geometry
incenter
ratio
area of a triangle
IMO
right triangle
IMO 1988

Lemma: Through the incenter $I$ of $\triangle{ABC}$ draw a line that meets the sides $AB$ and $AC$ at $P$ and $Q$, then:
\[ \frac{AB}{AP} \cdot AC + \frac{AC}{AQ} \cdot AB = AB+BC+AC \]
Proof of the lemma:
    Consider the general case: $M$ is any point on side $BC$ and $PQ$ is a line cutting AB, AM, AC at P, N, Q. Then:

$\frac{AM}{AN}=\frac{S_{APMQ}}{\triangle{APQ}}=\frac{\triangle{APM}+\triangle{AQM}}{\triangle{PQA}}=\frac{\frac{AP}{AB}\triangle{ABM}+\frac{AQ}{AC}\triangle{ACM}}{\frac{AP\cdot AQ}{AB \cdot AC}}=$

$=\frac{AC}{AQ}\cdot \frac{BM}{BC}+\frac{AB}{AP}\cdot \frac{CM}{BC}$

If $N$ is the incentre then $\frac{AM}{AN}=\frac{AB+BC+CA}{AB+AC}$, $\frac{BM}{BC}=\frac{AB}{AB+AC}$ and $\frac{CM}{BC}=\frac{AC}{AC+AB}$. Plug them in we get:
\[ \frac{AB}{AP} \cdot AC + \frac{AC}{AQ} \cdot AB = AB+BC+AC \]

Back to the problem
Let $I_1$ and $I_2$ be the areas of $\triangle{ABD}$ and $\triangle{ACD}$ and $E$ be the intersection of $KL$ and $AD$. Thus apply our formula in the two triangles we get:
\[ \frac{AD}{AE} \cdot AB + \frac{AB}{AK} \cdot AD = AB+BD+AD \]
and
\[ \frac{AD}{AE} \cdot AC + \frac{AC}{AL} \cdot AD = AC+CD+AD \]
Cancel out the term $\frac{AD}{AE}$, we get:
\[ \frac{AB+BD+AD-\frac{AB}{AK} \cdot AD }{AC+CD+AD- \frac{AC}{AL} \cdot AD }=\frac{AB}{AC} \]
\[ AB \cdot CD + AB \cdot AD - \frac{AB \cdot AC \cdot AD}{AL}=AC \cdot BD+ AC \cdot AD -\frac{AB \cdot AC \cdot AD}{AK} \]
\[ AB+AB \cdot \frac{CD}{AD}-\frac{AB \cdot AC}{AL}=AC+ AC \cdot \frac{BD}{AD} - \frac{AB \cdot AC}{AK} \]
\[ AB+AC - \frac{AB \cdot AC}{AL}=AB+AC - \frac{AB \cdot AC}{AK} \]
\[ \frac{AB \cdot AC}{AK} = \frac{AB \cdot AC}{AL} \]
So we conclude $AK=AL$.

Hence $\angle{AKI_1}=45^o=\angle{ADI_1}$ and $\angle{ALI_2}=45^o=\angle{ADI_2}$, thus $\triangle{AK_1} \cong \triangle{ADI_1}$ and $\triangle{ALI_2} \cong \triangle{ADI_2}$. Thus $AK=AD=AL$. So the area ratio is:
\[ \frac{E}{E_1}=\frac{AB \cdot AC}{AD^2} = \frac{BC}{AD} =\frac{BD+CD}{\sqrt{BD \cdot CD}}\geq 2 \]

I think we can be a bit quicker than that. Let X and Y be respective incenters of $ABD$ and $ACD$. Note that by spiral similarity taking $BDA$ to $ADC$, we have $XDY$ being a 45 degree rotation and dilation of both $BDA$ and $ADC$ (of course rotating in opposite directions). It follows then that $AK = AL$. Then $\angle AKX = 45 = \angle ADX \implies AK = AD$. Similarly, $AL = AD$. Then $\frac{[ABC]}{[AKL]}= \frac{AB \cdot AC}{AK \cdot AL}= \frac{AB \cdot AC}{AD^{2}}$. It's an easy finish from here; as shobber did, we can note that $BC \ge 2AD$, and so $\frac{AB \cdot AC}{AD^{2}}= \frac{BC}{AB}\cdot \frac{AB}{AD}= \frac{BC}{AD}\ge 2$.

[color=darkblue][b]Remark.[/b] Given are a triangle $ABC$ and a point $D\in [BC]$. Denote the incircles $C(I_{1},r_{1})$, $C(I_{2},r_{2})$ of the triangles $ABD$, $ACD$ respectively and the points $M\in AB\cap DI_{1}$, $N\in AC\cap DI_{2}$, $X\in AB\cap I_{1}I_{2}$, $Y\AC\cap I_{1}I_{2}$. Prove that $XY\parallel MN\Longleftrightarrow AX=AY$.[/color]

Remark by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=39796]28121941[/url]:

This was problem 5 at IMO 1988, proposed by Greece. As (ABC) (area) is greater or equal 2 times (AKL), the minimal value of the quotient is 2.
At Gazeta Matematica 1991(number 10), Neculai Roman published two generalizations of this problem:
1) Let ABC a triangle. Let k_1 the circle through A and B and tangent to AC; analogously, let k_2 the circle through A and C and tangent to AB. The second intersection of k_1 and k_2 is D. The line defined by the incenters of the triangles ABD and ACD meet the lines AB,AC in K and L, respectively. If S is the area of ABDC and T is the area of ACD, show that S is greater or equal to 2 times T.

2) Let ABC a triangle. with A>B, A>C. let D and D' two points of the segment BC such that angle CAD = angle ABC and angle BAD' = angle ACB. The line which join the incenters of ABD and ACD' intersect AB at K and AC at L.
If S = area(ABC) and T = area(AKL), then S is greater or equal to 4T*(sin(A/2))^2.

Here's a thoughtless bary bash[quote=orl]In a right-angled triangle $ ABC$ let $ AD$ be the altitude drawn to the hypotenuse and let the straight line joining the incentres of the triangles $ ABD, ACD$ intersect the sides $ AB, AC$ at the points $ K,L$ respectively. If $ E$ and $ E_1$ dnote the areas of triangles $ ABC$ and $ AKL$ respectively, show that
\[ \frac {E}{E_1} \geq 2.
\][/quote]
Set $ABC$ as reference triangle.Let $P=AI_1\cap BC$ and $Q=AI_2\cap BC$.Then clearly we have \[P=(0:b:a-b)\implies I_1=(a(a-b):bc:c(a-b))\]
\[Q=(0:a-c:c)\implies I_2=(a(a-c):b(a-c):bc)\]
Now let $K=(x:y:0)$.Then we have \[\left|\begin{array}{ccc}x  & y& 0 \\ a(a-b) & bc & c(a-b) \\ a(a-c) & b(a-c) & bc\end{array}\right|=0\implies K=(a-b:b:0)\]
Similarly $L=(a-c:0:c)$.Therefore \[\dfrac{[AKL]}{[ABC]}=\left|\begin{array}{ccc}1 & 0 & 0 \\ 1-\tfrac{b}{a} & \tfrac{b}{a} & 0 \\ 1-\tfrac{c}{a} & 0 & \tfrac{c}{a}\end{array}\right| = \dfrac{bc}{a^2}=\dfrac{bc}{b^2+c^2}\leq \dfrac{1}{2} \; \square\]
Thus we are done.


geometry
perimeter
trigonometry
limit
geometry open

This should be simple for you guys.

Basically, what is the relationship of the perimeters of regular polygons with a fixed area? (Preferrably more than it decreases as the number of sides increases)

Thanks!

$ \ell \equal{} BC$
$ n \equal{}$ number of sides.
$ S \equal{}$ area of the regular poligon.

We have $ \angle ADC\cong ADB \cong\frac {180^\circ}{n}$ , with $ n$ being the number of sides.

$ \tan \left(\frac {180^\circ}{n}\right) \equal{} \frac {\ell}{2h}\;\therefore\;h \equal{} \frac {\ell}{2\tan \left(\frac {180^\circ}{n}\right)}$

So the area is $ \;\;\frac {\ell}{2\tan \left(\frac {180^\circ}{n}\right)}\cdot\frac {\ell}{2}\cdot n\;\therefore\; S \equal{} \frac {\ell^2n}{4\tan \left(\frac {180^\circ}{n}\right)}$

Since $ S$ is fixed we have $ \ell \equal{} \sqrt {\frac {4S\tan \left(\frac {180^\circ}{n}\right)}{n}}$

And then the perimeter is

$ \boxed{nl \equal{} \sqrt {4Sn\tan \left(\frac {180^\circ}{n}\right)}}$

The perimeter increases as the number of sides increases. In fact:

$ \lim_{n\rightarrow \infty}\sqrt {\dfrac{4S\tan \left(\dfrac {\pi}{n}\right)}{\dfrac{1}{n}}} \equal{} \lim_{n\rightarrow \infty}\sqrt {4S\pi}$

Since the figure is a circle we have:

$ p \equal{} \sqrt {4\pi R^2\pi} \equal{} 2\pi R$ which is the perimeter of the circle, the maximum perimeter possible for a fixed area.


Prove that each member of the set of $ n\minus{}1$ consecutive integers

$ n!\plus{}2, n!\plus{}3,...,n!\plus{}n$

is divisible by a prime which does not divide any other member of the set

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=259996

Pierre.


LaTeX
quadratics
algebra unsolved
algebra

CAN YOU PLEASE HELP ME SOLVE THESE   FOR X ?
  HERE ARE THEY:

 A.(x-1)/(x-1)^2  - (x^3  + 1 )/(x+1)^2 = 6

 B. 1/(x) - 1/(x+6) =1(a) - 1(a+b)

A) $ 6\equal{}\frac{(x\minus{}1)}{(x\minus{}1)^2} \minus{} \frac{(x^3 \plus{} 1 )}{(x\plus{}1)^2}\equal{}\frac{1}{(x\minus{}1)} \minus{} \frac{(x^2\minus{}x\plus{} 1 )}{x\plus{}1}\equal{}$
$ \equal{}\frac{(x\plus{}1)\minus{}(x\minus{}1)(x^2\minus{}x\plus{} 1 )}{(x\minus{}1)(x\plus{}1)}\equal{}\frac{x\plus{}1\minus{}(x^3\minus{}x^2\plus{}x\minus{}x^2\plus{}x\minus{}1)}{x^2\minus{}1}$

$ 6x^2\minus{}6\equal{}x\plus{}1\minus{}x^3\plus{}2x^2\minus{}2x\plus{}1$
$ x^3\plus{}4x^2\plus{}x\minus{}8\equal{}0$

B) $ \frac{1}{x}\minus{}\frac{1}{x\plus{}6}\equal{}\frac{6}{x(x\plus{}6)}\equal{}\frac{b}{a(a\plus{}b)}\equal{}c$
$ cx^2\plus{}6cx\plus{}6\equal{}0$

hello, do you meant
A.$ \frac{x\minus{}1}{(x\minus{}1)^2}\minus{}\frac{x^3\plus{}1}{(x\plus{}1)^2}\equal{}6$
B.$ \frac{1}{x}\minus{}\frac{1}{x\plus{}6}\equal{}\frac{1}{a}\minus{}\frac{1}{a\plus{}b}$.
Sonnhard.

As far as $ \text{\LaTeX}$ is concerned, this may be extremely useful for a new member of ML:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5261[/url]

And concerning your problems, [b]mszew[/b] solved them: the first one reduces to a cubic equation, which doesn't have rational solutions, and it can't be trivially factored. Try Cardano or use a computer program (here's, for instance Mathematica's output:$ x_1 = 1.15276,\; x_2 = - 2.57638 - 0.549684 \;i,\; x_3 = - 2.57638 + 0.549684\;i$.
Second one is reduced to a quadratic equation. For explicit for of the solution just apply the formula for roots of quadratic equation.

[hide]Maybe a more suitable subforum for this type of problems would be intermediate topics?[/hide]

THANK YOU! BUT HOW DID WE OBTAIN [code]CX^2 + 6CX + 6[/code] ?

After multiplying both sides with $ x(x \plus{} 6)$ in the B), you get (assuming $ x(x \plus{} 6)\not \equal{} 0$)

$ x \plus{} 6 \minus{} x \equal{} cx(x \plus{} 6)$, hence $ cx^2 \plus{} 6cx \minus{} 6 \equal{} 0$.

Thanks , but can you do the sum 2 in details till the end ?

[b][color=darkblue]A.[/color]
$ x_1 \equal{} \minus{} \frac {4}{3} \plus{} \frac {1}{3}\sqrt [3]{62 \minus{} 3\sqrt {183}} \plus{} \frac {1}{3}\sqrt [3]{62 \plus{} 3\sqrt {183}}$[/b]
And two complex solutions.

[b][color=darkblue]B.[/color]

 $ x_1 \equal{} \frac { \minus{} 3c \minus{} \sqrt {3(2c \plus{} 3c^2)}}{c}$

$ x_2 \equal{} \frac { \minus{} 3c \plus{} \sqrt {3(2c \plus{} 3c^2)}}{c}$[/b]
where $ c\equal{}\frac{b}{a(a\plus{}b)}$


Alright, same rules as http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151259 , but we'll only have four people in this game (only sign up if you actually plan on playing, lol). It'll probably go quicker that way.  :)

Players:
1. [b]cincodemayo5590[/b]
2. [b]mnmath[/b]
3. [b]besttate[/b]
4. [b]Ignite168[/b]

[b]A fifth player can join, but we are going to start now.[/b] :) (Please ask if you'd like to join.)

IM IN

WE SHOULD HAVE 5

BAN Y FOR HEAD SUP

OK, let's start!  :)   Just copy what I have below, to keep score:

[quote="Round 1"]
[size=150]R[/size]

[u]Scores:[/u]

cincodemayo5590:
mnmath:
besttate:
Ignite168:
[/quote]

I guess to make it clearer, the letter that you guys are starting with is R.

[b]R[/b]

R[b]I[/b]

Scores:

cincodemayo5590:
mnmath:
besttate:
Ignite168:

RI[b]T[/b]

Scores:

cincodemayo5590:
mnmath:
besttate:
Ignite168:

RITU

Scores:

cincodemayo5590:
mnmath:
besttate:
Ignite168:

RITU[b]A[/b]

Scores:

cincodemayo5590:
mnmath:
besttate:
Ignite168:

Alright. The way the game works (sorry if this was unclear in the rules) is that, if you know you cannot continue the round without creating a full word, you should just say so, take the point, and begin a new round all in the same post. If someone takes too long to post (more than a couple days) we'll just give that person a point, and begin with the next person in order.

