [img]http://www.artofproblemsolving.com/Forum/images/smiles/mellow.gif[/img][img]http://www.artofproblemsolving.com/Forum/images/smiles/mellow.gif[/img][img]http://www.artofproblemsolving.com/Forum/images/smiles/mellow.gif[/img]

  :D If that's the case i also declare myself winner :D

dude you're not awesome and you're dead. also i didn't say so

Actually, by looking at the scoreboard, I see that [b]dragon96[/b] has [i]won[/i].  Because of [u]me[/u]! :D

well this is the [u][b][i][size=200][color=pink]100th[/color][/u][/b][/i][/size]post so I win

Nay, dragon96 does not necessarily win. Rules explicitly state that last person alive wins 3 pts.

Anyways this is dead. TIPS for future modders of this game of sorts - smaller map


Let a,b,c,d<1/3 such that a+b+c+d=1. Let also f(x)=(3-10x)/(x-3x 2 ). Prove that  \sum f(a)  \leq 32.

since $a,b,c,d<\frac 1{3}$
we let $x=\frac 1{3}-a, y=\frac 1{3}-b, z=\frac 1{3}-c, t=\frac 1{3}-d$
so $x+y+z+t=\frac 1{3}$ and $x,y,z,t>0$
let $x=\frac {m}{3S} , y=\frac {n}{3S} , z=\frac {p}{3S} , t=\frac {q}{3S}$
where $m,n,p,q>0$ and $S=m+n+p+q$
so your inequality will be equivalent to
$(m+n+p+q)(\frac 1{m} +\frac 1{n}+\frac 1{p}+\frac 1{q})+32 \geq 9(m+n+p+q)(\frac 1{m+n+p}+\frac 1{m+n+q}+\frac 1{m+p+q}+\frac 1{n+p+q})$ (*)
it's easy because we have
$m(\frac 1{n} +\frac 1{p} +\frac 1{q})\geq 9\frac{m}{n+p+q}$
write down similar inequalities and add them up we will have (*)
 :D


If $\sum_{k=0}^{n}\binom{3n}{3k}$
Prove that $\lim_{n\rightarrow \infty}\sqrt[3n]{S_{n}}=2$

[quote="aviateurpilot"]$U_{n}=\sqrt[3n]{S_{n}}$ and $V_{n}=\sqrt[3n]{2^{3n}/3}$ have the same limit $2$[/quote]

Why?  :D

A closed form for the sum can be found in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119683]this[/url] thread.


If the sides of a right triangle are $ 14$cm,$ 48$cm,and $ 50$cm. How many square centimeters are in its area?

This is a "copy" of the $ 7$-$ 24$-$ 25$ right triangle.

Thus, the answer is simply $ \frac{14 \times 48}{2}$, or $ \boxed{336}$.

Clearly, 14 and 48 are the legs.
Since a^2+b^2=c^2, the 2 smaller #'s are the legs.
Since the 14 and 48 legs form a right angle, you can use 1/2bh=1/2*14*48=7*48=336.

Also note that for a^2+b^2=c^2, and a,b,c form a right triangle, so do ka,kb, and kc.
k^2a^2+k^2b^2=k^2c^2.
k^2(a^2+b^2)=k^2(c^2) and divide by k^2.

Edit: Beaten.


[quote="jrrbc"]Theorem of S theory (to positive integer numbers)

$ t \equal{} r\Rightarrow(t.s \equal{} r.s)$

i want the proof

 :(

theorems not proved from the book [b]Introduction to Mathematical Logic [/b](Elliott Mendelson)[/quote]
It's pretty much the definition, ain't it? What theorems are you allowed to use to prove it?

Prove that $ a \equal{} b \Rightarrow a \plus{} n \equal{} b \plus{} n$ by induction (on $ n$). Then prove the desired property, also by induction.

Anyway, I think there is no meaning in doing this (probably because I am not understanding what we are doing) (I mean, in any formula, you can substitute $ a$ by $ b$ if $ a \equal{} b$). Also, this (at least the Olympiad Section) is not a good place to discuss logic or axiomatic theory...

Of course that is important for the olympiad section, you never doubt about the valuaty of the axiomatic mathematics? :P

About the question I dont know if those operations and relations $ \plus{}, . ,  \equal{}$ are defines as I know or if it some strange notation


inequalities
calculus
integration
inequalities proposed

[quote="Albanian Eagle"]my solution:
we have to prove that $\sum_{cyc}\frac{a^{2}}{1+bc}\geq \frac{1}{3(1+3abc)}$
or equivalently
$\sum_{cyc}\frac{a^{2}(1+3abc)}{1+bc}\geq \frac13$
and by using $a+b+c=1$ it is equivalent to
$\sum_{cyc}\frac{a^{2}(a+b+c+3abc)}{1+bc}\geq \frac13$
now
$LHS=\sum_{cyc}\left(a^{3}+\frac{a^{2}(b+c+2abc)}{1+bc}\right) =\sum_{cyc}\left(a^{3}+\frac{a^{2}b(1+ac)}{1+bc}+\frac{a^{2}c(1+ab)}{1+bc}\right)$
$=a^{3}+b^{3}+c^{3}+\sum_{cyc}\left(\frac{a^{2}b(1+ac)}{1+bc}+\frac{b^{2}a(1+bc)}{1+ac}\right)\geq a^{3}+b^{3}+c^{3}+2\sqrt{a^{3}b^{3}}+2\sqrt{b^{3}c^{3}}+2\sqrt{a^{3}c^{3}}$
$=(a^{\frac32}+b^{\frac32}+c^{\frac32})^{2}\geq \frac13 (a+b+c)^{3}=\frac13$
which is what we wanted.
what do you think of this solution?  :P[/quote]

Uhm, I spent some minutes trying to find your mistake but... can't find any thing. [b] Congragulation ![/b] Your sols is so elegant. Today is my lucky day now.
Let's see. Here is "ugly" one that I find it. Hope you can't solve :D , Albanian Eagle :roll:

Let $a,b,c$ are positive such that $a+b+c=1$ and denote:
$T= \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ Prove that:
$\frac{ab^{2}}{1+b}+\frac{bc^{2}}{1+c}+\frac{ca^{2}}{1+a}\geq \frac{1}{T(1+T)}$

[quote="Albanian Eagle"]Here is my solution:
$T+1=1+\sum_{cyc}\frac{a}{b}=\sum_{cyc}\frac{a(b+1)}{b}$
so
$(T+1)\left(\sum_{cyc}\frac{ab^{2}}{b+1}\right)\geq \left(\sum_{cyc}\sqrt{a^{2}b}\right)^{2}$
from Cauchy Schwartz
and in homogenous form we are left to show that
$\left(\sum_{cyc}\sqrt{\frac{a}{c}}\right)^{2}\left(\sum_{cyc}\frac{a}{b}\right)\geq \frac{(a+b+c)^{3}}{abc}$
after putting
$a=x^{2},b=y^{2},c=z^{2}$ and expanding the problem reduces to
$\sum_{cyc}\frac{x^{5}y^{2}}{z}+\sum_{cyc}\frac{x^{4}y^{4}}{z^{2}}+\sum_{cyc}x^{3}y^{3}+\sum_{x^{4}yz}\geq 3\sum_{cyc}x^{4}y^{2}+3x^{2}y^{2}z^{2}$
but this inequality can be reduced to summing the following inequalities
$\sum_{cyc}x^{4}yz\geq 3x^{2}y^{2}z^{2}$
$\frac{x^{5}y^{2}}{z}+\frac{x^{4}y^{4}}{z^{2}}+x^{3}z^{3}\geq 3x^{4}y^{2}$ cyclically
Hope it is clear enough! What is your solution?[/quote]

Not have time to type it, I am back to universityby now. But the idea is using integral( Harazi idea). Can you try it?

[quote="Gibbenergy"]I have no idea if this inequality hard or not. But it took me completely 2 pages to prove. Hope that I can see a smart solution:
Let $a,b,c$ are positive such that : $a+b+c=1$Prove:
$\frac{a^{2}}{1+bc}+\frac{b^{2}}{1+ac}+\frac{c^{2}}{1+ab}\geq \frac{1}{3(1+3abc)}$
 :D[/quote]

A shorter proof:
\[\sum_{cyc}\frac{a^{2}}{1+bc}=\sum_{cyc}\frac{a^{3}}{a+abc}\ge\frac{(a+b+c)^{3}}{3(a+b+c+3abc)}\]
where we use thw following well-known inequality:
\[\frac{a_{1}^{r}}{b_{1}^{s}}+\frac{a_{2}^{r}}{b_{2}^{s}}+\cdots+\frac{a_{n}^{r}}{b_{n}^{s}}\ge \frac{1}{n^{r-s-1}}\!\cdot\!\frac{(a_{1}+a_{2}+\cdots+a_{n})^{r}}{(b_{1}+b_{2}+\cdots+b_{n})^{s}}.\]

actually it's the same with what I did...
I just proved the well known inequality  :)


[quote="Meghrabi"]Can someone translate this sentence ?

"¿Se pueden distribuir los números del 1 al 16 en las casillas del tablero de modo que la suma de los números ubicados en tres casillas consecutivas sea siempre menor o igual que 24?"

Thanks :)[/quote]

Can you distribute the numbers from 1 to 16 in the squares of the board so that the sum of the numbers of three consecutive squares is always less or equal than 24?

How do you mean in the squares of a board ?

[quote="jaden"]casillas de un tablero = squares of a board

It's like a chess board, but in this case, with 4x4 squares, I suppose, because of the 16 numbers[/quote]

I thought it was an 1*16 board... don't know why.....

[quote="José"][quote="jaden"]casillas de un tablero = squares of a board

It's like a chess board, but in this case, with 4x4 squares, I suppose, because of the 16 numbers[/quote][/quote]

Yo pensé que era de 1*16, no sé porqué...


FTW
AMC
USA(J)MO
USAMO
MATHCOUNTS
AIME
AMC 10

Hello AoPS users. I am beginning the sequal for the tournament titled 'Another Tournament'. The signups begin now, and in the tournament. If you do good enough in one week, you don't have to answer the questions in the next week of the tournament. The tournament is single elimination, and when it reaches 2, 4, 6, 8, 12, 16, 24, or 32 contestants in the tournament there will be a FTW round.

