articles email MIT college Harvard AMC AIME

A friend on an email list for parents that I'm on just posted a link to an article "[url=http://www.dallasobserver.com/issues/2004-06-17/feature.html/1/index.html]Nice Kid, But . . .[/url]" about the randomness of the college admission process. What do the facts reported in the article make you think about the college admission process?
[/url]

When they mention the boy in the opening of the article, who had a good resume but was rejected by MIT, Harvard, etc..., I wasn't surprised. Like they say later in the article - there are probably 10000 people with good SAT scores (1550-1600) who also participate in various clubs and volunteer. His rank of 27 isn't exactly impressive either - is it somehow expected that all the people in front of him will also be accepted by elite schools? I'm starting to understand colleges more and more - there are thousand upon thousands of similar people with similar resumes who expect that they are good enough to make it anywhere, but only a few truly deserving and dedicated people that should be accepted.

College admissions is really unfair.  After senior year, it has made me really bitter about the whole process.  The whole thing is stupid.

[quote]Are you saying that just because you or someone close to you was not accepted into their choice college? Or for some broader reason?[/quote]

Both, but in this case I will say that I dislike the whole process (broader reasons).

I find the facts rather amusing.  Maybe I should get my mother to read that.  Maybe then she'd lay off a bit.  The public school thing is interesting.  My parents switched me out of a private school a few years ago, but now they want me to go to another private school.

The Texas State college 10% thing is pretty neat.  But that lets in a lot of people.  the top 10% can be huge some years.

They don't really know you, it is easy to make yourself look good on paper.

People lie and overemphasize their extracurricular activities.

It is hard for regular smart kids to get into decent colleges.

People become obsessed with "prestige" and have to get into an ivy league school or their life is over.

If you don't get into an ivy league school, people look down on you and think that you are not smart, and think that the people who did are far superior to you (this is just a general observation about parents in my town).

The best school the math team high scorer and AIME winner in my school last year got into was Boston University (this is not a "broader reason").

If you don't get in, you already didn't get it, and on top of that you have to deal with your guidance counselor and family thinking it is the end of the world.

It is not really about being smart or challenging yourself academically and excelling in hard classes.  It is not about how much you enjoy learning or how passionate you are about the subjects you learn.

They don't even consider who slacks off the second semester of senior year and who keeps on working hard.

There is a lot of ambiguity surrounding the whole process.

I can think of more if you like.

[quote]They don't really know you, it is easy to make yourself look good on paper.

People lie and overemphasize their extracurricular activities.
[/quote]

Come on, you can't blame the colleges for this. They have interviews - what else can they do? There are too many applicants for them to truly get to know each one. As for people lying - its nearly impossible to check every fact that every person writes on their application.

[quote]It is hard for regular smart kids to get into decent colleges.
[/quote]

I don't know what you mean either by regular smart kids or by decent colleges. Most of the kids from my school who are smart but not extraordinaly talented at something go to the University of Michigan, which is in fact, far better than a decent school. Also, when there is competition, you cannot blame the college admissions process for that.

[quote]People become obsessed with "prestige" and have to get into an ivy league school or their life is over.
[/quote]

So how are the colleges accountable for this? I mean, why would an elite college not want their institution to be prestigious? If people are obsessed, its their fault.

[quote]If you don't get into an ivy league school, people look down on you and think that you are not smart, and think that the people who did are far superior to you (this is just a general observation about parents in my town).
[/quote]

So what do you expect the colleges to do about this?

[quote]The best school the math team high scorer and AIME winner in my school last year got into was Boston University (this is not a "broader reason").
[/quote]

Again, so what? Admissions are based on national standards, and not on local high scores.

[quote]If you don't get in, you already didn't get it, and on top of that you have to deal with your guidance counselor and family thinking it is the end of the world.
[/quote]

That is the applicant's fault. No one is making one feel like the world is over. By the same token, I could blame say, the AMC, for not being accepted to the USAMO, since I feel it is the end of the world.

[quote]It is not really about being smart or challenging yourself academically and excelling in hard classes. It is not about how much you enjoy learning or how passionate you are about the subjects you learn.
[/quote]

How do you know its not? Perhaps what you think is being smart or challenging oneself differs from the standards of certain colleges. And if its not people who are passionate about the subjects they learn, who exactly gets in?

[quote]They don't even consider who slacks off the second semester of senior year and who keeps on working hard.

There is a lot of ambiguity surrounding the whole process.
[/quote]

They do consider people who slack off - your admission can be rejected!

As for the ambiguity, what do you expect? Naturally, they can't give you a checklist, or everyone will follow it.

She wasn't blaming the colleges, she was complaining about the process.  So pointing out that the colleges don't necessarily directly control some aspects doesn't speak to what she was saying.  She just thought the process is unpleasant, and I am inclined to agree with her (even though it worked out quite well for me).

[quote="JS1527"]Thanks JBL.  Tetrahedron, I am complaining about the whole college admissions process.  I mean, I am going to college, I am not against college!  I just think the whole process is unfair, stressful, and filled with confusion and ambiguity.[/quote]

So what would be an alternative process that you would prefer?

[quote]They do consider people who slack off - your admission can be rejected! [/quote]

This only happens in extreme cases.  They don't care if an A student gets Bs and Cs.  They care if you get an F, but that might not even get your admission rescinded.

[quote]So what would be an alternative process that you would prefer?[/quote]

I don't know, I guess I will not have to go through it again.  Sorry, I think there are many things that could be changed about the admissions process (like they should give reasons why they rejected you instead of having the whole thing shrouded in secrecy for 5 months, and then you get some letter in the mail, and they should base it more on your high school transcript) in terms of what goes on in the admissions office, but my complaits are with the process in general (for a HS student).

[quote]I don't know, I guess I will not have to go through it again. Sorry, I think there are many things that could be changed about the admissions process (like they should give reasons why they rejected you instead of having the whole thing shrouded in secrecy for 5 months, and then you get some letter in the mail, and they should base it more on your high school transcript) in terms of what goes on in the admissions office, but my complaits are with the process in general (for a HS student).[/quote]

I ask because I know many people who are discontent with the process, yet after thinking about it, I don't know how one would arrange a *better* one. I simply cannot imagine how some of the problems of the current system could be fixed.

