AMC
USA(J)MO
USAMO
AIME
geometry
number theory
AMC 10

alright.. can someone describe exactly how to get in the usamo?
i know something about getting an index score of 210 or something..
but.. i need to know exactly how it will be scored and stuff...
i will be a senior next year.. my chances are low.. but i and willing to work my butt off for this.. i think theres no better advantage to taking a test then knowing how its gonna be graded..
ya.. no one in my school has ever gone to the usamo level so i really want this..
thanks..
also.. if u guys got any material/books/websites/tips/strats.. anything to help me pass.. i'd love it..

thanks

kent

I believe that the USAMO floor is dependent on the year.  To get into USAMO you basically have to be in the top 250.  If for some reason 500 kids get a 200 index one year then the index will be raised.  So basically just aim to get as high a score as possible.  If you really want a good numerical goal I would say that a good goal for the USAMO would be a 130 on the AMC 12 and an 8 on the AIME.  That's a 210 index which should put you in in most circumstances.  The scoring is like what koopa said.  This website is the best site I know for preparing.  I suggest taking some of their classes and reading their books.  In Volume 1 they have a lot of problems that will help you with the AMC.  I haven't got to Volume yet but I'm thinking that it will be more of late AMC to AIME and possibly more.  You should also consider ordering past AMC and AIME test for practice.  [url]http://www.kalva.demon.co.uk/[/url] has past AIME s.  If you do most of the stuff I told you and you understand it you have a good chance.  Oh yeah don't memorize formulas or problem types.  It is a lot easier to just understand the concept and use that to figure a problem out.

[quote="kentliu"]alright.. can someone describe exactly how to get in the usamo? . . . thanks[/quote]

Start by reading the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11693]FAQ on the AMC contests[/url], and follow the links from there. Don't rely solely (or at all) on the Kalva site for the solutions to old exams; instead, buy and use the official AMC copies of old exams and study and think about their solutions after taking the tests for practice. Practice old AIMEs at least weekly between now and March 2005.

As the others have said, this AoPS site is the best Web site for AMC preparation. This site is raising the level of competition in the United States, so aim high to be one of the 120 or so test-takers after the AMC 12 and AIME next year.

Good luck.

So is there one cutoff score each year for all grades?  Or is there a score for each grade?
I'm going to be a 10th grader this year and was wondering what the cutoff score might be.  Also, the top 250 students are accepted in each grade nationwide or is it the top 250 students in all grades in a certain state?  I'm guessing it's the top 250 students in all grades nationwide that make it to the USAMO.  I live in Texas just to let you'll know, so maybe some of you know the requirements for Texas usually.

It's cool how if you get a 150 on the AMC you only have to get 5 questions right on the AIME (not that that is easy or anything, just fun to point out)

From the AMC website (http://www.unl.edu/amc):

Selection for the USAMO will be made according to the follwing rules:

The goal is to select about 250 of the top scorers from the prior AIME and AMC 12A, AMC 12B, AMC 10A and AMC 10B contests to participate in the USAMO.

Selection will be based on the USAMO index which is defined as 10 times the students AIME score plus the students score on the AMC 12 or the AMC 10.

The first selection will be the approximately 160 highest USAMO indices of students taking the AMC 12A or AMC 12B contest.

The lowest AIME score among those 160 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value.

The student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO.

To adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the (A & B) Contests.

The selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests.

Unfortunately, most years, even with 150 on the AMC 12, you will not make USAMO with a 5 on AIME. This year you needed at least 7 and last year at least 8.

Yes, that would be unfortunate, and I could be wrong, but what are the chances that a person who makes a 150 on the AMC 10 or 12 is going to make a 5 on the AIME?  It's possible I guess, but just not too likely, unless that person got really lucky guessing or something, which wouldn't help him/her on the AIME.  Now, I just got both AoPS volumes, and I have some serious studying to do.  Good day, and thanks for the help.

[quote="Rep123max"]So the floor is around 7 or 8? Does that mean since I will be a Freshmen next year and probably take AMC 10, if I qualify for AIME, my AMC 10 score wont matter, its just that I get whatever the floor was?[/quote]

basically yes, all you need is a aime score higher than the floor and an amc 10 score that aime-quals.  it's substantially more difficult when you're older tho...

oh man.. so AoPS is a new site?
man.. its gonna be tough for senior like me then..

[quote="koopa"]I think the cut off point is 200 points total in AMC and AIME.

Suppose your score for AMC = x, AIME = y. Then you need x + 10y >= 200.

Which is not that bad nowadays. AMC is a lot easier then in past. You should at least shoot for 120+ for AMC, and then 8+ should be pretty easy if you are strong in two of the four subjects: Algebra, Number Theory, Geometry, Combinatorics. By strong I mean you can do all(almost all) the problems in your area. Also, there ought to be some easy ones in your not so strong area also. So, it's not that bad. Getting into IMO is a complete different world...[/quote]

koopa, u said algebra, number theory, gemetry, and combinatorics..

may i ask what topics are involved in number theory and combinatorics?

Here's a description from a novice:

Number theory is like really advanced arithmetic.

The permutation and combination formulas you learn in Alg II are combinatorics.  There's a lot more, but that will give you an idea.

They say that all the problems can be done without any math above precal.  That s not to say that anyone who has completed precal can do good.  They pretty much take seemingly easy concepts and make them a whole lot more tricky.

Here 's a sample NT problem

 Find the largest integer n such that n + 10 divides n^3 + 100.

I would say that there is usually more algebra/ geometry on the AIME on average.

[hide]

Dividing n :^3: + 100 by n + 10 we get n :^2: - 10n + 100n + 10n :^2: - 900/(n+10).

For n + 10 to divide n :^3:  + 100, the result must be an integer, so n :^2: - 10n + 100n + 10n :^2: - 900/(n+10) must be an integer and thus 900/(n+10) must be also.

The largest number is obviously 890, when the fraction is 1[/hide]

[quote="tokenadult"][quote="Rep123max"]Does that mean since I will be a Freshmen next year and probably take AMC 10, if I qualify for AIME, my AMC 10 score wont matter, its just that I get whatever the floor was?[/quote]

If and only if the ninth-grade MOP program exists next year, and works the same way it did this year, AND if and only if there aren't a whole lot more ninth graders trying to get into MOP and succeeding at qualifying for AIME, then it will be as you suppose, because it was that way this year. This year people with USAMO scores of 0 went to ninth-grade MOP, and I think they got into USAMO in some cases mostly having a floor score on the AIME (which is not trivially easy to obtain) and whatever AMC 10 score got them into AIME in the first place (which anyone who has had algebra 1 and geometry ought to be able to obtain). But if next year is more competitive, all this may change, and I expect AoPS's very existence to make next year more competitive. And having ninth-grade MOP at all depends on funding, as indeed offering a broad eligibility for USAMO also does.

Good luck.[/quote]

IFF the program exists next year and works the same way?! The "if" part is certainly true, but I don't know about the "only if" part.

But if you think you'll be able to make a ~130 on AMC 12, and a ~10 on AIME, you're pretty safe, right?

The last three years have seen substantially easier contests and much more competition.  In 2002, the USAMO cutoff was an index of 210 (AIME floor of 6 or 7; I can't remember this one exactly). In 2003, the cutoff was a steep 226 (AIME floor of 8), but last year it came down to 210 (AIME floor of 7) again thanks mostly to slightly harder AMC-10/12 and AIME contests.

Previously, the AMC-12 had been much more difficult, with the AIME cutoff (through the AMC-12) dipping well below 100 just a couple years earlier.

Bottom line - I don't know if a 230 is particularly safe.  It all depends on what the folks in Lincoln have put together.  A 230 hasn't failed to reach USAMO yet, but it was pretty close in 2003.


Suppose you are given a graph of y = f(x).  How would you use that graph to sketch y = -f(x+1) + 2?

If you find Intermediate problems pretty easy, please leave this challenge problem for less experienced students to solve.

Source: Alabama State Contest

I haven't been here in a looooooong time...

[hide]

First, the graph of y=f(x+1) would be the graph of y=f(x) but moved to the left 1 unit.  y=-f(x+1) would be f(x+1) reflected over the x axis.  Finally, y=-f(x+1) + 2 would be the graph of -f(x+1) moved up 2 units.

[/hide]

[hide]Flip it over the x-axis, move it left one, and up 2.[/hide]


hi my name is rami i an a new member

any body want to say hi for me ?

 :blush:

[url=http://zzrz.com/english.htm]http://zzrz.com/english.htm[/url]

Hello rami, welcome to this site!
By the way, you should post it in the Game and Fun Factory, there is also a self-introduction thread! :D


What is the number of leap years between 1937 and 2005?

[quote="ToBe"][hide]$68/4=17$? i think. but thats assuming that 1938 is a leap year[/hide][/quote]

But 1938 wasn't a leap year...

[quote="nvidura"][b]it must be 16 since (2000-1937)/4 +1[/b][/quote]

would you please hide your answer next time, so other ppl can try. thx!

by the way, I got the same answer, same way.

how to find leap years:
[hide]to find out whether or not a year is a leap year, you use the last two digits of the year, and if it's divisible by $4$, it's leap. The exception is when the last two digits is $00$, in that case you divide all four digits by $4$[/hide]
answer:
[hide]I got $17$... i've got no clue how y'all did it: why would you do $2000-1937$? doesn't the question ask to find from $1937$ to $2005$? but you have to notice $1937$ isn't a leap year, so you have to start on $1940$, so it would be $2005-1940 = 65$, then you use $\frac{65}{4}  +1 = 16+1 = 17$.... but I might be completely wrong here...[/hide]
btw, Kristy, you aren't really suppose to double post... but I'm only saying... :lol:


geometry MATHCOUNTS algebra polynomial function AMC USA(J)MO

I got a 118.5 on AMC 12, and I really want to qualify for USAMO. That means I probably need a 10 or 11 on AIME. Right now I'm averaging 7-8, and I was wondering which topics you think are the easiest to improve on.

[quote="indianamath"]personally, Geometry was easiest to improve.  You just need to pinpoint the key points.  In one year, i went from a 16 on the AMC 8 :oops:    to a 138 on the AMC 10 :D   this year(which had lots of geometry).[/quote]
Actually, I'm going to have to say that it's number theory.
Number theory may look confusing, but once you get past the simple basics (mods and whatnot) it's smooth sailing.

[quote="indianamath"]number theory is easy to improve in, but AMCS's and MATHCOUNTS don't use Number Theory that much.[/quote]

are you kidding?

have you ever even SEEN an AMC?

it's 25 number theory questions

I heard polynomials and functions are used in questions toward the end of the AMC 10.

[quote="indianamath"]number theory is easy to improve in, but AMCS's and MATHCOUNTS don't use Number Theory that much.[/quote]
Number 25 on the AMC 10 was Number Theory (and Algebra).
I think Number 22 on the 10 was as well, I'm not sure though. (don't exactly remember the question, but I remember thinking that there were two number theory near the end).

[quote="indianamath"]those weren't really pure number theory problems. I thought you were talking about like finding the tens digit of (blahblahblah) and stuff like that.[/quote]
Mathcounts is actually very NT heavy.
Let's look at past nationals.
2005: 3 out of last 10
2004: 3 out of last 10
2003: 1 out of last 10
2002: 3 out of last 10

ok i'm wrong, just like a lot of my answers in MATHCOUNTS. :)

USAMO is very number theory heavy, which is why I can never do the problems when I look at one.

Personally, I think number theory is good too, because it provides a basis for a lot of the other things you'll need to do...

pressure with a capital P can bring out the worst in you

Tip 1: [u]Underline[/u] the question being asked. Put a square around important information that you might forget to take into account later into doing a question (such as a>b, x is an integer, y is positive, etc.)

Tip 2: Use scrap paper liberally. [b]Write big and neat.[/b] This will avoid errors such as turning + into -1, 4 into 9, dropping - signs, etc. Also, draw [i]large[/i] diagrams for geometry problems.

Also, when doing algebra, don't fail to write out intermediate steps just because it isn't cool. Don't write out the intermediate steps if you can see how to get there quickly, but you have to think hard about what's going on you might as well write it out. Things can get muddled in your head.

Tip 3: [b]Organize[/b] your scrap paper. Do not turn half-used scrap paper upside down and start working on another problem; if you have ample space draw a line under the last problem and start working on the next problem under it. [b]If in doubt get another piece of scrap paper.[/b] This tip will save you time when you go back and check over your answers.

Tip 4: Always allot time to check your work. Unless you are a master at avoiding silly errors because of the first three tips (or screwed up following tip 5) chances are you'll gain more points catching stupid mistakes than trying to solve more problems under time pressure.

