[quote="silouan"]Η επόμενη είναι αφιερωμένη σε έναν από τους καλύτερους γεωμέτρες που έχουν περάσει από τους ελληνικούς διαγωνισμούς τον staretak (Αν και θα την ξεφτιλίσει) .Η άσκηση είναι από προετοιμασία της Βουλγαρίας

Έστω $ M,N$ δύο σημεία μέσα σε κύκλο $ C(O,R)$ ώστε το  $ O$ να είναι το μέσον του $ MN$. Έστω $ S$ τυχαίο σημείο στον κύκλο και έστω  $ E,F$ οι τομές των ευθειών  $ SM,SN$ με τον κύκλο. Οι εφαπτόμενες στα $ E,F$ τέμνονται στο $ I$.
Νδο η μεσοκάθετος του  $ MN$ περνάει από το μέσον του  $ SI$.[/quote]

Silouane, arxika thanks polu gia ta kala sou logia :blush:

Kalo ergo ws moderator...

Loipon oso afora tin askisi, eutixws koitaksa to Forum simera (parolou pou edw exoun Pasxa kai genika oloi einai eksw..heh) opote eida kai tin askisi sou. Loipon, basika i askisi me ti lusi pou brika einai efarmogi ton polikwn:

Estw oti oi euthies NM, EF temnontai sto L. Tote epidi i poliki tou I dierxetai apo to L, i poliki tou L dierxetai apo to I. Etsi an LA, LB einai oi efaptomenes sto kiklo apo to L exoume oti A,B,I suneuthiaka (B sto tokso FE pou den anikei to S).  Tha deiksoume oti to B einai antidiametriko tou S.  Pragmati to i desmi A. LEBF einai armoniki opote kai i desmi S.AEBF einai armoniki (idies aksonikes gwnies) kai afou OM=ON exoume oti MN//SA opote SAB=90.  Twra, an T einai stin SI oste TO katheti stin MN tote O meso tou SB kai TO//SB dioti einai kai oi duo kathetes stins MN, opote to T einai meso tou SI, diladi to zitoumeno.

Tha epistrepsw me kapoia wraia porismata pou brika kathos tin elina (ta opoia einai polu wraia san ksexwristes askiseis...)

Silouane kai pali thanks :)  :)

Silouane, kathisa to prwi dioti ithela na brw mia lusi xwris polikes  kai genika xoris katholou logous etc.

Tha tin "postarw" se ligo... Apla kapoia idea pou eixa, tin exw balei sto diethnes forum sti geometria...have a look at it:

http://www.mathlinks.ro/viewtopic.php?p=1076897#1076897

[quote="staretak"] Tha deiksoume oti to B einai antidiametriko tou S.  Pragmati to i desmi A. LEBF einai armoniki opote kai i desmi S.AEBF einai armoniki (idies aksonikes gwnies) kai afou OM=ON exoume oti MN//SA opote SAB=90. [/quote]

Loipon meta apo tin poli swsti parembasi tou Nikou Rapanou, exw paraleipsei na balw apodeiksh sto gegonos oti to O anikei ston aksona-eutheia SB :o  (to opoio kai sinepagetai oti to B einai antidiametriko tou S) opote se prwti fasi i parapanw apodeiksei einai ellipis. Opote gia na deiksoume oti ta S,O,B einai sineuthiaka thewroume to antidiametriko B' tou B. Tote i desmi B'. AEBF einai armoniki kai an M', N' einai ta simeia tomis twn B'F, B'E me tin OL tote isxuei OM'=ON' (dioti $ B'A//OL\perp AB$). Omws, wlog ipothetoume oti OM'<OM opote tote kai ON'<ON opote to B' anikei sto esoteriko tou trigonou SFE (dioti i FB' anikei sto imiepido pou anikei kai to E os pros tin FS kai omoia i EB' anikei sto imiepipedo pou anikei to F os pros tin SE)  ,atopo afou einai eksoteriko tou logo tou kiklou. Ara S=B' kai ara i apodeiksi teliwnei opws sto arxiko post. Opote:

[color=red]Estw oti oi euthies NM, EF temnontai sto L. Tote epidi i poliki tou I dierxetai apo to L, i poliki tou L dierxetai apo to I. Etsi an LA, LB einai oi efaptomenes sto kiklo apo to L exoume oti A,B,I suneuthiaka (B sto tokso FE pou den anikei to S). Thewroume to antidiametriko B' tou B. Tote i desmi B'. AEBF einai armoniki kai an M', N' einai ta simeia tomis twn B'F, B'E me tin OL tote isxuei OM'=ON' (dioti $ B'A//OL\perp AB$). Omws, wlog ipothetoume oti OM'<OM opote tote kai ON'<ON opote to B' anikei sto esoteriko tou trigonou SFE (dioti i FB' anikei sto imiepido pou anikei kai to E os pros tin FS kai omoia i EB' anikei sto imiepipedo pou anikei to F os pros tin SE)  ,atopo afou einai eksoteriko tou logo tou kiklou. Ara S=B' kai an T einai stin SI oste TO katheti stin MN tote afou O meso tou SB kai TO//SB dioti einai kai oi duo kathetes stins MN, opote to T einai meso tou SI, diladi to zitoumeno.
[/color]

Niko thanks gia tin paratirisi... :oops:

ok stef!! apla eksairetikos!vale kai tin alli lysh!

@@silouan pos tin elyses?

Στέφανε πραγματικά εξαιρετικά ! :D

Λοιπόν nick έχω δύο λύσεις . Μία που δεν θα σας αρέσει καθόλου αλλά είναι πολύ σύντομη (complex numbers) (Just notice that $ i\equal{}\frac{2ef}{e\plus{}f}$ όπου με μικρά συμβολίζω τις εικόνες των μυγαδικών στο μιγαδικό επίπεδο και έχω θεωρήσει τον κύκλο μοναδιαίο ) (Μπορούμε και με πολικές συντεταγμένες να φτάσουμε αρκετά γρήγορα στο result)

Ο άλλος τρ΄πος που νομίζω ότι θα σας αρέσει
Έστω ότι η $ SO$  temnei τον  $ (O)$ στο  $ B$, και ότι η ευθεία που περνά από το $ S$ και είναι παράλληλη στην $ MN$ τέμνει τον $ (O)$ στο $ Q$
Τότε  $ S(MNOQ) \equal{} \minus{} 1$, άρα το $ EBFQ$ είναι αρμνικό τετράπλευρο . Άρα  $ I,B,Q$ συνευθειακά .

Ε μετά είναι πολύ εύκολο ... :)

Silouane thanks gia tis liseis sou,

Basika i 2i lisi sou einai idia me tin dikia mou (kata bathos... :roll: )

Loipon, grafw twra kai tin alli lisi pou brika pou einai pio stoixeiwdis:

Estw $ B$ to antidiametriko tou $ S$ (pali!).Feroume eutheia $ RBQ$ katheti stin $ IB$ me $ R$ stin $ SF$ kai $ Q$ stin $ SE$. Estw $ T$ to meso tou $ SI$. Afou $ TO//IB$, prepei kai arkei na deiksoume oti $ BR\equal{}BQ \Leftrightarrow \angle BRI\equal{}\angle BQI \Leftrightarrow \angle  BXI\equal{}\angle BYI \Leftrightarrow \angle FBX\equal{}\angle EBY$ (*) opou $ IX\perp SF, X\in SF$ kai $ IY\perp SE, Y\in SE$ logo ton cyclic tetrapleurwn $ RBIX$, $ QBIY$.Proxwrame stoixeiwdws gia na deiksoume tin (*).

Fernoume $ FW//\equal{}BZ \perp SB$ me $ FW\equal{}BZ\equal{}IF\equal{}IE$, opote to $ BFWZ$ einai parallilogrammo opote $ WZ \perp SF$ kai an $ G\equal{}WZ \cap SF$ tote to $ SGBZ$ einai eggrapsimo opote $ \angle ZSB\equal{} \angle ZGB\equal{} \angle GBF \equal{}\angle FBX$ (i teleutaia isxuei dioti $ FG\equal{}FX$ kai auto dioti $ \angle WFB\equal{}\angle FSB\equal{} \angle BFI$). Omoia deixnoume oti $ \angle EBY\equal{}\angle ZSB$ opote kai pernoume tin (*).

Simeiosh: Tin (*) mporoume na tin deiksoume poli pio eukola, apla deixnontas oti ta trigona $ FBX,EYI$ einai omoia :wink: .


For $ n$ a positive integer, $ n\geq 2$, $ n\equal{}p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$, denote $ a_n\equal{}\frac{1}{p_1}\plus{}\frac{1}{p_2}\plus{}...\plus{}\frac{1}{p_k}$.Prove that $ \sum_{k\equal{}2}^{\infty} a_2a_3...a_k<1$


The percentage of correct answers on each problem is displayed below. Note that problems 1-5 on A are 6-10 on B, but the data are not merged, e.g. the 23% in Division A Algebra problem 1 refers only to Division A participants.

[b]Division A[/b]
$ \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline
\text{Test} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\\hline
\text{Algebra} & 23.0\% & 25.3\% & 11.5\% & 1.1\% & 0.0\% & 1.1\% & 33.3\% & 21.8\% & 8.0\% & 0.0\% \\\hline
\text{Geometry} & 18.1\% & 32.5\% & 16.9\% & 24.1\% & 62.3\% & 0.0\% & 8.4\% & 6.0\% & 14.4\% & 4.8\% \\\hline
\text{Number Theory} & 16.9\% & 68.7\% & 13.3\% & 12.0\% & 14.5\% & 0.0\% & 14.5\% & 9.6\% & 10.8\% & 3.6\% \\\hline
\text{Combinatorics} & 33.7\% & 8.4\% & 20.5\% & 19.3\% & 0.0\% & 13.3\% & 44.6\% & 0.0\% & 8.4\% & 2.4\%\\ \hline
\end{tabular}$

[b]Division B[/b]
$ \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|} \hline
\text{Test} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\\hline
\text{Algebra} & 84.0\% & 40.0\% & 41.3\% & 2.7\% & 17.3\% & 8.0\% & 2.7\% & 8.0\% & 0.0\% & 0.0\% \\\hline
\text{Geometry} & 27.9\% & 23.0\% & 26.2\% & 3.3\% & 8.2\% & 3.3\% & 8.2\% & 3.3\% & 1.6\% & 26.3\% \\\hline
\text{Number Theory} & 80.0\% & 31.4\% & 0.0\% & 51.4\% & 2.9\% & 5.7\% & 51.4\% & 2.9\% & 8.6\% & 0.0\% \\\hline
\text{Combinatorics} & 37.2\% & 4.7\% & 4.7\% & 7.0\% & 27.9\% & 25.6\% & 2.3\% & 11.6\% & 4.7\% & 0.0\% \\\hline
\end{tabular}$

In Division A, the following problems were solved by no one:
Algebra #5
Geometry #6
Number Theory #6
Combinatorics #5, #8

The following problems were solved by just one person:
Algebra #4: ???
Algebra #6: ???
Algebra #10: Palmer Mebane (it's 0.0% above because he wrote the answer in a weird form and it was counted wrong)

In Division B, the following problems were solved by no one:
Algebra #9, #10
Number Theory #3, #10
Combinatorics #10

The following problems were solved by just one person:
Geometry #9: ???
Combinatorics #7: ???
Number Theory #5: ???
Number Theory #8: ???

If you or someone you know is one of the people who was the only one to solve a particular problem, post here and I'll add you/them to the list!

If we define a problem to be "easy" if over 50% of people got it right, and we define a problem to be "hard" if less than 10% of people got it right, then:

Division A had 2 easy problems and 18 hard problems out of a total of 40.

Division B had 4 easy problems and 24 hard problems out of a total of 40.

Will individual finals scores ever be released?

Oh yeah sorry!

Individual finals: The chart is points awarded, number of people who got that many points.

$ \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|} \hline
\text{Problem} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \text{avg}\\\hline
A1 & 9 & 6 & 0 & 1 & 0 & 0 & 0 & 0 & 0.56\\\hline
A2 & 11 & 4 & 0 & 0 & 0 & 0 & 0 & 1 & 0.69\\\hline
A3 & 15 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0.06\\\hline
B1 & 12 & 0 & 0 & 0 & 2 & 0 & 1 & 1 & 1.31\\\hline
B2 & 12 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 0.75\\\hline
B3 & 13 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0.88\\\hline
\end{tabular}$

People's totals:
Division A
7 Shaunak Kishore
3 Bryce Taylor
2 Matt Superdock, Gaku Liu, John Berman
1 Jacob Steinhardt, Haitao Mao, Andrew Ardito, Chris Olund, Wenyu Cao
0 Kevin Lee, Alex Anderson, Palmer Mebane, Hyunji Bae, Neel Patel, Tao Ran Chen

Division B
12 Preet Patel
10 John Chiarelli
7 Gene Katsevich
6 Troy Wu
4 Jeffrey Chao, Zev Rosengarten
2 Shawn Pan
1 Ed Zheng, Matt Myers
0 Raja Rajan Vel, Tong Chen, Rui Hu, Kevin Dong, Ian Harding, Surya Viswanathan, Michael Tan

[quote="MellowMelon"]Funny that no one got Algebra 5 in A. I know I wasn't the only one who "solved" that problem first on the test.[/quote]
I think everyone screwed up the negative values of $ a$. I didn't notice whether there was anyone who had essentially the right answer but got it wrong because of writing it incorrectly. The thing that's weird about it is that if the answer really was just $ a\equal{}2$ or whatever the positive solution was, then that problem is really easy compared to pretty much everything else... but no one got suspicious I guess... (maybe if we had done a better job of putting the problems in order of difficulty, people might have wondered why it wasn't A1/B6, but since other problems were out of order too it was harder to tell)

[quote="b-flat"]You mentioned that the results would be posted on the PUMaC website, but I cannot seem to find the results there...