[quote="JS1527"]They should make more spaces available in top schools.  I mean, more than 2,000 (just guessing) kids have MIT as their dream school, and they are all probably qualified and would be able to succeed there.  So why don't they deserve to go?  All of their hard work ends w/rejection.  I mean, so what if they were just a very hard working computer programmer (just an example) who did not save the world, climb mountains, or w/e.[/quote]

Well, the case has often been made that many, many students in the country are qualified to handle to workload at the most prestigious universities. However, the schools don't just want hardworking kids. If they did, they would simply look at test scores, grades, and other numerical indications of how smart a student is. Instead, they want a diversified and interesting group of students, so they focus a great deal on extracurriculars, achievements, etc... Also, if MIT expanded its student body size so much, the school would lose its prestige and thus less people would want to attend. It could also not be able to supply a large enough number of good professors to teach, thus decreasing the academic standards somewhat as well. Right now, if you go to MIT, you will be among the absolute smartest and most talented students within the country. If it expanded its student count by so much, things just wouldn't be the same.

A big problem is that as our population grows there will be more competition at every college.  The population will grow much faster than the list of elite colleges.  So naturally more qualified people must be rejected.  There is not much you can do about that.  Also, since there will be a greater number of qualified students in the future you can also expect there to be a greater number of top rank professors.  So I think that another problem might be that people should start making their lists bigger because there are going to be a lot of schools that are very close quality wise or at least as good as the best was 10-20 years ago.

It's also worth considering differences in the population going to college.  Even though at more prestigious universities the student bodies tend to come most heavily from the higher regions of the economic scale, far more middle- and lower-class people get higher education of some kind than has been true in the past.  Partly, this is due to the disappearance of good jobs (mainly in industry) that didn't require college education.*  College admissions were much easier before the days of financial aid packages.  So not only is there a problem of population growth, but there is the problem that larger sections of the population are interested in attending college, since more of our economy (and our jobs) rests on a quality education.

* As an aside, I would mention that a major irony of the civil rights movement in the North is that it fought for fairer percentages of minority workers to be hired into, for example, automobile factories -- just at the time when those jobs began declining in number and quality to be replaced by service-sector jobs.

Julie's comment which was most bothersome to me personally (as a parent) was the pressure she felt from counselors and parents to attend a prestigious, 'name' university - I agree that that is unfair added pressure to a very stressful process.

Everyone, keep your search patterns broad.  Ie there are three families of colleges that can satisy your academic needs:  Elite, famous univ. (Ivy, MIT, Stanford, etc);  top 25 liberal arts colleges (Williams, Amherst, etc); State research inst. (MI, CA, and nearly every other state) which offer excellent Honors programs and early access to grad level courses and research.

Since all three of these families of univ can lead to elite grad schools and excellent jobs, I strongly urge you to focus on the 4 undergraduate years - ie which univ. offers the academic, social, geography, and extracurricular blend that will  let you ENJOY yourselves with friends among the academic opportunities and pressures.

LOL

[quote]Also, if MIT expanded its student body size so much, the school would lose its prestige and thus less people would want to attend.[/quote]

See, this is exactly what I am talking about.  Do you want to go to MIT because of "presitge" or do you want to go because of higher quality of education, a chance to study at the best math/science/engineering school in the USA, meet new and interesting people, etc.

Ah, but you took this person's comment out of context. If you read on, (s)he also goes on to say that increasing the number of students would ALSO decrease the academic standards as well as the uniqueness of the students.

The main problem I have with your argument is that you are making it seem like an extremely hardworking student should be more desirable to a college than a hardworking student that also happens to have a [i]passion[/i] and is interesting. I mean, from what I can tell, colleges want kids that know where they are going. At MIT, they BOAST an "interesting" student body. Now, if all they had were a bunch of boring study nerds, it wouldn't be as cool of a school. But MIT students are known to work hard AND play hard. They have a passion. Colleges want students that dream big and try to achieve their dreams. Dreaming big does not mean just moping around going "oh I want to get into MIT, but I only have a 1570 on my SAT and that's not good enough, so I'll take it again." That's not making an impact. Colleges want [i]leaders[/i], that aren't afraid to be different, not just your standard- high-SATs,-top-10%-of-class,-AP/honors-classes,-volunteered-at-homeless-shelter/retirement home,-worked-at-Starbucks-and-played-the-piano kid. I mean, that's great, you know? But what on earth do these things tell a school about you? Nadda. I'm sure MIT sees a couple hundred of applications like that every year. You have to be original! Have fun and get points for it! If you like math- volunteer to teach underprivileged kids their multiplication tables. If you like animals, volunteer at an animal shelter or a zoo or a wildlife rescue center. FIND YOUR PASSION! If you like to skateboard or play basketball or ride horses, live it! Play a guitar in a band! Build a website for a non-profit organization (they will love you forEVER)! Paint, build, read, write, sing, dance... do whatever the hell inspires you and forget about college!

I'm going off subject, but I think you see what I mean? Retaking tests a million times to get that [b][i]perfect[/i][/b] score just isn't that impressive. You have to BE someone. Be unique and yourself. Don't just be a number.

Ok, I will get down off my soapbox now. I hope I didn't anger anyone?

[quote]It's what you have to do these days to avoid becoming another NBK--admissions jargon for "nice kid, but..."[/quote]
My thoughts on this are that someone needs a good proof-reader.

[i]nice[/i] confuted! i didnt notice that.

I was wondering about that when I skimmed the article.

Hmm...possibly playing a sport [i]other than tennis[/i] increases your chances of getting in too?

I was not surprised at all to read that that kid got rejected/deferred.  2nd try 1600s are not particularly uncommon, and nothing else about him made me think - wow, this kid is Harvard/Yale .. etc bound for sure.

Of course, top 10% in Texas is being able to read a chapter book.

[quote="JBL"]Fewer than 1% of students get 1600.  I'd call that uncommon.[/quote]
This is true, but you still have to take into consideration the huge amount of graduating seniors there are every year.
[quote="http://www.manhattan-institute.org/html/ewp_03.htm"]We estimate that there were about 1,299,000 college-ready 18-year-olds in 2000, and the actual number of persons entering college for the first time in that year was about 1,341,000.[/quote] It's probably more along the lines of 1,500,000 now, maybe more. So 1% of that is what, 15,000 kids? MIT enrolled 4,112 undergrads for the 2003-2004 year... I'm not trying to say that "less than 1%" is common, because it's not, but you still have to consider that fewer than 1% of 1.5 million kids is still a lot of warm bodies.