Example: Many people on the AIME make 2 stupid errors. Say you have gotten to question 12 already. Which do you think is more likely? Backtracking and finding those two errors, netting you 2 points, or solving two of the last three problems?

Tip 5: Don't get caught up in a hard problem if there are others that you haven't even read yet. The AMCs and AIMEs try, but are never in perfect difficulty order. Just because it's problem 15/25, doesn't mean you don't know how to do it.

[quote="tjhance"][quote="randomdragoon"]The easiest subject to improve quickly in is "Techniques for cutting back on stupid mistakes". I'm serious; if you master this between now and the AIME your score will go up by at least two points. (or twelve in AMC terms)[/quote]

I would really like to improve on this. Do you have any tips for doing it?[/quote]

Try reading this:

http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php

Number Theory is easy to improve in for MATHCOUNTS and the AMC, because most NT problems on those tests are mods and  . . . mods. :P  However, if you're looking at hard AIME and USAMO, number theory requires a lot of work to get good at.

For the AIME, I'd say the qickest to improve in would be complex numbers/trigonometry.  For some reason, the AIME often assumes that students don't know those subjects and has a problem with them at around #11-#13 that is about #5 in difficulty if you learn the basic ideas (see 2002 AIME I #12).  Unfortunately, the trig #12 on last year's AIME I was far from trivial, so maybe they've gotten away from this. :(

If you are geometrically inclined (I'm not), geometry is probably good too.


find  all the solutions of the eq.
\[cos(12x)=5sen(3x)+9tg^2x+ctg^2x\]
in the interval $[0,2\pi]$

Um, I'm just starting trig, so can someone help me with what $sen$ $tg$ and $ctg$ are?

this is probably just a terribly misguided guess, but i'm going to guess that sen=secant, tg=tangent, and ctg=cotangent.

i've never seen ctg, tg, and sn, before either >_<

It's a chinese way of saying it too.

Do you know what they mean? You should.

Reply, and I'll tell you.


algebra
difference of squares
special factorizations
number theory unsolved
number theory

Find all (x,y), given that
$ x,y \in Z \\
x^2(y\minus{}1)\plus{}y^2(x\minus{}1)\equal{}1$

[quote="NikolayKaz"]Find all (x,y), given that $ x,y \in Z, x^2(y \minus{} 1) \plus{} y^2(x \minus{} 1) \equal{} 1$[/quote]

Rewriting the equation we have $ \displaystyle xy \plus{}2\minus{}x\minus{}y\equal{} \frac {5}{x \plus{} y \plus{} 2} \in \mathbb{Z}$. So we are lead in a finite number of cases (only 4 because 5 is prime..)

Wow, how did you see that? What inspired you to do that? o_O

[hide="What I did :/"]Let $ u \equal{} x \plus{} y$ and $ v \equal{} xy$. We have $ x^2y \minus{} x^2 \plus{} xy^2 \minus{} y^2 \minus{} 1 \equal{} 0$, or $ uv \minus{} (u^2 \minus{} 2v) \minus{} 1 \equal{} 0$, or $ u^2 \minus{} uv \minus{} 2v \plus{} 1 \equal{} 0$. Multiplying by 4, we have $ 4u^2 \minus{} 4uv \minus{} 8v \plus{} 4 \equal{} 0$, or $ (2u \minus{} v)^2 \minus{} v^2 \minus{} 8v \plus{} 4 \equal{} 0$, that is, $ (2u \minus{} v)^2 \minus{} (v \plus{} 4)^2 \plus{} 20 \equal{} 0$, so $ (v \plus{} 4)^2 \minus{} (2u \minus{} v)^2 \equal{} 20$. We have a difference of squares, so we again only have finitely many cases to check.

Granted, many many more cases. Oh well. [/hide]

[quote="Zhero"]Wow, how did you see that? What inspired you to do that? o_O [...] or $ u^2 \minus{} uv \minus{} 2v \plus{} 1 \equal{} 0$. [...][/quote]

But it is the same equation, only try to explicit v and reduce the other fraction  :wink:

[hide]
First, putting y = x giving (x^2)(x-1) = 0.5 which is impossible, obviously putting unknown(s) to be zero gets no solution.
As x^2, y^2 >= 1,
so we first put one of the unknown be 1 obtaining the other be 2, so (1,2) or (2,1) are part of the solutions.
WLOG, assuming x>y,
with greater x, a negative y is needed for obtaining solutions.
e.g., putting x = 3,
y = -0.5 or -0.4. which neither are integers and both negative.
So putting x = a + 1, y = b + 1,
easily we get:
(1 + ab)(a + b)  + 4ab = 1, which L.H.S. is negative while the right is positive.
So the only solutions are (1,2) and (2,1).
[/hide]

What's about this problem : Find all pairs of integers $ (x,y)$  and positive integer n such that :
$ x^2(y^n\minus{}1)\plus{}y^2(x^n\minus{}1)\equal{}1$

[quote="TTsphn"]What's about this problem : Find all pairs of integers $ (x,y)$  and positive integer n such that :
$ x^2(y^n \minus{} 1) \plus{} y^2(x^n \minus{} 1) \equal{} 1$[/quote]
that's easy since u can notice $ xy(y^{n\minus{}1}x\plus{}x^{n\minus{}1}y)\equal{}1\plus{}x^2\plus{}y^2$ but it's known that if $ \frac{1\plus{}x^2\plus{}y^2}{|xy|}$ is an integer then it's equal to 3(the cases $ x\equal{}0$ or $ y\equal{}0$ are trivial).it means that $ |y^{n\minus{}1}x\plus{}x^{n\minus{}1}y|\equal{}3$ so we are left with 2 cases which are trivial.


In how many ways can 4 on/off switches in a row be set so that no two adjacent switches are on?

There are 8 ways for there to be at lest 2 switches on:

1100
1101
0110
0011
1011
1110
0111
1111

There are 2^4 - 8 = 8 ways for there to be no adjacent on switches

You can also just count the number of ways not to have two adjacent switches on:

All 4(0)s,
3(0)s and 1(1),
and

1001
1010
0101

for a total of $ 1 \plus{} 3 \plus{} 4 \equal{} \boxed{8}$ ways.

We can either have $ONON\implies 2!\times 2!=4\text{ ways}$ or $NONO\implies 4\text{ ways}$. So $4+4=\boxed{8\text{ ways}}$

If there is 2 on switch then there is 3 ways. If there is 1 on switch we have 4 ways. If there is 0 on switch there is 1 way. Hence $3+4+1=8.$


I can't figure out this question:

The radius of a particular circle is inscribed in a equilateral triangle is 2 units.  What is the perimenter of the riangle?  Express your answer in simplist radical form.

Thanks in advance. :lol:

Blah, I'll make a diagram later.

Basically, make a large triangle that uses 120 degrees of the central angle. Then divide it into a 30 60 90 triangle with one side as half of the side of the triangle. You can find that half, and then multiply by 6.

[quote="neurosurgeon1118"]couldn't you just draw the radius of the circle te be perpendicular to the traingle's base to form a 30 60 90 with 2 as the shorter leg the longer leg which is 1/2 of a side of a triangle to be 2root 3. So multiply by 2 and 4root 3 is the length of each side of the equilateral. the area of an equialeteral is A=(S^2root3)/4 so the area should be [b]12root3[/b]  :D

Am I right Adeesh[/quote]

No fingers, but it said perimeter.

[quote="happyme"]Thats exactly what I said...

 :huh:[/quote]

Don't copy of other people *cough karthik cough*

[quote="neurosurgeon1118"]sorry guys i just couldn't help myself, Adeesh was this on Warm Up 13[/quote]

Never give names out, surgeon, its denies netiquitte

Uh, its fine I guess?

I have no clue whats going on, this is getting slightly off-topic.

I have posted problems (and solutions) to most of the early Warm-Ups, so look in the past 2 or 3 pages, you'll get a lot of answers.

And didn't you just call the other guy Cartick or something?

o.o


integration
real analysis
real analysis unsolved

Suppose $ (a_n)_{n\equal{}0}^\infty$ is a sequence of positive numbers such that $ \sum_{n\equal{}0}^\infty a_n$ diverges. What can you say about
$ \sum_{n\equal{}0}^\infty \frac{a_n}{(a_0\plus{}\ldots\plus{}a_n)^2}$?

Let $ s_n \equal{} a_0 \plus{} \cdots \plus{} a_n$. Then $ 0 < s_{n \minus{} 1} < s_n$ and hence $ \frac {1}{s_{n \minus{} 1}} \minus{} \frac {1}{s_n} \equal{} \frac {a_n}{s_{n \minus{} 1} s_n} > \frac {a_n}{s_n^2} > 0$. Summing each sides shows that $ \sum_{n \equal{} 0}^{\infty} \frac {a_n}{s_n^2}$ is bounded, hence convergent.

Nice.  It seems (from the small assortment of examples I looked at) that perhaps the 2 in the exponent could be replaced with a smaller number and the result would still be true.  So, can anyone either prove this or find a sequence for which the 2 is best possible, i.e. $ \sum_{n \geq 0} \frac {a_n}{s_n^{2 \minus{} \varepsilon}}$ diverges for all $ \varepsilon > 0$?

$ \sum_{n \equal{} 0}^\infty \frac {a_n}{(a_0 \plus{} \ldots \plus{} a_n)^{1\plus{}\epsilon}} \approx \int_0^{\infty} \frac{f'(t)}{f(t)^{1\plus{}\epsilon}} dt$
where $ f(x) \approx \sum_0^x a_k$ is a positive  increasing function. Because $ \frac{1}{f^{\epsilon}(t)}$ is a primitive of $ \frac{f'(t)}{f(t)^{1\plus{}\epsilon}}$, that "shows" that for every $ \epsilon > 0$, the sum $ \sum_{n \equal{} 0}^\infty \frac {a_n}{(a_0 \plus{} \ldots \plus{} a_n)^{1\plus{}\epsilon}}$ converges. (you can easily make all this a proper proof, I think ... )

Very nice solutions [b]sos440[/b] and [b]alekk[/b]!


LaTeX
inequalities
inequalities unsolved

prove:$r>1$,$\frac{a}{ra+b+c}+\frac{b}{rb+a+c}+\frac{c}{rc+a+b}\leq \frac{3}{2+r}$
and also what about the general case

easy,but I can't use LaTeX,so can't write here,Who can tell me how to use LaTeX in this forum

we are are chinese.So if you like,tell me your qq number.I'll tell you how to use.(but a little) :D

Expansion works  :blush: .
Expanding all the terms we can know that this inequality is equivalent to:
\[6(r^2-r)abc \le (r-1)\displaystyle\sum_{sym}a^3+(r^2-2r+1)\displaystyle\sum_{sym}a^2b.\]But I think it is easy now.  :)

Edit: Jensen also works after homogenization using $a+b+c=1$.

thank you for your proof.
and here is my proof:
we have:$\frac{a}{ra+b+c}+\frac{b}{rb+a+c}+\frac{c}{rc+a+b}\leq \frac{b}{ra+b+c}+\frac{c}{rb+a+c}+\frac{a}{rc+a+b}$,$(1)$
$\frac{a}{ra+b+c}+\frac{b}{rb+a+c}+\frac{c}{rc+a+b}\leq \frac{c}{ra+b+c}+\frac{a}{rb+a+c}+\frac{b}{rc+a+b}$,$(2)$
add $(1)+(2)+r \frac{a}{ra+b+c}+\frac{b}{rb+a+c}+\frac{c}{rc+a+b}$
 :)  :)
and I'm sorry I find $AYMANE's$ solution doesn't work.check it please. :D  :D
Jensen won't be work(I think),maybe RCF theorem will work.

Ooops......you are right,isee my mistake :blush:

Finally, something non computational  :D . Let us prove that for $ r\geq 1$ and $a_1,...,a_n>0$ we have
  $\sum{\frac{a_1}{ra_1+a_2+...+a_n}}\leq\frac{n}{n+r-1}$. Put $S=a_1+...+a_n$ and $ x_i=\frac{a_i}{(r-1)a_i+S}$. Then we have $n+r-1=\sum{\frac{1}{1-(r-1)x_i}}\geq\frac{n^2}{n-(r-1)(x_1+...+x_n)}$ from where we get $x_1+...+x_n\leq\frac{n}{n+r-1}$

thank you for your nice proof :) $harazi$

My solution. Noting the given inequality is equivalent to
\[
\sum {(1 - r)(a - b)^2 (rc + a + b)}  \leqslant 0
\]
 :lol:


if $ A \subseteq {\mathbb{R}}^n$ be compact and has measure 0 then A has content 0.

The only difference between Lebesgue measure and Jordan content is that when you are finding the content, you use only [b]finite[/b] covers. The rest should be clear.