Also, how many people participated in each division?[/quote]
Results haven't been posted there yet. Our webmaster had a take-home exam due today...

There were 168 people in A and 107 in B, if I counted correctly.

By the way, only one team out of 36 solved T10, if I remember right. I think it was TJA (the question they missed was one of the ones that some other teams solved, maybe T8).

Not so. TJA had $ 1\plus{}\sqrt7$ for T10. Quagga was the one who got it (as far as I know)

Yeah, Quagga got that one after Gaku bashed it with Lagrange multipliers, and he wasn't even sure of his answer.


[quote="Chinaboy"]Just in case if you forgot the problem you were supposed to work on for next week:

If there are 5 boys and 4 girls and I want to choose a committee of 2 boys and 2 girls, how many different committees can I form?[/quote]

[hide="I think I have the answer"]I think it would be 5 C 2(boys) * 4 C 2(girls) = $ \ 10*6\equal{} 60$[/hide]

Um... abacadaea,

Thank you for telling me it's right, but I don't think this was your job to post here..

And I thought you live in Penn.

BUT NOBODY ELSE LIVES IN PENN!!!! *sniff sniff*. to compensate for that, i just post on other national/local communities.

have you seen anyone else from pennsylvannia on this forum besides me?

[quote="spiders_hate_chestnut"]i got the same thing as david[/quote]
And that is the right answer.(I asked you know who.. :lol: )


limit
LaTeX
real analysis
real analysis unsolved

[b]Definiton of f[/b]
$f: R^2 \to R$
$f(x,y)=\frac{xy}{x^2+y^2} \ for \ (x,y) \neq (0,0)$
$f(x,y)=0 \ for \ (x,y)=(0,0)$

[b]Problem[/b]

Study existence of partial derivates of f

[b]Part of Solution[/b]
$\frac{\partial f}{\partial x}(0,0)=\lim_{t\to 0} \frac{f((0,0) +t(1,0))-f(0,0)}{t}=$
$=\lim_{t\to 0} \frac{f(t,0)-f(0,0)}{t}=0$
why is this 0? isn't it an exception case like $\frac{0}{0}$ here ?
i dont understand this

Because for every $t$, $f(t,0)-f(0,0)=0$.  The thing you are taking the limit of evaluates to zero everywhere (except at the limit point), so the limit is certainly also zero.

[EDIT:i was wrong]
by definiton $f(t,0)=0$ and $f(0,0)=0$
so $f(t,0)-f(0,0)$ is always 0 ,algebricaly so its not an approximation
or an estimation,it's an exact value.
so now $\frac{f(t,0)-f(0,0)}{t}=0 \ \forall t\in R$
so the limit is obvious no ?

Yes -- I messed up my latex the first time, but it's fixed now.


Find limit h-->0 of (3^h-1)/h    .  Is there a way to manipulate this so that I can evaluate it?  I know it's the derivative of 3^x at x=0, but I want to be able to get it into a form where I can evaluate it?  Is there a way to?  Thanks.

What are you going to allow yourself to use? Can you use the fact that the derivative of ex is ex?

lim[size=75]x->0[/size] (3^h - 1) / h =
lim[size=75]x->0[/size] (3^h - 3^0) / h =
lim[size=75]x->c[/size] (f(c + h) - f(c)) / h

where c = 0, f(x) = 3^x, which should look familiar.

alright, thanks.  But what i was sort of looking for was a way to manipulate the expression so that i could get rid of h in the denominator and then just substitute in h=0.  But i guess that cant happen. thanks.

You can't because the limit is the expression for the derivative itself, which will always have a denominator approaching zero because, well, that's what it does.  That is, until you learn L'Hopital's rule.  But that won't be for a while yet.

Hello!:D

well we usually have that if xn-->0 then lim (a^xn-1)/xn=ln(a).

cheers! :D :D

he meant by first principles, evaluate the derivitive of y=3^x...
i tried this last time, it didnt work out too well, but i suggest if you must do this, use the definition of e=lim (n->oo) (1+1/n)^n
and evaluate the derivitive of f(x)=ln(x)... then you may use the chain rule if you wish...  another way is that the definition of e implies lim (h->0) (e^h-1)/h=1
but i am not exactly sure how this transforms into lim (h->0) (3^h-1)/h=ln(3)

If you are allowed to use (eh - 1) / h -> 1, then we can evaluate as follows:

(3h - 1) / h = (eh ln3 - 1) / h = ln 3 * (eh ln 3 - 1) / (h ln3) -> ln 3 * (1) = ln 3.

Also, if you don't mind using power series, then you can get rid of the denominator as you hoped.  For example, for nonzero h,
(eh - 1) / h = 1 + h/2! + h2/3! + ...
Because the right-hand side is continuous in h, you can subsitute h = 0 in the right-hand side to see that the limit is 1.


How do I get LaTeX to double-space? I'm also having problems with adjusting the margin. If I try to increase the text height, it will flow off of the page but still leave a huge gap on the top of the document. How do I fix this?

Play around with the following commands (It's obvious what they do)

\setlength{\topmargin}{.2in}
\setlength{\headheight}{0in}
\setlength{\headsep}{0in}

For double spacing, try this
[url=http://hajek.stat.ubc.ca/~webmaste/howto/editor/latex.html#doublespace]hajek.stat.ubc.ca/~webmaste/howto/editor/latex.html#doublespace[/url]


In my probability class, you have to have certain subjects complete, and although those are in my content area, those problems just don't come out, so I think it would be helpful if you were allowed to chose which topics you have problems on.

Hmmmmmmmm, good idea, Dojo, although I think Alcumus was already programmed to provide the algorithmically-based designation of subjects best for your level that was gauged in the diagnostic stage and additional questions afterward.

This idea is in development.  I'm not sure when we'll have it released to the public.


linear algebra
matrix
function
algorithm

I want to create a computer player for a 2-player board game,
but I don't know how to represent or to approach the problem.
I would like to know what game features and topics you think
are relevant.

[b]Game: The Crossing[/b]

o The board is a 13x13 matrix
o Player 1 and Player 2 take turn to place their pieces on empty cells
o Player 1 can place at most 4 consecutive horizontal pieces each turn
o Player 2 can place at most 4 consecutive vertical pieces each turn
o Player 1 wins if its pieces form a path from left edge to right edge
o Player 2 wins if its pieces form a path from top edge to bottom edge

[url=http://farm4.static.flickr.com/3616/3338965085_225a9e4d61_o.png][ Sample game runs with red going first (image) ][/url]

What do you think?

- Thanks

Can you give detailed info about the pieces, and also all the rest of the rules if there are any.

I assume your game engine is already working.

As for the AI, I think the best way to figure out what an AI would do is to play a bunch of games against yourself and see what strategies work best. Try and get some pseudocode from there and it should give you at least a rough AI, if not competent.

(I highly suspect that this game is deterministic if both players choose their best moves; since it is impossible to tie in this game, maybe one of the players has a winning strategy already? Then it wouldn't make for an entertaining game.)

EDIT> What language are you using anyways? I'm curious.

[quote="JBL"]Player 1 clearly has a winning strategy -- otherwise, she makes an arbitrary first play, then steals Player 2's assumed winning strategy.  But this argument doesn't tell us what such a strategy might be.[/quote]

Since the board is not infinite the arbitrary play will affect later gameplay, and if player 2's winning strategy relied on the position of player 1's move then the argument could be invalidated.

(See Sprouts: http://en.wikipedia.org/wiki/Sprouts_(game)

The argument doesn't work there.)

Re:

[b]Prototype[/b]

Hi, I am a bit surprised by the sudden replies. Thanks for replying. At the moment I have a prototype that you can try. You can tell that the AI is not very good, therefore I gave the AI some handicap by letting it start with some squares. Even with that, it is not hard to beat the AI. The prototype would show you the meaning of the game rules.

[url=http://albatross.x10hosting.com/page_viewer.php?page=1526&code=1384322218][ Crossing (Flash) ][/url]

Controls:
[ Arrows ] Move the Crosshair
[ Shift ] Clear the current ice block plus 3 on its right

Objective:
o Blast a pathway across the iceberg.

The submarine is not part of the puzzle. It is just background related
to the story.

[b]Q: Can the game have a draw?[/b]
Re: Yes. The game could result in a draw. With or without handicap.

[b]Q: What is your strategy when you play it yourself?[/b]
Re: I focus on blasting a section that has the highest chance to form a connected path from left to right. I would think about how the opponent could block my path, and place my segments at locations where the opponent cannot block my segments (if possible). I also make myself be aware of situations where the opponent could cut me off, and block the opponent before it makes me impossible to win or makes itself win.

[b]Q: How do you think that strategy would be translated and implemented in the code?[/b]
In brute force, this would be translated to making a move and simulating the opponent's moves, the simulate my move and so on. So the game would play all of the possible games and determine which move has the highest chance to succeed, or for each move, assume that the opponent (which is the player) would make the best move and select the best counter if the opponent does make the best move. But there are too many ways to place a segment and the complexity is high.

[b]Q: How does the current game do it?[/b]
The current AI does not simulate any future moves and does not calculate any chances. In each turn, for each cell, it estimates the minimum number of additional segments required to connect top-bottom and left-right if a segment is placed at that cell. For example, if a cell is the only missing cell required to connect left-right, then its left-right score is 0. This means that if the player places a segment there, the player will win. If the top-bottom score at a cell is 1, it means that the AI thinks that if it places a segment there, it would take just one more segment for itself to win. The AI then places a segment at the cell with the lowest total score, and extends that segment to a segment of 4 if there is no obstruction.

Wait, it should be impossible to draw since if one player cannot make it to the other side, then the other player must have a path blocking the first player's way.

EDIT> Oh wait, never mind. I was thinking of this:

http://nsg.upor.net/msx/basic/bridge.htm

That game must have a win state, per my argument above. This game is slightly different since a draw is forced when this configuration happens:

########NMNMN
NMNMNMNM#####

Bridge-it is analyzed easily (First player always wins with optimal play). So this game shouldn't be much different.

Anyways, nice GUI (Hopefully the mouse will have some function later). Any chance we can get our hands on the source, or is this intended for commercial release?

[quote="yanglunj"]Since the board is not infinite the arbitrary play will affect later gameplay, and if player 2's winning strategy relied on the position of player 1's move then the argument could be invalidated.

(See Sprouts: http://en.wikipedia.org/wiki/Sprouts_(game)

The argument doesn't work there.)[/quote]  Yes, strategy stealing relies on particular features of the game.  The game we're discussing does have one of these features, namely, it is monotonic.  That is, if we have two positions that differ only in that I have some additional squares in the first that I don't have in the other, I will be able to do at least as well in the former game as the latter.  So your objection was wrong and my original comment was right, except that this game can end in a draw.  This means that actually the first player has a non-losing strategy rather than a winning strategy.

In the context of the game, a "draw" is a "loss" because the submarine is still trapped by the iceberg. This is not really relevant to the math however.

[b]Q: Source code?[/b]

The full source code is long because the game itself has more than this puzzle.

This is the actual code section that runs that game stage:
[code]function S133_FPA(){
	var sn = 133;
	var sx = _root.COM_AssertStage(sn);
	var r,c,b,w;
	var board = sx.Board
	_root.Blaahval.WaterLevel = 0;
	if(_root.Blaahval.px > 30){
		_root.Blaahval.px = _root.Blaahval.ppx = 30;
	}

	_root.CharSwitchEnable=false;
	_root.PC1 = null;

	for(r=1;r<=board.Dim;r++){
		for(c=1;c<=board.Dim;c++){
		 	b = board["R"+r+"C"+c];

			if(b.Tile <> null and b.Tile <>undefined){
				w = b.Tile;
				if(b.Content=="X"){
					b.Tile._visible = false;
				}else{
					b.Tile._visible = true;
					if(b.Content=="O"){
						b.Tile.style = "TC_Rock";
					}else if(b.Content==0){
						b.Tile.style = "TC_Ice";
					}
					w.frame = w.style;
					_root.SO_AssertFrame(w);
					_root.SO_AssertLayer(w);
					_root.SO_Setxy(w);
				}
			}
		}
	}

	w = sx.Crosshair;
	w.tpx = sx.pxmin + 300 -30 + 20*w.C;
	w.tpy = sx.pymax - 70 - 20*w.R + 10;
	w.px += (w.tpx-w.px)/1.2;
	w.py += (w.tpy-w.py)/1.4;
	w.frame = w.style;
	_root.SO_AssertFrame(w);
	_root.SO_AssertLayer(w);
	_root.SO_Setxy(w);

	if(_root.KeyUPTime>0 and w.R > 1){ w.R--; }
	if(_root.KeyDOWNTime>0 and w.R < 13){ w.R++; }
	if(_root.KeyLEFTTime>0 and w.C > 1){ w.C--; }
	if(_root.KeyRIGHTTime>0 and w.C < 13){ w.C++; }
	if(_root.KeySHIFTReleaseTime>0 and sx.Board["R"+w.R+"C"+w.C].Content==0){
		if(board.NSConnected or (board.WEPathExist==false) or board.WEConnected){
			_root.TC_InitBoard(board);
			_root.COM_Dialog(_root.Blaahval,"Blank","Recycle","\n** Restart! **",20);
			return;
		}
		r = w.R;
		board.NumGroups ++;
		for(c=w.C;c<=w.C+3;c++){
			if(c> 13){break;}
			b=sx.Board["R"+r+"C"+c];
			if(b.Content<>0){break;}
			_root.COM_Dust(sn,b.Tile.px,b.Tile.py);
			b.Tile._visible = false;
			b.Tile.Invalid = true;
			b.Content = "X";
			b.Group = board.NumGroups;
		}
		w.C = c-1;
		_root.TC_MarkGroups(board);
		_root.TC_MapBoard(board);
		_root.TC_AddRocks(board,4);
		_root.TC_MarkGroups(board);
		_root.TC_MapBoard(board);
		if(board.WEConnected){
			_root.COM_Dialog(_root.Blaahval,"Blank","Heart","\n** Success! **",200);
		}else if(board.NSConnected or (board.WEPathExist==false)){
			_root.COM_Dialog(_root.Blaahval,"Blank","Frustration","\n** Blaahval is stuck! **",200);
		}
	}
}
[/code]

You can see that most of the code has nothing to do with the puzzle. They are just graphics and interface. The part that is related to the AI has this pseudo-code:

Do the following when it is the AI's turn:
1) TC_MarkGroup: Assign an unique group number to each continuous section
2) TC_MapBoard: Find the connectivity values and find the cell with the minimum score
3) TC_AddRocks: Places rocks for the AI player
4) TC_MarkGroup: Assign an unique group number again to each continuous section
5) TC_MapBoard: Find the connectivity values and find the cell with the minimum score
6) If it is detected that one side has connected, end the game. (The current code does not detect a draw, although it could easily be modified to do so.)

