geometry
rectangle
linear algebra
matrix
analytic geometry

Was that the quadrilatteral one? I made a careless error on that one... :(

Yeah-it was simple w/shoelace, but I also built a rectangle and subtracted 2 triangles, and split it into 2 triangles and used matrix determinants.

i also did it with subtracting 2 right triangles from a rectangle....but then i made a stupid mistake.  :oops:

oh well.

Before I did it, and had only read it, I was convinced it would be 2007, then, when I first read it on my calc, thought it sayed 2007. That was only for half a second though.

[quote=".:Logic:."]Shoelace REALLY helped me on that quadrilateral one :lol:[/quote]

that's because it was the intended method.

For some reason.... very few people besides me got it. When the solution was explained (the one without Shoelace) everyone was groaning and complaining.  :maybe:

Anyway, I got only 13th in Chapter, so I would have probably not made it in the top 16 if not for that one problem which I got, and very few others got.

*whew*  :lol:

yeah i know, my arrogant teammate was like "god, i'm so stupid. i should've known to use that."

and then he said that same thing about all the other problems he missed and concluded with "i'm going to beat you at state".  :roll:

Then at State i got 8th and he got around 60th.   :rotfl:  :rotfl:

This is a bit spammy but i have to complain about this annoying kid on my team. :P

[quote="Fanatic"]Is that kid johnz? Was he a 7th grader this time around???[/quote]

no it was Ray Qian.  he still thought i was as bad as i was last year.

[quote="perfect628"]Did anyone else get target 6 right and amaze others with your 3 methods? I was the only one in the surrounding area of tables(including last year's state champs) who could do it...[/quote]

I got really lucky on that one... I finished #5 fairly quickly, then spent a LONG time trying to find the area. In the last minute I realized you could subtract the two triangles, and I got the computation done about 5 seconds before they called time.  :D

[quote="Walk Around The River"][quote="binonunquineist"][quote="dragongrl6"]I probably sound sort of clueless...

but what's Shoelace? :D[/quote]

The shoelace method is just a way to find the area of a polygon on the coordinate plane given all of its coordinates. It's really complicated though and I don't like using it. Everyone at my school who used it missed it. :)[/quote]

i used it and i got it right.

i also had it programmed into my calculator, but i did it by hand to check[/quote]

Is shoelace that "listing the coordinates and multiplying, adding, and dividing by 2 at the end" method that looks like determinant?

[quote="Walk Around The River"][quote="binonunquineist"][quote="dragongrl6"]I probably sound sort of clueless...

but what's Shoelace? :D[/quote]

The shoelace method is just a way to find the area of a polygon on the coordinate plane given all of its coordinates. It's really complicated though and I don't like using it. Everyone at my school who used it missed it. :)[/quote]

i used it and i got it right.

i also had it programmed into my calculator, but i did it by hand to check[/quote]

O really? Haha one of my teammates also had it programmed into his calculator. He used his program and got the right answer, but then did it by hand and got a different answer. Apparently he trusted himself more than the calculator so he got it wrong :|

Shoelace worked for me- I finished the problem in under half-a-minute with it. It's not complicated at all, unless you get a 10+ sided figure, but if you did, it would still be the easiest way in most cases.

btw, to use shoelace it has to be convex.

and using determinants is pretty much exactly the same thing once you break it down, as shoelace, it's just easier to do on a calculator

[quote="LawOfSigns"]What exactly is the shoelace method?  I just cut the quadrilateral into 3 triangles.  My school did great this year...normally we get no recognition, maybe someone gets in the top 20 (of 150-180 or so) or maybe not.  This year we suprisingly got 3rd place team (just barely made it to state) and I got 3rd place individual (2nd-3rd by CD, with 4th place cutoff).  One school got 1st team, 1st ind, 2nd ind, and 3-4 other people in the top 10.  Our school is so dis-interested in mathcounts, that out of about 400 middle schoolers, only 5-9 [i]want[/i] to go to chapter...we have a 7th and 8th grade team, maybe an individual or two.[/quote]

arrange the vertices in either clockwise order in a matrix, multiply the pairs of numbers on the up diagonals, making sure you wrap around, multiply the pairs of numbers on the down diagonals, making sure you wrap around. Take the positive difference and divide by 2 and BAM

there's your area

[quote="jhredsox"]Ok, could you say that in a more detailed way? I tried looking it up, but nothing too good.

-jorian[/quote]

Oh, you'll enjoy this :lol:

http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf

Omg, I loooooooooooooooooooooooooooooooove you!!!!!!!!

Thank you soooooooooooooooooooooooooooo much!!!!!!!!!!

-Jorian Hoover

I did the quadrilateral by making 2 triangles, and adding the area. it took me a few seconds...

According to the website I'm looking at, Shoelace can be used on concave polygons as well.

My failure at Chapters is a warning about how I'm going to fail at States. Seriously, I think I'm the most overrated person on AoPS by far. Just look at the statistics, at least 17 people on AoPS beat me at Chapters, not counting people who didn't vote.

And Fanatic, that's really only the case when there's no one else near as good as you. If there are two people capable of consistently getting 44-46s, and one falls asleep, he's not going to be able to win, is he?

I have no clue what post you are responding to in your post...

Oopsies, I think that was in another thread. Or whatever. It was responding to the long thing jhredsox posted.

[quote="Ravi B"]The shoelace formula does hold even for non-convex polygons.

Whether the polygon is convex or not, you do need to make sure that the vertices are listed in cyclic order.[/quote]

i think you were responding to this, just you then got sidetracked.

can you unblock me ubemaya, i was jk with that.

-jorian

Walk Around The River... how did you program the shoelace (shoestring) method? I can only do it for 3 and 4 numbers.  Does anybody know how to program a TI-89  for an ambiguous number of points?    Thanks

Oooh. That quadrilateral one was my favorite one. I did the good old subtracting rectangles and got the answer ~3 seconds before they called time and scribbled it on paper. The guy let me rewrite it though

[quote="machack"]what are matrixes, determinants and all that.  ALso, what's shoelace?[/quote]

I have no clue what the 1st two are. :|

However, I can help you with the third one. :)

By shoelace, you are either referring to:

1) (n) A string or cord used for lacing and fastening shoes. Also called [i]shoestring[/i].

2) A wonderful Math Method which killed a problem on this year's Target Round at the Chapter level. You can educate yourself here:
http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf

[quote="1=2"]It's easier without shoelace.

[hide="solution"]We can graph those coordinates(as an estimate), and we can draw a rectangle which has coordinates at (1,1), (1, 2007), (2006, 1) and (2006, 2007). Plot the quadrilateral. Now you have a quadrilateral and 2 triangles with legs 2004, 2005 and 2004, 2006. We then compute

$\frac{2004(2005+2006)}{2}4019022$.

The area of the rectangle is 4022030, so the area of the quadrilateral is 4022030-4019022=3008.[/hide]

I think only I and another team mate of mine got this one right on my team. We both placed 1 and 2 on the whole thing, me being second.[/quote]

It's just my opinion, but shoelace is easier, sicne all you do is add and multiply, without actually plotting it, and coming op with all of that. It's just mindless :P

[quote="mathgeniuse^ln(x)"]So what is this formula???

It seems interesting.[/quote]

wow! 14th place at Nats guy doesn't know shoelace?

let's say you have the points in an easy shape you know the are is

(1,1)
(3,1)
(2,2)
(1,1) BE SURE TO REPEAT YOUR FIRST POINT

X  Take the sum of each slash and total each side.
X
X
X

Then take the positive difference of the two sums and divide by 2!!! that's your answer!

I divided the figure into a 1x2 rectangle and 2 triangles. one dimention of each triangles is obvious, and the other is jus perpendicular, so i can jus subtract the x or y values.
Then I add the areas.

This shoelace formula sounds interesting...

[b][size=200]Last time I am posting this link for all people interested in Shoelace:[/size][/b]

http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf

[quote="1=2"]It's easier without shoelace.

[hide="solution"]We can graph those coordinates(as an estimate), and we can draw a rectangle which has coordinates at (1,1), (1, 2007), (2006, 1) and (2006, 2007). Plot the quadrilateral. Now you have a quadrilateral and 2 triangles with legs 2004, 2005 and 2004, 2006. We then compute

$\frac{2004(2005+2006)}{2}4019022$.

The area of the rectangle is 4022030, so the area of the quadrilateral is 4022030-4019022=3008.[/hide]

I think only I and another team mate of mine got this one right on my team. We both placed 1 and 2 on the whole thing, me being second.[/quote]

yup. that's what i did.

checked it again with shoelace though

i did it with shoelace and checked it with shoelace. i'm surprised i got it right, though. i have a habit of screwing up on shoelace someway or another. :lol:

I made a stupid mistake on that one...

the answer was 3008, but I accidentally subtracted one so I got 3007...

yeah my friend got it because chess64 sent a file of the shoelace method and i didn't bother reading it  :ewpu:

I think it was 3008, but when I checked the first time, the answer key said 3009. The next day it said 3008

I'm going nuts I think


Hey folks,

I usually don't post in this forum but I just saw the trailer for Harry Potter and the Goblet of Fire. I'll give you a link if you didn't watch it on TV yet (even you did, just watch it  :P ).

http://movies.yahoo.com/shop?d=hv&id=1808475610&cf=trailer

Tell me guys. What do you think about this movie?

I personally think it's totally different style than other three and reminds me of Lord of Rings and etc. I think pure "magic" that appeared in Harry Potter 1 is defnitely gone. If Harry Potter 1 was designed for young children, this one is definitely for young adults.

Haha.. The trailer shortly shows who Harry likes.. I always wondered what kind of a girl Harry would go for..  :)

Comment please!  :lol:

"Mad Eye" Moody creeps me out :surrender:

Looks pretty good..but the movies are always reallly different from the trailers, so I wouldnt be surprised if its bad.

Wow, that were two really good trailers, but then again, all trailers for movies are put together beautifully.

The music already creeped me out. Music plays a large factor on the "scariness" and "mood" of a movie to me.

It seems like Dumbledore goes a little to much when he announces the things. Especially when he announces that Harry was in the Triwizard Tournament.

And I think its PG-13 because Peter Pettigrew cuts his hand off...  ;)  That's a little un PG ish.

I can't wait to watch it.

i would want to watch it, but i probably wouldnt be able to (my parents are evil, just ask drunken_math) :P

The trailer was certainly dramatic, but I don't buy the "darker" or more "mature" thing. If they pull it off well, it should be a good movie, but otherwise, it seems even more childish than before.

I read all of the books this summer, except for the last one of course.

And I LOVE HARRY POTTER!

I saw the trailer but it's not what i expected. Mad Eye Moody's eye is a bit different than what i expected. It looks more like a gadget. Even the dragon looked weird  ;) Also, Victor Krum does not look ugly in the movie compared to how the book characterized him. On the other side, Cedric Diggory is  :| .
The thing that really annoys me about the movie is Dumbledore...he's very jumpy and not so calm. The best actor for Dumledore would be Gandalf (Lord of the Rings).

I guess my hopes were too high for the book.

But, I'll book my tickets in advance as loyalty towards Harry Potter.  :heli:  :10:  :winner_first:  :lol:

[quote="G-UNIT"]It certainly does have a darker appeal.  Frankly, I thinks its better movies are made this way.  If any of you have seen "Batman Begins" you will agree it was the best Batman movie ever.  This was because the movie portrayed the dark and more mature side of batman comics, and it produced a significantly larger thrill than what was expected of it.  Likewise with Harry Potter.  J.K Rowling has explicitly stated the books begin to darken in storyline as the books come closer to an end (we have evidently witnessed this in the 6th book) so it makes sense that they make the transition from kiddy to mature, in the 4th book where its most needed.[/quote]

Maybe the last movie will be rated R?

Unlikely. The market lost would be too significant. Besides, I seriously doubt there will be something in the last book warranting a R-rating for the movie.

Can't wait for the movie! It better be as good as the book! I didn't know it was PG-13 I hope I won't have to drag my mom along...NOT FUN!

[quote="elston"]hm..pg-13 means a lot of ticket sales will be lost 'cause the liddle kiddies can't watch it anymore..

i prob shouldn't be saying this but hermoine seems to be getting more..um..."womanly" every movie.[/quote]the actors are well in their teens already...it's been quite a few years since the first movie came out.

Nice...None of the  harry potter clips are opening on my _ comp...it is releasing this year,right?

[quote="apusey"][quote]Haha.. The trailer shortly shows who Harry likes.. I always wondered what kind of a girl Harry would go for.. [/quote]
In the book Harry doesn't fall for Hermione.[/quote]I think he means Cho Chang, I think is her name.  Harry likes her in the 4th book.

To shady:  I think it comes out this November.


Similar topic has been posted before. Logarithmize the inequality and take Karamata.

[quote="stevenwang1992"]I am sorry that I have made a big mistake .The right hand of the inequality should be mutiply but not add[/quote]
I didn't see that. My solution is for multiplication.


http://www.math.fau.edu/web/mathday/mday08team.pdf

How would you solve #25? I didn't quite get the explanation.

