HMMT Princeton college calculus geometry inradius ARML

Good idea? Bad idea? I didn't think it really added anything, honestly I'd rather come in 10th among more intense competition than 1st or 2nd among lesser (by score) teams/individuals because we were deemed unworthy (well ok, we weren't deemed unworthy, but you get my drift) of competing at the next level. On the other hand, we (Lexington) did end up with significantly more awards...in fact we would have won zero in division 1 but won ten (I believe) in DIVISION TWO.

On another note, here is a win solution to Algebra #9.

[hide]Note that 14*4=56. So the whole thing factors as (x^2+ax+4)(x^2+bx+4) for some a and b (they turn out to be 7+/-rt5). So like, if you had to take z1 from one factor and z2 from the other, we would get an extremely ugly answer. So let's take z1 and z2 from the first factor. z1z2=4. That's pretty cool. Then z3z4=4. So z1z2+z3z4=8. There's a 1/3 chance that the answer is right. And it's an integer. Win.[/hide]

[quote="mathemonster"]i thought all A teams were div1? we were[/quote]

Nope. If you look on the website there are results for both Divisions 1/2, Team A/B.

But I thought Divisions 1/2 and Team A/B were an unnecessary confusion. I think whoever was handing out the plaques was also confused: my team placed for Div 1, Team B and we weren't supposed to get a plaque but we got one (but eventually returned it).

[quote="CatalystOfNostalgia"][...] because we were deemed unworthy (well ok, we weren't deemed unworthy, but you get my drift) of competing at the next level.[/quote]

How were divisions 1/2 even determined? I know it was supposed to be by the range that the students come from but honestly (1) how do you find that out (we never reported it and (2) does that really make things more fair? Our team was put into Division 1, most likely because the name of our school ends with "Math and Science Academy", but we don't have any selective admissions, nor is our math team very strong. We only met for a total of ~10 hours since the beginning of the year to work on problems.

Overall I think the idea of the different team contests is well-motivated but I don't see how the divisions thing is a good idea.

I guess it's to give smaller schools a chance to compete against each other...but isn't this the whole point of HMMT November? Like, if a school isn't serious about math, it's less likely they're going to make a trip from a place as close as even New York. For example it has crossed our (Lexington's) minds to send a team to Princeton, but we decided it's not worth it, and I daresay we're a pretty serious team. However, I still thought HMMT was more fun anyway...

Well here is how my solution above to Algebra #9 can be rigorized (sort of).

[hide]So we have our roots, z1, z2, z3, z4. We can do bounding or whatever to show that we indeed have 4 real roots (they probably would have specified complex magnitude or whatever when they said |z1z2+z3z4| anyway). The coefficients are all positive, so they all have to be negative. Then by rearrangement if z1<z2<z3<z4 we want z1, z4 to be paired, z2, z3 to be paired. Note that the 4 roots are going to be like a, 4/a, b, 4/b. wlog a<b, a<4/a, b<4/b. then, in order, they're going to be a, b, 4/b, 4/a. So the answer is 8.[/hide]

[quote="perfect628"]Bad idea, I mean really. If anything Division 2 made the teams/people that won awards feel kind of silly next to those that won in Division 1. Well I would imagine that is so, anyway.

Anyway was it just me or were the subject tests like...not as good as last year? (At least combo+alg, haven't looked in much detail at the others)[/quote]

Yeah, I guess silly kind of sums it up. When the 3rd place score for calculus was 32 and it was 11 in division 2, I was quite amused. However if this is the only year they did it, Lexington survives a hit to its ego because we would have been pretty far off from the top 10 in sweeps (maybe like 15th? but 6th in guts for the second year in a row haha).

As for the subject tests, I guess I kind of agree. There was a bit more computation (though last year had its fair share of computational problems). I thought the algebra test was decent; the geometry problems were pretty good (better than PUMaC...), but in a really weird order. For example I think #7 is the easiest problem on the whole test (probably because it's the only one I can get correct in under 30 minutes...though it did take me like 4 tries to get the computation correct). However there weren't any problems at the whole competition that jumped out as being really cool, like #10 on combinatorics or the tangent problem on algebra last year. And I probably would have liked the team round problems more had it...not been graph theory. Like, if I had wished against one topic, it would have been graph theory. What was with not breaking it up into multiple sets of problems like the last few years?

But overall, awesome competition. The guts round never fails to impress.

Is division 1 supposed to be harder than division 2?

Because:
Division 2 General Subject Test
Rank  Total  Name
1     70     Daesun Yim
2     69     Jason Steinberg
3     35     Christine Yan

I suppose it was a bit unfair also for the team test; we were an alternate team (base line), got shuffled into division B (by virtue of majority of alternates registered under B), and then won the team test. Though our individual scores still counted for division A. Edit: Wait, so I obviously totally misunderstood how the team test works; ignore.

By the way, what were tie-breaking rules? Apparently my geometry answers were exactly the same as the 3rd place individual for geometry, up to zeroes and blanks.

Unfortunately for us, we had a team of 5 people and thanks to the addition of alternates, we were bumped up to division 1. This caused me to go from top 5 (what I probably would have been in div 2) to 97th. I personally don't think that this is fair because we never chose to have extra people on our team, much less get moved to compete with the private schools. I do understand why we were moved (in case we had a super-genius individual added), but this was clearly not the case, so we were severely disadvantaged.

But overall, I agree that the division thing is completely unnecessary. Like someone said, the calculus test was heavily imbalanced between the two divisions, and it was actually quite embarrasing. Individuals' ability, besides, are quite independent of what kind of schools they are from, so there is no reason to segregate them.

[quote="236factorial"]But overall, I agree that the division thing is completely unnecessary. Like someone said, the calculus test was heavily imbalanced between the two divisions, and it was actually quite embarrasing. Individuals' ability, besides, are quite independent of what kind of schools they are from, so there is no reason to segregate them.[/quote]
I agree:  [b]individuals[/b] should only be ranked in aggregate.  It makes no sense that the 11th best student at the meet in most subjects didn't get an award, while the -- 80th?  100th? best student did.

On the other hand, [b]teams[/b] should be split into logical groupings:  public schools (e.g., Lexington, Nashoba, WW-P), magnet schools (TJ, IMSA, Boston Latin), private schools (Exeter, Deerfield, Lawrenceville), and non-school teams (Florida, Michigan, YP, Evil Geniuses).

This solves two issues:  first, it would provide a ranking of schools within each grouping.  My school is private, and I certainly don't mind losing to Exeter or Deerfield or whomever, but we can't reasonably be compared to TJ or AAST or the NYC Math Team.  By segregating teams, publics compete against publics, etc., which is inherently fair.

There should also be a global ranking of teams to declare the overall winner, which addresses Carl's concern.

Maybe team awards would go to top three publics, top three privates, top three magnets, and top ten overall.  My school still won't win a team award in this scheme, but at least we wouldn't end up in 43rd place, behind dozens of teams that have nothing in common with ours.

[quote="jvenezuela716"][quote="azjps"]By the way, what were tie-breaking rules? Apparently my geometry answers were exactly the same as the 3rd place individual for geometry, up to zeroes and blanks.[/quote]

Weren't you guys supposed to do the psychic test when it came to that? From the website it looks like they let the tie stand.[/quote]

They probably did it without me :) Still ridiculously happy with my results.

