AMC AIME email geometry USA(J)MO USAMO probability \/closed

The [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]AoPS class schedule[/url] for classes starting during the remainder of this year is now posted.

The only change that might take place is that we are considering offering the AIME Problem Series A (the same problem series we have run the past year and a half) in addition to the AIME Problem Series B (the new problem series that is the same level of material as the AIME Problem Series A, but with an almost entirely disjoint set of problems).

The Independent Study programs will be capped as follows:
Introductory A 15 students
Introductory B 15 students
Intermediate A 25 students
Intermediate B 25 students
Olympiad A 12 students

If we are able to expand our teaching resources, we will loosen these caps.

Enrollment in the Independent Study will take place on a first come, first served basis.

Audited enrollments in subject classes are no longer available.

[quote="LynnelleYe"]When I try to click on the "Are you ready?" for Intermediate Counting, my computer freezes. Is something wrong with that file?[/quote]

That doesn't happen to me.  Is anybody else having this problem?

I have no idea what might be causing such a problem.  You might as well try the usual tactics such as a cookies purge or going in through a different browser or even updating your PDF reader.

[quote="LynnelleYe"]When I try to click on the "Are you ready?" for Intermediate Counting, my computer freezes. Is something wrong with that file?[/quote]

It works fine for me.

[quote="LynnelleYe"]When I try to click on the "Are you ready?" for Intermediate Counting, my computer freezes. Is something wrong with that file?[/quote]

You might also try re-installing Acrobat reader.

Do you recommend me taking Intermediate Complex Nmbr and Intermediate Counting at the same time to prepare for AIME, given that although i did get 10 in AIME last yr, that was my only 3rd yr to take AIME (and other times I got 1 and 5). Also, consider the fact that last year's AIME was easier than some of the other years' in the past.

Also, should i take AIME Series B, A, or both???

(I think Mr. Rusczyk and Mr. Crawford knows my abilities, so I am asking you guys.... feel free to PM me or even email me. I would like extensive thoughts).

[quote="1234567890"]Why the lack of intro classes? I thought AoPS was trying to attract more middle schoolers.[/quote]

The Intro Geometry course will be offered in the Spring because Richard is finishing the Intro Geometry book and then he will revise the course a bit.

We're teaching Intro Counting/Probability this summer and we will teach it again in the Spring.

We would like to expand the schedule to include in Intro Algebra class, but that will probably have to wait until next year sometime.

Middle school resources are certainly a priority for us, but so are high school students who already make up the core of AoPSers.

[quote="xxreddevilzxx"]Do you recommend me taking Intermediate Complex Nmbr and Intermediate Counting at the same time to prepare for AIME, given that although i did get 10 in AIME last yr, that was my only 3rd yr to take AIME (and other times I got 1 and 5). Also, consider the fact that last year's AIME was easier than some of the other years' in the past.

Also, should i take AIME Series B, A, or both???
[/quote]

I think you're fine taking both of those Intermediate courses and an AIME Problem Series because you're at a level at which you'll struggle less with the Intermediate work than most Intermediate students.

I think taking both AIME Problem Series might simply put you in a time crunch, but you know your schedule better than I do.

[quote="joml88"]Will there be an Olympiad IS offered after the one in the fall/winter but before USAMO?[/quote]

Most of the Independent Study in the Spring will be before the USAMO, but it might be that only 80-90% of the work is complete by then.  It will probably start around the second week of January, though no schedule is firm yet.

[quote="joml88"]Will there be an Olympiad IS offered after the one in the fall/winter but before USAMO?[/quote]

Olympiad IS during the school year.... :P ....wow, I admire your organizational skills. I would probably fail all of my classes because of working on the IS problems!

well there's no pre/post test for the independent studies but they give ideas as to where you should be such as something like "this is probably right for you if you're not confident you can pass the AMC 10 or make National Mathcounts".  you can also get an idea of what topics will be covered in each independent study as there are two for each level except olympiad.

[quote="pkothari13"]is there some sort of pre-test/post-test or what not for the independent studies? how would we know which one is right for us?[/quote]

Judging from what I've seen of you in class, I think you'd learn a lot from either Intermediate Independent Study, though you're probably good at the Intermediate B level for a best fit.  But if you wanted to work through both Intermediate levels, that wouldn't be a bad fit either (during different semesters of course -- it's not a small amount of work).

[quote="Silverfalcon"]Just for curiosity, why no audited?

I like audited better because it was cheap. Now I'm not sure if I can take both AIME and Trig Complex Number class because that would be 200 dollars. I really want both classes since even I don't make to USAMO this year (hopefully I do), taking Trig Complex Number will help me a lot on later AIME problems.

 :( Why?

Just wondering..[/quote]

The reasons for no audited are stated [url=http://artofproblemsolving.com/Forum/viewtopic.php?p=266099#p266099]here[/url].

More than anything we've noticed that audited students spend far less time working on problems as a whole (with a few exceptions -- joml88 worked harder than other audited students in the Independent Study).  We have also found that many students are not advancing as quickly mathematically because they are not advancing their communication skills.  These link between these skills is almost universally underestimated.

[quote="pkothari13"]do you offer intermediate counting and probability during the second semester (Jan 2006 and onwards)?[/quote]

We plan to offer each class (except possibly AIME Problem Series A) exactly once during the coming school year.  It's either listed for Fall/Winter or we plan to list it for the Spring.

[quote="1234567890"]But these Olympiad classes are fixed, right? No more changes for them, am I correct?[/quote]

They are fixed unless something dramatic happens that forces a change of schedule.

[quote="WindSlicer"]Are the Intermediate A slots filling up quickly?[/quote]

I think we've had a couple of enrollments, maybe three at the most, but we've just listed it.  Over the first year of the program, that's been the most popular level and the one we would most expect to hit 25 enrollments.

[quote="sirorange"]Also, if a class has been taken off of the enrollment page, are we to assume it is full?[/quote]

So far the only reason we've ever had a class taken off is that it was past the enrollment deadline.  I suppose we will take a class off the enrollment page if it fills, though this is the first time we've really had to start thinking about what "filled up" means.

We in southern California have a rather liberal interpretation of 'Fall' :D

Ahh, that is why. For us in Mass. winter starts a week after summer ends :P Yeah, that "famed foliage" there isn't much of that!

[quote="sirorange"]Well, on the class description page, it says that Inequalities will be offered Fall 2005, and on the Enroll page it starts in December, so I was just wondering what was going on.[/quote]

What we really mean is that the class begins during the semester usually referred to as the "Fall semester".

Thanks for clearing that up :)  Just wanted to make sure a course I was still deciding whether or not to take was filling up.

Are the Intermediate A slots filling quickly? Because I plan on registering in a week or so. Will there still be enough space?


MATHCOUNTS
Princeton
college
geometry
AMC
AIME
AMC 8

It varies from state to state...I am going to try to find ignite's post so give you an idea of the cutoff.  But it really shouldn't matter, what happens at the competition is often very different from what happens at home.

edit: here it is: http://www.artofproblemsolving.com/Forum/post-492509.html

Go under state cutoffs under returning people

Yeah, I timed myself to the second (used an alarm clock) and no calculator for sprint - only target - but I couldn't even finish last 4 or 5 on sprint - ran out of time (not that I would've gotten those right anyway!)  :(

at the northern california state, what is considered a good score?

its a lot more competitive than other states, right?

in california 47's considered a good score

haha very funny. :rotfl:

thanks. i feel much better.  :wacko:

[quote="usaha"]With those scores, you couldnt even make it into the top 15 in Indiana, so it really does vary from state to state...[/quote]

Don't be so mean.

But it is true.

In Northern California, I have noticed that >42 is really good.

In Michigan, >40 is really good.

[quote="Walk Around The River"][quote="guyinPA"]I just took last year's State Sprint and Target Rounds - 2006 and I scored a total of 31 (21 on Sprint and 10 on Target) -- I feel as though this may not be a very good total score, but I honestly have no clue -- I KNOW it differs state to state, but can any of you shed light on minimum scores (maybe from last yr?) that moved people on from State to Nats? How much higher would I have needed on these tests last yr to move on?[/quote]

well, in florida, you woulda been somewhere from 18th-21st on tiebreakers.[/quote]
No, Kushal got a 32 and 25th ish place.
That'd probably put you at about 30th-40th place in FL.

yeah but with those scores in Maryland you can place really well..

wow apparently california is extremely competitive  :o

uh - probably doesn't matter much, but I just realized I did a stupid thing - my dad only gave me 30 - not 40 minutes when I took the test (don't know why I didn't realize) but I had taken an extra 8 minutes to do the last 6 problems just to see if I could do, and I got one of them right, so that gave me a 32 (22 + 10 target) with 2 minutes to spare! like I said, I bet 32 is like 31 - no diff at states, right?

Last year in MA the cutoff was low, 38. Most will be that or higher, imo.

[quote="vishalarul"][quote="7h3.D3m0n.117"][quote="guyinPA"]I just took last year's State Sprint and Target Rounds - 2006 and I scored a total of 31 (21 on Sprint and 10 on Target) -- I feel as though this may not be a very good total score, but I honestly have no clue -- I KNOW it differs state to state, but can any of you shed light on minimum scores (maybe from last yr?) that moved people on from State to Nats? How much higher would I have needed on these tests last yr to move on?[/quote]

Well with those scores im pretty sure you ahve a really good chance to get into nats or place within the top 10 but it depends on if you actually used the 40 minutes for sprint no calcs etc...[/quote]

Are you kidding?
With that score in California, you can't even get 30th.[/quote]

No way! With that score in [i]Northern[/i] California (not including South), you would get under 40th for sure. Including Southern CA, maybe 60th-70th.

[quote="#H34N1"]I thought bpms wasf rom Maryland, so how would he know about FL? :|[/quote]

He's from Florida.

[quote="Ubemaya"][/quote][quote="vishalarul"][quote="7h3.D3m0n.117"][quote="guyinPA"]I just took last year's State Sprint and Target Rounds - 2006 and I scored a total of 31 (21 on Sprint and 10 on Target) -- I feel as though this may not be a very good total score, but I honestly have no clue -- I KNOW it differs state to state, but can any of you shed light on minimum scores (maybe from last yr?) that moved people on from State to Nats? How much higher would I have needed on these tests last yr to move on?[/quote]

Well with those scores im pretty sure you ahve a really good chance to get into nats or place within the top 10 but it depends on if you actually used the 40 minutes for sprint no calcs etc...[/quote]

Are you kidding?
With that score in California, you can't even get 30th.[/quote][quote="Ubemaya"]

No way! With that score in [i]Northern[/i] California (not including South), you would get under 40th for sure. Including Southern CA, maybe 60th-70th.
[/quote]

oh not good. You know, maybe I should move to a different state. :D

ehee ... person - thanks a lot for the links - Very helpful!!!! And thanks to those that put together

[quote="guyinPA"]does it strike anyone else as sort of "unfair" that a person in one state with a lower score can go to nats when a person from a state with a higher score doesn't get to compete? Something sort of "undemocratic" about that -- uh, I mean, if you take 50 states X 4 people = 200 people -- then they should just let top 200 scorers across the US go -- I realize that some states won't be represented, but if there isn't a single student in that state worthy, why should they be?? (represented, that is) And IF there are 6 worthy kids in the smallest state (or largest) why SHOULDN'T all of THEM be given a chance?  :?:   Just food for thought.... I'm all for fairness... regardless of your zip code (the way it is, I'm glad I don't live in CA!!!)[/quote]

rcv posted this once:
"Yes, life1, I think you are completely and utterly naive about the fundamental philosophy of MATHCOUNTS.

I love MATHCOUNTS and the AMC contests. But they are different.

MATHCOUNTS is a grass-roots "coaching and competition" program. It starts at the local schools, with coaches (many are volunteers or teachers performing unpaid extra duty) working with kids to get them ready for a fun competition. The school is limited to taking a handful of students to the Chapter Competition, which fosters a competitive atmosphere in the school. Whether your are in a rural school with 200 low-income children of farm workers or from a suburban school with 1600 high-income students or from a magnet school with a high proportion of children of Princeton professors, a School MATHCOUNTS competition can really jazz your students.

At the next level, a volunteer Chapter Coordinator, along with a small army of volunteers hosts another competition, with the prospect that a fair proportion of students from that competition will advance to the next level. Often these volunteers donate, not just their time, but their own money to help fund the competition, the prizes, and the food.

At the State level, yet another volunteer, the State Coordinator, along with another small army of volunteers hosts yet another competition, with the guarantee that four students will advance to the National Competition to represent your State. Once again, the Engineers (and other volunteers) who run the contests donate time, money, food, and prizes.