[hide="Solution"]
Basic angle chasing yields the attached diagram. Since we are given that quadrilateral $ ABPC$ is [url=http://en.wikipedia.org/wiki/Cyclic_quadrilateral]cyclic[/url], we can let $ s\equal{}AB\equal{}BC\equal{}CA$ and apply [url=http://en.wikipedia.org/wiki/Ptolemy%27s_theorem]Ptolemy's Theorem[/url] to get $ BP\cdot AC\plus{}CP\cdot AB\equal{}BC\cdot AP\implies5s\plus{}20s\equal{}25s\equal{}AP\cdot s\implies AP\equal{}25$. Now, note that $ \angle PBD\cong\angle PAC$, so combined with $ \angle BPD\cong\angle APC$, we have $ \triangle BPD\sim \triangle APC$, so $ \frac5{AP}\equal{}\frac5{25}\equal{}\frac15\equal{}\frac{PD}{20}\implies PD\equal{}4\implies AD\equal{}AP\minus{}PD\equal{}25\minus{}4\equal{}\boxed{21}$.
[/hide]

Thanks a lot! My error was where I tried to use triangle ADC instead of triangle APC


Let $ ABC$ be a triangle and $ W$ its circumcircle.the circle $ W_a$ is tangent to $ BC$ at its midpoint and also tangent to $ W$ at the midpoint of arc $ BC$ which doesn't contain $ A$.$ W_b,W_c$ are defined similarly.The external common tangent of $ W_b,W_c$ which is the most distanced from $ A$ is called $ l_a$.similarly we define $ l_b,l_c$.prove that the three intersection points of $ l_a,l_b,l_c$ construct a triangle called $ A'B'C'$ which is homogeneous to $ ABC$.

from http://www.mathlinks.ro/viewtopic.php?search_id=163733052&t=214426
we can prove that Ta,Tb,T are collniear , where T is the exsimilarity center of Wa,Wb
tangent at Tb' // tangent at Ta => tangent at Ta // La  =>  tangent at Ta //BC//La
analously, AB//Lc AC//Lb


AMC
USA(J)MO
USAMO
AIME
ARML
analytic geometry
induction

[url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2008-ua/2008USAMOscores.pdf]They're up.[/url]

How did everyone do?

Failed more than I had thought. 2.

Oddly, I did better than or as well as around 145 people. Still, I sucked. But there's always next year.

9. pretty neat... near the score I predicted. prolly coulda got 12 if I caught and corrected something :)

Yay, I got 14, which is good enough for Red MOP!!!

And also, I predicted that I would get 14, so I am happy  :D  :lol:  :lol:

Hm well i guess i messed up #5 (which I had thought about but hoped not) so 15 for me.

Next year...

26! :)

One point higher than what I predicted, hurrah!

So I got a 7, missing red mop by half a problem.
I'm fairly depressed right now, considering that if I had been a freshman last year, I would have made Red MOP (I got a 9 on the USAMO last year), but since I was in 8th grade I wasn't eligible.
But oh well, there's always next year I suppose... :(

I got 15. No blue mop  :(  but meh still good  :)

I was hoping for a little more, but I know people tend to overestimate their scores  :wink:

... o.o. 9. I guess I can't write proofs T.T... bye bye mr. 5 other points.

HOLY CRAP!

Eric Larson didn't win...

Oh yeah and I got a 12  :mad:  :mad:

I knew I screwed up #4...

EDIT: [url]http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2008-ua/2008usamotop12.shtml[/url]

I got a 13, as expected. 7/0/0/6/0/0 most likely. More importantly, RED MOP!

I got an 11.  That was worse than I would have expected if I didn't know that first-time USAMO takers overestimate their scores, but better than I expected given that I knew that first-time USAMO takers overestimate their scores.  :P

I figure my distribution was probably 5/0/0/6/0/0.  I thought I would get 7 for #4, but that would require partial credit I didn't deserve or a 4 on #1, which wouldn't happen.

I got a 3.  :huh:

I guess I deserved it, since I overlooked about $ \infty$ things...

Well I got 28...

... except that according to the winner page it's a 27.5.

2.  T...o...t...a...l...p...h...a...i...l...u...r...e...

I don't know how I only got a 2 when I actually wrote a semi-legitimate proof with an eventually correct answer for #4.  Also, I thought I would get at least partial credit on #5 for proving the (maybe?) nontrivial case for all rational numbers, but I guess the proof sucked cuz I didn't consider the ar1+br2+cr3=0 thingomobobber.  And as for #3, I showed that there were a total of 2n^2 lattice points in any S_n, though, admittedly, the other lemmas in my proof were false ( :rotfl: ).  I thought I would get at least a 0/0/1 or 0/3 or 4/1 or 0/0=3-6 points, but I guess my proofs just sucked...

PS: Maybe they couldn't read my microscopic handwriting...Darn... :(
PPS: Congratulations to all of the HMs and top 12'ers for their achievement!  Especially to Evan O'Dorney.  I'm betting that he will be the youngest ever US IMOer to ever attend the competition.
PPPS: Anyone reading this will think I complain too much, which I do. :P

[quote="pythag011"][quote="JAM"]Umm... there have been 9th graders on the IMO team before...[/quote]

and a 8th grader I think.

NO!! ONLY 10 POINTS!![/quote]

Hmm...Is that possible?  I thought only 9th graders and above could attend MoP in the first place.  Or maybe it's changed since then.

I really had no idea about anything going into the USAMO. The only thing I really knew what to do was induction to prove things. The first olympiad level proof I did was the one on this year's USAMO. I ended up getting a 10 apparently, so I guess I'm going to Red MOP.

I'm not really all that excited about MOP. I probably won't know what's going on there in the classes, which I heard were pretty intense and it's a long 3 weeks.

I feel that other people deserve and would like to go to Red MOP more than I do. Don't even quite know how I made it in the first place. Oh well. I guess I'll try to make the most of MOP.

Many more 8th graders have attended MOP than have been on the IMO team. I checked some names from the late 1990's to 2000 (before there was such a thing as "red" MOP) and at the very least, Sasha Schwartz, Reid Barton, and Po Ruh Loh attended MOP in 8th grade or earlier - and there were others. Each of those three first went to an IMO as 9th graders.

[quote="Anaxerzia"]@Arcesso: Then you seem to be the target of Red MOP :)[/quote]

You do have a point there. Though I've heard cases of people going to Red MOP and becoming really discouraged after seeing so many geniuses there.

That is the one thing I realized that you just shouldn't do at math camps; take it as an inspiration rather than a discouragement.

[quote="b-flat"][quote="Math Geek"]PPS: Congratulations to all of the HMs and top 12'ers for their achievement!  Especially to Evan O'Dorney.  I'm betting that he will be the youngest ever US IMOer to ever attend the competition.[/quote]

Even O'Dorney is not on the IMO team, yet.  There is still the TST (Team selection test) to choose the actual team from among the winners (and Palmer Mebane).[/quote]

There are always great young talents.

For example, Kiran Kedlaya in the early 90's, Alex. K. on the 94 fream team, etc. More recently:

Reid Barton, made Blue as a 7th grader, black in 8th, followed by 4 IMO golad medals.

G.D.Carroll, made team as a 9th grader, followed by 2 golds and 1 silver. (Skipped the team in 11th grade, perfect score as a senior.)

Tiankai Liu, black MOSP 8th grader, followed by 3 gold medals. (Skipped team once.)

Paul Valliant, blue in the 8th, team member in the 9th, 2 gold and 1 silver.

Ricky Liu, Po-Ru Loh, and much more.

We should have a MOSP/USAMO hall of fame. It's time to pay respect to these wonderful students. I hope it will motivate people to do more!

Best,

Zuming

Most vividly in my memory, the 1998 team had three ninth graders (Barton, Carroll, Valiant). As a tenth grader that year in MOP, I found it a bit scary.

[quote="Anaxerzia"]Also, I think Brian Lawrence made MOP as a sevvie.[/quote]

Yeah, but that was also the expanded MOP year.


Sperner's lemma is well known. State and prove an analogous result for a tetrahedron divided into smaller tetrahedra.

Please refer to here:
http://en.wikipedia.org/wiki/Sperner's_lemma


[quote="Valentin Vornicu"]In a mathematical competition in which 6 problems were posed to be participants[/quote]
A little typo here, I think: "be" shouldn't be there.

so easy problem for 6th problem in IMO .It wasn't more than 1 minute.


[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ (a \plus{} b \plus{} c)^2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \geq\ \frac {27}{8}(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)$
 :)[/quote]
Try $ a\rightarrow\plus{}\infty$ and $ b\equal{}c.$ :wink:


Prove that if $a \in$Z then the only positive divisor of both $a$ and $a+1$ is $1$.

 :lol:

Euclidean Algorithm-

$\gcd(a,a+1)=\gcd(a,1)=1$

here's another method.

[hide]Let $(a,a+1)=k$. We must prove $k=1$. If $(a,a+1)=k$, then $k\mid a$ and $k\mid a+1$. Thus $a=kn$ for some integer n and $a+1=mk$ for some integer m. By sub, $kn+1=mk$ and $kn-mk=-1$. Factoring gives $k(n-m)=-1$ and thus $k\mid-1$ and $k=1$. QED.[/hide]


analytic geometry
geometry
3D geometry
tetrahedron

In a three dimensional rectangular coordinate system, find the total surface area of the solid defined by $ |x|\plus{}|y|\plus{}|z|\leq{1}$

[hide="Solution"]We have 8 symmetric octants, so it suffices to consider x,y,z>0. What we have is a tetrahedron with vertexes (0,0,0), (1,0,0), (0,1,0), (0,0,1). The volume of this can be calculated as the volume of a pyramid.

Let the first three points determine the base. The area of the base is $ \frac {1}{2} (1)(1) \equal{} \frac {1}{2}$.

The first and fourth points, then, determine the height to be $ 1$. The volume is $ \frac {1}{3} Bh \equal{} \frac {1}{3} \cdot \frac {1}{2} \cdot 1 \equal{} \frac {1}{6}$

Multiplying this by $ 8$ for our symmetric octants gives $ V \equal{} (8) \frac {1}{6} \equal{} \frac {4}{3}$
_________________________________________________________________________________[/hide]

[b]Follow-up:[/b] Find the volume contained by $ \big| x \plus{} y \big| \plus{} \big| y \plus{} z \big| \plus{} \big| z \plus{} x \big| \leq 1$

[b]Discussion question 1:[/b] Find the area contained by $ \big| ax \plus{} by \big| \plus{} \big| cx \plus{} dy \big| \leq 1$, where $ ad \minus{} bc \neq 0$

[b]Discussion question 2:[/b] Find the volume contained by $ \big| ax \plus{} by \plus{} cz \big| \plus{} \big| dx \plus{} ey \plus{} fz \big| \plus{} \big| gx \plus{} hy \plus{} kz \big| \leq 1$, where $ aek \minus{} afh \minus{} bdk \minus{} bfg \plus{} cdh \minus{} ceg \neq 0$

[b]Discussion question 3:[/b] What happens when $ ad \minus{} bc \equal{} 0$ in discussion Q1, or $ aek \minus{} afh \minus{} bdk \minus{} bfg \plus{} cdh \minus{} ceg \equal{} 0$ in discussion Q2?


If $\frac{x/4}{2} = \frac{4}{x/2}$ then $x =$

$\text{(A)} \ \pm 1/2 \qquad \text{(B)} \ \pm 1 \qquad \text{(C)} \ \pm 2 \qquad \text{(D)} \ \pm 4 \qquad \text{(E)} \ \pm 8$

This has already been posted (in the same forum too)  :D

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=72356]www.artofproblemsolving.com/Forum/viewtopic.php?t=72356[/url]


Let $ a,b,c>0$ such that $ 4abc\equal{}a\plus{}b\plus{}c\plus{}1$.Prove that
$ \frac{b^2\plus{}c^2}{a}\plus{}\frac{c^2\plus{}a^2}{b}\plus{}\frac{a^2\plus{}b^2}{c}$≥$ 2(ab\plus{}bc\plus{}ca)$

from the condition,we may assume that:$ a \equal{} \frac {x \plus{} y}{2z},b \equal{} \frac {y \plus{} z}{2x},z \equal{} \frac {x \plus{} z}{2y}$(1)
remember that $ LHS \geq 2\sum \frac {bc}{a}$
so we will prove that:$ \sum bc(\frac {1}{a} \minus{} 1)\geq 0$(2)
from 1 and 2,we have to prove that:
$ \sum (\frac {2x}{y \plus{} z})^2\geq \sum \frac {2x}{y \plus{} z}$
which is equivalent to
$ \box 2\sum x^6 \plus{} 3\sum xy(x^4 \plus{} y^4) \plus{} 4xyz(\sum x^3) \minus{} 4abc(\sum ab(a \plus{} b)) \minus{} 6a^2b^2c^2 \minus{} 2\sum a^3b^3 \geq 0$
but this is true ([6,0,0];[5,1,0];[4,1,1] >> [3,2,1],[2,2,2],[3,3,0])
Q.E.D  :)

[quote="SUPERMAN2"]Let $ a,b,c > 0$ such that $ 4abc \equal{} a \plus{} b \plus{} c \plus{} 1$.Prove that
$ \frac {b^2 \plus{} c^2}{a} \plus{} \frac {c^2 \plus{} a^2}{b} \plus{} \frac {a^2 \plus{} b^2}{c}$≥$ 2(ab \plus{} bc \plus{} ca)$[/quote]
[b]Another approach[/b]:
By $ AM \minus{} GM$, we obtain:
$ LHS\ge \frac {2bc}{a} \plus{} \frac {2ca}{b} \plus{} \frac {2ab}{c} \equal{} \frac {2}{abc}((ab)^2 \plus{} (bc)^2 \plus{} (ca)^2)$
But $ (ab)^2 \plus{} (bc)^2 \plus{} (ca)^2\ge \frac {1}{3}(ab \plus{} bc \plus{} ca)^2$
$ \Longrightarrow LHS\ge \frac {2}{3abc}(ab \plus{} bc \plus{} ca)^2$
Thus, it is enough to prove
$ \frac {2}{3abc}(ab \plus{} bc \plus{} ca)^2\ge 2(ab \plus{} bc \plus{} ca)$

$ \Leftrightarrow ab \plus{} bc \plus{} ca\ge 3abc$.
Indeed, from the condition, by $ AM \minus{} GM$, we obtain:
$ 4abc \equal{} a \plus{} b \plus{} c \plus{} 1\ge 4\sqrt [4] {abc}\Longrightarrow abc\ge 1$
$ \Longrightarrow a \plus{} b \plus{} c\ge 4abc \minus{} 1\ge 3abc$
But $ (ab \plus{} bc \plus{} ca)^2\ge 3abc(a \plus{} b \plus{} c)\ge 9a^2b^2c^2\Longrightarrow ab \plus{} bc \plus{} ca\ge 3abc$.
The result is lead as follow.