Is this what you want to see or do you want to see inside any of these function?

[quote="yanglunj"]Well, should we give it a go and try to find the said non-losing strategy?[/quote]  Well, I think it's probably very difficult to write down exactly what the strategy is, and the strategy-stealing argument is totally unhelpful for this.  An heuristic algorithm is probably the best we can realistically hope for.

A couple of suggestions for the AI that might lead to improvement: when it wants to connect to places, it seems to prefer to start at one of them and build towards the other.  My instinct is always to work from the middle outwards.  So, this may have something to do with how it chooses between points of equal value.

It would make more sense for the AI to add up the values of the four cells, rather than just choose a single high-value cell.  However, in this case, you should weight the values more heavily in favor of "winning right now" and "preventing my opponent from winning right now," i.e., use some sort of non-linear weighting function.

It should be possible to test improvement by personally playing against each version of the AI, or playing different versions of the AI against each other (which can generate a lot more data much more quickly, but may be mildly misleading in that beating an AI over and over again may not mean one can beat a human player).

Edit: The AI also makes moves that are plainly wrong, for reasons I don't understand.  For example, see the attached images -- the circles moves are the AIs most recent moves, and in both cases they are clearly suboptimal.

Editedit: Yes, I think it only looks for the first square of minimal value, or something.  It definitely should take into account all the squares that it will be blocking with a single move and sum their values.

Okay, here's a more sophisticated sort of improvement to think about: I just played a game in which I had three disconnected blocks I needed to join in order to win.  Between block 1 and block 2, there were several possible squares I could use to make the necessary connection, while between block 2 and 3, there was only a single possible square.  Your AI chose to (try to) prevent me from joining block 1 to block 2, even though I could still join them after its move, while letting me join block 2 to block 3, which it could have prevented.  A human player would never have made this mistake.  It's difficult to quantify this, unfortunately.

And one more: the AI seems singularly poor at estimating how easily I can join two blocks that are separated by a vertical (rather than horizontal) distance.  This is largely a feature of the poor weighting system, I think -- plain old linear just doesn't cut it.  Related to the above: perhaps the weighting scheme should take into consideration the number of ways in which I could join two blocks.  If we're talking about blocks at distance two moves, for example, it's important to know whether I can win in only a single way (in which case it's "non-urgent" and doesn't need to be blocked immediately) or if I can win in multiple ways (in which case it is worth making it a higher weight).  This amounts to a very limited "look one move ahead" approach.

Re: I think I know why the AI would do that in the last scenario:

[b]Scenario 3:[/b] Three disjoint sets with two junctions. Junction 1 can be blocked by one move; junction 2 can only be blocked in three moves.

From the view of the AI, the cells at those two junctions are the same because the LR-connectivity are both 1 (meaning that if a block is placed there, it would take just one more block to connect the left to the right). At the moment, there is nothing in the heuristic that identify the more important junction.

I will hardcode an instance of this scenario and show you the map. Perhaps there is a way from the existing maps that can tell the AI the more important junction.

[b]AI Upgraded:[/b]

[url=http://albatross.x10hosting.com/page_viewer.php?page=1616&code=2105734712][ The Crossing V2 (Flash) ][/url]

I found a bug in the code that made the AI do weird thing previously. I don't remember what it was.
I originally was just trying to post the maps, but in the process I kind of fixed the AI. The result is in
the link above. Now it is kind of hard to beat the AI so I took away the initial rocks.

Is it obvious what the non-losing strategy is?
( I don't know what it is. But I know that the AI could be beaten sometimes. )

It is not easy to state the non-losing strategy without using tons of illustrations, but here's my hint:

-Start at 1 tile above the bottom left corner. From there on if you play correctly you should win $ \frac {15}{16}$ of the time.

EDIT> The strategy is basically "Ignore what the opponent does, focus on getting your own path through the board." If that helps.

P.S. Alternative strategies COULD win in less moves, but they win less frequently. I also found that with the AI's thought process, you can go for straightforward wins like THIS:

Re: Straight win:

I didn't test that the AI would do that. Let me see whether the AI could be made to specifically not do that.

Yes, the AI is now a much better approximation of how a human might actually play.  I think actually putting a small number of random squares in might still be worthwhile to make it interesting -- in particular, I "got a feel" for how the AI plays fairly quickly, and now can beat it pretty consistently after a few initial losses.  The advantage of going first is fairly substantial.  It's probably worthwhile to play one or two dozen games against it yourself to debug (it was still making the occasional odd-seeming move when I played, and the computer makes another move after I win).  Would it be possible to create a version in which the computer makes the first move?

If you are going to add in some extra squares for the computer's benefit, you should be able to decide on a good number by having the AI play against itself a few thousand times or something -- the first player should still win a majority (and probably a substantial majority) of the time.  By the way, what happens if you play the AI against itself now, i.e., does the first player always win?

AI Upgrade: [url=http://albatross.x10hosting.com/page_viewer.php?page=1617&code=1802832887][ V3 ][/url]
In this upgrade I specifically targeted against the straight forward pattern above.

With random blocks: [url=http://albatross.x10hosting.com/page_viewer.php?page=1618&code=407336087][ V3 RB ][/url]
It seems that the random blocks confuse the AI. So it is actually easier to win with the random blocks (?)

AI starts first: [url=http://albatross.x10hosting.com/page_viewer.php?page=1619&code=272761355][ V3 OF ][/url]
This is the hardest of the three here.

There are some advantages patterns that the AI does not recognize but a human player would recognize. I haven't come across them enough to describe what they are to code the AI to use those patterns. There are also cases where the AI wastes a move trying to close a gap that the player could connect regardless. I think I will fix that next.

I can still consistently beat AI 3 since you can just apply the same general strategies; AI 1 is a lot harder because it makes some unintuitive moves that really get in the way (but still beatable with recognizing the basic win blocks: 1x2, 2x3, and 3x2.)

AI 2 is just plain evil. I don't see the AI being confused very much.

EDIT> I just realized that you could pit two "learning" AIs against each other and develop a perfect AI from there. (Or rather, two; one will consistently win while the other will play the most defensively). Same concept as a matchbox computer. You would need to read/write from an external file though.

AI Upgrade: [url=http://albatross.x10hosting.com/page_viewer.php?page=1620&code=1433551092][ V4 ][/url]

Re: Aristocrat

Which one is the best so far? I am confused.

Slight improvement, but...

(Sorry, I only played it once, I'll have more time to test this tomorrow.)

EDIT> When I say AI1, AI2, etc. I was erroneously referring to different versions of your third AI.

V3 RB is the hardest because random blocks screw over any potential winning strategies the player has.
V3 (first one) is unpredictable enough to be a challenge.

[url=http://albatross.x10hosting.com/page_viewer.php?page=1621&code=201841396][ V4 RB ][/url]: V4 with random blocks.


In Mathopolis, an adult is a person 21 years of age or older and a child is a person under 21 years of age.  Exactly half of the
adults in Mathopolis are female, and exactly half of the female
adults have exactly one biological child.  Nobody else has a
child, and there are no other children.  What percent of the
people of Mathopolis are children?

suppose that there are x number of children.
Then, there are 2x females and 2x males (adults.)
x/5x=1/5=20%


Bei mir sind gestern die Aufgaben der 2. Runde angekommen. Gibt es irgendein ordentliches Zeichenprogramm (möglichst 4 free), mit dem man Kreise und Sektoren zeichnen und die Sektoren dann noch unterschiedlich färben kann? Die Frage sollte erlaubt sein, da ich ja keine Frage zur Lösung der Aufgaben stelle, sondern nur nach einem Programm frage, mit dem ich meine Lösung darstellen kann.

Schön, dass du bereits eine Lösung hast. Ich habe bei den Aufgaben keine Ahnung :-D

Ich empfehle dir

http://geonext.uni-bayreuth.de/

Danke für die Empfehlung!
Ja, ich habe Lösungen für die Aufgaben 1 und 2. 3 und 4 gucke ich mir demnächst an.
Bis dann, der Blender

Zum Programm:
Cool, benutzerfreundlich und sogar idiotensicher (->me<-). Auch von mir ein thx :lol: !

... gerade mal ein Tag lang draußen, schon sind die Aufgaben gelöst  :)  ... (ich werde sie mir erstmal ordentlich durchlesen  müssen :oops: *pfeif*)
P.S: Gehe mal davon aus, dass eine mögliche Interpretation deines Namens [i]nicht[/i] stimmt :D  ;) ...

Viele Grüße und viel Spaß noch beim "Rechnen"  :D

Naja, ich habe den Namen deshalb gewählt, weil wir an dem Tag in der Schule über Blendertum (nicht im Unterricht) diskutiert haben. Also ist mein Name wirklich das, was er zu sein scheint. Mit Blender ist wirklich Blender gemeint... Das bezieht sich allerdings nicht auf mich, sondern war nur das erste, was mir eingefallen ist, als ich den Namen gewählt habe...
Also kannst du wirklich davon ausgehen, dass diese Interpretation nicht stimmt und du hast damit Recht...


inequalities
calculus
inequalities proposed

For $ x \in R$ ; find the minimum value of

$ \sqrt{x^2\plus{}8x\plus{}32}\plus{}\sqrt{x^2\minus{}4x\plus{}5}$

thanks :D

[quote="hungnsl"]For $ x \in R$ ; find the minimum value of

$ \sqrt {x^2 \plus{} 8x \plus{} 32} \plus{} \sqrt {x^2 \minus{} 4x \plus{} 5}$

thanks :D[/quote]
I think by Minkowski the minimum  value is $ 5$

:oops: i'm secondary student and i have just learnt inequalities, could you explain more?

For this problem, the expression has minimum value when $ x\equal{}0,8$ ( but i can't prove :(  )

Just by Calculus, we get min. value attains when $ 5x^2 \minus{} 24x \plus{} 16 \equal{} 0$, i.e. when $ x \equal{} 0.8$

Sorry, i do not understand your proof  :blush: , do u have any simple solution ?

$ \sqrt {(x\plus{}4)^2\plus{}(4)^2} \plus{} \sqrt {(x\minus{}2)^2\plus{}(\minus{}1)^2}$
two points (-4,4),(2,-1)
connect them have the line solve $ 0\equal{}\frac{\minus{}5}{6}(x\minus{}2)\minus{}1$
have the point (0.8,0)on  x-axis that is the answer 0.8

[quote="hungnsl"]For $ x \in R$ ; find the minimum value of

$ \sqrt {x^2 \plus{} 8x \plus{} 32} \plus{} \sqrt {x^2 \minus{} 4x \plus{} 5}$

thanks :D[/quote]
Consider two right triangles with sides $ 4, x\plus{}4$ and $ 1,2\minus{}x$. Then it is obvious that the minimum is $ \sqrt{5^2\plus{}6^2} \equal{} \sqrt{61}$. Reached when $ \frac{x\plus{}4}{4} \equal{} \frac{2\minus{}x}{1} \iff x \equal{} \frac{4}{5}$

Oh yeah , i see, it's an application of cauchy-schwarz

$ \sqrt{{a_1}^2\plus{}{b_1}^2}\plus{}\sqrt{{a_2}^2\plus{}{b_2}^2} \geq \sqrt{(a_1\plus{}a_2)^2\plus{}(b_1\plus{}b_2)^2}$

sorry for my foolishness  :blush:

thanks any way

[quote="hungnsl"]Oh yeah , i see, it's an application of cauchy-schwarz

$ \sqrt {{a_1}^2 \plus{} {b_1}^2} \plus{} \sqrt {{a_2}^2 \plus{} {b_2}^2} \geq \sqrt {(a_1 \plus{} a_2)^2 \plus{} (b_1 \plus{} b_2)^2}$

sorry for my foolishness  :blush:

thanks any way[/quote]
The inequality $ \sqrt {\sum_{i \equal{} 1}^n x_i^2 } \plus{} \sqrt {\sum_{i \equal{} 1}^n y_i^2 } \ge \sqrt {\sum_{i \equal{} 1}^n (x_i \plus{} y_i)^2 }$ is known as the "triangle inequality". This can be proved as you mentioned by Cauchy-Schwartz.