Now what happens if the magnet school is permitted to take a boatload of students to the Chapter Competition? Few other schools in the Chapter will be able to send anybody to the State Competition.

What if the advancement criteria is based on the statewide results, rather than Chapter Results? Then students in the rural part of the State will never advance to the State competition. Why will the rural Chapter Coordinator and his volunteers continue to invest their time and money? Why will the coaches in the rural part of the state continue to invest their time? Why even bother with MATHCOUNTS, except at the most elite schools in the State?

Because MATHCOUNTS has a 23-year history of a grassroots "coaching and competition" program. It is more than simply seeing who "wins" the contest. It's about learning. It's about teamwork. It's about physically getting students from different schools together, where they can mingle. Where their coaches can share ideas and strategies. It's about the experience and downright "fun" of the program!

In the end, each state *does* usually send four of their tip-top students to the National Competition. And yet, the rural coach can still have a student who is justifiably proud of their score of "3" on the Sprint.

For the high concentrations of great math students at high schools like TJ, Stuy, IMSA, and Exeter, the AMC contests allow them to shine. Likewise, the student in region "B" of your State can shine in the AMC contests, and will probably also have a much wider array of other opportunities in which he can shine than the student in region "A" of your State.

If you tried to change the rules of MATHCOUNTS to match those of the AMC, you would destroy MATHCOUNTS. Even if it didn't destroy MATHCOUNTS, why would you need two contests with the same rules. We could just do away with MATHCOUNTS, and rename the AMC-8 to be the AMC-8-MC. How delightful!"

Last year I was the cutoff for Wisconsin--37--25 sprint, 12 target. And assuming you're from PA, (by your username) I would think that it'd be a tad higher since Pennsylvania is bigger than WI :wink:

okay...I'm not so familiar with MC stuff, but there's only 8 questions in the target round. how can people score 10s or 12s?

or do each question count 2 points?

[quote="kevinchenxi"]
or do each question count 2 points?[/quote]

exactly

oh...

then when I attempted the 2006 state competition, I got

21 for sprint and 5*2 for target.

I got the exact same as the guy who started this thread.  :lol:

but how would that be for my state State Department?

[quote="guyinPA"]to ehee... not lookin' to argue, but why is it assumed student in "rural area" is less advantaged - or smarter  (or Dumber) than anyone else - I don't live in a "rural area" - more suburban-like, but I think that would insult me, if I did -- I mean, I know friends who have no "real" coach (just a parent or friend) - have no AoPS books, but just study really, really hard everything they can find and do VERY well on competitions -- and they don't live in rural areas - so if there are 10 of these kids who score high 30's say, versus someone in a small state scoring 20 - I still think it's unfair to the 30's kids who are 5th or 7th or 10th  and don't get to move on... just because their parents reside in a "less rural area" -- for ex, if you lived in CA and worked your hardest -- your very hardest and scored a 39 and didn't get to move on when kids in 47 other states do, for working much less, and scoring less, please tell me how that's fair? Hey, anyone in CA (or other top state) whose in that boat on this site?? Back me up, will ya?[/quote]
Others have had similar questions/concerns. You may want to read rcv's well written response touching this topic: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=757062#757062

Hey, I'm the guy who started this thread - AND I answered your question but it didn't post!! Weird ... anyway, yeah, 2 points x 8 max (8 questions) = 16 max points -- and I don't know why that is - it just is (maybe because target questions are supposed to be harder than sprint - or just take longer)

[quote="guyinPA"](maybe because target questions are supposed to be harder than sprint - or just take longer)[/quote]

that makes sense.

I'm in State Department, and it's the state with the U.S.-territory owned schools, and if I get a similar score to the one I just got, (31) would I be able to succeed to the nationals?

Uh - new here, but felt I had to add my 2 cents .. I totally see pa guy's pt of view -- I mean, other person in thread wrote in part "And yet, the rural coach can still have a student who is justifiably proud of their score of "3" on the Sprint." -- sure, kid could be Proud, but that doesn't mean he deserves to move on over other kids scoring higher!! sorry, but many of us disagree with that idea -- also, other person wrote or implied, in my opinion, that only the expensive magnet schools are capable of producing a kid great at math -- and while those kids may have certain advantages, I think that's nonsense --- did anyone here see Will Smith's Pursuit of Happiness? Poverty or lack of other advantages does not preclude being smart/doing well, and having everything (ie - certain "stars" in the news - just turn on tv - doesn't mean they are at all smart, in the classical sense - I'm not sure what the agendas are that some people are adament about keeping the rules the way they are (I guess they benefit certain states & places over others,huh?? And people think THAT is fair?)  :(  -   and to those that disagree, i think other's people's threads on here, incl. this one, are merely suggesting having open discussion/thinking about what's "most fair" -- and even the AMC constantly changes things like calculator rule, AIME cutoffs, etc -- so how 'bout some flexible thinking? I realize this may be a skewed forum, with many on here scoring at the top already, but please consider those kids on the next rung ..

[quote="kevinchenxi"]I'm in State Department, and it's the state with the U.S.-territory owned schools, and if I get a similar score to the one I just got, (31) would I be able to succeed to the nationals?[/quote]Gosh I don't think that anyone on this site knows much about the State Department. It would help if you knew approximately how many schools the State Department consisted of :)

EDIT: just looked it up: about 200 schools. No telling how many do MATHCOUNTS, so I'd say you might be in the top 10 with that score. :)

EDIT(again): 42 schools from State Department in MATHCOUNTS. You might have made nationals with that score.

... :lol:

if that score actually can get me close to the nationals,

[b]I'LL STUDY LIKE A BEAST[/b]

Ubemaya, what Chapter are you in?
Just wondering.

you can probably solve every mc problem you can ever find... :oops:


Let $F,G,H$ be polynomials with real coefficients such that $F(x)=G(H(x))$ and $H(0)=0$. Let $H_{0}(x)=H(x)+c$, where $c$ is a real number. Show that there exists another polynomial $G_{0}(x)$ such that $F(x)=G_{0}(H_{0}(x))$


IMO 2008
prime divisor
number theory
quadratics
IMO
IMO Shortlist

http://www.mathlinks.ro/viewtopic.php?search_id=550514793&t=126781   This says much more and the solution presented there gives infinitely many such numbers, not just one, as stated in that link. However, I guess the official solution runs along the ideas used in a problem given in a recent USAMO (proposed by Titu Andreescu and me). In any case, I doubt this is a good problem for IMO.

Sorry for double posting, but it's just as I thought: take $ p\minus{}1$ multiple of $ 4$ very large. Then we know there is $ n$ such that $ p|n^2\plus{}1$ and of course we may assume that $ n<p$ and even $ 2n\leq p$, simply because $ p$ also divides $ (p\minus{}n)^2\plus{}1$. Now, write $ p\minus{}2n\equal{}k>0$ and observe that $ p$ divides $ 4n^2\plus{}4\equal{}(p\minus{}k)^2\plus{}4$, thus $ p$ divides $ k^2\plus{}4$ and so $ k$ is at least $ \sqrt{p\minus{}4}$. This ismmediately implies $ p>2n\plus{}\sqrt{2n}$ if $ n$ is large enough, that is $ p$ is large enough.

Fix a prime $ p$ of the form $ 20m\plus{}1$. There are two solutions to the congruence $ n^2 \equiv \minus{}1 \pmod{p}$, one of them is less then $ \frac{p\minus{}1}{2}$ and one is greater. Let $ \frac{p\minus{}1}{2}\minus{}k$ be the smaller one ($ k>0$). Now we want find a lower bound for the $ k$, let's see what do we need. Take for simplicity $ \frac{p\minus{}1}{2} \equal{} a$. We want to have:
$ 2(a\minus{}k) \plus{} \sqrt{2(a\minus{}k)} \leq p\minus{}1$
or
$ a\minus{}k \plus{} \sqrt{\frac{a\minus{}k}{2}} \leq a$
which is equivalent to
$ a \leq 2k^2 \plus{} k$
and finally
$ p \leq 4k^2 \plus{} 2k \plus{} 1$.

Ok, now we see that $ (\frac{p\minus{}1}{2} \minus{} k)^2 \equiv \minus{}1 \pmod{p}$ is equivalent to $ (2k\plus{}1)^2 \equiv \minus{}4 \pmod{p}$. But $ p\minus{}4 \equiv 3 \pmod{5}$, so it's not a square. Of course $ (2k\plus{}1)^2 \not \equal{} 2p\minus{}4$ since the parity is different. Therefore $ (2k\plus{}1)^2 \geq 3p\minus{}4$, i.e. $ 4k^2\plus{}4k \plus{} 5 \geq 3p$ and it's clear that the desired inequality is satisfied.

[quote="harazi"]Sorry for double posting, but it's just as I thought: take $ p \minus{} 1$ multiple of $ 4$ very large. Then we know there is $ n$ such that $ p|n^2 \plus{} 1$ and of course we may assume that $ n < p$ and even $ 2n\leq p$, simply because $ p$ also divides $ (p \minus{} n)^2 \plus{} 1$. Now, write $ p \minus{} 2n \equal{} k > 0$ and observe that $ p$ divides $ 4n^2 \plus{} 4 \equal{} (p \minus{} k)^2 \plus{} 4$, thus $ p$ divides $ k^2 \plus{} 4$ and so $ k$ is at least $ \sqrt {p \minus{} 4}$. This ismmediately implies $ p > 2n \plus{} \sqrt {2n}$ if $ n$ is large enough, that is $ p$ is large enough.[/quote]

Very nice !

Seems like people who have seen the problem
Prove that there exists infinite $ n$ such that $ n^4\plus{}1$ has a prime divisor larger than $ 2n$

would have some advantages

Just a side remark. Even those who did not know that any $ p\equal{}4k\plus{}1$ divides some $ x^2\plus{}1$ could solve the problem like harazi did because all one needs is to have infinitely many primes that divide $ x^2\plus{}1$ for some $ x$ and the classical Euclid's proof of the infinitude of primes can be trivially ajusted to give this statement. The opening sentence in TomciO's solution, on the other hand, would tempt me to reduce a few points immediately if I were one of the graders because it is quite doubtful that the contestant could prove it if asked.

Anyway, the first day problems were sort of disappointing :(. Let's see what the second day will bring :)

Every prime that is congruent to $ 1$ mod $ 4$ appears exactly once as the divisor here, except for $ 5$ and $ 13$. Harazi's "large enough" threshold is $ p \equal{} 29$, and $ p \equal{} 17$ works easily because $ 17 \minus{} 4$ is not a perfect square.
If $ 2n \plus{} \sqrt {2n}$ were changed to $ 2n \plus{} \sqrt {2n} \plus{} 1$, we would need to exclude all primes of the form $ m^2 \plus{} 4$ as well. This still leaves infinitely many, but proving that is hard compared to the Euclid-style argument that there are infinitely many primes $ \equiv 1\mod 4$.

[Edit- fixed silly mistake]

Shouldn't 3rd problems be a little difficult in the sense that the solution should be a little long or at least involve some clever argument or trick?
I mean look at Harazi's post (#3)...
And then a gain P1 asking if you know Power Of a Point, and P2 being cross multiplication+multiplication+addition+completing the square...
Very disappointed.
Again the question is:
Poor choice of leaders or bad ISL?

Note that for any prime $ p \equiv 1 \pmod{4}$, we can always choose some $ \alpha \in \{ 0, \ldots, \frac{p\minus{}3}{2} \}$ such that $ \left(\frac{p\minus{}1}{2} \minus{} \alpha \right)^2 \equiv \minus{}1 \pmod{p}$. Denote $ m \equal{} \frac{p\minus{}1}{2} \minus{} \alpha$. This problem then becomes equivalent to choosing a suitable $ p$, with corresponding $ \alpha$, such that $ p > 2m \plus{} \sqrt{2m}$. We check that \[ p > p\minus{}1 \minus{} 2\alpha \plus{} \sqrt{p\minus{}1\minus{}2\alpha} \Leftrightarrow 2\alpha \plus{} 1 > \sqrt{p\minus{}1\minus{}2\alpha} \Leftrightarrow 4\alpha^2 \plus{} 6\alpha \plus{} 2\minus{}p >0.\] Solving this quadratic inequality, since we assumed $ \alpha \geq 0$, this is equivalent to \[ \alpha > \dfrac{\minus{}6\plus{}\sqrt{36\plus{}16(p\minus{}2)}}{8} \equal{} \dfrac{\minus{}3\plus{}\sqrt{4p\plus{}1}}{4}.\]

Hence, if $ \alpha > \dfrac{\minus{}3\plus{}\sqrt{4p\plus{}1}}{4}$, then we are done. Suppose not, then we have $ 0 \leq \alpha \leq \dfrac{\minus{}3\plus{}\sqrt{4p\plus{}1}}{4}$. Note that $ 4m^2 \equal{} (p\minus{}1\minus{}2\alpha)^2 \equiv (2\alpha\plus{}1)^2 \pmod{p}$. By the bounds on $ \alpha$, we get $ 0 \leq (2\alpha\plus{}1)^2 \leq \left( \dfrac{\sqrt{4p\plus{}1} \minus{} 1}{2} \right)^2$. Now, by assumption, $ 4m^2 \equiv \minus{}4 \pmod{p}$, so if $ \left( \dfrac{\sqrt{4p\plus{}1} \minus{} 1}{2} \right)^2 < p\minus{}4$, we get a contradiction. Solving this inequality, this is equilavent to $ p>20$.

