I love all the TV shows that start with STAR and continue with a 4-letter word, such as Star Trek, Star Wars (not quite a TV show, but anyway :D), Star Gate, Star Ship (Troopers - again not a TV show :)).

My favourite character is Col. O'Neil in StarGate SG-1 and his line "for crying out loud" :D ...

My favourite ship: USS Enterprise NCC 1701-E :)

My favourite aliens: the Goa'uld, runner up the Borg ...

My favourite episode of all: epsiode 326-401 of StarTrek TNG (it's when the borg gets Picard and tries to destroy earth, the battle of Wolf 349)

What about you?

I am not really a sci-fi fan,sorry [img]http://community.the-underdogs.org/smiley/happy/spin2.gif[/img]
I am more into mystery/thriller/drama
I liked the Babylon 5 series,i watched every single episode,but no star xxxx series [img]http://community.the-underdogs.org/smiley/happy/multicolour.gif[/img]

I can't believe you left out the original Star Trek series!! Granted, it is a little before the time of most of the people here (myself included), but it is the genesis of all of these shows!! If you like Star Trek Voyager, you would probably like the episodes of the original. The special effects and acting were not as good - but the plot lines and the stories were. I still prefer Voyager, I guess because that was my generation - and I'm not as distracted by the outdated colors and hairstyles. lol

You also should remember the Twilight Zone (not really a "Star" but close enough :) ). That was definitely my favorite of all of these shows. The story ideas were so creative and so thought provoking! They make the X-files look like garbage. Try and find some of the old original episodes if you can. I garauntee you will like it.

So overall, my favorite show was the original Twilight Zone series, followed closely by Star Trek Voyager. The Star Wars movies would be higher in my opinion if the last two episodes had been written better. But I loved the first three.

other people ? :) I can't belive only two of us like the StarTrek series ...

gauss202, I saw all episodes of TNG, Voyager, half of TOS and DS9 (because I didn't like them too much) all episodes of StarGate SG-1, half of the OuterLimits, almost all of Xfiles and some of Twilight Zone :) and I still think that TNG was the best.

so there aren't any sci-fi fans on AoPS / MathLinks? I just can't belive this :(

well anyway I'll add to my first post that actually I have
* my favourite good aliens: the asgard (stargate)
* my favourite bad aliens: the goa'uld (stargate), the wraith (stargate), the borg (star trek), the remans (star trek)

I've only seen StarWars and Stargate SG-1 out of those...and a lil' bit of StarTrek in Sunday School, but I really like SG-1. I watch it whenever I can, which is not that often, but o well.

I love Star Trek!!!   :lol:
But the old generation is definately far superior to all- I like next generation too, though.  I never really liked Deep Space 9, or Voyager.  The new Enterprise isn't too bad.

I think my favorite aliens are the Q in next generation.

[quote="hmmm"]But the old generation is definately far superior to all - I think my favorite aliens are the Q in next generation.[/quote]Picard & Co. are the [b]best[/b]. And ah .... yes the Q :) :P ... you know the Q ends up being a Goa'uld in StarGate SG-1 series 7? :P (the actor that is) ...

What happened to DS9  :( . I havent seen much of voyager to tell, but I personally preffer the continuity of DS9 over TNG. TNG may have more great espisodes, but it also as I alot more horroible espisodes. Even then, ive only seen the first 4 seasons of DS9, and two whole seasons+a bunch of random episodes of TNG because purchasing the seasons cost alot  :P . I find TOS very dry and borning compared to the rest of them, and havnt seen any Voyager nor much enterprize, but from the little ive sen of thoes, I still think id like DS9 better.

And its not "The Q" its just Q.  ;)

As for other Sifi I watch starwars, though it isnt my favorite, and havnt seen much else out side of a few episodes of Stargate-Altantis.

You [b]gotta[/b] check out StarGate, the movie, and the series. They are extremely entertaining, especially when Jack O'Neil does his numbers :D I personally think that the Goa'uld (as depicted in the original series) are the most scarry races of bad aliens ever made.

I probably will someday, but I dislike watching shows with commercials and also dislike watching random shows(as apposed to watching shows cronilogically). And thus as I dont even know what sations play SG reruns, let alone when they play the first one, I havnt watched any of them  ;) . BTW, for any other of you trekkie fans, ditl.org is a great resource. Contains anything and everything you ever wished to know about all the branches of trek  :D .

I say that the original Star Trek was better than the ones now.  My parents have all the old ones and I just absolutly love them. I have never seen any of the other ones listed.

[quote="JBL"]My girlfriend and I have been systematically viewing Voyager on DVD -- we finished season 4 before winter break, but then she brought some TNG instead of season 5.  I have to say that I think I prefer TNG -- something about Picard, maybe.  Good stuff.[/quote]When you finished all of TNG, all of DS-9 you find Voyager interesting just because it's something new in StarTrek, just as I find Enterprise now. Can't wait to see how Capt. Archer creates the UPF (aka the Federation).

Sorry - I can't stand StarTrek - Next Generation, voyeger etc and all such sci-fi episodes (including lost in space -  space 1999 and what have you ...)

Believe me, I have  tried, really tried ... very  very hard to sit through a full episode  but couldn't watch the whole episode. Just couldn't , I mean I found it more boring than my (then infant) son  found  Ronald Reagan's State of the Union speechs (Never failed, he used to fall asleep)

Only exception  is Star Trek  (The original) series. I have seen every episode of it about an average of  16.78 times... and absolutely  love it.

 :lol:
P.S. ( Even my sons  - only Star Trek  they have seen is the original one - No I am not making it up)

BTW, Is any one out here  a Inuyasha  (spelling?) fan?    I  don't know much about it but my sons  were home in their winter break and now my  computer's  hard disk  has about 30 GB of   backup files of their Inuyasha   episodes in the form of hundreds of avi files...

V. These are, I believe Japanese  [url=http://www.wot-club.org.uk/Inuyasha/]Cartoons[/url] . At least that is what I think. I know very little Japanese and my sons know even less .. still  looks like its a rage in some of the US  colleges right now... :)

P.S. [quote] I'm not alone in the Universe[/quote]

Glad there are two of you out there... :D

[quote="Valentin Vornicu"][quote="fredoniahead"]I agree. Stargate is a great sci-fi show.  I liked Farscape too. Did you know that Ben Browder(Crichton from Farscape) and Claudia Black(Aeryn Sun from Farscape) are going to be in Stargate next season?. Claudia is going to be 5 or 6 episodes and Ben is going to be cast in the series.  I saw this on Sci-fi channel website.[/quote]Finally a SG-1 fan on the site :P :) I'm not alone in the Universe :D
Can you give me the link where you read the above? Thanks.

@ Gyan, never heard of that series (or anything similar to the name), but you might have a chance to find out stuff my searching on [url=http://www.google.com]Google[/url] and/or [url=http://www.imdb.com]IMDB[/url].[/quote]

try these links:  http://scifiweekly.com/issue402/news.html ; http://scifiweekly.com/issue403/letters.html

i am of the opinion that the star wars movies can NOT be put under one tag, as the originals were of SUCH higher quality, in so many ways, than the new ones. the plot's better, the graphics are more believable (considering that computer generated stuff is just... so overdone, you can tell it- in the new ones), the people acting are sexier... yeah. gotta love the old ones, man. but the new ones? well, minus the Yoda-scene, and the pod-race? there isn't much substance.

I think the new ones are just much much much better (graphically speaking) than the old ones. You cannot possibly compare them. I love the battle where all the Jedi get caught in the middle of a huge colloseum, and Yoda appears in a space craft with the clones saying "Around the survivours [b]a perimeter[/b] create" :) Boy I love the way the little guy talks :D

There is this toy that you can buy that is like a combination of a Magic 8 Ball and a Yoda doll... you ask him a question, press his hand, and he answers... in Yoda-speak. It's hilarious and mine is NEVER wrong. But I agree, Yoda ROCKS!

[quote="Valentin Vornicu"][quote="fredoniahead"]try these links:  http://scifiweekly.com/issue402/news.html ; http://scifiweekly.com/issue403/letters.html[/quote]Oh, so she's Vala! I didn't knew that (I don't watch Farscape, so I could not recognize her).

They will probaby pull-out some jokes, as they previously did with O'Neill being called in the first season MacGuiver :) :D a couple of times. Naming Prometheus Enterprise (which I'd loved! :D) and other things like that (mentioning StarWars to Teal'c) are a proof that the writers will definitely not stop without introducing a little bit of Farscape in SG-1 :P[/quote]

No, Claudia wasn't Vala. She played Aeryn Sun, a Peacekeeper soldier.

[quote="fredoniahead"]No, Claudia wasn't Vala. She played Aeryn Sun, a Peacekeeper soldier.[/quote]No, I think that you misunderstood what I wrote. Claudia's Vala in SG-1 :) I have no idea who's she in Farscape.

[quote=Summer1980][quote=MathLion11]bruh how did you find this[/quote]

secrets :)[/quote]

huh

2 year bump
(whatever that means)


geometry
geometric transformation
projective geometry
geometry proposed

5. In the plane is given a circle and a point $P$ which doesn’t lie on that circle. Find the locus of the orthocenters of the triangles $PAB$, where $AB$  is any diameter of the circle. $\clubsuit$

A, B, P are not collinear, or they would not make a triangle. Hence, P cannot be the circle center. Let PA, PB meet the given circle again at E, D, respectively. Since AB is a circle diameter, the angles $\angle AEB = \angle BDA = 90^\circ$ are both right, so that AD, BE are 2 altitudes of the triangle $\triangle ABP$ and their intersection H is the triangle ortocenter. The locus of H is the polar of the point P with respect to the given circle (obvious, see Yaglom's Geometric Transformation III, pp. 66-67).

[quote="yetti"] ... The locus of H is the polar of the point P with respect to the given circle (obvious, see Yaglom's Geometric Transformation III, pp. 66-67).[/quote][color=darkred][b][u]Lemma[/u] (it is very usefully).[/b] Let $A,C,B,D$ be four collinear points in this order. Denote the middlepoint $M$ of the segment $[AB]\ .$ Then $\frac{CA}{CB}=\frac{DA}{DB}\ ,$ i.e. the points $C,D$ are harmonical conjugate w.r.t. the points $A,B$ if and only if $DA\cdot DB=DC\cdot DM\ .$[/color]

[u]Proof.[/u] Define $A(a)$, $B(b)$, $C(c)$, $D(d)$, $M(m)$, where $2m=a+b$.

$1.\blacktriangleright \boxed {\frac{CA}{CB}=\frac{DA}{DB}}\Longleftrightarrow \frac{\overline {CA}}{\overline {CB}}=-\frac{\overline {DA}}{\overline {DB}}\Longleftrightarrow$

$\frac{a-c}{b-c}+\frac{a-d}{b-d}=0\Longleftrightarrow\boxed {(a+b)(c+d)=2(ab+cd)}\ .$

$2.\blacktriangleright \boxed {DA\cdot DB=DC\cdot DM}\Longleftrightarrow \overline {DA}\cdot \overline {DB}=\overline {DC}\cdot \overline {DM}\Longleftrightarrow$

$(a-d)(b-d)=(c-d)(m-d)\Longleftrightarrow2(a-d)(b-d)=(c-d)(a+b-2d)\Longleftrightarrow$

$\boxed {(a+b)(c+d)=2(ab+cd)}\ .$

[color=darkblue][b][u]Proof of the proposed problem.[/u][/b] Denote: the fixed circle $w$ with the its mobile diameter $AB\ ;$ the second intersections $E,F$ of the lines $PA$, $PB$ with the circle $w\ ;$ the intersection $H\in BE\cap AF\ ;$ the intersection $R\in PH\cap AB\ ;$ the intersections $U,V$ of the line $PH$ with the circle $w\ .$ Then the point $R$ is the middlepoint of the segment $[UV]$ and the quadrilateral $AEHR$ is cyclically. Therefore, $PU\cdot PV=PE\cdot PA=PH\cdot PR\ ,$ i.e. $PU\cdot PV=PH\cdot PR\ .$ Using the above lemma results that the point $H$ is the harmonical conjugate of the point $P$ w.r.t. the points $U,V\ ,$ i.e. the geometrical locus of the orthocentre $H$ for the triangle $PAB$ is the polar-line of the point $P$ w.r.t. the given circle $w\ .$[/color]


algebra
polynomial
function
inequalities
algebra proposed

Find all polynomial functions  with coefficients in ${\mathbb Z}$ ,  of form  $p(x)=ax^2+bx+c=a(x-x_1)(x-x_2) \; ,\; a > 0,$  such that
  [b]1)[/b]  $0 < x_1 < x_2 < 1$,
  [b]2)[/b] for every $x \in (0,1)$ the inequality  $|p(x)|\sqrt{x(1-x)} \le \frac{1}{6}$ is verified   .

The answer is maybe $p(x)=a\left(x^2-\frac{1}{2}\right)\ (0<a\leq \frac{2}{3})$.

That polynomial is certainly not in $\mathbb{Z}[x]$

Find all polynomial functions  with coefficients in $ {\mathbb Z}$ , ?

I forgot this condition. :blush:

The condition 2) can be written like

$3|P(x)|\sqrt{1-(2x-1)^2}\leq 1$

change variable $y=2x-1$ we get $3|P(\frac{y+1}{2})|\sqrt{1-y^2}\leq 1$

Now apply the following
Lemma: R complex polynomial with degree $\leq n-1$ such that
$\forall y \in [-1,1] ,\; \sqrt{1-y^2}|R(y)|\leq 1$
Then we have for any $|y|\leq 1 , \; |R(y)|\leq n$

With $R(y)=3|P(\frac{y+1}{2})|$

we get $   |P(x)|\leq 1$

might be not difficult to find all polynomial with coefficient in $Z$ such that
$|a(x-x_{1})(x-x_{2})|\leq 1$ on $[-1,1]$ and $0<x_{1}<x_{2}<1$


In triangle $ABC$ the smallest angle is $A$. Let $D$ be a point on the shorter arc $AB$ of the circumcircle of triangle $ABC$ (radius $R$). The perpendicular bisectors of $AB,AC$ intersect $AD$ at $M,N$. Lines $BM,CN$ meet at $T$. Show that $BT+CT\leq 2R$.