[hide="Geo 9"]
Suppose the incircle of $ \triangle ABC$ is tangent to $ AB$, $ AC$ at $ D,E$ respectively; let $ AD \equal{} AE \equal{} x$, so $ BC \equal{} 21 \minus{} 2x$ and the inradius is $ \sqrt {16 \minus{} x^2}$. Then $ A \equal{} rs$, $ s \equal{} 21 \minus{} x$ gives
\[ \sqrt {x(21 \minus{} x)(5 \minus{} x)(16 \minus{} x)} \equal{} (21 \minus{} x)\sqrt {16 \minus{} x^2} \Longrightarrow \minus{} 16x \plus{} 21 \cdot 16 \equal{} 80x \Longrightarrow x \equal{} 7/2
\]
[/hide]

Also for algebra 9, after noting we can get $ z_1z_2 \equal{} 4$ and roots are negative, then AM-GM shows its $ \left|z_1z_2 \plus{} \frac{16}{z_1z_2}\right| \ge 8$, another way to guess the right answer.

having all these divisions is a bad idea

did anyone else notice the confusion when they were giving awards?

I personally don't mind divisions, seeing as my school was in D2. It allowed good teams to compete against good teams, and it allowed my school to sweep :P .

All the same, I think partitioning individual students and then giving them the same test is sort of pointless. To the best of my knowledge, D1 and D2 had the same tests - I think the D2 tests should be slightly easier.

Divisions are the best thing that the HMMT people could possibly have done. The people that got awards for getting an 11 on calculus aren't STUPID, they're some of the SMARTEST people that attend a public school. They deserve recognition.

And of course they should give the same individual tests to everyone. If they hadn't, I couldn't compare myself to D1 scores. But since they did, I know that I beat out any of their general test scores.

My school (NORTH) is NEVER going to be better than exeter or tj, but the competition gave us credit for what we've been for a long time: a dominant public school.

Personally, I think that the division system needs a major reworking.  We can't seperate schools by public and magnet: it turns out that some public schools, like Lexington, are insanely competitive, while some magnets and regional teams, like us, aren't so much.

Instead, maybe we could have a AA, A, and B system for the team and sweepstakes awards.  AA would be the best of the best, who have already placed well in previous competitions, like AAST amd TJ.  A would be for the less qualified magnets and regional teams, and B would be for smaller schools.  AA is invitation-only, but all teams get to choose between the A and B divisions.  They would all get different team rounds, and there would be a seperate sweepstakes for each division.  Sound fair(er)?

[quote="sunnyboy780"]Personally, I think that the division system needs a major reworking.  We can't seperate schools by public and magnet: it turns out that some public schools, like Lexington, are insanely competitive, while some magnets and regional teams, like us, aren't so much.[/quote]  If HMMT separated out public and magnet schools, then Lexington would rank as one of the top public schools, and yours wouldn't rank as one of the best magnet schools, in both cases matching reality.  Why separate the divisions any differently?

[quote="sunnyboy780"]Instead, maybe we could have a AA, A, and B system for the team and sweepstakes awards.  AA would be the best of the best, who have already placed well in previous competitions, like AAST amd TJ.  A would be for the less qualified magnets and regional teams, and B would be for smaller schools.  AA is invitation-only, but all teams get to choose between the A and B divisions.  They would all get different team rounds, and there would be a seperate sweepstakes for each division.  Sound fair(er)?[/quote]

If all the good teams were secluded into a separate AA division, then what would be the point of winning the A or B division? You would be the "best out of the not-as-good" which is essentially meaningless.

[quote="236factorial"]

If all the good teams were secluded into a separate AA division, then what would be the point of winning the A or B division? You would be the "best out of the not-as-good" which is essentially meaningless.[/quote]

A team that would win in that new A division wouldn't even have a chance at winning in the current A division, so at least they're getting something...

On the other hand, we could just scrap this division thing entirely and have a unified team round, with both easy and extremely challenging questions (like the ARML power round).  That way, no one will feel like they're the "best of the losers."  To compensate, HMMT could expand its November tournament, so anyone that feels as if the new system is too hard could go there.

I was on a frosh/prefrosh team; regardless of the fact that we come from a magnet school, we were in no way on the same tier as the magnet schools' top teams (including our own.) Divisions don't take this into account.

I don't really like the divisions system. It's got too many flaws.

[quote="Zhero"]I don't really like the divisions system. It's got too many flaws.[/quote]

Exactly. We should just compete. It's that simple.

[quote="sunnyboy780"]A team that would win in that new A division wouldn't even have a chance at winning in the current A division, so at least they're getting something...[/quote]I know local math leagues often have multiple divisions so the not-very-good schools can win team awards every now and then, but HMMT shouldn't be handing out awards to teams or individuals that have done worse than those leaving empty-handed (without good reason).

I am perfectly content to watch my team be pounded by Exeter.  They are a prep school, we are a prep school, and they are better -- so they [i]should[/i] be ranked above us.  Why would I want Exeter moved to a different division?  On the other hand, amalgamated teams are a very different beast than all-one-school teams.

In reference to the immediately preceding post, I strongly agree that everyone should be ranked together for sweeps (individuals and teams).  The only difference I suggest is that HMMT should provide awards to the top schools (not individuals!) of certain types:  public, magnet, private.  Merge magnets and privates if you want.  Under this year's system, public schools were isolated, so we found out that WW-P > Lexington > Belmont ....  But there's no need to isolate the publics to create this ranking.  Merge them into sweeps, but also give separate team awards.

It's easy enough to do.  For example, this year the top ten private schools were...

1. Phillips Exeter [3] -- 1333 points
2. Phillips Andover [8] -- 1119 points
3. Hopkins School [16] -- 865 points
4. Deerfield Academy [23] -- 796 points
5. Wheeler School [32] -- 608 points
6. Lawrenceville School [33] -- 607 points
7. Roxbury Latin [35] -- 590 points
8. Hotchkiss School [40] -- 474 points
9. Worcester Academy [43] -- 457 points
10. Portsmouth Abbey School [49] -- 417 points

So maybe give an award to PEA, PA, and Hopkins; WW-P, Lexington, and Belmont; and TJ, AAST, and Montgomery Blair.  Nine more team awards, 60 fewer individual awards.  That'll speed things up.

Rick

I guess the only problem with that is that it's kind of complicated. Like, quite a few people were already really confused about like, division 1, team test B versus division 2, team test A, etc. I think individuals should be ranked overall, not by division. As for dividing teams into divisions, it's a decent idea on the surface, but I don't know about non-flexibility (not being able to choose your division). Public schools should probably be given the choice to enter division 1 or 2, though I don't know what this would mean in terms of overcrowding division 1; there were already significantly more teams there. But like I said before, isn't the whole point of the November Tournament for the smaller schools/teams to compete at a less intense/easier level? And again, small public schools that don't take math competitions as seriously are less likely to come all the way here.

As for the AA, A, B idea, with AA being invitational, I think this would give rise to much controversy in that there is no good way to decide who gets invited into AA. And teams would complain, and people would have connections with the coordinators (I know Lexington does, we have several alumni who play some role in organizing the competition), etc.

Yay Hopkins with 7 people and less than 500 high schoolers... Is it fair that a private day school with <500 high schoolers has to compete with boarding schools though?


The initial exam looks easy enough, but since only 200 kids advance what is the cut off? Maximum one question wrong?

[quote="durt"]You're talking about USAPHO, right? The cutoff is actually suprisingly low. This year it was somewhere in the low to mid 60's (out of 150 possible points!).[/quote]

First of all it's out of 140.  Secondly, remember this year's was the hardest so far.


;)  i know it's gonna be so simple, but it's driving me mad!

ab + ac + cd = ea

each letter stands for a different digit.
work out which digit stands for which letter.

CAN ANYONE HELP ME PURLEASE!

Billy

wait i dunno how to solve these kinds of probs without having a number or something....

The problem is to figure out what the numbers are.

For example, it could be a = 4, b = 3, c = 2, d = 1, e = 7, but that would give:

43 + 42 = 74

Well, that's not true, so that's clearly not the right answer!  See if you can figure out the puzzle!