The question is correctly typed.  I need to find the value of
cos(180 + a)^0 or raised to the zero power.  Can you help me?

$a^0 = 1$ for all $a \neq 0$, that's why it doesn't make sense AT ALL.

Wait...$0^0\neq1$?  I always thought it was...

[quote="Arne"]What the -?[/quote]I agree with Arne.

Symmetry, if you're going to be making almost 11 posts per day, you should probably learn $\LaTeX$, which makes things much easier to read.

Also, let's not start another war on whether or not $0^0$ is defined.

post in getting started...these problems do not require any application of knowledge, there is just a way simple procedure that would be covered in a normal math class

I think those 0 exponents were meant to be degree symbols.  It is better to use ^o instead of ^0 for degrees.  Even better, use LaTeX, where the appropriate symbol is ^\circ.

Oh, it makes sense now, Ravi B :D Thanks.

I did not start any war.  I simple asked a question found in my math book.

[quote="ifai"][quote="symmetry"]If 1/(2sin x - 1) is undefined when x = a^0, what is the value of
cos(180 + a)^0?[/quote]
a^0 means $a^\circ$ and (180+a)^0 means $(180+a)^\circ$?[/quote]
If this is right then....
[hide="Sol"]
If $\frac{1}{2\sin{x}-1}$ is undefined so,
the denominator is $0$ and so,
$\sin{x}=\frac{1}{2}$
$x=30^{\circ}$
So,
$\cos{(210)}=\frac{-\sqrt{3}}{2}$
Correct?
[/hide]

Symmetry, learn [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]LaTeX[/url].  It will make your posts look much nicer and you will be able to avoid the confusion posed by your statement of the problem, i.e. the degree versus zeroth power confusion.

[quote="Rep123max"]Um, $\displaystyle a^0=\frac{a\cdot a\cdot a\ldots a}{a\cdot a\cdot a\ldots a}$ with the same number of a's. So if $\displaystyle a$ is 0, we are dividing by 0 (a big no-no).[/quote]

Another way to explain this is to say that $0^0=0^{a-a}=\frac{0^a}{0^a}=undefined$ iff $a \neq 0$.  In the case that $a=0, 0^a=1$.  Although this is mathematically correct, it is quite illogical to say that nothing will result into something.

Masoud Zargar

[quote="boxedexe"]Although this is mathematically correct, it is quite illogical to say that nothing will result into something.
Masoud Zargar[/quote]

No, it isn't, and please don't start the whole discussion about $0^0$ all over again.

There was a $0^0$ discussion in this forum, or are you just referring to the history of mathematics?  By the way, I am interested in knowing what is wrong with: $0^0=0^{a-a}=\frac{0^a}{0^a}=undefined$ iff $a\neq 0$. Please expound on why this is wrong.

Masoud Zargar

Please do that in an appropriate thread.

Okay, sorry.  I'll do that as soon as possible if need be.

Masoud Zargar


algebra
polynomial
superior algebra
superior algebra solved

This is probably very easy question, but I'm not familiar with field extensions. Is every finite extension of $\mathbb{Q}$ normal?

No, e.g. take $\mathbb{Q}[\sqrt[3] 2]$:
It's clearly an extension, but the polynomial $x^3-2$ has just one root in it.


let $x,y>0$ find the biggest $\alpha$ such that:
$(x+y)^4 \geq \alpha xy(x^2+y^2)$

The required biggest $\alpha$ is 8: We have

$\left(x+y\right)^4-8xy\left(x^2+y^2\right)=\left(x-y\right)^4\geq 0$,

and thus $\left(x+y\right)^4\geq 8xy\left(x^2+y^2\right)$. This is the particular case (n = 2) of [url=http://www.kalva.demon.co.uk/imo/isoln/isoln992.html]the IMO 1999 problem 2[/url].

  darij

$x>0,y>0\Longleftrightarrow x=r\cos \theta, y=r\sin \theta \left(0<\theta <\frac{\pi}{2}\right)$,

${\frac{(x+y)^4}{xy(x^2+y^2)}=\frac{2(1+\sin 2\theta )^2}{\sin 2\theta}=2\left(\sin 2 \theta+\frac{1}{\sin 2 \theta }+2\right)}\geq 8$


geometry
geometric transformation
homothety
ratio
parallelogram
circumcircle
geometry proposed

Let a $ \triangle {ABC}$,in a circle $ C(O,R)$ and let $ A_1,B_1,D_1$ the midpoints of $ BC,CA,AB$ respectively.Consider the circles $ C_1(A_1,\frac {R}{2}),C_2(B_1,\frac {R}{2}),C_3(D_1,\frac {R}{2})$.Prove that the circles $ C_1,C_2,C_3$ passing through a common point (let $ N$) and that the second common point of the circles are the midpoints $ A_2,B_2,D_2$ of $ OA,OB,OC$ respectively.Then prove that $ A_1A_2,B_1B_2,D_1D_2$ and $ ON$ are concurent.

The circle $ C_1 (A_1, \frac {R}{2})$ passes through $ B_2$ and $ D_2$, and that is obvious, a.s.o. Consequently, since there is a homothety centered $ G$ and ratio $ \minus{} \frac {1}{2}$ mapping $ A$ to $ A_1$, $ B$ to $ B_1$ and $ D$ to $ D_1$ and another homothety centered $ O$ and ratio $ \frac {1} {2}$ mapping $ A$ to $ A_2$, a.s.o., we come to  the conclusion that there is a homothety of ratio $ \minus{}1$ mapping $ A_1$ to $ A_2$, a.s.o. ( or, alternatively, see that $ A_2B_2A_1B_1$ a.s.o. are parallelograms ), consequently it is clear that $ A_1A_2 , B_1B_2$ and $ D_1D_2$ share the same midpoint.
Further, it's easy to see that $ O$, the circumcenter of $ \triangle A_2B_2D_2$ is the orthocenter of $ \triangle A_1B_1D_1$ and, consequently, the orthocenter of $ \triangle A_2B_2D_2$ is the circumcenter of $ \triangle A_1B_1D_1$.

As $ A_2B_2$ passes through the midpoint of $ OD_1$, a.s.o., $ A_1D_1$ will pass in turn through the midpoint of the segment joining $ B_2$ with the orthocenter of the $ \triangle A_2B_2D_2$ which, consequently, will belong to all three equal circles. Being equally apart of $ A_1, B_1$ and $ D_1$, it's $ N$, the 9-points circle of $ \triangle ABC$. Seeing that $ A_2NA_1O$ is a parallelogram, the problem is solved.

Best regards,
sunken rock


[quote="Algadin"]Let triangle $ ABC$. $ \widehat{A}\geq 90$
.$ AB \equal{} c,BC \equal{} a,CA \equal{} b$ Find the minimum of
$ P \equal{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc}$
 :roll:[[/quote]
We have $ a^2\ge b^2 \plus{} c^2$
$ P \equal{} 2 \plus{} (\frac {b}{c} \plus{} \frac {c}{b}) \plus{} a(\frac {1}{b} \plus{} \frac {1}{c}) \plus{} \frac {b \plus{} c}{a}\ge 4 \plus{} (\frac {b \plus{} c}{a} \plus{} \frac {2a}{b \plus{} c}) \plus{} \frac {2\sqrt {b^2 \plus{} c^2}}{b \plus{} c}\ge 4 \plus{} 3\sqrt {2}$
 :D


Prove that, for 0  :le: a,b,c  < 1 and a+b+c=2

abc/[(1-a)(1-b)(1-c)]  :ge: 8

zabelman wrote:[hide]First, note that all three numbers must be less than 1, because if one of them equaled 1 than the given fraction would be undefined. Therefore, the three numbers 1-a, 1-b, 1-c are positive. Substitute these three values as x, y, and z, so that x+y+z=1. The given fraction becomes (1-x)(1-y)(1-z)/(xyz). We have

(1-x)(1-y)(1-z)/(xyz)
=(x+y)(y+z)(z+x)/(xyz)
 :ge: (2 :sqrt: xy)(2 :sqrt: yz)(2 :sqrt: zx)/(xyz) (by AM-GM)
=8xyz/(xyz)
=8 QED.[/hide]

By the way, welcome to the forum!

Thanks for the welcome.  I wish that I had found this forum a long time ago, as even though I'm a Canadian high school student, I took part in the AIME, AMC, etc, and perusing this board is excellent preparation for the upcoming Canadian Mathematical Olympiad (nothing compared to the USAMO, although there are occasionally some good problems on it).

I actually had posted this more as an exercise than anything; I had arrived at that substitution for doing the question as well, although I must confess that it took me the better part of a day and a half, rather than minutes as it did for you.  I never seem to be able to make the correct substitution when solving inequalities of this type.

[quote] (nothing compared to the USAMO, although there are occasionally some good problems on it).
[/quote]

I COMPLETELY disagree.  true the questions are easier, but you only get three-hr to work on it instead of nine.  i suppose you meant that last year's CMO was exceptionally easy...  but nevertheless its still hard to hand in a perfect paper (well maybe you were one of them but...)  Anywayz, its rumoured that this year, the states will officially ask you to write the USAMO if your AIME + AMC score is high enough, even if you are only a Canadian citizen/landed immigrant.  The Canadian IMO selection team will also ask you to write USAMO (which is marked in Canada i believe) provided that you qualified for Winter Camp or did exceptionally well in COMC/CMO


calculus
integration
trigonometry
algebra
function
domain
calculus computations

For example when solving integrals by applying trigonometric substitutions: $ \int\sqrt{1\minus{}x^2}\,dx.$

Then one puts $ x\equal{}\sin u$ but the integrand (without the $ du$) apparently is $ \cos u,$ but, it's not always true that $ \sqrt{\cos^2u}\equal{}\cos u,$ hence one forgets that and keep going.

I was told that it doesn't matter, but it does matter when working with definite integrals. (That is clearly obvious to me.)

Here's why in the end it doesn't matter:

In order for this to make sense as a real integral, we must have $ x\in[\minus{}1,1].$ We usually write our substitution as $ x\equal{}\sin\theta$ but it might be best to think of it as $ \theta \equal{}\arcsin x.$ By that I mean we adopt the domain conventions of the arcsine, and say that $ \minus{}\frac{\pi}2\le\theta\le\frac{\pi}2.$ In those quadrants - the fourth quadrant and the first quadrant - we have $ \cos x\ge 0$ and hence $ \sqrt{\cos^2x}\equal{}\cos x.$

Make similar assumptions about quadrants for the other trigonometric substitutions.


Suppose BC is the longest side of triangle ABC.O is any point inside the triangle.P,Q and R are points on sides BC,CA and AB respectively.AP,BQ and CR meet at O.

Prove that OP+OQ+OR <BC.

If AP is the greatest among the three cevians AP,BQ CR prove further that

OP+OQ+OR<=AP.

Look at my post at http://www.mathlinks.ro/Forum/viewtopic.php?p=113711 (which I have updated in order to include the easy proof of the second inequality).

  darij


When it says "due on November 24", does that mean we can still submitt solutions on nov. 24th?

this is probably a really bad question

[quote="DPatrick"]Yes.  You can upload solutions on the website until midnight (Pacific Time) on Monday, and if you are mailing you must have a postmark on or before Monday.[/quote]
It's hard for me to read "midnight on Monday" know exactly which midnight it is.

Is it "the midnight after Monday" or "the midnight before Monday"?

I think some people have resolved this problem by saying "00:00" or "24:00" and distinguishing between those two times.

It's really 24:00 on Monday, that is, when the day ends. :D

I count 5 Mr. Green's in the above post.

Anyhow, part of the reason I was so rushed last round was because I thought "midnight of October 14th" meant after 11:59 on October 13th, so I (on my birthday) thought it was due by the end of October 13th and rushed unnecessarily, when it was really due at the end of October 14th.

A sad day for me... once I realized I had another day to do it. :( And I had already turned my (horrible) solutions in.

Well but isn't it also important to spread out your time per problem and don't wait till the last minute? (that's why they make the deadline that way to be due on midnight)

Oh and I count 0 Mr. Green's in the above post. :D  :D  :D

Yep, I'm definitely not making the same mistake this round.  :| :D

That's good...

maybe even the smartest kid in the world couldn't do all the problems in 1 hour and submit the solutions and still get a perfect score...or something close? :D  :maybe:  :wink:

Oh and I count 1 Mr. Green in the above post. :D

EDIT: the above post has a combo of smilies that don't make sense at all :rotfl:

lol, i need to  finish writing my solutions but i ran out of answer sheets, so i have to do it after school tomorrow. My printer is broken so i cant print any at home.


AMC
AMC 10
AIME
LaTeX
factorial
articles

You all know the rules (haha!). JUst this time there'es a slight twist: IF you break a rule you are done for that round, but if someone else can find a rule they post it. Whoever posts a rule unposted on for 3 days gets 1 point for every rule in the chain.

Rule 1: All rules must be in English.

Rule 1: All rules must be in English.
Rule 2: Rules must contain exactly $ \sqrt {38 \minus{} 13n}$ English words, where [i]n[/i] is the rule number.

Garyzx, one of the official rules is that you can't post an un-follow-up-able rule.

New round.