By the way, it seems like I smell homeworks here :lol:.


Find the sum of all the roots contained in the interval $ [0;2\pi]$

$ \sqrt {3}\tan x - \sqrt {3} \sin2x + cos2x = 0$

[hide="My Solution"]Dividing both sides by $ \sin 2x$
$ \sqrt {3}\cdot\frac {1}{2\cos^2x} - \sqrt {3} + \text{cot}2x = 0 \\
\therefore\;\sqrt {3}\cdot \frac { - \cos 2x}{2\cos^2x} + \frac {1 - \text{tan}^2x}{2\text{tan}x} = 0 \\
\therefore\;\sqrt {3}\cdot\frac {\cos2x}{\cos^2x} = \frac {\cos2x}{\cos^2x\cdot\text{tan}x} \\
\therefore\;\text{tan}x = \frac {\sqrt {3}}{3}\text{ or }\cos2x = 0$

$ \text{tan}x = \frac {\sqrt {3}}{3}\;\;x = \{\frac {\pi}{6}\,,\,\frac {7\pi}{6}\}$

$ \cos 2x = 0\;\;2x = \frac {\pi}{2} + k\pi\;\therefore\;x = \{\frac {\pi}{4}\,,\,\frac {3\pi}{4}\,,\,\frac {5\pi}{4}\,,\,\frac {7\pi}{4}\}$[/hide]

I solved it like this, but I was wondering if anybody had a more clever and faster solution for this problem. Thanks in advance.

hello, do you mean $ \sqrt{3}\tan(x)\minus{}\sqrt{3}\sec(2x)\plus{}\cos(2x)\equal{}0$?
Sonnhard.

hello, you can rewrite your equation int the form
$ 4\sec(x)\sin(\pi/6\minus{}x)\sin(\pi/4\minus{}x)\sin(\pi/4\plus{}x)\equal{}0$
Sonnhard.

hello, i have found the solutions
$ x\equal{}\frac{\pi}{6}$
$ x\equal{}\frac{\pi}{4}$
$ x\equal{}\frac{3\pi}{4}$
Sonnhard.

Let $ t \equal{} \tan x\Longrightarrow \sin 2x \equal{} \frac {2t}{1 \plus{} t^2},\ \cos 2x \equal{} \frac {1 \minus{} t^2}{1 \plus{} t^2}$,

The given equation can be written as $ \sqrt {3}t^3 \minus{} t^2 \minus{} \sqrt {3}t \plus{} 1 \equal{} 0$

$ \Longleftrightarrow (\sqrt {3}t \minus{} 1)(t \plus{} 1)(t \minus{} 1) \equal{} 0$, which follows Dr Sonnhard Graubner's solution. :lol:


I've often had this thought and was wondering if anyone else had too. Actually, I'm mainly just posting this because I thought it was interesting.

Often when you are trying to solve a hard problem, you try one method of attack for a while, and end up with a completely horrible equation to solve, say. You can see absolutely no way to solve this equation, you go back to the start and eventually solve the problem in a 'nice' way.
Now what I've thought is that after this, you could say "hey - what I have just done is a way to prove that horrible equation!"
Or to give a more specific example, you have a geometry problem, try a trig bash and, as long as you don't make any mistakes, you end up with a horrible trig equation that is probably "impossible" to prove. However, you could prove it by saying this trig equation is true if and only if this geometry problem is true, which I can prove in a nice way.

Of course, there'd be absolutely no way you'd be able to translate something horrible into a completely random geometry problem, but that idea always fascinated me..

[quote="mozilla"]Well, IMO 2001 problem 6 was a number theory problem with a horrible equation but you can solve it beautifully by using simple geometry.[/quote]

Really?  How?


While staying in a 15-story hotel, Polya plays the following game. She enters an elevator on the $ 6^{\mathrm{th}}$ floor. She flips a fair coin five times to determine her next five stops. Each time she flips heads, she goes up one floor. Each time she flips tails, she goes down one floor. What is the probability that each of her next five stops is on the $ 7^{\mathrm{th}}$ floor or higher? Express your answer as a common fraction.

There are 2^5 = 32 different ways to flip the coin 5 times.

Since Polya's stops must be at the 7th floor or higher, she must flip heads first. After that, she must flip head again because flipping tails will result in going back to the 6th floor.

So Polya must go in this order: 7th floor, 8th floor, ...

On the third flip, she can flip either heads or tails. If she flips tails, she will go back to level 7. The next flip must be heads. There are 2 choices for the last flip. They are 7,8,7,8,7 and 7,8,7,8,9.

If she flips heads for the 3rd flip, she will go in this order: 7,8,9, ...
The 4th flip can be either heads or tails.

Heads (two ways): 7,8,9,10,9 and 7,8,9,10,11

Tails (two ways): 7,8,9,8,7 and 7,8,9,8,9

There is a total of 6 ways to stay at the 7th floor or higher for each stop.
So there the probability is 6/32 which is simplified to 3/16.

[hide = Quicker Way]She starts at the 6th floor, so in order for the conditions to be satisfied she must go up on her first move, to the 7th floor.

On her second move, she must also go up, to the 8th floor. The probability of flipping two heads is $\frac{1}{4}.$

Now, she's on the 8th floor and has 3 moves left. If she goes up in any one of those moves, she won't be able to reach the 6th floor. The probability she flips at least one head (aka goes up at least once) is $1$ minus the probability she flips two tails, or $\frac{1}{4}.$

The answer is $\frac{1}{4} \cdot \frac{3}{4} = \boxed{\frac{3}{16}}.$[/hide]


How many pairs of vertical angles are formed by five distinct lines that have a common point of intersection?

For any pair of lines, there are $ 2$ unique pairs of vertical angles formed, so the answer is $ 2\binom52 \equal{} \boxed{20}$.

Edit: Darn, I seem to have forgotten what vertical angles are. :P

NOPE, good try

The answer is 5

(Obviously. 5 lines make 5 vertical angles)

The accepted solution seems to be incorrect. For example's sake have the $5$ lines intersect at the center of a circle, splitting it into $10$ congruent arcs. We count $5$ pairs of vertical angles of $36^{\circ}$, $72^{\circ}$, $108^{\circ}$, $144^{\circ}$, and $180^{\circ}$ each. Thus we have $5\cdot5 = 25$ pairs of vertical angles.

I disagree about the vertical angles measuring $180^{\circ}$. Vertical angles are formed by two lines, and you would only be able to form $180^{\circ}$ angles using two copies of the same line (which isn't allowed). If you discount them, you end up with the same answer as above.

I believe the currently accepted solution is $5$.

I reported it. They changed it from $5$.


geometry circumcircle incenter inradius inequalities trigonometry symmetry

Let [tex]ABC[/tex] be triangle. I is centre of triangle's incircle. [tex]AI[/tex] line cut circumcircle at [tex]A'[/tex]. Similarly, we have [tex]B',C'[/tex]. Prove that:
[tex]AA'+BB'+CC'\leq \frac{29}{6}R+\frac{7}{3}r[/tex].

Wlog, assume that A>=B>=C. (angles)
Note that AA' = 2R cos(B-C)/2
So it suffices to show that sum cos (B-C)/2 <= 29/12 +7/6 (r/R)
Finally taking into account that cosA + cosB +cosC= 1+ r/R we need to show that
[sum cos(B-C)/2] + 7/6 cos(B+C)-7/6(cos B +cos C) <=5/4
eliminating A, we need to show that :
cos(B-C)/2 [1-7/3 cos(B+C)/2] +7/6 cos(B+C) - 2cos(B-C)/4cos3/2(B+C) <= 5/4

[quote="keira_khtn"]Let [tex]ABC[/tex] be triangle. I is centre of triangle's incircle. [tex]AI[/tex] line cuts circumcircle at [tex]A'[/tex]. Similarly, we have [tex]B',C'[/tex]. Prove that:
[tex]AA'+BB'+CC'\leq \frac{29}{6}R+\frac{7}{3}r[/tex].[/quote]
Let $O, I$ be the triangle circumcenter and incenter, $R, r$ the circumradius and inradius and $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$. For an equilateral triangle $\triangle ABC$, $r = \frac R 2$, $AA' = BB' = CC' = 2R$ and the inequality becomes a trivial equality:

$AA' + BB' + CC'  = 6R$

${\frac{29}{6}\ R + \frac 7 3\ r = \frac{29 + 7}{6}\ R} = 6R$

Let $\triangle ABC$ be an arbitrary triangle. Consider the triangle $\triangle CAA'$. Since $AA'$ bisects the angle $\angle CAB$, the angle $\angle CAA' = \frac{\angle CAB}{2} = \frac \alpha 2$. Since the angles $\angle AA'C, \angle ABC$ span the same arc $AC$ of the circumcircle $(O)$, $\angle AA'C = \angle ABC = \beta$ and the remaining angle of this triangle is

$\angle ACA' = 180^o - \left(\frac \alpha 2 + \beta\right) = \gamma + \frac \alpha 2$

Using the sine theorem for this triangle,

$AA' = CA\ \frac{\sin{\left(\gamma + \frac \alpha 2\right)}}{\sin{\beta}} = 2R \sin{\left(\gamma + \frac \alpha 2\right)}$

Similar expressions could be obtained for $BB', CC'$, Hence,

$AA' + BB' + CC' = 2R \left[\sin{\left(\gamma + \frac \alpha 2\right)} + \sin{\left(\alpha + \frac \beta 2\right)} + \sin{\left(\beta + \frac \gamma 2\right)} \right]$

Let $a = BC, b = CA, c = AB$ be the triangle sides and $s$ the semiperimeter. The inradius is equal to

$r = \frac{|\triangle ABC|}{s} = \frac{abc}{4Rs} = 2R\ \frac{\sin \alpha \sin \beta \sin \gamma}{\sin \alpha + \sin \beta + \sin \gamma}$

Consider the triangle $\triangle ABC$ "rotating" according to Poncelet's porism. During this process, it passes through 2 different isosceles triangles, one acute and the other obtuse, 3 times through each one in one full "rotation". Let $\triangle A_1B_1C_1$ the acute isosceles triangle with $A_1B_1 = A_1C_1$. Then $A_1A_1' = 2R$ is maximum. When "rotating" the triangle counter-clockwise into the obtuse isosceles triangle $\triangle B_2C_2A_2$ with $B_2C_2 = B_2A_2$, the segments $A_1A_1', C_1C_1'$ both monotonously decrease into $A_2A_2', C_2C_2'$ and the segment $B_1B_1'$ monotonously increases into $B_2B_2' = 2R$. Because of symmetry of these isosceles triangles, $B_1B_1' = C_1C_1', A_2A_2' = C_2C_2'$ and obviously, $C_2C_2' < C_1C_1'$. Consequently,

$A_2A_2' + B_2B_2' + C_2C_2' = 2R + 2C_2C_2' < 2R + 2C_1C_1' = A_1A_1' + B_1B_1' + C_1C_1'$

becomes a minimum. In a subsequent "rotation" into the acute isosceles triangle $\triangle C_3A_3B_3 \cong \triangle A_1B_1C_1$ with $C_3A_3 = C_3B_3$, the sum $A_3A_3' + B_3B_3' + C_3C_3'$ again increases to its maximum. Therefore, it is sufficient to prove the desired inequality for an acute isosceles triangle $\triangle ABC$, for example, with $AB = AC,\ \alpha \le 90^o,\ \beta = \gamma < 90^o$. The desired inequality then becomes

$(?)\ \ 2R\left(1 + 2 \sin{\frac{3 \gamma}{2}}\right) \le \frac{29R}{6} + \frac{14R}{3}\ \frac{\sin^2{\gamma} \sin{2 \gamma}}{2 \sin \gamma + \sin{2 \gamma}} =$

$= \frac{29R}{6} + \frac{14R}{3}\ \frac{\sin^2{\gamma} \cos \gamma}{1 + \cos \gamma} = \frac{29R}{6} + \frac{14R}{3}\ (1 - \cos \gamma) \cos \gamma$

$(?)\ \ 4\sin{\frac{3 \gamma}{2}} \le \frac{17}{6} + \frac{14}{3} (1 - \cos \gamma) \cos \gamma$

Since both sides are obviously positive, we can square the inequality in question. The left side becomes

$16\left(\sin^2{\gamma} \cos^2{\frac \gamma 2} + 2 \sin \gamma \cos \gamma \sin{\frac \gamma 2} \cos{\frac \gamma 2} + \cos^2{\gamma} \sin^2{\frac \gamma 2}\right) =$

$16\left(\sin^2{\gamma} \frac{1 + \cos \gamma}{2} + \sin^2{\gamma} \cos \gamma + \cos^2{\gamma} \frac{1 - \cos \gamma}{2}\right) =$