Since 3 two digit numbers must equal another two digit number the number is 3, 2, or 1.  Then you use guess and check and modify when you need to.
[hide]24+23+35=82
         ab+ac+cd=ea[/hide]

elvenchamp777

but just couldn't get it together, now i see

miss 1 for a taking you to 2, double up for b = 4, then step back to odds c being 3 and d being 5.

so easy when you know, thanks so much! :)

Can I just point out that there are 42 solutions to this problem (42 being the meaning of life, etc..) :)

yeah..

There are definitely more solutions to this

(Theres not really much other than trial and error. I cheated - I used a computer :))

TripleM

did u program your computer to find out the answer?

[hide]ab+ac+cd=ea
This can be done without too much guess and check.
 ab
 ac
 cd
-------
 ea

Immediately, some things are clear.  A cannot equal 5 or higher.  However, this isn't enough.

10a+b+10a+c+10c+d=10e+a; 19a+11c+b+d=10e
Now, we solve for 10e.  We know that we can carry over at most 20, but we know that at most because the digits must be distinct, b+c+d=20 to 24.  24 can never work because 4+4+2 in the tens digit will create a hundreds digit.  23 can't work because 3+3+2=8 in the tens digit forces c to equal 1.  That doesn't work because 1+b+d at most only equals eighteen.  22 won't work because 2+2+2=6 which forces c=1 or 3.  Regardless of which number you put, b+d+c won't ever equal 22.  20 won't work because you can't have a digit number with 0 in the tens place.  21 however, will seem to work when c=4 or 5 which forces b and d which can have interchangable digits of 9,8 for 4 and 9,7 for 5.  However, it doesn't because if c=4, e=8 and if c=5, e=9, so it really doesn't work.  Now, we substitute 10e for 20a+10c+10 into the other equation as well as 20a+10c for 10e.  We find that the latter equation doesn't work because we get -a+b+c+d=0.  That's only true if a is at least 6 which would create a hundreds digit.  Therefore, we must use the other equation and we get c+b+d=10+a where all variables are distinct digits.  Here's one out of many answers 23+24+45=92.[/hide]

How did you program it?

did you use C?

or JAVA?

or should I just shut up?

here's another riddle (really easy solution... I got it pretty quickly)
A mathematician is walking along a farm road. He comes up to a house and starts talking to a farmer who is sitting on the porch. Later in the conversation the mathematician asks: "So, do you have any kids?" The farmer replies: "Yes, I have three in fact. The sum of their ages is my house number and the product of their ages is 72."
Knowing this the mathematician could not figure out the ages of the farmer's three kids. So he tells the farmer, who replies: "[i]By the way, my oldest one loves chocalate pudding[/i]" The mathematician now figured out the ages of the three kids... what are they?

Are you sure the solution is unique? I seem to have gotten two:

[hide]2,4,9
3,4,6[/hide]

EFuzzy, if you list out all of the possible ages, there are a LOT more than 2.

HINT:
[hide]Look for a relation between some of the possibilities of the children's ages[/hide]

I had this question and an MAO Invitational on a team round, we were the only team to get it in the fourth minute, and that made us tie for first...(IF WE HAD ONLY WRITTEN DOWN THE RIGHT ANSWER I HAD TOLD THEM FOR THE FIRST ONE!)

It is a little difficult at first..But the answer is really quite simple.

[i]"By the way, my oldest one loves chocalate pudding"[/i]

[hide="Hint 1"]Why do you think stating the oldest one would make a difference? What could possibly make that cause all the difference?
[/hide]

[hide="Hint 1 Exploited"]The only reason it should make a difference is if there were a set of twins...
[/hide]

[hide="Continued Explanation and Answer"]So we now transform $xyz=72$ to $xy^2=72$ So lets start listing out all possible values for $x$ and $y$, and let's find the sum (so that we can figure the house number).
The only two that exist are:
2, 6, 6 = 14
8, 3, 3 = 14

BUT WAIT!  Somethings fishy...

The sums are exactly the same...(Meaning the house number doesn't help to much) What did we miss?  remember how I said that there are two twins, and pointing out the oldest one gave it away?  If you look through all the values  (even all of the possible x, y, z)  you will see that this is the only moment in which the mathematician has to question, because the house number [i]didn't[/i] give it away, then when the farmer denotes the oldest child, the mathematician figures out the childrens ages.

The only case that there is an oldest child is when the ages are $\fbox{8,3,3}$.[/hide]

great solution... never thought of it done with saying: xyz=xy^2=72
here's the solution that i got:
[hide]call ages x,y, and z
so x+y+z = some house number and xyz=72

the line about the mathematician not knowing the ages with only this is key: that means that there must be two (or more) sets of numbers that have the same sum... so we could either list them out, or use other methods, but we eventually get the combinations:
6,6,2 and 8,3,3.
the reason mathematician couldn't figure the ages out was that he had to pairs of numbers which had same sum and product of 72. the line about the oldest one liking chocolate pudding means that there is only one child with largest age... so answer is 8,3,3[/hide]

If you have a moldy computer like mine, you could use QuickBasic to program it...


For every triangle ABC and every point N inside triangle ABC, there exists an affine transformation mapping the triangle ABC to a triangle A'B'C' and the point N to a point N' such that the point N' is the orthocenter of triangle A'B'C'.

$ AN$ cuts $ BC$ at $ D$. Let $ n$ be perpendicular to $ BC$ at $ D$ and $ A^* \not\equiv D$ arbitrary point on $ n.$ Parallel to $ AA^*$ through $ N$ cuts $ n$ at $ N^*.$ Transformation $ \mathcal T_1: (ABCN) \longrightarrow (A^*BCN^*)$ is affine. Let $ CN^*$ cut $ A^*B$ at $ F^*$ and let $ (M)$ be circle with diameter $ BC.$ Perpendicular to $ BC$ through $ F^*$ cuts $ (M)$ at $ F'$ on the same side of $ BC$ as $ A^*.$ $ BF', CF'$ cut $ A^*D$ at $ A', N'.$ Transformation $ \mathcal T_2: (A^*BCN^*) \longrightarrow (A'BCN')$ is also affine and $ N'$ is orthocenter of $ \triangle A'BC.$ Affine transformations form a group $ \Longrightarrow$ transformation $ \mathcal T \equal{} \mathcal T_2 \circ \mathcal T_1$ is also affine.


Nice inequality!

We have \[\frac{a}{b+c}-\frac{(b-c)^{2}}{4(b+c)^{2}}=\frac{4ab+4ac-b^{2}-c^{2}+2bc}{4(b+c)^{2}}= \frac{ab+bc+ca}{(b+c)^{2}}-\frac{1}{4}\] Hence we can rewrite the inequality to the form \[\sum_{cyc}\frac{ab+bc+ca}{(b+c)^{2}}\ge \frac{3}{2}+\frac{3}{4}=\frac{9}{4}\] which is the same as \[(ab+bc+ca)(\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}})\ge \frac{9}{4}\] which is the famous Iran 1996 inequality (see http://www.mathlinks.ro/Forum/viewtopic.php?&t=3547).


algebra
polynomial
function
complex analysis

Suppose that $ p(z)$ is a polynomial of degree $ n\geq 1$.
How will you show that there is $ R>0$ such that $ p$ never vanishes in $ \{z: |z|>R\}?$

Pick an $ R$ big enough so that the terms of $ p$ of degree less than $ n$ are trumped by the degree $ n$ term. Let $ g(z)$ be the lower order terms of $ p$ (i.e. everything except $ az^n$). Then on the circle of radius $ R$, $ |p(z)| > |g(z)|$ because $ z^n$ trumps the lower order terms. By rouche's theorem $ p(z)$ and $ z^n$ have the same number of zeros inside the circle of radius $ R$, so that means there are $ n$ zeros in there. But by the fundamental theorem of algebra, thats ALL the zeros of $ p(z)$, so outside the circle of radius $ R$ there are no zeros.