So, [b]mnmath[/b], you have one letter, G, in your score, and it is your turn to begin us with a new round.  :)

Oh. I was unclear about the rules.

Anyways,

[b]D[/b]

Scores:

cincodemayo5590:
mnmath: G
besttate:
Ignite168:

D[b]R[/b]

Scores:

cincodemayo5590:
mnmath: G
besttate:
Ignite168:

DRE

Scores:

cincodemayo5590:
mnmath: G
besttate:
Ignite168:

DRE[b]A[/b]

Scores:

cincodemayo5590:
mnmath: G
besttate:
Ignite168:

[quote="besttate"]you can fill in for me in 2 days when I leave if that's acceptable[/quote]

Yea, that's fine.  :)

And [b]mnmath[/b], it's your turn again. It's been two days already. [b]Samath[/b], maybe you should take [b]mnmath[/b]'s place instead if s/he doesn't reply soon.

Sorry. I was gone for 2 days. Now,

DREA[b]R[/b]

[b]B[/b]

Scores:

cincodemayo5590:
mnmath: GH
besttate:
Ignite168:


Gauss Duke college blogs geometry Mafia ARML

vote for anyone on aops except the admins, mods are okay, please do not vote for yourself, and do not vote more than once

[quote="gauss1181"]my vote goes for whoever picks me too :lol:[/quote]

new rule: you will not be allowed to copy any bright ideas that i have...

I am going to vote whoever made the first rule, or (^_^).

i vote for shirayuki because she is my best friend even though no one else here probably knows her 'cause even though she made an account she only went on aops like twice, so...does she still count???

I VOTE FOR MYSTICTERMINATOR PLZKTHXBAII.

(^_^)-1
iin77 - 1
shirayuki - 1
MysticTerminator-1

since i vote gauss, gauss automatically vote for me.

(^_^)-2
iin77 - 1
shirayuki - 1
gauss1181-1
MysticTerminator-1

since budi votes for winner, which is currently me, i get another vote.  :P

(^_^)-3
iin77 - 1
shirayuki - 1
gauss1181 - 1
MysticTerminator-1

[quote="crazychinesemaniac"][quote="mz94"]only one vote?
ok xoangieexo. it would be mystic, but hes a UNC fan  :mad:[/quote]

I suppose your a Duke fan. I vote for whoever reaches 15 first.[/quote]
 :clap2:  :clap2:

excellent detective work ;]

 :P

You might as well give a requirement as the "Top AoPSer." Like who's the smartest, etc.?

In this case, I vote for t0rajir0u.

I vote:

[hide]Wow this is hard. I think I'll choose MysticTerminator, isabella2296, Yongyi781, or (^_^) (just kidding)[hide]isabella2296[/hide][/hide]

[quote="stevenmeow"]both isabella and t0rajir0u spend alot of time making entries on their blogs about various areas of math
but i think t0rajirou's entries are better since they are more advanced although isabella's blog is better for the beginner

(hint i vote for t0rajir0u-- he also has the second most total posts)[/quote]
Maybe I would have voted for t0rajir0u if I knew who he was.

[quote="herefishyfishy1"]

 or (^_^) (just kidding)[/quote]

.......

[b]Vote:1=2[/b] because he is 1=2.

That's it.

no more.

I vote 1=2 because (^_^) already always win everything.

[b][i]The "I wanna play mafia so I'm making a vote count" vote count[/i][/b]
(^_^): 1=2, budi713, gauss1181, crazychinesemaniac
shirayuki: dreambold22
MysticTerminator: 27r, pythag011
iin77: kookamunga
gauss1181: (^_^)
Nerd_of_the_Ages: aanjohri
xoangieexo: mz94
t0rajir0u: 7h3.D3m0n.117, stevenmeow, TZF
isabella2296: herefishyfishy1
1=2: erabel, alexhhmun
Not voting: everyone else
With lots of people, it takes 40 votes for a lynch.

Unvote:

Vote: The person who will give me a freee copy of Mathematica  :P

STEVENMEOW. hes the smartest cat in the world

vote: 1=2 because he is a great ally.

To pythag: I can give you a free copy of Mathematica, unfortunately I won't guarantee that YOUR copy will be legal. [Yes I have my own legal copy, yay ARML!]

I'm not in this list? (Not that I deserve to be top AoPSer anyways...)


What did u think got the best out of the NBA draft? And who do you think will win next year's NBA championship? Which team will be a disappointment?

I just want to know your thoughts.

Celts got two steals in Green and Gomes.

Hakim Warrick was drafted so low in the draft. He's an all american first team and big east player of the year and he's taken 19th!!! i wonder why memphis of all teams picked him?

Raptors! Whooo! Go Charlie whats-his-face! Steal of the century!

Boy, did the Lakers screw up....drafting a high school kid to fill their most important role...good job ;)

[quote="furious"]i'm taking the suns and heat.[/quote]

 ;) Still trying to live up to those two over hyped teams...tsk...tsk...tsk...

Here's the next year's EAST and WEST finals..

EAST: Detroit vs. Indiana
WEST: San Antonio vs. Phoenix

I know SA will go to the Finals but it's a coin flip for the EAST

The reason I didn't pick Miami is because the Deisel is getting older and older and is in the downfall of his career. Also, here are my list of surprising teams for each conference(ordered from the biggest surprise and lower)

West:
GS (Baron Davis and Richardson are a great duo and did real good at end of season)
HOU (should've been in the 2nd round but the refs were on the Mavs side)
LAL (this team should make the playoffs especially since Phil Jackson is back)

East:
IND (Ron Artest is back and O'Neil and Jackson are great players)
PHI (Iverson will play just as well as he did this year and this team had great chemistry)
NJ (Watch out for this team, perhaps the best duo in the NBA?)

the heat forced a game 7 with detroit this year with numberous injuries (espically to shaq and wade at the end). shaq will have basically the same game as he did this year (i.e. no dramatic fall off) and wade will get better. i see them in the conference finals for sure.

[quote="noneoftheabove"]yes, but without the diesel dwayne wade can at least make it to the conference finals alone without shaq and shaq is still better than any other center out there even if he is getting worse.

Lebron will probably make it to the playoffs and i think the nuggets will go too.

Indiana??[/quote]

Without the Deisel, Wade can make the playoffs??? Did Kobe get to the playoffs without Shaq? Did any player get to the conference finals by HIMSELF in the last 3 years?? And yes, don't forget about Indiana, with Artest back, this team is perhaps better than 2004 when the won 61 games. Think about it, Steven Jackson, who's a great scorer, Jermaine O'Neil who is still young and will continue to get better, Jamaal Tinsley who's a superb PG, and they have a very good bench especially since they've demonstrated last year with the suspensions and injuries. May I also remind you that the Pacers this year beat Miami in 3 out of 4 games... Don't count this team out, this is one of the elite teams in the NBA.

[quote="colts18"][quote="noneoftheabove"] And who do you think will win next year's NBA championship? Which team will be a disappointment?[/quote]

I think that the Mavs will win next year. Possibly the Caveliers( I think i spelled it right). I think the Heat will be a big dissapointment.  :first:[/quote]

The Cavs should win close to 50 games next season (as long as all the players (Illgauskus, Mccinnis, etc.) don't quit like they did this season. Also, yes, the Mavs are a very dangerous team, the chemistry has done well and Jason Terry, Jerry Stackhouse, Michael Finley, etc. are great contributors, not to mention perhaps the best shooter in the game, Dirk Nowitzki.

The Pacers will definitely contend next year, unless Artest goes nuts again. The team made the playoffs as the 6 seed, despite having Artest for about 4 or so games, O'neal missing 10 or so, and Jackson missing a bunch more. True, Reggie's gone, but Granger, who was a terrific pick, should contribute. That team has a good chance for the finals.

[quote=Iversonfan2005]Without the Deisel, Wade can make the playoffs??? Did Kobe get to the playoffs without Shaq? Did any player get to the conference finals by HIMSELF in the last 3 years?? And yes, don't forget about Indiana, with Artest back, this team is perhaps better than 2004 when the won 61 games. Think about it, Steven Jackson, who's a great scorer, Jermaine O'Neil who is still young and will continue to get better, Jamaal Tinsley who's a superb PG, and they have a very good bench especially since they've demonstrated last year with the suspensions and injuries. May I also remind you that the Pacers this year beat Miami in 3 out of 4 games... Don't count this team out, this is one of the elite teams in the NBA.[/quote]

umm...well when shaq was with kobe wade made the playoffs by himself and now hes better then 2 years ago. And shaq is still the best center in the game even though hes isnt as good he can still help the team wade advanced to the conference finals when shaq was injured and sat out they swept everybody

Cavs need someone else besides James who can make plays. MJ didn't start becoming a championship collecting machine till he got some good role players (plus the triangle offense).

The Heat were really unlucky that Wade was out. Hopefully Shaq will be better and not be injured.

WARRIORS!!! The bay area team should do well in 2005-2006 with Davis and Richardson. The Warriors are a team that has lost so many good players (Jamison, Sprewell, Arenas, Hughes, Carter, Webber?) Hopefully they will make the playoffs. (Emphisis on Hopefully)

Kobe is stuck on a LA team with a new captain but not many new players. The question is, "will the triangle offense work for kobe's teammates?"

i dont think kobe is good enough to take it on single handedly with no supporting cast its basically like 1vs.5 every night except with people in the way its ridiculous

Predictions for various teams:

Cavs: Lebron James will play something like Jordan in his second year. Expect 45 and a 5-8 playoff berth. They lose first round, swept or in five games, with James averagin 40 points and scoring 60 or so in one of them.

Indiana: Start the season off badly, but end with a great record and one of the top four seeds in the East.

Detroit: Win somewhere between 48 and 52 games and lose in the Eastern Conference finals.

Miami: Eastern Conference champs, but it's all Wade. Shaq decides to take over the coaching job from Van Gundy; tired of running the floor.

well isnt it lebrons third year...well anyway i dont think lebron can score that many points if he had a supporting cast because then he wouldnt get the ball as much

[quote="tetrahedr0n"]Cavs: Lebron James will play something like Jordan in his second year. Expect 45 and a 5-8 playoff berth. They lose first round, swept or in five games, with James averagin 40 points and scoring 60 or so in one of them. [/quote]

You know that Jordan missed almost all of his second season with a broken foot, right?  And the East was so brutal that year (some things never change) that they won like 35 games and made the playoffs.  And if you've seen Lebron play, it's hard to think that he can score 60.  He's not an effortless scorer like, say, Kobe or McGrady - not yet, and he won't be next year either.  What I do expect is a season of 26-8-8 from him, which is still pretty darn respectable.  If they land a big free agent, I don't think the 3rd spot is unreasonable for the Cavs though, although I like Chicago and Washington more for that spot.  Indiana should win the East next year.

my west prediction

1. Spurs
2. Suns
3. Timberwolves
4. Mavs
5. Rockets
6. Nuggets
7. Grizzlies
8. Lakers

only reason Timberwolves are 3rd is because in NBA, division winners fill up first 3 seeds in the playoff regardless of their record against some of the 2nd ranked team in other divisions.

Conference Final : San Antonio vs. Houston

The biggest disappointment/bust: Seattle (not that they were good anyways)

The biggest improvement: Portland/Golden State

Why are we forgetting the Kings xxreddevilzxx?

I hope Rockets would really fly to the top but i know it is almost inpossible the next season. I just trust the Spurs.

Bobcats
they got 2 UNC dudes, should be great teamwork


Functional Analysis
advanced fields
advanced fields unsolved

Let $K$ be a compact subset of an infinite-dimensional Hilbert space $\mathcal H$ (real or complex; it doesn't matter). Prove that there is a compact subset $\widetilde K$ of $\mathcal H$ which contains uncountably many disjoint translates of $K$.

Edit:

Added the word "disjoint" :)

[quote="grobber"]Let $K$ be a compact subset of an infinite-dimensional Hilbert space $\mathcal H$ (real or complex; it doesn't matter). Prove that there is a compact subset $\widetilde K$ of $\mathcal H$ which contains uncountably many translates of $K$.[/quote]Let's say this is a real Hilbert space. Let $\mathcal H_1 = \mathcal H \times \mathbb R$, with the natural Hilbert space structure. Set $K_1 = K \times [0,1]$, which is compact. Pick $0 \ne y \in \mathcal H$ and define $F: \mathcal H_1 \to \mathcal H$ by $F(x,t) = x + ty$. This is continuous, and hence $\widetilde K = F(K_1)$ is compact. Then $\widetilde K = \cup_{0 \le t \le 1} \left(K+ty\right)$, a union of uncountably many translates of $K$.

Or did you want uncountably many [i]disjoint[/i] translates of $K$?  :|

[quote="grobber"]Yes, definitely disjoint. Sorry; I've edited it. :)[/quote]OK, I'll give it a try. Let $(e_k)_{k\ge 1}$ be an orthonormal set in $\mathcal H$, then it is well-known or an exercise that if $K \subset \mathcal H$ is compact then $K$ is closed, bounded, and
\[ \forall \epsilon \, \exists N \, \forall x \in K \, \sum_{k\ge N} |<e_k,x>|^2 \le \epsilon \]
i.e. $K$ is "equi-convergent", similar to Arzela-Ascoli. [i] If this is an orthonormal basis, this is equivalen t to the compactness of $K$, but we won't need this here.[/i]

Define the sequence of indices $n_1, n_2, \dots$ by the requirement that
\[ \forall x \in K \sum_{k\ge n_i} |<e_k,x>|^2 \le 2^{-2i} \]
Set $\mathcal H_1 = \mathcal H \times l^2$ and define $K_1 \subset \mathcal H_1$ as $K_1 = K \times \{y \in l^2 \big| |y_i| \le 2^{2-i} \, \forall i\}$. Then $K_1$ is compact. Now define $F: \mathcal H_1 \to H$ as $F(x,y) = x + \sum_{i \ge 1} y_i e_{n_i}$. This is a continuous map, so $\widetilde K = F(K_1)$ is compact. Clearly $K \subset \widetilde K$.