$8\left[(1 - \cos^2{\gamma})(1 + \cos \gamma) + 2(1 - \cos^2{\gamma}) \cos \gamma + \cos^2{\gamma}(1 - \cos \gamma)\right] =$

$= 8 (1 + 3 \cos \gamma - 4 \cos^3{\gamma})$

and the right side becomes

$\frac{1}{36} (289  + 952 \cos \gamma - 952 \cos^2{\gamma} + 784 \cos^2{\gamma} - 1568 \cos^3{\gamma} + 784 \cos^4{\gamma})$

Denote $x = \cos \gamma$. Since $0 < \gamma < 90^o$, $0 < x < 1$. Thus we have to show

$288 (1 + 3x - 4x^3) \le 289  + 952x - 168x^2 - 1568x^3 + 784x^4$

$784x^4 - 416x^3 - 168x^2 + 88x + 1 \ge 0$

Since we have an equality for an equilateral triangle, this most ugly quartic polynomial must have a root $x = \cos{60^o} = \frac 1 2$. Moreover, if a maximum is reached at $x = \frac 1 2$, this root has to be a double root. This can be verified by a long division:

$P(x) = 784x^4 - 416x^3 - 168x^2 + 88x + 1 = (2x - 1)^2 (196x^2 + 92x + 1)$

The roots of the remaining quadratic factor $x = \frac{-23 \pm 4\sqrt{30}}{98} < 0$ are both negative, i.e., not acceptable. The fact that the above quartic polynomial has a minimum for $x = \frac 1 2$ can be verified by taking the 2nd derivative:

$P'(x) = 8(392x^3 - 156x^2 - 42x + 11) = 0$ [color=white].[/color] for [color=white].[/color] $x = \frac 1 2$

$P''(x) = 48(196x^2 - 52x - 7) = 48 \cdot 16 > 0$ [color=white].[/color] for [color=white].[/color] $x = \frac 1 2$

Thus we have a single minimum in the interval $(0, 1)$ equal to zero for $x = \cos \gamma = \frac 1 2$, $\beta = \gamma = 60^o$ and $P(x) \ge 0$ elsewhere in this interval. As a result

$AA' + BB' + CC' \le \frac{29}{6}\ R + \frac 7 3\ r$

for an arbitrary triangle, with the equality only for an equilateral triangle.

Oh! It 's so long but true..
I have a short solution but have to use 2 lemma...A stronger result:
[tex]AA'+BB'+CC'\leq\ \sqrt{\frac{70}{3}}R+[12-2\sqrt{\frac{70}{3}}]r[/tex]
And find better [tex]k[/tex] constant st:
[tex]AA'+BB'+CC'\leq\ kR+[12-2k]r[/tex]
First inequality is particular case when:[tex]k=\frac{29}{6}[/tex]
And second inequality:[tex]k=\sqrt{\frac{70}{3}}[/tex]


Sea $ABCDEF$ un hexagono con todos sus lados opuestos paralelos tal que:
$-FC \cap AB= X_1, FC\cap ED= X_2$
$-AD\cap EF = Y_1 ,AD\cap BC =Y_2$
$-BE \cap CD = Z_1, BE\cap AF=Z_2$

Demuestre:
a)
$X_1, Y_1$ y $Z_1$ son colineales y $X_2, Y_2$ y $Z_2$ son colineales
b)
Las lineas $X_1Y_1Z_1$ y $X_2Y_2Z_2$ son paralelas

[quote="skuge"]Sea $ABCDEF$ un hexagono con todos sus lados opuestos paralelos tal que:
$-FC \cap AB= X_{1}, FC\cap ED= X_{2}$
$-AD\cap EF = Y_{1},AD\cap BC =Y_{2}$
$-BE \cap CD = Z_{1}, BE\cap AF=Z_{2}$

Demuestre:
a)
$X_{1}, Y_{1}$ y $Z_{1}$ son colineales y $X_{2}, Y_{2}$ y $Z_{2}$ son colineales
b)
Las lineas $X_{1}Y_{1}Z_{1}$ y $X_{2}Y_{2}Z_{2}$ son paralelas[/quote]
Sean:
G:$AB\cap DC$
H:$CD\cap FE$
I:$EF\cap BA$
los triángulos:BGC, EDH, IAF, IGH son semejantes, entonces si a,b,c son las longitudes de los segmentos:IG, GH, HI respectivamente; podemos decir que existen r,s,t tales que:BG=at, GC=bt, BC=ct, ED=ar, DH=br, AH=cr, IA=as, AF=bs, IF=cs. Con ayuda de esto, aplicando el teorema de menelao 3 veces(sobre IGH y una recta:AD, BE o  CF), podemos hallar las razones se necesitan para demostrar que $X_{1}, Y_{1}, Z_{1}$ son colineales(con el teorema recíproco de menelao, para el triangulo IGH)
Analogamente se demuestra que $X_{2}, Y_{2}, Z_{2}$ son colineales.
Para probar la parte b) es suficiente probar que $Y_{1}X_{1}$ y $X_{2}Y_{2}$ son paralelas, y ello se demuestra con semejanza(depende del grafico)

[quote="Chen241290"]
Sean:
G:$AB\cap DC$
H:$CD\cap FE$
I:$EF\cap BA$
los triángulos:BGC, EDH, IAF, IGH son semejantes, entonces si a,b,c son las longitudes de los segmentos:IG, GH, HI respectivamente; podemos decir que existen r,s,t tales que:BG=at, GC=bt, BC=ct, ED=ar, DH=br, AH=cr, IA=as, AF=bs, IF=cs. Con ayuda de esto, aplicando el teorema de menelao 3 veces(sobre IGH y una recta:AD, BE o  CF), podemos hallar las razones se necesitan para demostrar que $X_{1}, Y_{1}, Z_{1}$ son colineales(con el teorema recíproco de menelao, para el triangulo IGH)
Analogamente se demuestra que $X_{2}, Y_{2}, Z_{2}$ son colineales.
Para probar la parte b) es suficiente probar que $Y_{1}X_{1}$ y $X_{2}Y_{2}$ son paralelas, y ello se demuestra con semejanza(depende del grafico)[/quote]

Mi solución es similar usando homotesias y esto lo simplifica un poco:
usando la misma notación,
$FIA$ es homotesia de $CBG$ con centro $X_{1}$
$HED$ es homotesia de $FIA$ con centro $Y_{1}$
 Sabemos que el producto de dos homotesias resulta en una homotesia con centro colinear a los centros anteriores:
 $CBG$ es homotesia de $HED$ con centro $Z_{1}$
pero además es el producto de las dos homotesias anteriores entonces $X_{1},Y_{1}$ y $Z_{1}$ son colineares.
El resto de la solución es igual a la de Chen.

Se pueden evitar todos los cálculos echando mano de la geometría proyectiva. Sean $ U_{\infty},V_{\infty},T_{\infty}$ los puntos del infinito de $ FA,AB,BC$ y $ \tau_{\infty} \equiv U_{\infty}V_{\infty}T_{\infty}$ la recta del infinito del plano del hexágono. Entonces por reciproco del teorema de Pascal, $ A,B,C,D,E,F$ yacen en un misma cónica. Consiguientemente por el teorema de Pascal en el hexágono $ BADCFE,$ las intersecciones $ X_1 \equiv AB \cap FC,$ $ Y_1 \equiv AD \cap FE$ y $ Z_1\equiv EB \cap DC$ son colineales. De forma análoga en el hexágono $ EDAFCB,$ las intersecciones $ X_2 \equiv ED \cap CF,$ $ Y_2 \equiv DA \cap CB$ y $ Z_2 \equiv AF \cap BE$ son colineales. En el hexágono $ AFEBCD,$ las intersecciones $ P_1 \equiv FA \cap CB ,$ $P_2 \equiv EF \cap DC$ y $ P_3 \equiv EB \cap DA$ son colineales. Entonces los triángulos degenerados  $ \triangle Y_1Y_2T_{\infty}$ y $ \triangle Z_1Z_2U_{\infty}$ son homológicos $ \Longrightarrow$ $ P \equiv Y_1Z_1 \cap Y_2Z_2 \cap \tau_{\infty}$ $ \Longrightarrow$ $ P$ está en el infinito. Por tanto, las rectas $ X_1Y_1Z_1$ y $X_2Y_2Z_2$ son paralelas.


I would like to make one document with LaTeX in which there would be empty input boxes for name, surname, email and so on...

Is it possible to make an LaTeX document and then print it to PDF with these options?

Is it possible do make PDF in which you can put these text boxes so that other persons that get this PDF can input their name, surname and email and then print the document.

I know PDF isn't writeable, only readable document. But is it possible to make it only possible for someone to input this data and print out the document?

Thank you in advance.

You can use forms in PDF which will do exactly what you want. To achieve this use the hyperref package and see the [url=http://www.tug.org/applications/hyperref/manual.html#x1-160006]PDF and HTML forms[/url] section of the hyperref manual


vector
superior algebra
superior algebra unsolved

Let $n\geq 1$ be an integer. Find all rings $(A,+,\cdot)$ such that all $x\in A\setminus\{0\}$ satisfy $x^{2^{n}+1}=1$.

$1 = (-x)^{2^{n}+1}=-x^{2^{n}+1}=-1$, so $2 = 0$. Now, we note that $(t+1)^{2}= t^{2}+1$; applying this $n$ times yields $(t+1)^{2^{n}}= t^{2^{n}}+1$. We therefore find that if neither $x$ nor $x+1$ is $0$, $1 = (x+1)^{2^{n}+1}= (x+1)(x+1)^{2^{n}}= (x+1)(x^{2^{n}}+1) = x^{2^{n}+1}+x^{2^{n}}+x+1 = x^{2^{n}}+x$. Left-multiplying by $x$ yields $x = 1+x^{2}$, or $x^{2}+x+1 = 0$. Note that the condition that $x, x+1 \ne 0$ is equivalent to $x \ne 0, 1$. Note furthermore that this implies $x^{3}= 1$ for all nonzero $x$. Now, for $x, y$ nonzero elements, suppose $x, y \ne 1$. We claim that either $xy = 1$ or $xy^{2}= 1$, i.e. $x = y$ or $y^{2}$. Suppose not. Then $1 \ne xy^{2}= x(y+1) = xy+x$ and $1 \ne xy$, so we can apply the result that $t^{2}+t+1 = 0$ for $t \ne 1$ to find that $0 = (xy+x)(xy+x)+(xy+x)+1 = (xyxy+xy+1)+(xx+x+1)+(x^{2})y+xyx+1 = xy+y+xyx+1 = xy(x+1)+(y+1) = xyx^{2}+y+1$, so $xyx^{2}= y+1$ and right-multiplying by $x$ yields $xy = yx+x$, so $x = xy+yx$. Similarly, $y = xy+yx$, so $x = y$. This yields that besides $0, 1$, if we have any other element, the only other possible elements are its square, whereupon it is immediate that the only possible rings that satisfy this condition are the trivial ring with $0 = 1$, $\mathbb{F}_{2}$, and $\mathbb{F}_{4}$.

I'm interested in a full solution using vector spaces...I tried something similar in the contest itself, but forgot to prove one of my statements... :( ...got 2 f***ing points 4 it


geometry
incenter
geometric transformation
reflection
circumcircle
perpendicular bisector
geometry unsolved

[b]Construction[/b]. It is well-known that $ M_{a}E_{a}$ is a diameter of the nine-point circle; thus the midpoint $ N$ of $ M_{a}E_{a}$ is the nine-point center of $ ABC$. We can now construct the nine-point circle (we have its center and a point on it - $ E_{a}$ or $ M_{a}$). Now, the Feuerbach point is the tangency point of the incircle and the nine-point circle. This point obviously lies on the line $ IN$. Thus, prolong $ IN$ to intersect the nine-point center in two points; one of them is the Feuerbach point $ Fe$ (this part isn't quite exact, but at least it gives a Geogebra construction). Construct now the incircle (we have the incenter $ I$ and the $ Fe$, which we now that lies on it). Draw the tangents from $ M_{a}$ to the incircle. One of them is clearly the sideline $ BC$. Let the intersection points of the line $ BC$ with the incircle and nine-point circle be $ D$ and $ H_{a}$, respectively. Let $ M_{a}Fe$ intersect the incircle again at $ X$ and draw the perpendicular bisector of $ DX$ (this line is the bisector $ AI$). The intersection of this line $ AI$ with the perpendicular in $ H_{a}$ to $ BC$ gives the vertex $ A$. Draw the tangents from $ A$ to the incircle and intersect them with $ BC$. This gives the vertices $ B$ and $ C$.