That's a better way to compute $ R$, but you can just see this by noting that a degree $ n$ polynomial has at most $ n$ roots in any field, so $ R$ is trivially obtained as the maximum modulus taken over these finitely many roots.

Straight away you can contradict that the number of roots of a polynomial are finite hence they are bounded.the logic is same as Nukular but just framed differently


This is from the 2004 AMC 10 B.  # 13.

In the U.S, coins have the following thickness: penny, 1.55 mm; nickel, 1.95 mm; dime, 1.35mm; quarter, 1.75mm.  If a stack of these coins is exactly 14 mm high, how many coins are in the stack?

The solution says that the height in milimeter of any stack with an odd number of coins has a 5 in the hundreth place.  and that in order for a zero to be in the hundreds and the tenths place, the the number of coins must be a multiple of four.  Why does it have to be a multiple of four?
 :huh:
thanks    :)

Look at the decimal part of each number, ignoring the 1 in front of it.

$35\equiv15\text{ (mod 20)}$
$55\equiv15\text{ (mod 20)}$
$75\equiv15\text{ (mod 20)}$
$95\equiv15\text{ (mod 20)}$

If you multiply each by only 2, you come up with
$10\text{ (mod 20)}$, which will not create a whole dollar amount.  It will create an amout that is an even 10 cent, though.

$15\text{ (mod 20)}\cdot4\equiv60\text{ (mod 20)}\equiv0\text{ (mod 20)}\equiv0\text{ (mod 100)}$

I hope this helps you can see that you must have a multiple of 4.

[quote="myc"]what are modes?

Thanks  :)  :lol:[/quote]

Mods are basically remainders, except for negative numbers.

$N\equiv x\text{ (mod }y)$

Means that $y$ is an integer factor of $N-x$.


real analysis
calculus
integration
function
real analysis theorems

i need to answer the following, with examples

1- is any function that can be intefrable in the sence of riemann be lebesgue integrable (please explain why f(x)
must be nonnegative)

2- can you give me any unbounded function which lebesgue integrable

3- how can i determne whether the the function lebesgue integrable or not (bounded, contineous, monotone, .....)

4- what about functions which ultering its sign in the interval

1. Yes

3. By Caratheodory's criterion

4. Write $f$ as $f: =g-h$ where $g,h>0$.

1. I don't really understand the question - or at least, why you said anything about nonnegative.

2. Plenty of examples. $f(x)=\frac1{\sqrt{x}}$ on $(0,1),$ zero otherwise, would work.

3. and 4. Assuming measurability (and face it, you're just not going to construct a nonmeasurable function by accident), Lebesgue integrability is [b]absolute[/b] integrability. Think back to your calculus courses about all of the ways you determined the absolute convergence of a series. (Comparison, comparison, comparison - that's how you argue.)


I wanted to come back as captain but they wouldn't let me.


abstract algebra
geometry
geometric transformation
reflection
rotation
group theory
superior algebra

Is the following argument correct, thanks

I want to find 3 non-isomorphic groups of order 42.

1) C_42 th cyclic group of order 42 is  cyclic hence abelian.
2) D_21 Dihedral group of order 42, non-abelain.
3) C_3 \times D_7

C_42 is non-isomorphic to the others, since it is
abelian and the others are non-abelain.

Remains to show that D_21 and C_3 \times D_7 are non-isomorphic.

D_21 consists of 21 reflections of order 2 and 21 rotations of the form D^n.
The possible orders for the rotations are divisors of 21, i.e, 1,3,7, or 21.
So this group has no element of order 6.
Now consider C_3 \times D_7 , C_3 has a generator c which has order 3, and D_7 has a reflection S of order 2, so in C_3 \times D_7 there is an element (c,S) of order lcm(3,2)=6, hence the two groups are non-isomorphic.

That looks correct. You could also point out that $ D_{21}$ has $ 21$ elements of order $ 2$, whereas $ C_3 \times D_7$ only has $ 7$ such elements.


geometry circumcircle conics ellipse function trigonometry rectangle

[color=darkred]See http://www.mathlinks.ro/Forum/resources.php?c=60&cid=82&year=2006&sid=7bdfe1f609627201116641a1badef7b7
[b](France, TST - 2006, the first day, the first problem)[/b].

Let $ABCD$ be a square. Denote the intersection $O\in AC\cap BD$. Exists a positive number $k$ so that for any point $M\in [OC]$ there is a point $N\in [OD]$ so that $AM\cdot BN=k^{2}$. Ascertain the [u]geometrical locus[/u] of the intersection $L\in AN\cap BM\ .$[/color]

[quote="Ashegh"]is $O$, the circumcenter of square?[/quote]
I believe so.

It is no circumcenter, but circumcircle of the square !

Sorry, the point $O$ is the center of the square, $O\in AC\cap BD$.

i only find that the locus isnt line.

i hope it is circle, but i dont think so. :wink:

finally it is known that the locus is none of the line and circle.

it is a part of a circle and an ellipse and ...

[color=darkblue]I posted this "surprising" jewel-problem at [size=117][b]Jul 20, 2006.[/b][/size] Here is the my proof, [u]my dear Ashegh, Shobber and Amir.S !.[/u]

Denote w.l.o.g. $AB=1$. Let $f: [OC]\rightarrow [OD]$ be the given function, i.e. for any $M\in [OC]$ there is a unique point $N\in [OD]$ so that $AM\cdot BN=k^{2}\ .$ Particularly, $f(O)=U\ ,\ f(C)=V$ $\Longrightarrow$ $AO\cdot BU=AC\cdot BV=k^{2}$ $\Longrightarrow$ $BO\le BU=2\cdot BV\ ,\ \{U,V\}\subset [OD]$. But $BU\le BD=2\cdot BO\le 2\cdot BV$.
Therefore, $BU=BD$ and $BO=BV$, i.e. $U\equiv D$, $V\equiv O$ and $\boxed{\ k=1\ }\ .$ [b]Here is amazingly the all "background" of this problem ![/b]

$1.\blacktriangleright$ [b][u]The analytical method.[/u][/b] $A(0,0)\ ,\ B(0,1)\ ,\ C(1,1)\ ,\ D(1,0)\ ;\ M(m,m)\in [OC]\ ,\ N(n,1-n)\in [OD]$, where $\{m,n\}\subset \left[\frac{1}{2},1\right]\ .$
$AM\cdot BN=1\Longleftrightarrow \boxed{\ 2mn=1\ }\ .$ The point $L(x,y)$ is intersection of the lines $AN$, $BM$ with the equations $(n-1)x+ny=0$, $(m-1)x-my+m=0$ respectively. Thus, $m=\frac{x}{1+x-y}\ ,\ n=\frac{x}{x+y}\ .$ Therefore, $2mn=1\Longleftrightarrow$ $2x^{2}=(x+y)(1+x-y)$ $\Longrightarrow$ $x^{2}+y^{2}=x+y$ $\Longrightarrow$ $\boxed{\ \left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt 2}{2}\right)^{2}\ }$, i.e. [b]the geometrical locus of the point $L$ is the little $\mathrm{arcCD}$ of the circumcircle of the square[/b] $ABCD\ .$

$2.\blacktriangleright$ [b][u]The synthetical method.[/u][/b] $AM\cdot BN=AB^{2}$ $\Longrightarrow$ $\frac{AM}{BA}=\frac{BA}{NB}\ ,\ \widehat{BAM}\equiv\widehat{NBA}\ \ \mathrm{(S.A.S.)}$ $\Longrightarrow$ $\triangle AMB\sim \triangle BAN$ $\Longrightarrow$ $m(\widehat{ABL})+m(\widehat{BAL})=m(\widehat{ABM})+m(\widehat{BAN})=m(\widehat{BNA})+m(\widehat{BAN})=135^{\circ}$ $\Longrightarrow$ $m(\widehat{ALB})=45^{\circ}$ a.s.o.