You can easily prove that $C,O,T$ and $B$ are cyclic.Then $O$ is midpoint of arc $BC$ so $BT+CT \leq OA+OB=2R$

[quote="barasawala"][quote="cefer"]You can easily prove that $C,O,T$ and $B$ are cyclic.[/quote]

Really? I can't see  :blush:[/quote]

$M$ and $N$ are on perpendicular bisectors of sides $AB$ and $AC$ so $AM=AN$ and $AN=NC$ .So
$\angle BTC=\angle NTM=180-\angle ANC-\angle BMN=2\angle NAC-2\angle MAB=2\angle BAC=\angle BOC$
$\Longrightarrow B,T,O,C$ cyclic.


For integers $a$ and $b$, how many ordered pairs $(a,b)$ are there such that $ab$ is a multiple of $a+b$ if:

1) $0\le a\le b\le100$

2) $0\le a\le b\le1000$

3) $0\le a\le b\le x$

[hide]

$a+b \geq ab$

this does not happen when a>2 and b>2, so we look at the cases where $a,b\in(0,1,2)$

this leaves

$(0,0)$
$(1,1)$
$(2,2)$

as the only ordered pair solutions for any $x \geq 2$
[/hide]

[hide="Hint"]Note that the condition is equivalent to:

$(a-n)(b-n) = n^{2}$

where $n$ is an integer.

Note that the highest possible value for $a$ occurs when $a=n^{2}+n$ (and $b=1+n$). Note that the number of solutions $(a,b)$ for a particular value of $n$ is just the number of divisors of $n^{2}$.[/hide]

[hide="Yes, but..."] When $n^{2}+n > x$ we can still have smaller divisors of $n$ that produces solutions, but not all divisors work. [/hide]

Does anybody have a solution? I haven't been able to come up with one yet...

Do you have a source or did you think this problem up?

I made this problem up and posted it to see if anybody could find a nice solution.


I see there is an empty subforum [hide=http://www.artofproblemsolving.com/Forum/index.php?f=213]here[/url]. Why is it still there? Why isn't it deleted?

It's not USAMTS season right now. It has been active before, and will be active again; it's best to just leave the forum in place between those times.


Prove or disprove that :

a)exists not constants functions such that $ f(\sin \ x)\plus{}f(\cos \ x)\equal{}a\ ,\ (\forall)x\in \mathbb{R}$ with $ a \in \mathbb{R}^*$

b)exists not constants functions such that $ f(\sin \ x)\plus{}f(\cos \ x)\equal{}x^k\ ,\ (\forall)x\in \mathbb{R}$ with $ k \in \mathbb{N}^*$

[quote="Pain rinnegan"]Prove or disprove that :

a)exists not constants functions such that $ f(\sin \ x) \plus{} f(\cos \ x) \equal{} a\ ,\ (\forall)x\in \mathbb{R}$ with $ a \in \mathbb{R}^*$

b)exists not constants functions such that $ f(\sin \ x) \plus{} f(\cos \ x) \equal{} x^k\ ,\ (\forall)x\in \mathbb{R}$ with $ k \in \mathbb{N}^*$[/quote]

[b]a)[/b]A counterexample: $ f(k)\equal{}ak^2$ satisfies the condition.

[b]b)[/b]We have $ sin (x\plus{}2n\pi)\plus{}cos (x\plus{}2n\pi)\equal{}sin x\plus{}cos x$, $ \Rightarrow x^k\equal{}(x\plus{}2n\pi)^k$ for all $ n$ and $ x$, impossible.


I have this rather straightforward difference equation, but for some reason, I got stuck.

a(n+2) - 2*a(n+1) + a(n) = n, a(0)=0, a(1)=2

so the fist part is simple, repeated root x=1, a(n) for the 1st part = (B*n+C)*(1)^n=B*n+C

The second part is n, of the form: n*(k(0) + k*(1)*n)= n*k(0) + (n^2)*k(1)

Substituting this into the first part:

k(0)*(n+2) + k(1)*(n+2)^2 - 2*[ k(0)*(n+1) + k(1)*(n+1)^2)] + k(0)*n + k(1)*n^2 = n

The problem is when I expand this, it is reduced to:

2*k(1)=0

I don't seem to be able to find k(0)...

I have also tried a different method:

B*n+C + (1/2)*n*(n+1) from which I got B=0, C=1, so a(n) = n+(1/2)*n*(n+1)=(1/2)*n*(3+n).

I had a completely different result in Maple.

Any help would be appreciated. Max

[/i]

$ a_{n\plus{}2}\plus{}2a_{n\plus{}1}\plus{}a_{n}\equal{}n$. Let $ b_{n}\equal{}a_{n\plus{}1}\minus{}a_{n}$. Then $ b_{0}\equal{}2$ and $ b_{n\plus{}1}\minus{}b_{n}\equal{}n$. From here $ b_{n}\equal{}\frac{n(n\minus{}1)}{2}\plus{}2$. And from here we  get $ a_{n}\equal{}2n\minus{}\frac{n(n\plus{}1)}{4}\plus{}\frac{n(n\plus{}1)(2n\plus{}1)}{12}$

Thanks for this!

I have just checked your solution in Maple and the simplified form is: (11/6)*n + 1/6*n^3

If you do it through "rsolve", the answer is: (7/2)*n - (1/2)*n^2 + (1/6)*n^3.

I have just tried another method and got: n - (1/2)*n^2 + (1/3)*n^3.

I have got the coefficients wrong again...

simplify(2*n-1/4*n*(n-1)+1/12*n*(n-1)*(2*n-1)) = 7/3*n-1/2*n^2+1/6*n^3

This answer only works with (n-1).

Thanks for your help! Max


Vietnam olympiad mathematics 2004.
                           Day1
1.solve the system:
$x^3+x(y-z)^2=2$
$y^3+y(z-x)^2=30$
$z^3+z(x-y)^2=16$
3.let $A={1,2,...,16}$,find $k$ is integer be smallest such that any subset of include $k$ elements of $A$ ,there exists two distinct elements a and b satisfy $a^2 +b^2$ is prime.
                            Day2
1.let ${x_n}$ n=1,2,… be given a sequence defined by $x_1=$1 and
                 $x_{n+1}=\frac{(2cos2a)x_n+cos^{2}a}{(2-2cos2a)x_n +2-cos2a}$
  for all $n$=1,2,…
where $a$ is real.
Find all $a$  so that sequence {$y_n$} such that
                                        $y_n=\sum_{k=1}^n \frac{1}{2x_k+1}$
 Has finite limit when $n$ goes to infinite.
2.let $x,y,z$ is positive real such that  $(x+y+z)^3=32xyz$
  Find the maximal and minimal values of expression:
                       $P=\frac{x^4+y^4+z^4}{(x+y+z)^4}$

2^n +1 does not have a prime divisor of 8k+7?
It is Vietnam TST 2003.
You can on mathlinks.ro

[quote="mathubu"]If it is possible can you give  pdf version containing second day? I wonder that where the question 2^n +1 does not have a prime divisor of 8k+7 is published?[/quote]

see http://www.mathlinks.ro/Forum/viewtopic.php?p=268397#p268397  :)


limit
number theory proposed
number theory

Let the sequence $u_n$ define by:
$u_1=u_2=1$
$u_{n+2}=u_{n+1}^2-\frac{1}{2}u_n$
Prove that $u_n$ converges and find $\lim_{n\to \infty}u_n$


group theory
abstract algebra
superior algebra
superior algebra unsolved

If k is a field, K is its algebraic closure and f is an automorphism of K leaving
k fixed show that every finite extension of F (the fixed field of f) is cyclic.

I think this works:

As $ K$ is algebraically closed, it is a normal extension of any of its subfields. $ F$ is also the fixed field of $ \langle f\rangle$, the cyclic group generated by $ f$. In other words, the group of $ K/F$ is cyclic. Now, if $ L$ is any extension of $ F$, consider the chain of extensions:

$ F < L < K$

$ \operatorname{Gal}(K/L)$ is a subgroup of $ \operatorname{Gal}(K/F)$ and $ \operatorname{Gal}(L/F)$ is a quotient of $ \operatorname{Gal}(K/F)$.

As a subgroup and a quotient group of a given cyclic group is cyclic, the result follows.

[quote="puuhikki"]$ F$ is also the fixed field of $ \langle f\rangle$[/quote]
Why?
[quote="puuhikki"]Now, if $ L$ is any extension of $ F$[/quote]
Should $ L$ be an arbitrary finite extension of $ F$ instead of arbitrary extension?

If $ F$ is the fixed field of $ \langle f\rangle$ and $ K/F$ is Galois, I think $ \langle f\rangle$ should be finite. Is it?

You do not need infinite Galois theory: just take any finite extension $ L/F$ and replace $ K$ by the normal hull of $ L$. owk


What integer is tripled when nine is added to three-fourths of it?

$ 3x\equal{}9\plus{}\frac{3x}{4}$
$ x\equal{}3\plus{}\frac{x}{4}$.

So here it's pretty easy to see that $ \boxed{x\equal{}4}$

Let n be the number.

We have $ 3x\equal{}9\plus{}\frac{3}{4}x$, and we can divide by 3 and move $ \frac{1}{4}x$ to the left side. That leaves us $ x\minus{}\frac{x}{4}\equal{}3$. 3/4 x is 3, so x is 4.


If $ S$ is an orthonormal basis for an n-dimensional inner product space, and if
$ (u)_S \equal{} (u_1,u_2,...,u_n)$ and $ (v)_S \equal{} (v_1,v_2,...,v_n)$ ($ (u)_S,(v)_S$ are coordinates vectors), prove that

(a) $ \parallel{}u\parallel{} \equal{} \sqrt {u_1^2 \plus{} u_2^2 \plus{} ... \plus{} u_n^2}$
(b)$ d(u,v)\equal{}\sqrt{(u_1\minus{}v_1)^2\plus{}(u_2\minus{}v_2)^2\plus{}...\plus{}(u_n\minus{}v_n)^2}$
(c) $ < u,v > \equal{} u_1v_1 \plus{} u_2v_2 \plus{} ... \plus{} u_nv_n$

Hi mathwizarddude,
if you have a book read it and else buy one.


geometry
rectangle
analytic geometry
arithmetic sequence
geometry proposed

Consider  a rectangle  $ABCD$  with  $AB=\alpha$ and $AD=\beta$. The point  $E$  on  $AD$  is such that  $\frac{AE}{ED}=\frac{1}{2}$  and  $O$ is the center  of the  reactangle. Let $M$  be any  point on the line $OE$  interior  to the  rectangle. Find  the necessary  and  sufficient  condition  on $\alpha$ and  $\beta$  that the four  distances  from  $M$  to  lines  $AD, AB, DC, BC$  in this order  form an arithmetic  progression.

[quote="Kondr"][quote="nickolas"]Let $M$  be any  point on the line $OM$[/quote]
I'm afraid this is wrong. Also the point E is never used. So should it be "on the line OE"?[/quote]

Yes you were right. Im sorry  for my typo...

Put the coordinate origin at the point A and positive x-, y-axes in the direction of the rays AB, AD. The equation of the line OE passing through the points $E = \left(0,\ \frac \beta 3\right),\ O = \left(\frac \alpha 2,\ \frac \beta 2\right)$ is

$y - \frac \beta 3 = y - y_E = \frac{y_O - y_E}{x_O - x_E}\ (x - x_E) = \frac{\beta}{3 \alpha} x,\ \ y = \frac \beta 3 \left(\frac x \alpha + 1\right)$

The distances of any point M = (x, y) in the rectangle ABCD from the lines AD, AB, DC, BC are $x,\ y,\ \beta - y,\ \alpha - x$. To form an arithmetic sequence, it is required that

$y - x = (\beta - y) - y,\ \ (\beta - y) - y = (\alpha - x) - (\beta - y )$

$3y - x = \beta,\ \ 3y - x = 2 \beta - \alpha$

$\alpha = \beta,\ \ y = \frac{x + \alpha}{3} = \frac \alpha 3 \left(\frac x \alpha + 1\right)$

It follows that for an arithmetic sequence, it is necessary and sufficient that the rectangle ABCD is a square and the point M lies on the line OE in this square.


Not Number Theory!

Define $\tau = \sqrt[3]{108+12 \cdot \sqrt{93}}$.
The solutions are (thanks to MAPLE):
$\frac{-\tau}{6}+\frac{2}{\tau}$
$\frac{\tau}{12}-\frac{1}{\tau}+i \cdot \frac{\sqrt{3}}{2} \cdot \left( \frac{-\tau}{6}-\frac{2}{\tau} \right)$
$\frac{\tau}{12}-\frac{1}{\tau}-i \cdot \frac{\sqrt{3}}{2} \cdot \left( \frac{-\tau}{6}-\frac{2}{\tau} \right)$


inequalities
trigonometry
inequalities proposed

Prove that if $a,b,c$ are sidelengths of a triangle, then the inequality \[\frac{a}{\sqrt{2b^{2}+2c^{2}-a^{2}}}+\frac{b}{\sqrt{2c^{2}+2a^{2}-b^{2}}}+\frac{c}{\sqrt{2a^{2}+2b^{2}-c^{2}}}\geq \sqrt{3}\] holds.

[quote="mateivld"]$\sqrt{\frac{a^{2}}{2b^{2}+2c^{2}-a^{2}}\cdot\frac{1}{3}}\geq\frac{a^{2}}{a^{2}+b^{2}+c^{2}}$(GM-HM) sum up and we get the result
 :)[/quote]
Nice proof. Thanks.
I saw an very ugly proof (written by M. Bjelica to show how you can use [i]Mathematica 4.0[/i] in Mathematics) and tried to prove the inequality in some nice way. In my solution I used Jensen for $\frac{a}{\sqrt{k-3a^{2}}}$ and got $a^{2}+b^{2}+c^{2}\leq 4(ab+bc+ca)$ which can be proved by cosine theorem.