[quote="SUPERMAN2"]Let $ a,b,c > 0$ such that $ 4abc \equal{} a \plus{} b \plus{} c \plus{} 1$.Prove that
$ \frac {b^2 \plus{} c^2}{a} \plus{} \frac {c^2 \plus{} a^2}{b} \plus{} \frac {a^2 \plus{} b^2}{c}$≥$ 2(ab \plus{} bc \plus{} ca)$[/quote]
$ abc \ge 1$

By Chebyshev : $ LHS \ge \frac {2}{3}(a^2 \plus{} b^2 \plus{} c^2) \frac {ab \plus{} bc \plus{} ca}{abc} \ge \frac {2(4abc \minus{} 1)^2}{9abc}(ab \plus{} bc \plus{} ca) \ge 2(ab \plus{} bc \plus{} ca)$;

Another solution:
\[ LHS \geqslant RHS \Leftrightarrow \sum {ab({a^2} + {b^2}) \geqslant 2abc\sum {ab} }\]
\[ We\,have\,also\,: \,\sum {ab({a^2} + {b^2}) \geqslant 2\sum {{{(ab)}^2}\,so\,the\,inequality\,is\,equivalent\,to: } }\]
\[ \sum {{{(ab)}^2} \geqslant abc\sum {ab} } .\,We\,have: \,\,\,\sum {{{(ab)}^2}}  \geqslant \frac{1}
{3}(ab + bc + ac)(ab + bc + ac)\]
\[ so\,\,\,\,: \,\,\,\,\sum {{{(ab)}^2}}  \geqslant \frac{1}
{3}\sqrt {3abc(a + b + c)} (ab + bc + ac)\]
\[ \,\,\,\,\frac{1}
{3}\sqrt {3abc(a + b + c)} (ab + bc + ac)\, = \frac{1}
{3}\sqrt {3abc(4abc - 1)} \sum {ab}\]

\[ %Error. "eqalign" is a bad command.
{
  \& So\,it\,enough\,to\,show\,that\,\frac{1}
{3}\sqrt {3abc(4abc - 1)}  \geqslant abc \Leftrightarrow {(abc)^2} \geqslant abc\,wich\,is\,true  \cr
  \& because\,abc \geqslant 1.\,DONE! \cr}\]

[quote="SUPERMAN2"]Let $ a,b,c > 0$ such that $ 4abc \equal{} a \plus{} b \plus{} c \plus{} 1$.Prove that
$ \frac {b^2 \plus{} c^2}{a} \plus{} \frac {c^2 \plus{} a^2}{b} \plus{} \frac {a^2 \plus{} b^2}{c}$≥$ 2(ab \plus{} bc \plus{} ca)$[/quote]
Let $ 3u \equal{} a \plus{} b \plus{} c, 3v^2 \equal{} ab \plus{} bc \plus{} ca, w^3 \equal{} abc$
It's equivalent to:
$ \sum_{cyc} bc(b^2 \plus{} c^2) \ge 2(ab \plus{} bc \plus{} ca)abc \equal{} 2v^2w^3 \iff$
$ 27u^2v^2 \minus{} 3uw^3 \minus{} 18v^4 \ge 2v^2w^3 \iff$
$ \minus{} 18v^4 \plus{} v^2(27u^2 \minus{} 2w^3) \minus{} 3uw^3 \ge 0$
Fix $ u$ and $ w^3$, then the inequality is concave in $ v^2$ so we only have to check the cases where $ v^2$ are minimal or maximal, which is exactly when two of $ a,b,c$ are equal. ( [url=http://www.mathlinks.ro/viewtopic.php?t=278791]Because[/url] ) Assume wlog $ a \equal{} c$. Then we have to prove:
$ \frac {b^2 \plus{} a^2}{a} \plus{} \frac {a^2 \plus{} a^2}{b} \plus{} \frac {a^2 \plus{} b^2}{c} \ge 2(ab \plus{} ab \plus{} a^2)$ when $ 4a^2b \equal{} 2a \plus{} b \plus{} 1$ $ \iff$
$ a^3 \plus{} b^3 \plus{} a^2b \minus{} a^3b \minus{} 2a^2b^2 \ge 0$
But $ 4a^2b \equal{} 2a \plus{} b \plus{} 1 \iff b(4a^2 \minus{} 1) \equal{} 2a \plus{} 1 \iff$ $ b(2a \minus{} 1)(2a \plus{} 1) \equal{} 2a \plus{} 1 \iff b(2a \minus{} 1) \equal{} 1 \iff 2ab \equal{} 1 \plus{} b$.
Using that:
$ a^3 \plus{} b^3 \plus{} a^2b \minus{} a^3b \minus{} 2a^2b^2 \ge 0 \iff$
$ 8a^3b^3 \plus{} 8b^6 \plus{} 8a^2b^4 \minus{} 8a^3b^4 \minus{} 16a^2b^5 \ge 0 \iff$
$ (b \plus{} 1)^3 \plus{} 8b^6 \plus{} 2(b \plus{} 1)^2b^2 \minus{} b(b \plus{} 1)^3 \minus{} 4(b \plus{} 1)^2b^3 \ge 0$
$ 8b^6 \minus{} 4b^5 \minus{} 7b^4 \minus{} 2b^3 \plus{} 2b^2 \plus{} 2b \plus{} 1 \ge 0 \iff$
$ (b \minus{} 1)^2(8b^4 \plus{} 12b^3 \plus{} 9b^2 \plus{} 4b \plus{} 1) \ge 0$ which is obvious.

Equality iff $ a \equal{} b \equal{} c \equal{} 1$ :)

[quote="secrets"]I think you are arqady.[/quote]
Hehe :P I'm pretty sure that I'm not though. ;)


Find all functions $f:N \rightarrow N$ satisfying:
$\displaystyle f(2003mn+m+n)=2004f(mn)+f(m)+f(n)$

Let $g : 2003N+1 \rightarrow 2004N+1$ defined by $g(x) = 2004f( \frac {x-1} {2003}) + 1$.
It is easy to see that the correspondance from $f$ to $g$ is bijective, thus we only have to look for the functions $g$.
Now, it is easy too to verify that $g$ is completely multiplicative, that is $g(xy)=g(x)g(y)$ for all $x,y$.

Thus we only have to find all the completely multiplicative functions from $2003N+1$ onto $2004N+1$, but there is $g$ identically equals to 1, and otherwise $g$ is enterly defined by its values on the primes. Since, from Dirichlet's theorem there are infinitely many primes of the form  $2003k+r$ for all $r=1,2,...,2002$ and of the form $2004t+s$ where $s=1,3,5,...,2003$, there are too many such functions...
Are you sure that the problem is to find them all?

Pierre.

No thanks. I have other things to do...

Pierre.

[quote="pbornsztein"]Now, it is easy too to verify that $g$ is completely multiplicative, that is $g(xy)=g(x)g(y)$ for all $x,y$.[/quote]

How exactly do you verify that? (can't follow there :blush: )

pierre's statement at the end is not quite right...actually, we may arbritarily choose $g(2004)$ since 2004 has no divisors of the form $2003n+1$. the whole thing does not really depend on primes. i'm wondering how to solve that problem anyway...i don't see it at the moment. we did a similar problem in imo training, but i was sleeping... :D

Peter

[quote="Peter Scholze"]we did a similar problem in imo training, but i was sleeping... :D
[/quote]

The problem is adapted from the ISL1996. But our friend does not admit it as usual. Peter, have a look at my document[url=http://www.artofproblemsolving.com/Forum/orl/IMO_1996_shortlist.pdf]IMO 1996 ShortList[/url] and go to page 43/52.  :)

I think the problems requires several good ideas in order to get cracked. BTW Pierre worked at the problem before. Look [url=http://groups.google.fr/groups?hl=fr&lr=&ie=UTF-8&threadm=38C58666.4646%40club-internet.fr&rnum=10&prev=/groups%3Fq%3Dauthor:pbornszt%2540club-internet.fr%2Bimo%26hl%3Dfr%26lr%3D%26ie%3DUTF-8%26selm%3D38C58666.4646%2540club-internet.fr%26rnum%3D10]here[/url].  :)

[quote="pbornsztein"]Wow, what a spy!

Pierre.[/quote]

Hey Pierre,

I am from the KGB. And all your paths in the web will be traced back.  :D  But in order to be serious you know the old ML forum postings very well too.  :)


Find all pairs of nonnegative integers x and y satisfying the equation:
 3[y(x+1)-x-y]-2xy=2

[hide="It will help"]
$ 3(xy+y-x-y)-2xy=2$
$ xy-3x=2$
$ x(y-3)=2$
And remember that $ x,y\in\mathbb{Z}_{+}\cup\{0\}$[/hide]

[quote="sludgethrower"]Answer

$ 3(xy+y-x-y)-2xy=2$
$ xy-3x=2$
$ x(y-3)=2$

When $ x$ equals $ 2$ or $ 1$ (only positive divisors of 2) then $ y-3$ respectively equals $ 1$ and $ 2$

Hence pairs $ (x,\,y)$ are:

$ (2,\,4)$ and $ (1,\,5)$[/quote]
Your solution is correct and is the same as mine  :lol:


function
number theory proposed
number theory

$ n\in N$
$ n \equal{}\prod_{i \equal{} 1}^{r}p_{i}^{a_{1}}$
a)Find number of divisor of the form $ 4k\plus{}1$
b)Find the number of divisor of the form $ 3k\plus{}1$
Remark
Have some easy problem
1)Let $ n$ is an positive integer .Prove that number of divisor of the form $ 3k\plus{}1\geq$ the number of divisor in the form $ 3k\plus{}2$
2)Similar for divisor in the form $ 4k\plus{}1,4k\plus{}3$
Good.

More than we can find sum of all divisor in the form $ 4k\plus{}1$(Same method )

Why one of the $ m_{i}$ is odd then we has.
What do you do when all $ m_{i}$ even.
I has other method quite easy.

I have other method
$ f(d) \equal{} 1\text{if}d\equiv 1(\mod 4)$
$ f(d) \equal{}\minus{}1\text{if}d\equiv 3(mod 4)$
$ f(d) \equal{} 0\text{if}d\equiv 0(\mod 2)$
Then $ f(d)$ is multiple.
Lemma :
Let $ f(n)$ is multiple then
$ h(n) \equal{}\sum_{d|n}f(d)$ is multiple.
Now consider function
$ h(n) \equal{}\sum_{d|n}f(d)$
Let
$ x_{n}$ is the number of divisor in the form $ 4k\plus{}1$
$ y_{n}$ is the number of divisor in the form $ 4k\plus{}3$
Then
$ h(n) \equal{} x_{n}\minus{}y_{n}$
But we can find $ x_{n}\plus{}y_{n}$( number of odd divisor)
So we can find the general of $ x_{n},y_{n}$
Note : find $ h(p^{a})$ quite easy I give for you.
Now we consider function
$ g(d) \equal{} d.f(d)$
Then
$ g(d)$ is multiple
So you can find sum of all divisor in the form $ 4k\plus{}1$
Note :Same method for $ 3k\plus{}1$


Euler
geometry
circumcircle
geometry solved

In a triangle $ABC$, $H$ is the orthocenter, $O$ the circumcenter, and $w$ the circumcircle. The line $AO$ meets $w$ again at $A_{1}$, and the lines $A_{1}H$ and $AH$ intersect $w$ again at $A'$ and $A''$, respectively. The points $B'$, $B''$ and $C'$, $C''$ are defined similarly. Show that the lines $A'A''$, $B'B''$, $C'C''$ are concurrent at a point on the Euler line of triangle $ABC$.

[quote="hasan"]In a triangle $ ABC$, $ H$ is the orthocenter, $ O$ the circumcenter, and $ w$ the circumcircle. The line $ AO$ meets $ w$ again at $ A_{1}$, and the lines $ A_{1}H$ and $ AH$ intersect $ w$ again at $ A'$ and $ A''$, respectively. The points $ B'$, $ B''$ and $ C'$, $ C''$ are defined similarly. Show that the lines $ A'A''$, $ B'B''$, $ C'C''$ are concurrent at a point on the Euler line of triangle $ ABC$.[/quote]

What do you think [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=66285]circumcevian pingpong[/url] is for? ;)

The points H, O, H lie on the Euler line of triangle ABC; the triangles ABC, $ A_{1}B_{1}C_{1}$, A'B'C' are the circumcevian triangles of the points H, O, H with respect to the triangles A''B''C'', ABC, $ A_{1}B_{1}C_{1}$, respectively.

In order to prove circumcevian pingpong, try generalizing the proof of Lemma 4 in my [url=http://www.cip.ifi.lmu.de/~grinberg/Schroeder/Schroeder.html#7][Schroeder7][/url].

  Darij

[quote="barasawala"]Hey, there should be an easier solution without having to use these advanced knowledge  :maybe:[/quote]

Hey, I am using the Pascal theorem only, and I wouldn't quite call it advanced for a problem with loads of collinear points and points on a circle.

But if you want you can always "embed" the proof of the Pascal theorem into your proof, it just won't make much sense...

  Darij, from ZetaX's laptop


conics
ellipse
trigonometry
analytic geometry
parameterization
geometry

I think I might have posted this already, but I'm not sure...  :maybe: If I have, just delete this.

Consider the concentric circles with equations $ x^2\plus{}y^2\equal{}4$ and $ x^2\plus{}y^2\equal{}9$. A radius from the origin $ O$ intersects the inner circle at $ P$ and the outer circle at $ Q$. The line parallel to the $ x$-axis through $ P$ meets the line parallel to the $ y$-axis through $ Q$ at the point $ R$. Show that $ R$ lies on the ellipse $ \frac{x^2}{9}\plus{}\frac{y^2}{4}\equal{}1$.

[hide="With trig"]If the radius makes an angle of $ \theta$ with the $ x$-axis, then $ P \equal{} (2\cos\theta,2\sin\theta)$, $ Q \equal{} (3\cos\theta,3\sin\theta)$. Then since $ R$ has the same $ y$ coordinate as $ P$ and the same $ x$ coordinate as $ Q$, $ R \equal{} (3\cos\theta,2\sin\theta)$. And $ \frac{(3\cos\theta)^2}{9} \plus{} \frac{(2\sin\theta)^2}{4} \equal{} 1$ by the Pythagorean Identity.[/hide]

There is probably a much more elementary solution to this.

yeah thats how i did it too.  I don't think there is an easier way.