$3.\blacktriangleright$ [b][u]The trigonometrical method.[/u][/b] Suppose again $AB=1$ and denote $m(\widehat{OBM})=x$, $m(\widehat{OAN})=y$, where $\{x,y\}\subset \left[0,\frac{\pi}{4}\right]\ .$
$\frac{AM}{\sin (\widehat{ABM})}=\frac{AB}{\sin (\widehat{AMB})}$ $\Longrightarrow$ $AM=\frac{\sin(45+x)}{\cos x}\ ;\ \frac{BN}{\sin (\widehat{BAN})}=\frac{AB}{\sin(\widehat{ANB})}$ $\Longrightarrow$ $BN=\frac{\sin (45+y)}{\cos y}\ .$
$AM\cdot BN=1\Longrightarrow$ $\sin(45+x)\sin (45+y)=\cos x\cos y$ $\Longrightarrow$ $\cos (x-y)-\cos (90+x+y)=\cos (x-y)+\cox (x+y)$ $\Longrightarrow$ $x+y=45$ $\Longrightarrow$ $m(\widehat{ALB})=45$ a.s.o.

[b]Commentary.[/b] Denote the intersections $R\in CD\cap BM$ and $S\in CD\cap AN\ .$
$M\in [OC]$ $\Longrightarrow$ $MB=MD$ $\Longrightarrow$ $\widehat{SDM}\equiv\widehat{CBM}\equiv\widehat{MAS}$  $\Longrightarrow$ The quadrilateral $AMSD$ is inscribed  $\Longrightarrow$ $SM\perp AM$ $\Longrightarrow$ $\boxed{\ SM\parallel BD\ }\ .$
$N\in [OD]$ $\Longrightarrow$ $NA=NC$ $\Longrightarrow$ $\widehat{RCN}\equiv\widehat{DAL}\equiv\widehat{NBR}$ $\Longrightarrow$ $\Longrightarrow$ the quadrilateral $BNCR$ is inscribed $\Longrightarrow$ $RN\perp BD$ $\Longrightarrow$ $\boxed{\ RN\parallel BC\ }\ .$
$SM\parallel BD\ ,\ RN\parallel BC\Longrightarrow$ $MS\perp RN$, i.e. the quadrilateral $MRSN$ is orthodiagonally. Therefore, obtain the reference problem ([b]France, TST - 2006[/b]).[/color]

[color=darkred][b]Generalization.[/b] Let $ABCD$ be a rectangle. Denote the intersection $O\in AC\cap BD\ .$ Exists a positive real number $k$ so that for any point $M\in [OC]$ there is a point $N\in[OD]$ such that $AM\cdot BN=k^{2}\ .$ Ascertain the geometrical locus of the intersection $L\in AN\cap BM\ .$[/color]
[color=darkblue][b]Answer.[/b] The geometrical locus of the point $L$ is the little $\mathrm{arcCD}$ of the circumelypse for the rectangle $ABCD\ .$ If $A(0,0)$, $B(0,a)$, $C(b,a)$, $D(b,0)$, then the equation of the elypse is $\boxed{\ a^{2}x^{2}+b^{2}y^{2}-ab(ax+by)=0\ }$ (see the analytical method for the solution of the initial proposed problem). [/color]


Given a quadrilateral and a circle inscribed in it. Prove that the center of the circle and the 2 midpoints of 2 diagonal lines are collinear.

Let $ ABCD$ be the quadrilateral $ M,N$ the midlles of the diagonals $ AC$ and $ BD$, and let $ O$ be the center of the inscribed circle.
There are 3 hints(hope to be usefull)
[hide="1."]In any quadrilateral $ ABCD$ that is not a parallelogram , if $ O'$ lies on $ MN$ prove that:
$ [AO'B] \plus{} [CO'D] \equal{} [AO'D] \plus{}[BO'C]$[/hide]
[hide="2."]In a circumscribed quadrilateral $ ABCD$ if $ O$ is the center of the circle inscribed prove that:
$ [AOB] \plus{} [COD] \equal{} [AOD] \plus{} [BOC]$[/hide]
[hide="3."]From (1) and (2), $ O$ lies on $ MN$.[/hide]

PS: This is called Newton's line
PSS: Guess who's the Newton's point  :D


inequalities AMC USA(J)MO USAMO \/closed

[quote="Rep123max"]So is there definetly not going to be an Inequality course this summer, or is it up in the air that there might be one?[/quote]

It is not likely that we'll be offering that course this summer.  It would take significant student demand and you're the first to ask about it.

In that case, would it be offered during the fall? If, so I plan on enrolling...

Unfortunately we probably won't offer Olympiad Inequalities again until December or January.  There doesn't seem to be enough demand to warrant holding the class more than once a year at this point.  But at least that will give students time to take it before the 2006 USAMO (or most other Olympiads).

Oh okay, then i'll just wait until December or January :D


If $ x,y,z \in \mathbb{Z}$ then $ gcd(x,y,z)|gcd(x\minus{}y,y\minus{}z,z\minus{}x)$

This is a trivial consequence of [url=http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity]Bezout's lemma[/url].  That is:

- $ \gcd(a, b, c)$ is the smallest positive member of $ G(a, b, c) \equal{} \{ ax \plus{} by \plus{} cz : x, y, z \in \mathbb{Z} \}$,
- $ \gcd(a, b, c)$ divides every member of $ G(a, b, c)$, and
- Every element of $ G(x \minus{} y, y \minus{} z, z \minus{} x)$ is an element of $ G(x, y, z)$.


[hide]Let $ a_i \equal{} \minus{} n$ if $ 3 \nmid n$ and $ a_i \equal{} 2n \plus{} 1$ if $ 3 \mid n$. Then the sum of every 3 consecutive terms is $ 1$, and the total sum is $ 6(2n \plus{} 1) \minus{} 14n \equal{} 6 \minus{} 2n$. So if $ n > 3$, the second condition is satisfied.[/hide]


Let's see who can solve this one.

A train starts a trip at a uniform rate but after 1 hour must reduce its rate to $\frac 23$ of the original rate. It arrives at its destination 37 minutes late. If it could have traveled an additional 10 miles before reducing its rate it would have been only 32 minutes late. What was the original rate?

Since nobody's really replying...
[hide="Hint"]At a certain point, one train goes 10 miles five minutes faster than the other...[/hide]
Though that probably gave the whole thing away...
If nobody posts a solution, I will, so at least everybody will know 1 way to do the problem.

[hide] Make two equations. The first, in the case that the train is late 37 min., is r+(t-1)(2r/3)+(37/60)(2r/3)=tr, where t=time and r=rate of the train. This simplifies to r+(t-(23/60))(2r/3)=tr, which is also 1+(t-(23/60))(2/3)=t.
The second equation is r+10+(t-(28/60))(2r/3)=tr.
Solving for t in the first equation gives t=(67/30) hours, or 134 min.
Plug t in the second equation, which gives you r=180.