Rzeszut, see also  the last post from here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=m_a+%2Afrac&t=56265


Both parts in one go:

[hide]
Using Taylor series, we must have:
$ f(x) \equal{} f(0) \plus{} \frac {f'(0)x}{1!} \minus{} \frac {f(0)x^2}{2!} \minus{} \frac {f'(0)x^3}{3!} \plus{} ...$
since $ f''(x) \equal{} \minus{}f(x) \Rightarrow f^{(n \plus{} 2)}(x) \equal{} \minus{}f^{(n)}(x)$.
Using initial conditions as given, for part A: $ h(x) \equal{} cos(x)$.
for part B: The same argument (except replacing all the minus signs with plus signs) gives $ f(x) \equal{} e^x$.
[/hide]

[quote="spinos"]A. if for the function h(x) is"

h"(x)+h(x)=0 for all $ x\in R$
    and h(0)=1 , h'(0)=0,
    show that h(x)=cosx

B. if for the function f(x) is

f"(x)=f(x), for all $ x\in R$
    and f(0)=f'(0)=1
    find the function f(x)?[/quote]

A.
The characteristic equation is $ r^2\plus{}1\equal{}0$ so the roots are $ \pm i$.  Then by euler's we see that the roots are in the form $ A\sin x\plus{}b\cos x$.  Now plug in the initial conditions.

B.
The characteritic equation is $ r^2\minus{}1\equal{}0$ so the roots are $ \pm 1$.  Then the roots of the equations are in the form $ Ae^x\plus{}be^{\minus{}x}$.  Then plug in the initial conditions.

Doing it just manually and basically,
$ h(0) \equal{} \cos(0) \equal{} 1$
$ h'(x) \equal{} \minus{}\sin(x)$, so we now have $ h'(0) \equal{} 0$.
$ h''(x) \equal{} \minus{}\cos(x)$
Then we can see $ h(x) \plus{} h''(x) \equal{}  \cos(x) \minus{} \cos(x) \equal{} 0$.

We can see that if $ f''(x) \equal{} f(x)$, then $ f'(x) \equal{} f(x)$ as well.

Therefore we have
$ \frac{dy}{dx} \equal{} y$

$ \frac{dy}{y} \equal{} dx$

Solving this (with the initial value of $ f(0) \equal{} 1$ yeilds $ y \equal{} f(x) \equal{} e^x$.

[quote="Fermat's Little Turtle"]Doing it just manually and basically,
$ h(0) \equal{} \cos(0) \equal{} 1$
$ h'(x) \equal{} \minus{} \sin(x)$, so we now have $ h'(0) \equal{} 0$.
$ h''(x) \equal{} \minus{} \cos(x)$
Then we can see $ h(x) \plus{} h''(x) \equal{} \cos(x) \minus{} \cos(x) \equal{} 0$.

[color=red]We can see that if $ f''(x) \equal{} f(x)$, then $ f'(x) \equal{} f(x)$ as well.[/color]

Therefore we have
$ \frac {dy}{dx} \equal{} y$

$ \frac {dy}{y} \equal{} dx$

Solving this (with the initial value of $ f(0) \equal{} 1$ yeilds $ y \equal{} f(x) \equal{} e^x$.[/quote]

how did you get the line in red?

[quote="Fermat's Little Turtle"]Doing it just manually and basically,
$ h(0) \equal{} \cos(0) \equal{} 1$
$ h'(x) \equal{} \minus{} \sin(x)$, so we now have $ h'(0) \equal{} 0$.
$ h''(x) \equal{} \minus{} \cos(x)$
Then we can see $ h(x) \plus{} h''(x) \equal{} \cos(x) \minus{} \cos(x) \equal{} 0$.[/quote]
That proves that cos(x) satisfies the given condition, not that it is the [i]unique[/i] function which does so.

[quote]
We can see that if $ f''(x) \equal{} f(x)$, then $ f'(x) \equal{} f(x)$ as well.

Therefore we have
$ \frac {dy}{dx} \equal{} y$

$ \frac {dy}{y} \equal{} dx$

Solving this (with the initial value of $ f(0) \equal{} 1$ yeilds $ y \equal{} f(x) \equal{} e^x$.[/quote]
As the above poster mentioned, this doesn't even work. See $ f(x)\equal{}e^x\plus{}e^{\minus{}x}$ for an easy counterexample that doesn't satisfy $ f'(x)\equal{}f(x)$ but [i]does[/i] satisfy $ f''(x)\equal{}f(x)$.


Find all functions $f: \mathbb{Q}\to\mathbb{R}$ which fulfill the following conditions:

a) $f(1)+1>0$;

b) $f(x+y) -xf(y) -yf(x) = f(x)f(y) -x-y +xy$, for all $x,y\in\mathbb{Q}$;

c) $f(x) = 2f(x+1) +x+2$, for every $x\in\mathbb{Q}$.

First note that $f(0) = 1$ and $f(1) = -\frac{1}{2}$
Let $g(x) = f(x) + x$
We now have $g(x+y) = g(x)g(y)$ and $g(x) = 2g(x+1)$. The first equation (Cauchy) has solution $g(x) = 0$ or $g(x) = a^x$. However, $g(x)\neq 0$, so $g(x) = a^x$. It's now easy to see that $a = \frac{1}{2}$

[quote=reveryu]Do we must have a condition; f is continuous , monotone,or bounded to use cauchy equation?[/quote]

No, because $f: \mathbb{Q}\to\mathbb{R}$.
These conditions are needed for  $f: \mathbb{R}\to\mathbb{R}$.

 I guess I did the same thing as warut_suk, but decided to take the long route and not invoke Cauchy's f.e :)

So let P(x,y) = f(x+y)-xf(y)-yf(x) = f(x)f(y)-x-y+xy.

P(0,0) implies that f(0) = $f(0)^2$, so f(0) = 0 or 1.

$f(0) = 0 \iff f(x) = - x$ from P(x,0), which contradicts with a).

For f(0) = 1, you use induction on c) get f(n) = $\frac{1}{2^n} -n$. Then you use induction on b) to get f(mn) = $(f(n)+n)^m -mn$. Then you generalize to the rationales like you do in the proof of Cauchy F.E, and you're done.

(So the solution is f(x) = $\frac{1}{2^x} -x$ for $f: \mathbb{Q}\to\mathbb{R}$.

[quote=reveryu][quote=RagvaloD]No, because $f: \mathbb{Q}\to\mathbb{R}$.
These conditions are needed for $f: \mathbb{R}\to\mathbb{R}$.[/quote]
Thank you.
and does this function has range in $\mathbb{Q}$ also  ?[/quote]

Ofc nope :D $\frac{1}{2^r} \in \mathbb{Q}$ iff $r \in \mathbb{Z}$

 I am posting this behalf of my friend , he did this , but he  wants to know  did he miss any arguments.

here it is ,[hide]
Let $P(x,y),Q(x)$ be the assertion for $1,2$
$P(0,0) \Rightarrow f(0)=0$ or $f(0)=1$
If $f(0)=0$ then $Q(0) \Rightarrow f(1)=-1$ which contradicts $3$. so, $f(0)=1$
Now solving the recurrence $a_n =2a_{n+1}+n+2$ with $a_0 =1$ gives $a_n = \frac{1}{2^n} -n$(Recurrence solved by wolfram :mrgreen: but you can solve with generating functions or just by a non trivial guessing)
(Note: I didn't checked it please somebody check it)
Doubtful about this arguement: Now since the previous equation is satisfied for $\mathbb{Z}$ which is a sub set of $\mathbb{Q}$ so it follows all $x \in \mathbb{Q}$ also must follow $f(x)= \frac{1}{2^x} -x$ $\forall x \in \mathbb{Q}$ since the the $2$ is only satisfied by that for integers.[\hide]


[quote="tuandokim"]a,b,c>0
prove that:
$ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} abc}{4abc}\geq {(\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca})}^2 \plus{} \frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{8(ab \plus{} bc \plus{} ca)}$[/quote]
I haven't seen it before, so it might be new. :)
Anyways. It's pretty nice i think :) (And I have a very short proof :P ) There's equality iff $ a\equal{}b\equal{}c$..


Could someone please tell me how to know what conic section is being represented if given only the general formula?

Ah!  Thank you very much t0rajir0u.  I had a feeling that you would be my savior in learning this.  And what if I was asked to find the directrix, major axis, minor axis, foci, or eccentricity (for ellipse and hyperbola) when only given the general formula?

I'm not sure.  The difficult case is when $ b \neq 0$.  I would complete the square and perform an appropriate substitution to make $ b \equal{} 0$, and then the rest is straightforward.


geometry
MATHCOUNTS
USAMTS
geometric transformation
reflection
calculus
AMC

--Announcements for 8/24/03--

The [url=http://www.olemiss.edu/mathed/problem.htm]Ole Miss Problem of the Week[/url]is scheduled to resume on September 8th, 2003.  There are five problems posted each Sunday night or Monday morning from the five categories:
* Algebra in Action
* Middle School Madness
* Problem of the Week
* Elementary Brain Teaser
* Geometry Gambit

Students who submit the correct answer will, if they choose, have their name, school, and location published on the website.

--

In addition, the [url=http://mathforum.org/pow/]Math Forum @ Drexel Problem of the Week[/url]will be back on Monday, August 25th and Monday, September 1st, 2003.
The Geometry and Pre-Algebra Problems of the Week will resume on the former date, while the Math Fundamentals and Algebra problems will resume on the latter one.

Students who submit correct solutions may elect to have their names, schools, and ages published on the website; the best solutions will be displayed on the Solution to each problem.  [url=http://mathforum.org/calcpow/solutions/solution.ehtml?puzzle=75]Here is one of Simon's solutions[/url], for example.

--

As always, the MATHCOUNTS website has a Problem of the Week for you.  [url=http://www.mathcounts.org/Problems/2003Competitions.html]Here.[/url]

--

In the future, I will continue to post news on various math competitions online.  This information will generally come from the AMC, Mathcounts, Mandelbrot, USAMTS, etc. websites.

Whoa, that's really something!  Thanks for the interesting info!

-interesting_move

--News for 9/10/03--

Iiiiiiits official!

The Ole Miss Problem of the Week resumed this monday.  You may submit answers to problems from each of these categories:
* Algebra in Action
* Middle School Madness
* Problem of the Week
* Elementary Brain Teaser
* Geometry Gambit

Students who submit the correct answer will, if they choose, have their name, school, and location published on the website.
[url]http://www.olemiss.edu/mathed/problem.htm[/url]
--

The Math Forum @ Drexel Problem of the Week is also back.  The There are four categories:
* Geometry
* Pre-Algebra
* Math Fundamentals
* Algebra

Students who submit correct solutions may elect to have their names, schools, and ages published on the website; the best solutions will be displayed on the Solution to each problem.  [url]http://mathforum.org/pow/powstatus.html[/url]

--

As of 8/21/03, the Mandelbrot competition's website has been updated to reflect any changes for the 2003-2004 school year.  Online registration for the coming year is available, as are several sample contests, resources, etc.  [url]http://www.mandelbrot.org[/url]

--

The address and fax number for the USAMTS have not yet been announced.  You still have a little less than a month to complete your solutions and get them in.  [url]http://www.nsa.gov/programs/mepp/usamts.html[/url]

Enjoy!

--News for 9/15/03--

Blair Kelly, USAMTS Director, has updated the USAMTS [url=http://www.nsa.gov/programs/mepp/usamts.html]homepage[/url] to include new, year 15 cover sheets, entry forms, and the address/fax number that you should send solutions to.

- [url=http://www.nsa.gov/programs/mepp/usamts/mail.html]Where to mail/fax solutions[/url].
- [url=http://www.nsa.gov/programs/mepp/usamts/entry_form.pdf]Entry Form[/url]
- [url=http://www.nsa.gov/programs/mepp/usamts/cover_sheet.pdf]Cover Sheet[/url]

Enjoy!

--News for 10/8/03--

[url]http://gottagetyourown.com/[/url].  Fascinating.

This year, the [url=http://www.olemiss.edu/mathed/problem.htm]Ole Miss Problem of the Week Contest[/url] is being sponsored by Texas Instruments.  What does this mean?

It means that you now have the change to win a Ti-73, TI-83+ Silver Edition, or a TI-89!  Just submit a correct solution, and your name will be added to a list of correct answerers.  If your name is drawn in the random monthly drawing, you win a calculator!  Prizes are as follows:
Ti-73 for [url=http://www.olemiss.edu/mathed/brain/]Elementary Brain Teaser[/url]
Ti-73 for [url=http://www.olemiss.edu/mathed/middle/]Middle School Madness[/url]
Ti-83+SE for [url=http://www.olemiss.edu/mathed/algebra/]Algebra in Action[/url]
Ti-83+SE for [url=http://www.olemiss.edu/mathed/geometry/]Geometry Gambit[/url]
Ti-89 for [url=http://www.olemiss.edu/mathed/pow/]Problem of the Week[/url].

Enjoy!

--News for 10/27/03--

Hey you!  Yeah, you!

If you're in middle school, then try the Dositey.com Math Challenge at [url]http://www.dositey.com/problems/mathchall58.htm[/url].

Or try the ABACUS Internet Math Challenge at [url]http://www.gcschool.org/pages/program/Abacus.html[/url].  There are also years worth of past problems, so you'll have plenty of material to study for other stuff, too.

Enjoy!

[Note: cross-posted in the mathcounts board, since this is most pertinent to middle schoolers.]

--News for 11/8/03--

Check out [url]http://www.theproblemsite.com/[/url].  This site features several problems of the week, including [url=http://www.theproblemsite.com/mathhs/]Graeme's Problem of the Week[/url], [url=http://www.theproblemsite.com/maml/]The Maine Page[/url], and [url=http://www.theproblemsite.com/calculus/]Mr. T's Calculus Problem of the Month[/url]

Enjoy!

--News for 11/26--

USAMTS Year 15, Round 3 problems are now up.  View them at [url]http://www.nsa.gov/programs/mepp/usamts/problems/problems15R3.pdf[/url].  Or, [url=http://www.nsa.gov/programs/mepp/usamts/problems/problems15R3.pdf]click here for the main USAMTS website[/url].

Students should be recieving their results from Year 15, Round 1 in the mail shortly.  (I got mine today.)  Year 15, Round 2 solutions should have been sent in last week as the deadline was 11/23.