It remains to show that there are uncountably many disjoint translates of $K$ sitting somewhere in $\widetilde K$. Let $J \subset \mathbb N$ be an arbitratry set (there are uncountably many such sets), then let's set $K_J = K + \sum_{j \in J} 2^{2-j} e_{n_j} \subset \widetilde K$. In particular, $K_{\emptyset} = K$. Let's prove that $K_{I} \cap K_{J} = \emptyset$ if $I \ne J$:

Suppose $i \in I$ but $i \notin J$ and $k = n_i$. Then for any $x \in K_J$, $|<e_k,x>| \le 2^{-i}$ by construction. Now if $y \in K_I$, say $y = z + \sum_{j \in I} 2^{2-j} e_{n_j}$ with $z \in K$, then $|<e_k,y>| \ge 2^{2-i} - |<e_k,z>| \ge 2^{2-i} - 2^{-i} > 2^{-i}$, no matter  what $z$ is. Therefore $y \notin K_J$. :lol:

[quote="grobber"]Let $K$ be a compact subset of an infinite-dimensional Hilbert space $\mathcal H$ (real or complex; it doesn't matter). Prove that there is a compact subset $\widetilde K$ of $\mathcal H$ which contains uncountably many disjoint translates of $K$.[/quote]
Is this also true in a general Banach space?


I'm unsure about my answer to the following question. Anyone want to try the problem out and tell me what they got?

"It takes 852 digits to number the pages of a book consecutively. How many pages are in the book?"

[hide]
[i]* This is the correct solution, but not the original. Edited so future searchers may find what they're looking for more quickly.[/i]
Pages 1-9 total 9 pages and require the use of $1(9)$ digits.
Pages 10-99 total 90 pages and require the use of $2(90)$ digits.
Pages 100-999 total 900 pages and require the use of $3(900)$ digits.
Pages 100-N total $(N-99)$ pages and require the use of $3(N-99)$ digits.

Since pages 100-999 would alone require more than 852 digits we know the number of pages is in the hundreds. Since it takes $1(9)+2(90)=189$ digits to get to the hundreds we can see: $189+3(N-99)=852$ if we let N be the number of pages. Solving yields $\fbox{320}$ pages.
[/hide]

Look at this part one more time:
[quote="nigma"]Pages 10-99 then require an additional 178 digits ($2(99-10)$).[/quote]
Are you sure that pages 10-99 have $99-10=89$ pages?
[hide="Big hint"]
Pages 1-9 have $9$ pages, NOT $9-1=8$ pages ;) [/hide]

Thanks frt. Adjusting my logic I can setup the relationship $3x+189=852$ right?

[quote="nigma"]Thanks frt. Adjusting my logic I can set up the following relationship:
$3x+189=852$\\
$x=\fbox{221}$[/quote]
That's not the number of pages in the book.

[hide="Big Hints"]Your idea about using 1 digit from 1 to 9 is fine.
Your idea about using 2 digits from 10 to 99 is fine.
Define more clearly the variables, your x is not defined clearly. Of course you can use your x to find the number of pages in the book but I think you are already confused.

Think of it this way. Let $N$ be the number of pages in the book.

Ask the right questions:
How many digits are used from 1 to 9? $1\cdot(9-1+1)=9$. We used 9 digits so far from 1 to 9
How many digits are used from 10 to 99? $2\cdot(99-10+1)=180$. We used $9+180=189$ digits so far from 1-99
How many digits are used from 100 to 999? $3\cdot(999-100+1)=2700$. Since $852<2700+189$, $N$ is a 3-digit number.

How many digits are used from 100 to $N$? $3\cdot(N-100+1)=3(N-99)$

Set up the right equation: $9+180+3(N-99)=852$. Solve for $N$ and you got your answer.

Be specific when solving math problems. Problems are made to confuse but your job is to set things right.[/hide]

wow, I got pwned  :oops:

Thanks for the help & clarification.


how many of you guys are interested in doing a mathcounts countdown chat aol instant messenger? i figured it would be good practice especially with so much good competition!

Ooof, I would if I had AIM...when will this be on again?

im just trying to gather enough people to do it right now.

well we have to know your sn. anyone else?

what's up with AOL instant messaging i mean over here in utah its all MSN IM

I dont do mathcounts but Ill join. aalindsey100

who else has problems? i have some countdown problems...anymore people?

plus we need some like moderators to give the problems and keep count of scores. i figure someone will post the problem and it will run maybe like ciphering? give me your opinion.

one more thing...we should set a time each day. after school or later at night?

we need at least two moderators depending on how many people join...who wants to volunteer? if you do then im me.

one more thing...haha...we don't want too many people and really really advanced people doing it because then it'd be too easy for them so only people in 6-9th grade can participate.

this is going to be a daily thing until saturday feb 7th cuz thats when the chapter competition is.

so far...people we have are:

me: qwinn1190
ml2: centrocks3
theone853: reaperofthegrave
aalindsey100: aalindsey100
Rep123max: Rep345max
churchilljrhigh: kevinchen514

i'll get mathfiend to do it and any of white_horse_king88, HariSeldon888, pi^2 AlbinoSquirrel alc, CNRJames, and Mr. Ocax wanna do it? preferably lets keep this small.

each chat will run about a competition long which is about 70 questions or so? tell me what times all of you can do it but in the eastern coast time zone. we'll do this until saturday which is chapters for me.

we'll run the class like ciphering. rep you have problems am i correct? we'll do state level and we need a couple moderators. please help...we wanna get this going by wednesday!

for wednesday let's do 8 pm eastern time? rep will you supply problems and be moderator for the first day? if you do agree, then im me and i will tell you how it will all run. thanks all!

can i sign in late? (about 20 minutes). I have math circles on wednesday.

remember everyone....8:00 pm eastern time!

trinitrokirby (screen name 4 AIM) ... if my dad can let me miss swim team practice - ugh still trying to convince...

MathFiend21
trinitrokirby
kevinchen514
Rep345max
aalindsey100
reaperofthegrave
centrocks3
magnusmysterium
qwinn1190

i think thats a good amount of people...but if you want to join then post it. 8 pm eastern time.

for thursday:

handsome b good
MathFiend21
trinitrokirby
kevinchen514
Rep345max
aalindsey100
reaperofthegrave
centrocks3
magnusmysterium
confuted7
qwinn1190

moderators: qwinn1190 and centrocks3
time: 8-9:30PM eastern time

lol since the mathcounts counting class is going to go at the same time as the countdown chat i will do both and i will prepare for mathcounts at twice the efficiency (oh wait i'm too old for mathcounts)

lol it was sort of disaster. some people need a "good old fashion discipline" "it doesn't take a telepath to know what" they are thinking.

[quote="Rep123max"]rep345max ill do it, but if youre gonna do it use national countdown cause school and chapter are too easy.[/quote]
This is a kind of random comment but i personally think that chapter and school countdowns are easier than nats.

Mine is now whitehorseking88 because my old account got deleted or something.


I'm not sure whether you guys are already aware of this, but David Morin, one of the professors at Harvard for freshman mechanics, has published a full treatment of classical mechanics with many problems and exercises:

http://www.amazon.com/Introduction-Classical-Mechanics-Problems-Solutions/dp/0521876222

I already knew much of the material, having gone through the Physics Olympiad program and then through some college physics, yet I still found it very helpful for its rather challenging problems. It certainly reminded me of the AoPS books in the way it took a well-treated subject material and packaged it with good sets of hard problems. In terms of mechanics preparation for the physics olympiads, you will certainly get all you need and much more here.

Thanks a lot for the recommendation - I'll look into this.


induction
trigonometry
geometry unsolved
geometry

prove that $ \frac {1}{sin 2x}$+$ \frac{1}{sin 4x}$+.....+$ \frac {1}{sin 2^nx}$=$ cot x \minus{} cot 2^nx$

hello, i think induction is here the right way.
Sonnhard.

hello, for $ n=1$ we get
$ \cot(x)-\cot(2x)=\cot(x)-\frac{\cot(x)}{2}+\frac{\tan(x)}{2}=$
${ \frac{1}{2}\left(\cot(x)+\tan(x)\right)}=\frac{\sin^2(x)+\cos^2(x)}{2\sin(x)\cos(x)}=\frac{1}{\sin(2x)}$
Sonnhard.

hello, now we assume that
$ \frac{1}{\sin(2x)}\plus{}\frac{1}{\sin(4x)}\plus{}\cdots\plus{}\frac{1}{\sin(2^nx)}\equal{}\cot(x)\minus{}\cot(2^nx)$ and we have to show that
$ \frac{1}{\sin(2x)}\plus{}\frac{1}{\sin(4x)}\plus{}\cdots\plus{}\frac{1}{\sin(2^nx)}\plus{}\frac{1}{\sin(2^{n\plus{}1}x)}\equal{}\cot(x)\minus{}\cot(2^{n\plus{}1}x)$
comparing these both equations it is enough to prove that
$ \frac{1}{\sin(2^{n\plus{}1}x)}\equal{}\cot(2^nx)\minus{}\cot(2^{n\plus{}1}x)$
we get
$ \cot(2^nx)\minus{}\cot(2^{n\plus{}1}x)\equal{}\cot(2^nx)\minus{}\frac{1}{2}\cot(2^nx)\plus{}\frac{1}{2}\tan(2^nx))\equal{}$
$ \frac{1}{2}\left(\frac{\cos(2^nx)^2\plus{}\sin(2^nx)^2}{\sin(2^nx)\cos(2^nx)}\right)\equal{}\frac{1}{\sin(2^{n\plus{}1}x)}$
Sonnhard.


[hide="Click for solution"]
We can simply list:

The possible routes are MADCN, MACN, MBADCN, MBACN, MBCN, and MBN.  We know that this is all because the answer choices only go up to $ 6$, so the answer is $ \boxed{\textbf{(E)}}$.
[/hide]

Why are you answering you own questions?

For the Resources Section, worthawholebean said I was allowed to.


geometry
geometric transformation
reflection
analytic geometry
induction
number theory
relatively prime

Prove:

Given a rectangular billiard table with only corner pockets and sides of integer lengths m and n (with m and n relatively prime), a ball sent at a 45° angle from a corner will be pocketed in another corner after m+n-2 bounces.

I'm really not sure if that's possible to prove but if it can by elementary methods, I'd like to see =).

Are you sure it can be proven? Does the ball start at an integer coordinate location or something?

Yeah the ball starts at an integer coordinate location, but even so, I'm skeptical about this one.

INteresting..

Suppose m=1, n=2
Then you need 2 bounces to have the ball pocketed in another corner, instead of m+n-1=1

I may be wrong, but if that's true, it'd be a sufficient counterexample to disprove the statement..  :D

Your definitions are off by one; that's two steps, but only one bounce.

No, I think that only needs 1 bounce.. it bounces off the middle of the opposite long side, and then falls into the other corner..

It is possible to prove, and its not hard. I won't post a solution, but I think the subject gives it away:
[hide]
Billiard [b]reflection [/b]problem
[/hide]

I see.. I wasn't thinking about the bounce..  :blush:  :blush:  :blush:  :blush:
I'll try to solve it, now that I understand the problem properly..  :D

But then you could probably use induction on it, start with fixed m and n=1, and so on..

Suppose WLOG that $m \geq n$. We consider separately the horizontal and vertical movements of the ball.

Note that if the table were boundless, the ball would go from $(0,0)$ to $(d,d)$ where $d$ depends on how long it's been going. You set a restraint on it (the table boundaries) but it still moves the same distance, $d$ upwards and $d$ to the sides, except the path is not straight because the ball bounces.

We say the ball makes a distance $d$ when it moves $d$ in the x-coordinates and similarly in the y-coordinates (of course its coordinates may not actually change $d$, if there are bounces along the way). It is easy to see that after making a distance $d$, the ball goes from $(0,0)$ to $(a,b)$ where, if $d=2km+r$ with $r<m$, then $a=r$, and if $d=(2k+1)m+r$ with $r<m$, then $a=m-r$; $b$ is defined in the same way for $n$.

Then if the ball is pocketed it must end up at one of $(0,0), (m,0), (0,n), (m,n)$. The only way that can be achieved is if $r=0$ and similarly for $n$. That implies that $m|d, n|d$ so $mn|d$ because $m, n$ are coprime. It is trivial to see that $d=mn$ would pocket the ball, looking at the equations above, as well, and it's the smallest possible value. So we have to prove there are $m+n-2$ bounces when $d=mn$. We count separately the bounces on the m-sides and the n-sides. Note that the ball's "vertical" movement isn't altered when it hits a vertical side and the same goes for horizontal, so the bounces on the m-sides only depend on the distance made, and the same goes for the n-sides. It bounces on the m-sides every time it makes $m$, so since the path length is $mn$ it bounces $n$ times on them. Similarly, it bounces $m$ times on the n-sides. Since the last m-side and n-side bounce are on a pocket, those don't count , so in total it bounces $m+n-2$ times.

That looks right to me.

If you want to see the exact same thing geometrically:
[hide]
Draw a 'grid' of m by n billiard tables. Then, if we just consider each billiard table as a reflection of the one next to it, the balls path can be simply thought of as the line starting from the bottom left and going at 45 degrees without any bounces at all. Then we just need to find the first time the ball passes over a vertex - which happens at a multiple of m and a multiple of n, with smallest mn. Then if you just count the lines the ball passes over, its m+n-2 (subtracting two for the start and end.)
[/hide]


is from the Bay Area.  Congratulations to Bay Area IMO prticipants!

[quote="i_like_pie"]It would help if you mentioned their names...[/quote]
this is surprising, considering that 2/3 of the real moppers from BA were winners and were listed all over this spam forum


[quote="SUPERMAN2"]Let $ a,b,c,d > 0$ satisfying $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 4$.Prove that:
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} abc \plus{} bcd \plus{} cda \plus{} dab$≤$ 8$[/quote]
This inequality by PV THUAN,VN :) I know,he posted this inequality in mathlink


vector
linear algebra
linear algebra unsolved

If a linear transformation $\mathcal{A}$ on and $n$-dimensional vector space has $n+1$ eigenvectors such that any $n$ of them are linearly independent, does it follow that $\mathcal{A}$ is a scalar multiple of the identity?

Yes it does follow. There can be only one eigenvalue, and the eigenspace associated with that eigenvalue is $n$-dimensional, hence everything.

How did you determine that there could only be one eigenvalue?

The first $n$ eigenvectors $v_{i}$ form a basis in which $A$ is diagonal. The components of $v_{n+1}$ in this basis are all nonzero; otherwise there would be an $n$-tuple of dependent eigenvectors. Since $Av_{n+1}$ is a multiple of $v_{n+1}$, all diagonal entries of $A$ are the same.


abstract algebra
number theory
least common multiple
superior algebra
superior algebra unsolved

Let n be a positive integer. Prove that if an element of Sn has odd order, then it must be an even permutation.  Is every permutation of even order necessarily an odd permutation? Justify your claim

I really need help on this one. I would greatly appericiate a good explanation.