Set $ a\equal{}P_x$ and $ b\equal{}Q_y,$ then $ R\equal{}(a,b)$ and $ P_y \equal{} \frac{2b}{3}.$ So $ 4\equal{}P_x^2 \plus{}P_y^2 \equal{} a^2\plus{}\frac{4b^2}{9}\Rightarrow 1\equal{}\frac{a^2}{4}\plus{}\frac{b^2}{9}.$

I discovered both solutions, but I discovered bertram's first. It seems to me the trig solution is actually harder to find.

Well, I personally disagree. :P

Trig jumped out at me in this problem because there is only one degree of freedom in this problem, which is where the line is drawn. This means that the coordinates of $ P$ and $ Q$ each have to be expressible in one parameter total (that was $ \theta$ in my solution). And when you're dealing with points on a circle, using sines and cosines is natural.

I always try to work with coordinates first, which is probably why I can't solve any geometry problems....


An object moves $ 8$ cm in a straight line from $ A$ to $ B$, turns at an angle $ \alpha$, measured in radians and chosen at random from the interval $ (0,\pi)$, and moves $ 5$ cm in a straight line to $ C$. What is the probability that $ AC<7$?

$ \textbf{(A)}\ \frac{1}{6} \qquad
\textbf{(B)}\ \frac{1}{5} \qquad
\textbf{(C)}\ \frac{1}{4} \qquad
\textbf{(D)}\ \frac{1}{3} \qquad
\textbf{(E)}\ \frac{1}{2}$

Use two different areas of the triangle (Heron's and Sine Formula):
$ \dfrac{ab sinc}{2} \equal{} \sqrt {(s)(s \minus{} a)(s \minus{} b)(s \minus{} c)}$
$ 20sinC \equal{} 10\sqrt {3}$
$ c \equal{} \pi/3$
If the angle gets smaller, then $ AC$ decreases so you have $ \dfrac{\dfrac{\pi}{3}}{\pi} \equal{} \dfrac{1}{3}$

Btw, are these problems/solutions going to end up on the wiki?

[quote="worthawholebean"]An object moves $ 8$ cm in a straight line from $ A$ to $ B$, turns at an angle $ \alpha$, measured in radians and chosen at random from the interval $ (0,\pi)$, and moves $ 5$ cm in a straight line to $ C$. What is the probability that $ AC < 7$?

$ \textbf{(A)}\ \frac {1}{6} \qquad \textbf{(B)}\ \frac {1}{5} \qquad \textbf{(C)}\ \frac {1}{4} \qquad \textbf{(D)}\ \frac {1}{3} \qquad \textbf{(E)}\ \frac {1}{2}$[/quote]

First, we use the cosine rule

$ cos C \equal{} \frac{a^2 \plus{} b^2 \minus{} c^2}{2ab}$

$ cos C \equal{} \frac{8^2 \plus{} 5^2 \minus{} 7^2}{2(8)(5)}$

$ cos C \equal{} 1/2$

$ \therefore C \equal{} 60$ degrees

hence $ AC < 7$ for angles between $ 0$ and $ 60$, hence the probability is

P(<60) $ \equal{} 60/180$

$ \equal{} 1/3$

hence the answer is $ \textbf{(D)}\ \frac {1}{3}$

[b]Solution:[/b] By LoC, $49 > 89 + 80\cos\alpha \Rightarrow \cos\alpha < -\frac{1}{2}$. This implies that $\alpha > 120^{\circ}$. This gives us our answer of $\frac{60}{180} = \boxed{\frac{1}{3}}$.


[quote="Silverfalcon"]In the adjoining figure, the triangle $ABC$ is a right triangle with $\measuredangle BCA = 90^\circ$. Median $CM$ is perpendicular to median $BN$, and side $BC = s$. The length of $BN$ is

$\text{(A)} \ s \sqrt{2} \qquad \text{(B)} \ \frac{3}{2} s \sqrt{2} \qquad \text{(C)} \ 2s \sqrt{2} \qquad \text{(D)} \ \frac{s \sqrt 5}{2} \qquad \text{(E)} \ \frac{s \sqrt 6}{2}$[/quote]

[hide]Connect MN; $MN = s/2$. Then denote O as the intersetion of BN and CM

Due to similar triangles formed by connecting midpoints, $MO / OC = 1/2 = NO / BO$.

Time for some algebra, let $MO = x$ and $NO = y$. Plugging what we know, $BN = 3y$.

Now, since this is AMC, we arbitrarily let $x=1$.

Because CO is the attitude of the right $\triangle BCN$, $(2x)^2 = 2y(y) \Rightarrow 2x^2 = y^2$. Plugging in $x=1$, we have $y= \sqrt{2}$ and thus $BN = 3\sqrt{2}$.

Using Pathy on right triangle BOC, we have $s = 2\sqrt{3}$. Plug that into each of the choices, we see only $\boxed{E}$ gives $3 \sqrt{2} = BN$.

Lesson: Medians --> Midpoints --> Similar Triangles.
Alternatively, we note that medians intersect at a point that is 2/3 from the vertex and 1/3 from the side. [/hide]


Working 22 hours in the second week of June, Xenia was able to
earn $ \$$47.60 more than during the first week of June when she
worked 15 hours. If her hourly wage was constant, how many
dollars did she earn during the first two weeks of June? Express
your answer to the nearest hundredth.

Since she worked $ 7$ hours more, then her hourly wage is $ \frac{\$47.60}{7}\equal{}\$6.80$.
Her total work time is $ 15\plus{}22\equal{}37$, so her total pay is $ 37\times\$6.80\equal{}\boxed{\$251.60}$


function
real analysis
real analysis unsolved

Build a function $ f: R\to{R_{\plus{}}}$ with the property that for every $ x\in{Q}$ exist $ V$so that $ f(y)>f(x)$ for every $ y\in{V\cap{R}\minus{}\{x\}}$.
$ V$ is a neighborhood for $ x$.
I found $ f(x)\equal{}2^x$ for $ x\in{Q}$ and$ f(x)\equal{}2^x\plus{}1$ for $ x\in{R\minus{}Q}$. It is ok?
Can you give me another examples?

No, that's not OK though it is a good starting point. The problem is that $ f(y)$ in your example is not greater than $ f(x)$ if both $ x,y\in\mathbb Q$ and $ y<x$. On the other hand, the way you took care of irrationals is a good idea. Now all you need is to design a function defined on rationals that has a local minimum at each point. The fact that $ \mathbb Q$ is countable is crucial here ;)


inequalities
geometry
circumcircle
incenter
perpendicular bisector
geometry proposed

Given triangle $ABC$, let $h_{a},g_{a},w_{a},m_{a},n_{a}$ be length of orthocenter , Gergon point,inceter,centroid,Nagel point cevians from vertex $A$ similarly with vertex $B,C$ prove that
\[\sum h_{a}\le\sum g_{a}\le\sum w_{a}\le\sum m_{a}\le\sum n_{a}\]

Let $ M,D,V$ be the midpoint of $ BC,$ the orthogonal projection of $ A$ onto $ BC$ and  the foot of the A-angle bisector. Let $ X,Y$ be the tangency points of the incircle and the A-excircle with  with $ BC.$ It's enough to prove that $ AY \ge AM \ge AV \ge AX \ge AD.$ Perpendicular bisector $ \ell_a$ of $ BC$ cuts the arc $ BC$ of its circumcircle, not containing $ A,$ at its midpoint $ L.$ Since $ A,V,L$ are collinear,  $ AV$ is an internal ray of $ \angle DAM$ $ \Longrightarrow$ $ AM \ge AV$ and it is clear that $ AV \ge AX,$ due to $ IX \perp BC,$ where $ I$ is the incenter of $ \triangle ABC.$ On the other hand, $ M$ is also midpoint of $ XY$ $\Longrightarrow$ $ M$ is between $ Y$ and $ D$ $ \Longrightarrow$ $ AY \ge AM.$ Thus, $ AY \ge AM \ge AV \ge AX \ge AD.$


algebra
polynomial
number theory unsolved
number theory

Let $ m,n,p\in\mathbb{Z}$ and $ r\in\mathbb{R}$ such that
(i) $ \sqrt{p}\equal{}m\plus{}r$
(ii) $ m(m^2\minus{}nr)\plus{}r^3\equal{}0$

Prove that $ n$ is divisible by $ 9$

are you sure you didn't forget to add something?

Since $ x\equal{}r$ is irrational root of the polynomials: $ (m\plus{}x)^2 \minus{} p$ and $ x^3 \minus{} mn x \plus{} m^3$, the former polynomial must divide the latter one, implying that $ (m,n,p)\equal{}(2k,9k,6k^2)$ for some integer $ k$.


$ G, H, I, J, K, L$ are the midpoints, prove that $ GJ, HK, IL$ are concurrent $ < \equal{} >$ $ S(ACE) \equal{} S(BDF)$ .

This problems come from chinese expert teacher [b]Zhonghao Ye[/b]
The geometric way is here(6nd floor)：
[hide][url]http://bbs.cnool.net/topic_show.jsp?id=8342604&oldpage=1&thesisid=494&flag=topic1[/url][/hide]


A graph has 12k points. Each point has 3k+6 edges. For any two points the number of points joined to both is the same. Find k.

k=3, i will post my solution later when i will have more time.(yet i didn't construct example of this graph)


Find all functions $f: \mathbb R\to \mathbb R$ such that $f(x+f(y))=f(x)+x+f(x-y)$, $\forall x,y\in \mathbb R$.

(*) $f(x+f(y))=f(x)+x+f(x-y)$

Set $g(x): =f(x)+x$ for $x\in\mathbb{R}$.
Then
$g(x-y+g(y))=g(x+f(y))=x+f(y)+f(x+f(y))\\ \hspace*{2.88cm}=x+f(y)+f(x)+x+f(x-y)=g(x)+g(y)+g(x-y)$
and $x=y+z$ gives
(**) $g(g(y)+z)=g(y+z)+g(y)+g(z)$ for all $y,z\in\mathbb{R}$.

Replacing $z$ by $g(z)$ in (**), and then using (**) again, we get
$g(g(y)+g(z))=g(y+g(z))+g(y)+g(g(z))\\ \hspace*{2.7cm}=g(g(z)+y)+g(y)+g(g(z)+0)\\ \hspace*{2.7cm}=g(z+y)+g(z)+g(y)+g(y)+g(z+0)+g(z)+g(0)\\ \hspace*{2.7cm}=g(y+z)+2g(y)+3g(z)+g(0)$
and then switching $y$ and $z$ gives
$g(g(y)+g(z))=g(y+z)+2g(y)+3g(z)+g(0)=g(y+z)+3g(y)+2g(z)+g(0)$
and $g(y)=g(z)$ for all $y,z\in\mathbb{R}$.
So $g$ is constant and (**) gives $g\equiv 0$ and $f(x)=-x$ for all $x\in\mathbb{R}$.

my soluiton:
Let $c=f(0)$.putting $y=0$ we get $f(x+c)=2f(x)+x...(*)$
but putting $x=y=0$, $f(c)=2c$
putting $x=c$ in $(*)$, $f(2c)=2f(c)+c=5c$.
so $f(3c)=f(2c+c)=2f(2c)+2c=12c$, from $(*)$.
but $f(3c)=f(c+2c)=f(c+f(c))=f(c)+c+c=4c$.
hence $4c=12c\Rightarrow c=0$.
so from $(*)$, $f(x)=2f(x)+x$ and $f(x)=-x$ directly follows.$Q.E.D$
can you imagine how much pages i had to waste to make this problem? :P


inequalities
trigonometry
function
geometry proposed
geometry

[quote][color=darkred]Given is an acute $\triangle ABC$ with the inradius $r$ and the $A$- exinradius $r_{a}$ a.s.o. Denote :

- the radius $\beta_{a}$ of the circle which is tangent to the lines $AB$, $AC$ and [b]internally[/b] to the circle $w$ a.s.o. ;

- the radius $\gamma_{a}$ of the circle which is tangent to the lines $AB$, $AC$ and [b]externally[/b] to the circle $w$ a.s.o.

Prove the following inequalities : $\{\begin{array}{cc}\beta_{a}+\beta_{b}+\beta_{c}\ge 4r\\\\ \gamma_{a}+\gamma_{b}+\gamma_{c}\ge 4\cdot \frac{r_{a}+r_{b}+r_{c}}{3}\end{array}$

[b][u]A very nice question (dedicated to the my new friend April) :[/u][/b] $\boxed{\ \sum\gamma_{a}\ge 3\cdot \sum \beta_{a}+\frac{16}{3}\cdot (R-2r)\ }\ge 3\cdot\sum\beta_{a}$. Prove it ?!