The [url=http://www.olemiss.edu/mathed/problem.htm]Ole Miss Problem of the Week[/url] is on vacation until November 30th.

Oh, and [url=http://www.artofproblemsolving.com]AoPS[/url] is changing/has changed servers right now.

I did win a TI-83 silver from OLE MISS. It's more like a contest of luck. THe problems are easy and there are about 200 submissions so your chance of winning is slim to nont

-- News for 4/16/04 --

You may have heard of KoMaL, but did you know that it is online and you can participate for free?  Well, go to [url]http://www.komal.hu/info/bemutatkozas.e.shtml[/url].


Given non-constant linear functions $p(x), q(x), r(x)$. Prove that at least one of three trinomials  $pq+r, pr+q, qr+p$ has a real root.

[b]proposed by S. Berlov[/b]

This is, of course, a particular case of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20664]this problem[/url].


Generally (other than rational roots theorem and stuff) what are the ways to find the roots of a cubic, quartic, etc., equation?
 :huh:

Generally, I think it's much more useful to have information about the symmetric sums of the roots, aka Vieta's formula.

[quote="t0rajir0u"]The general methods are useless.

http://planetmath.org/encyclopedia/CubicFormula.html
http://planetmath.org/encyclopedia/QuarticFormula.html[/quote]

I wonder if there are contests that make you memorize these rather than pi.

All I know is that for equations of degree 5 and higher, there is [u]no[/u] formula that uses only the four basic operations, exponentiation, and roots that can be used to find all the roots of that equation.


yes thats very true mewto.KEEP PLAYING JASON SULLIVAN. DON'T REVEAL YOUR NAME!

Thanks for your support. :)

And that's not my name. nowhere near.  :P

[quote="mewto55555"][quote="JasonSullivan"]Hi,

I've just stumbled on this site and it looks AMAZING, I mean I love maths, really into it and I'm good at it in school etc. Just finished A levels (is this an american site? If so then that means I'm 18, just left high school etc). But wow you guys are fast! I just can't keep up in FTW! I freeze. Was anyone else like this when they first started or is it just me?

 :([/quote]

Hm, once you get used to it, you become faster and know were the numbers are. Also, there are a lmited number of questions on the site, so you begin to now the answer based on the first sentence, a phenomenon known as "memming"[/quote]
but memming is not reccommended  :P
sometimes you'll slip...

[quote="isabella2296"]I know of very few people(actually one) who actually mem [i]on purpose[/i]. What often happens is that you see the problem so many times, or you answer it or read the solution in the forum, and you suddenly realize that you know the answer upon seeing the problem! There are only so many problems on FTW, so naturally you will see quite a few repeated as you play more and more games.[/quote]

Hm the italicized portion makes one wonder whom izzy was referring to....

right okay. Is mem in reference to 'memory'?

And when exactly is chat allowed? merely from the point of view of replying to people who say hello.

Are the majority of ppl here American?

Yes, when most people on here say "mem," they mean they remembered the answer to the problem and had to do no work to get the answer.

It hasn't been divulged specifically what needs to be done to chat.  You need a sufficient number of games played and a sufficient number of productive posts.  Some people have said it's 50 games, but idk.  I don't know how many productive posts (if any) are needed, either.

Yes, the majority of people here are American.

Actually, people do come from all around the world but most are american or asian. :)

I don't know if this will really help anyone, but you do not NEED posts to be able to chat, although helping out isn't a bad thing (and will get people's attention more easily so they can unmute you).

Hmm...
I believe there was a little script that popped out saying in order to chat, you must have played a certain number of games and have had a certain number of productive posts on the AoPS forum or something like that.

[quote="isabella2296"]If you know the person, you'll know whom I'm talking about. This certain person mems glitch problems. Oh, and Mewto, no need to quote the entire post directly before you.

However, what I was saying is, it's not bad to mem the way most people do - you can't help it. I suppose you could wait 3 seconds before answering or whatever, but it doesn't make much of a difference.[/quote]

And that's why you get sub-2 mems :D

And are you referencing to me?  (earlier post)


Determine if the series converge

$ \sum_{n\equal{}1}^{\infty} \frac{(1 \plus{} sin (n^2))}{ \sqrt n}$

$ \sum_{n\equal{}1}^{\infty} \frac{(\frac{1}{3} \plus{} \frac{2}{3} sin(n))^n}{n}$

$ \sum_{n\equal{}1}^{\infty} \frac{(\frac{2}{3} \plus{} \frac{1}{3} sin(n))^n}{n}$

$ \sum_{n\equal{}1}^{\infty} \frac{(\frac{1}{3} \plus{} \frac{2}{3} sin(n))^n}{n^n}$

These are rather different from each other. The first is annoying, but obviously comparable to $ \sum \frac1{\sqrt{n}}$ and divergent. The last converges absolutely very rapidly. The other two are very hard, and are addressed in this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22093]legendary thread[/url]. They converge, but it requires a complicated and advanced argument to show that there aren't too many "big" terms.


geometry
algebra
polynomial
number theory
factoring polynomials
prime factorization
group theory

[quote="easyas3.14159..."]Why shouldn't it be? It has 2 factors, itself and 1.[/quote]

The answer to this question goes beyong ordinary middle school classroom math.

The fact is that primes are defined in different ways at different times.  When negative numbers come into play, 1 and -1 are considered units, while primes are integers like 2, -2, 3, -3, etc. that have 4 total divisors (positive and negative included).

Fortunately, there is no need to worry about this distinction early in your math education.  Unless you research into particular areas of number theory, you won't need to distinguish between these definitions of primes for a few years, if ever.

Though factorization becomes nonunique, the number of possibilities is still measurably limited.  This takes nothing substantive away from machinery of number theory.

I'm going to move this thread to a higher forum at this point.

When one gets around to generalizing the integers to a more general algebraic system (called a "ring"), we see the role that both $1$ and $-1$ play: they're called "units." In particular, $1$ and $-1$ are the only integers that have multiplicative inverses.

Given any nonzero integer, we can write it as a product of units times a product of primes. The list of primes (counting repetitions) in that factorization is unique; the list of units is not. We can include an arbitrary number of $1$'s in the factorization, and either an arbitrary even number or an arbitrary odd number of $-1$'s.

Consider the notion of factoring polynomials (say, over the rationals). Here, the units are the nonzero constants. And we would tend to see $(2x-1)$ and $\left(x-\frac12\right)$ as "essentially" the same factor. We could get from one to the other by multiplying by a unit.

[quote="keeyounglee"]A prime number (or prime integer, often simply called a "prime" for short) is a positive integer $p>1$ that has no positive integer divisors other than $1$ and $p$ itself. (More concisely, a prime number $p$ is a positive integer having exactly one positive divisor other than $1$.) Positive integers other than $1$ which are not prime are called composite numbers.

according to mathworld, a prime number is an integer greater than 1.[/quote]

That's not the only accepted definition.  There are branches of number theory that define primes in other ways, including negative primes.  In fact, I'm pretty surprised if mathworld doesn't mention that, because there are plenty of undergraduate and graduate classes that use that alternate definition of primes as a starting point.

[quote="10000th User"]But -1 is also divisible by $(-1)^n$, where $n$ is any positive odd integer, making factorization not unique and number theory will be boring :lol:[/quote]

LOL.  Of course, for those who actually take everything seriously, you know that $(-1)^n$ only results in either $-1$ or $1$.  You get the idea.

$-1$ and $1$ are not considered primes or non-primes, they're a special case. However, you can use it for prime factorisation: $17=1*17$.

[quote="t0rajir0u"]MCrawford and Kent Merryfield are correct.  The usual definition of primes is $\in \mathbb{N}$ - the natural numbers, $1, 2, 3, 4, ...$  And $-1$ is not in that set.

If you ask whether $-1$ is a prime in, say, $\mathbb{Z}$, which is $..., -3, -2, -1, 0, 1, 2, 3, ...$, then the answer is the same as it is for $1$.

[i]No[/i].  $-1$, like $1$, is unique.  It is a unit.[/quote]Not only that, there are even discrepancies on whether $0\in\mathbb{N}$ or not.

[quote="236factorial"]I don't know if this is a stupid question or not, but how would you prime factorize negative numbers like -144? There would have to be an odd number of negative prime numbers.[/quote]

Maybe like this: $-144=(-1)*144=(-1)*2^4*3^2$

[quote="t0rajir0u"]Unique prime factorization does not hold $\in \mathbb{Z}$.[/quote]

Right.... so there isn't a unique prime factorization for integrers numbers $k$, with $-1 \geq k$, and you can factorize it as you want, because there's no only one...............................................................

[quote="José"][quote="t0rajir0u"]Unique prime factorization does not hold $\in \mathbb{Z}$.[/quote]

Right.... so there isn't a unique prime factorization for numbers $k$, with $-1 \geq k$, and you can factorize it as you want, because there's no only one.....[/quote]

[i]Unique[/i] means that there is only [i]one[/i].

$-144 = (-3) \cdot (2)^{4} \cdot (3)^{1}$
$-144 = (-2) \cdot (2)^{3} \cdot (3)^{2}$

Here is a good definition of primality from group theory:  An element $p$ is prime iff $p | ab \implies p | a \vee p | b$.  For example, $5 | 20 \implies 5 | 10 \vee 5 | 2$.  However, $12 | 36$, which does not imply $12 | 4 \vee 12 | 9$.

Divisibility is still establishable $\in \mathbb{Z}$, and using this definition we can show $-2, -3, ...$ are primes $\in \mathbb{Z}$.

The conclusion we reach is that [i]there is more than one way to factorize a number $\in \mathbb{Z}$ into a product of primes.[/i]

[quote="t0rajir0u"][quote="José"][quote="t0rajir0u"][i]Unique[/i] means that there is only [i]one[/i].[/quote]

What I said.[/quote]

Oh.  You weren't being sarcastic?   :blush:[/quote]

Nonono, I'm sorry if I sounded sarcastic, I just wanted to help you...


I don't believe you're allowed to take both AIME's, but I could be wrong.

In theory, the AIME II is the same difficulty as the AIME I, but so far in practice, it's been more difficult. However, that's no guarantee for this year. In fact, it could be unintentionally easier.

[quote="123s"]My school has to take the AIME I, because the AIME II is during our vacation time. Is it possible for me to take the AIME II at a different school? Or do they only send a certain amount of tests according to how many qualified from each school?[/quote]You can only take one or the other, not both.

The AMC sends out exactly the number of tests necessary so that each qualifier gets 1 test.  No school will have "extra" copies of the AIME for extra students.

The majority of the people say the alternate AIME is much harder, but I for one, have performed much better on the AIME 2s.

[quote="123s"]Another question. Is it harder to qualify for the USAMO if you take the AIME ii?[/quote]In theory, no.  If it turns out that one version of the AIME is significantly harder than the other, then the AMC can use different indices for the students who took the different tests (and in fact did so last year).

No... I don't mean take both of them. What I mean is not take the AIME i, and instead take the AIME ii at a different school. If they only send a certain number of AIME tests to each school, then why is it possible that some students at one school take the AIME i, while others take the AIME ii? (If a student is sick for example). Do they send one of each test for each qualifier?
If I contact AMC, and ask them to send my test to a different school would it work?

If you have a specific reason that you [b]need[/b] to take the Alternate AIME at another school (i.e. your school cannot give the Alternate, since it'll be off, and you have to miss the regular 3/7 date for some legitimate reason), then you and/or your school should contact the AMC to make other arrangements.

Basically what happens if that a school has students which don't take the AIME on the regular date, then they can order the requisite number of copies of the Alternate AIME from the AMC, and they get shipped separately (for a fee).


calculus
integration
algebra
linear equation

Did anyone manage to solve this question without programs/cacluclators etc -- just using algebra? :maybe:

[quote="anirudh"][quote="Phelpedo"]Ok, then prove that there are infinite.[/quote]I think he is aksing a question[/quote]
Yes, I am.

[quote="Phelpedo"]Find all integral values of $x$ that make $x^{4}+6x^{3}+11x^{2}+3x+31$ a perfect square.[/quote]
at some point, we could say that its less than (x^2+3x+1)^2 and greater than (x^2+3x)^2 since a (x^2+3x+1)^2-f(x) is a positive linear equation (3x-30=0), which has solution x=10

[quote="i_like_pie"]I guess we are both correct. Are there $\infty$ solutions?[/quote]
you guessed wrong; I tested some of the values you give. They don't work, but their square roots are very close to integers, which may be why your calculator rounded. (WINDOWS CALCULATOR>TI-89 WOOT)

[quote="t0rajir0u"]What is with this sudden influx of posters using calculators and programs to "solve" problems?

Obviously getting the answer without understanding any of the concepts behind the solution is not the point of a forum which carries in its name the word [b]art[/b].[/quote]
Quoted and sigged for truthiness.

Er...okay...

That's why you should NEVER TRUST YOUR CALCULATOR.

And remember that the x can be negative too.

We can do the same thing I did above for negs, except it yields no solutions, so we have one solution, which is 10

[quote="Phelpedo"]And I think I answered it. (BTW, i_like_pie, that is libel :wink: )[/quote]
Definition of libel please?

[quote="Scrambled"][quote="i_like_pie"]I guess we are both correct. Are there $\infty$ solutions?[/quote]
you guessed wrong; I tested some of the values you give. They don't work, but their square roots are very close to integers, which may be why your calculator rounded. (WINDOWS CALCULATOR>TI-89 WOOT)[/quote]
I didn't use a Windows calculator, I made a program and used it.

[quote="t0rajir0u"]What is with this sudden influx of posters using calculators and programs to "solve" problems?

Obviously getting the answer without understanding any of the concepts behind the solution is not the point of a forum which carries in its name the word [b]art[/b].[/quote]
I understand that this forum is for "solving with art," but I don't see how that list of numbers can be found by paper and pencil.

[quote="i_like_pie"] I don't see how that list of numbers can be found by paper and pencil.[/quote]

Refer to [b]Scrambled[/b]'s post.  Your numbers are wrong.  $10$ is the only solution, and he even sketches a proof of why.