1. All rules must contain at least [b]one[/b] number (you can write it out or use the decimal form).

1. All rules must contain at least [b]one[/b] number (you can write it out or use the decimal form).
2. All rules must contain at most one number.
3. After this rule, there will be zero instances of the word "rule".

And I am assuming that the "2." doesn't count as a number...

[quote="james4l"]1. All rules must contain at least [b]one[/b] number (you can write it out or use the decimal form).
2. All rules must contain at most one number.
3. After this rule, there will be zero instances of the word "rule".
4. [color=green]New statements must be made in a new, unused color.  3 is a cool number.[/color]
5. [color=#00FFFF]Four words only, please.[/color][/quote]
6. [color=red]Don't exceed one "a".[/color]

1. All rules must contain at least [b]one[/b] number (you can write it out or use the decimal form).
2. All rules must contain at most one number.
3. After this rule, there will be zero instances of the word "rule".
4. [color=green]New statements must be made in a new, unused color.  3 is a cool number.[/color]
5. [color=#00FFFF]Four words only, please.[/color]
6. [color=red]Don't exceed one "a".[/color]
7. [color=orange]Zero quote tags showing.[/color]
8. [color=yellow]Further requirements must use a smiley in them, such as this one:  :ninja: [/color]

You're out, mew, you can only use 4 words.

New game:

1. All rules must have ONE smiley in them. =] (those count too)

1. All rules must have ONE smiley in them. =] (those count too)
2. [b]All rules from now on must be posted in black and be bold :D [/b]
3. [b]You must continue a quote stack in every post you make in this thread!  :evilgrin: [/b][/quote]

1. All rules must have ONE smiley in them. =] (those count too)
2. [b]All rules from now on must be posted in black and be bold :D [/b]
3. [b]You must continue a quote stack in every post you make in this thread!  :evilgrin: [/b]
4. [b]Conciseness: You must intentionally give your rule a quality such that it obviously breaks the word in front of it - here, the word is conciseness, for which, you can see, would be what a loquacious speaker might say, except more... or less... but still, you cannot, and I STRESS THIS POINT, obey that first word, delineated with a colon. Did I mention this? Well, you must have an adjective and a colon at the beginning of the rule, if I did not state that clearly enough. I see your faces. They must look like this:[/b]  :huh:
[/quote][/quote]

1- The required punctuation after the rule number cannot be a period (.) or parenthesis ()) and must be different from the punctuation from the previous rule.
2~ No quote stacks.
3% Three words max.
4; Three words minimum
5& No BB-Codes, alright?

[BB-Codes = 1 word]

1- The required punctuation after the rule number cannot be a period (.) or parenthesis ()) and must be different from the punctuation from the previous rule.
2~ No quote stacks.
3% Three words max.
4; Three words minimum
5& No BB-Codes, alright?
7> Prime rule numbers

Rule 4 breaks rule 7. You have to say, after this rule...

1. After this rule, the number of words in any two rules must be different. Further more, if rule i has a_i words and rule j has a_j rules, then |a_i - a_j| must be greater than i + j. Yes, i am evil!!!

1. After this rule, the number of words in any two rules must be different. Further more, if rule i has a_i words and rule j has a_j rules, then |a_i - a_j| must be greater than i + j. Yes, i am evil!!! (37)
2. No you are not evil yet. From now on append the numebr of words at the end (ok to be fair I will define "word" as a "non-mathematical" statement, so "|a_i - a_j|" is not a word). Got tath? The number of typos from now on must equate to C_n, where n is the rule number, and the C_n = the nth Catalan Number. Yes that is fairly evil too. (64)

Its not a typo if you mean to.

1. All rules must allow new rules to be made in less than 2 minutes with little thought.

[quote="aleph0"][quote="evilhamster"][quote="mathblitz"]1. All rules must allow new rules to be made in less than 2 minutes with little thought.
2. The word "after" cannot be used after the words "all rules."
3. After this rule all rule numbers must be consecutive composite numbers. (i.e. 4. blah blah blah 6. blah blah blah)[/quote]
4. All rules must have a quote stack.[/quote]
6. All rules must contain an emoticon like  :D .[/quote]

Rules involving emoticons eventually prevent further rules, because there is a smiley limit.

1. Rule number = x^x where x=actual rule number

4. Never include the previous rule. Also, no rule can cause a previous rule to break it.

That's 2 rules in 1. Choose one.

And james, must the next rule be 27, or can it be like $ \pi^{\pi}$?

27. You must have at least one "e" in your rule.

256. You must have either 7, 10, or 16 spaces in your rule so that it works.

3125. No rule after this one starts with the letter 'N'.


[quote="chien than"]Let $x;y;z>0$.Prove that:
$x+y+z+xy+yz+zx \geq \frac{-2-xyz}{2}$[/quote]

umm...what? move everything over the the left and you have all positive terms...also i don't see how there is equality

[quote="chien than"]Let $x;y;z>0$.Prove that:
$x+y+z+xy+yz+zx \geq \frac{-2-xyz}{2}$[/quote]
i think he didnt mean it .... can you check your problem plz , chien than


$ C(n, r) \equal{} \binom{n}{r}$ -- it's a binomial coefficient.  When $ r$ is maximal, this gives Fibonacci numbers.  Don't know about the partial sums, though.


If $ABCD$ is a cyclic quadilateral such that $AC$ is the diameter

(i)Does there exist a cyclic quadilateral such that all the four side are integer ?

(ii) If $AD=bc$ and either $AB$ or $BC$ is $\sqrt{b^2+c^2}$ for $b,c$ are positive integer . Does there exist an all integer-side quadilateral for this cyclic quadilateral?

[hide="Part i"]
Yes.

Since $AC$ is the diameter of the circle, we know that $\angle B=\angle D=90^\circ$.
So we can use Pythagorean triples.
For instance, if $AC=25$, we can have $AB=15$, $BC=20$, $CD=7$ and $DA=24$.[/hide]


geometry
vector
circumcircle
IMO Longlist

In the triangle $ABC$ let $D,E$ and $F$ be the mid-points of the three sides, $X,Y$ and $Z$ the feet of the three altitudes, $H$ the orthocenter, and $P,Q$ and $R$ the mid-points of the line segment joining $H$ to the three vertices. Show that the nine points $D,E,F,P,Q,R,X,Y,Z$ lie on a circle.

[hide="This is a well-known problem and the circle is known as the nine-point circle"]Just consider vectors, where the origin is the circumcenter of $ABC$. It's well-known that $\overrightarrow{H}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}$, and from this it's easy to prove $PD$, $QE$ and $RE$ have the same midpoint and the same length, therefore $D,E,F,P,Q,R$ belong to the same circle, with diameters $PD$, $QE$ and $RE$, and from this, it's obvious $X,Y,Z$ belong to that circle.[/hide]

Dear Mathlinkers,
for a proof issue of the Morley's circle and some histories,
http://perso.orange.fr/jl.ayme vol. 2  Les cercles de Morley, Euler...p. 3-5.
Sincerely
Jean-Louis

it is a populer theory that say this 9 points are lie in circle!!
sorry for my bad english


For $a_k$ and $b_k ( k=1,..,n)$ positive real numbers prove the ineq

$2(\sum\limits_{k=1}^n a_k)(\sum\limits_{k=1}^n b_k) \leq n[ \sum\limits_{k=1}^n (a_k)(b_k)+\sqrt {(\sum\limits_{k=1}^n (a_k)^2)(\sum\limits_{k=1}^n (b_k)^2)}]$

A very nice problem, generalizing a problem of Vasile Cartoaje from Kvant. A particular case appeared also in Mathlinks Contest. There is a three lines solution using Cauchy. Try to find it.

This three lines found my friend Victor Rojko in the classroom:
[∑ ai*(2(∑bi)/n - bi)] 2  ≤  (∑ ai 2 )*(∑(2(∑bi)/n - bi) 2 ) = (∑ ai 2 )(∑ bi 2 ).

Yes !
This was also my original solution.


inequalities
number theory proposed
number theory

Prove:If $ a\equal{}p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_{n}^{\alpha_n}$ is a perfect number, then
$ 2<\prod_{i\equal{}1}^n\frac{p_i}{p_i\minus{}1}<4$ ;
if moreover, $ a$ is odd, then the upper bound $ 4$ may be reduced to $ 2\sqrt[3]{2}$.

[quote="Albanian Eagle"]Prove:If $ a \equal{} p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_{n}^{\alpha_n}$ is a perfect number, then
$ 2 < \prod_{i \equal{} 1}^n\frac {p_i}{p_i \minus{} 1} < 4$ ;
if moreover, $ a$ is odd, then the upper bound $ 4$ may be reduced to $ 2\sqrt [3]{2}$.[/quote]
Because $ n$ is a perfect number so we have $ \prod_{i\equal{}1}^n \frac{p^{\alpha \plus{}1}\minus{}1}{p^{\alpha}(p\minus{}1)}\equal{}2$.
Because $ \frac{p^{\alpha\plus{}1}\minus{}1}{p^{\alpha}(p\minus{}1)}<\frac{p}{p\minus{}1}$ so the first inequality is followed.
By the condition we have $ (\prod_{i\equal{}1}^n \frac{p^{\alpha\plus{}1}\minus{}1}{p^{\alpha}(p\minus{}1)})^2\equal{}4$. In addition, it is not very hard to show that $ (\frac{p^{\alpha\plus{}1}\minus{}1}{p^{\alpha}(p\minus{}1)})^2> \frac{p}{p\minus{}1}$ for all $ \alpha>1$ or $ p>2$( this inequality is equivalent to $ p^{\alpha}>2\plus{}\frac{1}{p^{2.\alpha\plus{}1}}$. Even in case $ p\equal{}2$ and $ \alpha\equal{}1$ then $ a\equal{}6$, the inequality is obviously true.
When $ a$ is odd, it can be proved in the same way with the upper bound $ 2.\sqrt[3]{2}$

Well... if $ a$ is even we can use the fact that $ a=2^{k}(2^{k+1}-1)$ where $ 2^{k+1}$ is prime and the result is $ \displaystyle\prod{\frac{p}{p-1}=2+\frac{1}{2^k-1}}$. Both inequalities are obvius and we can't improve the left inequality because we don't know if there are infinitely many Mersenne's primes.

:)


If n is a composite number with n >4, then (n-1)! is congruent to 0 modn.

[hide] If n is composite it can be broken down into two smaller numbers, j and k, that are less than n.  They are thus included in (n-1)!, so (n-1)! is divisible by j and k and thus n. [/hide]

what about 4? since it is the square of a prime, the factors j and k would be the same, and both equal 2, but 2 only appears once in the factorization of 3!

E^(pi*i)=-1 - ;)  You make some mistakes.
Why don't you prove it by cases that n is perfect square and n is not perfect square
(which can prove by basically inquality)

Let  $p=\sqrt{n}$

Prove that $n-1 \geq 2\sqrt{n}$

By square
 $n(n-6) \geq -1$ with true when $n \geq 6$

And parrot your question is not true when $n =4$

Is that a statement related to Wilson's Theorem?
Maybe we can argue like this:

The stement is not true for n = 4.
Then for any n>4, in view of the prime factorization theorem,
we may let p be the smallest positive divisor of n other than 1.
Next, let n = pq, then we have q = n/p < n and q>2(with the assumption that $n \neq 4$.)
This means that $1<p<q<n$.
Of course p and q will appear in the list of 1,2,3,...,n-1.
Therefore, n = pq divides (n-1)!.

The above argument fails when n = 4 because we cannot assert that p<q!

[b]Opps! Forget the case when n is a perfect square of prime number.[/b] Need to think again!

The case of n = perfect square of a prime number is also true:
Let  $n=p^2$ and P>2.
Simply consider the list of numbers: $1,2,..,p-1,p,....,2p,......,p^2-1$.
Making use of p and 2p, then $n=p^2$ divides $1*2*3.....*(p^2-1)$.

Phew....


inequalities
real analysis
real analysis unsolved

Let $ \{u_i; i \geq 1\}$ be a sequence of positive real numbers.
(i) Show that if $ (u_i)$ decreases, then the convergence of $ \sum_i u_i$ is equivalent to the convergence of $ \sum_i (u_1 u_2 \cdots u_i)^{1/i}$.
(ii) Does (i) hold if $ (u_i)$ is not decreasing?

[hide="Hint for ii"]
Certainly not. Take $ u_i$ to be exponentially small except for a few, sparse instances where $ u_i\equal{}1$, say. Then $ \sum u_i$ obviously diverges but the geometric means are still tiny, and you can make sure they converge.
[/hide]

[hide="(i)"]One implication is trivial, since we have $ \sqrt[i]{u_{1}\cdot\ldots\cdot u_{i}} > u_{i}$. Another can be proven using [url=http://www.mathlinks.ro/viewtopic.php?highlight=inequality+carleman&t=5016]Carleman's Inequality[/url]. [/hide]


Let$ \mathbb{Q^\plus{}}$ denote the set of positive rational numbers.
Show that there exist one and only one function
$ f: \mathbb{Q^\plus{}}\rightarrow\mathbb{Q^\plus{}}$ satisfying the following conditions:
(i)If $ 0<q<1/2$ then $ f(q)\equal{}1\plus{}f(q/1\minus{}2q)$
(ii)If $ 1<q<\equal{}2$ then $ f(q)\equal{}1\plus{}f(q\minus{}1)$
(iii)$ f(q)*f(1/q)\equal{}1$ for all $ q\in \mathbb{Q^\plus{}}$

[hide]We can combine condition 3 with 1 and 2 to get conditions for $ q>2$ and $ \frac{1}{2}\le q <1$. So for any rational number $ q_1$, we can use the conditions we got or 1 and 2 to express $ f(q_1)$ in terms of the function of a smaller rational number $ f(q_2)$. By number 3, $ f(1)\equal{}1$. This recursion finishes with a finite number of steps, so the function is defined uniquely.[/hide]

[quote="Temperal"][hide]We can combine condition 3 with 1 and 2 to get conditions for $ q > 2$ and $ \frac {1}{2}\le q < 1$. So for any rational number $ q_1$, we can use the conditions we got or 1 and 2 to express $ f(q_1)$ in terms of the function of a smaller rational number $ f(q_2)$. By number 3, $ f(1) \equal{} 1$. This recursion finishes with a finite number of steps, so the function is defined uniquely.[/hide][/quote]

[hide="Hmm, can you explain how .. ?"]
I'm a bit doubtful of the validity of the above;

How do you know that the recursion finishes with a finite number of steps (or in particular that it can reach $ f(1)$)?