The article said there were about 900 perfect scores on the SAT I last year. So it is in deed rare, both percentage wise and in absolute terms. But keep in mind that there isn't much statistical significance between a score of 1500 and 1600.

[quote="JBL"][quote="blahblahblah"]2nd try 1600s are not particularly uncommon[/quote]

Fewer than 1% of students get 1600.  I'd call that uncommon.[/quote]

It's all relative.  Are 1600 applicants still 1% of MIT's applicant pool?


induction
inequalities
inequalities unsolved

Let $n$ be a positive integer. Show that if $0<x_1\leq x_2\leq x_n$, then\[\prod_{i=1}^n (1+x_i)\left(\sum_{j=0}^n \prod_{k=1}^j \frac{1}{x_k}\right)\geq 2^n(n+1)\]Determine when the equality occurs.

[quote="cezar lupu"]what about huygens inequality?[/quote]
why do you ask this? ;)

I assume you mean j = 1 in that sum.  Because otherwise the product is bad.

If n=1, then we have $(1/x)(x+1) \geq 4$ which is not true!

by the way I can prove Huygens inequality without induction just using AM-GM. Does anyone know any other proof?


[quote="magixter"][quote="13375P34K43V312"](reworded)

How many distinct sums can we get by adding together numbers, in which exactly four digits are used, namely 1, 2, 3, and 4. For example, 12+3+4, 1+2+3+4, and 342+1 are valid. 1234, 2341, etc., don't count.

At least I think that's how I went...it's pretty annoying casework as far as I know so I'm curious to see a nice solution.[/quote]

In the target problem, I think it said that 1234, 2341, etc. do count...[/quote]
Is the answer 85 or something if 2341 counts and stuff?

[hide="Is the answer"]46? (counting 1234 etc)[/hide]

All wrong so far.

[hide="hint"]
note that rearranging the units digit is irrelevant, and so forth[/hide]

Edit: I think this is right
[hide]Consider the cases a+b+c+d, ab+c+d, ab+cd, abc+d, abcd(does this count or not?).

Notice that except for the first one, we do not worry about carrying.

1) a+b+c+d=10, so 1 sum.

2) ab+c+d, Given a, we know b+c+d=10-a (uniquely defined), so there are 4 ways to choose a.

3) ab+cd, Given a+c, we know b+d=10-(a+c) (uniquely defined), and a+c can be any of 3,...,7= 5 sums.

4) abc+d, Given a and b, we know c+d=10-a-b (uniquely defined), so there are 4*3=12 sums.

5) abcd, 4!=24 sums.

So 1+4+5+12+24=46, or 22 is we don't count 1234 etc.

Edit: Oh, i overcounted the intersection of cases 2 and 3 didn't I?

Yeah, Cases 2 and 3 give the same for when it is 37 and 46 (when their tens digit are the same, case 2 takes on values 1,2,3,4 and case 3 takes on values 3,4,5,6,7 so I overcounted on when the tens digit is 3 and 4) so the answer is $46-2=\boxed{44}$[/hide]

[quote="13375P34K43V312"](reworded)

How many distinct sums can we get by adding together numbers, in which exactly four digits are used, namely 1, 2, 3, and 4. For example, 12+3+4, 1+2+3+4, and 342+1 are valid. 1234, 2341, etc., don't count.

At least I think that's how I went...it's pretty annoying casework as far as I know so I'm curious to see a nice solution.[/quote]

Er, in my opinion, I think that 1234, 2341, etc. do count.  :oops_sign:

counting 1234 and etc. case
4*3*2*1=24
counting the A+B+C+D case
1
counting AB+C+D case
its 4C2*2 (because AB can be BA)
counting AB+CD case
I think only 37,46,55,64,73 can be achieved..so 5
counting the ABC+D case
its 4C3*3!=24

Add them up and you get 24+1+12+5+24=66

BUT, there are overcounts like 43+1+2=34+12=46
and 34+1+2=37=13+24=37
(you can find these ones by fidning the maximum of AB+C+D, and the maximum is 46, while the second is 37, so there are only 2 that intersect with the AB+CD case)

Three digit cases should never intersect, I think..

SO 66-2
So $64$??I think this is completely wrong.

My edited solution gives the same answer as the one (official) posted in the other thread  :)

[quote="13375P34K43V312"](reworded)

How many distinct sums can we get by adding together numbers, in which exactly four digits are used, namely 1, 2, 3, and 4. For example, 12+3+4, 1+2+3+4, and 342+1 are valid. 1234, 2341, etc., don't count.

At least I think that's how I went...it's pretty annoying casework as far as I know so I'm curious to see a nice solution.[/quote]

[hide]1, 2, 3, 4, 5, 6, 7, 8, 9

          Number of (digits): 1    2    3
case
1: 4 0 0
2: 2 1 0
3: 1 0 1

ok, lemme try this

 :P

9C4

9C2/7C2

9C3/6C1

126 ways

36/21 ways

84/6 ways

wow...im kind of lost now[/hide]
maybe someone can continue with what i did...

sorry

[quote="13375P34K43V312"](reworded)

How many distinct sums can we get by adding together numbers, in which exactly four digits are used, namely 1, 2, 3, and 4. For example, 12+3+4, 1+2+3+4, and 342+1 are valid. 1234, 2341, etc., don't count.

At least I think that's how I went...it's pretty annoying casework as far as I know so I'm curious to see a nice solution.[/quote]

This problem makes no sense the way you worded it. First you say it's asking for all possible sums that can be gotten by adding numbers but then you imply it's only using 1,2,3 and 4.

[quote="rcv"]I haven't seen the original question, but if I interpret the question correctly, I break it down into four cases, based on the largest addend:

[hide]
With one exception, we note that the digits of the sum must add to 10.  So, the units-digit is uniquely determined when the higher digits are known.

4-digit addend:  Sum is ABCx, where A, B, C, are digits.  $4\times 3\times 2=24$

3-digit addend:  Sum is ABx, where A, B are digits.  $4\times 3=12$

2-digit addend:  Sum is 1x, 2x, 3x, 4x, 5x, 6x, or 7x.  $7$ choices.