Signups begin now;

Just to make it interesting.

Okay budi713 is the only person that has signed up.

Okay, so far we have two people that have joined;

budi713
ernie

meh.  i join but i might have to pull out.

I'll sign up, but also can drop out like AIME15.

I sign up, though I agree with Bugi and AIME15.

ok.

The message is too small. Please make the message longer before submitting.

The message is too small. Please make the message longer before submitting.

I'll join this one, but I am not going to join anymore.

budi713
ernie
AIME15
Ketvoman
Bugi
isabella2296
Bachukas
ThetaPi
alingaling

December, you should read the very first post of every new post you start posting on.

Please join. I will make the problems go up to USAMO lvl if we get enough people.

Well, if we get at least 16 people then it will be USAMO level at the end, but the first 8 rounds would be Mathcounts, then the next few rounds would be AIME, and the last few would be USAMO.

I'll join! Hopefully I'm able to participate in as many rounds as possible(I have a ton of homework in this year :().

[quote="isabella2296"]I don't suggest that - this is the [b]MathCounts[/b] forum, not the USAMO forum.[/quote]

But where else to make tournaments? I am also for USAMO like problems (I don't know what are these, but I hope that they are hard!!!)

They will get harder depending on the quantity of players, because I make each round harder than the previous.

You could probably make a USAMO tournament in the AMC forum. This is the MathCounts forum - I believe nebula42 said the level shouldn't go above AMC 10.

Most of the problems will be MOEMS lvl to AMC 10, it is only like the last 2 or 3 rounds that are AIME. I forgot to add ghjk to the list, but Bugi said that he quit, so here is the list;

budi713
ernie
AIME15
Ketvoman
ghjk
isabella2296
Bachukas
ThetaPi


$\triangle ABC: \ U\in AB,\ V\in AC,\ AU=AV;$
$XU\perp AB,\ XV\perp AC;\ YU\perp BX,\ YV\perp CX\Longrightarrow XY\perp BC.$

[u]Metoda 1[/u] (clasa a VII - a, avansat sau clasa a IX - a, mediu).

$\blacksquare \ XB^2-XC^2=(XB^2-XA^2)+(XA^2-XC^2)=$
$=(UB^2-UA^2)+(VA^2-VC^2)=UB^2-VC^2=$
$=(UB^2-UX^2)+(UX^2-VC^2)=(YB^2-YX^2)+(YX^2-YC^2)=$
$=YB^2-YC^2\Longrightarrow XB^2-XC^2=YB^2-YC^2\Longrightarrow XY\perp BC\ \square$

[u]Metoda 2[/u] (clasa a IX - a, avansat).

$\bigstar$ Notam cercul $w=C(X,r)$, unde $r=XU=XV$. Dreptele $UY,\ VY$ sunt polarele punctelor $B,\ C$ respectiv in raport cu cercul $w$, adica punctul $Y$ este intersectia polarelor punctelor $B,\ C$ in raport cu cercul $w$. Prin urmare, punctele $B,C$ apartin polarei punctului $Y$ in raport cu cercul $w$, adica $XY\perp BC.\ \bigstar$

[u]Observatie.[/u] Prima metoda este calitativ metrica iar cea de a doua metoda este instrumentata de cunostinte deosebite. Sa presupunem ca suntem in clasa a VII a. Cum s-ar rezolva aceasta problema cu cunostinte de nivel mediu ?

[u]Sugestie.[/u] Aici apare un subiect interesant pentru o dezbatere: odata cu propunerea unei probleme sa se precizeze si nivelul la care se situeaza solutia problemei fara a impiedica si solutiile care depasesc nivelul mentionat. Astfel, elevul va fi stimulat sa acceseze si cunostinte mai profunde din domeniul respectiv.


arithmetic sequence
combinatorics unsolved
combinatorics

prove that for natural number n>1 can divided the set {1,2,3,...,2^n} to the 2 set like A and B such that there is no arithmetic progression with leght 2n in the set A and B.


Find $\log_{3}10$ and $\log_{3}1.2$ in terms of $ x = \log_{3}4$ and $y = \log_{5}3$

 :sleeping: My brain isn't functioning right now

thanks bornformath and hmoonster

:wallbash:

i should have known that


trigonometry
inequalities proposed
inequalities

Let ABC be a triangle,prove that:

$ 2\sum\frac{\sin\frac{A}{2}}{\sin\frac{B}{2}\sin\frac{C}{2}} \geq \sum\frac{\cos{B}\plus{}\cos{C}}{\sin\frac{B}{2}\sin\frac{C}{2}} \geq \sum\frac{1}{\sin\frac{B}{2}\sin\frac{C}{2}}$
BQ

$ \cos{B}\plus{}\cos{C} \le 2\sin\frac{A}{2}$
a side is easy
BQ


a,b are two natural numbers such that (a,b)=1. Prove that (7a+3b,2a-b)=1.

It is not true, for example $ a\equal{}7,b\equal{}1$.


function
real analysis
real analysis unsolved

Hi,

My problem is that our professor did not go over an example like this in class, and I forgot my book in class one day and someone stole it, so I no longer have any references to help.

Here's the problem...

 Which of the following functions f has a removable discontinuity at a? If the discontinuity is removable, find a function g that agrees with f for x cannot = a and is continuous at a.

f(x)= (x^3-x^2-72x)/(x-9)    .... a=9

Now I know that this function has a removable discontinuity at a because 9 would make the function discontinuous. What I don't know is how to come up with a function, g, that agrees with f for x cannot = a and is also continuos at a. It confuses me. If anyone can help me I would appreciate it, Thank you.

p.s. I also attached the actual problem. I just figured that If i could get help with the one above, I could solve the other one too. Thanks again!

Josh

When $ x\neq 9$, $ f(x)$ can be written in a simpler form that does not look look like 0/0 for $ x\rightarrow 9$. To do this, factor the numerator of $ f$. You should get
[hide="this"]
\[ g(x)\equal{}x(x\plus{}8)\]
[/hide]
Also, note that not all discontinuities are removable. For example, $ f(x)\equal{}\frac{x}{|x|}$ is discontinuous at 0, but the behavior of $ f$ for positive numbers close to 0 is different than its behavior for negative numbers close to 0 (the function is 1 for all positive numbers, -1 for all negative numbers). $ f(x)\equal{}1/x$ also has a nonremovable discontinuity.


linear algebra
matrix
graph theory
combinatorics unsolved
combinatorics

Consider a rectangular matrix with 0's and 1's as entries. Show that each
set of r rows has 1's in at least r columns if and only if there is a 1 in each
row which are in different columns.

Consider a bipartite graph with $ n$ vertices $ A_1$, $ A_2$, ..., $ A_n$ on the left shore and $ m$ vertices $ B_1$, $ B_2$, ..., $ B_m$ on the right shore such that $ A_i$ is connected to $ B_j$ if and only if the entry of the matrix in the row $ i$ and column $ j$ is $ 1$. Apply the [url=http://en.wikipedia.org/wiki/Marriage_theorem#A_more_general_statement]generalized Hall theorem[/url].

  darij


If $9^{x}-9^{x-1}=216$ then what is the value of $2^{x}$?

Thats correct (and the same way I got it...)


linear algebra
matrix
linear algebra unsolved

First simplification:

Since $ \begin{pmatrix}0&0&1\\0&B&0\\1&0&0\end{pmatrix}^2\equal{}\begin{pmatrix}1&0&0\\0&B^2&0\\0&0&1\end{pmatrix},$ we have

$ A^{2009}\equal{}\begin{pmatrix}0&0&1\\0&B^{2009}&0\\1&0&0\end{pmatrix}.$

So it comes down to working with a $ 3\times 3$ matrix, not a $ 5\times 5.$

[b]Mr Kent Merryfield have a mistake [/b]($ \minus{} 1$)[b] but Suggest for me to think implies [/b]:
$ \begin{pmatrix}0 & 0 & \minus{} 1 \\
0 & B & 0 \\
1 & 0 & 0\end{pmatrix}^2 \equal{} \begin{pmatrix} \minus{} 1 & 0 & 0 \\
0 & B^2 & 0 \\
0 & 0 & \minus{} 1\end{pmatrix} \equal{} \begin{pmatrix} \minus{} 1 & 0 & 0 \\
0 & B & 0 \\
0 & 0 & \minus{} 1\end{pmatrix}$ then
$ A^{2009} \equal{} \begin{pmatrix}0 & 0 & \minus{} 1 \\
0 & B & 0 \\
1 & 0 & 0\end{pmatrix}^{2008}\begin{pmatrix}0 & 0 & \minus{} 1 \\
0 & B & 0 \\
1 & 0 & 0\end{pmatrix} \equal{} \begin{pmatrix} \minus{} 1 & 0 & 0 \\
0 & B & 0 \\
0 & 0 & \minus{} 1\end{pmatrix}^{1004}\begin{pmatrix}0 & 0 & \minus{} 1 \\
0 & B & 0 \\
1 & 0 & 0\end{pmatrix} \equal{} \begin{pmatrix}1 & 0 & 0 \\
0 & B^{1004} & 0 \\
0 & 0 & 1\end{pmatrix}\begin{pmatrix}0 & 0 & \minus{} 1 \\
0 & B & 0 \\
1 & 0 & 0\end{pmatrix} \equal{} \begin{pmatrix}1 & 0 & 0 \\
0 & B & 0 \\
0 & 0 & 1\end{pmatrix}\begin{pmatrix}0 & 0 & \minus{} 1 \\
0 & B & 0 \\
1 & 0 & 0\end{pmatrix} \equal{} \begin{pmatrix}0 & 0 & \minus{} 1 \\
0 & B^2 & 0 \\
1 & 0 & 0\end{pmatrix} \equal{} \begin{pmatrix}0 & 0 & \minus{} 1 \\
0 & B & 0 \\
1 & 0 & 0\end{pmatrix} \equal{} A$
 By $ Rank(B) \equal{} 1$ ,theorefore $ B^2 \equal{} Tr(B)B \equal{} ( \minus{} 7 \plus{} 4 \plus{} 4)B \equal{} B$ then $ B^{k} \equal{} B,\forall k \in \mathbb{N^{*}}$