Explanation:
[quote="I wrote"]
Let $ M_{a}Fe$ intersect the incircle again at $ X$ and draw the perpendicular bisector of $ DX$ (this line is the bisector $ AI$).[/quote]

This is a consequence of the following theorem due to Emelyanov.
[b]
Theorem[/b]. Let $ ABC$ be a triangle and let $ D$, $ E$, $ F$ be the tangecy points of the incircle with $ BC$, $ CA$, $ AB$, respectively. Let $ X$, $ Y$, $ Z$ be the reflections of $ D$, $ E$, $ F$ in $ AI$, $ BI$, $ CI$, respectively, and let $ M$, $ N$, $ P$ be the midpoints of $ BC$, $ CA$, $ AB$. Then, the lines $ MX$, $ NY$, $ PZ$ are concurrent at the Feuerbach point $ Fe$ of $ ABC$.

A synthetic proof can be viewed here:[url]http://pagesperso-orange.fr/jl.ayme/Docs/Symetriques%20de%20OI%20par%20rapport%20au%20triangle%20de%20contact.pdf[/url].

Do you have any simpler solutions on your mind, jrrbc? I find this somehow unnatural.

Dear Cosmin,
After constructing the incircle from Feuerbach point, we may solve in other way:
Draw tangent $ d$ from $ M_a$ to the incirlce to determine the line $ BC$, let $ D$ be the tangency point. The perp line $ d'$ from $ E_a$ to $ d$ is just the A-altitude. Let $ K$ be the orthogonal projection of $ I$ on $ d'$. Then the perp line to $ d$ at $ M_a$ (perp bisector of BC) meets $ KD$ at the midpoint $ M$ of the arc $ BC$ which doesn't contain $ A$ of the circumcircle (see my post in http://www.mathlinks.ro/viewtopic.php?t=256552 ). $ IM$ meets $ d'$ is just the vertex $ A$.
Maybe this solution has the same method with your solution but it doesn't need the consequence of Emelyanov's theorem.  :D


Do any of you know what President Garfield made an original proof for?

[hide="I do!"]The Pythagorean Theorem.[/hide]

[quote="joml88"][quote="ln(dx/dy)"]Darn. I thought I was the only one who knew that.[/quote]

Really?  This is a math website so it's a pretty reasonable to assume that for each bit of mathematical knowledge you have that at least one person on this sites also has that knowledge.

(Btw, I knew that Garfield came up with a proof for the pythagorean theorem also  :P )[/quote]

O, i really don't get your meaning..
Can you tell me who Garfield is?? Don't tell me the famour cat...

It says 'President Garfield' as the thread title and in the original post though...

[quote="eryaman"][quote="ln(dx/dy)"][quote="shobber"]O, i really don't get your meaning..
Can you tell me who Garfield is?? Don't tell me the famour cat...[/quote]
A president. How do you not know that?[/quote]

Doesn't shobber live in China?[/quote]

Well, i do live in China and i have never been to the USA......
So that means a president got an original solution of this theorem? That's really amazing!

[quote="tetrahedr0n"]Thats really cool that someone in that kind of office would have studied math (and more, have some original results in it!)[/quote]

Well, Michael Jordan was a math major! :D

...and Dan Quayle was dumb enough to spell "potato" wrong. Political figures are getting dumber and dumber. Especially George Bush and his famous Bushisms :P .

[quote="zscool"]Michael Jordan switched majors his junior year....

It is very cool to see people in office do math in their spare time...brings to mind Napolean and one of his servicemen Poncelet.[/quote]

Jordan?? He was a math major?? Incredible......
And for Napolean, do you mean the geometry theorem named after him??


[quote="indybar"]Prove for $x,y,z>0$:
a) $\frac{x^3}{x^2+xy+y^2}+\frac{y^3}{y^2+yz+z^2}+\frac{z^3}{z^2+zx+x^2}\geq \frac{x+y+z}{3}$
b) $\frac{x^3}{y^2-yz+z^2}+\frac{y^3}{z^2-zx+x^2}+\frac{z^3}{x^2-xy+y^2}\geq x+y+z$[/quote]
Too easy. I think you shouldn't post them here because they are too old.
a. Another solution:
$\frac{x^3}{x^2+xy+y^2}\ge\frac{2x-y}{3} \Leftrightarrow x^3+y^3\ge xy^2+x^2y$
b.$\frac{x^3}{y^2-yz+z^2}+\frac{y^3}{z^2-zx+x^2}+\frac{z^3}{x^2-xy+y^2}\geq \frac{(x^2+y^2+z^2)^2}{\sum(x^2y+x^2z)-3xyz}$
And we have:
$\frac{(x^2+y^2+z^2)^2}{\sum(x^2y+x^2z)-3xyz}\ge x+y+z \Leftrightarrow\sum x^4+\sum x^2yz\ge\sum(x^3y+x^3z)$


calculus
integration
analytic geometry
graphing lines
slope
vector
trigonometry

Does anyone know that how many questions do I have to get right to get a 5? Thanks in advance.

The curve for a 5 is approximately 65%. So, if you do about equally well on the FR and the MC, 65% of 45 questions is about 30.

I think it depends on the year.  If you actually understand the material well, you shouldn't have any trouble getting a 5.

Yea I think all AP tests are graded relative to everyone taking the examn, so if you pay 1 million dumb people to take the examn then a 5 is a lot easier to get. :D

hehe, for the easy 1 million dumb people. Well, to get a 5 in AB is defintely easy becuase I got it. Probably a bit tougher in BC, but not that tough, I think you can make a lot of mistakes and you dont need extremely advanced problem solving to solve the problems.

umm, you're on this site, which means you should know some math! getting a 5 on BC Calculus (at least in my school) seems to be much easier than a 5 on AB Calculus, and the curve is really big, so you should be fine

What?!?!?! Somehow that makes zero sense....

Yeah, I doubt that BC is easier then AB because BC covers infinite series and more integration techinques then AB.

[quote="MithsApprentice"]If you study well, a 100% on that exam is definitely not out of reach. (Of course, for those of us who just love making stupid mistakes, that may not always be true.)

I personally think there should be a 6 to recognize those students who achieve a perfect.[/quote]

*nods in agreement* I can get a 90% basically from doing the test as fast as I can... if I really took the time, I wouldn't be surprised to get 100% at least once in a while... not that I would ever actually take the time... ;)

I got a 5 in Calc AB last year and it was easy, I am not a top aopser. The only slope fields question was in free response and you had to draw a graph and solve the differential equation by the seperabhle differential equation method.

[quote="qweretyq"][quote="paladin8"]The BC curve is HUGE. If you go to the AP site, 40% of test-takers get 5 or above; compare that to about 25% for AB and 11% for like Physics.[/quote]

where can i take a look at these curves?
I'm shocked that AB 5's are only 25%, and with about 4-5 days left before the exam... I thought AB was very easy. I did practice tests from years before and I could answer almost all the questions.

Maybe this curve is just for last year since people didn't learn slope fields which was a new thing?[/quote]

Calc AB is pretty easy.  If you can do all the problems then you should be fine.

I'm a little worried about BC though...I'm self studying it this year and am still not very confident with my abilities to do vector, parametric and some of the trig stuff.  Well, one more weekend...

Really? I don't remember much except plugging it into my trusty TI-89 :D

Dont worry about Calc BC if you answered [b][u]most[/u][/b] of the questions. I mean there is a curve, and this is AOPS, so I can never tell if what you say "bombing" = I missed a free response part and some MC.

the polar one was pretty easy... ;)
i mean as far as i remember, all you had  to was to find the area of the under the curve with the integral formula, and for the other part, you had to let $dr/d\theta=0$.... ;)

i solved all of the MC except for one of them, which i was $70\%$ sure i knew the right answer but i didn't answer it...
i made one stupid mistake in the second part of the first question...(finding the volume of the solid...) ;)

I feel hella dumb
Because I didn't know how to derive or integrate that

And, on top of that, I didn't know how to use my calculator (89, also) for it

That's what I get for slackin off 2nd semester senior year....

oh well, I'm expectin at least a 4, but a 5 would be nice.

You cant get points off on free response like MC right?
You can only [i]gain[/i] points?

on MC, every wrong answer is 1/4 off.  on free response, you can only gain points.

[quote="joml88"]I was glad that they had a polar rather than a vector/parametric one.  It's just those stupid velocity/acceleration vectors that always throw me off.  I know how to do them but I often times screw them up :D .  For both parts of the free response I ended up working till the very end of the time limit.  Maybe my answers were more thorough than they had to be....I thought that a little more room to write the solutions would have been nice :lol: .  Guess I'm still on USAMO mode...[/quote]

Haha. We've been practicing grading our FR's in class a TON, so I pretty much know exactly how they grade; it's completely unrigorous and extremely easy to get points on, so I didn't bother to write a lot :D

I was hoping for a parametric one....

I'm just glad BC kids didn't get a related rates like AB had to do....

If it was up to me, free response would consist of 3 questions on parametric
and then 3 questions on rotating  areas and what not....

Ah you guys, I was HOPING for a parametric/vector FRQ or a related rates one.....

i did okay on the multiple choice but made some silly, and i mean SILLY errors on the FRQs. i messed up on parts of the polar and on parts of the taylor series one (i forgot to write down that 7 was the first term of the series! and that was given information.  :( ). but then even worse, i had:

$-1<\frac{x-2}{3}<1$

for the interval of convergence (which is correct) but i forgot my algebra and simplified it incorrectly! that sucks! i am still screaming at myself over that!

I HOPE I GET A $5$!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

I can't really remember my answers :!: Looking at his answers, though, I feel pretty good.  They have the questions up on collegeboard.com also: http://www.collegeboard.com/student/testing/ap/calculus_bc/samp.html?calcbc

oh man, i got half way through multiple choice and then suddenly realized that i had forgotten how to do related rates. i spent the rest of the time PRAYING that a related rate wouldn't be on the test. i was SO LUCKY.

[quote="gauss202"]For those who are interested, preliminary solutions to the 2005 AB and BC Free Response questions can be found here: http://clem.mscd.edu/~talmanl/APCalculus.html[/quote]
looking at the answers, of some of the FR, i got all of it right, except for one part of the first problem, finding the volume one...
and by the way..
[quote="joml88"]I know how to do them but I often times screw them up  . For both parts of the free response I ended up working till the very end of the time limit. Maybe my answers were more thorough than they had to be....[/quote]
that happened to me too, like in the last one, the series one the moment i finished the problem, time was up... ;) but my answeres were correct.... :P


AMC AIME trigonometry geometry LaTeX incenter search

Do you know any AIME Level Trig and/or Geometry Problem book. I don't want other topics but just those two concentrated.

Thanks.

Well here are are from AIME 1 2005

7. Quadrilateral ABCD has angle A=angle B=60 degrees (sorry i don't have latex)  BC=8, AD=10 and CD=12.  The length of AB can be expressed as m+sqrt[n] where m and n are natural numbers.  Find m+n.

15. In triangle ABC, AB=20.  The incircle of the triangle (not including the incenter) trisects the median CD.  If the are of the triangle is m+sqrt[n] where m and n are natural numbers, find m+n.  (I think that's the form you have to write it in...don't quite remember...if that isn't, just write the area)

But are there any good books or sets of problems online??

[quote="K81o7"]Don't know any, but i'm sure you could find one in any large bookstore.[/quote]
Impossible...I live in China... :(

You could probably search on the internet and find some...

For geometry, I would suggest Problems in Geometry (I am not sure if thats the right title). It is similar to Geometry Revisited but much easier.

Thanks I'm getting that one.