1-digit addend:  Sum is 10.  $1$ choice.  Note that this is the only choice where the digits of the sum don't add to 10, so this case is distinct from the other cases.

Total choices:  $1+7+12+24=\boxed{44}$
[/hide][/quote]

Very nice!, i'm a little confused with the 2-digit addend though...

If you could explain it that would be great,
thanks

[quote="rcv"][quote="jhredsox"]Very nice!, i'm a little confused with the 2-digit addend though...

If you could explain it that would be great,
thanks[/quote]
[hide]
There are two subcases for 2-digit addends:
A.  2-digits + 1-digit + 1-digit
B.  2-digit + 2-digits

For subcase A, the smallest possible sum is 19.  (One way of getting this sum is 12+3+4.)  The largest is 46.  (One way of getting this sum is 43+2+1.)  28 and 37 are also possible sums.  [Remember, the digits of the sum must add to 10, so there are no other sums.]

For subcase B, the smallest possible sum is 37.  (One way of getting this sum is 13+24.)  The largest is 73.  (One way of getting this sum is 42+31.)  46, 55, and 64 are also possible sums.

These two subcases overlap.  It isn't really necessary to consider every possible sum, if you can find one way of getting the smallest sum (19) and one way of getting the largest sum (73), and if you are sure you can get all possible sums between them.

Depending on the actual wording of the problem, I think this is a nice National Target question.  If you take the time to list all the combinations, you probably won't have time to do the other problem in the Target pair.  If you aren't careful with your casework, you'll miss something or overcount the 2-digit cases.  But, if you are skilled and guided by good intuition (or perhaps a little luck), I think this can be a 3-minute problem.[/hide][/quote]

Still, this proves how hard of a question this one was.  Nobody, not even a really smart moderator has gotten it.  Everyone is getting the answer that ourteam thought about protesting.

[quote="diophantient"][quote="rcv"][quote="jhredsox"]Very nice!, i'm a little confused with the 2-digit addend though...

If you could explain it that would be great,
thanks[/quote]
[hide]
There are two subcases for 2-digit addends:
A.  2-digits + 1-digit + 1-digit
B.  2-digit + 2-digits

For subcase A, the smallest possible sum is 19.  (One way of getting this sum is 12+3+4.)  The largest is 46.  (One way of getting this sum is 43+2+1.)  28 and 37 are also possible sums.  [Remember, the digits of the sum must add to 10, so there are no other sums.]

For subcase B, the smallest possible sum is 37.  (One way of getting this sum is 13+24.)  The largest is 73.  (One way of getting this sum is 42+31.)  46, 55, and 64 are also possible sums.

These two subcases overlap.  It isn't really necessary to consider every possible sum, if you can find one way of getting the smallest sum (19) and one way of getting the largest sum (73), and if you are sure you can get all possible sums between them.

Depending on the actual wording of the problem, I think this is a nice National Target question.  If you take the time to list all the combinations, you probably won't have time to do the other problem in the Target pair.  If you aren't careful with your casework, you'll miss something or overcount the 2-digit cases.  But, if you are skilled and guided by good intuition (or perhaps a little luck), I think this can be a 3-minute problem.[/hide][/quote]
The other problem in the pair was practically trivial for a number 7, I think it was meant to allow people to spend more time on 8.[/quote]

Indeed it was. I only wish that I had actually made some effort to check my (completely incorrect) answer. That would have put me up 12 places (21st, based on the fact that I would have completely lost the tiebreaker, instead of 33rd).

But, in order to make this relevant, let me say that the way I tried to do it was:

[hide]ABCD: 24 possibilities
ABC+D: 24/2=12 possibilities (D & C can be interchanged for the same sum)
AB+CD: 24/2/2=6 possibilities (D/B and A/C can be interchanged)
AB+C+D: 24/6=4 (D,C,B)
A+B+C+D: 1
Total: 47 (though I actually put down 53 because I forgot to divide case C by 2 for the A/C interchange)
However, you have to remove three possibilities:
42+1+3 (or equivalent)=32+14
32+1+4=23+14
21+34=42+13.

And there you are. 44.[/hide]

I got this
[hide] For 1234, there are 4! ways, or 24.
Next, for 123+4, there are 4! ways, BUT, there are repeats, so you divide by two which equals 12
Then, there's 1+2+3+4 which is 1 way.
After that, there is 12+3+4 which, since units digits are irrelevant, is 4 ways to choose the tens digit.
Finally, there is 12+34. This gives us 4 choose 2 ways, since you only count tens digits, so there are 6 ways.
24+12+6+4+1=47
however, numbers like 12+34 and 41+2+3 repeat! there are 2 of these so 47-2=45
i doubt if this is right though...   [/hide]


(1) Find the radius of the smallest circle whose interior and boundary completely contain the two circles with centers $ (0,0)$ and $ (24,7)$ and radii $ 3$, $ 4$, respectively.

(2) A circle has the same center as an ellipse and passes through the foci $ F_2$ and $ F_2$ of the ellipse.  The two curves intersect in four points.  Let $ P$ be any point of intersection.  If the major axis of the ellipse has length $ 15$ and the area of triangle $ PF_1F_2$ is $ 26$, compute the distance between the foci.

Thanks in advance!

[hide="(1)"]
Let $ A=(0,0)$ and $ B=(24,7)$. then the distance between their centers, $ AB$, is $ 25$ by the Pythaogrean Theorem. Also, since the longest line connecting a point on circle $ A$ and a point on circle $ B$ lies on the line connecting their centers, the diameter of the desired circle is just $ 25+3+4=32$, and the radius is therefore $ \boxed{16}$.
[/hide]

[hide="(2)"]
$ F_1F_2$ is the diameter of the circle, therefore $ \triangle PF_1F_2$ is a right triangle.
\begin{align*}
PF_1+PF_2&=2a=15 \\
(PF_1)(PF_2)&=2 \cdot 26 = 52 \\
F_1F_2&=\sqrt{PF_1^2+PF_2^2} \\
&= \sqrt{(PF_1+PF_2)^2-2(PF_1)(PF_2)} \\
&= \sqrt{225-104} \\
&= \sqrt{121} \\
&= \boxed{11}
\end{align*}
[/hide]


inequalities
function
inequalities proposed

$ a,b,c$are sides of a triangle! Prove that:

$ \frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{a+c-b}\geq\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

[hide="hint"]
Ravi substitution: $ x=a+b$ ,$ y=b+c$ ,$ z=c+a$ where $ a,b,c>0$
[/hide]

[quote="N.T.TUAN"]Use $ \frac{1}{x}+\frac{1}{y}\geq\frac{4}{x+y}\forall x,y>0.$ :D[/quote]]

Oh cool :D
and thanks for the IneQuality#5 "N.T.TUAN" that was a bad typo :D

[quote="sinajackson"][quote="navid"]Suppose: a>=b>=c  and then use Karamata Inequality for f(x)=1/x  .[/quote]

sorry but what is Karamata Inequality? :huh: its name is new to me :D[/quote]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=14975[/url]

also another similar one holds:
$ 1)\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\geq 3$
where $ a,b,c$ are the sides of a triangle...
which is very easy and simple to prove...  :D
and also this one which is more complicated than $ 1$:
$ 2)\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}\leq \sqrt{a}+\sqrt{b}+\sqrt{c}$
where $ a,b,c$ are the sides of a triangle...

for number 1 inequality you write when we suppose:
a+b-c=z ,b+c-a=x ,c+a-b=y  Then with Am-Gm Inequality we can easily prove it!
for number2 inequality as i Write in this  topic we can use Karamata Inequality for Concave function: F(x)=Sqrt(x).

another elementry approach for the problems that I gave above are the followings:
[hide]$ 1)$we know that $ a,b,c$ are the sides of a triangle so we can assume that $ a=x+y,b=y+z,c=z+x$ now the inequality is equivalent to:

$ \frac{x+z}{2y}+\frac{z+y}{2x}+\frac{y+x}{2z}\geq 3$

$ LHS=\frac{1}{2}(\frac{x}{y}+\frac{z}{y}+\frac{z}{x}+\frac{y}{x}+\frac{x}{z}+\frac{y}{z})$

but we know that $ A+\frac{1}{A}\geq 2$ for $ A>0$ so we get that:

$ \frac{x}{y}+\frac{y}{x}\geq 2$ and etc.

so we get that:

$ LHS\geq \frac{1}{2}.6=3$ as wanted...

$ 2)\boxed{\text{first solution}}$

for this one we claim that

$ \sqrt{a+b-c}+\sqrt{a-b+c}\leq 2\sqrt{a}$

as navid said put $ a+b-c=x,a-b+c=y$ so our claim is equivalent to:

$ \sqrt{x}+\sqrt{y}\leq \sqrt{2(x+y)}$

$ \Rightarrow x+y+2\sqrt{xy}\leq 2x+2y$

$ \Rightarrow (\sqrt{x}-\sqrt{y})^{2}\geq 0$

so our claim is proved,now similary we get that:

$ \sqrt{a+b-c}+\sqrt{a-b+c}\leq 2\sqrt{a}$

$ \sqrt{b+c-a}+\sqrt{b+a-c}\leq 2\sqrt{b}$

$ \sqrt{c+a-b}+\sqrt{c+b-a}\leq 2\sqrt{c}$

summing up all these inequalities gives us the desired result...

$ \boxed{\text{second solution}}$

like the last one we can assume that $ a=x+y,b=y+z,c=z+x$ so our problem is equivalent to show that:

$ \sqrt{2x}+\sqrt{2y}+\sqrt{2z}\leq \sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$

now square both sides,so its sufficient to show that:

$ 4\sqrt{xy}+4\sqrt{yz}+4\sqrt{zx}\leq \sum 2\sqrt{(x+y)(y+z)}$

but by couchy we know that $ 2\sqrt{(x+y)(y+z)}\geq 2\sqrt{(\sqrt{xy}+\sqrt{yz})^{2}}=2\sqrt{xy}+2\sqrt{yz}$

now agian summing all the same inequalities we get the desired result...  :) [/hide]


the problems here were getting a little too old, so i decided to post a new one--it's easy:

assume $a,b,c>0$ and $a+b=c$ what is the minimum value for $A$, if $A=\frac{a^4+b^4+c^4}{a^2b^2}$

if you mean $A$ by $i$, then yes.
how can it be less than zero, the numinator and denominator are both positive...

[hide="hint"]
this is  calling for Am-Gm.[/hide]

I don't think this is quite suited for Getting Started, but oh well:

[hide]
$a^4+b^4 \ge 2a^2b^2$
$(a+b)^4 \ge (2\sqrt{ab})^4 = 16a^2b^2$

So $A \ge \frac{18a^2b^2}{a^2b^2} = 18$.[/hide]

[hide]That is interesting considering that it happens at two early testing points; a=b=.5, c=1; amd a=b=1; c=2[/hide]

well, when i posted the problem i was thinking of an easy AM_GM, but i'm sorry if it is too hard for Getting Started. :oops:
(Am-Gm is too hard for Getting Started??)


Hi, I encountered another 2 questions that I'm not quite sure how to do.

1.) Given that f and g are continuous functions on [a,b], and that f(a)<g(a) and

g(b)<f(b), show that there exists at least one number c ∈ (a,b) such that f(c)=g(c)

2.) Suppose f(x)<g(x)<h(x) for all x near a, and

 lim f(x) and
x->a

lim h(x) both exists, does it follow that
x->a

 lim g(x) exists? Justify the answer with an appropriate proof or counterexample.
x->a

Thank you so much!!!  :)  :)

1) Apply the intermediate value theorem to $f-g$.

2) For definiteness, say $f(x)=-1$ for all $x$ and $h(x)=1$ for all $x$.  Fiddle around with things from there.


inequalities
function
calculus
calculus computations

I am stuck up in the following multivariable problem:

[quote]
Prove that for any p > 1 and x, y, z >= 0

[(x+y+z)^p]/3  <=  (z^p + y^p + z^p)/3
[/quote]

Any help would be appreciated.

Thank you.

You probably meant $ \frac{(x\plus{}y\plus{}z)^p}{3^p}\le\frac{x^p\plus{}y^p\plus{}z^p}{3}$

yes. sorry for the typo. can you please help me out?