[b]Remark.[/b] $\sum\gamma_{a}\ge 3\cdot\sum\beta_{a}+\frac{16}{3}\cdot (R-2r)\ge 12r+\frac{16}{3}\cdot (R-2r)=\frac{4}{3}\cdot (4R+r)=\frac{4}{3}\cdot\sum r_{a}$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=152882 [/color][/quote]

[hide="For April and Co."][color=darkblue][b][u]The proof of the "my dedication".[/u][/b] I"ll use the following remarkable identities $\{\begin{array}{c}\frac{1}{\cos^{2}x}=1+\tan^{2}x\\\\ \frac{r_{a}}{p}=\frac{r}{p-a}=\tan\frac{A}{2}\end{array}$ , $\{\begin{array}{c}r_{a}r_{b}+r_{b}r_{c}+r_{c}r_{a}=p^{2}\\\\ r_{a}+r_{b}+r_{c}=4R+r\end{array}$

and the following well-known inequalities : $\{\begin{array}{ccc}\{a,b,c\}\subset (0,\infty ) & \implies & 3\cdot(a^{3}+b^{3}+c^{3})\ge (a+b+c)\cdot (a^{2}+b^{2}+c^{2})\\\ \\{a,b,c\}\subset\mathcal R }& \implies & a^{2}+b^{2}+c^{2}\ge ab+bc+ca\end{array}$. Therefore,

$\frac{\beta_{a}}{r}=\frac{\gamma_{a}}{r_{a}}=\frac{1}{\cos^{2}\frac{A}{2}}=1+\tan^{2}\frac{A}{2}=1+\frac{r^{2}_{a}}{p^{2}}$ $\implies$ $\{\begin{array}{c}\beta_{a}=r+\frac{rr^{2}_{a}}{p^{2}}\\\ \gamma_{a}=r_{a}+\frac{r^{3}_{a}}{p^{2}}\end{array}\|$ $\implies$ $\sum\gamma_{a}-3\cdot\sum\beta_{a}=$ $\sum r_{a}+\frac{1}{p^{2}}\cdot\sum r^{3}_{a}-9r-3r\cdot\sum \frac{r^{2}_{a}}{p^{2}}=$

$(\sum r_{a}-9r)+\frac{1}{p^{2}}\cdot(\sum r^{3}_{a}-3r\cdot\sum r^{2}_{a})\ge$ $4(R-2r)+\frac{1}{p^{2}}\cdot(\frac{1}{3}\cdot \sum r_{a}\cdot\sum r^{2}_{a}-3r\cdot\sum r^{2}_{a})=$ $4(R-2r)+\frac{1}{3p^{2}}\cdot\sum r^{2}_{a}\cdot(\sum r_{a}-9r)=$

$4(R-2r)\cdot(1+\frac{1}{3p^{2}}\cdot\sum r^{2}_{a})\ge$ $4(R-2r)\cdot(1+\frac{1}{3p^{2}}\cdot\sum r_{b}r_{c})=$ $4(R-2r)\cdot(1+\frac{1}{3})=$ $\frac{16}{3}\cdot (R-2r)$ and [size=167][b]it is ready !![/b][/size][/color][/hide]

[color=darkblue]From the topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=152882, we have:

$1.\blacktriangleright\sum\frac{1}{\beta_{a}}=\frac{4R+r}{2Rr}\implies \sum\beta_{a}\ge\frac{9}{\sum\frac{1}{\beta_{a}}}=\frac{18Rr}{4R+r}\ge\frac{16Rr+4r^{2}}{4R+r}=4r$

$2.\blacktriangleright 2\sum\frac{1}{\gamma_{a}}\ge\sum\frac{1}{\theta_{a}}=\frac{2(R+r)}{R^{2}}$
$\implies\sum\gamma_{a}\ge 4\cdot\frac{\sum r_{a}}{3}\iff\frac{9R^{2}}{R+r}\ge 4\cdot\frac{4R+r}{3}\iff 11R^{2}\ge 20Rr+4r^{2},$ what is truly.

[b]Remark.[/b] $\left\{\begin{array}{c}\beta_{a}=\frac{bc}{p(p-a)}\cdot r,\;\textrm{a.s.o.}\\ \sum\beta_{a}\ge 4r\end{array}\right\|\implies{\sum\frac{bc}{p(p-a)}\ge 4}$ [b]Is it new and hard?[/b][/color]

[color=darkblue][b][u]Proof 1.[/u][/b] $\{\begin{array}{c}bc=p(p-a)+(p-b)(p-c)\\\ r_{a}r_{b}+r_{b}r_{c}+r_{c}r_{a}=p^{2}\\\ \tan\frac{A}{2}=\frac{r_{a}}{p}\end{array}\|$ $\implies$ $\sum\frac{bc}{p(p-a)}=$ $\sum [1+\frac{(p-b)(p-c)}{p(p-a)}]=$

$3+\sum \tan^{2}\frac{A}{2}=$ $3+\frac{1}{p^{2}}\cdot \sum r^{2}_{a}\ge 3+\frac{1}{p^{2}}\cdot\sum r_{b}r_{c}=4$ $\implies$ $\sum\frac{bc}{p(p-a)}\ge 4$ $\implies$ $\sum\beta_{a}\ge 4r$.

[b][u]Proof 2[/u][/b] ( quickly, but I don't like it ! ). $\sum\frac{bc}{p(p-a)}=\sum\frac{1}{\cos^{2}\frac{A}{2}}=\sum(1+\tan^{2}\frac{A}{2}))=3+\sum\tan^{2}\frac{A}{2}\ge 4$

because the function $\tan^{2}: (0,\frac{\pi}{2})\rightarrow (0,\infty )$ is convexe and apply the Jensen's inequality : $\sum\tan^{2}\frac{A}{2}\ge 1$.

[b]The April's remark.[/b] $\sum\tan^{2}\frac{A}{2}\ge \sum\tan\frac{B}{2}\tan\frac{C}{2}=1$[/color]

Nice equality Virgil Nicula, I have remark
$r_{a}^{2}+r_{b}^{2}+r_{c}^{2}=(4R+r)^{2}-2p^{2}$ and $r_{a}^{3}+r_{b}^{3}+r_{c}^{3}=(4R+r)^{3}-12p^{2}R$ thus we get
$\sum\beta_{a}=3r+r\frac{\sum r_{a}^{2}}{p^{2}}=r+r(\frac{4R+r}{p})^{2}\quad (1)$ and
$\sum\gamma_{a}=\sum r_{a}+\frac{\sum r_{a}^{3}}{p^{2}}=4R+r+\frac{(4R+r)^{3}-12p^{2}R}{p^{2}}=r-8R+\frac{(4R+r)^{3}}{p^{2}}\quad (2)$
Now by $4R+r\ge \sqrt{3}p$ we get $\sum\beta_{a}\ge r+r\frac{\sqrt{3}(4R+r)}{p}\ge 4r$ and $\sum\gamma_{a}\ge r-8R+3(4R+r)=4(R+r)$ this one is weaker but I mean with the equalities with $r,R,p$ we can make them strong as we want.


In triangle $ ABC$ $ AK$ $ BL$ and $ CM$ are altitudes  and $ H$ is orthocenter. $ P$ is the midpoint of $ AH$. Lines $ BH$ and $ MK$ intersect in point $ S$. Lines $ LP$ and $ AM$ intersect at $ T$.
Prove that line $ TS$ is perpendicular to $ BC$.

[hide="Hint"]1.Use the nine-point circle
2.Prove that LSMT is inscribed quadrilateral[/hide]

Could You post some usefull properties connected with  nine-point circle?

It's easy to solve the problem when we know that $ LSMT$ is cyclic but I can't see how to prove it.

I was just wondering if someone could give me a link with  sample problems which can be solved by using 9-point circle.

I'm rather poor in geometry and I'd like to learn it.

$ \angle SMC\equal{}\angle LBC\equal{}\angle CAK\equal{}\angle TLA$ -> LSMT is cyclic. Hence $ \angle LPK\equal{}\angle LMK\equal{}\angle LTS$ Q.E.D.

Dear Mathlinkers,
a collection of proofs can be now ....
see also    http://www.mathlinks.ro/Forum/viewtopic.php?t=218868
Sincerely
Jean-Louis


As shown in the figure below, a circular park consists of an outer-ring path for walkers (white) and a ring-shaped flower garden (gray) surrounding a central circular fountain (black). The walking path is six feet wide in all places, the garden ring is eight feet wide in all places, and the fountain has a diameter of 10 feet. What is the diameter, in feet, of the circle that is the outer boundary of the walking path?

[asy]import graph;
size(101);
draw(Circle((0,0),19));
filldraw(Circle((0,0),13),gray(.6),black);
fill(Circle((0,0),5),black);
draw("$8'$",(0,5)--(0,13));
draw("$6'$",(13,0)--(19,0));[/asy]


inequalities
trigonometry
geometry
inradius
circumcircle
trig identities
Law of Cosines

I'm really out of it right now, so my apologies if I did anything really stupid.

[hide="Solution"]By the Law of Cosines, $ b^2 \plus{} c^2 \minus{} a^2 \equal{} 2bc \cos A$. It suffices to show that:

$ \sum_{cyc} \frac {1}{2bc \cos A} \geq \frac {1}{d^2} \equal{} \frac {4K^2}{a^2 b^2 c^2}$

$ \iff \frac {a^2bc}{4\cos A} \plus{} \frac {ab^2 c}{4\cos B} \plus{} \frac {abc^2}{4\cos C} \geq 2K^2$

$ \iff \frac {ab}{2} \cdot \frac {ac}{2} \cdot \frac {1}{\cos A} \plus{} \frac {ab}{2} \cdot \frac {bc}{2} \cdot \frac {1}{\cos B} \plus{} \frac {ac}{2} \cdot \frac {bc}{2} \cdot \frac {1}{\cos C} \geq 2 K^2$

However, $ \frac {ab}{2 \cos A} \geq \frac {ab}{2} \geq \frac {ab}{2} \sin C \equal{} K$, and symmetric, thus we have:

$ \frac {ab}{2} \cdot \frac {ac}{2} \cdot \frac {1}{\cos A} \geq K^2 \implies \sum_{cyc} \frac {ab}{2} \cdot \frac {ac}{2} \cdot \frac {1}{\cos A} \geq 3 K^2 > 2K^2$[/hide]

[b][color=red]Edit:[/color][/b] See below  :rotfl:

Ya, r is the inradius, not the circumradius, so your first line is wrong.

[u]Solution.[/u] Notice that the cosine law implies that $ \sum \frac {1}{b^2 \plus{} c^2 \minus{} a^2} \equal{} \frac {1}{4K}\sum\tan A\geq \frac {3\sqrt {3}}{4K}$, where $ K \equal{} [ABC]$.  The inequality follows from the identity $ \sum\tan A \equal{} \prod\tan A$ and the Arithemetic-geometric means inequality or by Jensen.  Therefore, if we prove that $ \frac {3\sqrt {3}}{4K}\geq\frac {1}{4r^2} \equal{} \frac {s^2}{4K^2}$, i.e. $ 12K\sqrt {3}\geq (a \plus{} b \plus{} c)^2$, then we are done.  But that inequality is equivalent to $ 12K\sqrt {3}\geq \sum a^2 \plus{} 4K\sum\frac {1}{\sin A}$.  This is true since $ \sum a^2\leq 4K\sqrt {3}$ (this can be proved using Heron and AM-GM) and $ 4K\sum\frac {1}{\sin A}\leq 8K\sqrt {3}$ (follows by Jensen's inequality or the identity $ \sum\cos^2 A \plus{} 2\prod\cos A \equal{} 1$ with AM-GM). $ \Box$

Virgil Nicula, this is a nice inequality. :)

[quote="Virgil Nicula"] [color=darkred]Prove that in an acute triangle $ ABC$ there is the relation $ \sum \frac {1}{b^2 \plus{} c^2 \minus{} a^2}\ge \frac {1}{4r^2}$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b][/color] $ \sum \frac {1}{b^2 \plus{} c^2 \minus{} a^2} \equal{} \frac {1}{2S}\cdot\sum\frac {\sin^2A}{\sin 2A}\stackrel{C.B.S.}{\ \ \ge\ \ }\frac {1}{2S}\cdot\frac {\left(\sum\sin A\right)^2}{\sum\sin 2A} \equal{} \frac {1}{4r^2}$ .

[quote="rofler"]This is true since... is false

Weitzenbocks inequality is reversed.[/quote]

Oops, you're right.  I should've made sure it was true.

[quote="minsoens"][quote="Virgil Nicula"]$ \frac {1}{2S}\cdot\frac {\left(\sum\sin A\right)^2}{\sum\sin 2A} \equal{} \frac {1}{4r^2}$ [/quote]please, how do you prove this? thank you[/quote]
$ a \plus{} b \plus{} c \equal{} 2p$ , $ \frac {a}{\sin A} \equal{} \frac {b}{\sin B} \equal{} \frac {c}{\sin C} \equal{} 2R$ , $ S \equal{} pr$ and $ abc \equal{} 4RS$ . Thus,

$ \sum\sin A \equal{} \sum\frac {a}{2R} \equal{} \frac pR$  and $ \boxed {\ \sum\sin 2A \equal{} 4\prod\sin A\ } \equal{} 4\prod\frac {a}{2R} \equal{}$ $ \frac {abc}{2R^3} \equal{} \frac {4RS}{2R^3} \equal{} \frac {2S}{R^2}$ .


Find an equation of the tangent to the curve at the given point

y = (1/8)x^4 , (2,2)

Let $f(x) = \frac{1}{8}\cdot x^{4}$. Then $f'(x) = \frac{1}{2}\cdot x^{3}$ and $f'(2) = 4$. So the equation of the tangent is $y = 4 \cdot (x-2)+2$


Hello all,

           I will be taking the usnco local section in march. I can get about a 55/60 right now, but competition at my school is fierce and so I am aiming for a 60. The questions I find difficult are the questions about random facts of elements (I.e which metal is commercially created by electrolysis: Al or Fe). I was wondering if anyone knew how to prepare for those types of problems?

Those facts are not random. In the case of your posted question, "electrochemical series" are the key words.

Can someone perhaps give me a list of those series/facts that they test? I could also use names of basic household products and their chemical names, I.e baking soda (sodium bicarbonate).

Thanks.

Baking Powder is not sodium bicarbonate. $ \ce{NaHCO3}$ is commonly known as $ \text{Baking}$ [i]$ \ext{Soda}$[/i].

Yea, I realized after I wrote it... But school comps are just so slow that it wasn't worth fixing (at the expense of missing calc :D )

As someone who got a 54 (though I should have had a 57) on the local exam last year with bad prep, I can tell you one word: Atkins. It has everything you need. Things like "what is bronze an alloy of?" (answer: Copper and Tin) are found in that book. Read it for 3 weeks straight, combined with 4 - 5 practice tests and you should be good (although I recommend just doing national exams since they "overprepare" you for the difficulty of the local exam).

Regarding a list of things you need to know, this summer I categorized all of the National problems from 2000-2007.  In the next day or two, I'll be posting them at http://www.scientificendeavors.com/chemistry/olympiad.php.

Wow, Greg, I must say that I'm quite impressed with your progress so far on this project.  I'll definitely be watching it closely, it looks very promising.  I just wish you had written it a year earlier so I'd have a better shot at making the team ;)  Keep up the awesome work, I'm looking forward to reading your future work!