Implies :
Characteristic equation of  $ A$ is : $ (\lambda^2\plus{}1)\det\begin{pmatrix}\minus{}7\minus{}\lambda&5&3\\\minus{}5&4\minus{}\lambda&2\\\minus{}9&6&4\minus{}\lambda\end{pmatrix}\equal{}(\lambda^2\plus{}1)(\lambda^2\minus{}\lambda)\equal{}0$ have 4 roots are $ \minus{}i\plus{}i,0,1$
Then the exist $ T \in GL_4{(\mathbb{C})}$
 Such that : $ A\equal{}T\begin{pmatrix} i&0&0&0&0\\
0&i&0&0&0\\
0&\minus{}i&0&0&0\\
0&0&0&0&0\\
0&0&0&0&1\end{pmatrix}T^{\minus{}1}\equal{}Tdiag(i,\minus{}i,0,1)T^{\minus{}1}$
$ A^{2009}\equal{}Tdiag(i^{2009},(\minus{}i)^{2009},0,1^{2009})T^{\minus{}1}\equal{}Tdiag(i,\minus{}i,0,1)T^{\minus{}1}\equal{}A$


[quote="Pain rinnegan"]Let $ a,b,c > 0$. Prove that:
\[ 3 \plus{} \frac {a(c \plus{} a)}{b(b \plus{} c)} \plus{} \frac {b(a \plus{} b)}{c(c \plus{} a)} \plus{} \frac {c(b \plus{} c)}{a(a \plus{} b)}\ge 4\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\]
[hide]This inequality is strengthening the following one :
\[ \frac {a(c \plus{} a)}{b(b \plus{} c)} \plus{} \frac {b(a \plus{} b)}{c(c \plus{} a)} \plus{} \frac {c(b \plus{} c)}{a(a \plus{} b)}\ge 2\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\]
for $ a,b,c > 0$, which is almost obvious by AM-GM.[/hide][/quote]

WLOG, assume $ a\equal{}\max\{a,b,c\},$ after expanding:

$ \iff \left( ac \left( a\minus{}b \right) ^{2}\plus{}a{c}^{2} \left( 3a\plus{}b \right) \plus{}2b{c}^{3}\plus{}a \left( b\plus{}c \right)  \left( b\minus{}2c \right) ^{2}\plus{}3{b}^{4}\minus{}5{b}^{2}{c}^{2}\plus{}4{c}^{4} \right)  \left( a\minus{}b \right)  \left( a\minus{}c \right) \plus{} \left( b\plus{}c \right)  \left(  \left( 4a\minus{}3c \right) {b}^{2}\plus{} \left( 4a\minus{}3b \right) {c}^{2}\plus{}3abc \right) \left( b\minus{}c\right) ^{2} \ge 0.$

clearly

Let $ \frac{a}{b+c}=\frac{m.l}{k} , \frac{b}{a+c}=\frac{k.m}{l} , \frac{c}{a+b}=\frac{k.l}{m}$ and $ (k,l,m >0) \Longrightarrow$

$ k^2+l^2+m^2+2.k.l.m=1  \Longrightarrow$

$ k=cos\alpha , l=cos\beta , m=cos\theta$

$ \alpha,\beta,\theta$ are angles of a triangle.

Inequality equivalents to

$ cos^4{\alpha}.cos^2{\theta}+cos^4{\beta}.cos^2{\alpha}+cos^4{\theta}.cos^2{\beta}+3cos^2{\alpha}.cos^2{\beta}.cos^2{\theta} \geq
4cos{\alpha}.cos{\beta}.cos{\theta}(cos^2{\alpha}.cos^2{\beta}+cos^2{\beta}.cos^2{\theta}+cos^2{\alpha}.cos^2{\theta})$

By Schur's inequality;

$ cos^4{\alpha}.cos^2{\theta}+cos^4{\beta}.cos^2{\alpha}+cos^4{\theta}.cos^2{\beta}+3cos^2{\alpha}.cos^2{\beta}.cos^2{\theta} \geq
2.\sqrt[3]{cos^2{\alpha}.cos^2{\beta}.cos^2{\theta}}(cos^2{\alpha}.cos^2{\beta}+cos^2{\beta}.cos^2{\theta}+cos^2{\alpha}.cos^2{\theta})$

And $ cos{\alpha}.cos{\beta}.cos{\theta} \leq \frac{1}{8}$

Solution is done.  :wink:

[quote="Pain rinnegan"]Let $ a,b,c > 0$. Prove that:
\[ 3 \plus{} \frac {a(c \plus{} a)}{b(b \plus{} c)} \plus{} \frac {b(a \plus{} b)}{c(c \plus{} a)} \plus{} \frac {c(b \plus{} c)}{a(a \plus{} b)}\ge 4\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\]
[hide]This inequality is strengthening the following one :
\[ \frac {a(c \plus{} a)}{b(b \plus{} c)} \plus{} \frac {b(a \plus{} b)}{c(c \plus{} a)} \plus{} \frac {c(b \plus{} c)}{a(a \plus{} b)}\ge 2\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\]
for $ a,b,c > 0$, which is almost obvious by AM-GM.[/hide][/quote]
It is known that
$ 4\sum \frac{a}{b\plus{}c} \le \sum \frac{b\plus{}c}{a}.$
Therefore,
$ 4\sum \frac{a}{b\plus{}c} \le 2\sqrt{\left(\sum \frac{a}{b\plus{}c}\right)\left(\sum \frac{b\plus{}c}{a}\right)}.$
It suffices to prove that
$ \sum \frac{a(c\plus{}a)}{b(b\plus{}c)}\plus{}3 \ge 2\sqrt{\left(\sum \frac{a}{b\plus{}c}\right)\left(\sum \frac{b\plus{}c}{a}\right)}.$
Setting $ x\equal{}\frac{a}{b\plus{}c},$ $ y\equal{}\frac{b}{c\plus{}a}$ and $ z\equal{}\frac{c}{a\plus{}b},$ this inequality becomes
$ \frac{x}{y}\plus{}\frac{y}{z}\plus{}\frac{z}{x}\plus{}3 \ge 2\sqrt{(x\plus{}y\plus{}z)\left( \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\right)}.$
We will prove that this inequality holds for any positive real numbers $ x,$ $ y,$ $ z.$ This is a good application of the AM-GM Inequality:

Due to cyclicity, we may assume that $ y$ is between $ x$ and $ z.$ In this case, using the AM-GM Inequality, we get
$ 2\sqrt{(x\plus{}y\plus{}z)\left(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\right)} \le \frac{x\plus{}y\plus{}z}{y}\plus{}y\left(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\right),$
and hence, it is enough to check that
$ \frac{x}{y}\plus{}\frac{y}{z}\plus{}\frac{z}{x}\plus{}3 \ge \frac{x\plus{}y\plus{}z}{y}\plus{}y\left(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\right).$
This inequality reduces to
$ \frac{(x\minus{}y)(y\minus{}z)}{xy} \ge 0,$
which is obviously true.

Remark. There is another solution using Schur's Inequality. I will post it soon.

[quote="manlio"]Pain wrote:

......This inequality is strengthening the following one :
\[ \frac {a(c \plus{} a)}{b(b \plus{} c)} \plus{} \frac {b(a \plus{} b)}{c(c \plus{} a)} \plus{} \frac {c(b \plus{} c)}{a(a \plus{} b)}\ge 2\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\]
for $ a,b,c > 0$, which is almost obvious by AM-GM.

Sorry I don't see why it is obvious by AM-GM. :blush:

Thank you very much.[/quote]

Of course, manlio.

\[ \frac{2a(c\plus{}a)}{b(b\plus{}c)}\plus{}\frac{b(a\plus{}b)}{c(c\plus{}a)}\ge 3\sqrt[3]{\frac{a^2(c\plus{}a)(a\plus{}b)}{bc(b\plus{}c)^2}}\ge \frac{6a}{b\plus{}c}\]

then sum the similars and it's done.

[quote="Pain rinnegan"]Let $ a,b,c > 0$. Prove that:
\[ 3 \plus{} \frac {a(c \plus{} a)}{b(b \plus{} c)} \plus{} \frac {b(a \plus{} b)}{c(c \plus{} a)} \plus{} \frac {c(b \plus{} c)}{a(a \plus{} b)}\ge 4\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\]
[hide]This inequality is strengthening the following one :
\[ \frac {a(c \plus{} a)}{b(b \plus{} c)} \plus{} \frac {b(a \plus{} b)}{c(c \plus{} a)} \plus{} \frac {c(b \plus{} c)}{a(a \plus{} b)}\ge 2\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\]
for $ a,b,c > 0$, which is almost obvious by AM-GM.[/hide][/quote]
And this is the second proof:

Similar to the above proof, we see that it suffices to prove that
\[ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \plus{} 3 \ge 2\sqrt {(x \plus{} y \plus{} z)\left(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\right)}\]
for $ x,$ $ y,$ $ z > 0.$

Using the third degree Schur's Inequality, for any nonnegative real numbers $ u,$ $ v,$ $ w,$ we have
\[ u^3 \plus{} v^3 \plus{} w^3 \plus{} 3uvw \ge (u^2v \plus{} v^2w \plus{} w^2u) \plus{} (uv^2 \plus{} vw^2 \plus{} wu^2).\]
Also, by the AM-GM Inequality,
\[ \begin{aligned} (u^2v \plus{} v^2w \plus{} w^2u) & \plus{} (uv^2 \plus{} vw^2 \plus{} wu^2) \ge \\
& \ge2\sqrt {(u^2v \plus{} v^2w \plus{} w^2u)(uv^2 \plus{} vw^2 \plus{} wu^2)}.\end{aligned}\]
Therefore,
\[ u^3 \plus{} v^3 \plus{} w^3 \plus{} 3uvw \ge 2\sqrt {(u^2v \plus{} v^2w \plus{} w^2u)(uv^2 \plus{} vw^2 \plus{} wu^2)}.\]
Setting $ u \equal{} \sqrt [3]{\dfrac{x}{y}},$ $ v \equal{} \sqrt [3]{\dfrac{y}{z}}$ and $ w \equal{} \sqrt [3]{\dfrac{z}{x}}$ in this inequality, we get
\[ \begin{aligned} \sum\frac {x}{y} \plus{} 3 & \ge 2\sqrt {\left(\sum\sqrt [3]{\frac {x^2}{y^2}}\sqrt [3]{\frac {y}{z}}\right)\left(\sum \sqrt [3]{\frac {x}{y}}\sqrt [3]{\frac {y^2}{z^2}}\right)} \equal{} 2\sqrt {\left(\frac {\sum x}{\sqrt [3]{xyz}}\right)\left(\frac {\sum xy}{\sqrt [3]{x^2y^2z^2}}\right)} \\
& \equal{} 2\sqrt {\left(\sum x\right)\left(\sum \frac {1}{x}\right)}.\end{aligned}\]


Euler
geometry
circumcircle
ratio
trigonometry
incenter
similar triangles

A(1)B(1)C(1) is the orthic(pedal) triangle of an acute non-equilateral triangle ABC

The incircle of triangle A(1)B(1)C(1) touches its sides at A(2), B(2), C(2)

Prove triangles ABC and A(2)B(2)C(2) have the same Euler line.