1. Set $ f(p) \equal{} \left(\frac {x^p \plus{} y^p \plus{} z^p}{3}\right)^\frac {1}{p}$
2. Compute $ f'$
3. ???
4. Profit

[edit] fixed a typo in the first formula.

Another approach: (1) it suffices to consider the case $ x\plus{}y\plus{}z\equal{}1$. (2) Use Lagrange multipliers to find the minimum of $ x^p\plus{}y^p\plus{}z^p$ on the set $ x\plus{}y\plus{}z\equal{}1$, $ x,y,z\ge 0$.

[color=darkblue][b][u]Proof.[/u][/b] You can use the well-known inequality

$ \left\{\begin{array}{c} a\le b\le c \\
\ x\le y\le z\end{array}\right\|$ $ \implies$ $ (a \plus{} b \plus{} c)(x \plus{} y \plus{} z)\le 3(ax \plus{} by \plus{} cz)\ .$

Since $ 0 < a\le b\le c\implies$ $ a^k\le b^k\le c^k$ for any $ k\in \mathcal N^*$ obtain :

$ \left\{\begin{array}{c} (a \plus{} b \plus{} c)(a \plus{} b \plus{} c)\le 3\left(a^2 \plus{} b^2 \plus{} c^2\right) \\
 \\
(a \plus{} b \plus{} c)\left(a^2 \plus{} b^2 \plus{} c^2\right)\le 3\left(a^3 \plus{} b^3 \plus{} c^3\right) \\
 \\
\minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \\
 \\
(a \plus{} b \plus{} c)\left(a^{n \minus{} 2} \plus{} b^{n \minus{} 2} \plus{} c^{n \minus{} 2}\right)\le 3\left(a^{n \minus{} 1} \plus{} b^{n \minus{} 1} \plus{} c^{n \minus{} 1}\right) \\
 \\
(a \plus{} b \plus{} c)\left(a^{n \minus{} 1} \plus{} b^{n \minus{} 1} \plus{} c^{n \minus{} 1}\right)\le 3\left(a^n \plus{} b^n \plus{} c^n\right)\end{array}\right\|\ \bigodot$ $ \implies$

$ (a \plus{} b \plus{} c)^n\le 3^{n \minus{} 1}\left(a^n \plus{} b^n \plus{} c^n\right)$ $ \implies$ $ \left(\frac {a \plus{} b \plus{} c}{3}\right)^n\le \frac {a^n \plus{} b^n \plus{} c^n}{3}\ .$

[b][u]Remark.[/u][/b] Otherwise, you can use the Jensen's inquality for the convex function $ f(x) \equal{} x^n$ , $ x > 0\ :$

$ f\left(\frac {a \plus{} b \plus{} c}{3}\right)\le \frac {f(a) \plus{} f(b) \plus{} f(c)}{3}$ $ \implies$ $ \left(\frac {a \plus{} b \plus{} c}{3}\right)^n\le \frac {a^n \plus{} b^n \plus{} c^n}{3}\ .$[/color]


inequalities
function
algebra
binomial theorem

For $ x \in R$ and $ x \ge 1$, prove that:
$ \sqrt{x\minus{}1}(\sqrt{x\plus{}1}\plus{}1)<x\sqrt{x}$

Squaring both sides gives:

$ (x\minus{}1)(x\plus{}2\plus{}2\sqrt{x\plus{}1}) < x^3$

Expanding the left side gives:

$ x^2\plus{}x\plus{}2x\sqrt{x\plus{}1}\minus{}2\sqrt{x\plus{}1}\minus{}2 < x^3$

I think there is some elegant way to proceed from here with A.M. > G.M. inequality, but I can't figure it out

Hello
[hide=hint]
Mulitply both sides by $ \sqrt {x\plus{}1} \minus{}1$
[hide]

[quote="Unidranix"]For $ x \in R$ and $ x \ge 1$, prove that:
$ \sqrt {x \minus{} 1}(\sqrt {x \plus{} 1} \plus{} 1) < x\sqrt {x}$[/quote]

I'm going to replace the $ x$ with $ x^2$ for computational ease:

For $ x^2 \in R$ and $ x^2 \ge 1$, prove that [the left hand-side is
less than the right-hand side:

$ \sqrt {x^2 \minus{} 1}(\sqrt {x^2 \plus{} 1} \plus{} 1)$ vs.  $ x^2\sqrt {x^2}$ becomes

$ \sqrt{x^4 \minus{} 1} \plus{} \sqrt{x^2 \minus{} 1}$ vs. $ x^3$

For $ x^2 \equal{} 1$, the original statement is true.

Using a square root approximation method such as the binomial theorem,

$ \sqrt{x^4 \minus{} 1} \equal{} x^2 \minus{} \frac{1}{2x^2} \minus{} ...$

Then $ \sqrt{x^4 \minus{} 1} < x^2 \minus{} \frac{1}{2x^2}$

---------------------------------------------------------------

$ \sqrt{x^2 \minus{} 1} \equal{} x \minus{} \frac{1}{2x} \minus{} ...$

Then $ \sqrt{x^2 \minus{} 1} < x \minus{} \frac{1}{2x}$

$ \sqrt{x^4 \minus{} 1} \plus{} \sqrt{x^2 \minus{} 1} < (x^2 \minus{} \frac{1}{2x^2}) \plus{} (x \minus{} \frac{1}{2x})$ vs. $ x^3$

$ x^2 \plus{} x \minus{} \frac{1}{2x} \minus{} \frac{1}{2x^2}$ vs. $ x^3$

$ x^2 \plus{} x \minus{}\frac12(\frac{x \plus{} 1}{x^2})$ . . . (*)

If the left-hand side (just above) can be shown to be less than
the right-hand side, then the original given statement is true.

Noting earlier the case for $ x^2 \equal{} 1$, let's look at the values for
$ x^2 > 1$.

In (*), the expression $ \frac{x \plus{} 1}{x^2}$ starts off at a value of
$ 2$ for $ x \equal{} 1$, and it gets smaller for $ x > 1$.  It is decreasing
function for $ x > 1$.

Then $ x^2 \plus{} x \minus{} \frac12(\frac{x \plus{} 1}{x^2}) \le x^2 \plus{} x \minus{} 1$ vs. $ x^3$ . . . (**)

Let $ x \equal{} 1 \plus{} \epsilon$, for $ \epsilon > 0$, and $ \epsilon$ belongs to the real numbers.