[quote="i_like_pie"][quote="Phelpedo"]And I think I answered it. (BTW, i_like_pie, that is libel :wink: )[/quote]
Definition of libel please?[/quote]
I think he's referring to the part where someone asked phelpedo and you replied. He's joking, of course, so I wouldnt take it as something malicious

[quote="i_like_pie"][quote="Scrambled"][quote="i_like_pie"]I guess we are both correct. Are there $\infty$ solutions?[/quote]
you guessed wrong; I tested some of the values you give. They don't work, but their square roots are very close to integers, which may be why your calculator rounded. (WINDOWS CALCULATOR>TI-89 WOOT)[/quote]
I didn't use a Windows calculator, I made a program and used it.
[/quote]
Thats why WINDOWS TRUMPS TEXAS INSTRUMENTS(the calculator brand im assuming u made a program with) because i used it to figure out ur list of numbers don't work

Also, you should notice when looking at your data that there is a stark lack of numbers between 10 and 185544, while the others aren't very far apart at all. This is important when dealing with really big numbers who's square roots will round off very closely on a calculator display screen.

And be alert of rounding on calculators


a) Determine the injective functions $ f:\mathbb{R}\rightarrow\mathbb{R}$ such that

\[ f(3^{x}\minus{}1\plus{}f(y))\equal{}\log_{3}(2x\plus{}1)\plus{}f(y),\mbox{ for any }x\in[0,\infty),\ y\in\mathbb{R}.\]

b) Find with proof if there exist bijective functions $ f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[ f(3^{x})\minus{}f^{2}(2x\plus{}1)\ge\frac{1}{4},\mbox{for any }x\in\mathbb{R}.\]

[hide="a"]Put $ x \equal{} 0$, cancel stuff, and get $ f(f(y)) \equal{} f(y)$. Since $ f$ is injective, $ f(y) \equal{} y$.[/hide]

[hide="b"]Put $ x \equal{} 0$. Rearrange things to get $ 0\geq f^{2}(1)\minus{}f(1)\plus{}\frac{1}4 \equal{}\left(f(1)\minus{}\frac{1}2\right)^{2}$, so $ f(1) \equal{}\frac{1}2$. Put $ x \equal{} 1$, do the same thing to get $ f(3) \equal{}\frac{1}2$. So no bijective (or even injective) functions $ f$ exist.[/hide]


Nice, interesting and beautiful. I will prove the stronger inequality:
$(a^{2}+2)(b^{2}+2)(c^{2}+2) \geq 3(a+b+c)^{2}.$
  I will use the fact that if x,y have the same sign, then (1+x)(1+y)>=1+x+y.
  Write the inequality as $\prod \left(\frac{a^{2}-1}3+1\right) \geq \frac{1}{9}(a+b+c)^{2}.$
   If a,b,c>=1 then $\left(\frac{a^{2}-1}3+1\right) \geq 1+\sum \frac{a^{2}-1}3= \frac{1}{3}\sum a^{2}\geq \frac{1}{9}(a+b+c)^{2}.$
   If two of them are at least 1, let them be a and b. Then the LHS is at least $( 1+\frac{1}{3}(a^{2}-1+b^{2}-1) ) (c^{2}+2)= \frac{1}{9}(a^{2}+b^{2}+1) (1^{2}+1^{2}+c^{2}) \geq (a+b+c)^{2}$  by Cauchy.
   If one of them is at least 1, let it be a, then we use the same argument as in the precedent case.
   If a,b,c<=1 then by Bernoully inequality we have $\prod \left(1+\frac{a^{2}-1}3\right) \geq 1+\sum \frac{a^{2}-1}3= \frac{1}{3}\sum a^{2}\geq \frac{1}{9}(a+b+c)^{2}.$

There is a simple lower-power solution using strictly AM-GM:

productcyc(a^2+2)
=((abc)^2+1+1) + sumcyc(a^2) + 3sumcyc(a^2) + 2sumcyc((ab)^2+1))
>=3(abc)^2/3 + sumcyc(a^2) + 3sumcyc(ab) + 4sumcyc(ab)
>=sumsym(a^4/3*b^2/3) + 7sumcyc(ab) (expanding the well known inequality xyz>=productcyc(x+y-z), where x=a^2/3, y=b^2/3, x=c^2/3)
>=2sumcyc(ab)+7sumcyc(ab)
=9sumcyc(ab), proving the required.  Equality obviously holds iff a=b=c=1.

Of course, even though it's simpler, Harazi's result is more elegant in another sense because it is stronger.

Particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=19666 for k = 1.

  darij

Does anyone have the *OFFICIAL* solutions to this question on the APMO?

I remembere there were 2 nice solutions, one with vectors, one with trig.

[quote="mathboy_cntHDVN"]@Michael Lipnowski :please edit it
i don't   understand[/quote]
  $\prod_{cyc}(a^{2}+2) =((abc)^{2}+1+1)+\sum_{cyc}(a^{2})+3 \sum_{cyc}(a^{2})+2 \sum_{cyc}((ab)^{2}+1))$
$\geq 3(abc)^\frac{2}{3}+\sum_{cyc}(a^{2})+3 \sum_{cyc}(ab)+4 \sum_{cyc}(ab) \geq \sum_{sym}(a^\frac{4}{3}{b^\frac{2}{3})+7 \sum_{cyc}(ab)}$  (expanding the well known inequality $xyz\geq \prod_{cyc}(x+y-z)$, where $x=a^\frac{2}{3}, y=b^\frac{2}{3}, x=c^\frac{2}{3})$
$\geq 2 \sum_{cyc}(ab)+7 \sum_{cyc}(ab) =9 \sum_{cyc}(ab)$, proving the required. Equality obviously holds iff $a=b=c=1$.
  @mathboy_cntHDVN : This 's posts of Michael Lipnowski in latex  :lol:

My solution:
Put $ t \equal{} \frac {b \plus{} c}{2}$
By AM-GM we have:
$ (b^2 \plus{} 2)(c^2 \plus{} 2) \geq 3 \plus{} 2bc \plus{} 2(b^2 \plus{} c^2) \geq 3 \plus{} 6t^2$ and $ t^2 \geq bc$
We need prove that:
$ (a^2 \plus{} 2)(3 \plus{} 6t^2) \geq 9(2at \plus{} t^2)$
<=>$ (a \minus{} t)^2 \plus{} 2(at \minus{} 1)^2 \geq 0$
This inequality is true
Equality occurs when  $ a \equal{} b \equal{} c \equal{} 1$

[quote="Arne"]Prove that the inequality
$ \left(a^{2} \plus{} 2\right)\left(b^{2} \plus{} 2\right)\left(c^{2} \plus{} 2\right) \geq 9\left(ab \plus{} bc \plus{} ca\right)$
holds for all positive reals $ a$, $ b$, $ c$.[/quote]
I will prove the stronger inequality:
$ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \ge 3(a \plus{} b \plus{} c)^2$
I have:
$ (a^2 \plus{} 2)(b^2 \plus{} 2) \equal{} (a^2 \plus{} 1)(b^2 \plus{} 1) \plus{} a^2 \plus{} b^2 \plus{} 3 \ge (a \plus{} b)^2 \plus{} \frac {1}{2}(a \plus{} b)^2 \plus{} 3$
$ \equal{} \frac {3}{2}[(a \plus{} b)^2 \plus{} 2]$
$ \equal{} > (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \ge \frac {3}{2}[(a \plus{} b)^2 \plus{} 2](c^2 \plus{} 2) \ge \frac {3}{2}[ \sqrt {2}(a \plus{} b) \plus{} \sqrt {2}c]^2$
$ \equal{} 3(a \plus{} b \plus{} c)^2$

$ \equal{}>(a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2) \ge 3(a \plus{} b \plus{} c)^2 \ge 9(ab \plus{} bc \plus{} ca)$

Can anyone please explain in more detail how Bernoulli's inequality is used in Harazi's solution?

Thank you in advance! :)

Take
$ x=a+b+c $ ,   $ y=ab+bc+ca $ ,   $ z=abc $.
Then given inequality is equivalent to  \[ z^{2}+2y^{2}-4xz+4x^{2}-17y+8\geq 0 \]
It is equivalent to
 \[ (z-\frac{x}{3})^{2}+\frac{8}{9}(y-3)^{2}+\frac{10}{9}(y^{2}-3xz)+\frac{35}{9}(x^{2}-3y)\geq 0 \]

It is easy to prove \[ y^{2}-3xz\geq 0 \]  and \[ x^{2}-3y\geq 0  \]

I took it from official solutions. There were another two solutions. One of them is  by Jensen ,  other is by vectoral product of vectors.

[quote="mudok"]Take
$ x=a+b+c $ ,   $ y=ab+bc+ca $ ,   $ z=abc $.
Then given inequality is equivalent to  \[ z^{2}+2y^{2}-4xz+4x^{2}-17y+8\geq 0 \]
It is equivalent to
 \[ (z-\frac{x}{3})^{2}+\frac{8}{9}(y-3)^{2}+\frac{10}{9}(y^{2}-3xz)+\frac{35}{9}(x^{2}-3y)\geq 0 \] [/quote]

May anyone explain to me how would one think of grouping the addends like that? I mean, there is no formula nor anything obvious in this particular grouping right? So how can you master the craft of grouping, even if it works only in some cases? And, what "shows" you how to group the elements in this problem?
Thank you in advance! :)

\begin{align*} &\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9\left(ab+bc+ca\right) \\ &\iff a^2b^2c^2+\sum_{cyc} 2a^2b^2+\sum_{cyc} 4a^2+8 \geq \sum_{cyc} 9ab \\ &\iff \sum_{cyc} 2a^2b^2+6 \geq \sum_{cyc} 4ab.\end{align*} Trivial by Schur and AM-GM.

Here is my solution to Japanese Kids those who haven't learnt Schur's inequality yet , which is beyond my ability, because I can't memorize it even now.

Here goes my solution :

From $\frac{a^2+2}{(a+b+c)^2} =\{(b+c)^2+2\}\left(\frac{1}{a+b+c}-\frac{b+c}{(b+c)^2+2}\right)^2+\frac{2}{(b+c)^2+2}\geq \frac{2}{(b+c)^2+2},$

Equality is achived when $\frac{1}{a+b+c}=\frac{b+c}{(b+c)^2+2}\Longleftrightarrow a =\frac{2}{b+c}>0\cdots [1]$

$\therefore \frac{(a^2+2)(b^2+2)(c^2+2)}{(a+b+c)^2}\geq \frac{2(b^2+2)(c^2+2)}{(b+c)^2+2}=\frac{2(bc-1)^2+(b-c)^2}{(b+c)^2+2}+3\geq 3,$

Equality is achived when $bc=1$ and $b=c>0\cdots [2], $  by using the well known inequality $(a+b+c)^2\geq 3(ab+bc+ca),$ yielding

$(a^2+2)(b^2+2)(c^2+2)\geq 3(a+b+c)^2\geq 9(ab+bc+ca).$ From [1], [2] Equality is achived when $a=b=c=1.$

[quote=sqing][quote=Arne]Prove that the inequality \[\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9\left(ab+bc+ca\right)\] holds for all positive reals $a$, $b$, $c$.[/quote]
[b]WLOG [/b]$(a^2-1)(b^2-1)\geq 0, $ then
$$(a^2+2)(b^2+2)(c^2+2)\ge3(a^2+b^2+1)(1+1+c^2)\ge3(a+b+c)^2\ge9(ab+bc+ca)$$[/quote]

Very nice way of proving it. Congratulations on finding it!

Let  $a, b, c$ be positive reals.  Prove that $$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 4\left(a^2+bc+ca\right)$$
$$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 4\left(a^2+bc+c^2\right)$$

$8<9$  :(

We first e x p a n d which yields the equivalent inequality
$$a^2b^2c^2+2\sum_{\mathrm{cyc}} a^2a^2+4\sum_{\mathrm{cyc}} a^2-9\sum_{\mathrm{cyc}}ab+8=a^2b^2c^2+2\sum_{\mathrm{cyc}} (ab-1)^2+4\sum_{\mathrm{cyc}} a^2-5\sum_{\mathrm{cyc}} ab+2\geq 0.$$
Motivated by the $a=b=c=1$ equality case we discard the $(ab-1)^2$ terms, so it remains to prove that
$$4\sum_{\mathrm{cyc}} a^2+a^2b^2c^2+2\geq 5\sum_{\mathrm{cyc}}ab \impliedby \sum_{\mathrm{cyc}} a^2+3a^{\frac{2}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}}\geq 2\sum_{\mathrm{cyc}} ab$$
by AM-GM. By Schur on $a^{\frac{2}{3}}$, etc., we have
$$\sum_{\mathrm{cyc}} a^2+3a^{\frac{2}{3}}b^{\frac{2}{3}}c^{\frac{2}{3}}\geq \sum_{\mathrm{sym}} a^{\frac{4}{2}}b^{\frac{2}{3}},$$
from which the conclusion is clear by AM-GM. $\blacksquare$

Here is an alternate finish from the second-to-last centered inequality.
The inequality there is homogeneous and symmetrical, so WLOG suppose that $a\geq b \geq c$ and $ab=1$. We will prove that the inequality is strongest when $a=b=1$. The inequality in this case is equivalent to
$$\left(a+\frac{1}{a}\right)^2-2c\left(a+\frac{1}{a}\right)+c^2+3c^{\frac{2}{3}}-4\geq 0.$$
Viewing the LHS as a quadratic in $a+\tfrac{1}{a}$, its minimum occurs when $a+\tfrac{1}{a}=c$ and is decreasing on $[c,\infty)$ and thus on $[2,\infty)$ since $a\geq b\geq c$. This proves the desired claim, since $a+\tfrac{1}{a} \implies a=1$. The inequality then reduces to $c^2-4c+3c^{\frac{2}{3}}\geq 0$, which is just AM-GM on $c^2$ and three copies of $c^{\frac{2}{3}}$. $\blacksquare$

[quote=Arne]Prove that the inequality \[\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9\left(ab+bc+ca\right)\] holds for all positive reals $a$, $b$, $c$.[/quote]
2023 Shandong High School Mathematics Competition [b]Q13[/b]

Solution is attached.
Although a bit lengthy, I found this solution to be very natural. :)

[b]Version 2- 4_Variables[/b]
Let $a,b,c,d,f$ be positive reals. Then prove that

$$(a^2+2)(b^2+2)(c^2+2)(d^2+2)(f^2+2)\geq \dfrac{27(a+b+c+d+f)^2}{4}$$

[b]Generalization 1[/b]
Let $a_{1},a_{2},\cdots,a_{n}$ ($n\geq 2$) be positive reals. Then prove that

$$\prod_{i=1}^{n}{(a_{i}+2)}\geq \dfrac{3^{n-2}}{2^{n-3}}\left(\sum_{cyc}{a_{1}}\right)^2$$

These generalizations are proving the stronger one.