Therefore, if $ p$ is sufficiently large (i.e. $ p>20$) and $ p\equiv 1 \pmod{p}$, we can always find some $ m$ such that $ p|(m^2\plus{}1)$ and $ p > 2m \plus{} \sqrt{2m}$.

[quote="harazi"]Sorry for double posting, but it's just as I thought: take $ p \minus{} 1$ multiple of $ 4$ very large. Then we know there is $ n$ such that $ p|n^2 \plus{} 1$ and of course we may assume that $ n < p$ and even $ 2n\leq p$, simply because $ p$ also divides $ (p \minus{} n)^2 \plus{} 1$. Now, write $ p \minus{} 2n \equal{} k > 0$ and observe that $ p$ divides $ 4n^2 \plus{} 4 \equal{} (p \minus{} k)^2 \plus{} 4$, thus $ p$ divides $ k^2 \plus{} 4$ and so $ k$ is at least $ \sqrt {p \minus{} 4}$. This ismmediately implies $ p > 2n \plus{} \sqrt {2n}$ if $ n$ is large enough, that is $ p$ is large enough.[/quote]

Similarly I got to the step that $ p$ divides $ k^2 \plus{} 4$, obviously if $ \frac{k^2\plus{}4}{p}>1$ we will done. So p=$ k^2$+4. So this will lead to that any prime congruent to 1 mod 4 need also be congruent to 5 mod 8.(otherwise p-4 congruent to 5 mod 8 which cannot be $ k^2$) This is clearly a contradiction by Dirichlet's Theorem!

P.S Is there an elementary way to show that there are infinitely many prime of form 8K+1?

yes. one way for example is considering divisors of $ 2^{2^m} \plus{} 1$  :wink:

Author of this problem is [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=4906]Kęstutis Česnavičius, Lithuania.,[/url] from [url=http://math.jacobs-university.de/]Jacobs University Bremen[/url] which is the host of the [url=http://www.imo2009.de/index.html]50th International Mathematical Olympiad 2009[/url] in Bremen Germany. It will also be the 20th anniversary after the last IMO in Braunschweig, Germany in 1989. Formerly the German Democratic Republic (GDR) was supposed to host the IMO 1992 but due to the unification with the Federal Republic of Germany (FRG) the event was cancelled and organized by Moscow, Russia, as a substitute host in 1992.

If solution of the problem consists only of two simple steps – it isn’t third.
[b]My solution. [/b] Consider the set of numbers $ S\equal{}${$ p \in \mathbb{P} | p \equiv 1 \mod 4, p \not \equal{} 5,13$}.
For every $ p \in S$ let us define set of positive integers $ N_p$ such that $ n \in N_p \leftrightarrow p|n^2\plus{}1$ and $ n_p$ - the minimal of them.
It’s easily seen that $ n_p<p$ and also $ n_p<\frac{p}{2}$ else we can take $ p\minus{}n_p$ instead of  $ n_p$.
Let $ k_p$ be $ p\minus{}2n_p$. Then $ k_p^2\plus{}4\equal{}(p\minus{}2n_p)^2\plus{}4 \equiv 4(n_p^2\plus{}1) \equiv 0 \mod p$, hence $ k^2\plus{}4>\equal{}p$.
Let us notice that $ k_p>4$. Really, if this isn’t true then there are only two variants: $ k_p\equal{}1$ and $ k_p\equal{}3$ since $ k_p$ is odd. If $ k_p\equal{}1$ then $ p|1^2\plus{}4$ and $ p\equal{}5$ - contr; if $ k_p\equal{}3$ then $ p|3^2\plus{}4$ and $ p\equal{}13$ -contr.
So we have $ k_p^2\plus{}k_p>k_p\plus{}4 \ge p$, therefore $ p\minus{}k_p\plus{}\sqrt{p\minus{}k_p}<p$ or $ 2n_p\plus{}\sqrt{2n_p}<p$. Since $ |S|\equal{}\infty$ then there are infinitely many different integers $ n_p$ such that $ p \in S$, so we are done.
Evidently it’s possible to estimate this value better. For example if we take the set $ T\equal{}${$ p \in \mathbb{P} | p \equiv 1 \mod 4, p \equiv 1 \mod 5$} then for every $ p \in T$ $ p\minus{}4$ isn’t square of integer and hence $ k_p^2\plus{}4 \ge 5p$.

[quote="harazi"]\we may assume that $ n < p$ and even $ 2n\leq p$, simply because $ p$ also divides $ (p \minus{} n)^2 \plus{} 1$. .[/quote]
vhy???  :?:  :maybe:

I realized that I had written this solution up around five months ago, why not post it; the same solution as some other on the thread except with more of the details provided for motivating the solution, similar to [hide=problem I solved since this write-up]USAMO 2006, P3[/hide].
[hide=Solution]
$\textbf{Lemma 1:}$ \\
It suffices to find infinitely many primes $p$ such that there exists some $n \in \mathbb{N}$ such that $p \mid n^2+1$ and $n < \frac{p-1}{2}-\sqrt{\frac{p}{4}}$. \\
$\textbf{Proof)}$ \\
Notice that for each prime $p$, there is at most one $n \in \mathbb{N}$ satisfying the condition because $x^2+1$ has at most two roots in $\mathbb{F}_{p}[x]$ which come in a pair $n,p-n$, one of which is at least $\frac{p}{2}$ and consequently does not satisfy the condition. \\
Therefore, if we have that there are infinitely many primes $p$ with $n \in \mathbb{N}$ satisfying the condition for said prime and take $A$ to be the set of all such primes, then we can consider the map $f: A \to \mathbb{N}$ such that $p \mid f(p)^2+1$ and $f(p)< \frac{p-1}{2}+\sqrt{\frac{p}{4}}$. \\
Moreover, $f$ is injective because if $f(p) = f(q)$ with $p \neq q$, $$n^2+1 \geq pq > (2n+\sqrt{2n})^2$$ which is not possible. \\
As $A$ has infinitely many elements the image of $f$ must be infinite, then taking any element of the image, notice that $$p>2(\frac{p-1}{2}-\sqrt{\frac{p}{4}})+\sqrt{2(\frac{p-1}{2}-\sqrt{\frac{p}{4}})}>2f(p)+\sqrt{2f(p)}$$ meaning that each element of the infinite image of $f$ satisfies the condition as $f(p)^2+1$ always has a prime divisor that is at least $2f(p)+\sqrt{2f(p)}$, as desired. $\blacksquare$ \\

We shall now prove that there are indeed infinitely many prime numbers $p$ such that there is an $n \in \mathbb{N}$ satisfying the condition which by $\textbf{Lemma 1}$ imply the problem. \\

Take a prime $p$ such that $p \equiv 1 \pmod{4}$ and $p \equiv 2 \pmod{5}$ of which there are infinitely many by Dirichlet's Theorem. \\

$\textbf{Lemma 2:}$ \\
There exists some $n \in \mathbb{N}$ such that $p \mid n^2+1$ \\
$\textbf{Proof)}$ \\
Let $g$ be a primitive root $\pmod{p}$ and let $n = g^{\frac{p-1}{4}}$. \\
Then $n \in \mathbb{N}$ and $$n^2 \equiv (g^{\frac{p-1}{4}})^2 \equiv -1 \pmod{p}$$ which immediately gives us the existence of an integer $n$ such that $p \mid n^2+1$. \\

$\textbf{Lemma 3:}$ \\
There exists some $n \in \mathbb{N}$ such that $p \mid n^2+1$ and $p > 2n+\sqrt{2n}$ \\
$\textbf{Proof)}$ \\
Notice by $\textbf{Lemma 2}$ that there exists some $n \in \mathbb{N}$ such that $p \mid n^2+1$, similarly, notice that $p \mid (p-n)^2+1$, therefore we can in fact find some $n \in \{0,..,\frac{p-1}{2}\}$ such that $p \mid n^2+1$. \\
Naturally, we may write $n = \frac{p-1}{2}-k$ for some $k \in \{0,...,\frac{p-1}{2}\}$. \\
Then $$-1 \equiv (\frac{p-1}{2}-k)^2 \equiv k^2+k+\frac{1}{4} \pmod{p}$$
Therefore $$p \mid k^2+k+\frac{3p+5}{4}$$ because $\frac{5}{4} \equiv \frac{3p+5}{4} \pmod{p}$ and $4 \mid 3p+5$ as $p \equiv 1 \pmod{4}$. $\blacksquare$\\

$\textbf{Claim:}$ $k > \sqrt{\frac{p}{4}}$ \\
$\textbf{Proof)}$ \\
Assume for the sake of contradiction that $k \leq \sqrt{\frac{p}{4}}$. \\
Then, notice that $$2p> p+\frac{5}{4}+\sqrt{\frac{p}{4}} =  (\sqrt{\frac{p}{4}})^2+\sqrt{\frac{p}{4}}+\frac{3p+5}{4}\geq k^2+k+\frac{3p+5}{4}$$
meaning that $p = k^2+k+\frac{3p+5}{4}$ as $p \mid k^2+k+\frac{3p+5}{4}$. \\
Therefore we have that $p = 4k^2+4k+5$, yet,
$$2 \equiv p \equiv -k^2-k \equiv 0,3,4 \pmod{5}$$ which is a clear contradiction, as desired. $\blacksquare$ \\

Then, notice that $$n = \frac{p-1}{2}-k < \frac{p-1}{2}-\sqrt{\frac{p}{4}}$$
which by $\textbf{Lemma 1}$ suffices. $\blacksquare$ \\
[/hide]

Let $p\equiv 1\pmod{8}$ be a large prime, let $0 < n < p$ be the smaller solution to $n^2\equiv -1\pmod{p}$, and let $a = \frac{p-1}{2} - n$. Then we have

$-1\equiv \left(\frac{p-1}{2}-a\right)^2\equiv \left(-\frac{1}{2}-a\right)^2\equiv a^2 + a + \frac{1}{4}\pmod{p}\implies a^2 + a\equiv -\frac{5}{4}\pmod{p}$.

Since $p\equiv 1\pmod{8}$, $\frac{p-5}{4}$ is odd and hence cannot be of the form $a^2 + a$. It follows that $a^2 + a\ge \frac{5p-5}{4}$, which gives the very weak bound of $a > \frac{\sqrt{p}}{2}$ for sufficiently large $p$. Then $2n = 2\left(\frac{p-1}{2} - a\right) < p - \sqrt{p}$ which is enough to imply $2n + \sqrt{2n} < p - \sqrt{p} + \sqrt{p}  = p$, as desired.

......

Pick some prime $p \equiv 1 \pmod{4}$. There are two solutions to $n^2 \equiv -1 \pmod{p}$ that add up to $p$, hence one must be less than $\tfrac{p-1}{2}:=a$. Let the smaller one be $a-k$. Note that $(a-k)^2 \equiv -1 \pmod{p}$ translates to $(2k+1)^2 \equiv -4 \pmod{p}$. This means
$$(2k+1)^2 \geq p-4 \iff p \leq 4k^2+4k+5.$$
On the other hand, if $p\leq 2n+\sqrt{2n}$, then
\begin{align*}
p &\leq 2(a-k)+\sqrt{2(a-k)}\\
2a+1 &\leq 2(a-k)+\sqrt{2(a-k)}\\
2k+1 &\leq \sqrt{2(a-k)}\\
4k^2+6k+2 &\leq 2a+1\\
p &\geq 4k^2+6k+2,
\end{align*}
which is a contradiction for sufficiently large $k$. On the other hand, if $k$ stays bounded then $p$ stays bounded as well, since $(2k+1)^2+4$ has finitely many prime divisors for each $k$, which is impossible by Dirichlet. Thus for all sufficiently large $p$, picking $n$ as described will make $n^2+1$ have a prime divisor (namely $p$) greater than $2n+\sqrt{2n}$. $\blacksquare$

Consider large primes $p\equiv 1\pmod 4.$ Then exists $n<\tfrac p2$ such that $p\mid n^2+1.$ This implies $p\mid (p-2n)^2+4.$ Clearly, $p-4>2n$ and so $p-2n\geq \sqrt{p-4}>\sqrt{2n}.$

[quote=CantonMathGuy]Is this really this easy? To give an indication, I was able to solve it before AoPS went to sleep...