I wonder what about trig. Trig is something that surely appears on AMC and if I'm good at it, it usually makes geometry problem easier. :ninja:


Let $a,b,c$ are positive such that $abc=1$. Prove
$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}+2(a+b+c) \geq 3 (\frac1a+\frac1b+\frac1c)$

[hide="rather messy, but it works"]
Substitute $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$. After clearing denominators, it reduces to
\[x^{3}+y^{3}+z^{3}+2(x^{2}z+y^{2}x+z^{2}y)\geq 3(x^{2}y+y^{2}z+z^{2}x)\]
But we have $x^{3}+y^{2}x-2x^{2}y=x(x-y)^{2}$, and
\begin{eqnarray*}x^{2}y+y^{2}z+z^{2}x-x^{2}z-y^{2}x-z^{2}y&=&x^{2}(y-z)-y^{2}(x-z)+z^{2}(x-y)\\ &=&(x^{2}-y^{2})(y-z)+(z^{2}-y^{2})(x-y)\\ &=&(x+y)(x-y)(y-z)-(y+z)(x-y)(x-z)\\ &=&(x-y)(y-z)(x-z)\\ \end{eqnarray*}
so the inequality is equivalent to
\[x(x-y)^{2}+y(y-z)^{2}+z(z-x)^{2}\geq (x-y)(y-z)(x-z)\]
WLOG let $z=\min \{ x,y,z \}$. Then let $p=x-z$ and $q=y-z$, so that $x-y=p-q$ as well. Replacing $(x,y,z)$ with $(p,q,0)$ does not change any of the differences, so the right hand side is left unchanged. But as $x\geq p$, $y\geq q$, $z\geq 0$, the left hand side decreases:
\[x(x-y)^{2}+y(y-z)^{2}+z(z-x)^{2}\geq p(x-y)^{2}+q(y-z)^{2}=p(p-q)^{2}+q^{3}\]
It remains to prove that $p(p-q)^{2}+q^{3}\geq pq(p-q)$. If $q=0$, the inequality is trivial. Otherwise, let $r=\frac{p}{q}$ so that it becomes equivalent to
\[r(r-1)^{2}+1\geq r(r-1)\]
But this follows from
\[r(r-1)^{2}+1-r(r-1)=r^{3}-3r^{2}+2r+1=1+(r+2)(r-1)^{2}\geq 0\]
[/hide]


Hi friends,
                 I am planning to join for MSc mathematics course in India. I would like to know about the the opportunities for PhD abroad.
1. What else exam other than GRE should I write ?
2. What percentage of my expense will be covered by the scholarship, where do I apply to them ?
3. Do I have to write entrance exam ie., math paper for each University separately ?
4. What about the recommendation letters ?
5. What else I can do during my MSc years like MTTS to improve my math skills and better my CV ?
6. Most Important of all- Is this foreign PhD really worth, what do I get out of it ? ( question is intended both materialistically and intellectually)

Hoping for help,
Regards,
Srikanth

[quote="pardesi"]Hi srikanth ..
well i think you will get good replies if you repost it [url=http://www.mathlinks.ro/index.php?f=143]here[/url] :)[/quote]

Don't you mean
[url]http://www.artofproblemsolving.com/Forum/index.php?f=143[/url]
 :maybe:

[quote="pardesi"][quote="vishalarul"]Don't you mean
[url]http://www.artofproblemsolving.com/Forum/index.php?f=143[/url]
 :maybe:[/quote]
yes i think that's obvious and that's where u r lead into when u click on "here" link[/quote]
No, when you click that link, you are lead to
[url]http://www.artofproblemsolving.com[/url]

no it doesn't happen to me and i suppose it does't happen to srikanth since he posted in the college forum just few hours after i asked him to do so :wink:

Not necessarily. I get lead to the home page as well.

I shouldn't even be in the forum, as I'm not Indian.  :|

wait are u all posting from Aops or mathlinks

Hope the answers are useful.

Disclaimer: Accurate answers can be gotten from the universities, use these as guidelines:
1. What else exam other than GRE should I write ?
>>> Most  require or strongly suggest  you take GRE.   The marks are generally used as a "cutoff"  but other than that not much weightage.

>>>2. What percentage of my expense will be covered by the scholarship, where do I apply to them ?

Generally full tuition, room and board are covered in virtually all graduate assistance. You may have to teach (TA)  (15-20 hours a week)

3. Do I have to write entrance exam ie., math paper for each University separately ?

>>> Most likely no.
4. What about the recommendation letters ?

>>.>Very important.
5. What else I can do during my MSc years like MTTS to improve my math skills and better my CV ?

>>> publishing papers in peer reviewed journals.
6. Most Important of all- Is this foreign PhD really worth, what do I get out of it ? ( question is intended both materialistically and intellectually)

>>> Depends on you..for many   the subject,  faculty  (who is the adviser .. what field .. how good your work is etc) makes one place more worthy than others...


I wrote an article with 5 sections. I want that each displayed equation be numbered according to that section. Like (1.1), (1.2) etc for the first section, then (2.1), (2.2) etc for the second section and so on.

1. How can I make Latex to know when a section starts and ends?
2. How can I make the equations be numbered as above in the output (final) document?
3. How can I make the refferences to those equations change according to the above numerotation?

Apologize for my silly questions but I'm exhausted after typing 14 pages.

Thanks

Assuming you have amsmath loaded (which you should do anyway) then it's been made really simple for you. Just put \numberwithin{equation}{section} at the start of your document and it's done!


set theory
combinatorics proposed
combinatorics

The elements of the set $F$ are some subsets of $\left\{1,2,\ldots ,n\right\}$ and satisfy the conditions:
i) if $A$ belongs to $F$, then $A$ has three elements;
ii)if $A$ and $B$ are distinct elements of $F$ , then $A$ and $B$ have at most one common element.
Let $f(n)$ be the greatest possible number of elements of $F$. Prove that $\frac{n^{2}-4n}{6}\leq  f(n) \leq  \frac{n^{2}-n}{6}$

i proved the RHS! (for all n, $f(n)\leq \frac{n(n-1)}{6}$).

to prove this, see that all elements of the family F have 3 diferent pairs of elements, and this pais occurs exactly one time in F, because the intersection of every two elements of F have at most 1 element!.

so, $3f(n) \leq C(n,2)$, and we get the result!

The LHS isn't all that hard as well . Let $K_{i}$ consist of ${(a,b,c)|a+b+c \equiv i \bmod n, a,b,c\in{1,2,\ldots n}}$ Now we see very clearly that all K_i's are disjoint.
  We also show that in any $K_{i}$ any two 3-sets can have atmost one element in common (as if they had two then by the congruence condition and the fact that our elements are chosen from ${1,\ldots,n}$ the third also would be fixed). The fact that every set has 3 elements is obvious (If you don't see this .. :rotfl: ) So our set $K_{i}$ satisfies the conditions that $F$ does. Now, it is clear that all the $K_{i}$'s are non-intersecting and their union is the set of all 3-subsets which has size ${n \choose 3}$. Since we have n such $K_{i}$'s we can say that there exists a set of size atleast$\frac{{n\choose 3}}{n}= \frac{(n-1)(n-2)}{6}$ So we choose that $K_{i}$ which satisfies our condition. Now,
   $\frac{(n-1)(n-2)}{6}\leq|K_{i}| \leq f(n)$ thus giving our condition. Now since $\frac{n-1)(n-2)}{6}\geq \frac{n^{2}-4n}{6}$ we are done. Note this gives us a stronger condition - have I gone wrong anywhere?(The idea of partitioning it $\bmod n$ is a general heuristic and I usually try it on any nice set theory problem.)
  Ashwath


Do any of you know about any programs where one can be mentored by a mathematician personally? All the greats had opportunities like that. I think it could really help. Sadly, none of my teachers can solve the problems we discuss in these forums. :(  I need constant contact with a person who has alot of experience and can break hard problems down face to face. Thank you. Happy Holidays :lol:

There is no problem.
Just contact via PMs with persons you want to teach you.

Maybe one of the independent studies programs listed at http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php would meet your needs?

If you want an in-person mentor, that may be harder to find depending on where you are located.   You might try a local college math department, if there is one near you, or look for programs that provide mentors from local industry.

go to like a college near ur area and talk to the math professors.. try to get a private college or at least a good state one..


geometry
calculus
topology
summer program
Mathcamp
Stanford
college

All the math I know since freshman year geometry is self taught, as I skipped algebra II to take a C++ course and then continued on to self teach myself precalc and calc BC. Going into my senior year, I have a 5 on the AP Calc BC test and a decent knowledge more advanced math, i.e. Multivariable Calculus, ODE's, The calculus of differential forms, some real analysis. Right now I'm working on some point-set topology. My question is, how do I emphasize to a college to which I am applying that I have [u]self-studied[/u] all my high school math and beyond? For the stuff that's beyond, should I send my notes/exercises?

An example of a book that I'm working through:

[url=http://www.amazon.com/exec/obidos/tg/detail/-/0817637079/]Advanced Calculus : A Differential Forms Approach[/url]

Excellent book, recommend it to anyone with a knowledge of basic calculus.

I'd recommend at the very least writing about it as one of your essays.

That is an interesting book, and you should definitely mention in your application that you do that kind of reading for "fun."

[quote="curl"]All the math I know since freshman year geometry is self taught, as I skipped algebra II to take a C++ course and then continued on to self teach myself precalc and calc BC. Going into my senior year, I have a 5 on the AP Calc BC test and a decent knowledge more advanced math, i.e. Multivariable Calculus, ODE's, The calculus of differential forms, some real analysis. Right now I'm working on some point-set topology. My question is, how do I emphasize to a college to which I am applying that I have [u]self-studied[/u] all my high school math and beyond? For the stuff that's beyond, should I send my notes/exercises? [/quote]

Your question isn't clear.   What information do you want to convey, and why?

For all but the most elite US universities, differential forms, real analysis etc are at the very limit of what an applicant might know, and having studied them in any form places you at the top of the applicant pool (in the absence of severe disqualifying "issues" such as a criminal record, or worse, a lack of money).

At universities where a math professor may read your application and you will not be the only applicant to have seen such material, they know that both course-taking and self-study can vary widely in the results, and that people can cite whatever they want in an application, including reams of notes and exercises.  If they really need to know the details (probably not) they can invite you for a conversation with a professor, or ask to see the notes and exercises that you allude to.  I doubt it would ever be that crucial.

There is also a difference, that savvy admissions officers will be aware of, between self-study added to your existing course work, and self-study where your advanced mathematical interests are assigned some course credit in lieu of regular high school courses.  Both demonstrate aptitude and motivation, but if it's a situation where you free up 1-2 hours during your school day for an accredited hobby, it is exaggerated to celebrate the self-study aspect (as opposed to just the learning of advanced material, by whatever means) and people familiar with such arrangements may not necessarily be bowled over by it.

I want to convey the effort that I have put forth to cover all of high school mathematics on my own (which is not at all a great feat, by the way), as well as my curiosity and fascination with math.

It almost goes without saying that learning a serious programming language, differential forms and real analysis within the first $n-1$ years of high school involves a lot of self-study.   Discussion of the effort as such, beyond just stating the fact of self-study, can backfire for many reasons, and is mostly irrelevant.  More pertinent would be to quantify or concretely indicate the mathematical level reached, (e.g., "calculus, analysis and topology at about the third-year undergraduate level", or "solved all exercises in Spivak's book on manifolds").  Others can judge the effort from such information.

my son was in a similar situation, needing to somehow document a bunch of stuff beyond Calc BC that he learned on his own. I don't think that sending copies of primary material like notes etc will be considered "proof" that you mastered the subjects you have studied. Anybody can assert anything, and few colleges have the resources to sort through massive amounts of primary material to assess the rigor. My son documented multi and linear alg by taking distance learning courses thru http://www.utexas.edu. They are self paced, so if you already know those subjects you could work thru them pretty fast and get grades on an official transcript in time to submit it with your apps. He also audited a bunch of classes at the local uni. Administration would not allow him to register, but the profs did allow him to submit all the homework and exams for grading. At the end, the profs each wrote a note with what his grade would have been if he had been allowed to register, and he included copies of those with his apps. That still left a lot of stuff that he somehow "knew" which could not be documented (5 years of Mathcamp will do that). But there was plenty of documented stuff to make it obvious that he is a strong math person. Success in competitions can also be viewed as documentation that you have learned a significant body of mathematical knowledge outside a standard curriculum.

[quote="mathclass"]you would have to show that you know the material well, and you didn't just skim through for the sake of saying you did "Ind study". I would guess there aren't AP tests for Real analysis (am I right? :)) so maybe you could go to your local community college & ask to take the Real Analysis, etc, finals. and see what you get, above a 90% or 95% I bet colleges would take that as proof that you understand the material.[/quote]

.. If your local community college offers Real Analysis (mine doesn't)

[quote="blahblahblah"]Worse, taking "real analysis" at a community college is hardly going to be at the level of the analogous course at a decent university, much less an elite one.[/quote]

Entirely true that it's hardly likely that a real analysis course at a community college, where such a thing exists, is anything at all like a real analysis course at a top university. But for the top university admissions officer sorting through applications, it is enough that a student who took the most advanced courses at his local community college showed more ambition than the student who took only high school courses. The top universities all have first-year courses that introduce challenging material even to very well prepared new admittees.