Sorry My New Zealand friend but I'm really in such a hurry... ;)
I say the 3 triangles are the big,medium and small ones. :P
You need to prove that the sides of the two triangle are correspondingly parallel(Angle chasing...Using the cyclic points), then their Euler lines must be parallel(if not meet), as the circumcenter of small triangle is the ... of the big one(sorry for my poor English), the answer is obvious now...

Denote $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$ the triangle internal angles. Let $D \in BC, E \in CA, F \in AB$ bet the altitude feet of the triangle $\triangle ABC$ forming the orthic triangle $\triangle DEF$. The triangles $\triangle AEF \sim \triangle DBF \sim \triangle DEC \sim \triangle ABC$ are all similar by SAS, for example, the triangles $\triangle AEF \sim \triangle ABC$ have the angles $\angle EAF \equiv \angle CAB$ common and the adjacent sides in the same ratio:

$\frac{EA}{AB} = \frac{AF}{CA} = \cos \alpha,\ \ \frac{EA}{AF} = \frac{AB}{CA}$

Thus the remaining corresponding angles of these triangles are also equal:

$\angle FDB = \angle EDC = \angle CAB = \alpha$
$\angle DEC = \angle FEA = \angle ABC = \beta$
$\angle EFA = \angle DFB = \angle BCA = \gamma$

Consequently,

$\angle ADF = \angle ADE = 90^\circ - \alpha$
$\angle BED = \angle BEF = 90^\circ - \beta$
$\angle CFE = \angle CFD = 90^\circ - \gamma$

and the altitudes AD, BE, CF bisect the internal angles $\angle D, \angle E, \angle F$ of the orthic triangle $\triangle DEF$. Therefore, the orthocenter H of the triangle $\triangle ABC$ is the incenter of the orthic triangle $\triangle DEF$. Let P, Q, R be the tangency points of the incircle (H) of the orthic triangle with its sides EF, FD, DE. Since the tangent lengths to the incircle (H)  from the vertices D, E, F are equal, DR = DQ, EQ = EP, FP = FR, the triangles $\triangle DQR, \triangle EPQ, \triangle FRP$ are isosceles, which means that the bisectors DA, EB, FC of the angles $\angle QDR \equiv \angle FDE,\ \angle REP \equiv \angle DEF,\ \angle PFQ \equiv \angle EFD$ are perpendicular to their bases QR, RP, PQ. Since $QR, BC \perp AD,\ RP, CA \perp BE,\ PQ, AB \perp CF$, the sides $QR \parallel BC,\ RP \parallel CA,\ PQ \parallel AB$ of the triangles $\triangle PQR, \triangle ABC$ are parallel. This means that these 2 triangles are centrally similar, the similarity center S being the concurrency point of the lines AP, BQ, CR connecting their corresponding vertices. But then all lines connecting the corresponding points of the centrally similar triangles $\triangle PQR \sim \triangle ABC$ pass through the similarity center S. Let O be the circumcenter of the triangle $\triangle ABC$ and X the orthocenter of the triangle $\triangle PQR$. Obviously, the orthocenter H of the triangle $\triangle ABC$, being the incenter of the orthic triangle $\triangle DEF$, is the circumcenter of its contact triangle $\triangle PQR$. Since the triangles $\triangle ABC, \triangle PQR$ are not equilateral, the points O, H, X are different from each other and consequently, also from the similarity center S. The lines OH, HX connecting the circumcenters O, H and the orthocenters H, X of the triangles $\triangle ABC, \triangle PQR$, respectively, both pass through the similarity center S. Hence, the 4 points O, H, X, S are collinear, which means that the Euler lines OH, HX of the centrally similar triangles $\triangle ABC, \triangle PQR$ are identical.

[b]Note:[/b] If the triangle $\triangle ABC$ has all angles acute, the orthocenter H is the incenter of the orthic triangle $\triangle DEF$ and the vertices A, B, C are the excenters against the vertices D, E, F. If the triangle $\triangle ABC$ has an obtuse angle, say $\angle C > 90^\circ$, then the vertex C is the incenter of the orthic triangle $\triangle DEF$ and the vertices A, B and the orthocenter H are the excenters against the vertices D, E, F. Then the contact triangle $\triangle PQR$ of the excircle against the vertex F, centered at the orthocenter H, is centrally similar to the original triangle $\triangle ABC$ and these 2 triangles have the same Euler line.

It is an old problem. Suppose $A'B'C'$ is the triangle formed by the tangency points of the incircle with $AB, BC, CA$. Then the Euler line of $A'B'C'$ is the line $OI$ (Inversion $(I,r^2)$)


geometry
calculus
AMC
AIME
number theory
relatively prime
AIME I

In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at $50$ miles per hour along the highways and at $14$ miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of this region is $m/n$ square miles, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

[hide="Thrilling..."]Let the roads be the $x$ and $y$ axes. The situation is symmetric across each of these, and also across $y=x$ and $y=-x$, making a total of $8$ parts. I will only consider the part between $x$ axis and $y=x$.

The furthest point horizontally that can be reached is $5$ miles along the highway and no distance along the prairie.

Consider a point $(x,y)$. Let $x-w$ be the distance along the highway traveled. The time to get there is then $\frac{x}{50}-\frac{w}{50}+\frac{\sqrt{w^{2}+y^{2}}}{14}\;\text{hours}$

For given $x$, $\frac{x}{50}$ is constant and can be ignored.
$-\frac{w}{50}+\frac{\sqrt{w^{2}+y^{2}}}{14}$ is then optimized in terms of $w$, which occurs when:

$-\frac{1}{50}+\frac{w}{14\sqrt{w^{2}+y^{2}}}=0\implies w=\frac{7}{24}y$

So now we have the most optimal way to reach $(x,y)$ from the origin: travel along the highway until $(x-\frac{7}{24}y,0)$ and then travel along the prairie the rest of the way.

Now we substitute and have:
$\text{time}=\frac{1}{10}=\frac{1}{50}x+\frac{12}{175}y$

This intersects $y=x$ at $x=y=\frac{35}{31}$.

Our region is therefore triangular. The height is $\frac{35}{31}$ and the base is $5 \implies \text{area}=\frac{175}{62}$

There are $8$ such regions, and multiplying by $8$, our total area is $\frac{700}{31}$.

This gives $m+n=\boxed{731}$.[/hide]

Is there an easy way to maximize that expression w/o calculus (or for that matter, is there an easy way to do this problem without calculus?)

if you know snell's law, then just use that...or this is just a certain inequality...i don't know the name, but i am sure somebody does

Does anyone have a solution that does not use optimization?  Thanks!

This is a past Aime problem...

[url=http://www.artofproblemsolving.com/Wiki/index.php/2000_AIME_I_Problems/Problem_13]2000/I/13[/url]

Does anyone know a way to solve this problem without using optimization and/or calculus?

Let $x$ be the distance traveled on the highway and $r$ be the distance traveled on the prairie.  By $d=rt$, $$\frac{x}{50}+\frac{r}{14}=\frac{6}{60}\implies r=\frac{7}{25}(5-x).$$  We toss the problem on a coordinate plane and focus on the first quadrant only(and multiply by 4 at the end) and at $(x,0)$, we can travel an additional $\frac{7}{25}(5-x)$ miles on the prairie. Hence, draw a circle at $(x,0)$ with a radius of$\frac{7}{25}(5-x)$ and this is  the distance that we can go in the prairie.  As $x$ ranges from $0$ to $5$, the circles are homothetic or can be dilated with respect to the point $(5,0)$.  To prove this fact, let $B$ be $(5,0)$ and let $O_x$ be the center of the circle with center $(x,0)$ and radius $r_x$ of $\frac{7}{25}(5-x)$.  Hence, $\frac{O_xB}{r_x}=\frac{25}{7}$ for all $x$ from 0 to 5 so they are indeed homothetic.  Since the circles are homothetic, the max distance traveled on the prairie is the lines to the tangent of these circles(which would be tangent to all of the circles by the definition of homothety).  For the tangent in the first quadrant, Let $O$ be $(4,0)$, $B$ be $(5,0)$ and $C$ be the tangent with respect to the $(4,0)$ circle(note that we could have chosen any point on the tangent).  The hypotenuse of right triangle $OBC$ is 1 and $OC$ has length $\frac{7}{25}$.  Hence, $\tan CBO=\frac{7}{24}$ which implies that the slope of the tangent is $-\frac{7}{24}.$  To intuitively see this, draw a perpendicular from $C$ to the x-axis and note that $\tan CBO$ is magnitude of the slope because the slope is negative.  Hence, the line is $y=-\frac{7}{24}(x-5)$ and similarly, the tangent line with respect to the circles homothetic with $(0,5)$ is $y=-\frac{24}{7}x+5$ and the intersection is $(\frac{35}{31},\frac{35}{31})$ and the area in quadrant 1 is a square and two right triangles which is $\frac{35}{31}\cdot\frac{35}{31}+2\cdot\frac{1}{2}\cdot\frac{35}{31}\cdot (5-\frac{35}{31})=\frac{175}{31}$ so the total area is $\frac{700}{31}=\boxed{731}.$ $\Box$