A general and abstract way of seeing it:
We have a homomorphism $f: S_n\mapsto \mathbb{Z}_2$, with kernel $A_n$. If $x^k=e$, then $(f(x))^k=e$ in the image. Since the image of an odd permutation has order 2, the order of the permutation itself must be a multiple of 2.

A more concrete view: look at the cycle decomposition. The order of a permutation is the lcm of the lengths of its cycles, and every cycle of odd length is an even permutation, so an odd permutation has some cycle of even length.

For the second: try to find a counterexample.

how do you know that every cycle of odd lenght has an even permutation?  Please explain in more depth.

man I'm super confused can anyone help me figure out this problem.  is there something about permutations that i am missing i don't have much information on permutations and i'm trying to wing this one the best i can.  google isn't providing much info about permutation and i really could use some defined help.

There is the canonical homomorphism $sgn: S_n \rightarrow \{ -1 , 1 \} , \pi \mapsto \prod_{1\leq i<j\leq n} \frac{\pi(i) - \pi(j)}{i-j}$.
Now let $x \in S_n$ be an element of odd order $o$. We want to prove that $sgn(x)=1$.
Assume that $sgn(x)=-1$, then $1=sgn(e)=sgn(x^o)=sgn(x)^o = (-1)^o=-1$ which is clearly not true.
For the second question: take $(12)(34)$.

chill out gap, here's the most fundamental stuff on permutations on {1,2,...,n} :

-every element can be written as a product of cycles

-every cycle can be written as a product of transpositions
in this way (a1..... an) = (a1 a2)(a1 a3)....(a1 an)

-therefore every permutation is a product of transpositions

-when  a permutation is multiplied with a transposition, the number of cycles goes up with one, or down with one

- definition : a permutation is even iff the number of transpositions in its cycle is even (by previous fact, this is well defined)

- the order of a permutation is the greatest common multiple of all the cycle lengths

That should do it , some examples :

n=4   (123) =(12) (13) has order three, and is even permutation
n= 2 (12) has order two and is odd permutation
n= 4 (12)(34) has order two and is even permutation

When the order is odd, all cycle lengths are odd, and thus all those cycles are a product of an even number of transpositions!


Calculate:

$\lim_{n\to\infty} \frac{\binom{2n}{n}}{2^{2n}}$

[hide]$\frac{\binom{2n}{n}}{2^{2n}}=\frac{(2n)!}{(n!)^22^{2n}}$. Distributing the $2n$ factors of 2 in the denominator by giving one to each factor of each $n!$, this is $\frac{(2n)!}{(2*4*6*\cdots*2n)^2}$.

Cancelling out the even factors from $(2n)!$, the fraction is $\frac{1*3*5*\cdots*(2n-1)}{2*4*6*\cdots*2n}=\frac 12 *\frac 34 *\frac 56 *\cdots*\frac{2n-1}{2n}$.

It turns out that $\frac 12 *\frac 34 *\frac 56 *\cdots*\frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}$, with equality only when $n=1$. The base case is $\frac 12$ on both sides.

The inductive step involves proving that $\frac{1}{\sqrt{3n+1}}*\frac{2n+1}{2n+2}< \frac{1}{\sqrt{3n+4}}$. Squaring and cross-multiplying, this is equivalent to $(3n+4)(2n+1)^2< (3n+1)(2n+2)^2$.

The left side expands to $12n^3+28n^2+19n+4$ while the right side is $12n^2+28n^2+20n+4$. So the inequality is true.

$\frac{\binom{2n}{n}}{2^{2n}}\leq \frac{1}{\sqrt{3n+1}}$, so $\lim_{n\rightarrow \infty} \frac{\binom{2n}{n}}{2^{2n}}=\boxed{0}$.[/hide]

My solution:
$a_n=\frac{1}{2}\cdot\frac{3}{5}\cdot\cdots \frac{2n-1}{2n}\cdot$
Let $b_n=\frac{2}{3}\cdot \frac{4}{5}\cdot\cdots\frac{2n}{2n+1}\cdot$
For $n\in\mathbb{N}$ the inequality holds:
$\frac{n}{n+1}<\frac{n+1}{n+2}$, so $a_n<b_n$. Notice, that $a_nb_n=\frac{1}{2n+1}$

$a_n^2<\frac{1}{2n+1}$
$0<a_n<\frac{1}{\sqrt{2n+1}}$

$\lim_{n\to\infty}a_n=0$

By the way. How do we call in English the following theorem:

We're given three sequences:$a_n, b_n, c_n$ such that:
$\lim_{n\to\infty}a_n=g$
$\lim_{n\to\infty}c_n=g$

and for all natural $n$ the inequality holds:
$a_n\leq b_n\leq c_n$
Then $\lim_{n\to\infty}b_n=g$

[quote="dondigo"]By the way. How do we call in English the following theorem:

We're given three sequences:$a_n, b_n, c_n$ such that:
$\lim_{n\to\infty}a_n=g$
$\lim_{n\to\infty}c_n=g$

and for all natural $n$ the inequality holds:
$a_n\leq b_n\leq c_n$
Then $\lim_{n\to\infty}b_n=g$[/quote]

I think it's called Pinching Theorem, Squeezing Theorem, Sandwich Theorem or (my personal favorite) Two Policemen Theorem.

[quote="Gyan"][
Okay,  just to make it a  little more interesting ,   can you show the value (when n is large) is about $\frac 1 {\sqrt{\pi n }}$

Or to prove that:

$\sqrt{\frac 2 {(2n+1) \pi}} < \frac{\binom{2n}{n}}{2^{2n}}< \sqrt{\frac 2 {(2n) \pi}}$[/quote]

I'm afraid I don't know how. But I'd like to see the solution :)

There is a theorem that says that if $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=l$ and if$l>1$ then $a_n\to\infty$ else if $l<1$then$a_n\to 0$ and finally if $l=1$ we cannot pronounce ouserselfs on the limit. This theorem can be enhanced by the following one $l=\lim_{n\to\infty}(\frac{a_{n+1}}{a_n})^n$ the statements about the positioning of $l$ remain the same.
So let $a_n=\frac{\binom{2n}{n}}{2^{2n}}$.
 Thus we have $\frac{a_{n+1}}{a_n}=\frac{2n+1}{2n+2}$ from where we get $(\frac{a_{n+1}}{a_n})^n=(\frac{2n+1}{2n+2})^n=(1-\frac{1}{2n+2})^n=[(1-\frac{1}{2n+2})^{-(2n+2)}]^{-\frac{n}{2n+2}}$ so $\lim_{n\to\infty}(\frac{a_{n+1}}{a_n})^n= e^{-\frac{1}{2}}=\frac{1}{\sqrt{e}} <1$. So according to that theorem we have that $a_n \to 0$.

dondigo:
You may like to look  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=walli&t=15144]here[/url]  a similar problem discussed before.
 Also look for "stirling approximation" for $n!$ and it is straight-forward  if you know calculus.. otherwise it is some what hard.
You can look for stirling approximation in google , for example
Also [url=http://hyperphysics.phy-astr.gsu.edu/hbase/math/stirling.html]here.[/url]

Basically useful formula is $n! \approx \sqrt{2\pi n } (n/e)^n$


Given real numbers $a, x_0$ and $y_0.$ Let us define two sequences $(x_k)$ and $(y_k)$ recursively for $n = 0, 1, 2, \ldots$:

\[x_{n+1} = a(x_n - y_n)\]

\[y_{n+1} = a(x_n + y_n).\]

Determine dependently on $x_0$  and $y_0$ all numbers $a$ such that the sequence $(x_k)$ is periodic.

One answer is right. I don't see any proper case distinction. Where do you show that these solutions are correct or that no further solutions exist? Try again.  :)


any one have? want to check my answers..thank u!

The questions are all up in the Contests section (button is in the top toolbar). The answers are in another thread in this forum.

but as i click the contest, it shows:

MathLinks Contest - Edition 1
MathLinks Contest - Edition 2
MathLinks Contest - Edition 3
MathLinks Contest - Edition 4
MathLinks Contest - Edition 5
MathLinks Contest - Edition 6
MathLinks Contest - Edition 7

it is not contest, but resources...
anyway, try this...
http://www.mathlinks.ro/resources.php?c=182&cid=44&year=2009

It's "Contests" under the AoPS theme and "Resources" under MathLinks.


a wonderful site for olympiad
http://www.grupoteorema.mat.br/links.html :roll:

Wow, thank you very much! I must vote for you 6*  :wink: But what is language here ?  :maybe:


ratio
number theory
prime numbers
number theory proposed

Given 4 consecutive terms of a progression (the arithmetical ratio is different from 0 ) . They are co-prime numbers . 4 numbers are not perfect squares at the same time but the product of them is a perfect square . Prove that : The product of them is divisible by $2520^2$

Do you mean an arithmetic progression or a geometric one?

Numbers in geometric progression wouldn't be coprime ;)

This problem is very strange . If it's true , it'll be very nice

Are that numbers coprime all together $i.g. (x,y,z,t)=1$ or pairwise and is it $none-of-them-is-square$ or $they-aren't-all-squares$?


Please folks, help me on this problem.

Let X be a vector space and U, V subspaces of X. Show that

(U+V)/U and U/(intersection of U and V) are isoform.

Thnaks!

That can't be right (just let $U$ be one-dimensional, $V$ two-dimensional, and the intersection trivial.) I assume you meant that

$(U+V)/U$ is isomorphic to $V/(U\cap V).$

You've got the statement wrong; that should be $V/(U\cap V)$.

This isn't restricted to vector spaces- the isomorphism is canonical, and works just as well in many other algebraic structures. Map $(u+v)+U$ to $v+U\cap V$. The inverse maps $v+U\cap V$ to $v+U$.

yes, yes! You guys were right. I typed it wrong...sorry!
How to type the math like you do? I am new hehe at the forum....

But still, how to prove the problem? How to show that it is a injection and surjective?

thanks again...


algebra
polynomial
modular arithmetic
search
number theory proposed
number theory

The unsolved problem :

If n is an integer larger than 1, must there be integers x, y, and z, such that 4/n=1/x+1/y+1/z?
A number of the form 1/x where x is an integer is called an Egyptian fraction.
Thus, we want to know if 4/n is always the sum of three Egyptian fractions, for n>1.

 if 4/n=1/x + 1/y + 1/z, then there exists a third degree polynomial x^3 + ax^2 + bx + c where 4/n=b/c for integers b and c. This is because 1/x + 1/y + 1/z=(z(x+y) +xy)/xyz. And the polynomial (a+x)(a+y)(a+z)=a^3 + (x+y+z)a^2 + (z(x+y)+xy)a +xyz, which proves the statement when b=z(x+y) +xy and c=xyz.

Your thoughts?

[quote="Christine"]The unsolved problem :

If n is an integer larger than 1, must there be integers x, y, and z, such that 4/n=1/x+1/y+1/z?
A number of the form 1/x where x is an integer is called an Egyptian fraction.
Thus, we want to know if 4/n is always the sum of three Egyptian fractions, for n>1.
[/quote]

Hello !

I don't have a complete answer but an intermediate one : decomposition $ \frac{4}{n}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is possible at least for any $ n>1$ containing one prime divisor equal to $ 2\pmod{3}$ or such that $ n\neq 1\pmod{24}$

$ 1).$  It's impossible for $ n=1$ (obvious)

$ 2).$ It is possible for any positive integer having a divisor equal to $ 2\pmod{3}$ :
$ n=pq$ with $ q=2\pmod{3}$ implies $ \frac{4}{n}=\frac{1}{n}+\frac{1}{p\frac{q+1}{3}}+\frac{1}{pq\frac{q+1}{3}}$
So it is possible, for example, for any even $ n$.

$ 3).$ It is possible for any positive integer equal to $ 3\pmod{4}$ :
$ n=3\pmod{4}$ implies $ \frac{4}{n}=\frac{1}{\frac{n+1}{2}}+\frac{1}{\frac{n+1}{2}}+\frac{1}{n\frac{n+1}{4}}$

$ 4).$ It is possible for any positive integer equal to $ 13\pmod{24}$ :
$ n=13\pmod{24}$ implies $ \frac{4}{n}=\frac{1}{\frac{n+3}{4}}+\frac{1}{\frac{n(n+3)+8}{12}}+\frac{1}{n\frac{n+3}{8}\frac{n(n+3)+8}{12}}$

So, it remains all positive integers containing only prime divisors equal to $ 1\pmod{3}$ and such that $ n=1\pmod{24}$
I search again.

Thanks pco,

I've just realized that this is the Erdős–Straus conjecture, you can find it at
http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Straus_conjecture

At the bottom of the "Erdos-Straus conjecture" page  in wikipedia, you see
Webb, William A. (1970). "On 4/n = 1/x + 1/y + 1/z". Proceedings of the American Mathematical Society 25 (3): 578–584. MR0256984.

i don't have access to that. Does anyone have the access?


[quote="ChemicalWarfare"]I need the problem written as a polynomial or a mixed expression.  I dont understand how to solve it with long devision.  Which is what my teacher wants.
The problem is
y 2 -2y+5 / y+1

The book says the answer is y-3 + 8\y+1

I have to solve it with long devision, can somone show me the process?[/quote]
Well,  you only need to expand the numerator in powers of $y+1$, I mean
\[
\frac{y^2-2y+5}{y+1}=\frac{(y+1)^2-4(y+1)+8}{y+1}=\frac{(y+1)^2}{y+1}-\frac{4(y+1)}{y+1}+\frac{8}{y+1}
\]
then is equal to
\[
= (y+1)-4+\frac 8{y+1}=y-3+\frac{8}{y+1}.
\]
And that's all.  U understood ?

RX


Let $x_1,...,x_m,y_1,...,y_n$ be positive integers such that $x_1+...+x_m = y_1+...+y_n<mn$. Prove that in the equality $x_1+...+x_m = y_1+...+y_n$ one can cancel some terms (not none, and not all)

[quote="pleurestique"]seems to be false...