Also, since you never used any of the function definitions, would that hold true for all functions defined on the same domain?
[/hide]

[hide="Oh, I see"]We can find $ f\left(\frac{a_1}{b_1}\right)$ in terms of $ f\left(\frac{a_2}{b_2}\right)$ with $ a_1,a_2,b_1,b_2$ all integers and $ a_1\plus{}b_1>a_2\plus{}b_2$. This guarantees a finite number of steps, so the function is uniquely defined.[/hide]


Support AMC USA(J)MO USAMO USAMTS AIME geometry

There were like a hundred people invited to MOSP last year but only thirty this year.  What happens at MOSP and what does it take to make it into MOSP?  Are they ever going to invite that many people like they did last year?  :roll:

Well, they only invited a huge number of people to MOSP last year because they received a large donation that year.

Fierytycoon

To get into MOSP you need to place in the top 24 to 30 in the USAMO.  It is unlikely that MOSP will get large donations like last year frequently, but it might happen again.

I was at MOSP this year, and we heard that the Akamai Foundation is actually giving us money again (yay!), so MOSP is going to be expanded to 50-60 people next year. Thus is will be a lot easier for non-seniors to get in next year, though not automatic like 2002. Seniors, however, still have to make top 12 to go.

[quote]I was at MOSP this year, and we heard that the Akamai Foundation is actually giving us money again (yay!), so MOSP is going to be expanded to 50-60 people next year. [/quote]

Oh yeah! Last year I have a chance to make it as a non-senior.  :D

Fierytycoon

meh. me too.
:(

Well, right now I'm at PROMYS....heh.

[quote]I was at MOSP this year, and we heard that the Akamai Foundation is actually giving us money again[/quote]

so does that mean Akami Foundation gave the large donation last time also

[quote="ComplexZeta"]Yes. In 2002, Akamai gave AMC a very large donation to support the centralized USAMO event in Boston and the expanded MOP.[/quote]

Of course, now that they have the download thing down pat (well, mostly), it is likely that to minimize costs, they will not have a centralized USAMO event for a while.   :cry:   Boohoo.  However, this means that they probably will be able to afford a larger MOP program as long as they keep getting the donations.

about how much money did akami give to mosp a few years ago?

[quote="mathfanatic"]Of course, now that they have the download thing down pat (well, mostly), it is likely that to minimize costs, they will not have a centralized USAMO event for a while.[/quote]

I did not qualify for USAMO 2002, so I did not attend the central event. However, the impression I got from those I talked to at MOP was that it was totally unnecessary. Also, the concept is rather ridiculous- the test is exactly the same whether you download it or fly somewhere to take it, and it seems downright irresponsible to spend so much money to flew so many people in just to take a test. Not to mention it disrupts the schedules of students who are probably in the midst of getting ready for AP exams and such concerns.

[quote="ComplexZeta"]At least it was fun.[/quote]

Well certainly it was, and I do truly wish I could have been there. But it did spend a whole lot of Akamai money, nonetheless.

the only problem is it was a gigantic commercial for akamai and mit.

However, it was not required for students to go to Boston for the USAMO 2002. Therefore, theoretically, those who wanted to cheat could stay home and take it at schools. But how many cheaters do you think there are? I think (and hope) that there are not very many.

[quote]But how many cheaters do you think there are? I think (and hope) that there are not very many.[/quote]

If there are any, they are certainly not cheating very well. The top USAMO scorers are almost always on the IMO team, and cheating wouldn't be very possible on the TST since it is held at MOP.

Fierytycoon

The other thing about cheating on the USAMO is it would be pretty hard to pull off. You have to have a general understanding of a problem to write a solution, even if someone tells you. Not to mention, there are very few people who can even do USAMO problems, certainly there isn't a teacher in my school that could stand a chance, much less help me cheat. I seriously doubt there is much, if any, cheating on the USAMO. If cheating were to happen anywhere, it would be on USAMTS, but I'm not even sure if that happens very much.

[quote="ComplexZeta"]At least it was fun.[/quote]

It sure sounded fun.  You lucky...first year I get to USAMO, and I end up having to spend 6 hours in the principal's office [taking the test].  Not to mention all the annoying little 6th graders passing by and making faces at me...

[quote="hello"]But it is a good idea that they made the USAMO 2002 in Massachusettes. Because if everybody comes to one place you can't cheat because there are proctors. If you take the test at your school you can cheat. Your teacher might help you, you never know. Also people can easily cheat on the amc/aime because the teachers have answers and solution pamphlets so if they aren't fair they can give answers to students. That is certaintly possible.[/quote]

True, there are probably people who cheated on the AMC or AIME.  The AMC committee does its best to keep cheating down to a minimum by having the principal sign a form, etc.  But does it really matter?  The people who cheat on the AMC almost certainly won't make it past AIME, which is the big "weeder-outer" since 10,000 go in and only 250 come out....

And besides, what will someone accomplish by cheating?  They probably won't make it to the following level, since if they could have, they wouldn't be cheating.  And if someone somehow cheated on the USAMO (I can't imagine how...), they might make it to MOP.  But they wouldn't make it to IMO - even a really good cheater can't beat Tiankai or Po-Ru, right?

[quote="JBL"][color=cyan]I think you all probably overestimate the amount of cheating that goes on.  Particularly on USAMO.  Even probably on USAMTS not very much if any happens.

Also, I didn't really think the big USAMO was so fun, even though I knew many of the people who went.  I mean, it was nice.  But not that nice.  I never minded the sitting in a room for 6 hours before then.  Also, the accomidations for the Friday testing were really awful, resulting in many people having to take the test under much worse conditions than having little sixthgraders make faces at them.[/color][/quote]

Okay, but it's always nicer to have a free trip than not, right?

In some sense I agree with Joel (gasp!) on this issue. The Boston trip was scheduled for Thursday through Sunday. I got to the Boston airport at 8PM, and no one picked me up there until 10. I was considering trying to find a taxi, but that would have been incredibly expensive for an allegedly free trip (which really wasn't because of all the long-distance phone calls I had to make because my parents were wondering if I'd ever make it to the hotel). We (my friend who was on the same flight and I) decided to look for some place to get dinner when we arrived at the hotel around 10:30, but it was pouring with rain, and all the restaurants in the area were already closed by then. So that wasn't exactly the nicest way to start off the trip.

However, it was rather nice to meet lots of other people from around the country who also like to do math. Furthermore, I got to see some people who were at SUMaC 2001, and I had not seen some of them since then, so it was nice to see them again. The lectures they had were also fun. Hearing someone as famous (at least in the math world) as Ronald Graham speak tends to be a good experience.

As for being in a room with no other people rather than many other people, I should say that my three experiences in the USAMO were radically different. In 2001, I was the only one from my school taking the USAMO, so no one could care less if I were to talk to myself about the problems (which is a habit I have while doing math). In Boston in 2002, I was in a room with many other people, so talking to yourself about the problems was out of the question. In 2003, there was one other person from my school taking USAMO, so I had to be a bit careful of what I said (or, more precisely, how loudly I said it). But we had a very big room, and so with only two people in it, we were far apart, so we had a lot of space to move around, which was nice. Furthermore, our teacher spoiled us and gave us whatever food we wanted, since he figured that most people eat lunch some time between 9:30 and 2, which was our window for taking the USAMO on the west coast.

All in all, I enjoyed the trip to Boston, and it didn't hurt that I got to skip one extra day of the most painful class I have ever taken (and hopefully will ever take).


geometry
ratio
LaTeX
analytic geometry
similar triangles
geometry unsolved

Let $A_1$, $A_2$, $A_3$ be three points in the plane, and for convenience, let $A_4= A_1$, $A_5 = A_2$.  For $n = 1$, $2$, and $3$, suppose that $B_n$ is the midpoint of $A_n A_{n+1}$, and suppose that $C_n$ is the midpoint of $A_n B_n$. Suppose that $A_n C_{n+1}$ and $B_n A_{n+2}$ meet at $D_n$, and that $A_n B_{n+1}$ and $C_n A_{n+2}$ meet at $E_n$.
Calculate the ratio of the area of triangle $D_1 D_2 D_3$ to the area of triangle $E_1 E_2 E_3$.

Hello.
I think i solved it.
First show that A1,D1,E2 are on a same line ;
then prove that D1D3 ll E2E3 ll A2A3 ;
and then show that D1D3 / E2E3 = 11/14 .

So, here is my solution for this problem:

First we will show that $A_{n}A_{n+1}\parallel E_{n}E_{n+1}\parallel D_{n+1}D_{n+2}$.
Let $G$ be the center of gravity of $A_{1}A_{2}A_{3}$, thus the intersection of the lines $A_{n}B_{n+1}$. By Menelao's Theorem on $A_{1}GB_{1}$ we have: $ \frac{A_{1}E_{1}}{E_{1}G}\cdot \frac{GA_{3}}{A_{3}B_{1}}\cdot\frac{B_{1}C_{1}}{C_{1}A_{1}}=-1$, by midpoints and properties of the gravicenter $ \frac{A_{1}E_{1}}{E_{1}G}=\frac{3}{2}$. So every $E_{n}$ divides each $A_{n}G$ in such proportion, and by Thales each pair $A_{n}A_{n+1}\parallel E_{n}E_{n+1}$. Similar resoning gives us each pair $A_{n}A_{n+1}\parallel D_{n+1}A_{n+2}$, but in  now each $D_{n}$ cuts $B_{n}G$ in proportion $ \frac{B_{n}D_{n}}{D_{n}G}=\frac{3}{4}$.

So the three triangles are similar, and furthermore, homothetic. Therefore $G$ is also the center of gravity of $D_{1}D_{2}D_{3}$ and $E_{1}E_{2}E_{3}$. Finally, we know that the proportion of two similar triangles is the proportion of one of their corresponding lines, and we have, because of our previous observations on the medians :

$ \frac{D_{1}D_{2}D_{3}}{A_{1}A_{2}A_{3}}=\frac{4}{14}$, $ \frac{E_{1}E_{2}E_{3}}{A_{1}A_{2}A_{3}}=\frac{2}{5}$, which finally gives:

$ \frac{D_{1}D_{2}D_{3}}{E_{1}E_{2}E_{3}}=\frac{5}{7}$, and since the proportion of the areas is the square of the proportion of the sides we have:

$ \frac{|D_{1}D_{2}D_{3}|}{|E_{1}E_{2}E_{3}|}=\frac{25}{49}$

Yay!! My first LaTeX proof!!

one can use barycentric coordinates to trivialize this problem

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1989Problem3

Vo Duc Dien

By ratio lemma, $\frac{B_nD_n}{A_{n+2}D_n} = \frac{1}{6}$, so $\triangle D_1D_{2}D_{3}$ consists of $\triangle A_1A_2A_3$ scaled by $-\frac{2}{7}$ at the centroid. Similarly, $\frac{A_nE_n}{B_{n+1}E_n} = \frac{2}{3}$, so $\triangle E_1E_2E_3$ consists of $\triangle ABC$ scaled by $\frac{2}{5}$ at the centroid. Hence $\frac{[D_1D_2D_3]}{[E_1E_2E_3]} = \left(\frac{\frac{2}{7}}{\frac{2}{5}}\right)^2 = \frac{25}{49}$.


algebra
polynomial
irrational number
number theory open
number theory

Is $\pi ^ e$ rational or not? I know that an irrational number to an irrational power is rational (ex. :sqrt:$2$<sup>:sqrt:$2$</sup>) , but can anyone prove that this holds in this case? Good luck!

[quote="bomb"]Isnt irrational to power of irrational irrational transeendatal even

Bomb[/quote]

I don't think so : $e^{i \pi} = -1.$

Pierre.

I think the most general result known for this type of problem is what's called the Gelfond-Schneider theorem:

If $b$ is an (edit: algebraic and) irrational number, and $a$ is any number other than 0 and 1 which is [i]algebraic[/i] (meaning that there is a polynomial with integer coefficients to which $a$ is a root), then $a^b$ must be transcendental (meaning not algebraic)

For example, taking $a=b=\sqrt{2}$ gives that $\sqrt{2}^{\sqrt{2}}$ must be transcendental.

However, the theorem doesn't apply with $a=\sqrt{2}^{\sqrt{2}}$ and $b=\sqrt{2}$ because $a$ wouldn't be algebraic), and indeed taking $a^b=\sqrt{2}^2=2$.