trigonometry integration function limit algebra polynomial vector

[color=blue][b]12th Grade[/b][/color]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=134082]Problem 1[/url]
Let $f : \left( 0, \infty \right) \to \mathbb R, \, f(x) = \frac{\sin x}{x}$.
a) Prove that $f \in C^\infty \left( 0, \infty \right)$ and that
\[f^{(n)}(x) = \frac1{x^{n+1}}\cdot \int_{0}^{x}t^{n}\cdot \cos \left( t+\frac{n \pi}2 \right) \, dt, \ \forall n \in \mathbb N^\ast, \ \forall x > 0 . \]
b) Prove that $\left| f^{(n)}(x) \right| \leq \frac1{n+1}, \, \forall n \geq 1, \, \forall x > 0$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=134083]Problem 2[/url]
Let $f : \left[ 0, 1 \right] \to \mathbb R$ be a continuous function. For any $n \in \mathbb N^\ast$ define function
\[g_{n}: \left[ 0, 1 \right] \to \mathbb R, \ g_{n}(x) = \int_{0}^{1}\min \left\{ x, t^{n}\right\}f(t) \, dt . \]
a) Prove that $g_{1}$ is twice differentiable on $\left[ 0, 1 \right]$ and that $g_{1}^{\prime \prime}(x) =-f (x), \, \forall x \in \left[ 0, 1 \right]$.
b) Give an example of a function $f$ for which $g_{2}$ is not twice differentiable in $0$.
c) Compute $\lim_{n \to \infty}g_{n}(x), \, \forall x \in \left[ 0, 1 \right]$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=134081]Problem 3[/url]
Let $\left( G, \cdot \right)$ be a finite group with the following property: the set of the orders of the elements of $G$ is composed of $n$ consecutive numbers, where $n \geq 2$.
a) Prove that $G$ is commutative if and only if $n=2$.
b) Prove that if $n \geq 3$, then the only element $a \in G$ such that $ax = xa, \, \forall x \in G$ is $e$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=134080]Problem 4[/url]
Let $\left( A,+, \cdot \right)$ be a commutative ring with $n \geq 6$ elements, that is not a field.
a) Prove that $u : A \to A$, $u(x) = \begin{cases}0, \ \ \ \ x \neq 0 \\ 1, \ \ \ \ x = 0 \end{cases}$ is not a polynomial function.
b) Prove that $P$, the number of polynomial functions $f : A \to A$, verifies
\[n^{2}\leq P \leq n^{n-1}. \]

[color=red][b]9th Grade[/b][/color]

[hide="CLICK TO SEE THE 9th GRADE PROBLEMS"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=137452]Problem 1[/url]
Let $a,b,c$ be rational numbers. For each $n \in \mathbb Z$ we consider the equation
\[E_{n}: (a+n) x^{2}-2 (b+n) x+c+n = 0 . \]
(a) Prove that if $a,b,c$ are in arithmetic progression, then the equations $E_{n}$ have a common solution.
(b) Prove that if every equation $E_{n}$ has real roots, then $a,b,c$ are in arithmetic progression and each equation has rational solutions.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=137454]Problem 2[/url]
Let $M$ be the midpoint of the side $[BC]$ of the triangle $ABC$. The parallel to $BC$ through a point $G \in (AM)$ intersects $AB$ in $X$ and $AC$ in $Y$; $CX$ and $GB$ intersect in $Q$, $CG$ and $BY$ in $P$. Denote $\frac{AG}{AM}$ by $k$.
(a) Prove that $\frac{GP}{PC}= \frac{k}2$.
(b) Express the vector $\overrightarrow{CP}$ as a function of $\overrightarrow{AB}$, $\overrightarrow{AC}$ and the scalar $k$.
(c) Prove that if $G$ is the centroid of $ABC$, then $MPQ$ is similar to $ABC$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=137453]Problem 3[/url]
We divide a square into $n^{2}$ congruent squares by drawing parallels to the sides.
(a) Find the number of squares of different dimensions.
(b) If we draw the diagonals in every small square, then how many isosceles right triangles of different dimensions are there?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=137456]Problem 4[/url]
(a) Prove that for all $n \in \mathbb N^\ast$, by dividing $3^{2^{n}}: 2^{n}$ with remainder, one obtains an even quotient.
(b) Prove that the sequence $\left\{ a_{n}\right\}_{n \geq 1}$ given by
\[a_{n}= \left\lfloor \frac{3^{n}}{n}\right\rfloor, \, \forall n \in \mathbb N^\ast \]
has an infinity of even elements, an infinity of elements divisible by $3$ and an infinity of elements divisible by $5$.[/hide]

[color=green][b]10th Grade[/b][/color]

[hide="CLICK TO SEE THE 10th GRADE PROBLEMS"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=137457]Problem 1[/url]
(a) Solve in $\mathbb C$ the system $|z| = \left| \frac{a}{z}\right| = |z-a|$, where $a \in \mathbb R, \, a>0$.
(b) Solve the equation $\log_{x}\left( x+10.25 \right) = \lg 10.5$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=134730]Problem 2[/url]
Find all functions $f : \mathbb R \to \mathbb R$ that have the following property: for any distinct points $A(p,f(p)), \ B(q,f(q))$, the line $AB$ intersects $Oy$ in $(0,-pq).$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=137458]Problem 3[/url]
(a) Prove that there is a convex hexagon with congruent angles and sides of lengths $1,2,3,4,5,6$ (not necessarily in this order).
(b) Prove that there are exactly two types of hexagon that fulfill the conditions in (a).

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=137459]Problem 4[/url]
Let $n \in \mathbb N^\ast$ and $\mathcal P(A)$ be the power set of $A = \left\{ 1, 2, \ldots, n \right\}$. A function is called [i]ascending[/i] if $f : \mathcal P(A) \to \mathcal P(A)$ and, for all $X,Y \subset A$, we have that $X \subset Y \, \Longrightarrow \, f(X) \subset f(Y)$.
(a) Prove that if $f$ is an ascending function, then there is $B \subset A$ such that $f(B) = B$.
(b) Prove that for all $k \in \overline{1,2^{n}}$ there is an ascending function $f_{k}$ such that the equation $f_{k}(X) = X$, $X \subset A$, has exactly $k$ solutions.[/hide]


It's clear, since from $13 = 3(x^2 - xy + y^2)$ we get that 3 must divide 13, but it doesn't :)

[quote="riddler"][quote="shyong"][quote="Fermat -Euler"]resolve this equation

[color=red]13(x+y)=3(x^3+y^3)[/color]
x and y are integers[/quote]

[hide]
$13(x+y)=3(x+y)(x^2+y^2-xy)$ .

From here it is clear that the only answer is $x+y=0$ so $(x,y)=(k,-k)$ for any integer $k$ .

[/hide][/quote]
 :huh:  :huh:  :huh:  :huh:   please be more clear
[hide="solution"]
$13(x+y)=3(x+y)(x^2+y^2-xy)$
$13=3(x^2+y^2-xy)$
$13=3x^2+3y^2-3xy$
$3x^2+3y^2-3xy-13=10$
$3x(1-y)-3(1-y^2)=10$
$(3x-3-3y)(1-y)=10$
$y=1$ or other things.... why is this wrong? :(  :?
[/hide][/quote]

Dear riddler , your 4th step should be $=0$ instead of $10$ . And your 5th step should be $3x(x-y)$ instead of $3x(1-y)$  ;)

Actually , you dont needa multiple it out in your 3rd step .
[hide="why"]
Because there is a factor of $3$ on your RHS but there is no factor of $3$ on LHS . This is enough to say that there are no integral solution for it .
 :) [/hide]

urrrrrrggghhhh :wallbash:  :wallbash_red:  :ewpu:

[quote="shyong"]From here it is clear that the only answer is $x+y=0$ so $(x,y)=(k,-k)$ for any integer $k$ .
[/quote]

Now WHY did you post that, stephenyamqc?

Stephen, you make the others feel like that you copied others' answer.


$ 1^2\times\sum_{i \equal{} 1}^n i \equal{} 210 \iff n^2 \plus{} n \minus{} 420 \equal{} (n \plus{} 21)(n \minus{} 20) \equal{} 0 \iff n \equal{} \minus{} 21, 20$.
So, $ 20\times4 \equal{} \boxed{80 \ \text{cm}}$.

could someone explain this so it's easier to understand?

Sorry, i still don't understand how you solved this.

gah...

The sum of the first $ n$ numbers is $ 210$. Rewritten, $ \frac {n(n \plus{} 1)}{2} \equal{} 210\implies n^2 \plus{} n \minus{} 420 \equal{} 0\implies (n \minus{} 20)(n \plus{} 21) \equal{} 0$. The solutions for $ n$ are $ 20$ and $ \minus{} 21$, but it's impossible to have a [i]negative[/i] number of layers, so there are $ 20$ layers.
We can rearrange the blocks to form a right "triangle" with side length $ 20$. The perimeter is thus $ 2(20 \plus{} 20) \equal{} 2(40) \equal{} \boxed{80}.$

does that make sense [b]NOW[/b]?

Sorry, I still don't understand how you're doing this problem. Doesn't a triangle have three sides and what do you mean layers? Why are you using that equation? I think you'll need to start from the beginning and make it more clear.

hm. ill give this a shot.

we know that there are 1+2+3+4+...n=210 blocks. so we need to find n, and we can find that this is 20 (you can figure how to do that.). now, we can rearrange the blocks, instead of having them in the shape that we have them, as a right triangle. that leads us to roflcopts and everyone else's solutions. get it NOW?

ok i get how you found the number of layers. But i don't get how you found the perimeter. Because a triangle has three sides.

shape the blocks into a "right-triangle-like" shape. The "hypotenuse" is a staircase. The perimeter is thus $ 20\plus{}20\plus{}20\plus{}20\equal{}\boxed{80}$.
Draw it out yourself and see.

But if it's a right triangle, then why are you adding 4 20's for something that has 3 sides?