$1+1+1+3 = 2+2+2 < 4\cdot 3$

but there are no common terms on the two sides to cancel out. or do you mean something else by cancelling out? do you mean that some subset of $x$'s has a sum equal to that of some subset of $y$'s and these can be cancelled?[/quote]

The problem means that some subset of the $x$'s has a sum equal to the sum of some subset of the $y$'s. So like $1+1 = 2$ or $1+3 = 2+2$ in your example. This problem is in Math Olympiad Treasures, btw. The solution given in the book uses induction.


Let $(C)$ be a conic and $ABC$ is a triangle such that the sides of $BC, CA, AB$ intersect the conic at $M_{1}$ and $M_{2}$, $N_{1}$ and $N_{2}$, $P_{1}$ and $P_{2}$. Suppose there exist the double elements $X_{1}$ and $X_{2}$ of the involution formed by $(B, C)$ and $(M_{1}, M_{2})$ and define $Y_{1}, Y_{2}, Z_{1}, Z_{2}$ similarly. Show that the six points $X_{1}, X_{2}, Y_{1}, Y_{2}, Z_{1}, Z_{2}$ are the vertices of a complete quadrangle.

Suppose an involutions formed by conjugate pairs of points $(A, A'), (B, B'),....$. Double elements $M$ and $M'$, if exist, are self conjugate points $(M, M), (M', M')$. I hope it is clear.  :)

It's known that double points in an involution are harmonic conjugates WRT two pairs of homologous points, so WLOG assume that $X_1,Y_1,Z_1$ are on $\overline{BC},$ $\overline{CA},$ $\overline{AB},$ respectively. $X_2,Y_2,Z_2$ are then on the extensions of $BC,CA,AB.$ By property of the involution $B \mapsto C,$ $M_1 \mapsto M_2,$ $X_1 \mapsto X_1,$ we obtain

$\frac{\overline{BM_1} \cdot \overline{BM_2}}{\overline{CM_1} \cdot \overline{CM_2}}=\frac{\overline{X_1B} \cdot \overline{X_1B}}{\overline{X_1C} \cdot \overline{X_1C}}= \frac{\overline{X_1B}^2}{\overline{X_1C}^2}$

Similarly, we have the expressions

$\frac{\overline{CN_1} \cdot \overline{CN_2}}{\overline{AN_1} \cdot \overline{AN_2}}=\frac{ \overline{Y_1C}^2}{\overline{Y_1A}^2} \ , \ \frac{\overline{AP_1} \cdot \overline{AP_2}}{\overline{BP_1} \cdot \overline{BP_2}}=\frac{ \overline{Z_1A}^2}{\overline{Z_1C}^2} \Longrightarrow$

$\frac{\overline{X_1B}^2}{\overline{X_1C}^2} \cdot \frac{ \overline{Y_1C}^2}{\overline{Y_1A}^2} \cdot \frac{ \overline{Z_1A}^2}{\overline{Z_1C}^2}=\frac{\overline{BM_1} \cdot \overline{BM_2}}{\overline{CM_1} \cdot \overline{CM_2}} \cdot \frac{\overline{CN_1} \cdot \overline{CN_2}}{\overline{AN_1} \cdot \overline{AN_2}} \cdot \frac{\overline{AP_1} \cdot \overline{AP_2}}{\overline{BP_1} \cdot \overline{BP_2}}.$

By Carnot's theorem for the conic $\mathcal{C}$ intersecting the sidelines of $\triangle ABC$ at 6 points, the RHS of the latter expression equals 1 $\Longrightarrow$

$\frac{\overline{X_1B}}{\overline{X_1C}} \cdot \frac{\overline{Y_1C}}{\overline{Y_1A}} \cdot \frac{ \overline{Z_1A}}{\overline{Z_1C}}=-1.$

Hence, by Ceva's theorem $AX_1,BY_1,CZ_1$ concur at a point $P$ $\Longrightarrow$ $X_2,Y_2,Z_2$ lie on the trilinear polar of $P$ WRT $\triangle ABC$ $\Longrightarrow$ $X_2 \in Y_1Z_1,$ $Y_2 \in Z_1X_1$ and $Z_2 \in X_1Y_1,$ i.e. $X_1,X_2,Y_1,Y_2,Z_1,Z_2$ are the vertices of the complete quadrilateral bounded by the sidelines of the cevian triangle of $P$ and its trilinear polar.


you can find the solution in the magazin of the "bashgahe danesh pajohan javan" and it has been proved by "Hesame mahdavifar". :)

Posted few times already - use search button

[quote="yash_17"]If n,a,b is natural & (a,b)=(a,n)=(b,n)=1
	prove that:
	$\frac{1}{a}+\frac{1}{a+b}+\frac{1}{a+2b}+...+\frac{1}{a+nb}$
	[b]is not natural.[/b][/quote]
 I can only prove in case $b=1$ ! Who have a generation solution  :maybe:

See here for a solution:
https://artofproblemsolving.com/community/u257113h1331945p8387502


Prove that

\[ \sum{\frac{x}{\sqrt{kx^{2}\plus{}y^{2}}}\ge{\frac{1}{\sqrt{k}}\]

holds $ \forall (x,y,z)\in{R_{\plus{}}^{3}}$ if and only if $ k\ge{\frac{1}{8}}.$

[quote="fjwxcsl"]Prove that
\[ \sum{\frac {x}{\sqrt {kx^{2} \plus{} y^{2}}}\ge{\frac {1}{\sqrt {k}}\]
holds $ \forall (x,y,z)\in{R_{ \plus{} }^{3}}$ if and only if $ k\ge{\frac {1}{8}}.$[/quote]

[quote="fjwxcsl"][quote="hedeng123"][quote="fjwxcsl"]Prove that
\[ \sum{\frac {x}{\sqrt {kx^{2} + y^{2}}}\ge{\frac {1}{\sqrt {k}}}}\]
holds $ \forall (x,y,z)\in{R_{ + }^{3}}$ if and only if $ k\ge{\frac {1}{8}}.$[/quote][/quote]

Very nice proof ,hedeng123!

In addition,let $ {x = y = z = 1},$ we can get the necessary condition $ k\ge{\frac {1}{8}.}$[/quote]

thank you  :lol:

[quote="hedeng123"][/quote][quote="fjwxcsl"][quote="kuing"][quote="fjwxcsl"]Prove that
\[ \sum{\frac {x}{\sqrt {kx^{2} + y^{2}}}\ge{\frac {1}{\sqrt {k}}}}\]
holds $ \forall (x,y,z)\in{R_{ + }^{3}}$ if and only if $ k\ge{\frac {1}{8}}.$[/quote]
\[ \sum{\frac {x}{\sqrt {kx^{2} + y^{2}}}\ge{\frac {3}{\sqrt {k + 1}}}}\]
holds $ \forall (x,y,z)\in{R_{ + }^{3}}$ if $ {\frac {1}{8}} \ge k \ge 0.$[/quote]

How nice!

How to prove?[/quote][quote="hedeng123"][/quote]

Nice answer,hedeng123!  But

$ \frac {x^2}{\sqrt {kx^2 + y^2z^2}},\cdots$

 should be changed

 $ \frac {x^2}{\sqrt {kx^4 + y^2z^2}},\cdots$

[quote="fjwxcsl"][/quote][quote="hedeng123"][/quote][quote="fjwxcsl"][quote="kuing"][quote="fjwxcsl"]Prove that
\[ \sum{\frac {x}{\sqrt {kx^{2} + y^{2}}}\ge{\frac {1}{\sqrt {k}}}}\]
holds $ \forall (x,y,z)\in{R_{ + }^{3}}$ if and only if $ k\ge{\frac {1}{8}}.$[/quote]
\[ \sum{\frac {x}{\sqrt {kx^{2} + y^{2}}}\ge{\frac {3}{\sqrt {k + 1}}}}\]
holds $ \forall (x,y,z)\in{R_{ + }^{3}}$ if $ {\frac {1}{8}} \ge k \ge 0.$[/quote]

How nice!

How to prove?[/quote][quote="hedeng123"][/quote][quote="fjwxcsl"]

Nice answer,hedeng123!  But

$ \frac {x^2}{\sqrt {kx^2 + y^2z^2}},\cdots$

 should be changed

 $ \frac {x^2}{\sqrt {kx^4 + y^2z^2}},\cdots$[/quote]

 :oops: I have changed,thank you.

[quote="hedeng123"][/quote][quote="fjwxcsl"][/quote][quote="hedeng123"][/quote][quote="fjwxcsl"][quote="kuing"][quote="fjwxcsl"]Prove that
\[ \sum{\frac {x}{\sqrt {kx^{2} + y^{2}}}\ge{\frac {1}{\sqrt {k}}}}\]
holds $ \forall (x,y,z)\in{R_{ + }^{3}}$ if and only if $ k\ge{\frac {1}{8}}.$[/quote]
\[ \sum{\frac {x}{\sqrt {kx^{2} + y^{2}}}\ge{\frac {3}{\sqrt {k + 1}}}}\]
holds $ \forall (x,y,z)\in{R_{ + }^{3}}$ if $ {\frac {1}{8}} \ge k \ge 0.$[/quote]

How nice!

How to prove?[/quote][quote="hedeng123"][/quote][quote="fjwxcsl"]

Nice answer,hedeng123!  But

$ \frac {x^2}{\sqrt {kx^2 + y^2z^2}},\cdots$

 should be changed

 $ \frac {x^2}{\sqrt {kx^4 + y^2z^2}},\cdots$[/quote][quote="hedeng123"]

 :oops: I have changed,thank you.[/quote]

$ \sqrt {kx^4+x^2y^2z^2+ky^4+x^2y^2z^2+kz^4+x^2y^2z^2},.........$

 should be changed

 $ \sqrt {kx^6+x^2y^2z^2+ky^6+x^2y^2z^2+kz^6+x^2y^2z^2},.........$

Notice $ .......$!


Chloroform is a liquid once used for anesthetic.  What is the volume of $ 5.0$ g of chloroform if the density of chloroform is $ 1.49$ g/mL?

Is there something missing from the problem?  I think I need the mass...

Thanks in advance!

"Does this rag smell like chloroform to you?" --Pick-up joke

Density = mass / volume. Thus, you could find the volume if you have mass and density, and you are given both parameters.


Prove that, for any positive integer $ n$, there exists a polynomial $ P(x)$ of degree at most $ n$ whose coefficients are all integers such that, $ P(x)$ is divisible by $ 2^n$ for every even integer $ x$, and $ P(x)\minus{}1$ is divisible by $ 2^n$ for every odd integer $ x$.

For example \[ P(x)\equal{}\prod_{i\equal{}1}^{2^{n\minus{}1}\minus{}1} (x\minus{}2i).\]

I'm not sure what you mean, Rust, since your polynomial doesn't seem to have degree at most $ n$...


[quote="MCrawford"]Palindromic primes are numbers that are prime and which read the same backwards and forwards; for example, 11 is a palindromic prime.  What is the sum of all palindromic primes between 100 and 200?[/quote]

[hide]the numbers possible are
101,111,121,131,141,151,161,171,181,191

the only ones that work are 101,131,151,181,191

therefor the sum is 755[/hide]


Αφού μελετήσαμε τη διαίρεση πολυωνύμων στη Β Λυκείου, σκέφτηκα "γιατί να μην έχουμε πρώτα πολυώνυμα?"
Υπάρχει τίποτα σχετικό ή να αναπτύξω κάποιες προτάσεις?

Yparxei i algevra twn poliwnimwn opou kai ekei iparxoun "prwtoi arithmoi" pou einai ta anagwga poliwnima. Opws kai stous akeraious iparxei i eukleideia diairesi pou pire to onoma tis akrivws epeidi iparxoun antistoixes idiotites. Omoia iparxei i prwtogenis analisi kathe poliwnimou se anagwga poliwnima (opws kai stous fisikous i analisi se prwtous paragontes) ktl ktl...

Ean anevoume ligo pio panw, milame gia ton daktilio twn poliwnimwn diladi ena sinolo efodiasmeno me kapoies prakseis prosthesis kai pollaplasiasmou pou exei kapoies idiotites omoies me autes tis thewrias arithmwn gia ton Daktilio twn akeraiwn arithmwn :) Ola loipon auta genikeuontai akoma pio panw se diaforous allous daktilious me kapoies idiotites!

Thelei ligo proxwrimeni algevra alla mporeis na paikseis poli omorfa me ola auta siga siga... Me tetoia asxoleitai i klassiki eisagwgiki algevra (oxi h grammiki algevra) pou ginetai sta prwta 1-2 xronia tou mathimatikou kai apo kei kai pera anevaineis kai pas...

Alexandros

στους μηχανολόγους μηχανικούς και μηχανικών υπολογιστών έχει τίποτα?? :roll:

xm... exw tin entipwsi oti sto programma spoudwn den tha exei tin sigekrimeni algevra dioti den efarmozetai amesa!Auta pou tha deis sigoura einai grammiki algevra (pou den exei sxesi me auta pou leme, ektos ean anaferei o kathigitis merika pragmata ap'eksw - ap'eksw) kai PARA polles analiseis (oloklirwmata, paragwgous, dipla - tripla oloklirwmata kai genika diaforiko kai oloklirwtiko logismo se 1 kai perissoteres diastaseis kai diaforikes eksiswseis), giati auta sas xreiazontai para poli.

Tipote omws den se empodizei apo tin gnwsi... Mporeis kalista na psakseis se selides panepistimiwn (akoma kai ellinikwn), kai na psakseis eisagwgikes simeiwseis panw stin algevra kai idiaitera na periexoun thewria Omadwn kai Daktiliwn. Ean eidika vreis kanena mathima afierwmeno sta poliwnima tote einai o,ti kalitero. Auto pou tha psakseis esi tha einai koinws "Introduction to Algevra" 'h "Group and Ring Theory".

Auta...
Alexandros

thanks! BTW μου αρέσει ΠΑΡΑ ΠΟΛΥ η ανάλυση (ένας λόγος ακόμη που διάλεξα αυτή τη σχολή)!!!