Answer: 180 mph [/hide]

My answer doesn't agree, but I will post my solution just in case you are right.
[hide]Basically it's telling you that in Case 2, the train went 10 miles 5 minutes faster than in Case 1. Since time is equal to distance over rate, we right an equation: $\frac {10}{\frac {2}{3}r} - \frac {10}{r} = \frac 1{12}$. Solving, we get $\fbox {r = 60}$.[/hide]


Hey,I will give you the first one.
$x,y,z>0$Prove:
$\frac{(y+z)(z+x)(x+y)}{xyz}\ge 8+72\sum_{cyc}\frac{(y-z)^{2}}{(7x+2y+2z)^{2}}$
And guess when will the equality happen :D (Try $x=9,y=2,z=2$)
This means Sept 2 :P  :D

(It is like the VASC's one)

I am not so experienced so probably my problems are not the most interesting, but at least the spirit is the same  :lol: :

$\frac{a^{3}+b^{3}}{c(a+b)^{2}}+\frac{b^{3}+c^{3}}{a(b+c)^{2}}+\frac{c^{3}+a^{3}}{b(a+c)^{2}}\ge \frac{3}{2}$

edit: sorry need to fix second one :P

[quote="Xevarion"]I am not so experienced so probably my problems are not the most interesting, but at least the spirit is the same  :lol: :

$\frac{a^{3}+b^{3}}{c(a+b)^{2}}+\frac{b^{3}+c^{3}}{a(b+c)^{2}}+\frac{c^{3}+a^{3}}{b(a+c)^{2}}\ge \frac{3}{2}$

[b] Solution. [/b]

  From the well-known inequality \[a^{3}+b^{3}\geq ab(a+b)\] we only need to prove that \[\frac{ab}{c(a+b)}+\frac{bc}{a(b+c)}+\frac{ca}{b(c+a)}\geq\frac{3}{2}.\] Now, by setting $x=bc, y=ca$ and $z=ab$ the inequality becomes \[\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\geq\frac{3}{2}.\]

[quote="Xevarion"]edit: sorry need to fix second one :P[/quote]

$(a+b+c)^{3}+3\sqrt{3}\left(\sqrt{a^{2}+b^{2}+c^{2}}\right)^{3}\ge (1+3\sqrt{3})(a^{3}+b^{3}+c^{3})+(51-9\sqrt{3})abc$

:D

And another ;) (possibly this is known before me)

$a+b+c \ge \sqrt{\frac{2a^{2}+2b^{2}+2c^{2}}{3}}+(3-\sqrt{2})\sqrt[3]{abc}$

[quote="pvthuan"]Sept 2, 1945 is Vietnamese independence day. Please, Vietnamese guys and foreign friends propose some inequalities to celebrate our national day.

 :lol:[/quote]
Do you have any one?
Yes. This is my problems but I think is trival for you.
Let a,b,c be the lenght of three side in a triangle. Prove that
$\sum_{cyc}\left( \frac{a+b-c}{c}\right)(a^{2}+b^{2}) \ge \sum_{cyc}ab \left( \frac{a+b}{c}\right)$
Viet Nam ! Ho Chi Minh ! Forever!

Come on, boys. I have NOT seen any REMARKABLE and ORIGINAL inequalities for this big event.

 :lol:

With all of my pride, I post this problem and shout loudly that I am a Vietnamese!  :roll:
Here is my problem:
Let a,b be positive number and $a+2b=3$.
Find the minimum value of the following expression:
$A=\frac{1}{a^{2}}+\frac{2}{b^{2}}+8ab+4b^{2}$

It is not a difficult problem with all of you, but it have many meaning on today. And I am very glad to see a so nice answer!  :D
So, let try it~!

Here is a nice inequality created today - the national independence day:

Let $a,b,c,d$ be nonegative real numbers such that \[a^{2}+b^{2}+c^{2}+d^{2}=1,\] prove that \[\frac{abc}{1-abc}+\frac{bcd}{1-bcd}+\frac{cda}{1-cda}+\frac{dab}{1-dab}\leq\frac47.\]

The following stronger inequalities hold for $a^{2}+b^{2}+c^{2}+d^{2}=1$: \[\frac{abc}{1-3abc}+\frac{bcd}{1-3bcd}+\frac{cda}{1-3cda}+\frac{dab}{1-3dab}\leq\frac 4{5}\] and still \[\frac{abc}{1-4abc}+\frac{bcd}{1-4bcd}+\frac{cda}{1-4cda}+\frac{dab}{1-4dab}\leq\ 1.\]

Only for the second, because it is very strong.

I use  lemma with the help of Zhaobin for the first one.


calculus
derivative
function
real analysis
real analysis unsolved

Let  $ f$ be a real function, three time differentiable on $ [\minus{}1;1]$, such that $ f(\minus{}1)\equal{}f(0)\equal{}0$, $ f(1)\equal{}1$, $ f'(0)\equal{}0$. Show that $ f'''(x) > 3$ for some $ x \in (\minus{}1;1)$.

You've probably meant that there exists point $ x$ where $ f'''(x)\ge 3$, otherwise $ f(x) \equal{} \frac {x^2}{2} \plus{} \frac {x^3}{2}$ provides a counterexample. Anyway, just consider Taylor expansion of the function $ f$ at the point $ 0$.
You'll get that $ f(x) \equal{} \frac {x^2 f''(0)}{2} \plus{} \frac {x^3 f'''(a(x))}{6}$. Now taking $ x \equal{} 1$ and $ x \equal{} \minus{} 1$ in last formula we can easily get that $ f'''(a(1)) \plus{} f'''(a( \minus{} 1)) \equal{} 6$ and we are done :)


calculus
integration
function
geometry
calculus computations

A cubic function $ y \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d\ (a\neq 0)$ touches a line $ y \equal{} px \plus{} q$ at $ x \equal{} \alpha$ and intersects $ x \equal{} \beta \ (\alpha \neq \beta)$.
Find the area of the region bounded by these graphs in terms of $ a,\ \alpha ,\  \beta$.

[hide="is it....."]$ |(\beta\minus{}\alpha)^4(\frac{a}{4}\minus{}\frac{\alpha}{3})|$ ? [/hide]

[quote="kunny"]incorrect. :([/quote]
let us assume $ \alpha<\beta$

$ A\equal{}\bigg|\int_\alpha^\beta\;(ax^3\plus{}bx^2\plus{}cx\plus{}d\minus{}px\minus{}q)\;dx\bigg|$
$ \equal{}\bigg|\int_\alpha^\beta\;a(x\minus{}\alpha)^2(x\minus{}\beta)\;dx\bigg|$
[since the curve touches the line at $ x\equal{}\alpha$]

$ A\equal{}\bigg|\int_\alpha^\beta\;a(x^3\minus{}(2\alpha\plus{}\beta)\,x^2\plus{}(\alpha^2\plus{}2\alpha\beta)\,x\minus{}\alpha^2\beta)\;dx\bigg|$
$ \equal{}\bigg|a\,\left(\frac{x^4}{4}\minus{}(2\alpha\plus{}\beta)\,\frac{x^3}{3}\plus{}(\alpha^2\plus{}2\alpha\beta)\,\frac{x^2}{2}\minus{}\alpha^2\beta\,x\right)\bigg|_\alpha^\beta\;\bigg|$
$ \equal{}\frac{|a|}{12}(\beta\minus{}\alpha)^4$

That's right.  :)

$ \int_{\alpha}^{\beta } (x \minus{} \alpha )^2(x \minus{} \beta )\ dx$

$ \equal{} \int_{\alpha}^{\beta } (x \minus{} \alpha)^2\{x \minus{} \alpha \minus{} (\beta \minus{} \alpha)\}\ dx$

$ \equal{} \int_{\alpha }^{\beta } \{(x \minus{} \alpha )^3 \minus{} (\beta \minus{} \alpha )(x \minus{} \alpha )^2\} dx$

$ \equal{} \left[\frac {(\beta \minus{} \alpha )^4}{4} \minus{} \frac {\beta \minus{} \alpha }{3}(x \minus{} \alpha )^3\right]_{\alpha }^{\beta }$

$ \equal{} \minus{} \frac {1}{12}(\beta \minus{} \alpha )^4.$

You can use integral by parts directly to the definite integral.


AMC
AIME
geometry
USA(J)MO
USAMO
trigonometry
analytic geometry
\/closed

Is it a good idea to take both the AIME prob series at once or separately? Also, should AIME Problem Series A be taken first or B?