(**) becomes:

$ {(1 \plus{} \epsilon})^2 \plus{} (1 \plus{} \epsilon) \minus{} 1$ vs. $ {(1 \plus{} \epsilon)}^3$

$ 1 \plus{} 2\epsilon \plus{} {\epsilon}^2 \plus{} 1 \plus{} \epsilon \minus{} 1$ vs. $ 1 \plus{} 3\epsilon \plus{} 3{\epsilon}^2 \plus{} {\epsilon}^3$

$ 1 \plus{} 3\epsilon \plus{} {\epsilon}^2$ vs. $ 1 \plus{} 3\epsilon \plus{} 3{\epsilon}^2 \plus{} {\epsilon}^3$

Subtract the terms on the left-hand side from each side:

$ 0$ vs. $ 2\epsilon \plus{} {\epsilon}^3$

Then $ 0 < 2\epsilon \plus{} {\epsilon}^3$, because the right-hand side is
composed of only positive terms.

Therefore,  for $ x \in R$ and $ x \ge 1$,
$ \sqrt {x \minus{} 1}(\sqrt {x \plus{} 1} \plus{} 1) < x\sqrt {x}$ is true.

I'm sincerely sorry that i bothered you guys.
It should be:
$ \sqrt {x \minus{} 1}(\sqrt {x \plus{} 1} \plus{} 1) \le x\sqrt {x}$
And the equality holds when $ x \equal{} \frac {\sqrt {5} \plus{} 1}{2}$
Sorry

$ \sqrt{x\minus{}1}(\sqrt{x\plus{}1}\plus{}1)\le x\sqrt{x}$
$ \sqrt{x\minus{}1}(\sqrt{x\plus{}1}\plus{}1)(\sqrt{x\plus{}1}\minus{}1)\le x\sqrt{x}(\sqrt{x\plus{}1}\minus{}1)$
$ \sqrt{x\minus{}1} \le \sqrt{x}(\sqrt{x\plus{}1}\minus{}1)$
$ x\minus{}1 \le  x(x\plus{}2\minus{}2\sqrt{x\plus{}1})$
$ x^2\plus{}x\plus{}1 \ge 2x\sqrt{x\plus{}1}$
$ x^4 \plus{}2x^3\plus{}3x^2\plus{}2x\plus{}1 \ge 4x^3\plus{}4x^2$
$ x^4 \minus{} 2x^3\minus{}x^2\plus{}2x\plus{}1 \ge 0$
$ (x^2\minus{}x\minus{}1)^2 \ge 0$

Equality holds when $ x \equal{}\frac{\sqrt{5}\plus{}1}{2}$

Thank you RobRoobiks for your nice solution, although this inequality is ugly.
 :lol:


[quote="Silverfalcon"]Calculate:

$\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)\cdots\left(1-\frac{1}{2005}\right)$[/quote]

[hide]
$\prod_{n=3}^{2005} \frac{n-1}{n}$
Major cancellation:
$\frac{(n-1)(n)(n+1)(n+2) \cdots (n+2001)}{n(n+1)(n+2)\cdots(n+2002)}$
$\frac{n-1}{n+2002}=\boxed{\frac{2}{2005}}$
What is SAMO?  :?
[/hide]

[hide=" Answer with Explanation"]Simplify the first fraction, and the last, and everything in between cancles out, so you get $\frac{2}{2005}$[/hide]

[hide]Simplify to get $\frac{2}{2005}$[/hide]

[hide]the first and the last can be simplified and the middle cancels out so the answer is 2/2005[/hide]

Nice job.  :)

SAMO stands for South Afrian Math Olympiad. This problem came from 2nd round of Junior Division this year which is for students around 8th grade to 9th grade. I thought this problem willl fit in mathcounts forum since I saw this type of problem often.


Let $ A$ be a real square matrix with spectral radius $ \rho < 1$. Find a norm $ \|\cdot\|$  on $ \mathbb{R}^n$ such that $ \forall x\in \mathbb{R}^n  \, \|Ax\| \le (1\minus{}\delta)\|x\|$ for some $ \delta > 0$.

Let $ |x|$ stand for some norm in $ \mathbb R^n$. Choose $ m$ such that the corresponding matrix norm of $ A^m$ is less than $ 1$ and put $ \|x\|\equal{}\sum_{k\equal{}0}^{m\minus{}1} |A^kx|$


analytic geometry
function
linear algebra
matrix
vector
trigonometry
geometry

Please give a precise statement of the "right hand rule".  (Something that would be suitable for an analysis textbook.)

The precise version is the coordinate definition of the cross product.

Now, why the rule in the first place? Well, everything would work just as well if we used the left-hand rule instead. It's just a convention, and we only need it for internal consistency. It is a choice between always using the right-hand rule and always using the left-hand rule, as mixing the two would create unwanted discontinuities.

Thank you for your reply.

Shouldn't there be a more geometrical definition, though?

I was thinking something like this.  Define a "coordinate frame" to be a $ 3$ by $ 3$ orthogonal matrix.  (You can think of the first column as corresponding to the first axis of a coordinate frame, the second column as corresponding to the second axis, and the third column as corresponding to the third axis.)  Define a coordinate frame $ R_0$ to be "right-handed" if and only if there exists a function $ R: [0,1] \to \mathbb{R}^{3 \times3}$ such that:

1) $ R$ is continuous
2) $ R(t)$ is orthogonal for all $ t \in [0,1]$
3) $ R(0)$ is the identity matrix
4) $ R(1) \equal{} R_0$.

This is an attempt to capture my intuitive idea of the right hand rule, which involves taking the standard coordinate axes (represented by my right hand) and moving them so that they coincide with the coordinate frame in question.

Next, define an ordered triple of unit vectors $ (a,b,c)$, with $ a,b,c \in \mathbb{R}^3$, to be "right handed" if and only if there exists a right-handed coordinate frame $ R_0 \equal{} [x_0,y_0,z_0]$ such that:

1) $ x_0 \equal{} a$
2) The component of $ b$ on $ y_0$ is positive, and the component of $ b$ on $ z_0$ is $ 0$
3) The component of $ c$ on $ z_0$ is positive.