[hide = Solution]
Let $p \equiv 1 \pmod{4}$ be a prime (of which there are infinitely many), and let $n$ be the smallest positive integer with $p \mid n^2 + 1$. Clearly $0 \le n < p$, and in fact, $n < \frac{p}{2}$ (otherwise, take $n \to p - n$).

Now, suppose that $k$ is the smallest (odd) integer with $p \mid \left(\frac{p - k}{2}\right)^2 + 1$. We have $p \mid (p - k)^2 + 4$, or $p \mid k^2 + 4$, so $k \ge \sqrt{p - 4}$. Since $n = \frac{p - k}{2}$, we have
\[n = \frac{p - k}{2} \le \frac{p - \sqrt{p - 4}}{2},\]
so $\sqrt{p - 4} \le p - 2n$, or $p - 4 \le (p - 2n)^2$. This becomes $p^2 - (4n + 1)p + 4(n^2 + 1) \ge 0$. For the left hand side to be nonnegative, we must have $p$ equal to or larger than the larger root of $x^2 - (4n + 1)x + 4(n^2 + 1) = 0$, or
\[p \ge \frac{(4n + 1) + \sqrt{(4n + 1)^2 - 16(n^2 + 1)}}{2} = \frac{4n + 1 + \sqrt{8n - 15}}{2} \ge 2n + \frac{1}{2} + \sqrt{2n - 4},\]
which is larger than $2n + \sqrt{2n}$ for $n$ large enough. We are done. (If our $n$ is too small, pick a larger $p$; the small $n$ will only consume finitely many $p$.)
[/hide][/quote]

what is the motivation to let $p \equiv 1 \pmod{4}$ be a prime and let $n$ be the smallest positive integer with $p \mid n^2+1$? Also why does we must have $p$ equal to or larger than the larger root of $x^2 - (4n + 1)x + 4(n^2 + 1) = 0$?

First solve of the year!

Let $p\equiv 73\pmod{76}$ be a large prime (there exist infinitely many such primes by Dirichlets).

Since $p\equiv 1\pmod 4$, we have $-1$ is a QR mod $p$. So there exists a positive integer $n< p$ satisfying $p\mid n^2 + 1$. We may assume $2n< p$ because we also have $p\mid (p-n)^2 + 1$ and $p$ is odd. We wish to show that $p-2n > \sqrt{2n}$. Let $p-2n = k$. Then we have \[p\mid  4(n^2 + 1) = (2n)^2 + 4,\]which implies $p\mid k^2 + 4$, so $k^2 \ge p-4$. Note that $p\equiv 16\pmod{19}$, and since $12,13,14,15$ are not quadratic residues modulo $19$, $k^2$ cannot be equal to $p-4,p-3,p-2,p-1$. Thus, $k^2 \ge p$, which implies $k^2 > 2n$, so $k> \sqrt{2n}$, as desired.

Note: Since $n^2 + 1$ has finitely many prime divisors for any fixed $n$, we must generate infinitely many $n$.

[quote=delegat]Prove that there are infinitely many positive integers $ n$ such that $ n^{2} \plus{} 1$ has a prime divisor greater than $ 2n \plus{} \sqrt {2n}$.

[i]Author: Kestutis Cesnavicius, Lithuania[/i][/quote]

Wait I solved an IMO #3 thats $>$ year $2000$ in less than 15 minutes? Im so Proud :)
___________________________________________________________________________________________________________________________________________________
First we set out to simplify the problem statement.

We see that $ 2n + \sqrt {2n} < n^{2} + 1$ for all $n > 2$.
Thus, if there exists infinite primes of the form $k = n^{2} + 1$, then the problem is solved, since the maximum prime divisor of $k$, which is itself, is necessarily higher than the bound given.

We rewrite the problem as follows: Prove that there exist infinite prime numbers of form $n^{2} + 1$ which $n \in Z^{+}$

[b]Claim: [/b]There exist infinite prime numbers of form $n^{2} + 1$ which $n \in Z^{+}$

If $n^{2} +1$ is prime then $n^2$, and $n$, are even. Thus we can consider the expression mod $4$, to assert that $n^{2} + 1$ is of form $4r + 1$ for positive integers $r$.

Simply proving there are infinite primes of form $4r + 1$ will suffice. This is well known. The proof for that is as follows

[quote = Proof]
We proceed by way of contradiction. Suppose there are only finite primes of form, labeled $p_{1}, p_{2}, .... p_{n}$.
$N=(p_{1}...p_{n})^{2}+1$; $N>p_{i}$ for $1 \le i \le n$. Hence $N$ can not be prime. Any number of the form $a^{2}+1$, has, except possibly the factor $2$, only prime factors of the form $4m+1$ .Since dividing $N$ by each prime factor of the form $4j+1$ leaves a remainder $1$, $N$ can not be composite

[b][u]CONTRADICTION[/u][/b].
Thus the statement is proven
[/quote]

Accordingly, we are done
___________________________________________________________________________________________________________________________________________________

EDIT: Wait this is 30M No way...

Okay!

[quote=bryanguo]The above proof fails (the proof of the claim is incorrect). It is one of Landau's unsolved problems.

See [url=https://en.wikipedia.org/wiki/Landau%27s_problems#Near-square_primes]this[/url].[/quote]

Wait how come?

I 99% expected that since I'm not very good at math, but why doesn't this work:

Claim: There exist infinite prime numbers of form $n^{2} + 1$ which $n \in Z^{+}$

If $n^{2} +1$ is prime then $n^2$, and $n$, are even. Thus we can consider the expression mod $4$, to assert that $n^{2} + 1$ is of form $4r + 1$ for positive integers $r$.

Simply proving there are infinite primes of form $4r + 1$ will suffice. This is well known. So done

Well consider the following invalid argument.
[quote=Variant]
 If $n^{2} +1$ is prime then $n^2 + 1$ must be odd for $n \ge 2$. Thus we can consider the expression mod $2$, to assert that $n^{2} + 1$ is of form $2r + 1$ for positive integers $r$.

Simply proving there are infinite primes of form $2r + 1$ will suffice. This is well known. So done
[/quote]

Oh I see...

But still, why doesn't it work? Like why isn't mod 4 good enough? It usually works with similar diophantines so im confused

Are there other circumstances where this is true as well?

Let's take [b]YaoAOPS[/b]'s example and push it even further.

[quote=Variant]
 If $n^{2} -1$ is prime then $n^2 - 1$ must be odd for $n \ge 2$. Thus we can consider the expression mod $2$, to assert that $n^{2} - 1$ is of form $2r - 1$ for positive integers $r$.

Simply proving there are infinite primes of form $2r - 1$ will suffice. This is well known. So done
[/quote]
But here the claim is false, because $n^2 - 1$ factors!

Do you see what's going on here?  You've expanded the search space too much.  An analogous "real life" version of your argument would be as follows:

[quote=Variant of a variant]Let's prove that over 1 billion people live in Mongolia.  To do this, note that if someone lives in Mongolia, they must live in Asia.  Thus, it suffices to prove that over 1 billion people live in Asia, which is true.[/quote]

Oh!!!!!!!! I see! Thank you so much

Basically for my proof to work I have to prove all primes of form 4n+1 are also of form x^2 + 1 which is false

tysm

For each sufficiently large prime $p \equiv 1 \pmod{4}$, let $n$ be the minimal root of $n^2 + 1 \equiv 0 \pmod{p}$. I claim that this $n$ has $p > 2n + \sqrt{2n}$, which will finish. Indeed, clearly $n \le \frac{p-1}{2} = a$, so let $n = a - k$. Now
\begin{align*}
  n^2 + 1 \equiv 0 \pmod{p} &\implies (a - k)^2 + 1 \equiv 0 \pmod{p} \\
  &\implies (2k+1)^2 + 4 \equiv 0 \pmod{p} \\
  &\implies p \le 4k^2 + 4k + 5.
\end{align*}
Putting this back in terms of $n$ gives
\begin{align*}
  p \le 4k^2 + 4k + 5  &\implies p \le (p - 2n)^2 + 4 \\
  &\implies p^2 - (4n+1)p + (4n^2+4) \ge 0 \\
  &\implies p \ge \frac{4n + 1 + \sqrt{8n - 15}}{2}
\end{align*}
which implies the bound for sufficiently large $n$.

[hide=A small remark on motivation]
I think the main motivation here is that getting direct bounds on $p$ is pretty annoying (how do we get $2n + \sqrt{2n}$ directly from $n^2 + 1$? This would be extremely unnatural), so this is not worth trying out. Instead, considering $p$ first and trying to bound the smallest value of $n$ that follows is much more reasonable. In addition, the bound $n \le \frac{p-1}{2}$ is immediate but not what we want, so we expect this is only sharp for finitely many $n$. Indeed,
\begin{align*}
  \left( \frac{p-1}{2} \right)^2 + 1 &\equiv 0 \pmod{p} \implies p = 5 \\
  \left( \frac{p-1}{2} - 1 \right)^2 + 1 &\equiv 0 \pmod{p} \implies p \le 13
\end{align*}
which gives the idea to generalize to $n = \frac{p-1}{2} - k$ pretty easily.
[/hide]

Let $p$ be some arbitrarily large prime that is $\equiv 1\pmod{4}$. Then let $n$ be the smallest $n$ that satisfies $n^2 \equiv -1\pmod{p}$. Notice that $n \leq \frac{p-1}{2}$ since there are at most two distinct solutions to $n^2 \equiv -1\pmod{p}$ that add up to $p-1 \pmod{p}$ (you can prove this with primitive roots). So then let $k = \frac{p-1}{2} - n$.
Expanding $(\frac{p-1}{2} - k)^2 + 1 \pmod{p}$ gives $4k^2 + 4k + 5 \equiv 0\pmod{p}$.
Then plugging in $k = \frac{p-1}{2} - n$ again gives $(p - 2n)^2 + 4 \geq p$. Expanding gives $p^2 - (4n + 1)p + 4n^2 + 4 \geq 0 \implies p \geq \frac{4n + 1 + \sqrt{8n - 15}}{2}$ which is enough as $\sqrt{8n - 15} \geq \sqrt{2n}$ for sufficiently large $n$.


geometry
circumcircle
geometric transformation
reflection
homothety
geometry unsolved

Let $ ABC$ be a triangle, and let $ AM$ and $ AN$ be the trisectors of angle $ A$, with $ M$ between $ B$ and $ N$. Let the extensions of $ AM$ and $ AN$ intersect again the circumcircle of triangle $ ABC$ at $ P$ and $ Q$, respectively. Let $ R$ and $ S$ be the intersections of $ AP$ with $ BQ$, and $ AQ$ with $ CP$, respectively. Let $ D$ and $ E$ be the intersections of $ BQ$ with $ CP$, and $ MS$ with $ NR$, respectively. If $ F$ is the reflection of $ E$ on $ BC$, prove that $ A, D$ and $ F$ are collinear.

$ \angle RAN\equal{}\angle MAN\equal{}\angle NAC\equal{}\angle QAC\equal{}\angle QBC\equal{}\angle RBN\Rightarrow A,B,R,N$are concyclic.
By same way, $ A,M,S,C$are concyclic, too.
We have $ \angle FNM\equal{}\angle RNB\equal{}\angle MAB\equal{}\angle BCP\equal{}\angle QAC\equal{}\angle QPC\Rightarrow BC\parallel PQ$.
And by same way, $ \angle FNM\equal{}\angle FMN\equal{}\angle DPQ\equal{}\angle DQP\equal{}\angle MAN$
$ \Rightarrow FM\parallel DP,FN\parallel DQ,MN\parallel PQ$, by homothety, we have $ PM,QN,DF$are concurrent.
$ \Rightarrow A,D,F$are conlinear.

Sorry but I'm not good at English.


calculus
integration
geometry
3D geometry
sphere
analytic geometry
trigonometry

Find the volume of the solid bounded by the cone x^2 + y^2 = z^2 and the spheres x^2 + y^2 + z^2 = 2 and x^2 + y^2 + z^2 = 8 containing the point (0,0,3/2).

This question made me a bit confused. Okay, here we use spherical coordinates and we take the upper half of the cone. But i couldn't figure the bounded part between 3 solids? Can you help me with this?