[b]Generalization 2[/b]
Let $a_{1},a_{2},\cdots,a_{n},k$ $\left(n\geq 2\right)$ be positive reals. Then prove that

$$\prod_{i=1}^{n}{\left(a_{i}^2+k\right)}\geq \dfrac{3^{n-2}.2^4}{k^{n+1}}\left(\sum_{cyc}{a_{1}}\right)^2$$

Another solution. Again expand but notice $a^2b^2c^2+\frac13(a^2+b^2+c^2)+2\ge\frac23(a^2bc+b^2ca+c^2ab)+2\ge\frac43(a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab})$ and combined with $a^2b^2+1\ge 2ab$ we want to show $4(a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab})+11(a^2+b^2+c^2)-15(ab+bc+ca)\ge 0.$ This follows from \begin{align*}&\frac72\left((a-b)^2+(b-c)^2+(c-a)^2\right)+4(\sqrt{ab}(\sqrt a-\sqrt b)^2+\sqrt{bc}(\sqrt b-\sqrt c)^2+\sqrt{ca}(\sqrt c-\sqrt a)^2)\\&+4\left(a(\sqrt a-\sqrt b)(\sqrt a-\sqrt c)+b(\sqrt b-\sqrt a)(\sqrt b-\sqrt c)+c(\sqrt c-\sqrt a)(\sqrt c-\sqrt b)\right)\\&\ge 0.\end{align*} The last part again by Schur.


limit
logarithms
calculus
calculus computations

$\lim_{n\to\infty} \left(1+\frac 4n\right)^{\frac n{10}}=$

It finally works!!!!!!!!!!!!!!!!!

I am so happy!!!!!!!!!!

Thanks :roll:

There are lots of ways, but this is probably the most natural.

$\lim_{n\to\infty} \left(1+\frac xn\right)^n=e^x$

It follows that

$\lim_{n\to\infty} \left(1+\frac 4n\right)^{\frac n{10}}=\left(\lim_{n\to\infty} \left(1+\frac 4n\right)^n\right)^{\frac 1{10}}=e^{\frac 25}$

How come I can't print my question in boldface?  This is so complicated.  I'll spent some time and  figure that out tomorrow.  Have to go to sleep now.

A general principle: if a limit of $a^b$ where $a$ and $b$ are both variable is giving you trouble, then take the logarithm. What this does for $1^{\infty},0^0,$ and $\infty^0$ indeterminate cases is to convert them into product or quotient indeterminate cases: $\frac00,\frac{\infty}{\infty},$ or $0\cdot\infty.$ These are still indeterminate, but more familiar and you have more tools.

In this case, let $L=\lim_{n\to\infty}\left(1+\frac4n\right)^{\frac{n}{10}}$

Then $\ln L=\lim_{n\to\infty}\frac{n}{10}\ln\left(1+\frac4n\right).$

We have a number of tools to deal with this. Another appearance-changer that I like: make an appropriate substitution. In this case, try letting $u=\frac1n.$ (I'm not spending any time worrying about the distinction between a continuous variable and a discrete variable - I could justify it, if I had to.) We now have:

$\ln L=\lim_{u\to0^+}\frac{\ln(1+4u)}{10u}.$

Now you should have plenty of ways to proceed.

Oh - and you don't want to print your question in boldface. You want to print it in $\text{\LaTeX}.$ Hold your mouse over or click on any of the formulas in blablahblah's post or my post to see how those formulas are constructed.

Whenever the limit
$\lim_{x\to{k}} {(f(x))}^{g(x)}$
is of the form $1^{\infty}$ use the following transformation.

Change the limit to $\lim_{x\to{k}} {e^{{(g(x))}.{(f(x) - 1)}}}$

It is proved by the same way as beginner solved the problem.


geometry
geometric transformation
reflection
function

Question:

I am provided the list "cat, monkey, dog, mule." Therefore, I generalize the next entry is spider.

If you discovered the rationale for the generalization, then you should be able to enter a numeric sequence that reflects the rationale.

  _,_,_,_,_?   Any help?

What are you trying to say? I can generalize that the next number in the series 0, 0, 0, 0, ___ is 69 because I was thinking $f(x) = (\frac{23}{8})(x-1)(x-2)(x-3)(x-4)$.

I can generalize cat, monkey, dog, mule, homosexuality. The first one has an odd number of letters, the second one has an even number of letters, the third one has an odd number of letters, and the fourth one has an even number of letters, so the fifth one must have an odd number of letters.

[quote="Allan Z"][quote="Treething"]What are you trying to say? I can generalize that the next number in the series 0, 0, 0, 0, ___ is 69 because I was thinking $f(x) = (\frac{23}{8})(x-1)(x-2)(x-3)(x-4)$.

I can generalize cat, monkey, dog, mule, homosexuality. The first one has an odd number of letters, the second one has an even number of letters, the third one has an odd number of letters, and the fourth one has an even number of letters, [b]so the fifth one must have an odd number of letters[/b].[/quote]

o.O? spider.. 6 letters.. how is that odd[/quote]

What he's trying to say is that one cannot "generalize" anything from only a finite set of information. There will always be different ways of fitting a function or a relation to a set of data, so heck "spider" might as well be "homosexuality". To induct (either mathematically or just conceptually, as it is here) you need a condition which requires the following data or words, whatever, to logically fit with what already is known.

Ummm.... has anyone here ever looked at the local "Puzzles and Brainteasers" forum where this belongs? Of course there is no strictly mathematical solution to these things; just like minimizing a function, you are in essence considering the set of all solutions which is infinite and presenting the most logical and easy to see solution, which can be subjective at times (although the point of a good puzzle is that there is one unconstestable nicest solution). Bearing this in mind, I wouldn't scorn this problem would try to solve it, which I honestly can't right now.

( :rotfl: Why did I just axiomatically describe something of common sense?? )

[quote="wilbur"]I am provided the list "worm, man, dog, ant." Therefore, I generalize the next entry is spider.
If you discovered the rationale for the generalization, then you should be able to enter a numeric sequence that reflects the rationale.

Enter the sequence of numbers,......[/quote]
[quote="anyone"][hide]
Its based on the number of legs of the animal. So the the series is 0 (# of worm's legs), 2 (# of mans legs), and so on. So numerically it's just the sequence of even numbers beginning with 0. So the sequence is 0, 2, 4, 6, 8, 10...[/hide][/quote]

This is something to compare it to ( :o outdone by getting started forum)


Find $ m$ such that the follow  equation has root:
$ \sqrt [4]{1 \plus{} x} \plus{} \sqrt[4]{1 \minus{} x} \equal{} m$

[quote="thuyanh158"]Find $ m$ such that the follow  equation has root:
$ \sqrt [4]{1 \plus{} x} \plus{} \sqrt [4]{1 \minus{} x} \equal{} m$[/quote]

$ f(x)\equal{}\sqrt [4]{1 \plus{} x} \plus{} \sqrt [4]{1 \minus{} x}$ is an even function defined only for $ x\in[\minus{}1,\plus{}1]$. This function is increasing over $ [\minus{}1,0]$, maximal at $ 0$ and decreasing over $ [0,\plus{}1]$.

So the answer is $ m\in[f(1),f(0)]$, that's to say $ m\in[\sqrt [4]2,2]$

[quote="Victory.US"][quote="pco"][quote="thuyanh158"]Find $ m$ such that the follow  equation has root:
$ \sqrt [4]{1 \plus{} x} \plus{} \sqrt [4]{1 \minus{} x} \equal{} m$[/quote]

$ f(x) \equal{} \sqrt [4]{1 \plus{} x} \plus{} \sqrt [4]{1 \minus{} x}$ is an even function defined only for $ x\in[ \minus{} 1, \plus{} 1]$. This function is increasing over $ [ \minus{} 1,0]$, maximal at $ 0$ and decreasing over $ [0, \plus{} 1]$.

So the answer is $ m\in[f(1),f(0)]$, that's to say $ m\in[\sqrt [4]2,2]$[/quote]

in invert, if  $ m\in[\sqrt [4]2,2]$ then do we have to prove that the equation has roof,? don't have to :?:[/quote]

You're right. We must just add that $ f(x)$ is continuous over $ [0,1]$ and so $ m\in[f(1),f(0)]$ $ \implies$ $ f(x)\equal{}m$ always have at least one root.


is there any team in the RI or MA that would want a member or would want members for next year?

Looking at last year's results: there were several teams called either "Massachusetts or "Eastern Massachusetts" - I think that's all the same organization. There's a multi-team "Western Massachusetts" organization, and there's a multi-team "Connecticut" organization. I don't see anything from Rhode Island.

My suggestion is to contact Bryan Sullivan, the President of ARML (you can find his email on the ARML site). He has long been associated with the Western Massachusetts team, and probably has some idea of the activities in the area.


Suppose $a$ and $b$ are positive and $n$ is natural. Proove that $\sum_{i=1}^{n}\frac{1}{a+ib} < \frac{n}{\sqrt{a(a+nb)}}$

[quote="Beat"]Suppose $a$ and $b$ are positive and $n$ is natural. Proove that $\sum_{i=1}^{n}\frac{1}{a+nb} < \frac{n}{\sqrt{a(a+nb)}}$[/quote]
I think you mean$\sum_{i=1}^{n}\frac{1}{a+ib} < \frac{n}{\sqrt{a(a+nb)}}$ :rotfl:

If I am right,
It can be proved by:
$\frac 1 {a+kb}+\frac 1{a+(n-k)b} \le \frac{2}{\sqrt{a(a+nb)}}$
and then obvious the all the n equalities can't be hold :)


abstract algebra
induction
group theory
superior algebra
superior algebra unsolved

Hello, I try to demonstrate the structur theorem which says that an abelian finite group $G$ without torsion elements is a product of cyclics groups

By induction, I take $x\in G$ and I say that $\frac{G}{<x>}$ is a product of cyclics groups $\prod_i G_i$, after that I consider the morphism (since $G$ is abelian) $\phi: G \rightarrow \frac{G}{<x>}$.
Now I don't know how to prove correctly that I can do that:
\[ G = \phi^{-1}\Big(\frac{G}{<x>}\Big) = \phi^{-1}(\prod_i G_i) = \prod_i \phi^{-1}(G_i) = \prod_i (G_i<x>) = (\prod_i G_i)<x> \]

I think I'm OK with first equal (since $\phi$ is a morphism, is it sufficient ?), second equal it's just definition, third I think it's not well written, fourth it's also not very clear for me, last equal is ok.

Thanks !

You can not take any $x$, e.g. $G=\mathbb{Z} / 4 \mathbb{Z}$:
when $x=2 \mod 4$, then you would get that $\mathbb{Z} / 4 \mathbb{Z}$ is ismorphic to $\mathbb{Z} / \mathbb{Z} ^2$ which is simply wrong.

The $x$ with the highest order should do.

i'm sorry for doubting you, but i don't remember that theorem being that easy
so either  my proof was way too long, or there is tiny problem here

what exactly do you do in the disputed equality
does your = stand for isomorphism or really equality
and does (product groups) <x >
stand for direct product of (product groups) and <x> or
really the set of all products ab with a in (product groups) and b in <x>

One of the two proves I know uses one important lemma (hidden, since I'm not sure if Rodolphe2005 want's to try for himself):
[hide]
Let $G$ be an abelian group.
Let $x$ be an element of maximal order in $G$.
Lemma: every cyclic subgroup of $G/<x>$ generated by $y$ has a representant $y' \in G$ with $ord_G(y')=ord_{G/<x>} (y)$.[/hide]

I've find a better way regarding $p$-Sylows of $G$.
fredbel6: It's isomorphisms and product ab and last equals it's a direct product but what I've written is not very well written and there is some erros I'm sorry ...
Zetax I would be interested to see one of your proof or a link to it. Thx

Forgive me for nagging one, it's only because I am interested
A friend of mine once suggested that too, proving the abelian group theorem with Sylow
however, isn't there a subtlety there?
You need Cauchy's theorem to prove Sylow
and if you prove Cauchy's theorem you can always use induction on the order, except if the group is abelian
that case is to be considered easy because by classification of abelian groups it is okay too then
But you do need it

You see my point?  I'm afraid Sylow makes use of the abelian group theorem, therefore you would be running in a circle

Or can anyone avoid my problem?


Let

$f(x)=LCM{\left(x,y\right)}$

Where x is some number, and y is it's digits reversed. If x has a final digit of zero, then we ingore the 0. So, for example,

$f(24)=LCM{\left(24,42\right)}$

and

$f(20)=LCM{\left(20,2\right)}$

Compute

f(f(f(12)))

Changing the two numbers to their prime factorizations helps a lot  :wink:


greatest common divisor
number theory
relatively prime

1)Let $x,y,z$ be positive integers such that
\[\frac{1}{x}-\frac{1}{y}= \frac{1}{z}\]
Let $h$ be the greatest common divisor of $x,y,z$. Prove that $hxyz$ and $h(y-x)$ are perfect squares.

2) Let $k$ be an odd positive integer. Prove that
\[(1+2+...+n)|(1^{k}+2^{k}+3^{k}...+n^{k}) \]
for all positive integers $n$.

3) Consider the following two-person game. A number of pebbles are lying on a table. Two players make their moves alternately. A move consists of taking $x$ pebbles off the table, where $x$ is a perfect square. The player who is unable to make a move loses. Prove that there are infinitely many initial situations in which the player who goes second has a winning strategy.