[hide="easier to understand?"] So if there are a total of n rows, there will be a total of $ 1 \plus{} 2 \plus{} ... \plus{} n \equal{} \frac {n(n \plus{} 1)}{2}$ squares. Setting this equal to 210, we find n=20.

Now imagine that you're looking at this from the top, like a birds-eye-view. You will see just twenty squares with side length 1, giving a perimeter of 20.

From the bottom looking up, you'd see twenty squares again, giving 20 more.

From both sides, since there are twenty rows, you'd see the sides of twenty squares, giving twenty for each side. Since there are two sides, this gives 40.

In all, you have 20+20+40=80.[/hide]

Also I get the feeling that you're trolling. If you are, stop.

GAHHH...

  [geogebra]75743cc7f6bcab1022730f805b355e5e8c039913[/geogebra]

[geogebra]b0180764bf9831dab968fdc77996ec20b09ccf99[/geogebra]

Do you see how the first formation can have the blocks all shifted to the right to form the second formation?

Since it has 20 layers, its perimeter is $ 20\plus{}20\plus{}(\text{all those stairstep segments})\equal{}20\plus{}20\plus{}2(20)\equal{}\boxed{80}$.

Do you see why the stairstep segments add up to $ 2(20)?$ that is because we can look at it from one point of view, and see the same length at the opposite side, which is 20 because it has 20 layers. Similarly, the other "dimension" is also 20. So the perimeter is $ \boxed{\boxed{80}}$

Or if you didn't see the second part, you could just count out the perimeter 1 by 1, but that would be very slow.


A bug starts at the origin. It crawls in the following pattern: 1 inch due east, 1/2 inch due north, 1/4 inch due west, 1/8 inch due south, and continues crawling like this indefinitely.
a. Does the bug ever get to one point?
b. If so, what are the coordinates of this point?
c. What is the total distance that the bug crawls?

In our class (Adv. Analysis II) we are learning about infinite series. When I first saw the problem, I said to myself that it will never reach a point thinking of the worm problem (A worm is 1 ft in a hole and it goes up half the distance everyday. When will he get to the top?)

But then it didn't fit in with the topic of infinite series. According to number 23 on this site [url]http://64.233.161.104/search?q=cache:aVsImqqWK2MJ:www.math.unl.edu/link/outreach/mathday/probe/i/problems/ProbeI-2002.pdf+%22a+bug+starts+at+the+origin%22&hl=en&ie=UTF-8[/url] none of the choices include "Will never reach"

So I would appreciate it if someone could help me. Which is right? This is an extra-credit problem and I need to get it right. So if you can, please help.[/url]

One good way to think about the problem is with complex numbers.

Shouldn't this be in another forum, like Intermediate or something?

wow, you can just use what we learned in class
if you can't get this yourself, i pity you

apparently we have lots of people from AAST here... :\

silly underclassmen :)

Jhyon8149 wrote:But then it didn't fit in with the topic of infinite series. According to number 23 on this site (omitted) none of the choices include "Will never reach"

So I would appreciate it if someone could help me. Which is right? This is an extra-credit problem and I need to get it right. So if you can, please help.

Note that in the site you linked to, the question asked what the "ultimate destination" of the bug is, i.e. what point its path tends to lead it towards.  That doesn't mean that the bug will necessarily ever reach that point, just where it heads toward as it moves an infinitely many steps.

But I think the question is to find that point that it tends toward, regardless of the fact that it will never actually reach it.  (You might mention this.)  In that case, the only hint I'll offer is [hide]consider horizontal and vertical motion separately[/hide].


Hi all,

Can someone tell me how to install LaTex on Mac OS X?
I've been a long time user of Latex - TexnicCenter on PC, and I'd like to setup something similar on my Mac for college.
Basically, I want to know what the best thing to install is...

Thanks a lot.

I have seen good reports about TeXShop; see [url=http://en.wikipedia.org/wiki/TeXShop]here[/url] and [url=http://www.uoregon.edu/~koch/texshop/]here[/url].


Somebody help me with this problem?
Given a positive integer a.
Find a smallest positive integer k such that.
k^2 - 4ka - 4 is a perfect square.

$ (k\minus{}2a)^{2}\minus{}(4a^{2}\plus{}4)\equal{}m^{2}$
$ (k\plus{}2a\minus{}m)(k\plus{}2a\plus{}m)\equal{}4(a^{2}\plus{}1)$
So there exist $ p,q$ such that
$ k\plus{}2a\minus{}m\equal{}p,k\plus{}2a\plus{}m\equal{}q$ where $ pq\equal{}4(a^{2}\plus{}1) and p(mod2)\equal{}q(mod 2)$
So $ k\equal{}\frac{p\plus{}q\minus{}2a}{2}$
Easy to check
$ p\plus{}q>2a$ so $ k>0$
Contine....

Exactly $ k\equal{}2a\plus{}p\plus{}q$, were $ a^{2}\plus{}1\equal{}pq$ and p is maximal divisor $ a^{2}\plus{}1$, suth that $ p\lе a.$
For example a=1 give $ p\equal{}1,q\equal{}2,k\equal{}5$.


[quote="iamagenius"]i'm currently reading ch. 4 of aops vol. 1 about proportions. the example in the starting paragraph is

$5\,\, apples = 39\,\, cents$.

then it says if $a$ is the number of apples and $c$ is the cost then $a/c=5/39$.

but shouldn't it be $a/c=39/5$ from the first equation?[/quote]

no because a is the number of apples....and according to ur statement there are 5 apples...not 39

[quote="mathamatix"]5 a=39c

dividing by 5

a=39c/5
dividing by c

a/c=39/5

as you said

or am i being totally dense?[/quote]
yeah thats what i thought too.

[quote="k8reindeer"]think about it logically for a second. If you have 5 apples for 39 cents, shouldn't each apple cost more than 1 cent? so cents per apple should be more than one. $\frac{a}{c}$ is apples per cent (not %) and it's the reciprocal of cents per apple. so it should be less than one.[/quote]
so whats wrong with the equations.

sorry

i misread the question i now agree with everyone else.

[quote="mathamatix"]5 a=39c

dividing by 5

a=39c/5
dividing by c

a/c=39/5

as you said

or am i being totally dense?[/quote]

You're multiplying by a and c. Those are your units.

ya i had something at first but then i thought it made sense because it seemed like he was trying to explain the apple to cents ratio but ya the more i think about the more it doesnt make sense..

here is how i see it,

if [b]5 apples[/b] cost [b]39 cents[/b] then:

[b]1 apple[/b] costs [b]7.8 cents[/b]

so the Total Cost would be:

$c=7.8a$, where $a$ is the number of apples and $c$ is the cost in cents

$\frac{1}{7.8}=\frac{a}{c}$

$\frac{1}{\frac{39}{5}}=\frac{a}{c}$

$\boxed{\frac{5}{39}=\frac{a}{c}}$

Here, how about a more explicit equation:

$1 dollar=100 cents$
If you treat those as variables:
$\frac{dollars}{cents}=\frac{100}{1}$
Suddenly, money seems more than it really is.

[quote=".:Logic:."][quote="rnwang2"]Hmm...I think the problem here is that

$5\,\,\text{apples}=39\,\,\text{cents}$

is not a valid equation, because it is really saying $5=39$.[/quote]

It is a valid equation, because you treat apples and cents as variables.

The equation $5a = 39c$ definitely works :)[/quote]
that's what i had. so from this equation a/c = 39/5.

but a/c = 5/39 according to the book.

[quote=".:Logic:."][quote="rnwang2"]Hmm...I think the problem here is that

$5\,\,\text{apples}=39\,\,\text{cents}$

is not a valid equation, because it is really saying $5=39$.[/quote]

It is a valid equation, because you treat apples and cents as variables.

The equation $5a = 39c$ definitely works :)[/quote]
But you can't just replace "apples" with the variable $a$ can you?

$a$ represents the number of apples.

Say $a=39$ (you buy 39 apples). That would cost 5 cents?

by saying 5 apples is equal to 39 cents, and then saying 5a=39c they are different things. in order for the first to be true, when you substitute 5 in for a, 39 needs to come out for c, which doesnt work with 5a=39c. a and c are units, not variables. (the equation would work if it were 5c=39a, whic gives $\frac{a}{c}=\frac{5}{39}$

The book is right. These guys know what they're talking about.

5 apples cost 39 cents. Therefore, one apple costs 39/5 cents. The number of apples is a, and the total cost is c. Therefore, cost/number of apples = 39/5 so the number of apples/cost = 5/39. Nowhere does it say "apples = a and  cents = c".


Can anybody plz.. post an article on [b][u]KVPY interview[/u][/b] it's urgent u c... :?:  :maybe:

you can feel the interview to be very easy if you are seriously preparing for IIT JEE and are almost halfway thru ur syllabus by now. anyway this year for mumbai we had Prof arvind kumar(director HBCSE),shri MA prasad who has actively been involved in training students for the olympiad and a sadist chemistry prof(a bald one){he was there for me too last year.

if ur basic concepts in physics,primarily mechanics are clear then AK wont cause much trouble.MA prasad asks simple and standard olympiad problems in combinatorics(such as those on PHP,invariance,parity etc.)
the chem one is really a sadist and he derives utmost pleasure out of proving u wrong.he asks routine questions on bonding (thats what he asked me).
and finally dont try to pretend as if u know something  u dont actually know,or else you are done.
concepts of work energy and rotational dynamics should suffice in physics.

THIS IS JUST TO GIVE YOU A ROUGH IDEA OF WHAT IT ACTUALLY IS THERE.THE ACTUAL INTERVIEW MAY BE DIFFERENT,WHO KNOWS.

I appeared for the the interview this year. Do you when the results are usually declared?

The results are generally out by the March or April.


With the problem you said, I got $\frac{813617 + 54241\sqrt{15}}{19609 + 2801\sqrt{7}}$. I'd give the full solution but I think you mistyped something somewhere. Take a look at the problem...


What is the smallest number that is divisible by four different primes?