Use the relations for spherical coordinates:

$ x=\rho \sin \phi \cos \theta$
$ y=\rho \sin \phi \sin \theta$
$ z=\rho \cos \phi$

Then your spherical equations look like $ \rho^{2}=2$, $ \rho^{2}=8$, and $ \rho^{2}\sin^{2}\phi = \rho^{2}\cos \phi \hspace{1 mm}\rightarrow \sin^{2}\phi = \cos^{2}\phi$.  Note that all three equations are $ \theta$ invariant.  What values of $ \phi$ work in the above?

From here it should not be too difficult to set up the triple integral and solve.

Using the point $ (0,0,3/2)$ in the description specifies which of several possible solids bounded by these surfaces we're talking about. So we know were in the upper half space, $ z>0.$

Since we're in the upper half space, we can write the equation of the cone as $ z=\sqrt{x^{2}+y^{2}}$ or (in cylindrical coordinates), $ z=r.$

Yes, this would be simplest in spherical coordinates. What you need is to write the equations of the three surfaces in spherical coordinates.

$ x^{2}+y^{2}+z^{2}=2$ becomes $ \rho=\sqrt{2}$

$ x^{2}+y^{2}+z^{2}=8$ becomes $ \rho =\sqrt{8}$

and $ z=\sqrt{x^{2}+y^{2}}$ becomes $ \phi=\frac{\pi}4.$

(OK, I'm mostly repeating what edoo said.)

Thanks for your help. Now i have done with expressing the integral.


AMC
USA(J)MO
USAMO
trigonometry
inequalities
geometry
pigeonhole principle

I'm trying to make an archive of the solutions to the TJ USAMO problems. Why don't you contribute by solving the problems in the following pdf?

http://www.tjhsst.edu/~tmildorf/math/2004TJUSAMO2.pdf

[hide="#1"]

The largest possible sum of a subset is $90+91+\cdots 100 = 1045$, and the smallest is 1.  But in a set of 11 numbers there are $2^{11}= 2048$ subsets.  Since there are only 1045 possible sums, two subsets must have the same sum.  If they are not disjoint, simply remove all common elements; those that are left also must have equal sums (and they are subsets).

[/hide]

[hide="#2 - Very nice problem"]We'll substitute $a_{i}=\tan\phi_{i}$. For every real $a_{i}$ there's uniquely determined value $\phi_{i}\in(-{\pi\over 2},{\pi\over 2})$. Then we have

$\frac{a_{i}-a_{j}}{1+a_{i}a_{j}}=\frac{\tan\phi_{i}-\tan\phi_{j}}{1+\tan\phi_{i}\tan\phi_{j}}=\tan(\phi_{i}-\phi_{j})$

and

$\sqrt{2}-1=\tan{\pi\over 8}$

(From $\tan x=\sqrt{2}-1$ it can be derived $\cos x=\sqrt{\frac{2+\sqrt{2}}{4}}=\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}}$, and that's obviously the value of $\cos\frac{\pi/4}{2}$)

Therefore our condition becomes

$k\pi<\phi_{i}-\phi_{j}<k\pi+{\pi\over 8}$

for some integer $k$, where $\phi_{i},\phi_{j}\in(-{\pi\over 2},{\pi\over 2})$. Because $-\pi<\phi_{i}-\phi_{j}<\pi$, the possible values for $k$ are only $-1$ and $0$. Now multiply the inequality by ${8\over\pi}$ to turn it into

$8k<{8\phi_{i}\over\pi}-{8\phi_{j}\over\pi}<8k+1$

and substitute $b_{i}={8\phi_{i}\over\pi}$. Because of the restriction on $\phi_{i}$, we have $b_{i}\in(-4,4)$

Since $k\in\{-1,0\}$, we must have either $-8<b_{i}-b_{j}<-7$ or $0<b_{i}-b_{j}<1$. The first possibility can take place only if $b_{i}<-3$ and $b_{j}>3$, and of course we can choose arbitrarily many numbers $b_{k}$ without fulfilling that requirement. The second possibility means that in the interval of length $8$ we must have two points that are less than $1$ apart. Obviously that will always happen if we choose $\boxed{n=9}$ points. (Eight points aren't enough, e.g. $(-3.5,-2.5,-1.5,-0.5, 0.5, 1.5, 2.5, 3.5)$)[/hide]

I printed out all 13 of these...going to be working on them over a period of time, and I'll give you the solution i get if you want.

Help with four and five please? I can't get them, and I've tried intermittently for a few days.

Question: for #3, is a single digit considered a string?  Also, is 0 a perfect square?

[hide="hint on #4"]Think what condition would HAVE to happen if you had a region of area greater than 4.  Think mods...[/hide]

[hide="another hint on #4"] Suppose you had two points in the plane such that one is the other plus something of the form (2m, 2n) where m and n are integers.  Then what?  When does this necessarily have to happen?[/hide]

By the way, the generalization of #4 to any lattice is called Minkowski's Theorem.

[quote="Phelpedo"]#4 is one of those questions that is quite obvious but tricky to argue rigorously...[/quote]
Yeah, which is why I need a proof :(

[hide="why no one posted the answer to 0 yet?"]Assume the opposite for contradiction: no two know the same number. Thus there are n people who know 0,1,2,3,...n-1 people respetively (you can't know n people because you don't count as knowing yourself). But then someone knows all n-1 other people, contradicting the guy who knows nobody. Contradiction, $\mathbb{Q. E. D}$[/hide]

[quote="K81o7"]i can put my solutions for 4 and 5, but i have to go now...so i'll put them when i get back in around 2 hours...

Ah heck...i'll outline them [hide="#4"] Suppose you have two points $(x, y)$ and $(x+2m, y+2n)$, s.t. $m$ and $n$ are not both 0.  If the area exceeds $4$, this follows by a pigeonhole type argument...Then the midpoint of (-x, -y) and (x+2m, y+2n), namely (m, n) is in the set.  Done.[/hide]

[hide="hint on 5"]Basically, first show that if a point is outside a triangle, then there exists a line through that point which does not intersect the triangle.[/hide]

[hide="5"]Okay...so draw a line through the center of the square, so that the triangle is on some side of it.  You can prove this has to be the case by proving the above hint with two simple cases.  Then draw the perpendicular bisector on the side where the triangle is of the line segment you just made through the center of the square.  You have two small regions there now.  Each is a cyclic quadrilateral whose circumradius is less than $\frac{1}{2}$ by a pythagorean theorem argument.  Use the pigeonhole principle: three points and two regions means two points in one region, so two points with distance less than 1.  Then a little work considering points on the segment dividing the two regions to be thorough, and you're done.[/hide][/quote]
I don't follow the solution for 4...

I'm sure you can lookup a proof of Minkowski's somewhere that goes into more detail than him...

This is k81o7's solution, worded more rigorously:
[hide]
We must have some pair of points $(x,y), (x+2m,y+2n)$, with $m,n$ integers not both 0, in the figure. Otherwise, the length and width of the figure are both less than 2, so the area is less than 4. Since the figure is centrally symmetric, $(-x,-y)$ are also in the figure. We can take the midpoint of $(-x,-y)$ and $(x+2m, y+2n)$, which is $(m,n)$. By our definition of $(m,n)$, it is a lattice point which is not the origin.
[/hide]

Well, for the first part of #4, [hide] you couldn't use a width and height argument.  But you could consider that square with area 4, and consider positions of things inside that...ie a "mod" argument.[/hide]

Can't you do #4 using Pick's theorem? :maybe:

If it were a lattice polygon, we would be quite finished :lol:

[quote="Phelpedo"]I don't quite understand what you mean. If the figure is less than  2 in each direction, it can be contained in a 2X2 square, so its area is less than 4. Thus, it has to be at least 2 in at least one dimension.[/quote]

Oh...I see what you're saying now...

Width=maximum width, ie maximum amount taken up of any horizontal line?

Pretty cool proof for #3
[hide] We will prove that the smallest possible $n$ is $16$.

Let the terms of the sequence be $a_{1}, a_{2}, ...a_{1}6$. Let $a_{i}$ be the first term of the string and $a_{j}$ be the last one.

Now fix $i=1$ and consider the $16$ different string we obtain if we let $j=1,2,3...16$. Each product can be factorized as $2^{a}3^{b}5^{c}7^{d}$. If, at any stage, all of $a,b,c,d$ are even, then we are done because the product is a perfect square. Therefore, assume that they are not all even. There are then $2^{4}-1=15$ possible combinations of odd and and even, so by the Pidgeonhole principle, at least 2, $a_{m}$ and $a_{n}$, yield the same parity for each of the exponents. We then let $i=a_{m}$, $j=j_{n}$, so the exponents in the prime factorization of the product are the difference of the exponents of $a_{1}a_{2}...a_{m}$ and $a_{1}a_{2}...a_{n}$. Since all of the exponents have the same parity, the differences will all be even, so the product is a perfect square.

It is clear by this construction that any sequence $a_{i}*...*a_{j}$ can be obtained by dividing $a_{1}a_{2}...a_{j}$ by $a_{1}a_{2}...a_{i}$. If $n<16$, then it is possible for all the exponents to have a different odd-even sequence, with not all of them even.
[/hide]

EDIT: Ravi, that is a good counterexample :blush: If someone can offer a proof of that first part, I would be much obliged.

[quote="Ravi B"]Phelpedo, I don't understand your argument.  First, you never define length and width.  Second, you seem to be saying that if the length or width is at least 2, then there must be two points that are the same mod 2.  But consider the triangle with vertices $(0, 0)$, $(1, 0)$, and $(100, 1)$.  I would say that the length of this triangle is bigger than 2 (although I don't know your definition of length), but it doesn't have two points that are the same mod 2.[/quote]

Well, by the definition I was assuming he was using, that triangle would have width 1...

But still, it's true that by that method, you in fact have to butcher up the figure and move things around mod 2...

Here's the way I did it (but didn't have time to outline above)...
[hide]
Consider the square defined by (0, 0), (2, 0), (0, 2), (2, 2).  Take the region inside this square union two adjacent sides union a vertex.  This is the set of all points mod 2 (ie any point in the plane is congruent to some point in this square mod 2, and in this case, no two points in the square are congruent mod 2).

If there are no two points congruent mod 2 in a region, then that means we can create a function which takes every point in the region to the point congruent to it mod 2 in the square, and no two points in the region will map to the same point in the square (because no two are congruent mod 2).  Thus, for each point in the square, there is at most 1 corresponding point in the region.  Then the area of the region is less than or equal to the area of the square (okay, you can make infinity arguments to make this rigorous, but i'm not going to do that here...), which is less than or equal to 4.  Thus, if a region in the plane has no two points which are congruent mod 2, then it has maximum area 4.  The contrapositive statement is what we wanted.[/hide]

If I understand it correctly, your area argument seems a bit dangerous. You claim that because there is a bijection between points in the square and points in the region, the areas are equal, but this argument isn't true. Example: suppose I have a square with side 1 and a square with side 2, both centered around the origin. The function $f(x)=x/2$ establishes a bijection between points in the big square and the little square. However, their areas are not equal.

Yes, that's true...that's why I said it wasn't rigorous...
But the residue class argument makes sense.  I don't know how to make it completely rigorous, but it's clear as to why it should be true...
I shouldn't have said something about number of points, but rather about a particular residue class.  It is convex, you could make an argument about forming the lattice of even points, say, $\mathbb{Z}_{2}$ in the complex plane, and then using the division algorithm over $\mathbb{C}$ to talk about non-integer remainders in that lattice, but even then there's other stuff you need to prove...I don't know if there's a very elementary way to do it if you want to be completely rigorous...

Phelpedo, you are right that K81o7's argument needs some care.  Fortunately, it all works out in an elementary way.  The key point is that mod is locally a translation.

Here is the argument.  Let $P$ be the given region.  Suppose that it does not contain two points $(x, y)$ and $(x+2m, y+2n)$ where $m$ and $n$ are integers, not both $0$.  We will show that the area of $P$ is at most $4$.