πραγματι η αλγεβρα των πολυωνυμων ειναι φοβερη,και πολυ χρησιμη.ενα βιβλιο εκτος ηλεκτρολογων θα ειναι το καλυτερο.απο οτι ξερω δεν διδασκεται αυτη η αλγεβρα στους ηλεκτρολ. αλλα πρεπει να ξερεις τι σου αρεσει περισσοτερο ,αναλυση ή πολυονυμα? :)

[quote="cretanman"]
Tipote omws den se empodizei apo tin gnwsi... Mporeis kalista na psakseis se selides panepistimiwn (akoma kai ellinikwn), kai na psakseis eisagwgikes simeiwseis panw stin algevra kai idiaitera na periexoun thewria Omadwn kai Daktiliwn. Ean eidika vreis kanena mathima afierwmeno sta poliwnima tote einai o,ti kalitero. Auto pou tha psakseis esi tha einai koinws "Introduction to Algevra" 'h "Group and Ring Theory".
[/quote]

Θα δώσω λοιπόν άλλη μια φορά ένα link που μπορεί να βοηθήσει

[url]http://mathstuff.rootforge.org/gr/downloads.php[/url]  :)

πολυ καλο link. που τα βρισκεις. :)   :coolspeak:


let $M,N$ be the modpoints of the bases of trapezoid $ABCD$.show that the line $MN$,passes through the intersetion point

$O$ of the diagonals and the point S where the extensions of the legs intersect.

[hide]
First, we can show that $S$, $O$ and $N$ are collinear by using Ceva's theorem.

(In my solution, $AB$ and $DC$ are the bases such that $M$ is the midpoint of $AB$ and $N$ is the midpoint of $DC$.)

Since $AB\parallel DC$, we have $\frac{SA}{AD}=\frac{SB}{BC}$

So $\frac{SA}{AD}\cdot \frac{DN}{BC}\cdot \frac{CB}{BS}=1$

$\therefore$ by Ceva's theorem, $S$, $O$, and $N$ are collinear. $(1)$

There are a couple of ways to show that $S$, $M$, and $N$ are collinear, and one of such ways is to use vectors.

$\overrightarrow{SM}=\frac{\overrightarrow{SA}+\overrightarrow{SB}}{2}$
$\overrightarrow{SN}=\frac{\overrightarrow{SC}+\overrightarrow{SD}}{2}$

(These are not very hard to prove. For instance, in order to prove the first one, express $\overrightarrow{SM}$ in two ways, using $\overrightarrow{AM}=-\overrightarrow{BM}$.)

$\overrightarrow{MN}=\frac{\overrightarrow{AD}+\overrightarrow{BC}}{2} =\frac{\left(\overrightarrow{SD}-\overrightarrow{SA}\right)+\left(\overrightarrow{SC}-\overrightarrow{SB}\right)}{2}$

(This can be proven by taking points $P$ and $Q$ on $DC$ such that $AD\parallel MP$ and $BC\parallel MQ$.)

So $\overrightarrow{SM}+\overrightarrow{MN}=\overrightarrow{SN}$

Therefore, $S$, $M$, and $N$ are collinear. $(2)$

From $(1)$ and $(2)$, we can conclude that the four points $S$, $M$, $O$, and $N$ are collinear.

[/hide]


Ring Theory
number theory
relatively prime
advanced fields
advanced fields unsolved

Let K be a number field with ring of integers O.
Prove there are infinitely many prime ideals in O.

Each maximal ideal is prime. For each ordinary integer prime $ p$, there is some maximal ideal containing $ p$- and any two of these are relatively prime, hence distinct.

how do we know there is some maximal ideal containing p?

and why does two p and q being relatively prime imply that the two ideals are distinct?

thanks so much.

oh ok, i understand more now.

But new question,

why is it that if p is an ordinary integer prime --> <p> is a maximal ideal in O.

Why is it? It isn't. For example, look at the ideal generated by $ 5$ in the ring of Gaussian integers. It's not maximal, because the ideals generated by $ 2\plus{}i$ and $ 2\minus{}i$ respectively each contain it. Those two ideals are maximal.


[quote="27r"]This is kinda creepy, so stop making threads about her. We all know you love her but you don't have to express it here.[/quote]
lol
well said

look on google

well, i can't find it, so help! and jessica i DON'T love her.

You don't [u]HAVE[/u] to contact her, you know...

[quote="pythag011"]I know her email and phoen number. If you give me a good reason why you want to contact her, I'll give you her phone number.[/quote]

you need a gold medal at imo, not just usamo winner, before you can go around spamming, david.

maybe youll see her at MAO competitions next year?

seems the florida forum in general has died...RIP.

jihmmy is stalking me also :|
he keeps on pming me and seems to know both my emails....

btw nice username plushieteng :wink:


inequalities
calculus
inequalities proposed

Let $n$ be real and $k\geq-1$
Show that $[n+k+2,0,0]+[n,k+1,1]\geq[n+k+1,1,0]+[n+1,k+1,0]$
Where the denotions are the usual ones for symmetric sums.

Follows directly by Vornicu Schur :)
Why don't you replace $k+1\mapsto k$ ? :maybe:  :roll:

yeah i meant $[n+k+1,0,0]+[n,k,1]\geq [n+k,1,0]+[n+1,k,0]$ for $k\geq-1$   :oops:
can you post your solution. I also used Vornicu-Schur (or SOS, whatever ... :P )

$LHS-RHS=\sum_{cyc}(x-y)(x^{k}-z^{k})x^{n}+(x-z)(x^{k}-y^{k})x^{n}$ $=: \sum_{cyc}S_{x}(x-y)(x-z)$.
For natural $k$, we see easily $S_{x}\ge S_{y}$ (assuming $x\ge y\ge z$).

OK of course the prblem with this sort of things is always to find a good argument (i.e. without much calculus) why this holds also for [i]non integer[/i] $k$ :blush: Do you have one? And for $-1\le k\le 0$, I have to think about  :maybe:

ah!... I messed it up again! it should be $k\geq 0$ .
and yes i have an argument withut much calculus, though you should 've found it now! :)


Ring Theory
superior algebra
superior algebra unsolved

Is the ideal genereted by $ (5-\sqrt{3})$ prime in $ \mathbb{Z}[\sqrt{3}]$?

No, it isn't.
$ 5\minus{}\sqrt{3} \equal{} (1 \plus{} \sqrt{3})(\minus{}1 \plus{} 2\sqrt{3}),$
but
$ 1\plus{}\sqrt{3}, \minus{}1\plus{}2\sqrt{3} \not \in (5\minus{}\sqrt{3}),$
since
$ N(5 \minus{} \sqrt{3}) \nmid N(1 \plus{} \sqrt{3}), N(\minus{}1 \plus{} 2\sqrt{3}),$
where:
$ N: \mathbb{Z}[\sqrt{3}] \in a \plus{} b\sqrt{3}\mapsto a^{2} \minus{} 3b^{2}\in \mathbb{Z}.$

Simpler and more generally:

prime $ \pi|\pi\pi'\in\mathbb{Z}\implies \pi|p$ prime in $ \mathbb{Z} \implies \pi'|p'\equal{}p \implies \pi\pi'|p^2$

In fact number rings are UFD $ \iff$ PID $ \iff$ irreducibles have norms associate to prime powers.


geometry
geometric transformation
reflection
analytic geometry
rotation
trigonometry

I was working on a problem concerning the reflection of images in coordinate planes; just for curiosity, I tried to derive some sort of "formula" for reflection, it is:

If $ (x_2,y_2)$ is the reflected image of $ (x_1,y_1)$ over line $ l$, then

$ (x_2,y_2) \equal{} (x_1\cdot \cos2\alpha \plus{} (y_1 \minus{} y_h)\cdot \sin2\alpha, \minus{} y_1\cdot \cos2\alpha \plus{} (x_1 \minus{} x_h)\cdot \sin2\alpha)$

where $ \alpha$ is the angle of $ l$ from $ x$-axis counterclockwise, $ x_h, y_h$ are x - and y - intercepts, respectively.

Is this correct?

Furthurmore, my derivation is quite brute-forcy (Construct perpendicular line and solving linear system) and so the anwer is ugly, too. But then I found it can be easily written into nicer form (above) by using trig. identity; and I regard it an implication that this formula might be derived in a better way. If this is true, can anybody give me the actual name and introduction of it? Thanks in advance.

A reflection about an arbitrary line $ l$ can be represented as the composition of a translation, a reflection about $ y \equal{} 0$, a counterclockwise rotation by an appropriate angle, and another translation.

A reflection about $ y \equal{} 0$ takes $ (x, y)$ to $ (x, \minus{}y)$.

A counterclockwise rotation by $ \theta$ takes $ (x, y)$ to $ (x \cos \theta \minus{} y \sin \theta, x \sin \theta \plus{} y \cos \theta)$.  This can be proven using polar coordinates.

Finally, a translation takes $ (x, y)$ to $ (x \plus{} h, y \plus{} k)$.

Once you figure out the appropriate angle to rotate and the appropriate translations, you will be able to derive your formula in a more systematic way.  In fact, all of these transformations are much better represented as matrices.


Polynomial P(x)=$ x^3 +ax^2 +bx +c$ have three different real roots.
                 Q(x)=$ x^2 +x+2001 $
Polynomial P(Q(x)) have no real root.
Prove P(2001)>$\frac{1}{64}$

The minimum of $Q(x)$ is for $x=-\frac{1}{2}$ and it is $Q(-\frac{1}{2})=2001-\frac{1}{4}$

So $Q(x)>2001-\frac{1}{4},\ \forall x \in R$ and $Q(x)$ goes to infinity when $x$ does.

Let $x_0=2001-\frac{1}{4}$

There is no root of $P(x)$ greater than (or equal to) $x_0$.
If $y \geq x_0$ then from continuity of $Q(x)$ we get that there is a $x$ such that $Q(x)=y\Rightarrow P(y)=P(Q(x)) \neq 0$.

Let $x_1<x_2<x_3$ the three roots of $P(x)$.

They are all less than $x_0$

Now, we will find the sign for $P(x_0), P'(x_0), P''(x_0), P'''(x_0)$

The coefficient of $x^3$ in $P(x)$ is $1>0$, hence
$\forall x>x_3, P(x)>0 \Rightarrow $
$P(x_0)>0$

$P'(x)=3x^2+2ax+b$ (3>0)

From Rolle Theorem we have that, $P'(x)$ has a root in $(x_1, x_2)$ and a root in $(x_2, x_3)$

And after the second root (for greater values of $x$), $P'(x)>0$ (because 3>0). So,
$P'(x_0)>0$

$P''(x)=6x+2a$

$P''(x)$  is an increasing line, and it has a root at the midpoint of roots of $P'(x)$.

Since $x_0$ is greater that all of them, we have $P''(x_0)>0$

Finally, $P'''(x_0)=6$

Now, we take Taylor around of $x_0=2001-\frac{1}{4}$

$P(x_0+h)=P(x_0) + \frac{P'(x_0)}{1!}h + \frac{P''(x_0)}{2!}h^2 + \frac{P'''(x_0)}{3!}h^3 $

If we set $h=\frac{1}{4}$ we have

$P(x_0)>0$

$\frac{P'(x_0)}{1!}h>0$

$\frac{P''(x_0)}{2!}h^2>0$

$\frac{P'''(x_0)}{3!}h^3 = \frac{1}{1}h^3=(\frac{1}{4})^3=\frac{1}{64}$

Finally $P(x_0+h)=P(2001)>\frac{1}{64}$

Since P(x) has three roots ,say p,q r ,then P(x)=(x-p)(x-r)(x-q)
P(Q(x))=(Q(x)-p)(Q(x)-r)(Q(x)-q)
Since Q(x) has no real roots so the discriminant of of Q(x)-p is negative it means
p+1/4 < 2001
Same for r and q.
We have P(2001)=P(Q(0))=(2001-p)(2001-r)(2001-q) and using the inequalities above we get
1/64<P(2001)


[quote="dinoman"]When the Worm family went on vacation, Dr. Worm drove 4/9 of the distance at 40 mph, 7/18 of the distance at 70 mph and the final 30 miles at 60 mph. If there were no stops, the entire trip took them ____  hours.[/quote]

[hide]the total distance is 180 miles
80 miles was driven at 40 mph
70 miles was driven at 70 mph
30 miles was driven at 60 mph

t=d/r
2+1+1/2=3 1/2 hours[/hide]

I'll give you a hint....

it's called magic circle

draw a circle

divide it in half horizontally....

you will have a half circle and another half circle...

on the bottom half circle, divide it in half which is two one fourth circle...

now, lable them...

the top half circle Distance
the left fourth circle Times
the last rigth fourth circle Speed...

now it's a magic circle....

if you want to know about speed, cover speed section with your finger and look at other two sections....
the line in the middle ( the horizontal one ) would mean division... the line that's dividing Times and Speed will mean multiplication...

magic circle...


calculus
derivative
function
complex numbers
complex analysis
complex analysis unsolved

I need a sanity check, please.

I'm to prove or provide a counterexample of the following:

[b]
If $f(x+yi)=u(x,y)+iv(x,y)$ is differentiable everywhere, then so is
$g(x+yi)=v(x,y)+iu(x,y)$[/b]

Okay, so the Cauchy-Riemann equations for $f(x+yi)$ are:

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and
$\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$

If I take $g(x+yi)$ (from above) and find $g'(z)$ where $z=x+yi$ using the limit
definition of a derivative, I get the following:

$\frac{\partial v}{\partial x}=\frac{\partial u}{\partial y}$ and
$\frac{\partial u}{\partial x}=-\frac{\partial v}{\partial y}$

which are not the Cauchy-Riemann equations.

Am I done at this point (I proved it false)?  What would be a counterexample?

Thanks,
Bailey[/b]

$u(x,y)=x, v(x,y)=y$ should do since the first function is $f(z)=z$ whereas the second one is $g(z)=y+ix=i \cdot (x-iy) = i \cdot \bar{z}$.


[quote="randomdragoon"]A robot starts at the origin, and travels to the point $(3,3)$. It can only change direction at a lattice point, and it can only go north, east, or northeast. How many different routes are possible?[/quote]
Ooooh, very clever twist.
[hide]
[code]
1,1,1,1
7,5,3,1
25,13,5,1
63,25,7,1
[/code]
63[/hide]

To anyone who doens't get bpms's code (like I didn't at first), it is a diagram of all 16 points and each number represents the number of ways to get to (3,3) from that point.

Here's an extension that I don't know how to do, so it might be impossible but I strongly strongly doubt it.
How many paths are there where you can go north, east, and northeast in an n by k grid?
[hide="something I noticed"]The grid resembles pascal's triangle[/hide]

[quote="mathgeniuse^ln(x)"][quote="randomdragoon"]A robot starts at the origin, and travels to the point $(3,3)$. It can only change direction at a lattice point, and it can only go north, east, or northeast. How many different routes are possible?[/quote]

[hide]I think one can use recursion in this.

Let $F(m,n)$ be the number of paths to reach $(m,n)$.