Let $ n\equal{}p_{1}^{a_{1}}\cdots p_{r}^{a_{r}}$, where the $ p_{i}$ are distinct primes raised to the $ a_{i}$th power.  Our condition requires that $ r\equal{}4$ and since we want smallest, we want $ a_{i}\equal{}1$.  Then just pick the 4 smallest primes-2, 3, 5, 7.  So $ n\equal{}2\times 3\times 5\times 7\equal{}210$.

the 4 1st primes are

2,3,5,7

we multiply them all together

2*3*5*7=210

so 210 is the answer

rate my post please thanks !! :)


Can someone please outline the proof of Hess's law in particular the assumptions that go behind it probably by taking the example of a reaction?
also when it is written that for a compound $AB$  $\Delta_{f}(AB)^{0}$ is the heat released when the
compound $AB$ is formed from it's constituent elements in their standard states is it assumed that elements are in the same temp. as that of $AB$ :?:

The Hess law is based on the fact that the enthalpy is a state function, and so the value of its variation between two states depends only on that states and not on the path taken from one to the other.
By definiton, the standard enthalpy of formation for a given compound is variation in enthalpy that occurs when 1 mol of the compound forms from its elements (or their elementary substances) in their most stable form, with [i]all substances in question being in the standard state[/i].
The standard state corresponds to a temperature of 25ºC, a pressure of 1 bar for gases, a concentration of 1 M for solutions, pure state for solids and liquids, and the most stable alotrop at 25ºC for elementary substances.


a fair coin is tossed repeatedly until there's a run of an odd number of heads followed by a tail.determine d expected no. of tosses.

Game states:

I. First toss, number of consecutive heads before this toss is even, or previous toss was tails.
-$ \frac {1}{2}$ probability of landing on heads and transitioning to game state II.

-$ \frac {1}{2}$ probability of landing on tails, game state not changed.

II. Previous consecutive numbers of heads is odd.
-$ \frac {1}{2}$ probability of landing on heads and transitioning to game state I.

-$ \frac {1}{2}$ probability of landing on tails, ending the game.

Expected number of tosses to end the game = $ x$.

After 3 tosses:
$ x \equal{} \frac {1}{4}\cdot 2 \plus{} \frac {1}{2}\cdot (x \plus{} 2) \plus{} \frac {1}{8}\cdot (x \plus{} 3) \plus{} \frac {1}{8}\cdot 3$
EDIT> Note:
-1/4 chance game ends in 2 tosses.
-1/2 chance game state is reverted to I in 2 tosses. Expected number of tosses for this scenario is x+2.
-1/4 chance game state is II after 2 tosses.*

*This leads to:
-1/8 chance game ends in 3 tosses.
-1/8 chance game reverts to state I in 3 tosses. Expected number of tosses for this scenario is x+3.

$ x \equal{} \frac {5}{8}x \plus{} \frac {9}{4}$

$ x \equal{} \boxed{6}$


polynomial
quadratics
number theory unsolved
number theory

Find all complete numbers x,y for which:

x^2 + 3y^2= 1998x

Happy New Year!

Which is complete number?
This problem not very hard but we need an computer  :)

Whole numbers (i.e. integers).

Let $ z \equal{} x \minus{} 999$; the given can be written

$ z^2 \plus{} 3y^2 \equal{} 999^2$

Let $ \omega \equal{} \minus{} \frac {1}{2} \plus{} i \frac {\sqrt {3}}{2}$ be a primitive third root of unity; then we are working in the Eisenstein integers $ \mathbb{Z}[\omega]$.  Now, $ \sqrt { \minus{} 3} \equal{} 1 \plus{} 2 \omega$ so $ z \plus{} y \sqrt { \minus{} 3} \equal{} (z \plus{} y) \plus{} 2y \omega$.

$ z^2 \plus{} 3y^2 \equal{} 999^2$ is just the norm.  We therefore seek to factorize $ 999^2 \equal{} 3^6 37^2$ in the Eisenstein integers.

Now, $ \sqrt { \minus{} 3} \equal{} 1 \plus{} 2 \omega$, so $ 3 \equal{} (1 \plus{} 2 \omega)(1 \plus{} 2 \omega^2)$ and $ 3^6 \equal{} (1 \plus{} 2 \omega)^{6} (1 \plus{} 2 \omega^2)^{6}$.  So much for that.

$ 37 \equiv 1 \bmod 3$, so it is the product of two Eisenstein primes.  In particular,

$ 37 \equal{} 25 \plus{} 3 \cdot 4 \equal{} (5 \plus{} 2 \sqrt { \minus{} 3})(5 \minus{} 2 \sqrt { \minus{} 3}) \equal{} (7 \plus{} 4 \omega)(7 \plus{} 4 \omega^2)$

so that finally we have

$ ((z \plus{} y) \plus{} 2y \omega)((z \plus{} y) \plus{} 2y \omega^2) \equal{} (1 \plus{} 2 \omega)^6 (1 \plus{} 2 \omega^2)^6 (7 \plus{} 4 \omega)^2 (7 \plus{} 4 \omega^2)^2$.

The rest is computation.

Edit:  In particular,

$ (z \plus{} y) \plus{} 2y \omega \equal{} (1 \plus{} 2 \omega)^a (1 \plus{} 2 \omega^2)^{6 \minus{} a} (7 \plus{} 4 \omega)^b (7 \plus{} 4 \omega)^{2 \minus{} b}, a \equal{} 0, 1, 2, ... 6, b \equal{} 0, 1, 2$

appart from trivial solutions $ (0,0)$ and $ (1998,0)$ we can assume without loss of generality that $ x,y>0$. if $ (x,y)$ is a solution than obviously $ (x,\minus{}y)$ is to.

from $ x^2\plus{}3y^2\equal{}1998x$ we get $ z^2\plus{}3y^2\equal{}999^2$ ($ z$ is a natural number) by analising quadratic equasion. By divisibility of $ 3$ we can get an equivalent equation $ a^2\plus{}3b^2\equal{}37^2$ with $ z\equal{}27a$ and $ y\equal{}27b$

now analizing the equation $ a^2\plus{}3b^2\equal{}37^2$ by modulo $ 4$ we come to the conclusion that $ b$ must be even. therefore $ a^2\plus{}12c^2\equal{}37^2$. where $ b\equal{}2c$. obviously this inequallity must hold: $ c<\sqrt\frac{37^2}{12}<11$ therefore we must check only possibilities $ c\leq10$.

out of this possibilities only $ c\equal{}10$ gives a solution. (i hope i didnt miscalculate something. :lol: )
therefore $ a\equal{}13$ and $ b\equal{}20$ which yields $ y\equal{}540$ and $ x_{1,2}\equal{}648;1350$

all solutions therefore are:
$ (x,y)\equal{}(0,0);(1998,0);(648,540);(648,\minus{}540);(1350,540);(1350,\minus{}540)$


Does there exist a function f : N -> N such that f(f(n)) = n ?

Obviously I know how this is done (since Arne mentioned in F.E. 2 that this question comes from Singapore )... but I shouldn't say  :D

This is very nice! :D

We just try and build a function that verifies this functional eqs! :D :D
lets choose f(0)=0, f(1)=1 and the eqs is verified for n=0 and n=1. now lets choose f(2)=3 and f(3)=4 and the eqs is verified for n=2. now we choose f(4)=9 and put f(5)=10, f(6)=11, f(7)=12, f(8)=13,f(9)=f(f(4))=16 and the eqs is verified for n=3 and n=4. so we can see how to build this function:

1) if there exists n in {0,1,2,...,p} with f(n)=p+1 we chose f(p+1)=n^2
2) but if for any  n in {0,1,2,...,p} we have  f(n)<> p+1 we chose f(p+1)=f(p)+1.

all done!!
cheers!!!!! :D


it is very easy!!
Begin from memorizing
Multiply table
2x2=4
.....
....

[quote="vidz"]I really need help on doing division! I was really good when i was little but I suck know! Is there any easy way to do it?[/quote]

Example: $12 \div 4$

well the way to think in your head is to think: [i]What number multiplies with 4 to get 12?[/i] eventually you'l start to see easy patterns and then you can memorize the easy ones...it gets a tiny bit harder when you move on up to long division...

on another note this should be in classroom math

[quote="7h3.D3m0n.117"][quote="usaha"]What type of division:Dividing polynomials?[/quote]

ah upon reading the complete OP i realized he sucks now but he was good before........and here I thought I was responding to a fourth grader...no offense...

if dividing by polynomials here are some tips

1. Factor
2. try to avoid signs like $+$ or $-$ as they will affect your division...

for example

$\frac{(x+1)(x+6)}{(x+6)}$

the whole problem is multiplication and you can cancel out the $x+6$.[/quote]
um, no I think he means numerical division.

[quote="bpms"]
um, no I think he means numerical division.[/quote]

yeah then this should be in classroom math...the only reason why i thought it was something harder was what usaha wrote and why it could be in a high school forum

learning divison in high school?
I wish I was then I could slack off
http://en.wikipedia.org/wiki/Division_(mathematics)

[quote="7h3.D3m0n.117"]for example

$\frac{(x+1)(x+6)}{(x+6)}$

the whole problem is multiplication and you can cancel out the $x+6$.[/quote]

Incorrect, you still need a domain restriction of $x\not=-6$


i am taking calculus ab this year and plan on taking the ap test in may and i was wondering based on your experience or what you have heard, which one does a better job of preparing someone for the ap test?

Thanks in Advance

Tokenadult, I'm moving this to College, but I'm not sure this is the right forum.

This is a fine question for right here.

I don't have an opinion about AP calculus AB review books, but I hope someone who does have an opinion will speak up.

The Princeton Review is supposed to be better than Barrons for Calculus BC, so I'd assume that the same goes for AB.

i think for barron's its actually the same book for both BC and AB (its combined)


I found a problem, but a relative fast algorithm is needed. It goes as follows: I have been given a number $A\in [100,10000]$. Determine the smallest positive integer $B$ such that the digits of $AB$ are either in increasing or decreasing order. Brute force seems to be too slow for some of those integers.

Hm, I'd say it looks exactly like one of those brute force problems - perhaps some optimization would help?

Also one question about the problem statement - are the resulting digits supposed to be in [i]strictly[/i] increasing / decreasing order or not?