The theorem doesn't apply directly to $e^{\pi}$ ($e$ isn't algebraic), but we can write $e^{\pi}=(e^{\pi i})^{-i}=(-1)^{-i}$, and now the theorem does hold and we get $e^{\pi}$ is transcendental.  However, I think the $\pi^e$ case is still open

(editted due to error pointed out by Grobber...thanks!)

In the Gelfond-Schneider theorem $b$ must also be algebraic (so it's algebraic and irrational, not just irrational, as you wrote, kevinatcausa).

Hmmm ... Never knew about Gelfond-Schneider theorem. Alright well it's still open for whoever wants it. Thanks guys.


function
logarithms
limit
algebra unsolved
algebra

Find all continious functions such that $ f: [0,\plus{}\infty)\to [0,\plus{}\infty)$ and
\[ f(xy)\equal{}f(x)^{y^{\beta}}f(y)^{x^{\beta}},\]
where $ \beta\in \mathbb{R}$ is some constant.

Solution...
[hide]
First of all, we must say that $ \beta \geq 0$. If $ \beta < 0$ then right side doesn't exist if $ x \equal{} 0$.
So, let $ \beta \geq 0$.

It's easy to prove that if $ x \neq 0$ and $ f(x) \equal{} 0$, then $ f(x) \equal{} 0$ for all $ x$.
Suppose $ f(x) > 0$ for all $ x > 0$.

Taking $ \ln{}$ for both sides, we will have:
$ \ln f(xy) \equal{} y^{\beta}\ln f(x) \plus{} x^{\beta}\ln g(y)$.

Let $ g(x) \equal{} \ln f(x)$, so we will have:

$ g(xy) \equal{} y^{\beta}g(x) \plus{} x^{\beta}g(y)$;

If $ x \equal{} e^s$ and $ y \equal{} e^t$ ($ t,s \minus{} real$), then we will have:

$ g(e^{s \plus{} t}) \equal{} (e^{\beta})^t g(e^s) \plus{} (e^{\beta})^s g(e^t)$.

Let $ e^{\beta} \equal{} a$ and $ h(x) \equal{} g(e^x)$, so:

$ h(s \plus{} t) \equal{} a^t h(s) \plus{} a^s h(t)$;

$ \frac {h(s \plus{} t)}{a^{s \plus{} t}} \equal{} \frac {h(s)}{a^s} \plus{} \frac {h(t)}{a^t}$.

Last step: $ q(x) \equal{} \frac {h(x)}{a^x}$, so:

$ q(s \plus{} t) \equal{} q(s) \plus{} q(t)$ - for all real $ s,t$ and $ q$ - continuous.
It's well-known that $ q(x) \equal{} kx$, where $ k$ - some real number.

So, $ h(x) \equal{} a^{x} (kx)$, $ g(x) \equal{} a^{ln x} (k\ln x)$, $ f(x) \equal{} e^{g(x)} \equal{} e^{e^{\beta \ln{x}}({k\ln{x}})} \equal{} e^{ x^{\beta}k\ln{x} } \equal{} x^{k x^{\beta}}$

If $ \beta \equal{} 0$ then we have $ f(0) \equal{} lim_{x\to 0} f(x) \equal{} 0$.

If $ \beta > 0$ then we have $ f(0) \equal{} lim_{x\to 0} f(x) \equal{} 1$, so $ f(0) \equal{} 1$.

So, we have answers:

If $ \beta < 0$ - there no such function.
If $ \beta \equal{} 0$ then functions:

$ f(x) \equal{} 0$;
$ f(x) \equal{} 1$ ($ k \equal{} 0$ - so, $ f(x) \equal{} 1$ and $ f(0) \equal{} 1)$;
$ f(x) \equal{} x^k$, $ f(0) \equal{} 0$, $ k \neq 0$.

If $ \beta > 0$ then functions:
$ f(x) \equal{} 0$
$ f(x) \equal{} x^{k x^{\beta}}$ and $ f(0) \equal{} 1$.
[/hide]

A can only add, that $ e^{kx^{\beta}\ln x}\equal{}x^{k x^{\beta}}.$

[quote="Yuriy Solovyov"]A can only add, that $ e^{kx^{\beta}\ln x} \equal{} x^{k x^{\beta}}.$[/quote]
Thank you, I added it answer to solution, but there is some problem... $ \lim_{x\to\ 0} x^{k x^{\beta}} \equal{} 1 \neq f(0)$.

Put $ y\equal{}0$, get: $ f(0)\equal{}f(0)^{x^{\beta}}$, so, I think $ f(0)\equal{}1$ because if $ f(0)\equal{}0$ then if put $ x\equal{}0$, we get $ f(0)\equal{}1.$ :)


He took the shot when it was 97, but i got up to 102.

So i assume that this was multiple people making guests and trying to get 100?

I know there were at Least 2.  Hint hint.

Take a look at what files he has open on the bottom bar. He has 1 Modzilla Firefox (the window that is open) and 6 IE windows open that all start with "AoPS For T..."

Hmmm...

THis should be in the other thread....y'know, the spammy one.  The FTW funny  sights


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geometry
analytic geometry
parallelogram
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AMC
AIME

Let $\mathcal{T}$ be the set of ordered triples $(x,y,z)$ of nonnegative real numbers that lie in the plane $x+y+z=1.$  Let us say that $(x,y,z)$ supports $(a,b,c)$ when exactly two of the following are true: $x\ge a, y\ge b, z\ge c.$  Let $\mathcal{S}$ consist of those triples in $\mathcal{T}$ that support $\left(\frac 12,\frac 13,\frac 16\right).$  The area of $\mathcal{S}$ divided by the area of $\mathcal{T}$ is $m/n,$  where $m$ and $n$ are relatively prime positive integers, find $m+n.$

[quote="minsoens"][hide]

Imagining a 3 dimensional coordinate system, the graph of $x+y+z=1$ is an equilateral triangle $XYZ$ with vertices at $X(1,0,0)$, $Y(0,1,0)$, and $Z(0,0,1)$
Thus it has a length of $\sqrt{2}$

If only one of $x \ge \frac{1}{2}$, $y \ge \frac{1}{3}$, $z \ge \frac{1}{6}$ is false,
then there are three cases:
#1 $x < \frac{1}{2}$, $y \ge \frac{1}{3}$, $z \ge \frac{1}{6}$
#2 $y < \frac{1}{3}$, $x \ge \frac{1}{2}$, $z \ge \frac{1}{6}$
#3 $z < \frac{1}{6}$, $x \ge \frac{1}{2}$, $y \ge \frac{1}{3}$

In case #1, a line is drawn $\frac{1}{2}$ of the way up from base $YZ$ on $\triangle XYZ$ and parallel to it.
In case #2, a line is drawn $\frac{1}{3}$ of the way up from base $XZ$ on $\triangle XYZ$ and parallel to it.
In case #3, a line is drawn $\frac{1}{6}$ of the way up from base $XY$ on $\triangle XYZ$ and parallel to it.
The overlapping regions are where two of the original cases are false, therefore it is not what we want.

By now the equilateral triangle should be divided into $6$ sections: $3$ small eq. triangles and $3$ parallelograms. We want the area of the $3$ triangles, because if the point is inside a parallelogram, two of X, Y, or Z are less than $\frac{1}{2}$,$\frac{1}{3}$,$\frac{1}{6}$, respectively.

The eq. triangles have side lengths of $\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{3},\frac{\sqrt{2}}{6}$. The areas of the small eq. triangles are: $\frac{\sqrt{3}}{72}$, $\frac{\sqrt{3}}{18}$, $\frac{\sqrt{3}}{8}$

The fraction is $\frac{\frac{14\sqrt{3}}{72}}{\frac{2\sqrt{3}}{4}}=\frac{7}{18}=\frac{m}{n}$

$m+n=\boxed{25}$

[/hide][/quote]

How did you relize that they were equilateral triangles ( the small ones, not the big ones ) ?

EDIT: whoops, read the inequalities wrong. You are correct minsoens.

Remark, this problem is similar to 1984 AIME, #3 :)

I think this is on some AIME, like number 9-12ish.  Draw accurate diagram really helps.  But...how do you do it fast??? (It took me 18-20 min to do this the first time)

[quote="pkerichang"][quote="Iversonfan2005"][quote="pkerichang"] But...how do you do it fast??? (It took me 18-20 min to do this the first time)[/quote]

What do you mean? :? It's not like you even have to calculate the areas of all the equilateral triangles, it's simply side ratios squared and you know the sides. I think that's pretty fast :P[/quote]

You're right...now I look back on it, I think 10 min. is spent on just drawing the diagram (being in Algebra 2 at that time, I can't get too pro with 3-d coordinates...).  And I don't realize the parallel-ness of the problem so I hard calculate all three sides of all three triangles... Now I should do better.[/quote]

You guys play with 3-d coordinates in Algebra2?.....
Anyway, if we know that the area x<1/2, y>=1/3 and z >=1/6 are what we want how can we prove that in the 3-d equilateral triangle that all x<1/2 space will satisfy that y>=1/3 and z>=1/6?

[quote="cdymdcool"][quote="minsoens"][hide]

Imagining a 3 dimensional coordinate system, the graph of $ x \plus{} y \plus{} z \equal{} 1$ is an equilateral triangle $ XYZ$ with vertices at $ X(1,0,0)$, $ Y(0,1,0)$, and $ Z(0,0,1)$
Thus it has a length of $ \sqrt {2}$

If only one of $ x \ge \frac {1}{2}$, $ y \ge \frac {1}{3}$, $ z \ge \frac {1}{6}$ is false,
then there are three cases:
#1 $ x < \frac {1}{2}$, $ y \ge \frac {1}{3}$, $ z \ge \frac {1}{6}$
#2 $ y < \frac {1}{3}$, $ x \ge \frac {1}{2}$, $ z \ge \frac {1}{6}$
#3 $ z < \frac {1}{6}$, $ x \ge \frac {1}{2}$, $ y \ge \frac {1}{3}$

In case #1, a line is drawn $ \frac {1}{2}$ of the way up from base $ YZ$ on $ \triangle XYZ$ and parallel to it.
In case #2, a line is drawn $ \frac {1}{3}$ of the way up from base $ XZ$ on $ \triangle XYZ$ and parallel to it.
In case #3, a line is drawn $ \frac {1}{6}$ of the way up from base $ XY$ on $ \triangle XYZ$ and parallel to it.
The overlapping regions are where two of the original cases are false, therefore it is not what we want.

By now the equilateral triangle should be divided into $ 6$ sections: $ 3$ small eq. triangles and $ 3$ parallelograms. We want the area of the $ 3$ triangles, because if the point is inside a parallelogram, two of X, Y, or Z are less than $ \frac {1}{2}$,$ \frac {1}{3}$,$ \frac {1}{6}$, respectively.

The eq. triangles have side lengths of $ \frac {\sqrt {2}}{2},\frac {\sqrt {2}}{3},\frac {\sqrt {2}}{6}$. The areas of the small eq. triangles are: $ \frac {\sqrt {3}}{72}$, $ \frac {\sqrt {3}}{18}$, $ \frac {\sqrt {3}}{8}$

The fraction is $ \frac {\frac {14\sqrt {3}}{72}}{\frac {2\sqrt {3}}{4}} \equal{} \frac {7}{18} \equal{} \frac {m}{n}$

$ m \plus{} n \equal{} \boxed{25}$

[/hide][/quote]
How did you draw those three inequalities on the three dimensional plane?  I think the region we are really looking of should be the area in side the equilateral triangles excluding 3 small one.[/quote]I think I looked at the area like you did. It's really hard to visualize 3D inequalities :ninja:


$ x_{1}\equal{} 1$
$ x_{n\plus{}1}\equal{} x_{n}\plus{}{x_{n}}^{\frac{1}{3}}$
Prove that there exist a,b different from 0 and
$ lim\frac{x_{n}}{a.n^{b}}\equal{} 1$

These problem solved many times. For your case $ a\equal{}(\frac{2}{3})^{3/3},\ b\equal{}\frac{3}{2}$.


The ecuation $ \frac{1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{x} \equal{} \frac {1}{a \plus{} b \plus{} x}$ has a finite number of solutions. Find the sum of this solutions.

On solving the equation
the simplified form obtained is
(a+b)x^2+(a+b)^2 x +ab(a+b)=0
dividing by a+b we get
x^2+(a+b)x+ab=0
so the sum of roots of equation = -(a+b)

I suppose this is what joshisunny16 wanted to say.
On simplification, the equation becomes
$ x^2 \plus{} (a \plus{} b)x \plus{} ab \equal{} 0$ and the result follows by Viete's theorem.
But here we need to assume that $ a \plus{} b \ne 0$

Also @balan razvan it is equation, not ecuation.


geometry
parallelogram
inequalities
trigonometry
geometric transformation
reflection
algebra

Let $O,A,B,C$ be variable points in the plane such that $OA=4$, $OB=2\sqrt3$ and $OC=\sqrt {22}$. Find the maximum value of the area $ABC$.

[i]Mihai Baluna[/i]

You can get the estimation for the area from Ptolemey's inequality. Try to build some triangles up to parallelograms to get convenient quadraliteral.

I think this was used in some Romanian TST (1999, if I am not mistaken). Anyway, the main idea is that the maximal area is achieved when $O$ is the orthocenter of $ABC$. After that, it does become rather ugly, from what I can remember.