Let $S$ be the square $[0, 2) \times [0, 2)$.  We have the following bound on the area of $P$: \begin{eqnarray*}|P| & = & \bigl| \bigcup_{m, n \in \mathbb{Z}}P \cap [S+(2m, 2n)] \bigr| \\ & \le & \sum_{m, n \in \mathbb{Z}}\bigl| P \cap [S+(2m, 2n)] \bigr| \\ & = & \sum_{m, n \in \mathbb{Z}}\bigl| [P-(2m, 2n)] \cap S \bigr| \\ & = & \bigl| \bigcup_{m, n \in \mathbb{Z}}[P-(2m, 2n)] \cap S \bigr| \\ & \le & |S| \\ & = & 4 \, . \end{eqnarray*}

[quote="Ravi B"]
Let $S$ be the square $[0, 2) \times [0, 2)$.  We have the following bound on the area of $P$: \begin{eqnarray*}|P| & = & \bigl| \bigcup_{m, n \in \mathbb{Z}}P \cap [S+(2m, 2n)] \bigr| \\ & \le & \sum_{m, n \in \mathbb{Z}}\bigl| P \cap [S+(2m, 2n)] \bigr| \\ & = & \sum_{m, n \in \mathbb{Z}}\bigl| [P-(2m, 2n)] \cap S \bigr| \\ & = & \bigl| \bigcup_{m, n \in \mathbb{Z}}[P-(2m, 2n)] \cap S \bigr| \\ & \le & |S| \\ & = & 4 \, . \end{eqnarray*} [/quote]

Sorry, but I don't understand these computations, probably because I have no clue what the $\bigcup$ is.

it's union of all such sets...that is, the area of $P$ is the area of the union of its intersections with each tranlation of $S$ (ie if you use $S$ to make the lattice $2\mathbb{Z}$, the area of $P$'s intersections with all of those copies of $S$...namely because those copies of $S$ all add up to the plane, which is where $P$ entirely lies anyways.

The second step is self-explanatory...they're technically equal from an area standpoint, but meh, doesn't matter.

The next step is the different between a translation of $S$ one way or a translation of $P$ the other way.

The step after that is just going back

And the last step is true because we have no points in $P$ which "repeat" throughout all the sets we're taking the union of.  I guess that might have to be cleared up to make this really rigorous, but Ravi B puts it well.

So why is there a less than or equal sign in the second step? Is that an identity or a typo?

You both are right: The $\le$ on the second step could just as well have been an equals sign.  I was being unnecessarily conservative.

In words, here is what is going on.  It's a proof by "tangrams".  To show that the area of our region $P$ is at most $4$, we will break $P$ up into pieces, and then move those pieces to fit into a $2 \times 2$ square.

Specifically, let's tile the plane with $2 \times 2$ squares.  That effectively breaks $P$ into pieces.  Let's translate all of those pieces of $P$ into one particular square, say $[0, 2) \times [0, 2)$.  (Translation preserves area.)  By the "no two points" property, those translated pieces will all be disjoint.  So now we have a bunch of disjoint pieces inside a $2 \times 2$ square.  It follows that the sum of their areas must be at most $4$.  That implies that the area of $P$ is at most $4$.

In the spirit of this theme, I'll add another problem, closely related to this one (if you have read solution before, please don't give it away): prove that ANY figure of area greater than $n$ can be translated to cover at least $n+1$ lattic points.


For each plumbing repair job, Mr. Wrench charges $ N$ dollars for coming out to the house plus $ x$ dollars per hour that he works at the house. He charged $ \$97$ for a one-hour repair job and $ \$265$ for a five-hour repair job. What is his charge for a two-hour repair job?

From the given, we have a system of equations

$ N\plus{}x\equal{}97$

$ N\plus{}5x\equal{}265$

Subtracting the first equation from the second, we have

$ 4x\equal{}168 \implies x\equal{}42$. Substituting this into our first equation, we get that $ N\equal{}97\minus{}42\equal{}55$.

Thus, the charge for two hours is $ 55\plus{}2(42)\equal{}\boxed{139}$.


geometry
perimeter
algebra
system of equations
perpendicular bisector

Can anyone solve these? They're extra credit problems for my precalc class, and I think if I do them I could boost my grade up to an A....

[b]Find the sum of the reciprocals of two real numbers, given that these two numbers have a sum of 50 and a product of 25.[/b]

I have no clue on this one. I tried using a system of equations, but it didn't really work out to suit what the problem is asking.

[b]A typical telephone number in the United States used to be 7 digits.  Recently, several areas of the country have started using ten-digit telephone numbers.  Columbus, Georgia and Orlando, Florida are two such cities.  If the entire country goes to a ten-digit telephone, exactly how many different ten-digit telephone numbers will be available such that the first digit cannot be a 0 or 1 and the fourth digit cannot be a 0 or a 1?[/b]

I think this can be solved using combinatorics or permutations...one of them. I'm not completely sure though.

They should be easy....I'm just retarded with math.  :D

[hide="1"] We have two numbers $a, b$.  We are given that $a+b = 50, ab = 25$.  Then

$\frac{1}{a}+\frac{1}{b}= \frac{a+b}{ab}= \frac{50}{25}= \boxed{2}$. [/hide]
[hide="2"] The first and fourth digits can be a digit from $2$ to $9$; there are $8$ such digits.  The second, third, fifth, sixth, seventh, eighth, ninth, and tenth digits can be a digit from $0$ to $9$; there are $10$ such digits.  Hence there are a total of

$8^{2}\cdot 10^{8}= \boxed{ 6400000000 }$

Such phone numbers. [/hide]

I already figured out the first one soon after posting, but I didn't know the second one.

Thanks!

Ok, I'm trying to do more of these.

[b]Suppose a country only issued two types of coins to be used as currency: one 7 unit coin and one 11 unit coin. This would cause a dilemma since certain prices could not be paid exactly, such as 10 units. If this was the case, what would be the highest price that could not be paid with any combination of the two coins?[/b]

I'm not good with these type of problems at all.

[b]Given a triangle with side lengths of 15, 20, and 25 units, what is the length of the shortest altitude?[/b]

I used Heron's formula....would it be 12?

[b]Points A, B, C, and D are all on segment AD, in the order listed.  Segment AB is congruent to segment CD.  Segment BC equals 12 units in length. Also, an additional point E exists that is not on segment AD with segment BE being congruent to segment CE.  Segment BE equals 10 units in length.  If the perimeter of triangle ADE is twice the perimeter of perimeter of triangle BCE, find the length of segment AB.[/b]

I don't know where to go with this one....

[hide="the first one you just posted"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30040

Suppose a country only issued two types of coins to be used as currency: one 7 unit coin and one 11 unit coin. This would cause a dilemma since certain prices could not be paid exactly, such as 10 units. If this was the case, what would be the highest price that could not be paid with any combination of the two coins?

I think the rest is easy after you know this.[/hide]

let's fix that link real quick: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30040[/url]

triangle one: note that $15: 20: 25 = 3: 4: 5$ which means it's a right triangle with hypotenuse 25 and legs 15,20.  this means area is 150.  using $A = \frac{ah_{a}}{2}$ we find that $h_{a}= \frac{2A}{a}$.  using $a = 25$ we get the longest altitude to be 12.  gj, but no need to use heron.

geometry one:  [hide]since $AB = CD$ it should be obvious that $E$ lies on the perpendicular bisector of $AD$ as well.  we have that the perimeter of $BCE$ is 12+10+10 = 32 units.  hence the perimeter of $ADE$ is 64.  Let $M$ be the midpoint of $BC$ (and of $AD$)  it is obvious that $EM \bot AD$.  it is obvious by pythagoras that $EM = 8$

Let $AM = x$.  We have $AE = DE = \sqrt{64+x^{2}}$ and $AD = 2x$.  hence the perimeter of $ADE$ is 64.  so we have

\[2\sqrt{64+x^{2}}+2x = 64\]
\[\sqrt{64+x^{2}= 32-x}\]
\[64+x^{2}=1024-64x+x^{2}\]
\[0-960-64x\]
\[x = 15\]

hence $AD = 30$[/hide]

In case you didn't follow the links, the first one uses something called the chicken mcNugget Theorem.
That states that for positive integers m and n then the largest number you cannot make with a linear combination of m and n is $mn-m-n$
The name originally came from the boxes of chicken mcnuggets sold at McDonalds.


A chemistry student mixed some 30%-copper sulfate solution with some 40%-copper sulfate solution to obtain 100 mL of a 32%-copper sulfate solution. How much of the 30%-copper sulfate solution did the student use in the mixture?

I used to know how to do this, but I totally forgot and I have no idea what to search for on google to find out the procedure. It's something simple, I know, but I can't seem to figure it out.

Thanks.

[quote="brainomega"].3*a+.4*b=.32*100
a+b=100

Solve the system of equations.[/quote]
Not sure how to do that. I ended up getting something crazy when I tried to isolate .4b on one side and solve for B, then use that to solve for A. I got 79.25 for B and 1 for A.

Thanks. I knew it was easy, I just couldn't figure it out for some reason.


complex analysis
function
topology
complex analysis unsolved

This should be simple, but I'm blanking on how to go about it: Does there exist an entire holomorphic function taking only real values on the boundary of the unit circle?

There are no such non-constant functions.

Assume that the function is a non-constant function. Consider the image of the (open) unit disk. Denote it by $S$. Then we have:
i) $S$ is an open set.
ii) By the condition, $\partial S \subset \mathbb{R}$.
iii) $S$ is bounded.

Clearly, there is no set that satisfies properties i-iii.

Alternative approach: apply the Maximum Modulus Principle to the functions $e^{i f(z)}$ and $e^{-i f(z)}$.

I'm sorry, but why is it a tautology to say that there's no set corresponding to a non-constant function that satisfies the first three properties (an open, bounded set, with a real boundary).

With respect to the maximum modulus approach: With respect to both functions, |f(z)| on the boundary is 1.  So either e^(I*f(z)) or e^(-I*f(z)) is constant, since e^(I*f(z)) takes its maximum on the boundary <=> 1/e^(i*f(z)) takes its maximum in the circle. Is that what you're getting at?

[quote="gka"]I'm sorry, but why is it a tautology to say that there's no set corresponding to a non-constant function that satisfies the first three properties (an open, bounded set, with a real boundary).
[/quote]
There is no open set that has properties i-iii. This is a geometrically obvious statement. This statement is also very easy to prove. Hint: Project the set to the $y$-axis. The projection is a connected bounded open set, i.e. it is an interval $(a,b)$. So the projection has $2$ boundary points. But the projection of the boundary of $S$ (i.e. a subset of the $x$-axis) to the $y$ axis is just one point.

[quote]
With respect to the maximum modulus approach: With respect to both functions, |f(z)| on the boundary is 1.  So either e^(I*f(z)) or e^(-I*f(z)) is constant, since e^(I*f(z)) takes its maximum on the boundary <=> 1/e^(i*f(z)) takes its maximum in the circle. Is that what you're getting at?[/quote]
Yes. One of them is a constant, so $f(z)$ is a constant.

Here's another approach. Let $f=u+iv$, $u,v$ being harmonic functions. Let $\Omega$ be the open unit disk, then by data
\[\begin{cases}\Delta v = 0 & \text{ in }\Omega, \\ v = 0 & \text{ on }\partial \Omega \end{cases}\]
yielding that $v=0$ in $\Omega$, hence in $\mathbf C$.

Thus $f=u$, hence is constant.


abstract algebra
superior algebra
superior algebra unsolved

Is this easy , difficult or unsolved?

Let $W$ be a group, is there a group $G$ such that $G/Z(G)\cong Inn(G)\cong W$

I found that $W$ cannot be cylic (unless when trivial)
Can you help me?

What do you mean by "for every perfect $W$"?

Simply, that if $W$ is perfect, $W$ is isomorphic to an inner automorphism group of some group, but it is actually isomorphic to its own inner automorphism group

[quote="fredbel6"]I found that $W$ cannot be cylic (unless when trivial)[/quote]Well, moreover it cannot be abelian, right?

Why not, consider the quaternion group?
Its centre is -1 and 1
the inner automorphism group has four elements and thus is abelian, and nontrivial?

Isn't $G/Z(G)\cong Inn(G)$ always true (i.e for any group $G$) ?
Just consider the operation of $G$ on $G$ by conjugation.
The kernel of this operation is $Z(G)$ and the image is $Inn(G)$.
Thereby by the isomorphism theorem we get the claim.

{x} yes that is true, the 'exciting part' is the second isomorphism sign, as the first one always holds
i just wrote that down to make it easier for everyone to find it

i still have no solution
all remarks welcome!

Why can't we set $W=G$ ? I don't quite get the problem :(
Also what is a perfect group ?

I am beginning to have serious doubts about the perfect group statement
I am afraid i mixed it up with another problem!

so {x}thx
this is the problem now

I give you a group W
can you give me a group G such that Inn(G) is isomorphic with W?
Why do you think you can set G=W?


search
topology
real analysis
real analysis unsolved

Is the metric space $(\mathbb{R}\setminus\mathbb{Q}, d)$, where $d(a,b)=|b-a|$ just a standard metric, a separable space? My thoughts are no. You can't get a countable subset of $A\subset\mathbb{R}\setminus\mathbb{Q}$ such that $\text{Cl}(A)=\mathbb{R}\setminus\mathbb{Q}$. I don't really have a contradiction, so it is only intuition I'm relying on. A yea or ney on separable would be fine, since it is in search of a different counterexample.

A countable dense set: $\sqrt{2}+\mathbb{Q}$.