Number 3:
[hide]Assume to the contrary.  Call n the largest such possible case where player 2 wins.  Obviously, $n+k^{2}$ must also win for player one since he can just take $k^{2}$ pebbles and force player 2 to lose (remember, player 2 wins when there's n only if it's player 1's turn.  If it's his turn, the roles are reversed.)  Now, if $n+k^{2}+c$ pebbles were on the table where 0<c<2k+1, then player one cannot take less than or equal to $k^{2}$ pebbles because of the initial assumption and the rule of role reversal.  (In other words, since player 1 wins if the number of pebbles exceed n when it is his turn, player 2 wins if the number of pebbles exceed n when it is player 2's turn.)  However, since we can make k arbitrarily large, $n+k^{2}+c<(k+1)^{2}$ pebbles would force player 1 to take an undesirable amount and make player 2 win which is a contradiction.[/hide]
Is there solution that involves divisibillity?


geometry circumcircle incenter trigonometry ratio conics hyperbola

This problem is consequence of $ I_1J_1,I_2J_2,CP$ are concurrent, but I haven't proved it, yet  :P

Denote $ X = (J_1)\cap AB, Y = (J_2)\cap AB, X' = (I_1)\cap AB, Y' = (I_2)\cap AB,$ $ J_1J_2\cap AB = T, I_1I_2\cap AB = T'$ and $ \angle BPC = \phi$
From Casey's theorem we have: $ AX.BC + BX.AC = (CP + PX).AB\Leftrightarrow (AP - PX).BC + (BP + PX).AC$ $ = (CP + PX).AB\Leftrightarrow$ $ PX = \frac {AP.BC + BP.AC - CP.AB}{AB + BC - AC}$
Analogically $ PY = \frac {AP.BC + BP.AC - CP.AB}{AB + AC - BC}$

We find easily $ PX' = \frac {PA + PC - AC}{2}, PY' = \frac {PC + PB - BC}{2}$
$ TX = \frac {XY.J_1X}{J_2Y - J_1X}(\triangle TXJ_1\sim \triangle TYJ_2)$ $ T'X' = \frac {X'Y'.I_1X'}{I_2Y' - I_1X'}(\triangle T'X'I_1\sim \triangle T'Y'I_2)$
We have $ XJ_1 = PX.tan\frac {\phi}{2}, YJ_2 = \frac {PY}{tan\frac {\phi}{2}},X'I_1 = \frac {PX'}{tan\frac {\phi}{2}}$, $ Y'I_2 = PY'.tan\frac {\phi}{2}$
So $ TX + XP = \frac {(PX + PY).PX.tan\frac {\phi}{2}}{\frac {PY}{tan\frac {\phi}{2}} - PX.tan\frac {\phi}{2}} + XP$ $ = \frac {PX.PY(tan^2\frac {\phi}{2} + 1)}{PY - PX.tan^2\frac {\phi}{2}}$
Analogically $ TX + XP = \frac {PX'.PY'(tan^2\frac {\phi}{2} + 1)}{PY'.tan^2\frac {\phi}{2} - PX'}$. It's sufficient to prove that $ TX + XP = TX' + X'P\Leftrightarrow \frac {PX.PY(tan^2\frac {\phi}{2} + 1)}{PY - PX.tan^2\frac {\phi}{2}}$ $ = \frac {PX'.PY'(tan^2\frac {\phi}{2} + 1)}{PY'.tan^2\frac {\phi}{2} - PX'}\Leftrightarrow tan^2\frac {\phi}{2} = \frac {PX'.PY(PX + PY')}{PY'.PX(PY + PX'}$ $ \Leftrightarrow tan^2\frac {\phi}{2} = \frac {(PA + PC - AC)[2(PA.BC + BP.AC - PC.AB) + (AB + BC - AC)(PC + PB - BC)]}{(PC + PB - BC)[2(PA.BC + BP.AC - PC.AB) + (AB + AC - BC)(PA + PC - PB)]}$
$ tan\frac {\phi}{2} = \frac {S_{PBC}}{\frac {BP + PC - BC}{2}.\frac {PB + PC + BC}{2}}, tan\frac {\phi}{2} = \frac {\frac {AP + PC - AC}{2}.\frac {PA + PC + AC}{2}}{S_{PAC}}$ $ \Rightarrow tan^2\frac {\phi}{2} = \frac {PB(PA + PC + AC)(AP + PC - AC)}{PA(PB + PC + BC)(BP + PC + BC)}$
So it's sufficient to prove that $ \frac {PB(PA + PC + AC)(AP + PC - AC)}{PA(PB + PC + BC)(BP + PC - BC)} =$ $ \frac {(PA + PC - AC)[2(PA.BC + BP.AC - PC.AB) + (AB + BC - AC)(PC + PB - BC)]}{(PC + PB - BC)[2(PA.BC + BP.AC - PC.AB) + (AB + AC - BC)(PA + PC - PB)]}$ $ \Leftrightarrow \frac {PB(PA + PC + AC)}{PA(BP + PC + BC)} =$ $ \frac {(2(PA.BC + BP.AC - PC.AB) + (AB + BC - AC)(PC + PB - BC)}{(2(PA.BC + BP.AC - PC.AB) + (AB + AC - BC)(PA + PC - PB)}(1)$
Denote $ BC = a, AC = b, PA = c, PB = d, PC = e$
$ (1)\Leftrightarrow \frac {d(c + e + b)}{c(d + a + e)} = \frac {ac + bd - ce - de + ab + cd + ae - be + d^2 - a^2}{ac + bd - ce - de + ab + cd + be - ae + c^2 - b^2}$
After some calculations(believe me they are not hard) we get:
$ cd^3 - cda^2 - c^2e^2 + ace^2 - bce^2 + a^2c^2 + a^2bc + c^2ad + d^2ca - a^3c$ $ = bde^2 - d^2e^2 - ade^2 + c^3d - b^2dc + b^2d^2 + b^2ad + d^2bc + dbc^2 - db^3 (2)$
From Stewart's theorem we have: $ a^2c + b^2d - cd(c + d) = e^2(c + d)$
We get that:
$ (*)a^3c + ab^2d - acd(c + d) = e^2(ac + ad)\Leftrightarrow a^3c + ab^2d - ac^2d - acd^2 = e^2ac + e^2ad$
$ (**)a^2bc + b^3d - cdb(c + d) = e^2(bc + bd)\Leftrightarrow a^2bc + b^3d - c^2db - cd^2b = e^2bc + e^2bd$
$ (***)(a^2c + b^2d - cd(c + d))(c - d) = e^2(c + d)(c - d)\Leftrightarrow a^2c^2 + b^2dc + cd^3 + e^2d^2 =$ $ c^2e^2 + c^3d + a^2cd + b^2d^2$
From $ (*),(**),(***)$ we see that $ (2)$ is true, so $ (1)$ is true too. DONE!

Hi nice proof, I have an idea from your proof, If you can prove
\[ \tan^2\frac {\phi}{2} = \frac {PX'.PY(PX + PY')}{PY'.PX(PY + PX')} \Rightarrow \frac {1}{\tan\frac {\phi}{2}}(\frac {1}{PX'} + \frac {1}{PY}) = \frac {1}{\cot\frac {\phi}{2}}(\frac {1}{PX} + \frac {1}{PY'})
\]
This mean
\[ \frac {1}{r_{I_1}} + \frac {1}{r_{J_2}} = \frac {1}{r_{I_2}} + \frac {1}{r_{J_1}}
\]
This really nice indenity, more we can see
\[ \frac {1}{r_{J_1}} - \frac {1}{r_{I_1}} = \frac {1}{r_{J_2}} - \frac {1}{r_{I_2}}\quad (1)
\]
This is sufficient to show $ I_1J_1,I_2,J_2$ and $ CP$ are concurrent, indeed Let $ I_1J_1\cap CP = S$ and $ (I_1), (J_1)$ touch $ PC$ at $ M,N$, resp, we have
\[ \frac {SM}{r_{I_1}} = \frac {SN}{r_{J_1}} = \frac {MN}{r_{J_1} - r_{I_1}} = \frac {PM + PN}{r_{J_1} - r_{I_1}}\quad (2)
\]
So
\[ SP = SN + NP = \frac {r_{J_1}(PM + PN)}{r_{J_1} - r_{I_1}} + NP = \frac {r_{J_1}PM + r_{I_1}PN}{r_{J_1} - r_{I_1}} = \frac {\frac {PM}{r_{I_1}} + \frac {PN}{r_{J_1}}}{\frac {1}{r_{J_1}} - \frac {1}{r_{I_1}}} = \frac {\tan\frac {\phi}{2} + \cot\frac {\phi}{2}}{\frac {1}{r_{J_1}} - \frac {1}{r_{I_1}}}
\]
Similarly, let $ S' = I_2J_2\cap CP$ then
\[ S'P = \frac {\tan\frac {\phi}{2} + \cot\frac {\phi}{2}}{\frac {1}{r_{J_2}} - \frac {1}{r_{I_2}}}\quad (3)
\]
From $ (1),(2),(3)$ derive $ S\equiv S'$ more, if $ I_1J_1,I_2,J_2$ and $ CP$ are concurrent at $ S$ let $ I_1J_1,I_2J_2$ intersect $ BC$ at $ P,Q$, easily seen $ (I_1J_1PS) = (I_2J_2QS) = - 1$ it is sufficient to show $ I_1I_1,J_1J_2,PQ$ are concurrent.

[quote="yetti"]$ P$ is arbitrary point on the side $ AB$ of a $ \triangle ABC$ with circumcircle $ (O)$ and $ CP$ cuts $ (O)$ again at $ D.$

$ (I_1), (I_2)$ are ordinary incircles of $ \triangle APC, \triangle BPC$ with incenters $ I_1, I_2.$

$ (J_1), (J_2)$ are Thebault circles with centers $ J_1, J_2,$ tangent to the rays $ PD, PA$ resp $ PD, PB$ and internally tangent to the circumcircle $ (O).$

Prove that the center lines $ I_1I_2, J_1J_2$ intersect on the line $ AB.$[/quote]

$ Q$ at infinity when $ P$ is tangency point of  $ C$-excircle and $ AB$. :D

[hide="For encyclopedia"] Take away the circle (O) - $ (I_1), (I_2)$ are not ordinary incircles and $ (J_1), (J_2)$ are not Thebault incircles. AB, CD are just common tangents intersecting at P. Let CD cut $ I_1I_2, J_1J_2$ at K, L. Assume $ I_1I_2, J_1J_2$ intersect at $ Q \in AB.$ and let $ I_1J_1$ cut CD at R.

$ PI_1, PI_2, PJ_1, PJ_2$ are angle bisectors $ \Longrightarrow$ the cross ratios $ (Q, K, I_1, I_2) \equal{} (Q, L, J_1, J_2) \equal{} \minus{} 1$ are harmonic. From $ (Q, K, I_1, I_2) \equal{} \minus{} 1,$ the pencil of lines $ (RQ, RI_1 \equiv RJ_1, RK \equiv RL, RI_2)$ is harmonic. Together with $ (Q, L, J_1, J_2) \equal{} \minus{} 1,$ this makes $ RI_2$ cut the line $ QJ_1LJ_2$ at $ J_2.$

Alternately, Q is the external similarity center of $ (I_1), (I_2)$ and $ R$ the external similarity center of $ (I_1), (J_1)$ $ \Longrightarrow$ the external similarity center S of $ (I_2), (J_1)$ is on QR. Q is also the external similarity center of $ (J_1), (J_2)$  $ \Longrightarrow$ the external similarity center $ R' \equiv I_2J_2 \cap CD$ of $ (I_2), (J_2)$ is on SQ. Therefore, $ R' \equiv SQ \cap CD \equiv RQ \cap CD \equiv R.$[/hide]

Incircle (I) of $ \triangle ABC$ touches AB at F. $ I_1I_2$ is tangent at $ T \in PC$ of a hyperbola $ \mathcal H$ with foci C, F passing through A, B - see [url]http://www.mathlinks.ro/viewtopic.php?p=1353036#1353036[/url] (1st solution).
Parallel to AB through C cuts (O) again at E. $ J_1J_2$ is tangent at $ S \in PD$ of an ellipse $ \mathcal E$ with foci C, E passing through A, B - see [url]http://www.mathlinks.ro/viewtopic.php?p=1371790#1371790[/url] (2nd page).

AB is the common chord of the ellipse $ \mathcal E$ and hyperbola $ \mathcal H$ with one common focus C, such that A, B are on the same side of the directrices $ e, h$ of $ \mathcal E, \mathcal H$ with respect to C. Otherwise, these 2 conics are quite special: Their common chord AB is parallel to the ellipse major axis line CE and the other hyperbola focus F is on their common chord AB. Disregarding all this except for the one common focus C, the general theorem is:

[color=blue]Let $ \mathcal G, \mathcal H$ be 2 arbitrary intersecting conics with one common focus C. Let $ g, h$ be the directrices of $ \mathcal G, \mathcal H$ with respect to C. Let AB (and possibly A'B') be their common chord(s), such that A, B are either both on the same sides of $ g, h,$ or both on opposite sides of $ g, h.$ (The same condition for A', B'.) Let arbitrary line through the common focus C cut $ \mathcal G$ at S, S' and $ \mathcal H$ at T, T'. Then the tangents of $ \mathcal G$ at $ S, S'$ cut the tangents of $ \mathcal H$ at T, T' on AB or A'B'.[/color]

[hide="Better identification of the common chords"] Let (A), (B) be circles with centers A, B and radii AC, BC. Assume A, B are both on the same sides of $ g, h$ (always true for ellipses). Since $ \frac {AC}{d(A, g)} \equal{} \frac {BC}{d(B, g)},$ $ g$ goes through the external similarity center $ X \in AB$ of (A), (B). Likewise, $ \frac {AC}{d(A, h)} \equal{} \frac {BC}{d(B, h)}$ and $ h$ goes through X. Therefore, AB (and possibly A'B') go through the intersection X of the directrices $ g, h.$ Assume A, B are both on the opposite sides of $ g, h$ (possible only for hyperbolas). For similar reasons, both $ g, h$ go through the internal similarity center $ Y \in AB$ of (A), (B). On the other hand, assume A, B are on the same side of $ g$ and on the opposite sides of $ h.$ Then $ g$ goes through the external similarity center X of (A), (B), while $ h$ goes through the internal similarity center Y of (A), (B) - we get nothing.[/hide]

Well known lemma: Let $ ABC$ be a triangle. $ (P)$ is circle that pass $ B,C$ and cut segment $ AB,AC$ at $ D,E$, respectively. $ (Q)$ is circle that touch $ BD,CE$ and externally tangent to circle $ (P)$ at $ M$. Bisector of $ \angle BMC$ pass through incenter of triangle $ ABC$.