I've posted about 5-10 subjects on the above link.  Still a few more to go; I'll update when I get a chance.

Ok, for everyone interested: at [url=http://scientificendeavors.com/chemistry/olympiad.php]scientificendeavors.com/chemistry/olympiad.php[/url] I have a beta version of a guide to the Chemistry Olympiad.  I hope that you will find it interesting, informative, and useful.


[quote="quykhtn-qa1"]Let $ a,b,c$ be nonegative numbers,no two of which are zero such that:$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$.Prove that:
$ \frac {a^4}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {b^4}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {c^4}{c^2 \minus{} ca \plus{} a^2} \ge 1 \plus{} \frac {2abc(a \plus{} b \plus{} c)}{(a^3 \plus{} b^3 \plus{} c^3)^2}$
 :)[/quote]
Nice result,quykhtn-qa1.This is my solution for this one:
We have $ \sum {\frac{{{a^4}}}{{{a^2} \minus{} ab \plus{} {b^2}}}}  \equal{} \sum {\frac{{{a^4}\left( {a \plus{} b} \right)}}{{{a^3} \plus{} {b^3}}}} \equal{} \left( {a \plus{} b \plus{} c} \right)\sum {\frac{{{a^4}}}{{{a^3} \plus{} {b^3}}}}  \minus{} \sum {\frac{{{a^4}c}}{{{a^3} \plus{} {b^3}}}}$
Use this lemma: $ \sum {\frac{{{a^4}}}{{{a^3} \plus{} {b^3}}} \ge \frac{{a \plus{} b \plus{} c}}{2}}$
We get that
$ \sum {\frac{{{a^4}}}{{{a^2} \minus{} ab \plus{} {b^2}}}}\ge\frac{(a\plus{}b\plus{}c)^2}{2}\minus{} \sum {\frac{{{a^4}c}}{{{a^3} \plus{} {b^3}}}}\equal{}\frac{{{{\left( {a \plus{} b \plus{} c} \right)}^2}}}{2} \minus{} \left( {ab \plus{} bc \plus{} ca} \right) \plus{} abc\sum {\frac{{{b^2}}}{{{a^3} \plus{} {b^3}}}}$
$ \equal{}1\plus{}abc\sum {\frac{{{b^2}}}{{{a^3} \plus{} {b^3}}}}$
So we must to prove that $ \sum {\frac{{{b^2}}}{{{a^3} \plus{} {b^3}}}}  \ge \frac{{2\left( {a \plus{} b \plus{} c} \right)}}{{{{\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right)}^2}}}$
By Cauchy-Schwarz inequality we have:
$ \sum {\frac{{{b^2}}}{{{a^3} \plus{} {b^3}}}}  \ge \frac{{{{\left( {a \plus{} b \plus{} c} \right)}^2}}}{{2\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right)}}$
It suffices to prove that:
$ \begin{array}{l}
 \frac{{{{\left( {a \plus{} b \plus{} c} \right)}^2}}}{{2\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right)}} \ge \frac{{2\left( {a \plus{} b \plus{} c} \right)}}{{{{\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right)}^2}}} \\
  \Leftrightarrow \left( {a \plus{} b \plus{} c} \right)\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right) \ge 4 \\
 \end{array}$
It's true by $ CS$with $ a^2\plus{}b^2\plus{}c^2\equal{}2$.
Our proof is completed. :)

[quote="leedt26"][quote="quykhtn-qa1"]Let $ a,b,c$ be nonegative numbers,no two of which are zero such that:$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$.Prove that:
$ \frac {a^4}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {b^4}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {c^4}{c^2 \minus{} ca \plus{} a^2} \ge 1 \plus{} \frac {2abc(a \plus{} b \plus{} c)}{(a^3 \plus{} b^3 \plus{} c^3)^2}$
 :)[/quote]
Nice result,quykhtn-qa1.This is my solution for this one:
We have $ \sum {\frac {{{a^4}}}{{{a^2} \minus{} ab \plus{} {b^2}}}} \equal{} \sum {\frac {{{a^4}\left( {a \plus{} b} \right)}}{{{a^3} \plus{} {b^3}}}} \equal{} \left( {a \plus{} b \plus{} c} \right)\sum {\frac {{{a^4}}}{{{a^3} \plus{} {b^3}}}} \minus{} \sum {\frac {{{a^4}c}}{{{a^3} \plus{} {b^3}}}}$
Use this lemma: $ \sum {\frac {{{a^4}}}{{{a^3} \plus{} {b^3}}} \ge \frac {{a \plus{} b \plus{} c}}{2}}$
We get that
$ \sum {\frac {{{a^4}}}{{{a^2} \minus{} ab \plus{} {b^2}}}}\ge\frac {(a \plus{} b \plus{} c)^2}{2} \minus{} \sum {\frac {{{a^4}c}}{{{a^3} \plus{} {b^3}}}} \equal{} \frac {{{{\left( {a \plus{} b \plus{} c} \right)}^2}}}{2} \minus{} \left( {ab \plus{} bc \plus{} ca} \right) \plus{} abc\sum {\frac {{{b^2}}}{{{a^3} \plus{} {b^3}}}}$
$ \equal{} 1 \plus{} abc\sum {\frac {{{b^2}}}{{{a^3} \plus{} {b^3}}}}$
So we must to prove that $ \sum {\frac {{{b^2}}}{{{a^3} \plus{} {b^3}}}} \ge \frac {{2\left( {a \plus{} b \plus{} c} \right)}}{{{{\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right)}^2}}}$
By Cauchy-Schwarz inequality we have:
$ \sum {\frac {{{b^2}}}{{{a^3} \plus{} {b^3}}}} \ge \frac {{{{\left( {a \plus{} b \plus{} c} \right)}^2}}}{{2\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right)}}$
It suffices to prove that:
$ \begin{array}{l} \frac {{{{\left( {a \plus{} b \plus{} c} \right)}^2}}}{{2\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right)}} \ge \frac {{2\left( {a \plus{} b \plus{} c} \right)}}{{{{\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right)}^2}}} \\
\Leftrightarrow \left( {a \plus{} b \plus{} c} \right)\left( {{a^3} \plus{} {b^3} \plus{} {c^3}} \right) \ge 4 \\
\end{array}$
It's true by $ CS$with $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$.
Our proof is completed. :)[/quote]
Very nice solution :lol: .This is also my solution :)

[quote="quykhtn-qa1"]Let $ a,b,c$ be nonegative numbers,no two of which are zero such that:$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$.Prove that:
$ \frac {a^4}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {b^4}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {c^4}{c^2 \minus{} ca \plus{} a^2} \ge 1 \plus{} \frac {2abc(a \plus{} b \plus{} c)}{(a^3 \plus{} b^3 \plus{} c^3)^2}$
 :)[/quote]
After CS we need to prove that
$ \frac{2}{\sum(2a^2\minus{}ab)}\geq\frac{1}{a^2\plus{}b^2\plus{}c^2}\plus{}\frac{abc(a\plus{}b\plus{}c)}{(a^3\plus{}b^3\plus{}c^3)^2},$ which is obviously true after expanding.

[quote=quykhtn-qa1]Let $ a,b,c$ be nonegative numbers,no two of which are zero such that:$ a^2\plus{}b^2\plus{}c^2\equal{}2$.Prove that:
$ \frac{a^4}{a^2\minus{}ab\plus{}b^2}\plus{}\frac{b^4}{b^2\minus{}bc\plus{}c^2}\plus{}\frac{c^4}{c^2\minus{}ca\plus{}a^2} \ge 1\plus{}\frac{2abc(a\plus{}b\plus{}c)}{(a^3\plus{}b^3\plus{}c^3)^2}$
 :)[/quote]

stronger
[img]http://s7.sinaimg.cn/middle/006ptkjAzy75RNBPQhg56&690[/img]


Solve the equation in $R$.  $(\sqrt 2-x)(x+\sqrt{x^2+4})=3$

i did it in kind of a weird way...but anyway

[hide]first i multiplied each side by the sort of conjugate, so that gave

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

note that $x\sqrt{2}-x^2+\sqrt{2}\sqrt{x^2+4}-x\sqrt{x^2+4}=3$

so substituting that in, i got

$4x^2-8=3(3+2x^2-2\sqrt{2}\sqrt{x^2+4})$ rearranging, we get

$2x^2+17=6\sqrt{2}\sqrt{x^2+4}$  squaring, and rearranging, we get

$4x^4-4x^2+1=0=(2x^2-1)^2$

and checking both roots, we get $\boxed{-\frac{\sqrt2}{2}}$ as the only suitable answer[/hide]


Show that if $p_k,\,q_k,\,r_k,\,s_k\,\in \mathbb{R},\,1\le k\le n,$ then
\[
\left (\displaystyle\sum_{k=1}^n p_kq_kr_ks_k\right)^4\le\left(\sum_{k=1}^np^4_k\right)\left(\sum_{k=1}^nq^4_k\right)\left(\sum_{k=1}^nr^4_k\right)\left(\sum_{k=1}^ns^4_k\right).
\]

Denote:
$\sum_{k=1}^np^4_k=A$
$\sum_{k=1}^nq^4_k=B$
$\sum_{k=1}^nr^4_k=C$
$\sum_{k=1}^ns^4_k=D$

and

$x_i=A^{- \frac{1}{4}}p_i$
analogically $y_i,z_i,t_i$ then  we have:

$\sum x_i ^4=\sum y_i ^4=\sum z_i ^4 = \sum t_i ^4=1$

now use AM-GM for sums of $x,y,z,t$ and you get the result

In exactly the same way we can prove ineq for $n$ variables

[quote="ZetaX"]$\left( \sum_{k=1}^n p_k q_k r_k s_k \right)^4 \leq \left( \sum_{k=1}^n p_k^2 q_k^2 \right)^2 \left( \sum_{k=1}^n r_k^2 s_k^2 \right)^2 \leq \left(\sum_{k=1}^n p^4_k\right)\left(\sum_{k=1}^nq^4_k\right)\left(\sum_{k=1}^nr^4_k\right)\left(\sum_{k=1}^ns^4_k\right)$ with Cauchy-Schwarz[/quote]
Yes. But, actually, it is a Holder inequality. You can easily generalize it for case of different exponents $t_1,t_2,...,t_l>1$ s.t. $1/t_1+...1/t_l=1$.


Let $ (X_1,d_1),(X_2,d_2)$ metric spaces and $ \mathcal{B}_1,\mathcal{B}_2$ Borel $ \sigma$-algebra of $ X_1$, that is $ X_2$. Let $ \mathcal{B}$ Borel $ \sigma$-algebra on product $ X_1\times X_2$. Is $ \mathcal{B} \equal{} \mathcal{B}_1 \times \mathcal{B}_2$? If not, give an example. Under what conditions they are the same?

I'd guess that separability is involved here.


Let $p \geq 5$ be a prime number. Prove that there exists a positive integer $a < p-1$ such that neither of $a^{p-1}-1$ and $(a+1)^{p-1}-1$ is divisible by $p^{2}$ .


Circle P with radius 1 is tangent to Circle Q with radius 2, and both are tangent to Circle R with radius 3. Circle S is constructed, which passes through the centers of the other circles. What is the length of the radius of Circle S?

If that's not challenging enough, here's something: What is the area of the region enclosed by the first 3 circles' tangent points?

O1,O2 and O3 are the center of circles.
obvious that is a 3,4,5 triangle.
this trianlge is one right triangle. ==> R = 5/2

It's a 3-4-5 right triangle.  So r = h /2 = 5/2

Area of the triangle is just 6


[size=150]If p is prime, show that
  1²* 2²* 3² * .... (p - 2)² is congruent to (-1)^((p + 1)/2) mod p[/size]
 and 2² * 4² * 6² * ... * (p-1)² is congruent to (-1)^((p + 1)/2) mod p

The first one seems to be wrong.
And please don't write in that textsize.

[size=150][size=150][/size]Sorry, itś the following:
  1² * 3² * 5² * ... * (p - 2) is congruent to (-1) ^((p + 1)/2)
and  2² * 4² * 6² * ... * (p - 1) is congruent to (-1) ^((p + 1)/2)
 for any p prime of the type: p = 2m + 1.[/size]


Let $ \{p_1 < p_2 < \cdots\}$ be set of all primes. Prove that there exists a positive integer number $ K$ such that $ p_{i \plus{} 1}^2 \minus{} 1 < p_i\cdot p_{i \plus{} 2}\,\forall i > K$.

This implies $ 2 p_{i\plus{}1} \leq p_i \plus{} p_{i\plus{}2}$ which is wrong for infinitely many $ i$.

It even implies $ 2 p_{i\plus{}1} < p_i \plus{} p_{i\plus{}2}$ if one looks closer.
Now assume for some integer $ K$ we always have $ 2 p_{i\plus{}1} < p_i \plus{} p_{i\plus{}2}$ for all $ i>K$. This means $ p_{i\plus{}2} \minus{} p_{i\plus{}1} > p_{i\plus{}1} \minus{} p_i$. As primes (with one exception) are odd integers, we get $ p_{i\plus{}2} \minus{} p_{i\plus{}1} \geq  p_{i\plus{}1} \minus{} p_i \plus{} 2$. Define $ q_i : \equal{} p_{i\plus{}1}\minus{}p_i$ for $ i \geq 1$ and $ q_0 : \equal{} 2$, we already saw that $ q_{i\plus{}1} \geq q_i \plus{} 2$ for all $ i>K$.
For $ k>K$ we have
\[ p_k \equal{} \sum_{i\equal{}0}^{k\minus{}1} q_i \equal{} p_K \plus{} \sum_{i\equal{}K}^{k\minus{}1} q_i \equal{} p_K \plus{} \sum_{i\equal{}0}^{k\minus{}K\minus{}1} q_{i\plus{}K} \geq \\
\geq p_K \plus{} \sum_{i\equal{}0}^{k\minus{}K\minus{}1} q_{K}\plus{}2i \equal{} p_K \plus{} (k\minus{}K)(k\minus{}K\minus{}1) \plus{} (k\minus{}K) \cdot q_K.\]
This grows quadratically in $ k$, but $ p_k$ is well known not to do this, contradiction.