[quote="HilbertThm90"]You can't get a countable subset of $A\subset\mathbb{R}\setminus\mathbb{Q}$ such that $\text{Cl}(A)=\mathbb{R}\setminus\mathbb{Q}$.[/quote]
You were probably thinking of $\text{Cl}(A)$ as the closure of $A$ in $\mathbb R$. But it is the closure of $A$ in $\mathbb R\setminus \mathbb Q$ that is relevant here. For the set given by jmerry, $\text{Cl}_{\mathbb R}(A)=\mathbb R$ and $\text{Cl}_{\mathbb R\setminus \mathbb Q}(A)=\mathbb R\setminus\mathbb Q$.

try to prove that if $X$ is a separable space, then any subset of $X$ is separable too.


[hide]I'm guessing you meant degrees instead of taking it to the zeroth power, which would be trivial.
$\cot{x}=\cos{x}/\sin{x}=\cos{x}/\cos{(90-x)}=y$
Two complementary angles are reciprocals here.
$1/(1+y)+1/(1+1/y)=(1+y+1+1/y)/(1+1+y+1/y)=1$
Thus, every angle adds up with its complement and forms one.  Only 45 degrees doesn't have a match since it's its own complement.
$\cot{45}=1$, so the answer is 4*3+1+0.5=13.5[/hide]

A latex hint: a perfect circle (good for function composition or the degree symbol) can be had with \circ: $f\circ g$ or $360^\circ$


group theory
abstract algebra
linear algebra
matrix
modular arithmetic
number theory
greatest common divisor

Prove that for fixed n, GL_n(Z) has just a finite number of subgroups, any two not being isomorphic. Great problem!

You mean that it has a finite number of isomorphism classes of subgroup: the subgroup generated by $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ in $GL_2(\mathbb{Z})$ has infinitely many subgroups, all of which are isomorphic to the integers.

could you give us a hint harazi ? it is a very interesting pb  :)

Yes, reduce the elements of each subgroup modulo different prime numbers. I think we discussed some other problems with this idea.

Ok, I admit this is a terrible problem and a hint is more than useless. Try to solve the second problem from the file.

hi Harazi,
I've just read the paper you mentioned in your previous post.
There is something I can't understand in exercice 2.d.
Why $(I+p^rB)$ does belongs to $Gl(n,Z)$ ?? For instance,
$M=I+p^rI$ does not since $M^{-1}$ is not in $GL(n,Z)$? Where is
my mistake ?

I also thought about this, but I think they refer for this particular case at $ M_n(Z)$ and not $GL_n(Z)$; I should also add that I've just realized my argument for concluding there are a finite number is idiot and thus I have no solution for the last question. Do you have one?

prove that $GL_2(Z)$ has infinitely many subgroups, 2 of them not being isomorphic !!

:?  :? Excuse me if I'm wrong  :blush: , but isn't it obvious that every group $G$ which is not finitely generated has infinitely many isomorphism classes of subgroups?!!??!! What's wrong with this: take an arbitrary $g_1 \in G$; because $G \neq \langle g_1 \rangle$, there exists an element $g_2 \in G \setminus \langle g_1 \rangle$; because $G \neq \langle g_1,g_2 \rangle$, there exists an element $g_3 \in G \setminus \langle g_1, g_2 \rangle$. Construct the infinite sequence $g_1,g_2,\ldots$ inductively, using at each step that $G$ is not finitely generated. Then, the subgroups $P_1 = \langle g_1 \rangle, P_2 = \langle g_1,g_2  \rangle, \ldots, P_k = \langle g_1,\ldots,g_k \rangle, \ldots$ are pairwise non-isomorphic, because the minimal number of generators of $P_k$ is precisely $k$.

That doesn't work, vess.
Example: let the group be $(\mathbb{Q},+)$. Every finitely generated subgroup is isomorphic to $\mathbb{Z}$.

I am not sure this is entirely correct. Think about $\mathbb Z_2\times \mathbb Z_3$ (direct sum of additive groups). The element $b=(0,1)$ does not belong to $\langle a=(1,0)\rangle$, and yet $\mathbb Z_2\times\mathbb Z_3=\langle a,b\rangle=\langle (1,1)\rangle$, so even though it's generated by two elements, it can have fewer than $2$ generators. In other words, with your notation, we can have $P_1\cong P_2$, can't we? :?

Or did I misunderstand you? :? If so, I apologize.

Edit: jmerry got here first :)

I see, I'm sorry for spamming the topic. I had tacitly assumed something that is criminally false: that $G$ has a minimal set of generators (silly me  :) ). Anyway, a group $G$ must have infinitely many isomorphism classes of subgroups provided that it has an infinite set $X$ of generators that is minimal in the following (non-usual) sense: for every finite subset $S \subset X$, the minimal number of generators of the group $\langle S \rangle$ is $|S|$. How about the converse? What do you think, jmerry & grobber?

Come to think of it, I think I can show that $\text{GL}_n(\mathbb{Z})$ has only finitely many isomorphism classes of finite subgroups. First of all, if $p \geq 3$ is any prime, then the set of elements of $\text{GL}_n(\mathbb{Z})$ that are congruent to $1$ (the identity, or $I_n$) modulo $p$ form a torsion-free subgroup of $\text{GL}_n(\mathbb{Z})$. (Indeed, if we assume the contrary, then there exists a matrix $M \in \text{M}_n(\mathbb{Z})$ not divisible by $p$ such that $1 + p^rM$ has prime order $\ell$ for some positive integer $r$. Hence, $\ell p^rM = - \sum_{2 \leq i \leq \ell} \binom{\ell}{i}p^{ri}M^i$, which is easily seen to be a contradiction by comparing the powers of $p$ in both sides. ) Hence, every finite subgroup of $\text{GL}_n(\mathbb{Z})$ maps injectively, for every prime $p \geq 3$, into ${\text{GL}_n(\mathbb{Z}_p)}$, a group of finite order $(p^n-1)(p^n-p)\cdots (p^n-p^{n-1})$. Hence, every finite subgroup of $\text{GL}_n(\mathbb{Z})$ has bounded order, so these subgroups are only finitely many up to isomorphism (and, furthermore, we may compute the l.c.m. of their orders as the g.c.d. of the $(p^n-1)(p^n-p)\cdots (p^n-p^{n-1})$).

for the infinite subgroups: try to find $A$ and $B$ such that $(A, B)$ generates a free group.

[quote="vess"]Hence, $\ell p^rM = - \sum_{2 \leq i \leq \ell} \binom{\ell}{i}p^{ri}M^i$, which is easily seen to be a contradiction by comparing the powers of $p$ in both sides. ) Hence, every finite subgroup of $\text{GL}_n(\mathbb{Z})$ maps injectively, for every prime $p \geq 3$, into ${\text{GL}_n(\mathbb{Z}_p)}$, a group of finite order $(p^n-1)(p^n-p)\cdots (p^n-p^{n-1})$. Hence, every finite subgroup of $\text{GL}_n(\mathbb{Z})$ has bounded order, so these subgroups are only finitely many up to isomorphism (and, furthermore, we may compute the l.c.m. of their orders as the g.c.d. of the $(p^n-1)(p^n-p)\cdots (p^n-p^{n-1})$).[/quote]

hi Vess, could you the end of your proof ? I can't understand what you mean .
 thanx

Hi Alekk, sorry for the bad explanations. The idea is pretty simple, actually  :) . To show --- without aiming at good upper bounds --- that the finite subgroups of $\text{GL}_n(\mathbb{Z})$ have bounded order (and thus are finitely many), it is not necessary to reduce modulo all primes, but just modulo some $p \geq 3$ --- say, modulo $3$. We know that the subgroup $\Gamma_3 := \langle X \in \text{GL}_n(\mathbb{Z}) \mid X \equiv I_n \pmod{3} \rangle < \text{GL}_n(\mathbb{Z})$ is torsion-free, i.e. it has no elements $x \neq I_n$ of finite order (and the same things holds modulo every odd prime $p$). Now, let $H$ be a finite subgroup of $\text{GL}_n(\mathbb{Z})$. Then, for any two distinct elements $A,B \in H$, $AB^{-1} \neq I_n$ is an element of finite order, and hence cannot belong to the torsion-free group $\Gamma_3$, which means that $A \not\equiv B \pmod{3}$. Therefore, reduction modulo $3$ yields an embedding of $H$ into the finite group $\text{GL}_n(\mathbb{Z}_3)$, which implies that $|H|$ is bounded by its order $(3^n-1)(3^n - 3) \cdots (3^n - 3^{n-1})$. (or we could just remark that there are only finitely many residue classes of $n \times n$ integer matrices modulo $3$). QED

I suppose the fact that $\Gamma_p := \langle X \in \text{GL}_n(\mathbb{Z}) \mid X \equiv I_n \pmod{p} \rangle$ is torsion-free for every prime $p \geq 3$ is clear: in comparing the exponents of $p$ in the two sides of the equation I wrote, we just distinguish two cases with respect to $\ell = p$ or $\ell \neq p$, and easily obtain a contradiction in both cases.

Actually, if we want to obtain better upper bounds for the l.c.m. of the orders of the finite subgroups of $\text{GL}_n(\mathbb{Z})$ (this is what I meant in the brackets), the same argument shows that we can embed, for all primes $p \geq 3$, every finite subgroup $H$ of $\text{GL}_n(\mathbb{Z})$ into the finite group $\text{GL}_n(\mathbb{Z}_p)$ of order $(p^n-1) (p^n - p) \cdots (p^n - p^{n-1})$, which shows that $|H|$ must divide this number. Hence, the l.c.m. of the orders of all finite subgroups does not exceed the quantity
\[
\text{GCD}_{p \geq 3} \, \big\{ (p^n-1) (p^n - p) \cdots (p^n - p^{n-1}) \big\}.
\]

thanx for your explanations  :)  very nice proof !


combinatorial geometry
combinatorics solved
combinatorics

We are given a set of 2004 points in the plane $A_i$ and another point $P$, no three of which are collinear. Let $F$ be the number of triangles $A_iA_jA_k$ containing $P$ in its interior. Prove that $F$ is even.

I think $F$ is even for any number of points which is divisible by 4. First of all, you can see that this is true for 4 points. Then, if we consider 8 points, we divide them into 2 groups of 4 points each. There is an even number of triangles in each group which contains $P$, and we have to prove that the number of triangles which contain $P$ and which have 2 vertices in one group and 1 in the other one is also even.

Let's denote the sides and diagonals of the first quadrilateral $a_1, a_2, a_3, a_4, b_1, b_2$, and the ones of the second quadrilateral $c_1, c_2, c_3, c_4, d_1, d_2$. Now we form quadrilaterals with the extremities of the following pairs of segments: $(a_1,c_1), (a_2,c_1), (a_2,c_2), (a_2,c_1), (a_3,b_3), (a_3,b_4), (a_4,b_4), (a_4,b_3), (b_1,d_1), (b_1,d_2), (b_2,d_2), (b_2,d_1)$. We can see that every triangle which has 2 vertices in one group of 4 points and 1 vertex in the other group has its vertices among the vertices of one of the quadrilaterals formed like this, and also this is true for only one quadrilateral (check it out). But for each quadrilateral the number of triangles which contain $P$ is even, so we have our conclusion for 8 points. We can do something similar for 12 points ( divide them in 3 groups of 4 points etc.). Because there is an even number of triangles which contain $P$ and which have 2 vertices in one group and 1 vertex in another group, then there is an even number of triangles which contain $P$ and have their vertices in the 3 different groups of 4 points.

$2004 = 4 \times 501$, so we divide 2004 in 501 groups of 4 points. A triangle can: have all 3 vertices in one group, and in this case there is an even number of triangles which contain $P$, it can have 2 vertices in one group and 1 vertex in another group, and again we have an even number of such triangles which contain $P$, or it can have its vertices in 3 different goups, but the conclusion is still true, because it's true for 12 points. Check it out yourself for 12 points (because I'm not sure it's a good solution :D ), and tell me what you think, will you?

Bye!

Seems correct!
The main thing is to notice that in any set of 4 points there is an even number of triangles that contain the point P.

Actually, there's a much simpler way to arrive at the conclusion once you realize that the problem holds for every 4 points (which actually contains two cases, depending on whether there is a point inside the convex hull or not). Just consider [i]all[/i] possible groups of 4 points; a little double-counting and you're done.