[quote="texas137"]my son was in a similar situation, needing to somehow document a bunch of stuff beyond Calc BC that he learned on his own. I don't think that sending copies of primary material like notes etc will be considered "proof" that you mastered the subjects you have studied. Anybody can assert anything, and few colleges have the resources to sort through massive amounts of primary material to assess the rigor. [/quote]

Few colleges have the [i]need[/i] to assess the rigor.   At the non-elite schools, it doesn't matter what the details of the study program are, just that it substantially exceeds high school.   At the elite schools, for purposes of admission the fine details of the self-study program are not important either; if it is consistent with other information (such as competitions or summer mathematical programs as you mention) the claims will probably be taken at face value as an indication of interest, ambition and high, but not precisely calibrated, ability.
For purposes of course placement or credit once at the university, the content of the self-study doesn't matter because beyond the standard year 1-2 placement exams, a student can generally enroll in whatever courses they want, and universities rarely accredit academic studies without institutional documentation.    Apart from competitions for large scholarships, the need to investigate the exact material studied will rarely arise, and if it does, the school can simply have a math professor contact the applicant.

It's interesting to note that at most of the top math schools there is a professor of mathematics on the admissions committee, or else a committee of the mathematics faculty to whom the main admissions committee can refer the files of applicants who appear to have strong math backgrounds. A former Stanford admissions officer, in her book Questions and Admissions, describes how Stanford adopted the latter system after noting that it was losing out on strong math students to other colleges that were better staffed to evaluate the credentials of strong math students who applied.

Most departments will have a liaison to the admissions office if not a fixed spot on the admissions committee.  Mathematics is unusual in that it is one of the few subjects where a (relatively) large number of applicants claim specific achievements outside the standard curriculum.

[quote="fleeting_guest"]
Few colleges have the [i]need[/i] to assess the rigor.   At the non-elite schools, it doesn't matter what the details of the study program are, just that it substantially exceeds high school.   At the elite schools, for purposes of admission the fine details of the self-study program are not important either; if it is consistent with other information (such as competitions or summer mathematical programs as you mention) the claims will probably be taken at face value as an indication of interest, ambition and high, but not precisely calibrated, ability. [/quote]

We're basically in agreement. Courses being used to support an application for admission do not need to be detailed to the same extent as courses being used for placement or credit. But I would only expect colleges to take a limited amount of bald assertion on face value. If a student has self-studied a lot of post-secondary math, it is in their best interest to document what they can to make the whole package plausible. They can document with grades, letters from profs who know their work, or by success in competitions. I think it is reasonable to assume that someone from the math department where a student applies will review that sort of thing, and accept a certain amount of additional, undocumented but plausible stuff based solely on a student's description of what they did. I do not think it is reasonable to assume that someone from the math department is going to sift through (hundreds of?) pages of the OP's "notes/exercises...from Multivariable Calculus, ODE's, the calculus of differential forms, real analysis, and point-set topology", or quiz them, and come up with some opinion about whether or not the student has actually mastered any of it. I also think it is unreasonable to assume that the OP can simply assert that he has covered those subjects with no documentation of anything beyond AP calc. It is the student's responsibility, not the math department's of every school where s/he is applying, to make his/her application a winning one. If they can document some of that stuff with a grade on a transcript or a short letter from a prof that anyone in the math dept. can understand and believe without spending a lot of time on it, I think that's the student's best route.

Well, again, I just want to communicate the fact that I am a very motivated student and that I am very devoted to math, not necessarily that I am a master of the entire undergraduate curriculum.

That being said, I think that an admissions officer would take my word on the level of math I have reached as much as he or she would take my word on a personal story or a life-changing event. Everything written in a college essay could be cooked up, yet officers do seem to accept what is in most people's essays because it seems reasonable and honest. I am hoping that the same logic will extend to my independent study. As a precaution, though, I will take the advice of some people here and ask a teacher that knows about my progress for a recommendation.

I am actually a bit rusty in some of the areas I studied; for example I studied ODE's summer before my junior year, and now that it's my senior year, I have forgotten the bulk of the material. I may or may not try to refresh my memory on these topics and go on to take a final at a local college to get the certified credit.

Either way, thanks for the advice folks.

Do taking courses off EPGY Stanford seem credible to universities?

Not if the classes themselves cover the same material....

A class being "harder" than another just means that the prof. assigns more work, or just requires more from his/her students by traveling at high speed through material, or perhaps with more rigorous proofs. So to answer your question, no, it wouldn't look bad if you took classes at Columbus instead of OSU so long as the courses you plan on taking cover the same material as they would have at OSU. College admissions officers aren't going to judge you based off which college you take classes at; the fact that you have taken classes at the college level at all will be impressive to them.

If you can only get into classes at one place and not the other, you did the best you could by getting into the place you could get into. A lot of high school students try to increase their academic challenges by self-studying or attending part-time classes at some college near their home, so do what you can, and be satisfied that at least you are challenging yourself beyond the high school curriculum.

[quote="h_s_potter2002"]Off topic, but I have a question.  I live around a lot of colleges, like OSU, Captiol, Otterbein, and Columbus State.  Obviously OSU is going to have harder classes that Columbus State, so would it look bad If I took college classes at Columbus State instead of OSU?[/quote]

I think they'll be impressed in general that you were taking classes at the college.  But.  More information is necessary.  For example, the University of Illinois at Chicago, hereafter to be referred to as UIC, is about a 25 minute train ride away from me.  It is, to put it nicely, not a very good school.  The University of Chicago, one of the strongest math schools in the country (but I'm biased, I'm actually starting my freshman year there in 6 days :)), is about an hour and a half away by public transportation.  Therefore, I went to UIC.  And evidently, the University of Chicago wasn't at all offended.

There could be other equally valid reasons why you would go to Columbus State instead of OSU.  OSU is more expensive, OSU doesn't like high school students in its classes, and so on.  Even if you just though Columbus State would be easier, which would be the worst from the point of view of a college to which you are applying, it's entirely understandable.  You're in high school still, and balancing a college class on top of that is very time consuming.  To take the easier isn't shameful.

[quote="Magnara"] For example, [b]the University of Illinois at Chicago[/b], hereafter to be referred to as [b]UIC[/b], is about a 25 minute train ride away from me.  It is, to put it nicely, not a very good school.  The University of Chicago, one of the strongest math schools in the country (but I'm biased, I'm actually starting my freshman year there in 6 days :)), [/quote]

Sure, UIC is not as good as Univ. of Chicago. However, UIC's math department is by no means weak. UIC's math department is very comparable to the math department at the Ohio State University in Columbus, OH.

Are you thinking of the University of Illinois at Urbana Champaign (UIUC)?

[quote="qvc122"]No, the University of Illinois at Urbana Champaign (UIUC) is better in math and engineering than both UIC and Ohio State University (OSU). However, the math departments at both UIC and OSU are not bad at all.[/quote]

OSU is a sort of "geometric mean" of UIC and UIUC; in quantity and level of researchers, it bears roughly the same relationship to UIC as the relationship between UIUC and OSU.  All three are large departments with a few people in (or near) almost any field that a student could be interested in, but whether or not there are very strong people or a research group, will depend on the school and the subject.

At undergraduate level what will make more the difference is the level of other students, the amount of funding, the quality of life, and so on; one will not be in a position to really detect the level of the faculty.


Currently, I have tried many free word document to pdf converters which i find unsatisfactory.

Every time I try one, the $ \text{\LaTeX}$ images I added are blurred and it is barely legible.

Does anyone know a good pdf converter that produces clear images?

Thanks.

Have you tried this?
[url]http://sourceforge.net/projects/pdfcreator/[/url]

If you have, you can google "pdf printer" and get a lot of programs...

eh LaTeX documents aren't very hard to write..

I know......

I can use the text command, right??

but MikeTeX is screwed up on my computer for some reason and even with the fancydhr (or something like that) it can't export as a pdf file.

that's why i need a good pdf converter......

Did you look at the quality settings before you converted?

Or an alternative, print it out and scan it into a pdf.

[quote="USAMTS Rules"]The solutions will be printed and/or photocopied in black-and-white for grading. [/quote]

my latex comes out blurry too when i convert it to pdf, but it comes out ok when i print it.

but eventually graders are going to print it so no problem.


The title is self explanatory.
Thanks, and see everyone soon.

There will be computers available. It's been 4 years since I was at Colorado, but at that time we had a group go over to the library to use the computers on a nightly basis during free time. They do not recommend bringing laptops as they can easily get broken, especially in a summer camp setting with lots of people running around.


calculus
derivative
function
calculus computations

Has f(x) = exp(x) / 3x any inflection point(s)?

At an inflection point, the second derivative of a funciton is zero.  $\frac{d^2}{dx^2}(\frac{e^x}{3x})=\frac{e^x}{3x}-\frac{2e^x}{3x^2}+\frac{2e^x}{3x^3}=0$
This can be rearranged as $e^x(x^2-2x+2)=0$  Neither of these factors has any real roots and therefore, there is no point of inflection. :D


function
induction
algebra proposed
algebra

find all functions such that:($f: \mathbb{Q}^+\to\mathbb{Q}^+$)
$f(x+1)=f(x)+1$ and $f(x^2)={f(x)}^2$

I'm assuming the "Q+" means positive rational numbers.

[hide]
$f(1) = f(1)^2 \rightarrow f(1) = 1$ or $0$, but $0 \notin \mathbb{Q}^+$, therefore $f(1) = 1$.  From there, an easy induction using the first property of the function gives $f(x) = x$ as the only possible solution.

[/hide]

$f(x)=x$ for any natural namber.
but problem wants to find the she solution of this equation in $\mathbb{Q}^+$

Damn.  I'm looking at $[0,1]$, seeing if it gets me anywhere

[quote="justdudxit"]-.  I'm looking at $[0,1]$, seeing if it gets me anywhere[/quote]
Its not as easy as you wrote i think.

[quote="sqrt4"]As already said $f(1)=1$
and by Induction f(x)=x $\forall x \in N$
so $f(nx)=nx=nf(x)$     $\forall x,n \in N$
Now let $x=\frac{a}{b}$
then $x*b=a*1$
and $f(b*x)=f(a*1)$ using $f(nx)=nf(x)$
we get $f(x)=\frac{a}{b}*f(1)=\frac{a}{b}=x$

Correct?[/quote]
I don't think so.
we can't get $f(nx)=nx=nf(x)$easily by induction based on the condition$f(x+1)=f(x)+1$

[quote="sqrt4"]Well I do.
By Induction:$f(x)=x$ $\forall x \in N$
$nx$ is also natural so f(nx)=nx is correct
and then nx=x+x+x.....+x=f(x)+f(x)+....+f(x)=nf(x)

I don´t see the mistake.[/quote]
well,I see.
It's right that$f(nx)=nf(x),x \in N$
but In your first post ,you wrote"Let $x=\frac{a}{b}$
butl$\frac{a}{b}$is not a integer,except $b=1$
Be careful when trying to solve the problems.

[quote="Rust"]Let $f(\frac{p}{q})=a$, then $f(\frac{p}{q}+q)=a+q$.
We have \[f((\frac{p}{q})^{2})=a^{2},(a+q)^{2}=(f((\frac{p}{q})^{2}+2p+q^{2})=f((\frac{p}{q})^{2}+2p+q^{2})=2p+q^{2}+a^{2}\Longrightarrow 2aq=2p\Longrightarrow a=\frac{p}{q}.\] [/quote]
Sorry,I can't understand your solution.But it seem to be like my solution and there seems to be  some typo in it.


Any of your school math tecahers hate you?

my math teacher hates me

in 5th grade i really liked my teacher.
then in 6th grade my teacher was really mean so i hated her and she hated me
in 7th and 8th grade my teacher was mean too and I hate her so she hates me


Thought it might be entertaining for you guys:

 Find the restrictions on $ x$ if:

#1. $ |x\minus{}5| \leq |2x\minus{}3|$

#2. $ |2x| \leq |x\minus{}3|$

#3. $ |x| > |2x\minus{}6|$

i like to deal with absolute value problems by squaring :D
since these inequalities are preserved while squaring(both sides are positive)

1.$ x^2 \minus{} 10x \plus{} 25 <\equal{} 4x^2 \minus{} 12x \plus{} 9, so
3x^2 \minus{} 2x \minus{} 16 >\equal{}0$
. the solution to the equation is -2,8/3, so x<-2 or x>8/3

the rest can be solved similarly

I am not good at these kinds of problems, but I will still give it a go.

#2. Square it to get 4x^2 < or = x^2 - 6x + 9

         3x^2 < or =  9 - 6x

         3x^2 + 6x - 9 < or = 0

         Solve the equation and the solutions are 1, and -3, so the solution is;
         x < -3 , and x > 1

I let somebody else solve number 3.