[hide="not brute force"]
There are at most $9$ digits in the product $AB$. Now, assuming that non-decreasing and non-increasing is okay, you can take all such numbers (if it isn't okay, there are less numbers). There are $2\binom{18}{9}$ such numbers, which can be seen if you digits as one type of object and increases (or decreases) in digit as another type of object, and simply arrange them in an arbitrary manner, then multiply by two because we have two cases (either increasing or decreasing). Actually we are off by a little bit due to double-counting and leading zeroes, but we have the same order of magnitude at the very least. This gives us about $100,000$ such numbers. Now just go through the divisors of each of them, and you are good to go. (The number of divisors of a number is bounded by $\log(n)$, so this is $O(100,000\log(100,000)) = O(2,000,000)$ which is fine.
[/hide]

Yes, this would work if the sequence has to be strictly monotonous.

The sequence does not have to be strictly monotonous.

I've just written a program which works reasonably instantly for all n in that range. Its basically just a DP.

Let f(m,k,c) be true or false depending on whether there is a c-digit increasing-digit number congruent to m mod A with first digit >= k.
Then f(m,k,c) is true if f((m-i*10^(c-1)) mod A, i, c-1) is true for some k<=i<=9, where you have i as the first digit.
I precalculate the powers of 10 mod A, up to some reasonably large number (I don't know how big these numbers can get, but I couldn't find much that didn't end up with a 10 digit product or more. I chose 100 digits maximum to be safe!)
Base case is c=0, which is true only if m=0.
Calculate the others from c=1 up until you find f(0,k,c) is true for some k,c. You can then do the typical reverse-the-dp to find the solution.

Repeat that for decreasing numbers, and choose the best answer.

Could you send me the program? I have not yet studied computer science very much and I don't know what is DP.

This prints out the smallest increasing-digit multiple of the input. I'll leave the other half to you.
[code]/*ans[m][k][c] = exists a c-digit increasingdigit number congruent to m mod N with first digit >= k*/
#include <iostream>
#define NUMDIG 100
int main() {
    int N;
    while (true) {
        scanf("%d",&N);
        static bool ans[10000][10][NUMDIG];
        for (int i=1; i<N; i++)
        for (int j=0; j<10; j++) ans[i][j][0]=false;
        for (int j=0; j<10; j++) ans[0][j][0]=true;

        int pows[NUMDIG];
        pows[0]=1;
        int gc, gm, gk;
        for (int i=1; i<NUMDIG; i++) pows[i] = (pows[i-1]*10)%N;
        for (int c=1; c<NUMDIG; c++) {
            for (int m=0; m<N; m++) {
                for (int k=1; k<=9; k++) {
                    ans[m][k][c] = false;
                    for (int i=k; i<=9; i++) {
                        // first digit is i
                        if (ans[((m-i*pows[c-1])%N+N)%N][i][c-1]) {
                            ans[m][k][c]=true;
                            if (m==0) {
                                // found a solution
                                gc=c;
                                gm=m;
                                gk=k;
                                goto out;
                            }
                            break;
                        }
                    }
                }
            }
        }
        out:
        while (gc>0) {
            for (int i=gk; i<=9; i++) {
                if (ans[((gm-i*pows[gc-1])%N+N)%N][i][gc-1]) {
                    printf("%d",i);
                    gm = ((gm-i*pows[gc-1])%N+N)%N;
                    gk = i;
                    gc--;
                    break;
                }
            }
        }
        printf("\n");
    }
}[/code]


More info needed here: what units is $t$ in? I'll assume it is months in this solution.

So we have $f(t) = Ab^{t}$ representing the number of thefts each month. Given in the problem is that $2f(t) = f(t+4)$. So $2Ab^{t}= Ab^{t+4}$. Cancelling out some things, $2 = b^{4}$. So $b = \sqrt[4]{2}$, and just get a calculator to find that to 2 decimal places.

[quote="sharkman"]The rate of television theft is doubling every four months. Determine to
two decimal places, the base b for an exponential model y = Abt of the rate
of television thefts as a function of time[/quote]
take the derivative


algebra
polynomial
number theory unsolved
number theory

Suppose that $n\geq 2$ be an integer number and $P(x)=x^n+a_{n-1}x^{n-1}+...+a_1x+1$ be a polynomial with integer coefficient and for every $k$ ($k=1,2,...,n-1$)  $a_k=a_{n-k}$. Prove that there exist infinite ordered pair $(x,y)$ in integer number s.t. $x\mid P(y)$ and $y\mid P(x)$

if you take $a_k=C_n^k$ the polinomial would be $P(x)=(x+1)^n$ and $y=x+1$ and things are good i think...


projective geometry
geometry
angle bisector
geometry proposed

[b][size=100][color=DarkBlue]A quadrilateral $ ABCD$ is given and let $ P,\ Q$ be, two points on its side-segments $ BC,\ AD$ respectively, such that $ \frac {BP}{PC} \equal{} \frac {AQ}{QD} \equal{} \frac {AB}{CD}$. Let $ P'$ be, the harmonic conjugate of $ P,$ with respect to $ B,\ C$ and also let $ Q'$ be, the harmonic conjugate of $ Q,$ with respect to $ A,\ D.$ Prove that $ P'Q'\perp PQ$.[/color][/size][/b]

Kostas Vittas.

PS. This result has been mentioned by [b]yetti[/b] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=116208][b][color=red]Here[/color][/b][/url].

If we take the points $ X,Y$ on $ BD,AC$ such that $ \frac {_{\overline{BP}}}{^{\overline{PC}}} \equal{} \frac {_{\overline{BX}}}{^{\overline{XD}}} \equal{} \frac {_{\overline{AY}}}{^{\overline{YC}}} \equal{} \frac {_{\overline{AQ}}}{^{\overline{QD}}},$ then $ PXQY$ is a rhombus.  Taking the harmonic conjugates $ P',Q',X',Y'$ of $ P,Q,X,Y$ in the respective segments, we figure out easily that the homologous lines are parallel to each other and we still get analogous relations:

$ \frac {\overline{P'Y'}}{\overline{AB}} \equal{} \frac {\overline{P'C}}{\overline{BC}} \ , \ \frac {\overline{P'X'}}{\overline{CD}} \equal{} \frac {\overline{P'B}}{\overline{BC}} \ \Longrightarrow$

$ \frac {\overline{P'Y'}}{\overline{P'X'}}\cdot \frac {\overline{CD}}{\overline{AB}} \equal{} \frac {\overline{P'C}}{\overline{P'B}}$ $ \Longrightarrow \ \overline{P'X'} \equal{} \overline{P'Y'} \ \Longrightarrow$ $ P'X'Q'Y'$ is a rhombus.

Therefore, $ P'Q'$ is the angle bisector of $ \angle X'P'Y'$ whose sides are parallel to the sides of the supplementary angle defined by lines $ AB,DC.$ Hence, their bisectors are parallel as well and the conclusion follows.

I think that it is enough to post here only the schema, because of the proof is familiar to the one I posted in the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=116208][b][color=Blue]Parallels[/color][/b][/url] of [b]conejita[/b].

Kostas Vittas.


geometry
3D geometry
MIT
college
articles

$1367631$
What's special about this number?

Let your imagination run wild - yes, it is an integer, an odd integer
too, and is divisible by $3 \cdots$

Let's see who can come up with the most interesting property of this number.

It's once divisible by the sum of its digits...

First you notice many powers of 7 are  (For example $7^{1},7^{3},7^{4},7^{6}$ )  are Palindromic.

So this number $7^{9}=1367631$ happens to be palindromic number too.  Big deal.!

(In other words, it is palindromic, and so is its cube-root,  and cube-root of its cube root) !!!

[hide="Of course"](Of course, I am talking about writing the numbers in base 18 ("18", here  is written in decimal base)   , but that's  just a minor point :) ) [/hide]

The digits of 1367631 sum to 27 and it has $7^{2}$ prime factors

Okay here are some more! :)  (Hope one of them is what xxxyyyy had in  his mind)

[b]This  is a cube of a number, and still a cube of a number if you reverse its digits[/b]  : )

(Okay, no big deal for a number which happens to be palindromic, but still it sounds nice, not that many numbers have   this property)
[b]
The number is a perfect cube  not only in decimal system but also in Octal [/b] (Base 8 )  $1367631_{8}= 389017 =73^{3}$) (system – for those who write computer code)

[b]The number is a perfect cube in Hexadecimal (Base 16)  system[/b]. ($1357631_{16}= 20346423=273^{3}$

Okay: [b]The number is a perfect cube in any  base syste[/b]m (greater than 7)  :) :)

 (Remember in [b]base 18 it is also[/b] $7^{9}$)

Even more – It is a perfect cube in any base (even less than 7 – if you  allow the digits larger than the value of the base)

(That’s  someone above said: $1^{6}+3\cdot 1^{5}+6\cdot 1^{4}+7\cdot 1^{3}+6 \cdot1^{2}+3 \cdot 1+1 = 27$  is a perfect cube)
[hide]
I am sure  every one knows by now $(1+a+a^{2})^{3}= a^{6}+3a^{5}+\dots$ [/hide]

Getting back to base 10 systems …

[b]It is the smallest palindromic number which is a cube of a  composite number.[/b]
 (And in fact  not that many palindromic cubes are known which are not cubes of primes)

Numbers like 25 are rare, because you can write it as $25=5^{2}$ (using digits 2 and 5)

(Only  operations one can use are (+,-,*,/,^ and  parenthesis) if a number  can be written by  reusing it’s own   digits – It is really a special number!)

How many other numbers you can think which has this property??

Well , you guessed it [b]the number above has this property![/b]

We   have $(117-6)^{\frac{6+3}{3}}= 16367361$ (And this is true in any  base >7)

And last but not the least - For any physicist ....
[b]Look at this beautiful picture[/b]:

[url=http://www2.warwick.ac.uk/fac/sci/csc/images/wf111.jpg]link of the picture[/url]

Beautiful, Isn't it - Picture from MIT of eigenstate of a system with $1367631$ atoms: :)

From vwiki Article:
http://en.wikipedia.org/wiki/Metal-insulator_transition

Everything I was going to say has been said, but also, there are 7 digits, and there is a 7 in the middle of the number. Interesting.