Both are pretty much the same difficulty.

And for your first question...

It depends on how good you are and how much money you are willing to pay...

These classes cost money, but for a lot of people they are really worth it...

[quote="Blazer123"]Should Precalculus come first or should Algebra 3 come first?[/quote]
Algebra 3 is first. Precalculus uses skills in Algebra 3 for trigonometry/complex numbers/vectors topics not introduced in Algebra 3.
[quote="Ihatepie"]I'm taking precalculus in class this year. I know that the aops book covers much more than my curriculum, but do you think i'd get more out of taking the precalculus class and an aime class, or both aime classes while picking it up along the way?[/quote]
This depends, because the AIME class covers most of all fundamentals in AIME, while Precalculus goes deeper into Trigonometry/Complex Numbers (leaving off Combinatorics to Interm CP, Geometry to Olympiad Geo, etc.) If you don't have a lot of time and you are preparing for AIME-level contests, two AIME series would be a better choice.

[quote="Ihatepie"][quote="rrusczyk"]It also depends on how much time you have and what your goals are.  If you don't have a lot of time, just take one.  If you have tons of time, you might be better off taking a subject class, like Precalculus, in place of the second AIME class.

You can take the classes in either order.[/quote] I'm taking precalculus in class this year. I know that the aops book covers much more than my curriculum, but do you think i'd get more out of taking the precalculus class and an aime class, or both aime classes while picking it up along the way?[/quote]

You'll get way more with Precalc + AIME than with 2 AIME classes.  Our Precalc class will cover a lot that your school won't cover (and what we cover in common with your class will be covered better in ours :) )

[quote="Blazer123"]Should Precalculus come first or should Algebra 3 come first?[/quote]remy is correct; Algebra 3 before Precalc.

Precalculus has AMC12/AIME level trigonometry/vectors, right? Can USAMO problems be solved with AIME-level trigonometry?

Yes, and it has tools that will be useful on some USAMO problems.

If you're looking for extra AIME practice, sure.

[quote="simof"]would a person doing woot get much out of the aime problem series?[/quote]
I would assume to the extent that everyone gets better by doing problems.

I'm taking precalculus this December, are there concepts from the book, Intermediate Algebra, that will be used in the class? Also, what would be better for AIME-prep...A or B

[u]Choice A[/u]
1. Intermediate Algebra (book)
2. Precalculus (online class + book)
3. Doing past AIME tests on my own

[u]Choice B[/u]
1. Precalculus (online class + book)
2. AIME problem solving series

Algebra 3 is mostly just nontrig/vector precalc.

How good are you at that kind of Algebra? If not so good, Choice A. If you've mastered it, I suggest Choice B + past AIME tests.

Also, as there's a lot of Counting and Geometry in AIME, you might want to do Intermediate Counting.

I only know AoPS, sorry. But:

[quote="rrusczyk"]Finishing Intro to Geo should help with most of the non-trig geometry on AIME[/quote]

Also, do analytic geometry

[quote="AIME15USAMO"]is introduction to geometry enough geo for the AIME? If not, could you suggest something to improve my geometry?[/quote]

That was all the geometry I knew for the AIME this year, and I was able to solve 3 out of 5 of the geometry problems on the AIME I...

[quote="AIME15USAMO"]I am taking analytic geometry in school next semester, is that good enough or should I do something else, in addition to that?[/quote]

I think good enough...

Or if you feel like you need more geo practice, there's a lot of recommended books, like Geometry Revisited or geo parts of ACoPS.

Whoa, I think ACoPS is going a little far

Art and Craft of Problem Solving, by Paul Zeitz.
http://www.amazon.com/Art-Craft-Problem-Solving/dp/0471135712

Yeah, I guess it's more for Olympiads than AIME. People usually work on it during or after Intermediate AoPS books/classes.

[quote="isabella2296"]Should Interm C&P be taken before an AIME course? What about Interm NT?[/quote]

Either order is fine.


I can't seem to solve this problem, and it really is bothering me:

Let $n \geq 3$.  Find the greatest $c_n$ with the following property:  If $x_1, \ldots, x_n, y_1, \ldots, y_n$ are nonnegative real numbers satisfying $x_1 + \cdots + x_n = y_1 + \cdots + y_n$, then there exists a proper nonempty subset $J \subset \{1,2,\ldots,n\}$ such that \[c_n \sum_{j\in J} x_j \leq \sum_{j\in J} y_j \leq \frac{1}{c_n} \sum_{j\in J} x_j.\]

I really do not wonder! As far as I remember this is classified as hard problem by Gabriel Carol...

That is very true, it is a extremely hard problem, but it would be nice if a solution was found (if it has a solution)... :?


Of course , I knew dot product , but I have one to ask you : Can you show me way what ancient scientists discovered this . This made me a bit of confused :(

[quote="JRav"]What makes you confused?  You're implying that you already have this information but you don't understand the information.[/quote]
Nope , I imply that I don't know how to discovery it ~O)

I'm not sure how accurate all of this information is, but it's interesting nontheless [url=http://jeff560.tripod.com/matrices.html]Earliest Uses of Symbols for Matrices and Vectors[/url] (see Symbols for Vectors)
(Main Page - [url=http://jeff560.tripod.com/mathsym.html]Earliest Uses of Various Mathematical Symbols[/url])

[quote="t0rajir0u"]It's a strange story, and I'm not sure if I believe it, but to my knowledge the dot and cross product were simultaneously discovered by [url=http://en.wikipedia.org/wiki/William_Rowan_Hamilton]Hamilton[/url] in the form of the quaternions.  It was only later that vector language was adopted and widely used (since it applied in any number of dimensions).  Of course, there are some very good and deep reasons that the quaternions only exist in four dimensions.

Edit:  in slightly more detail, if you have two vectors $ \left < a, b, c \right > \equal{} ai \plus{} bj \plus{} ck$ and $ \left < d, e, f \right > \equal{} di \plus{} ej \plus{} fk$, then

$ (ai \plus{} bj \plus{} ck)(di \plus{} ej \plus{} fk) \equal{} \minus{} \left < a, b, c\right > \cdot \left < d, e, f\right > \plus{} \left < a, b, c \right > \times \left < d, e, f \right >$

in the quaternions.[/quote]

All right , I've read it in historical book . Thank t0rajir0u


function
analytic geometry
trigonometry
calculus
derivative
linear algebra
matrix

Let $ f: D\subset \mathbb{R}^2 \to \mathbb{R}^2$, cont. on $ D\equal{}\{y: \parallel{}x\minus{}y\parallel{}\leq r\}$, is diff. on $ int D\equal{}\{y: \parallel{}x\minus{}y\parallel{}<r\}$, and $ f(\overline{D})\equal{}\textbf{0}\equal{}(0,0)$, where $ \overline{D}\equal{}\{y: \parallel{}x\minus{}y\parallel{}\equal{}r\}$. There exits $ \alpha$ such that $ \alpha \in int D$, and $ f'(\alpha)\equal{}\textbf{0}$?

Counterexample in polar coordinates: $ f(R,\theta)\equal{}(\sin(\pi R/r),\theta)$. The derivative matrix $ f'$ has rank 1 along the circle $ R\equal{}r/2$, but it never vanishes.