Andrea inscribed a circle inside a regular pentagon, circumscribed a circle around the pentagon, and calculated the area of the region between the two circles.  Bethany did the same with a regular heptagon (7 sides).  The areas of the two regions were $ A$ and $ B$, respectively.  Each polygon had a side length of $ 2$.  Which of the following is true?

$ \textbf{(A)}\ A\equal{}\frac{25}{49}B\qquad \textbf{(B)}\ A\equal{}\frac{5}{7}B\qquad \textbf{(C)}\ A\equal{}B\qquad \textbf{(D)}\ A\equal{}\frac{7}{5}B\qquad \textbf{(E)}\ A\equal{}\frac{49}{25}B$

[hide]
We show that the area is always the same no matter how many sides the polygon has.  Draw a radius of length $ r$ of the incircle to the midpoint of one of the sides, and a radius of length $ R$ of the circumcircle to an endpoint of that side.  These form a right triangle with legs $ 1$ and $ r$, and hypotenuse $ R$.  Then

$ 1\plus{}r^2\equal{}R^2\Rightarrow \pi\equal{}R^2\pi \minus{} r^2\pi$

which is the difference in area.  So this difference is always equal to $ \pi$ and the answer is $ C$.
[/hide]

[hide="Click to reveal hidden content"]
Let the radii of the larger and smaller circles be R and r, respectively.  We then see that two radii make a right triangle with one vertex at the vertex of the polygon, another at the center of the circles, and the last at the midpoint of a side.  Thus, $ R^2 \minus{} r^2 \equal{} \left( \frac{s}{2} \right)^2$, where $ s$ is the length of a side.  Since $ s$ is the same for the pentagon and the heptagon, we see that the desired areas are equal: $ \pi (R_p^2 \minus{} r_p^2) \equal{} \pi (R_h^2 \minus{} r_h^2)$.
[/hide]

[hide]For n sides, the area between is $ \pi (1/\sin^2(\pi/n)\minus{}1/\tan^2(\pi/n))\equal{}(1\minus{}\cos^2(\pi/n))/\sin^2(\pi/n) \equal{} \pi$.[/hide]

[hide="comment"]I was about to embark on a trig bash but I stopped and realized that the Pythagorean Theorem is cool.[/hide]


function algebra domain geometry inequalities search

I just HAD to revive this thread, since I have all kinds of new problems to post for you guys.

The first one was inspired by one of the "HARD FUNCTIONAL EQUATIONS" given to us by Dorette at training camp.  There are a couple of ideas in it, but hopefully it won't be too difficult to solve!

1) Let f be a continuous, real-to-real function (ie its domain is the real numbers, and its range is the real numbers or some subset of them).  Find all possible functions f such that both of the following two statements are true:

a) 2f(x+y+z) = f(x+y) + f(y+z) + f(x+z) + 2005
b) f(2005) = 2005

Have fun!

[quote="manuel"]I have an original problem and is still unsolved in the combinatorics unsolved problems!
im not from canada, but im going to post it here!!!

 Does there exists a configuration of an infinite amount of lines in the plane with the following properties:
i) there are no 2 lines parallel
ii) no 3 concurr
iii) the area of every minimal polygon (polygons without segments inside) is equal
????????[/quote]

Wow... that sounds like a very interesting problem.  By unsolved, do you mean that you know the solution but nobody else has found it, or do you mean that even you don't know the solution?

I can prove that no pentagons can be minimal polygons, but I don't think that helps, as there are lots of ways to divide up the plane using quadrilaterals and triangles only.  This problem seems quite hard.

Elyot ! i haven't found any solution!! and a lot of people stronger than me have thought  about it but still no solution!! hope in the future i can write a mini paper on this problem ! when i became a professional mathematician! :)
but maybe you can help me to attack it.. my idea was to add lines inductively identifying the locus where the conditions hold everytime you  add a line... or i had another idea: if $A(P_i)$ denotes the area of polygon $P_i$, prove that $c < A(P_i)<k$ for two real constants $c$ and $k$. ;)

I see, I see.  Normally in here, we post problems that we have found the solution to, not research problems that are unsolved... I'm sure guys like Yufei will enjoy having a crack at this though.

S'ok... geez this forum is so dead without David and Yufei posting...

Any luck on #1?  I can give hints... ;)

heheh... that's a start.  Can you finish it?  It's not that hard...

[quote="Arne"]Define $g(x) = f(x) + 2005$ then $g(x)$ satisfies the Cauchy equation ;) (that is $g(x + y) = g(x) + g(y)$) and since $f$ and $g$ are continuous, there exists a constant $k$ such that $g(x) = kx$ and $f(x) = kx - 2005$. And since $f(2005) = 2005$ we get $k = 2$ and $f(x) = 2x - 2005$.[/quote]

Correct!  My exact solution.  The substitution is enough to throw most people off... But this was probably easy for you ;)

Do you have another problem to post?  I might post another one if I have time...

The problem $g(x+y)=g(x)+g(y)+\alpha$, with $g$ satisfying a condition in which all additive functions satisfying said condition are linear, has been discussed elsewhere on the forums before.

Yes Elyot, it was pretty straightforward to me and I know the general solution to the equation blahblahblah mentioned above :)

Do you mean I should post an original problem?

Well, if $a$, $b$ and $c$ belong to $[-3, + \infty)$ and $a + b + c = 6$ prove that \[\frac{3a - 4}{a^2 + 4} + \frac{3b - 4}{b^2 + 4} + \frac{3c - 4}{c^2 + 4} \leq \frac{3}{4}.\]

By the way, that's certainly not my best problem, but my other good problems have been posted before here (or some of them will be IMO proposals next year.)

But this inequality is a bit too similar to another inequality which appeared in some olympiad, can't remember which one, about 10 years ago, so it won't make IMO :)

It's not an IMO difficulty problem if I can solve it in under half an hour. :D

I claim that

$\frac {3a-4}{a^2+4}-\frac {a}{4}+\frac {1}{4}\leq 0$

Indeed, it's equivalent to

$\frac {-(x-2)^2(x+3)}{4(x^2+4)}\leq 0$

which is obvious.

But then

$\sum \frac {3a-4}{a^2+4}\leq \sum \frac {a}{4}-\frac {3}{4}=\frac {3}{4}$

as desired.

Of course, this is distilling a bunch of work into a neat solution, so here is the reasoning.  I tried convexity and a bunch of other crap and it didn't work, so I tried to show that

$\frac {3a-4}{a^2+4}\leq \frac {a}{x}+y$ for some $x$ and $y$

Obviously, I needed equality at $a=2$, which gave me $8xy-2x+16=0$.  But I needed a double root at two, so I differentiated the relation with respect to $a$, and set $a=2$ again, to get the relation $4xy-3x+16=0$.  This gave me an $x$ and $y$, and I just needed to check that they worked, which they did, by the above factorization.

[quote="blahblahblah"]It's not an IMO difficulty problem if I can solve it in under half an hour. :D [/quote]

I don't know, I realized recently that IMO problems really don't need to be that difficult - look at Q4 this year, and Q1 and Q4 last year... I mean, they are not trivial but I did each of these problems in less than half an hour. Remember that there are some weak countries at IMO's as well and that we need some problems which are not out of reach for those countries.

[quote="blahblahblah"]
Of course, this is distilling a bunch of work into a neat solution, so here is the reasoning.  I tried convexity and a bunch of other crap and it didn't work, so I tried to show that

$\frac {3a-4}{a^2+4}\leq \frac {a}{x}+y$ for some $x$ and $y$

Obviously, I needed equality at $a=2$, which gave me $8xy-2x+16=0$.  But I needed a double root at two, so I differentiated the relation with respect to $a$, and set $a=2$ again, to get the relation $4xy-3x+16=0$.  This gave me an $x$ and $y$, and I just needed to check that they worked, which they did, by the above factorization.[/quote]

I know, that's the idea the problem is based on, but if you haven't seen this trick before than you might have a hard time. Actually, IF this inequality would have been an IMO problem, I don't think more than half of the contestants would have solved it.

The only thing about the problem is that it does not lend itself to a variety of solutions - I really suspect that you can't solve it by any method except for that one.  Basically, all of the MathLinkers would get 7s, and most of the rest would be $0$s.  :D

Mathlinkers are famous because of their extraordinary abilities to solve inequalities :D

A quote from the report on IMO 2005 by Geoff Smith, leader of the UK team:

[i]The harder questions are selected next, and since there is a good combinatorics question, the obvious thing is to select what has been classified as the hardest algebra question. Many leaders don’t know the technique known as Muirhead’s inequality, and some of those who do seem to think that it is an advanced exotic technique. In fact many of the IMO students are networking to great effect, partly with the help of the twin internet sites Mathlinks and AoPS – the Art of Problem Solving, and they are sharing ideas and techniques independent of their national training regimes. The IMO problem selection committee and jurors will have to monitor these sites to see what ideas the students are sharing. It turns out that Muirhead’s inequality might be better known among the students than parts of the jury.[/i]

The whole report is available at [url]http://www.imo-register.org.uk/2005-report.html[/url]

That's funny.  The markers are just lucky that Muirhead was the most advanced tool used to solve the inequality - marking papers which used 'IPT', 'n-1 EV principle', or 'Vornicu-Schur' would be a real mess.

I've always wondered about that (only in the academic sense, as I'm too old for IMO now and didn't qualify while in high school).  I think that Vornicu-Schur might be alright, because of Valentin, and because it's in his book, and because the proof is quite straightforward.  I don't think the other ones would be alright, because I don't even know if you can find the proofs on ML.  I guess the reality is that the kind of elementary inequalities treated in the IMO are not really a genuine area of mathematical research, and the only places in which you can find some of the techniques proved are in problem books or on ML, and even then you might have difficult finding a given proof in an arbitrary language (for example, has anyone seen a proof of Muirhead's?  Kiran Kedlaya's notes don't contain a full proof, but everyone has tried it for enough small cases to believe it).

Hm, I believe Vornicu-Schur would be fine, too - Valentin is sort of an authority in math olympiad land :)

Another authority is Hojoo Lee. (I guess you know him, he proposed IMO 2001 Q1 and Q2, 2002 Q2, 2004 Q2 and Q4, 2005 Q3) He is writing a set of notes right now, mainly focusing on IMO inequalities!

The set of notes (unfinished) is available at http://my.netian.com/~ideahitme/tin050525.pdf

Note that he proves Muirhead for three variables, but he doesn't prove the general case :D

[quote="Arne"]Imagine someone uses inflection points or the Vornicu inequality (by the way, Valentin was a coordinator for that question 3 at this IMO). I guess the markers wouldn't have a clue what that particular student is talking about.

By the way, you can use such theorems if they are "well-known", but some reference should be available if the theorem used is sort of... sketchy, or let's say, less well known.

Do you think the website would be sufficient as reference? Or should the theorem really be in some kind of publication/book?[/quote]Haha, I really thought about people trying to use Vornicu-Schur inequality ( :D ) or Myth-Vasc inflection points theorems :) I think those papers, if given correct proofs, would have received full marks, especially if they were in my pile :D :D

Can you introduce something about the Myth's Inflection point theorem and Vornicu's inequality?

Shobber - use the search function on this forum!  :o

That's the easiest for everybody here - we don't need to repost it and you don't need to wait until someone makes the post.

By the way, these are advanced techniques, do you know the basics well enough?

I used it but I only found Myth's Inflection Point Theorem. So where is Valentin's?

Here's a nice presentation from Darij: http://www.mathlinks.ro/Forum/viewtopic.php?p=156212#p156212

Thanks for the links.  :)
And yes, Arne is right. I cannot understand a single word about the two theorems.  :(

OK... time limit to solve this one is TEN MINUTES!
First one to solve it gets to post the next one!

ABC is a triangle such that cos(B) = cos(C) = x.  Let f(x) be the following multipart function:
f(x) = -1 if the centroid of ABC lies outside the incircle of ABC
f(x) = 0 if the centroid of ABC lies on the incircle of ABC
f(x) = 1 if the centroid of ABC lies inside the incircle of ABC

If x is a uniformly distributed random real number greater than 0 and less than 1, calculate the average value of f(x).


geometry
analytic geometry
rectangle
geometry unsolved

The following problem was asked by one of my professors at school:

 Given a L-shaped polygon show how to bisect the area of the L-shape using only a straight edge. If the meaning of L-shape is not clear view it as the polygon with the following points as corners (in case of further ambiguity take this as a polyline in the order given): $(0,0), (p,0), (p,1), (1,1), (1, q), (0,q),(0,0)$, $p,q \geq 1$.
 What about the case when the two legs of the L-shape need not have same width?

Let's place the polygon L into Cartesian system of coordinates, such that y-axis of the system is not parallel to any edges of L. Now let's split L with line p parallel to y. As moving p in the direction of  x-axis, the diference $D=S_1-S_2$ between areas of the two parts continuously changes. Because D can obviously be both negative and positive and changes continuously for some position of p holds D=0 and the areas are equal.

The problem asks us to give a "construction" using a straight edge only. It is trivial to see that there exists some line which bisects the area (in fact in any direction, or through any point), the problem is to construct one such  line using a straight edge only.

In a rectangle any line through the center will cut it in 2 equal area parts.

   So if you find centers of the whole piece and the hole separately and connect them

   this will be one of your professor's lines.  There are others also constructible

   by edge and pencil, but this one is the quickest to do.

   BTW, a rectangular hole (or any other divisible in two equal areas by center line)

   could have been  anywhere in your  original  rectangle and at any angle.

   Still, a well balanced coordinate system could be handy sometimes, so keep it in your

   tool-box  just in case.

   Thank you.

   M.T.

That's a  nice solution. By the way the problem according to the prof. was originally posed to him by the renown (Turing award winner) computer scientist Michael O Rabin.


number theory
superior algebra
superior algebra solved

Hello, I am working on a number theory problem, it is as follows:

Show $\sum_{x=0}^{p-1} ((ax+b)|p)=0$ given p does not divide a.