Minor variation: Let $ ABC$ be a triangle and $ (P)$ is circle that pass $ B,C$ and cut ray $ AB$ beyoond $ A$ and ray $ AC$ beyond $ A$ at $ D,E$, respectively. $ (Q)$ is a circle that touch $ AD,AE$ and internally tangent to circle $ (P)$ at $ M$. Bisector of $ \angle BMC$ pass through incenter of triangle $ ABC$.

Problem:  Let $ (J_1),(J_2)$ touch $ (O)$ at $ M_1,M_2$,resectively and let $ K,L$ be midpoint of arc $ BC,CA$, respectively. $ KM_1$ is bisector of $ \angle CM_1B$. By lemma, $ KM_1$ pass through $ I_2$ and similarly, $ LM_2$ pass through $ I_1$. By applying Pascal theorem for $ A,K,M_1,B,L,M_2$, we get $ AK\cap BL$, $ KM_1\cap LM_2,$ $ AM_2\cap BM_1$ are collinear, or equivalently, triangle $ I_1AM_2$ and $ I_2BM_1$ are perspective. Hence, By Desargues theorem, $ I_1I_2,AB,M_1M_2$ are concurrent. $ M_1$ and $ M_2$ are external similitude center of $ (J_1),(O)$ and $ (J_2),(O)$, respectively. By Monge-D'Alembert theorem applying for $ (J_1),(J_2),(O),$ $ M_1M_2$ pass through $ X$, where $ X\equal{}J_1J_2\cap AB$ is external similitude center of $ (J_1),(J_2)$. Hence $ AB,J_1J_2,M_1M_2$ are concurrent. Together with fact that $ I_1I_2,AB,M_1M_2$ is concurrent, we get the result.

Dear Mathlinkers,
for this difficult and nice problem, you can see

http://perso.orange.fr/jl.ayme    vol. 20 "Shape and Mouvement" p. 17

Sincerely
Jean-Louis

The solution to this nice problem can be quite short ... if we are allowed to quote some famous results :

Let $ T_1 ~,~ T_2 $ be the tangency points of $ (J_1) ~,~ (J_2) $ with $ (ABC) $ ,
$ M_1 ~,~ M_2 $ the midpoints of minor arcs $ AC ~,~ BC $ respectively .

1. $ T_1T_2 ~,~ AB ~,~ J_1J_2 $ are concurrent  .

Proof: Let $ X = J_1J_2 \cap AB $ , then $ X $ is also the intersection point of the two external common tangents . Consider the inversion at $ X $ that preserves $ (ABC) $ , $ (J_1) $ and $ (J_2) $ are sent to each other , we see that $ X $ sends $ T_1 $ to $ T_2 $ so these three lines are concurrent at $ X $ .

2.  $ M_1 ~,~ I_1 ~,~ T_2 $  as well as  $ M_2 ~,~ I_2 ~,~ T_1 $ are collinear .

Proof: omitted

Therefore , we just need to prove that $ I_1I_2 ~,~ AB ~,~ T_1T_2 $ are concurrent , but it can be done immediately because of Pascal , by proving $ I_1 ~,~ I_2 ~,~ AB \cap T_1T_2 $ are collinear .

Another proof, using polarity transformation with center at the common focus of 2 conics and mapping them into 2 circles,
is at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=369005&p=2033012#p2033012[/url].


conics
ellipse
geometry
geometric transformation
reflection
quadratics
rectangle

Given four circles $\left(O_1\right)$, $\left(O_2\right)$, $\left(O_3\right)$, $\left(O_4\right)$. Construct a square ABCD whose sides AB, BC, CD, DA are tangent to these circles $\left(O_1\right)$, $\left(O_2\right)$, $\left(O_3\right)$, $\left(O_4\right)$, respectively.

Another problem:

Let M and N be given points and let d be a given line on the plane; also, let k be a given length. Find a point A on the line d such that AM + AN = k.

Rewording the second problem:

Given the foci of an elipse and its large axis ($M,N$ are foci and $k$ is the length of the large axis) and a line $d$, construct the intersection points between $d$ and the ellipse.

I have a construction, but I'll let others enjoy it for now :).

[b]Solution 1:[/b]

The locus of points with the constant sum of the distances $k = AK + BK$ from the points $A, B$ is an ellipse with major axis $a = \frac k 2$ and minor axis $b = \sqrt{a^2 - \left( \frac{AB}{2} \right)^2}$ (an easy construction using Pythagorean theorem). Let $M$ be the midpoint of the segment $AB$ and $MC$ a normal to $AB$ at the point $M$ such that $MC = b$. Project the ellipse to a circle $(M)$ centered at the point $M$ and with radius a (just draw the circle). All that is necessary is to project the line $l$ into a line $l'$ intersecting the circle $(M)$ at points $P', Q'$ and than drop normals to the line $AB$, intersecting the original line $l$ at the desired points $P, Q$. Let $X$ be the intersection of the line $l$ with the line $AB$. This point is one point of the line $l'$. Let the line $l$ intersect the normal $MC$ to $AB$ at the point $Y$. Construct the point $Y'$ on the normal $MC$, such that $\frac{MX'}{MX} = \frac a b$. $XY'$ is the projection $l'$ of the line $l$. In case the lines $l$ and $AB$ are parallel, the line $l'$ is also parallel to $AB$ and passes through the point $Y'$. In case the lines $l$ and $AB$ are perpendicular, project the ellipse into a circle $(M)$ with radius $b$ by rotating it around the minor axis, rather than around the major axis as before.

[b]Solution 2:[/b]

Draw a circle $(A)$ centered at the point $A$ and with radius $k$. Reflect the point $B$ in the given line $l$ into the point $C$. Construct 2 circles $(P), (Q)$ centered at points $P, Q$ on the line $l$, passing through the points $B, C$ and touching the circle $(A)$. Their centers $P, Q$ are obviously the solutions of the problem. To construct the circles $(P), (Q)$, draw an arbitrary circle $(R)$ centered at some point $R$ on the line $l$, passing through the points $B, C$ and intersecting the circle $(A)$ at points $E, F$. The line $BC$ is the common radical axis of the circles $(P), (Q), (R)$ and the line $EF$ is the radical axis of the circles $(A), (R)$. The lines $BC, EF$ intersect at the common radical center $X$ of the circles $(A), (P), (Q), (R)$. The radical axes of the circle pairs $(A), (P)$ and $(A), (Q)$ are their single common external tangents, which must pass through the radical center $X$. Construct tangents to the circle $(A)$ from the point $X$. Let $S, T$ be the tangency points. The lines $SA, TA$ intersect the given line $l$ at the desired points $P, Q$, the centers of the circles $(P), (Q)$. This is because the angles $\measuredangle ASX, \measuredangle ATX$ are both right.

[b]Solution 3:[/b]

Start as in solution 2, i.e., draw the circle $(A)$ centered at the point $A$ and with radius $k$ and reflect the point $B$ in the given line $l$ into the point $C$, but construct the circles $(P), (Q)$ with the help of an inversion. For example, invert the circle $(A)$ into a line $a'$ in the circle $(O)$ centered at the intersection $O$ of the line $BC$ and the circle (A) and with radius $OB$. The problem is then reduced to constructing the circles $(P'), (Q')$ passing through the inverted points $B', C'$ and touching the line $a'$. Finally, invert the tangency points $S', T'$ of the constructed circles $(P'), (Q')$ with the line $a'$ back into the tangency points $S, T$ of the circles $(P), (Q)$ with the circle $(A)$ and draw the lines $SA, TA$ intersecting the given line $l$ at the desired points $P, Q$. Alternately, invert the circle $(A)$ into a circle $(A')$ in the circle $(B)$ centered at the point $B$ and with radius $BC$. The unknown circles $(P), (Q)$ are inverted into lines $p', q'$ tangent to the inverted circle $(A')$ and passing through the point $C$, which is on the inversion circle $(B)$ and as such, it stays in place. Invert back the tangency points $S', T'$ of the lines $p', q'$ with the circle $(A')$ into the tangency points $S, T$ of the circles $(P), (Q)$ with the circle $(A)$ and draw the lines $SA, TA$ intersecting the given line $l$ at the desired points $P, Q$.

[quote="bolzano"]Erect a square ABCD with AB,BC,CD,DA respectively tangent to (O1),(O2),(O3),(O4)(They are given circles)[/quote]

Assume that the quadrilateral $O{}_1O{}_2O{}_3O{}_4$ is convex. Let $r_1, r_2, r_3, r_4$ be the radii of the circles $(O_1),\ (O_2),\ (O_3),\ (O_4)$,  $d{}_1{}_3 = O{}_1O{}_3,\ d{}_2{}_4 = O{}_2O{}_4$ the segment lengths between the opposite centers, $X$ the intersection of the center lines $O{}_1O{}_3, O{}_2O{}_4$ and $\phi = \angle O_1 X O_2$.  Assume that the problem has been solved and denote $a = AB$ the length of the square side. Let $T_1,\ T_2,\ T_3,\ T_4$ be the tangency points of the circles $(O_1),\ (O_2),\ (O_3),\ (O_4)$ with the square sides $AB,\ BC,\ CD,\ DA$ and $S_1,\ S_2,\ S_3,\ S_4$ the intersections of the lines $O{}_1T{}_1,\ O{}_2T{}_2,\ O{}_3T{}_3,\ O{}_4T{}_4$ with the opposite square sides $CD,\ DA,\ AB,\BC$. Obviously, $S{}_1O{}_1T{}_1 \parallel S{}_3O{}_3T{}_3  \parallel BC \parallel DA$ and $S{}_2O{}_2T{}_2 \parallel S{}_4O{}_4T{}_4  \parallel AB \parallel CD$. Denote the angles $\alpha = \angle O{}_1O{}_3S{}_3 = \angle O{}_3O{}_1S{}_1$ and $\beta = \angle O{}_2O{}_4S{}_4 = \angle O{}_4O{}_2S{}_2$. We have

$\pm r_1 + d{}_1{}_3\ cos(\alpha) \pm r_3 =  \pm r_2 + d{}_2{}_4\ cos(\beta) \pm r_4 = a$
$\beta - \alpha + \phi = \frac \pi 2$

Considering various combinations of the plus/minus signs, this leads to several quadratic equations for $cos(\alpha)$, each of which may or may not have real roots:

$\pm r_1 + d{}_1{}_3\ cos(\alpha) \pm r_3 = \pm r_2 + d{}_2{}_4\ sin(\phi - \alpha) \pm r_4 =$
$= \pm r_2 + d{}_2{}_4\ [sin(\phi)\ cos(\alpha) - cos(\phi)\ sin(\alpha)] \pm r_4$

$[\pm r_1 + d{}_1{}_3\ cos(\alpha) \pm r_3 - (\pm r_2 + d{}_2{}_4\ sin(\phi)\ cos(\alpha) \pm r_4)]^2 =$
$= [d{}_2{}_4\ sin(\phi)]^2 (1-cos^2(\alpha))$

These quadratic equations can be solved with compass and straightedge. Once the angle $\alpha$ is constructed, the direction of the lines $S{}_1O{}_1T{}_1 \parallel\ S{}_3O{}_3T{}_3 \parallel\ BC \parallel\ DA$ is known and it is sufficient to draw tangents to the circles $O_1,\ O_3$ parallel to this direction and tangents to the circles $O_2,\ O_4$ perpendicular to this direction. A rectangle with a given ratio of width and length can be erected on the 4 given circles in a similar way.


conics
geometry
projective geometry
geometry unsolved

Prove that any (projective) involution ($f$) from a conic to itself has the following property: $A f(A)$ passes through a fixed point, where $A$ varies through all points of the conic.

I tried to prove this, but I don't think I have the knowledge to do it. Maybe you'll enlighten me  :surf:

oh boy. if we finally start trying to understand all of grobber's "well-known" results in projective geometry, this could get very long ;)

does this mean i won't see a proof of this during my life time ? (or after...)

In general, for a point $U$ on th conic, let $U'$ be its image through the involution. Take any two points $A,B$ on the conic s.t. $A\ne B,B'$, and put $P=AA'\cap BB'$. The map taking $X$ (on the conic) to the second inetrsection between $PX$ and the conic is clearly an involution, and any involution is determined by the images of just two points, which do not coincide and are not the image of one another, such as $A$ and $B$ above, so these two involutions must coincide.

In order to prove that an involution is determined if we know the images of two points $A,B$ (with $A\ne B,B'$), we could do this:

For any other point $C$, we have $(A,B;B',C)=(A',B';B,C')$, so the image of $C$ is uniquely determined.

Are we all happy now, perfect_radio? :)

yes, we (me, my dogs, the spiders which reside in my house, the rats from the basement) are all happy now

and, strangely enough, i understood your solution


[quote="_"]What is the sum of the first 100 odd natural numbers?[/quote]

[hide]The sum of the first $n$ odd natural numbers is $n^2$.

So, the first 100 would be $100^2$, or 10000.[/hide]

[hide]the answer is 10000, as you can see sometimes on the little picture boxes on the left of the page. lol. cheaters...like me! =D[/hide]

whoops, i did mess up in my $\LaTeX$ a bit. :D ill edit it right away.  What do you mean by the little boxes that you can see solutions?  I can't see them :? its kind of cheap to just look at someone elses answer and then post it with no explination or anything :(

i think he means the box that has all the cool formulas.  Its only there if you have the AoPS skin.

[quote="BOOYA"]the answer is 10000, as you can see sometimes on the little picture boxes on the left of the page. lol. cheaters...like me! =D[/quote]

That's a bit cheap you know...

Anyway here's my answer

[hide]To find the answer we need to do this:

100^2 or           1+199
                        ----------- *100  =  10000
                             2[/hide]

[quote="risinglegend"]10000 is the answer[/quote]
Can you please stop posting just the answer and not hiding it. It's getting really annoying! ;)