Notational remark: $ \overline{D}$ normally means the closure of $ D$. Write $ \partial D$ for the boundary.


calculus
integration
function
real analysis
real analysis unsolved

I will sketch the proof of the first problem. We will consider the generating functions of the left hand side, and of the right hand side.  Then we will prove that they are equal. It is well known that \[\arcsin t = \sum_{k=0}^{\infty}\frac{(2k-1)!!}{(2k)!!\cdot (2k+1)}t^{2k+1},\] and \[\frac{1}{\sqrt{1-t^{2}}}= (\arcsin t)' = \sum_{k=0}^{\infty}\frac{(2k-1)!!}{(2k)!!}t^{2k}.\] Therefore, the generating function of the left hand side is $\frac{\arcsin t}{\sqrt{1-t^{2}}}$: \[\frac{\arcsin t}{\sqrt{1-t^{2}}}= \sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{(2k-1)!!(2n-2k-1)!!}{(2k)!!(2n-2k)!!}\cdot \frac{t^{2n+1}}{2k+1}.\] Now let us find the generating function $f(t)$ of the right hand side: \[f(t) = \sum_{k=0}^\infty \frac{(2k)!!}{(2k+1)!!}t^{2k+1}.\] Observe that \[1+t(t f(t))' = f'(t) = \sum_{k=0}^\infty \frac{(2k)!!}{(2k-1)!!}t^{2k}.\] We get a differential equation for $f(t)$: $1+t(t f(t))' = f'(t)$. It is equivalent to \[(1-t^{2}) f'(t) = 1+t f(t).\] We have \[(\sqrt{1-t^{2}}f(t))' = \sqrt{1-t^{2}}f'(t)-\frac{tf(t)}{\sqrt{1-t^{2}}}= \frac{(1-t^{2})f'(t)-t f(t)}{\sqrt{1-t^{2}}}= \frac{1}{\sqrt{1-t^{2}}}.\] Therefore, \[\sqrt{1-t^{2}}f(t) = \arcsin t,\] and \[f(t) = \frac{\arcsin t}{\sqrt{1-t^{2}}}.\] We proved the first identity. The proof of the second identity is very similar, except that it corresponds to the following equality of generating functions: \[\arcsin t \times \arcsin t = (\arcsin t)^{2}.\]

:coolspeak:  :first: .
Thank you very much


AMC
USA(J)MO
USAMO
geometry
rotation
3D geometry
octahedron

Derek wrote:Because it all sums up to 42.

[hide]  [/hide]

Haha. Nice one.

Because it's the fourth prime number. And they chose to use the fourth prime number because four is the square of the smallest prime number.

I think it's because there are seven steps to solving a USAMO problem:

1. Read problem
2. Identify the subject area
3. Identify what values you need to find
4. Assign variables
5. Note possible useful formulas
6. Work on it for a couple hours
7. Give up; read solution on kalva

Just kidding!  :lol:

But 7 is a nice number.  It's: number of days in a week, number of continents, prime, etc.

[quote="xxreddevilzxx"]Why is a full scoring for a USAMO or IMO problem a 7??

is there any reason??? just interested....[/quote]The answer is obvious: because 7 is the largest prime digit.

[quote="white_horse_king88"]Because 7 rotated is L, the Roman Numeral for 50. 6 * 50 = 300. 300 = AMC + AIME. ? :?[/quote]The USAMO grades problems with 7 points because the IMO grades problems with 7 points (way ahead in time before the USAMO was "born"). The IMO grades problems with 7 points because it is the largest known prime digit :D ;)

No, the IMO grades problems with 7 points beacuse if you take 1 and then you add 1 six times for each of the six problems, you get 1+1+1+1+1+1+1, and the sum is 7.  LOL, jk, why do we care, just asking?

Or try this one.  the IMO grades problems with 7 points because if you take anynumber in the whole universe, multiply it by 2, and 6, multiply the result by 3, add 24, and divide that whole thing by 6, and then subtract the number you started with, you get 7! OMG! go figure, AoPSers, how did that happen?

Incidentally, 7 happens to be a very common number.   When someone says, I'm thinking of a number between 1 and 10, most people respond, "7!" although the answer is rarely 7.

Because God took 7 days to create the world

[quote="darij grinberg"]since any number < 7 would leave the graders the possibility of throwing a dice, what should be prevented (really? :D )...[/quote]

An octohedron! 0-7 is 8 numbers, so it's perfect!  :D

[quote="darij grinberg"]since any number < 7 would leave the graders the possibility of throwing a dice, what should be prevented (really? :D )...[/quote]

Well that means you can't get a 0 since 0 is not a number on the dice  :D(which is nice!)

Well the expected value from throwing a dice would be $\frac{1}{6}(1+2+3+4+5+6)=\frac{7}{2}$ per problem. Thus, the total expected score is $6\cdot\frac{7}{2}=21$=MOP Qualification. Hence any USAMO participant is [i]expected[/i] to qualify for MOP :P

[quote="paladin8"][quote="WindSlicer"]Because it's 1 less than my favorite number! I can't believe none of you knew that.. pshh.. heh

And bubala.. err that doesn't work..[/quote]

Yes it does.

$\frac{3(2x+6)+24}{6}-x = 7$.[/quote]

Oh, I thought by "multiply it by 2 and 6" means multiply it by 12..

We really need to start saying "die" when we are using singular.  This looks bad.

An alternative way to do expected value would be $E(x) = \frac {42}2 = 21$  ;)  which nevertheless sets us all up with high expectations.

I remember that from Countdown prep. Always with the fastest way possible tsk.

[quote="Zyloch"]Back to the question of why 7, it's obvious that $6\cdot7=42$ and $42$ is the answer to life, the universe, and everything :)[/quote]Why is 42 the answer to life, the universe, and everything?

Why don't you watch the movie "Hitchhiker's Guide to the Galaxy" to find out... :P It looks pretty cool.

It's 7 because I said so. End of discussion.

Since most solutions get 0, 1, 2, 6, or 7, then are they not clearly saying that one full solution is better than 3 or more partial solutions?

[quote="MithsApprentice"][quote="Magnara"]All that AMC Director guy ever does is talk about stupid things, like when we can and cannot talk about tests.  Where is he when we have a really important question?[/quote]

The fair administration of the AMC exams is not what I would call "stupid things". Generally your questions are already answered in AMC published materials. If not, (like experience-based questions) then plenty of members on this forum can answer them. In my opinion, there's just so much more important work to be done at the AMC office than answering questions here.[/quote]

I assume that was sarcasm, since I can't see anyone considering "Why is a full scoring for a USAMO or IMO problem a 7?" a "really important question".  Nonetheless, Magnara, please watch what you say, as it can offend others.  We appreciate all the hard work that Mr. Dunbar and the rest of the AMC office do.

I edited it out because I wasn't sure. Either way, I got complaints from users so try to show whether or not u really mean something.

This is just more evidence that there should be a <sarcasm> tag :).  Oh yeah, and 7 might be a good number for the MOP qualification indices.

[quote="rrusczyk"]I believe Magnara was being facetious.[/quote]

Yeah...I didn't think he would say something like that, but eh it didn't sound like he was joking from here. Anyways, I think we should have sarcasm tags as well ^_^.

[quote="rrusczyk"][quote="MithsApprentice"][quote="Magnara"]All that AMC Director guy ever does is talk about stupid things, like when we can and cannot talk about tests.  Where is he when we have a really important question?[/quote]

The fair administration of the AMC exams is not what I would call "stupid things". Generally your questions are already answered in AMC published materials. If not, (like experience-based questions) then plenty of members on this forum can answer them. In my opinion, there's just so much more important work to be done at the AMC office than answering questions here.[/quote]

I believe Magnara was being facetious.[/quote]

It's true.  I was being facetious.

It's hard enough having a dry sense of humor in real life.  It's even harder on the internet.  Especially with a bunch of people who take everything way too seriously.  Either that, or you guys just think I'm a complete moron.

[quote="Zyloch"]Back to the question of why 7, it's obvious that $6\cdot7=42$ and $42$ is the answer to life, the universe, and everything :)[/quote]

What do you mean? $42=6\cdot9$, not $6\cdot7$. :)

Not to worry, I understood it was facetious.  In my spare time, I do have a sense of humor.

Good luck tomorrow and Wednesday to all participants.

Steve Dunbar
AMC Director