Notice the following

$F(m,n)=F(m-1,n)+F(m,n-1)+F(m-1,n-1)=3F(m-1,n-1)+F(m-2,n)+F(m,n-2)$

I can't find a general, but you can find any number, given the first three which are $F(0,0)=1, F(0,1)=1, F(1,0)=1$[/hide][/quote]
This is definitely recursion.
Notice the symmetry.
I conjecture that where, in pascals triangle the sum of each row is $2^{n}$, the sum of the terms in this triangle $3^{n}$.


I mean NOT from the teacher
any other ways??
coz I know I did very bad but still want to know^^

http://www.cmc.uwaterloo.ca/english/contests/open.shtml

Check the results; your name might be on the Honour Roll.

If it isn't, then I guess you'll have to ask your teacher.


$\ (p-1)!+1=p^n$
Prove that this equality isn't real with p is a prime number, p>5,n belong to N :oops:

Math is my life :!:

[quote="draperp"]Does [u]any[/u] integer for $p$ work?[/quote]
Does this question make any sense¿ :?

But posted: http://www.mathlinks.ro/Forum/viewtopic.php?t=76606

Ah,I got it.Thanks :blush:

Math is my life :!:


calculus integration logarithms limit algebra polynomial function

In light of the proposer's response (following), this is re-edited to take out spoiler and put in TeX.

The answer is $e^e$.

There's a lot of clutter here, but the integrand is $n + f(x)$ where $f(x)$ is differentiable at $1$ and $f(1) = e$. That means we can write the integral as
$I_n = \int_1^{1+\frac1n} n + e + O(x-1) dx$. From this,
$I_n = 1 + \frac{e}n + O(\frac1{n^2})$
$\ln(I_n) = \frac{e}n + O(\frac1{n^2})$
$\ln(I_n^n) = n\ln(I_n) = e + O(\frac1n)$, which goes to $e$ as $n \to\infty$.

Thus $I_n^n \to e^e$ as $n\to\infty$.

To Mr.Kent Merry field, thank you for your reply.You are right.This is very tricky ,because I made this problem in an artificial way.I will show you my answer.
Since we have      $ \frac{d}{dx}e^{xe^{xlogx}}=e^{xe^{xlogx}}(xe^{x}logx)'=[(x+1)logx+1]e^{x(e^{x}logx+1)}$,we obtain $\  I_n=\left[e^{xe^{x}logx}+nx\right]_1^{1+\frac{1}{n}}\
                             = e^{(1+\frac{1}{n})e^{1+\frac{1}{n}}log(1+\frac{1}{n})}$.  Therfore $\lim_{n\to\infty}I_n^{n}=e^{(1+\frac{1}{n})e^{1+\frac{1}{n}}log(1+\frac{1}{n})^{n}}=e^{e}$

I would add that I see Taylor polynomials, Taylor series, power series, and estimates of the size of functions as all lying close to the heart of calculus as a powerful problem solving technique. The question of which functions do or don't have closed-form antiderivatives is more of a peripheral and incidental question.

The "big-O" and "little-o" notation is quite a useful shorthand. Users of calculus to solve problems should make themselves familiar and comfortable with it.

Thank you again!You may be right.In Japan,senior high school 'student  at the age of 17 ~18,usually study Limits and continuity (except ε－δ,de l'Hopital,Taylor expansion etc ),Differentiation and Integration. that is why  I make problems according to this principle.Sorry for my poor English.


Postati cineva probleme pe ML, parca olimpiada a fost azi, dar nimic nu se aude.
Care sait oficial olimpiadei ?

Uite ce am gasit eu: http://www.olimpiade.ro/informatii.php?etapa=3&materie=2&editie=5 .

Da mai si eu sunt putin uimit. Valentin mi-a zis ca o sa posteze problemele imediat. :?


A sequence $ (a_n)$ is defined by $ a_0 \equal{} a_1 \equal{} 0, a_2 \equal{} 1$ and $ a_{n \plus{} 3} \equal{} a_{n \plus{} 1} \plus{} 1998a_n$. Prove that equality $ a_{2n \minus{} 1} \equal{} 2a_na_{n \plus{} 1} \plus{} 1998a_{n \minus{} 1}^2$ holds for every positive integer $ n$.

[quote="tdl"]A sequence $ (a_n)$ is defined by $ a_0 = a_1 = 0, a_2 = 1$ and $ a_{n + 3} = a_{n + 1} + 1998a_n$. Prove that equality $ a_{2n - 1} = a_na_{n + 1} + 1998a_{n - 1}^2$ holds for every positive integer $ n$.[/quote]

Unless you have missed typing somethinig when defining the problem, the equality does not hold;
Counter example:

$ a_0 = 0\\
a_1=0\\
a_2 = 1\\
a_3 = 0\\
a_4 = 1\\
a_5 = 1998\\
a_6 = 1\\
a_7 = 2*1998\\
a_7=a_{2*4-1}=a_4a_5 + 1998a_3^2=1*1998 + 1998*0=1998 \neq 2*1998$

Sorry, I have repaired it: Prove that $ a_{2n\minus{}1}\equal{}2a_na_{n\plus{}1}\plus{}1998a_{n\minus{}1}^2$ :oops:

[hide="My idea"] Prove by induction that $ a_{2n\minus{}1}\equal{}2a_na_{n\plus{}1}\plus{}1998a_{n\minus{}1}^2$ and $ a_{2n}\equal{}2a_{n\plus{}2}a_n\minus{}a_n^2\plus{}a_{n\plus{}1}^2$[/hide]


function
geometry
advanced fields
advanced fields unsolved

w(x,y,z)=x^2dy/\dz + y^2dz/\dx -2x(x+y)dx/\dy
What is the degree of this differential form ?
Prove that w is closed
Find A,B :R^3 --> R C^(oo) functions such that
q = Adx + Bdy verify w=dq

well this form is in fact not closed it woulb be if we change -2x(x+y) to -2z(x+y) and by the way it's standard problem don't know why have you posted it here as an unsolved any third (and some second) year student could solve it

Can you moved this problem to College->Geometry-> proposed problems
thx


linear algebra
matrix
calculus
calculus computations

Hi,
I am detailing a specific problem of finding the maximum of a double summation. hope some one can turn up with some solution.

think of a matrix of size $ M \times N$  filled with values $ C_{j,k}$  where $ j$ represents the row number ($ j\equal{}[1,M]$) and  $ k$ represents the column number ($ k\equal{}[0,N\minus{}1]$). (matrix figure attached)

now my problem is to find an expression for:
$ \max_{k}(\max_{j}(\sum_{p\equal{}1}^{N}C_{j,p}\sum_{p'\neq
p}^{}\overline{C_{j,p'}}e^{2 \iota \pi \frac{k(p\minus{}p')}{N}}))$

here  $ C_{j,p}$ can be any value from the set [$ \minus{}1\minus{}1 \iota$, $ 1\plus{}1 \iota$ , $ 3\plus{}3 \iota$  , $ \minus{}3\minus{}3 \iota$  ] and remember  $ j$ ranges from $ 1$ to $ M$  while $ k$ ranges from $ 0$ to $ N\minus{}1$  .

Thanks for ur efforts!

have you any results ?
I can show that upper bound will be $ \geq 18N(N \minus{} 1)$ but I'm not sure about: will be $ 18N(N \minus{} 1)$ this value best upper bound . :oops:

yes i agree with ur upperbound but this is a very big upperbound compared to instantaneous results of the expression and i am looking for un upperbound which is quite close to the instantaneous results... hope we get something better than this... thanks for ur effort anyways...


Bill is old enough to vote but not old enough to receive Senior Citizen discounts. x years ago, Bill's age was the square root of his age x years from now. If his age is a perfect square, how old is he now?

LoL... yah i did... cause the algebriac way was... ewww.....

I had (B-x)^2 = B+x

so then I had B^2 -2Bx + x^2 = y+x...

[quote="Syntax Error"]i assume you used guess and check then. OU INTERCEPTS! RENALDO WORKS RUNS IT IN FOR A TOUCHDOWN. WOOOOOOOOOOT[/quote]
What the...
a.) Is "OU" a team or is it Franais? :lol:
b.) Who's Renaldo? I thought your name was Neal...
c.) Aren't you a moderator? :lol:
d.) Last time I checked this was a math forum :lol:  :lol:

"Ewww" indeed :P

ou=oklahoma university
renaldo=renaldo works, one of the tailbacks for oklahoma university
yes i am a mod
im  pretty sure this is a math forum

i couldnt hold in my excitement from the ou vs texas game in college football

Oy!, can't believe you care about a college football game when there's Red Sox vs. Yankees Best 4 out of 8 going on! :lol:


USAMTS probability geometry algebra polynomial calculus integration

1. Find all solutions of the following system of equations:
[tex]x+y+z+w = 10[/tex]
[tex]x^2 + y^2 + z^2 + w^2 = 30[/tex]
[tex]x^3 + y^3 + z^3 + w^3 = 100[/tex]
[tex]xyzw = 24[/tex]

2. A basketball player starts shooting baskets. He hits his first shot and misses his second. Thereafter, the probability that he hits a shot is equal to the percentage of shots that he has already made. Thus, his probability of hitting the third shot will be 1/2. What is the probability that, if he shoots 2004 shots, he makes exactly 1002 of them?

3. Find the area of an equilateral triangle containing in its interior a point P, with distances from the vertices of 3,4, and 5.

4. Find the positive integer n < 100 for which the number [tex] A_n =  7^1 + 7^2 + ... + 7^n [/tex] ends with the maximum number of equal digits.

5. Prove that the following polynomial in x is linear for all n:
[tex]
\displaystyle\sum_{i=0}^{n}i \times \binom{n}{i}x^i(1-x)^{n-i}[/tex]

Thanks to JBL for providing problems!

The rules are basically the same as the real USAMTS, don't get help from anyone, etc... The problems are due at the end of the weekend 2 weeks from this one. Solutions can be PMed to me. At the end of the time period, the best solutions will be posted, as well as a list of high scorers. Suggestions and inquiries are welcome.

My guess would be that this is in fact much harder than a USAMTS round -- you're probably going to need an unusually large amount of that two weeks :)

And just to clarify, they are due on or before Sunday, June 6.

Umm, so, for number 5, am I correct in assuming that for all n means for all integral n greater than or equal to 1?

How much calculator use is appropriate?  I managed to reduce number three to a quartic.  Should I try to find a more elegant solution, or just set my 89 on it?

Calculators may assist in the solving process in any way you see fit; however, know that full solutions are required for credit. Simply plugging in and returning an answer will get you nothing in this competition.

[quote="The Official USAMTS Rules"]Participants may use resources such as books or the internet to do mathematical research to try to solve the problems.[/quote]

You may use any non-live resource to solve these questions.  That is, Dr. Math is not acceptable, but Mathworld is (unless you e-mail the guy who runs mathworld to ask for the answers to specific questions).  However, you should be able to explain and justify every step you make.  If you use a computer or calculator to complete a step that it is not clear you could have done by hand or which you cannot explain, you may be penalized.  However, if you write your own computer program and can demonstrate its correctness, that's worth credit.

[quote="The USAMTS FAQ"]Can I use a computer to help solve a problem?

Using a computer to answer a question is ok. You can still receive full credit of 5 points for a computer solution to a problem. However, it is unlikely that a grader will give a "commendation" for such a solution.

Can you think of a way to answer a problem without a computer? Non-computer answers usually give more "insight" into a problem than computer solutions. Sometimes using a computer to get an answer is helpful in devising non-computer solutions. (It helps to know the answer!)

Computer solutions should be accompanied by an outline of the program as well as a printout of the outcome. In no case should you submit a floppy disk or CD with your solutions.[/quote]

While this does not make itself entirely clear, I believe what is indicated is that computer algebra systems are looked down upon while computer programing is acceptable.

Since Tetrahedon is grading this, I'll leave it to him to be more specific about his grading policies.

Hmmm, so if I program the quartic formula onto my 89 I'm in good shape...

:-)

Those are nice problems, tetrahedr0n and JBL! I especially like 2 and 5. Are any of them original creations?

Hehe, 2 and 5 are my additions.  2 is definitely not original -- I was originally shown it by my mentor for my Intel project but I would imagine that it is quite old.  Number 5 is my own invention, and I will tell the story of how I came across it after the round is over in my solution to it (which I will add on to whatever nice solutions other people get).

I'm not sure, but I think they do.  Judging by this year and a few other stray rounds I've seen, they definitely do, but I've only seen about 6 or so rounds.

[quote="Magnara"]I don't really know where to ask this question, this seems as good a place as any.  Do USAMTS traditionally increase in difficulty from 1 to 5?[/quote]

They try. The first is usually easier than the other 4, but for 2 through 5 it's more a matter of which you like more. I often find myself struggling with #3 and solving #5 in 10 minutes. (and by 'often' i mean it happened once last year :))

Can someone draw up Zabelman's proof for the equilateral triangle?  I can't get it to work.

Here is the diagram after the rotation.  Since we've rotated 60 degrees, the angle between the two segments of length 3 is 60 degrees, which means that the dashed segment is of length 3.  Then we have, with the rotated segment of length 4 and the unrotated segment of length 5, a 3-4-5 triangle.  Thus, the angle between the (rotated, and thus also unrotated) bits of length 3 and 4 is 150 degrees, and the law of cosines finishes us off.

Ok thanks... BTW, that was a good proof.

Yep.  I think this problem often is brought up when teaching how to solve geometry problems by rotation because it's so nice and simple.

I got for #3, 9 + 25 :rt3: /4.  I had the idea of using a rotation but the whole time I rotated it around a different vertex, I just found out it was one in particular you need to rotate the triangle around for it to work.  Good problems!

[quote="beta"][quote="Magnara"]
A_39 :equiv: 3999 mod 10000
A_59 :equiv: 5999 mod 10000
A_79 :equiv: 7999 mod 10000
A_99 :equiv: 9999 mod 10000
[/quote]

Can you explain how you calculated that(looking for pattern in powers of 7)?[/quote]

Calculator :-)

It's actually not that hard to do by hand -- just look for the pattern, like always, mess around with equations a little bit.

There are not-so-hard algorithms for taking moduli of large numbers, so long as they are expressed as a^b, because you can just reduce the mod for a^1, a^2, as you go, meaning you never end up with a number larger than a times the modulus.

I did that with Binomial Theorem (50-1)^n

[quote="beta"]How can a calculator deal such a big number like A_79 or A_99???[/quote]

The 89 can handle:
mod((7^100-7)/6,10000) with little difficulty.