In detail: Let $\mathcal{X}$ be the set of 3-point subsets of $\{A_i\}$ that form a triangle with $P$ in the interior, and let $\mathcal{Q}$ be the set of 4-point subsets of $\{A_i\}$. What we know is that, for any $S \in \mathcal{Q}$, an even number of its 3-point subsets are in $\mathcal{X}$. In other words, the cardinality of the set \[ \mathcal{Y}_S = \{ T \mid T \subset S, T \in \mathcal{X} \} \] is even for any $S \in \mathcal{Q}$. So the cardinality of the set \[ \mathcal{Y} = \{ (T, S) \mid T \subset S, T \in \mathcal{X}, S \in \mathcal{Q} \} \] is even since $|\mathcal{Y}| = \sum_{S \in \mathcal{Q}} |\mathcal{Y}_S|$. But for every triangle $T \in \mathcal{X}$, it is contained in exactly 2001 such 4-point subsets (there are 2001 remaining points to choose from), which means that \[ |\mathcal{Y}| = 2001|\mathcal{X}|, \] implying that $|\mathcal{X}|$ is even as well.

I posted a different solution here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3670126#p3670126


Pretty Easy Problem.

What is the remainder when 10988065 is divided by 2103?

Okay. We found a way to do this very easily with a calculator, but we (as in our class) want to know a
quicker way we can do WITHOUT a calculator or long division...

[hide="Answer"]1993 hehe. the year[/hide]

wow...um well...*cough*  *sigh*  *cough* since it is the target you use your calculator, although i have no idea why it would be like #7!!!!!!FAST WAY--I have no idea

Long Division is very quick... so. er. Do it.

hmm....yes long diision is kinda quick but lets see its target and calc is sooooooo much faster...im not saying that you should become totaly calculator dependent (me in ALGEBRA II) but sometimes its ok.

Guys, it's a target round, you can use calculators in the target round.
Just use the INT division.

Guys, it's a TARGET round, you can use calculators in the TARGET round. =D

10988065 *division sign* 2103 *equals*
round it down very quickly in your head.
ans-*whatever you jjust got* equals
ans -> frac

um sorry but that was kinda hard to understand...how about words or even better latex...unless it was some kinda joke...it wasnt very funn though...sarcasm doesnt travel over the internet!!!

Divide, and subtract the number that is greater than 0, and then multiply by divisor.

i.e.

10988065 is divided by 2103:

$ 10988065/2103\equal{}5224.947694...$

(subtract 5224)

$ .947694...\cdot2103\equal{}\boxed{1993}$

Since these numbers are relatively prime, their's no "quick" way to do this without a calculator. I think the problem writers had a calculator in mind when writing this problem :)

with smaller numbers, I can do it in my head, but this... no

[quote="mathcountsfreak"]Well that's a pretty ridiculous problem. Just punch it in on your calculator? No thinking required.
I guess its a school round though.
Its not likely for a problem like that to be on a higher level though. Might be a countdown problem to see who is quicker at long division.[/quote]

Whoever can do it in 45 seconds is pretty beast.

mentally, yes, but with 45 seconds and with pencil and paper, it's not too bad, since the digits of 2103 are small


The number $ 101$ is the smallest three-digit palindromic prime. What is the second-smallest one?

It asks for prime palindromes, so the answer would be 131

[quote="Ebob"]the next smallest would also be in the 100's, making it end in 1.
The next lowest would therefore be $ \boxed{111}$[/quote]

$111$ is divisible by 3, $3*37$, Since $1+1+1=3$, and if it does add up to 3 or anything divisible by 3, it will be divisible by 3, according to the divisibility rules.

121 isn't prime but 131 is prime so 131 is answer.


geometry
analytic geometry
college
parallelogram
geometry proposed

Prove Varignon's Theorem [u]using mass points[/u]:
If the midpoints of consecutive sides of a quadrilateral are connected, the resulting quadrilateral is a parallelogram.

Taken from Tom Rike's speech on Mass Points and Barycentric Coordinates (Berkeley Math Circle).

Are mass points even rigorous when not used on triangles? (I would expect that given three points, all other points in the plane could be expressed as some combination of those 3 points.)

I apologize if this is slightly off-topic. I have attempted an approach in which I assigned a mass of 1 to each of the vertices, resulting in a mass of 2 for each of the midpoints; I was stuck there.

Yes, I realize that.  However, I would like to know how it is done using mass points.

[hide="Solution using mass points"]Let the parallelogram have vertices A, B, C, and D.  Let the midpoints of the sides of the parallelogram be E, F, G, and H, in that order.  Assign a weight of 1 to A, B, C, and D.  Then E, F, G, H will all have weights of 2.  Now draw in segments EG and FH, the diagonals of quadrilateral EFGH.  These points intersect at some point P.  Since E and G have weights 2, and since F and H have weights 2, P will have weight 4.  Thus, P bisects EG and FH.  This means the diagonals of quadrilateral EFGH bisect each other, and hence EFGH must be a parallelogram.[/hide]


I have a question about the Chi Square test and confidence intervals:

How does one know which confidence interval to look up in a chi-square table?  I saw this tutorial on rolling dice (http://www.advancinghordes.com/newsletters.php/faq_item/2)  and after rolling a 6-sided dice 21 times, their chi-square value was 5.57.  Looking at p=.05, they concluded that since 5.57 < 11.07, the die was fair. Specifically, they were 95% sure it was fair.  However, if you look at the chi-square table, 5.57 is also less than 15.09 and 20.52, meaning one could be 99.9% sure it is fair... am I misunderstanding something?  It seems that the lower the p-value, the easier it is to be more confident in one's data.

It's been awhile since AP stat, but I'll try to make an intuitive appeal.

Contrary to what the article said, a Chi-Square test is used to disprove something, not to prove something. (When a die passes the test, we can only say we are confident that the die is fair.)
Even for a fair die, isn't there a possibility that in those trials, you get all 6's (or something else outlandish)? That would inflate the chi-square value, possibly past the value listed in the table. If you are working with a 95% confidence interval, you would reject the die and say it's not fair, even though it is in fact fair. That is why the values for 99% and 99.9% are higher. They encompass a greater range of chi-square values. Therefore, when the die fails the test, you can be more confident in saying the die is not fair.
So the die actually passes the 99% test, but you're only saying that you're more confident that if the die had failed the test, you could be more sure that it was in fact not fair.

I had the exact same problem last year. The confidence interval didn't make sense. So I asked about it here, and it was explained perfectly; it all makes sense in a non-intuitive kind of way. Here's the post that did it for me: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=37629

I am still somewhat hazy on this topic, so would you mind if I posted a sample problem (I just made it up right now)?

Let's say I have a 6 sided die that I roll 600 times.  Each face is expected to come up 100 times if fair.  Let's say I get the following results:  face1 = 87, face2 = 89, face3 = 108, face4 = 122, face5 = 90, face6 = 103.

When I do the chi-square test, I get X2 = 8.44.  When consulting a chi-square table(http://www.ento.vt.edu/~sharov/PopEcol/tables/chisq.html), I know the degree of freedom is 5, but which confidence interval do I use?  It seems I can be up to 99.9% confident this die is fair...

[quote="pkothari13"]It seems I can be up to 99.9% confident this die is fair...[/quote]

That is not what the test tells you. You use the test to decide (based on the experiment results) whether the proposed hypothesis (that the die is fair) makes sense. When you use the test, if the chi-square value is higher than the value given in the table (the die fails), you can reject the hypothesis.
However, there is always a possibility that the hypothesis was correct, but through pure chance, the experiment yielded some unlikely results, like rolling all 6s. If your confidence interval is too small, you could reject the hypothesis because the results give a chi-square value higher than the table's listed value.
The confidence interval tells you how certain you are that, in the case you do reject the hypothesis, the die is not fair. It is not an indication of how sure you are that the die is fair.
That is why the numbers for 99% and 99.9% are progressively larger. They include more of the chi-square values, so when the die does fail the test, you can be more certain that it is indeed not fair. Using higher confidence intervals means that you are less likely to say a die is not fair when it actually is.
On the other hand, if the die passes the test, you still have no way of knowing for sure whether the die is fair or not. The test does not tell you if the die is fair. It only tells you if it is safe to say that the die is not fair.

Hikaru: I'm glad to hear that I was of help.

[quote="Bictor717"]Hikaru: I'm glad to hear that I was of help.[/quote]
I completely forgot to thank you in that other thread (it was last year, I don't remember why :(), so I'd just like to say thanks now for your patience with me over there :) You did clear it up for me, and your explanations led to some interesting class discussions the next few days. Thanks, Bictor717!  :oops:


Use twice your previous [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=904772#904772]problem[/url]


Let $ S\equal{}\{x_0;x_1;..x_n\} \subset [0, 1]$ be a finite set of real numbers with $ x_0\equal{}0, x_1\equal{}1,$ such that every distance between pairs of elements occurs at least twice, except for the distance $ 1$. Prove that all of the $ x_i$ are rational.

The condition that all points are distinct is, clearly, missing.

I think it has been here long enough to post the (sketch of a) solution. Assume that some of the numbers are irrational. Write the linear system representing the equality of distances and consider it as a system for $ x_2,\dots,x_{n}$. Since it has a real solution and all the coefficients are rational, it also has a rational solution. Thus, it has a whole line of solutions. Now go far away along this line so that $ [0,1]$ is no longer the greatest distance and all points are distinct (the latter condition is generic and the first will eventually hold if at least one point moves at all). Then the largest distance between the points will be unique, but the linear system still guarantees that it'll be repeated somewhere. Thus our assumption is false and there can be only one solution, which is rational.


inequalities
real analysis
real analysis unsolved

I'm trying to prove the following:

Let $ Y$ be a closed subspace of a normed space $ X$. Let $ x_0 \in X\Y$. Prove that there exists an $ f \in X^*$ such that $ f(x_0) = 1$ and $ f(y) = 0$ for all $ y \in Y$.

I'm trying to use the following theorem:

Let $ C$ be a closed convex subset of a normed space $ E$. Let $ x_0 \in E, x_0 \notin C$. Then there exists an $ f \in E^*$ and $ \alpha \in \mathbb{R}$ such that $ Re f(c) \leq \alpha < Re f(x_0) \forall c \in C$.

How I'm going to use above theorem is as follows:
Consider X/Y, that is X quotient Y. In X/Y, Y is just a point, it's convex so Y is our C. x_0 is x_0 as in theorem. So according to theorem, there exists $ f \in (X\Y)^*$ and $ \alpha$ such that $ Re f(y) \leq \alpha < Re f(x_0) \forall y \in Y$. This looks promising right?

Now, the problem I'm having and where I need your help is that I need to establish that $ x_0$ is sent to precisely $ 1$ and $ y$ is sent to $ 0$.  but all we have above is that there exists some alpha such that $ x_0$ is sent to values greater alpha and $ y$ is sent to values less than or equal to alpha. Do you see the problem I'm having?

So I thought about doing the following:
we established the existence of alpha such that
$ Re f(y) \leq \alpha < Re f(x_0) \forall y \in Y$
so what if we switch the role of alpha and Y so that this time we get second inequality
$ Re f(x_0) < \alpha' \leq Re f(y) \forall y \in Y$

I'm getting little dodgy here and say:
$ \alpha < Re f(x_0) < \alpha'$ and
$ \alpha' \leq Re f(y) \leq \alpha$

Can we somehow squeeze $ Re f(x_0)$ to be $ 1$ and $ Re f(y)$ to be $ 0$ ?????
But as soon as I rescale alpha to meet the need in one of the inequalities, the other inequalities become nonsense....  :(

I would appreciate your help on this.  :)

[quote]In X/Y, Y is just a point[/quote]
$ Y$ is not just a point -- it's a very special point.

If $ x_{o}\in X \minus{} Y$ then it's true by Lemma Urisona. because we have two closed subset in normed set such as they are noncrossing.
and if $ x$ can be in $ Y$ then your problem is not true because I can choose  $ y\in Y$ such as $ y \equal{} x_{0}$ and we have $ f(1) \equal{} x_{0} \equal{} y \equal{} f(0)$


Is it just me, or do the rag-tag group of 15-20 people on this form, make the canadian national form one of the buisiest national forms on aops? Whenever I'm just surfing/trolling aops, and scroll past the 'north america' section, I alway see a canadian who has posted most recently, out of all the places in NA.

Do we pwn that hard?

Well, we definitely aren't normal.  And we post an amazing amount of times, lol.  :D

Thats true, we always have something on our minds. :jump:

[quote="rzsolt"][quote="rem"]Yeah. This forum is way busier than any other national forum.
Great job, Canadians!

PS. Actually one of the reasons why it's so active is cause we have several ppl online at the same time, so that they make several posts at that period of time, which increases the number of posts.[/quote]

funny, how canada is spread across so many time zones, but we still have so many people with messed up scheduals on at the same time...[/quote]

LOL... That just means that people will post at even stranger than usual hours during the day.

We're the forum with the most number of posts  :D

definitely tells you something.