According to the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=24803]Triangle inequality[/url], for an arbitrary point $O$ inside of a triangle $\triangle ABC$, the following inequality holds (with equality when $O$ is the triangle orthocenter):

$OA \cdot OB \cdot BC + OB \cdot OC \cdot CA + OC \cdot OA \cdot AB \ge AB \cdot BC \cdot CA$

Let $a = BC, b = CA, c = AB$, $R, R_A, R_B, R_C$ the circumradii of the triangles $\triangle ABC, \triangle OBC, \triangle OCA, \triangle OAB$. Using the formula $|\triangle ABC| = \frac{abc}{4R}$, etc, the inequality can be written as

$OA \cdot OB \cdot a + OB \cdot OC \cdot b + OC \cdot OA \cdot c \ge abc$

$R \cdot |\triangle ABC| \le R_A \cdot |\triangle OBC| + R_B \cdot |\triangle OCA| + R_C \cdot |\triangle OAB|$

$|\triangle ABC| \le \frac{R_A}{R} \cdot |\triangle OBC| + \frac{R_B}{R} \cdot |\triangle OCA| + \frac{R_C}{R} \cdot |\triangle OAB|$

The area $|\triangle ABC|$ is maximum at equality, i.e., when $O$ is the orthocenter. If this is the case, then $R_A = R_B = R_C = R$ and the inequality turns into a trivial equality:

$|\triangle ABC| = |\triangle OBC| + |\triangle OCA| + |\triangle OAB|$

Thus the problem is to find the triangle area given the distances of the orthocenter $O$ from the vertices $A, B, C$. Let $\alpha, \beta, \gamma$ be the angles at the vertices $A, B, C$. The distances $OA, OB, OC$ are equal to

$OA = 2R \cos \alpha,\ \ OB = 2R \cos \beta,\ \ OC = 2R \cos \gamma$

$\frac{OA}{OB} = \frac{\cos \alpha}{\cos \beta} = \frac{4}{2 \sqrt 3} = \frac{2}{\sqrt 33},\ \ \frac{\cos^2{\alpha}}{\cos^2{\beta}} = \frac 4 3$

$\frac{OC}{OA}= \frac{\cos \gamma}{\cos \alpha} = \frac{\sqrt{22}}{4} = \frac{\sqrt{11}}{2 \sqrt 2},\ \ \frac{\cos^2{\gamma}}{\cos^2{\alpha}} = \frac{11}{8}$

Since $\alpha + \beta + \gamma = \pi$, $\cos \gamma = -\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$.

$\frac{\cos^2{\alpha} \cos^2{\beta} - 2 \cos \alpha \cos \beta \sin \alpha \sin \beta + \sin^2{\alpha} \sin^2{\beta}}{\cos^2{\alpha}}  = \frac{11}{8}$

Substituting $\sin \alpha = \sqrt{1 - \cos^2{\alpha}},\ \sin \beta = \sqrt{1 - \cos^2{\beta}}$, denoting $x = \cos^2{\alpha}$ and substituting $\cos^2{\beta} = \frac 3 4\ \cos^2{\alpha} = \frac{3x}{4}$, this becomes

$\frac{3x^2}{4} - x \sqrt{3(1 - x)\left(1 - \frac{3x}{4}\right)} + (1 - x)\left(1 - \frac{3x}{4}\right) = \frac{11x}{8}$

$12x^2 - 25x + 8 = 4x \sqrt{3(3x^2 - 7x + 4)}$

$144x^4 + 625x^2 + 64 - 600x^3 + 192x^2 - 400x = 144x^4 - 336x^3 + 192x^2$

$264x^3 - 625x^2 + 400x - 64 = 0$

Thus we obtained a cubic equation, as expected (see the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=27049]Triangle edges and orthocenter[/url]). We can test this cubic equation for rational roots with the help of the rational root theorem: Possible rational roots are $\frac p q$, where $p$ divides $64 = 2^6$ and $q$ divides $264 = 2^3 \cdot 3 \cdot 11$. Or we can solve numerically and factor out any rational roots we spot. The above equation has a rational root $x = \frac{8}{33}$. Factoring it out yields

$264x^3 - 625x^2 + 400x - 64 = (33x - 8)(8x^2 - 17x + 8) = 0$

and the remaining 2 roots are $x = \frac{17 \pm \sqrt{33}}{16}$. The root with the plus sign is greater than 1, i.e., not acceptable for a cosine. The remaining irrational root is also not acceptable, because for  $x = \frac{17 - \sqrt{33}}{16}$,

$12x^2 - 25x + 8 = \frac{12(289 + 33 - 34 \sqrt{33})}{256} - \frac{25(17 - \sqrt{33})}{16} + 8 =$

$= \frac{483 - 850 + 256 - (51 - 50) \sqrt{33}}{32} = \frac{\sqrt{33} - 111}{32} < 0$

i.e., it is not a root of the original equation before we squared it. Consequently, it must be $x = \cos^2{\alpha} = \frac{8}{33}$, $\cos \alpha = \pm \frac{2 \sqrt 2}{\sqrt{33}}$. The minus sign is not acceptable, because the angle $\alpha$ would be obtuse, the orthocenter outside of it, and it obviously would not have maximum area for the given lengths $OA, OB, OC$. Therefore,

$\cos \alpha = \frac{2 \sqrt 2}{\sqrt{33}},\ \ \cos \beta = \frac{\sqrt 3}{2} \cos \alpha = \frac{\sqrt 2}{\sqrt{11}},\ \ \cos \gamma = \frac{\sqrt{11}}{2 \sqrt 2} \cos \alpha = \frac{1}{2 \sqrt 3}$

$\sin \alpha = \frac{5}{\sqrt{33}},\ \ \sin \beta = \frac{3}{\sqrt{11}},\ \ \sin \gamma = \frac{\sqrt{11}}{2 \sqrt 3}$

$R = \frac{OA}{2 \cos \alpha} = \frac{OB}{2 \cos \beta} = \frac{OC}{2 \cos \gamma} = \sqrt{\frac{33}{2}}$

$a = \frac{2R}{\sin \alpha} = \frac{33 \sqrt 2}{5},\ \ b = \frac{2R}{\sin \beta} = 11 \sqrt 2,\ \ c = \frac{2R}{\sin \gamma} = 6 \sqrt 2$

$|\triangle ABC| = \frac 1 2\ bc \sin \alpha = 10 \sqrt{33}$

I saw this problem back then, and liked it- I've proposed versions of it myself. The answer requires the solution of a cubic equation, most conveniently for $2R$. If $OA=a,OB=b,OC=c$,
$(2R)^3-(a^2+b^2+c^2)2R-2abc=0$

Then the area is $\frac{ab\sqrt{4R^2-c^2}}{4R}+\frac{ac\sqrt{4R^2-b^2}}{4R}+ \frac{bc\sqrt{4R^2-a^2}}{4R}$.

In this case, that cubic is $(2R)^3-(16+12+22)(2R)-16\sqrt{66}=0$, or $2R^3-25R-4\sqrt{66}=0$. There aren't many possible reasonable roots, and we find that the unique positive root is $R=\sqrt{\frac{33}{2}}.$

It's trivial to show that the maximum occurs when $O$ is the orthocenter- look at what happens if you fix two vertices and let the other vary.

It's nice to rig it backward from a nice triangle: I like $a=25,b=33,c=39$.

$O$ must be the orthocenter in order to have a critical point of the oriented area function. For $A,B$ fixed and $C$ variable, the tangent line at $C$ must be parallel to $AB$ at a critical point $(A,B,C)$, which means $OC$ is perpendicular to $AB$.

If one point $A$ is fixed, the critical points come in pairs under reflection in $OA$, which exchanges max/min. [([b]Edit:[/b] <deleted some stupid remarks about number and type of critical points>].

So my guess is that we want the case when $O$ is orthocenter and it is inside $ABC$, and this is the unique maximum of unoriented area.

So much effort on the trivially easy part!

Fix $B$ and $C$; the maximum area clearly occurs when $A$ is as far away from $BC$ as possible, which occurs when $OA$ is perpendicular to $BC$ and $A$ is farther from $BC$ than $O$.

Therefore, if a maximum exists, it must come when $O$ is the orthocenter and $O$ is inside $ABC$. The configuration can be specified by two angles $\alpha=AOB$ and $\beta=BOC$, with compact domain $\alpha\ge0$, $\beta\ge0$, $\alpha+\beta\le2\pi$. The area is the continuous function $\frac12|ab\sin\alpha+bc\sin\beta+ac\sin(2\pi-\alpha-\beta)|$, which must take its maximum by the Extrame Value theorem.

Actually, any exterior orthocenter possibilities would be saddle points.

Thanks, that remark about the saddle points makes it clear (and the argument for $O$ being the orthocenter seems identical to the one before it).
So the critical points of the function $f(B,C)$ = oriented area of $ABC$ occur when $O$ is the orthocenter and are of the following types:

1 maximum of the oriented area when $O$ lies inside $ABC$ and $ABC$ is positively oriented
2 saddle points when $O$ lies outside $ABC$
1 minimum when $O$ lies inside $ABC$ and $ABC$ is negatively oriented.

This 1-2-1 pattern is good for cohomological reasons, because on $S^1$ (fixed $A,B$ and variable $C$) we expect 1 max and 1 min ($(1-t)$, and on $S^1 \times S^1$ we expect the distribution of critical point indices like $(1-t)^2$ = 1 min, 2 saddle, 1 max.
The saddle points and the max/min are mirror images with respect to reflection in $OA$.

So the consistency of mathematics is upheld, and only the nontrivial part of the problem remains.
I should ask whether this problem in general involves solving an irreducible cubic or quartic.

[quote="fleeting_guest"]I should ask whether this problem in general involves solving an irreducible cubic or quartic.[/quote]
See my first post for that- the equation is the trig identity $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ for $A,B,C$ the angles of a triangle.

[quote="jhaussmann5"][quote="yetti"]$|\triangle ABC| \le \frac{R_A}{R} \cdot |\triangle OBC| + \frac{R_B}{R} \cdot |\triangle OCA| + \frac{R_C}{R} \cdot |\triangle OAB|$

The area $|\triangle ABC|$ is maximum at equality, i.e., when $O$ is the orthocenter.[/quote]

I don't quite see this.  It's true that there is equality when $O$ is the orthocentre, but the right-hand side is not constant, as far as I can see.  So, you can't conclude that the area is maximal simply when there is equality.[/quote]

The argument I put forward is either incorrect or insufficient. I do not see whether or how it can be fixed at the moment. Thanks to jmerry for and fleeting_guest for a correct argument.

Yetti


[quote="danee 1nonly"]Is it legal to submit a solution done with computer programming?[/quote]
Yes.
$\\$


Given a nonnegative real number $x,$ let $\langle x\rangle$ denote the fractional part of $x;$ that is, $\langle x\rangle=x-\lfloor x\rfloor,$ where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x.$  Suppose that $a$ is positive, $\langle a^{-1}\rangle=\langle a^2\rangle,$ and $2<a^2<3.$  Find the value of $a^{12}-144a^{-1}.$

p3: Thanks [b]peace09[/b] for assistance (I needed to look at a part of his solution :() with the simplification.

First, note that the floor of $\frac{1}{a}$ is going to be $0$ and the floor of $a^2$ is going to be $2$, so $\frac{1}{a} = a^2-2$. This means $a^3-2a-1=0$. Factoring out $a+1$ gives $a^2-a-1=0$; this means our solution is $a = \frac{1+\sqrt5}{2}$ (This doesn't turn out to be necessary, but it's nice to find). Next, we concentrate on $a^{12} - \frac{144}{a}$. We will make use of (1) $a^3 = 2a+1$ (this is where peace09 helped; I didn't think of using this) and (2) $a^2 = a+1$. (1) gives $a^12 = (2a+1)^4$, so we reduce to $(2a+1)^4 - \frac{144}{a}$. Expanding $(2a+1)^4$ gives $16a^4+32a^3+24a^2+8a+1$. Using (1) and (2), we simplify to $16(a+1)^2+32(2a+1)+24(a+1)+8a+1 = 16a^2+128a+73$. Using (2) we further simplify to $16(a+1)+128a+73 = 144a+89$. Our expression is now $144a+89-\frac{144}{a} = \frac{144a^2+89a-144}{a}$. Using (2), the numerator simplifies to $\frac{144a+144+89a-144}{a} = \frac{233a}{a} = \boxed{233}$.


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number theory solved
number theory

See also http://www.mathlinks.ro/viewtopic.php?t=150370 .

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=4542]Related problem by Kiran Kedlaya[/url]:

Let $ x$, $ y$, $ z$ be positive integers. Prove that $ \left(xy \plus{} 1\right)\left(yz \plus{} 1\right)\left(zx \plus{} 1\right)$ is a perfect square if and only if $ xy \plus{} 1$, $ yz \plus{} 1$, $ zx \plus{} 1$ are all perfect squares.

  darij

[quote="ttd"]Find infinitely many triples $ \left(a,b,c\right)$ of positive integers such that the integers $ a$, $ b$, $ c$ are in arithmetic progression and such that $ ab \plus{} 1$, $ bc \plus{} 1$, $ ca \plus{} 1$ are perfect squares.[/quote]

Sorry to revive the thread, but I don't see a solution posted

- A similar problem is posted in "pre-Olympiad"  thread
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159494

(The reason I found this old post was ,  I did the search when I saw the above  thread)
The solution  is quite simple, if you know it:

[hide]$ 2m \minus{} n , 2m, 2m \plus{} n$ where $ 3m^{2} \minus{} n^{2} \equal{} 1$  there are infinite solutions for this Pell equation. [/hide]

Sorry if already posted somewhere else...