I did that problem with some mod 10000 arithmetic and a calculator, but I didn't need the super power of an 89, I just needed to multiply 4 digit numbers a few times (ok, closer to 10, which isn't a 'few' ).

Here's my solution in latex:

[hide]By the geometric series formula, we have . Note that the  satisfy the recursion

since

Thus, , and so the last 2 digits of  are the same as the last 2 digits of . Since

the last 2 digits of  for any  must be 07, 56, 99, or 00. So, for  to end with 2 or more identical digits, it must end in 9's or 0's, i.e. it must be one less than, or equal to, a multiple of a power of 10.

If  ends with  0's, then

\begin{eqnarray*}
\tfrac{7}{6}\left(7^n-1\right) &\equiv& 0 \mmod{10^k}\\
7^{n+1} &\equiv& 7 \mmod{10^k}\\
7^n &\equiv& 1 \mmod{10^k}.
\end{eqnarray*}

(We could cancel the 7's since 7 is relatively prime to .) Likewise, if  ends with  9's, then

\begin{eqnarray*}
\tfrac{7}{6}\left(7^{n-1}-1\right) &\equiv& -1 \mmod{10^k}\\
7^n-7 &\equiv& -6\mmod{10^k}\\
7^n &\equiv& 1 \mmod{10^k}
\end{eqnarray*}

Thus,  and  end with (at least)  9's and 0's respectively if and only if . I will prove that the order of 7 in modulo 10000 is 100, which will prove that  is the first of the 's to end with at least 4 identical digits.

\lskip

The proof that  is mainly just alot of arithmetic mod 10000. (Note:  denotes the order of  modulo . Also, from now on, any congruence is in mod 10000 unless otherwise specified.) We have

Thus, , and so for  to end in -01,  must be a multiple of 4. We then have

\emph{This} proves that , and so for  to end in -001,  must be a multiple of 20. Next, we have

Thus (finally),  is the smallest positive integer that satisfies . So, as explained before,  ends with at least four 9's, and it is the smallest of the 's to end with 4 or more identical digits. Since  is the only integer less than 100 which gives  4 identical ending digits,  is the answer.[/hide]

[quote="zabelman"]
Here's my solution in latex:

[/quote]

It's not really in latex...

Sure it is! It just isn't compiled. To be more specific, it's Latex source code.


limit
algebra
polynomial
calculus
derivative
calculus computations

Since $\arctan 1=\frac{\pi}4$ and $\frac{d}{dx}\arctan x|_{x=1}=\frac12,$ we have the Taylor polynomial (really, this is just the definition of the derivative):

$\arctan(1+y)=\frac{\pi}4+\frac y2+o(y).$

So, as $x\to\infty,$

$\arctan(\frac{x^2}{x^2-1})= \arctan(1+\frac1{x^2-1} )=\frac{\pi}4+\frac1{2(x^2-1)}+o(x^{-2})$

So $6x^2(\frac{\pi}4-\arctan(\frac{x^2}{x^2-1})) = \frac{-6x^2}{2(x^2-1)}+o(1)$

The limit as $x\to\infty$ is $-3.$

As I've said here many times, it's not my gut instinct to start with L'Hôpital's rule. I find the language of Taylor polynomials and big-O, little-o, to be much more easily controlled and manipulated.

I like your magic.  What does the big-o and little-o notation mean?  Is Kent Merryfield a calculus master?  Nice solution.


hi
can we request roommates?
i have a friend who went last year and we want to be roommates

I requested my roommate. All we did was email Dr. T before camp and ask to be roommates.
Yeah, I think they usually assign roommates by age.

[quote="jhredsox"]umm....a little off topic from this topic, but i don't want to start a new one just for this

for the qualification quiz, do you have to fully understand every question you answer on there, or is it ok to not have a full understanding?

jorian[/quote]
Show as much work as you can come up with.

Ok, sorry, I had worded that badly. I asked leet what I really meant and he was nice enough to give me a long and good answer.
-jorian

[quote="13375P34K43V312"]hmmm, who would that friend be? just wondering.

You can probably e-mail Dr. T. in advance, but generally, they assign roomates based on birthday (my birthday was 8/11, I was paired with someone whose birthday was 8/30). However, there were I believe two rooms where the two kids were from the same school, I don't know if that was requested or not.

Ask Ubemaya, he was one of those people. He got paired with someone from his school, in his grade.[/quote]
ok i emailed Dr. T and its fine
if you want to know, his AoPS username is whitesoxfan

Whos? D. T or your friend because I didn't think D. T likes baseball very much...


trigonometry
geometry
rectangle
quadratics
ratio
complex numbers
algebra

[quote="mathgeniuse^ln(x)"][quote="bpms"]Find a radical value of $\sin 18$.[/quote]

[hide]$\frac{-1+\sqrt{5}}{4}$[/hide][/quote]
With proof.

[quote="mathgeniuse^ln(x)"]Well, I can prove it with complex numbers, but then I can't display it in this forum.[/quote]
I didn't know complex numbers were beyond HSB, but now you can display them.

[hide="Geometric Solution"]

Consider a regular pentagon $ABCDE$ with side length 1. Now, construct lines through $A$ and $B$ perpendicular to $AB$. Draw a line passing through $C$ and $E$. Let the intersection of line $CE$ with the line through $A$ be $F$ and the intersection with the line through $B$ be $G$. The perpendicular lines create 18 degree angles. By definition of sine, $EF=CG=\sin{18}$. Because quadrilateral $AFGB$ is a rectangle, $FG=AB=1$. Therefore, $CE=1+2\sin{18}$. However, $CE=\sqrt{2-2\cos{108}}$ also by applying the Law of Cosines on triangle $CDE$. Equating the two expressions results in:

$1+2\sin{18}=\sqrt{2-2\cos{108}}$

We know that $\cos{108}=-\cos{72}=\sin{18}$. Squaring both sides and substituting in the identity just established yields:

$4\sin^{2}{18}+4\sin{18}+1=2+2\sin{18}$

$4\sin^{2}{18}+2\sin{18}-1=0$

Using the Quadratic Formula shows that $\sin{18}=\frac{-1 \pm \sqrt{5}}{4}$. However, $\sin{18}$ is in the first quadrant and is therefore positive, so the only possible value for $\sin{18}$ is $\boxed{\frac{-1+\sqrt{5}}{4}}$.[/hide]

[hide="the complex solution"]
Consider the 5th roots of unity,excluding 1.
Then $1+x+x^{2}+x^{3}+x^{4}=0$
where $x$ is a 5th root of unity
Taking real parts of both sides,
$1+\cos72+\cos144+\cos216+\cos288=0$
$1+\sin18-\cos36-\cos36+\sin18=0$
$1+2\sin18-2(1-2\sin^{2}18)=0$
$4\sin^{2}18+2\sin18-1=0$
Use quadratic formula, choose the value in the first quadrant, and get the same answer as above.
[/hide]

slightly more elementary:

we look at the equation:
$\sin(3x)=\cos(2x)$     $x=18$ is a solution; as is $x=90$

using basic trig IDs:

$\sin(3x)=\cos(2x) \iff \sinh\cos2x+\sin2x\cosh = 1-2\sin^{2}x \iff\sinh(1-2\sin^{2}x)+2\sinh\cos^{2}x=1-2\sin^{2}x \iff\sinh(1-2\sin^{2}x)+2\sinh(1-\sin^{2}x)=1-2\sin^{2}x \iff 4\sin^{3}x-2\sin^{2}x-3\sinh+1=0$

so $\sin(x)=1$ is a solution ($x=90$), factor it out, you have a quadratic, and the result comes

Oh, yeah, I remember the solution for $\cos{3x}=\sin{2x}$.

Thanks

Just to note, this problem is beyond the High School Basics level. ;)

Pentagon! Golden Ratio!

The triangle with angles $36^\circ, 72^\circ, 72^\circ$ directly gives answer.


logarithms
real analysis
real analysis unsolved

Solve differential equation $ \frac{2}{3}xyy'\equal{}\sqrt{x^6\minus{}y^4}\plus{}y^2$.

$ y\equal{}x^{\frac{3}{2}}$ fits, but I wouldn't expect uniqueness here either.

Let $ y\equal{}u x^{3/2}$. Now $ \frac{2}{3}xu \frac{du}{dx}\equal{}\sqrt{1\minus{}u^4}$.
$ \frac{2u du}{\sqrt{1\minus{}u^4}}\equal{}\frac{3dx}{x}.$
$ \arcsin \frac{y^2}{x^3}\equal{}\ln C x^3$
$ \arcsin \frac{y^2}{|x|^3}\equal{}\ln C|x|^3$, $ y^2\equal{}|x|^3$, where $ C>0.$


Slove this equation: $ sin^3(x\plus{}\frac{\pi}{4})\equal{}\sqrt{2}sinx$

When you post trig problems like these, they look much prettier if you use

[code]\sin^3 x[/code] to get $ \sin^3 x$

instead of

[code]sin^3 x[/code] which gives you $ sin^3 x$

[quote="queen"][hide="hint"]
just take the cosx to otherside and factor it .
[/hide][/quote]

You can post full solution?

[hide="GUESS AND CHECK"]Let $ \sin{x}\equal{}\sqrt{2}/2$, because of the rt2 on the RHS. Therefore $ x\equal{}\dfrac{\pi}{4}$, and then 1=1. YAY[/hide]

Can someone give a better solution?

I thought 1=2!!!!

 :rotfl:  :rotfl:  :rotfl:

This is my solution:

Equation $ \Longleftrightarrow sin(x \plus{} \frac {\pi}{4}).sin^2(x \plus{} \frac {\pi}{4}) \equal{} \sqrt {2}sinx$

$ \Longleftrightarrow \frac {1}{\sqrt {2}}(sinx \plus{} cosx).\frac {1 \minus{} cos(\frac {\pi}{2} \plus{} 2x)}{2} \equal{} \sqrt {2}sinx$

$ \Longleftrightarrow \frac {1}{\sqrt {2}}(sinx \plus{} cosx).\frac {1 \plus{} sin2x}{2} \equal{} \sqrt {2}sinx$

$ \Longleftrightarrow (sinx \plus{} cosx)(1 \plus{} sin2x) \equal{} 4sinx$

$ \Longleftrightarrow (sinx \plus{} cosx)^3 \equal{} 4sinx$

$ \Longleftrightarrow sinx^3 \plus{} cosx^3 \plus{} 3sin^2xcosx \plus{} 3cos^2xsinx \equal{} 4sinx$

$ \Longleftrightarrow \frac {sinx^3 \plus{} cosx^3 \plus{} 3sin^2xcosx \plus{} 3cos^2xsinx}{cos^3x} \equal{} \frac {4sinx}{cos^3x}$

$ \Longleftrightarrow tan^3x \plus{} 1 \plus{} 3tan^2x \plus{} 3tanx \equal{} tanx(1 \plus{} tan^2x)$

$ \Longleftrightarrow 3tan^3x \minus{} 3tan^2x \plus{} tanx \minus{} 1 \equal{} 0$

$ \Longleftrightarrow 3tan^2x(tanx \minus{} 1) \plus{} (tanx \minus{} 1) \equal{} 0$

$ \Longleftrightarrow (tanx \minus{} 1)(3tan^2x \plus{} 1) \equal{} 0$

$ \Longleftrightarrow tanx \equal{} 1 \Longleftrightarrow x \equal{} \frac {\pi}{4} \plus{} k\pi,k\in Z$.

[quote="thanhnam2902"]This is my solution:

Equation $ \Longleftrightarrow sin(x \plus{} \frac {\pi}{4}).sin^2(x \plus{} \frac {\pi}{4}) \equal{} \sqrt {2}sinx$

$ \Longleftrightarrow \frac {1}{\sqrt {2}}(sinx \plus{} cosx).\frac {1 \minus{} cos(\frac {\pi}{2} \plus{} 2x)}{2} \equal{} \sqrt {2}sinx$

[/quote]
shouldnt it be:
$ \frac {1}{\sqrt {2}}(sinx \plus{} cosx).\frac {1 \boxed{\plus{}} cos(\frac {\pi}{2} \plus{} 2x)}{2} \equal{} \sqrt {2}sinx$

Oh! No.
Note: $ \boxed{cos2x\equal{}1\minus{}2sin^2x \Longleftrightarrow sin^2x\equal{}\frac{1\minus{}cos2x}{2}}$

I got something strange,
but I'm almost certain that
all my steps are legal.

$ \sin^3(x\plus{}\frac{\pi}{4}) \equal{} \sqrt2\sin x$

Let $ y \equal{} x \plus{} \frac{\pi}{4}$

Thus $ \sin^3y \equal{} \sqrt2 \sin(y\minus{}\frac{\pi}{4})$

       $ \sin^3y \equal{} \sqrt2 (\sin y \cos \frac{\pi}{4} \minus{} \cos y \sin \frac{\pi}{4})$

$ \sin^3y \equal{} \sin y \minus{} \cos y$

$ \cos y \equal{} \sin y ( 1 \minus{} sin^2y)$

$ \cos y \equal{} \sin y (cos^2y)$

$ 1 \equal{} \sin y * \cos y$

$ 2 \equal{} 2 * \sin y * \cos y$

$ 2 \equal{} \sin 2y$

Yet we know this is impossible, and there
is no value for y.
But since $ y$ is $ x\plus{}\frac{\pi}{4}$, x cannot
exist either.

What did I do wrong?

[quote="Math4tots"]$ \cos y \equal{} \sin y (cos^2y)$

$ 1 \equal{} \sin y * \cos y$
[/quote]
What if $ \cos y\equal{}0$?

[quote="queen"][hide="hide"]
$ sinx(sin^2(x \plus{} \frac {\pi}{4}) \minus{} \sqrt {2}) \equal{} 0$
$ sinx \equal{} 0$  then  $ x \equal{} 0$ also
$ sin(x \plus{} \frac {\pi}{4}) \equal{} \sqrt {\sqrt {2}}$
$ \frac {\sqrt {2}}{2}(sinx \plus{} cosx) \equal{} \sqrt {\sqrt {2}}$
$ sinx \plus{} cosx \equal{} \frac {2\sqrt {\sqrt {2}}}{\sqrt2}$         (A)
$ A^2$ gives:

$ sin2x \equal{} 2\sqrt {2} \minus{} 1$
dont know what to do now  :blush:
[/hide][/quote]

$ 2\sqrt{2}\minus{}1>1$

Thank every one. This problem is easy if let $ t\equal{}x\plus{}\frac{\pi}{4}$