(Note:  $(a|p)$ is the Legendre Symbol)

Now after a bit of unsightly work I managed to have this expressed as a large sum and simply took this expression modulo b and ended up with

$\sum_{a=1}^{p-1} (a|p)$  which we know is zero due to an equal number of
residues and non residues etc.

My instinct, though, says this is not the most elegant method to approach this and I was wondering if anyone has a suggestion for a different path of attack.

Thanks,
JNG

The Legendre symbol $u|p$ depends only on the residue of $u$ mod $p$. $ax+b$ for $0\le x\le p-1$ form a complete set of residues mod $p$. That gets us to your last line; we have only one version of the sum to calculate.


limit
analytic geometry
function
calculus
calculus computations

I am rather puzzled about wht you are asking? If these limits exist, then certainly one can find them by almost trivial means. Use the fact that for any sequence tending to $ (0,0)$ we must have the sequence tending towards the limit. So in the first case just set $ y \equal{} 0$. In the second set $ x \equal{} y$ then take the limit as $ x \to 0$...

[quote="sludgethrower"] If these limits exist[/quote] And have you shown that they do?

Of course I haven't. But I should put on my pure mathematics hat...

The second one is $ < \frac{e^{\minus{}\frac{1}{x^2}} \cdot 3^{\minus{}\frac{2}{y^2}}}{3^{\minus{}\frac{2}{y^2}}} \equal{} e^{\minus{}\frac{1}{x^2}}$

This certainly tends to zero as $ x \to 0$ independently of $ y$.

The first one doesn't actually exist:

To see this we take any coordinates $ (x,0)$ with $ x \to 0$. This would give limit $ 0$.

Now take $ x \to 0$ with coordinates $ (x,e^{ - \frac {1}{x^2}})$. These coordinates would certainly tend to $ (0,0)$. But substituting in we actually get:

${{ \frac {e^{ - \frac {1}{x^2}}y}{e^{ - \frac {2}{x^2}} + y^2}}} = \frac {1}{2}$

So the limit in the two cases are not consistent. Hence the limit cannot exist.

I am sorry to say m2006123, that you were wrong in both cases...

Actually, there are issues of definition here- what do we do where $ x$ is exactly zero in the first, or where either $ x$ or $ y$ is exactly zero in the second? To me it makes the most sense to remove the removable singularity in those exponentials, so that these functions are defined on a punctured neighborhood of zero (and are zero on the axes).

Yes, these questions are usualy given with seperate values allocated to those points where the function cannot be properly defined. In my experience usually such that the function is continuous.


there is a list (and pictures for some of them) of the perfect scorers on the amc

i also believe there might be another thread on how AOPSers did on the aime and a poll.

My name is Peter Ralli, from Lincoln-Sudbury Regional High School in MA.  On the AIME, I managed a four. :(

[quote="juggle7"]I believe this thread was meant for those that scored 150 on AMC.[/quote]

Peter Ralli did score 150 on AMC.

I think Juggle7 was replying to Calculus, not blerpiez.

[quote="juggle7"]I believe this thread was meant for those that scored 150 on AMC.[/quote]

In truth, I don't know what I was thinking.

[quote="Fountainhead"]You might get a few more responses if
A. There were more perfect scorers
B. You didn't title the thread as "Aime scores for genius people". People might be more inclined to post here if the framework of the thread didn't sound so self-congratulatory.[/quote]

i'm sorry it sounded like that, that wasn't what i meant. my own amc score sucked pretty bad. but there were some people i'd heard a lot about, like the mathcounts people and the usamts people and the perfect score people, and i was just curious about their scores.


algebra
function
domain
trigonometry
advanced fields
advanced fields theorems

Solve the following initial value problem. Describe carefully the domain of the solutions:

$ z_x \plus{}z_y \equal{} z \; ; z \equal{}\cos t$  on the initial curve C: x=t, y=0 , $ \minus{}\infty <t < \infty$

Working through the problem a number of times I arrive at:

$ z\equal{}\cos(y\minus{}x)e^{y}$ and the book answer is $ z\equal{}\cos(x\minus{}y)e^{y}$ realizing $ \cos \theta \equal{} \cos\minus{} \theta$ I get the book answer.

However, the first time I worked through the problem I found (as a result of a typo), $ z \equal{}(y\minus{}x)e^y$ and substituted into the PDE I found this satisfies the PDE as well.

The theorem in the text indicates that the solution should be unique. I'm not suggesting the theorem is wrong, I'm wondering what piece I don't understand?
My guess is it is something to do with the initial curve?

Actually any function $ z(x,y)\equal{}f(x\minus{}y)e^{y}$ will satisfy the equation but only the one with $ \cos$ will satisfy the given initial condition as well. Note that the uniqueness theorem doesn't say that the solution of the PDE is unique; it only states that the solution of the initial value problem (which includes both the PDE and the initial condition) is unique.


function
complex analysis
complex analysis unsolved

Let [i]D[/i] ={Z$\in$ $\mathbb{C}$: Re(z) $\gneq$ 0}. For each [i]n[/i]
$\in$ $\mathbb{N}$ define $f_{n}$ : D $\to$ $\mathbb{C}$ by

$f_{n}$ (z) = z+ $\frac{1}{1+nz}$

Find the positiv limit of {$f_{n}$} and show that this function is not continuos at z=0 . Hence or otherwise deduce that{$f_{n}$} is not uniformly convergent.

I presume that "positiv limit" should be "pointwise limit"; call it $f(z)$. We have $f(z)=z$ for any $z\ne 0$, because $1+nz\to\infty$ as $n\to\infty$. Only the point $z=0$ stands out: $f_{n}(0)=1$ for any $n$, hence $f(0)=1$ as well. Thus $f$ is discontinuous at $0$.

Not uniformly convergent... because the difference $f_{n}(z)-f_{n}(0)$ is not very small when $z=1/n$...


The sum of positive real numbers $ a,b,c,d$ is $ 1$. Prove that:

$ \frac{a^2}{a\plus{}b}\plus{}\frac{b^2}{b\plus{}c}\plus{}\frac{c^2}{c\plus{}d}\plus{}\frac{d^2}{d\plus{}a} \ge \frac{1}{2},$

with equality if and only if $ a\equal{}b\equal{}c\equal{}d\equal{}\frac{1}{4}$.

[quote="moldovan"]The sum of positive real numbers $ a,b,c,d$ is $ 1$. Prove that:

$ \frac {a^2}{a \plus{} b} \plus{} \frac {b^2}{b \plus{} c} \plus{} \frac {c^2}{c \plus{} d} \plus{} \frac {d^2}{d \plus{} a} \ge \frac {1}{2},$

with equality if and only if $ a \equal{} b \equal{} c \equal{} d \equal{} \frac {1}{4}$.[/quote]

$ 1\minus{}(\frac{ab}{a\plus{}b}\plus{}\frac{bc}{b\plus{}c}\plus{}\frac{cd}{c\plus{}d}\plus{}\frac{da}{d\plus{}a})\equal{}(a\minus{}\frac{ab}{a\plus{}b})\plus{}(b\minus{}\frac{bc}{b\plus{}c})\plus{}(c\minus{}\frac{cd}{c\plus{}d})\plus{}(d\minus{}\frac{da}{d\plus{}a})\equal{}\frac {a^2}{a \plus{} b} \plus{} \frac {b^2}{b \plus{} c} \plus{} \frac {c^2}{c \plus{} d} \plus{} \frac {d^2}{d \plus{} a} \ge \frac{1}{2}$

$ \Leftrightarrow 1 \ge \frac{2ab}{a\plus{}b}\plus{}\frac{2bc}{b\plus{}c}\plus{}\frac{2cd}{c\plus{}d}\plus{}\frac{2da}{d\plus{}a}$ which is true as by AM-HM,

$ (x\plus{}y)^2 \ge 4xy \Rightarrow 1\equal{}\frac{a\plus{}b}{2}\plus{}\frac{b\plus{}c}{2}\plus{}\frac{c\plus{}d}{2}\plus{}\frac{d\plus{}a}{2}$
$ \ge \frac{2ab}{a\plus{}b}\plus{}\frac{2bc}{b\plus{}c}\plus{}\frac{2cd}{c\plus{}d}\plus{}\frac{2da}{d\plus{}a}$

Equality holds $ \Leftrightarrow a\equal{}b,b\equal{}c,c\equal{}d,d\equal{}a \Leftrightarrow a\equal{}b\equal{}c\equal{}d\equal{}\frac{1}{4}$

well, the form of BCS we use here to prove it is the Andreescu form with equality for;

$ \frac{a}{a\plus{}b}\equal{}\frac{b}{b\plus{}c}\equal{}\frac{c}{c\plus{}d}\equal{}\frac{d}{d\plus{}a} \equal{} \frac{a\plus{}b\plus{}c\plus{}d}{2(a\plus{}b\plus{}c\plus{}d)}\equal{}\frac{1}{2}$

so $ a\plus{}b\equal{}2a$ $ \Leftrightarrow$ $ a\equal{}b$ etc. and we get $ a\equal{}b\equal{}c\equal{}d$, plugging that in the condition we get $ 4a\equal{}1$ and so

$ a\equal{}b\equal{}c\equal{}d\equal{}\frac{1}{4}$

[quote="stephencheng"][quote="crazyfehmy"]It's directly consequence of Cauchy Schwarz inequality.  :wink:[/quote]

Yes , it is . But how to solve for the equality case then.[/quote]
$ \sum a \equal{} 1$
Use the Cauchy-Schwarz inequality
$ \sum \frac{a^2}{a\plus{}b} \ge \frac{ (\sum a)^2}{2 \sum a } \equal{} \frac{ \sum a}{2} \equal{} \frac{1}{2}$

[quote="enigmatic"][quote="stephencheng"][quote="crazyfehmy"]It's directly consequence of Cauchy Schwarz inequality.  :wink:[/quote]

Yes , it is . But how to solve for the equality case then.[/quote]
$ \sum a \equal{} 1$
Use the Cauchy-Schwarz inequality
$ \sum \frac {a^2}{a \plus{} b} \ge \frac { (\sum a)^2}{2 \sum a } \equal{} \frac {\sum a}{2} \equal{} \frac {1}{2}$[/quote]

This is also a part of the question:

[quote="moldovan"]
Prove that:
......
[b]with equality if and only if[/b] $ a\equal{}b\equal{}c\equal{}d\equal{}\frac{1}{4}$
[/quote]

I showed the equality case of his solution in my post above.

Dear friend, if you use an inequality, then you can ALWAYS determine the case of equality. In this case, we use Cauchy in Titu's Lemma form; and equality hods for $ a \equal{} b \equal{} c \equal{} d \equal{} \frac {1}{4}$. :)

if the word "[b]if & only if [/b]" worries you, then first you get that equality holds from Cauchy; and then you [i]insert[/i] the values into the inequality to see if equality holds. :rotfl:

Titu's Lemma gives $\sum \frac{a^2}{a+b} \ge \frac{(a+b+c+d)^2}{2(a+b+c+d)}=\frac{1}{2}$.  Equality holds iff $\frac{a+b}{a}=\frac{b+c}{b}=\frac{c+d}{c}=\frac{a+d}{d} \implies \frac{b}{a}=\frac{c}{b}=\frac{d}{c}=\frac{a}{d}.$ This results in $a^2=b^2=c^2=d^2 \implies a=b=c=d$. Since $a+b+c+d=1$, we know that $ a=b=c=d=\frac{1}{4}$.


Anyone have any more ARML problems they want to post?  I need practice if I wnat to make Penn State next year! :D

You can find all of this year's questions [url=http://www.arml.com/]here[/url].

Here are all of this year's questions with solutions:

http://www.arml.com/2005contest.htm

If you're trying to work your way up to being able to solve ARML problems a good place to start is with AHSME/AMC12 problems. Especially the ones between, say, 13 and 25.

A few practice AMC12 problems can be found here:

http://www.math.uncc.edu/~hbreiter/#American_Mathematics_Competitions

If you have questions the problems, feel free to post them here!

[quote="furious"]here's one:
Determine all integer values of $\theta$ with $0\le\theta\le90$ for which $(\cos\theta+i\sin\theta)^{75}$ is a real number.[/quote]

Yay, first post for Carolina's section for 2006!

$(\cos\theta+i\sin\theta)^{75}=\cos{75\theta}+i\sin{75\theta}$, so we need the $i$ to cancel out, so $\sin{75\theta}=0$, so $75\theta=180$, so $\theta=\frac{12}{5}$

EDIT: Oops..forgot we wanted integer values, so obviously $\theta$ has to be all multiples of $12$ including $0$, so there are a total of $\boxed{8}$ values.


algebra
polynomial
number theory proposed
number theory

Suppse that $n$ be a positive integer. Find the number of integer polynomials $p(x)$ such that the coefficients belong to the set ${\{0,1,2,3}\}$ and $p(2)=n$

[quote="tµtµ"]Before posting my (longuish) solution : is it $[n/2] + 1$ ?[/quote]

I don't think it is true. We can write: $P(x) = x.g(x^2) + h(x^2)$
So $n = 2.g(4) + h(4)$. So there is $[\frac{n-1}{2}] +1$

I had a very similar argument.
But $P(x) = 1$ is the only solution for $n=1$, $P(x)=2$ and $P(x)=x$ are the solutions for $n=2$. Since $[1/2]=0$ and $[2/2]=1$ I'm a bit confused  :o

[quote="tµtµ"]Before posting my (longuish) solution : is it $[n/2] + 1$ ?[/quote]

yes it's true   ;)


Yeah, I'm curious too. For the Eastern team, if that makes any difference.

Ask your Math teacher or Math department head to arrangve for you to tryout.

If your teacher doesn't know anything about ARML, have the teacher contact Martin Badiaon of Canton High School. He's the Coach of Eastern Mass ARML. Practices are held at Canton.

How far away is Canton from the Lunenburg/Fitchburg/Leominster  Worcester County Area?

Lol, wow, I didn't expect it to be that far away.

[quote="panda888"]Doesn't Worcester County participate on the Western Mass team?

Practices for the Western Mass team are at Hudson HS.[/quote]

Yes, it does.