3. For x<3, it becomes
x>-2x+6.
3x>6
x>2

For x>0, it becomes
x>2x-6
x<6

Then, answer is $ (2,6)$ (I guess :|)

In this case it's fine, since the thread is not [i]that[/i] old, and the problems were not all finished yet (so people may still be looking for feedback on their answers / solutions).


trigonometry
inequalities proposed
inequalities

Let x,y,z is possitive real number satisfies xy+yz+zx=3 . Prove this ineq
$ \sqrt{x^2\plus{}3}\plus{} \sqrt{y^2\plus{}3} \plus{} \sqrt{z^2\plus{}3} \geq 3\plus{}x\plus{}y\plus{}z$

[quote="eagerkill"]Let x,y,z is possitive real number satisfies xy+yz+zx=3 . Prove this ineq
$ \sqrt {x^2 \plus{} 3} \plus{} \sqrt {y^2 \plus{} 3} \plus{} \sqrt {z^2 \plus{} 3} \geq 3 \plus{} x \plus{} y \plus{} z$[/quote]
See here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133953

Hi,arqady,Nice to see you :) Hope my solution is right and not so ugly.
By change variables $ x \equal{} \cot A,y \equal{} \cot B,z \equal{} \cot C$,Where $ ABC$ is an acute triangle.
$ \frac {1}{\sin A} \plus{} \frac {1}{\sin B} \plus{} \frac {1}{\sin C} \ge \sqrt {3} \plus{} \cot A \plus{} \cot B \plus{} \cot C$
But $ \frac {1}{\sin x} \minus{} \cot x \equal{} \frac {1 \minus{} \cos x}{\sin x} \equal{} \tan \frac {x}{2}$
So it suffices to prove:
$ \tan \frac {A}{2} \plus{} \tan \frac {B}{2} \plus{} \tan \frac {C}{2} \ge \sqrt {3}$
But this is obvious by $ \tan x$ is convex on $ [0,\frac{\pi}{2})$.

[quote="zhaobin"]Hi,arqady,Nice to see you :)[/quote]
Mutually! :lol:


A natural number is a [i]palindrome[/i] when one obtains the same number when writing its digits in reverse order. For example, $481184$, $131$ and $2$ are palindromes.

Determine all pairs $(m,n)$ of positive integers such that $\underbrace{111\ldots 1}_{m\ {\rm ones}}\times\underbrace{111\ldots 1}_{n\ {\rm ones}}$ is a palindrome.

I think it 's $m < 10$ or $n < 10$

When doing the multiplication like in small classes, if $m \geq 10$ or $n \geq 10$ there is an extra carry which introduces a malicious $8$ on the right part.

I would find it difficult to write a full and complete solution in competitions ....

It's not so hard to write a full solution...

Note that $N = \underbrace{111\ldots 1}_{m\ {\rm ones}}\times\underbrace{111\ldots 1}_{n\ {\rm ones}}$ has exactly $m + n - 1$ digits, since $\underbrace{111\ldots 1}_{m\ {\rm ones}}\times\underbrace{111\ldots 1}_{n\ {\rm ones}} < 2\cdot 10^{m-1}\times2\cdot 10^{n-1} = 4 \cdot 10^{m+n-2}$. If $m, n > 9$, then considering the tenth leftmost digit, there would be a carry, thus the number consisting of the first nine digits of $N$ is bigger than the number formed by the last nine digits of $N$, in reversed order. Then if $m,n > 10$ the number $N$ is not a palindrome.

Would that be seen as a full solution ???


In an office during the day the boss gives the secretary a letter to type each time putting the letter on the top of the secretary's inbox.When there is time the secretary takes the top letter off the pile and types it.If there are five letters in all and the boss delivers them in the order 1 2 3 4 5 which of the following could not be the order in which the secretary types them?
A, 12345
B, 24351
C,32415
D,45231
E,54321

Can anyone help me to solve this problem?

Is this assuming you can't just grab one out of the middle?

[hide]It would have to be D.

You wait until the 4th is delivered. Do it and then wait til 5th is delivered. You take that but you can't do the 2nd right after because you have to do it in order so it would be the 3rd.[/hide]

This problem definitely doesn't belong in this forum.

Here is a thread for which this same problem was included in the Problem of the Day.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=362143#p362143

Yes.  Please look through the forums to determine the forum in which the problems you post are most relative in difficulty.


Hi africans.
Can some one send the last test of african MO?
By the way,what was the results?

It's also hard to find Belarussian MO, which contains many interesting problem?

[quote="shobber"]http://www.mathlinks.ro/Forum/viewtopic.php?t=48794
They are the pan aftrican MO problems. Don't know if they are the ones you want.[/quote]Actually shobber here's the link with all the problems: http://www.mathlinks.ro/Forum/resources.php?c=1&cid=28 ;)


linear algebra
matrix
vector
algebra
polynomial
combinatorics proposed
combinatorics

A graph will be called [i]even[/i] if each of its vertices has an even degree.

Let $ V$ be the set of vertices of an even graph $ G$ such that $ \left|V\right|$ is an even number. Prove that the set $ V$ can be partitioned into two non-empty subsets $ V_{1}$ and $ V_{2}$ such that the induced subgraph of $ G$ on each of these two subsets $ V_{1}$ and $ V_{2}$ is an even graph.

  darij

I don't know if the problem is new.  Here is a related problem from Lovasz's book [i]Combinatorial Problems and Exercises[/i].

Let $ G$ be an arbitrary graph with vertex set $ V$.  Then $ V$ can be partitioned into two subsets $ V_{1}$ and $ V_{2}$ such that the induced subgraphs are even.  (This problem allows empty subsets.)

As a corollary, $ V$ can be partitioned into two sets such that one induced subgraph has all-even degrees and the other has all-odd degrees.

Lovasz credits Gallai (unpublished), Chen (SIAM J. Appl. Math 1971), and Posa.

Darij tells me to post the following in his name:

The problem with $ G$ being an arbitrary graph and $ V_{1}$ and $ V_{2}$ not having to be empty is known to him (see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=59434[/url]), but he thinks his problem requires some other idea.

It does need a little more work, but the algebraic proof in that other thread (see the link in NoName's message) can be adapted.

Let's start just as in the topic mentioned above: let $ A$ be the adjacency matrix of the graph, regarded as an endomorphism of $ \mathbb F_{2}^{n}$ (with $ n$ being the number of vertices of $ G$). We want to prove that there is a non-zero vector $ x\in\mathbb F_{2}^{n}$, different from the all-$ 1$ vector (this, together with $ x\ne 0$, says that the sets $ V_{i}$ are both non-empty) such that $ Ax \equal{} 0$. In other words, what we want to show is that the rank of $ A$ cannot be $ n\minus{}1$; we already know that it's at most $ n\minus{}1$ because $ G$ is even and hence the sum of all the columns of $ A$ is zero.

After casting out the $ i$'th row and column for any $ i\in\overline{1,n}$, the problem is reduced to showing that a symmetric $ m\times m$ matrix $ M$ over $ \mathbb F_{2}$ having $ 0$ on the main diagonal can't be invertible if $ m$ is odd. The simplest proof I could think of is this: write the determinant of $ M$ in the usual way, as a sum of products of elements of $ M$. The products which contain an element on the main diagonal are zero, and the other ones can be partitioned (because $ m$ is odd and $ M$ is symmetric) in equal pairs which are the transpose of one another. It follows that the sum of all the products must be $ 0$, i.e. $ \det M \equal{} 0$.

I don't have much to say because your solution is exactly the same as mine, Grobber. Only one remark:

[quote="grobber"]the problem is reduced to showing that a symmetric $ m\times m$ matrix $ M$ over $ \mathbb F_{2}$ having $ 0$ on the main diagonal can't be invertible if $ m$ is odd.[/quote]

This, in fact, holds for any field with characteristic $ 2$ instead of $ \mathbb{F}_{2}$, and is a particular case of the following theorem:

[b](*)[/b] An alternating $ m\times m$-matrix over any field cannot be invertible if $ m$ is odd.

Here, a matrix $ A$ is called [i]alternating[/i] if it satisfies $A^{T} = -A$ and all of its entries on the main diagonal are $0$. Over fields with characteristic $ \neq 2$, a matrix $A$ is alternating if and only if it is antisymmetric (i.e., it satisfies $A^T = -A$), but this does not hold over fields of characteristic $ 2$. Instead, over fields of characteristic $ 2$, a matrix is alternating if and only if it is symmetric and all of its entries on the main diagonal are $ 0$.

You can prove the assertion [b](*)[/b] by first showing it for fields of characteristic $ \neq 2$ (trivially: $ \det A = \det A^T = \det\left(-A\right) = \left(-1\right)^{m} \det A = -\det A$, and thus $2 \det A = 0$, so that $ \det A = 0$) and then concluding that it must hold for fields of characteristic $ 2$ as well by a polynomial identity argument. The latter argument works because the determinant of an alternating matrix is a universal polynomial in its entries below the main diagonal and these entries are completely independent.  (However, the same cannot be said of an antisymmetric matrix, so that this argument would not work anymore if we replaced "alternating" by "antisymmetric" in [b](*)[/b].)

  Darij


$ \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\ge{\frac{3}{4}}$

can be rewritten as follows

$ 4[a(c+1)+b(a+1)+c(b+1)] \ge 3(a+1)(b+1)(c+1)$

$ 4ab+4bc+4ca+4a+4b+4c \ge 3(abc+ab+bc+ca+a+b+c+1)$

$ ab+bc+ca+a+b+c \ge 6$ since $ abc=1$

$ abc=1 \Rightarrow  ab=\frac{1}{c}$

Now, $ \frac{1}{c}+c\ge2$ since $ (c-1)^{2}\ge0$

Then $ ab+c\ge2$, $ bc+a\ge2$, $ ca+2\ge2 \Rightarrow ab+bc+ca+a+b+c \ge 6$ is true.

[quote="Litlle 1000t"]$ a,b,c$ are positive reals such that $ abc=1$. Prove that
$ \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\ge{\frac{3}{4}}$[/quote]

[b] HINT[/b]

 Put

$ a=\frac{x}{y}, b=\frac{y}{z}, , c = \frac{z}{x}$ ,

 clear the denominators and it is sufficient to prove that

   $ 4[xy(x+y)+yz(y+z)+zx(z+x)] \geq 3(x+y)(y+z)(z+x)$ ,

 which easy becaoms

 $ x(y-z)^{2}+y(z-x)^{2}+z(x-y)^{2}\geq 0$

 We have equality only if      $ a=b=c=1$

[u] Babis [/u]

I have another solution  :oops:

$ \frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\geq\frac{3}{4}$

$ \Leftrightarrow4[a(c+1)+b(a+1)+c(b+1)]\geq3(a+1)(b+1)(c+1)$ $ (1)$

Setting $ p=a+b+c$ and $ q=ab+ac+bc$ we have :

$ (1)\Leftrightarrow4(p+q)\geq3(p+q+2)\Leftrightarrow p+q\geq6$

But,

$ p+q=(a+b+c)+(ab+ac+bc)\geq6\sqrt[6]{(abc)^{3}}=6$ from AM-GM.

Done.

EDIT : Oh, it's the same than stergiu  :oops_sign: , sorry, I didn't see your solution

see also [url]http://www.mathlinks.ro/viewtopic.php?p=521094#p521094[/url]

This problem also was in Czech Slovak M.C. (beffore then it appeared in France)

Ohh, this problem comes from French TST  :o


While accessing some pages,I get a message saying it might be unsafe to let an ActiveX control to interact with other parts of the page.
I'm using IE 6.0

What is this problem ?

Can you give us a link to the problem pages? Thanks.

I got the message while I accessed this page.
BTW I accessed this Page from http://www.artofproblemsolving.com/Forum/search.php?search_id=egosearch
i.e. I was viewing my posts and then accessed this page.

Here's the exact message :

I usually get the message when accessing my post from http://www.artofproblemsolving.com/Forum/search.php?search_id=egosearch

Else there is no problem.

I think your firewall may be a little too sensitive.


It is once again that time of year for MATHCOUNTS.  Chapter competitions have already taken place across the nation.

Due to the popularity of the format for our last MATHCOUNTS Math Jam we will again be running a countdown round style Math Jam on Wednesday, Feb. 18.  We will keep a tally of points based on the first three students to submit correct answers as well as the best explanations of the solutions offered.  To see how this worked last time, check out the transcript:

http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=20

There will be a slight change in the way the game will be run.  No split points will be awarded for best solutions.  If two or more solutions are equally nice, the student who answered the problem the quickest will be awarded the extra point.

Come and challenge some of the top MATHCOUNTS students in the nation!

The questions will come from all levels.  I think they will be appropriate practice for the upcoming competitions.

darn.  the math jam is four days after our chapter competition.

[quote="frost13"]I just checked the Math Jams Page and it doesn't list the MATHCOUNTS Jam on the 11th, but on the 18th.  Was that a typo, or have you added one and just haven't listed it yet?

thanks[/quote]

Thank you for noticing my mistake.  The Math Jam will indeed by on the 18th.  I will edit my post.