If anyone would like to critique this definition, or say that it is unnecessary, or that it is overly-complicated, please do so!  I am eager to hear it.  It has always bothered me that some definition of the right-hand rule such as this is not given in textbooks.

Clearly, everything you want is determined by your original choice of how to show the coordinate axes. If you define your coordinate axes in a right-handed fashion, then your cross product obeys a right hand rule, and if you define your coordinate axes in a left-handed fashion, then your cross product obeys a left hand rule.

What you you communicate about this by words or concepts alone? We can that there are two possible orientations; a determinant may be either positive or negative. But I claim that you cannot get someone else to agree which of those possibilities should be called right and which should be called left without exchanging some physical object, or some reference to a physical object that is in both experiences. What kind of physical object? A drawing. A glove. A sample of alanine or glucose or DNA or any of a million other biologically active chiral compounds. A beam of circularly polarized light. The way three particular stars are related in the night sky as seen from Earth. We can use the notions of summer and winter to define "north," sunrise to define "east" and gravity to define "up," but we're still relying on a physical object: the Earth itself.

What if I try to say something like, "on your computer screen, draw the $ y$ axis going the right, the $ z$ axis going up, and the $ x$ axis pointing out of the screen toward you" we're still relying on that word "right" being meaningful - which is it if you're a member of the human community and understand from the humans you have met what you mean by your right hand. That still defends on a physical object - a typical human body.

[quote="Kent Merryfield"]We can that there are two possible orientations; a determinant may be either positive or negative.[/quote]
This is the definition I gave to a friend who asked me how to characterize an ordered triplet of right-handed vectors (their triple scalar product is positive).  As the Merryfields have hopefully made clear, this is more or less a matter of convention.

Thanks everyone.

Here's a related question.  If I can answer this, I think I will be satisfied.

Suppose we define a "coordinate frame" to be an ordered orthonormal basis of $ \mathbb{R}^3$.  That is, an ordered triplet of mutually orthogonal unit vectors in  $ \mathbb{R}^3$.  (I prefer this to my earlier definition of a coordinate frame as an orthogonal matrix in $ \mathbb{R}^{3 \times 3}$.)

Define two coordinate frames $ a$ and $ b$ to be "related" if and only if there exists a continuous function $ f: [0,1] \to \mathbb{R}^9$ such that $ f(t)$ is a coordinate frame for all $ t \in [0,1]$, and $ f(0) \equal{} a$, and $ f(1) \equal{} b$.  It is clear that this is an equivalence relation on the set of all coordinate frames.  How do you prove that the number of equivalence classes induced by this equivalence relation is two?  In other words, how do you prove that every coordinate frame is related to either $ (e_1, e_2, e_3)$ or $ (e_1, e_2, \minus{} e_3)?$  (Here $ e_1, e_2,$ and $ e_3$ are the standard basis vectors in $ \mathbb{R}^3$.)

Here is an objection I have to using the cross product (defined in terms of its coordinates) to define the right hand rule.  Suppose I find two orthogonal unit vectors $ v_1$ and $ w_1$ in $ \mathbb{R}^3$ such that the coordinate frame $ (v_1, w_1, v_1 \times w_1)$ is related to $ (e_1, e_2, e_3)$ (by the relation I defined above).  Suppose then that I find two other orthogonal unit vectors $ v_2$ and $ w_2$ in $ \mathbb{R}^3$ such that the coordinate frame $ (v_2, w_2, v_2 \times w_2)$ is related to $ (e_1, e_2, \minus{} e_3)$.  Is that possible?  Do I need to worry about that?

[quote="Singularitarian"]Define two coordinate frames $ a$ and $ b$ to be "related" if and only if there exists a continuous function $ f: [0,1] \to \mathbb{R}^9$ such that $ f(t)$ is a coordinate frame for all $ t \in [0,1]$, and $ f(0) \equal{} a$, and $ f(1) \equal{} b$.  It is clear that this is an equivalence relation on the set of all coordinate frames.  How do you prove that the number of equivalence classes induced by this equivalence relation is two?  In other words, how do you prove that every coordinate frame is related to either $ (e_1, e_2, e_3)$ or $ (e_1, e_2, \minus{} e_3)?$  (Here $ e_1, e_2,$ and $ e_3$ are the standard basis vectors in $ \mathbb{R}^3$.)[/quote]
It's a pity you gave up the orthogonal matrix definition because what you are talking about is the statement that [url=http://en.wikipedia.org/wiki/Orthogonal_group#Over_the_real_number_field]$ O_n(\mathbb{R})$ has two connected components.[/url]  (What's going on here is that the determinant is continuous, so it is impossible to pass from a matrix with determinant $ 1$ continuously into a matrix with determinant $ \minus{}1$.  On the other hand, it is intuitively likely that you can continuously deform two matrices with determinant $ 1$ into each other - think of continuously deforming two parallelpipeds of the same signed volume into each other.)

[quote="Singularitarian"]Suppose then that I find two other orthogonal unit vectors $ v_2$ and $ w_2$ in $ \mathbb{R}^3$ such that the coordinate frame $ (v_2, w_2, v_2 \times w_2)$ is related to $ (e_1, e_2, \minus{} e_3)$.  Is that possible?  Do I need to worry about that?[/quote]
No.  The triple scalar product of two vectors and their cross product is always positive because of the way the cross product is defined.

Thank you, that was helpful.

Let's say I want to prove that if $ R \in O_3(\mathbb{R})$ and $ \det (R) \equal{} 1$ then there is a path in $ O_3(\mathbb{R})$ from $ R$ to the $ 3 \times 3$ identity matrix .  I believe I could do this by writing down an explicit formula, involving sines and cosines, for a continuous function $ f: [0,1] \to O_3(\mathbb{R})$ such that $ f(0) \equal{} R$ and $ f(1) \equal{} I$ (the identity matrix).  Does that sound like a reasonable approach?

How difficult is it to prove that $ O_n(\mathbb{R})$ has two connected components for any positive integer $ n$?  I didn't see a proof on Wikipedia---can you recommend a good book that has a proof?

I haven't actually gone through the proof, but you can take advantage of the significant restrictions that the form of an orthogonal matrix has imposed on it.  For example, in two dimensions every orthogonal matrix is some combination of rotation and reflection (the pure rotations form one connected component and the rotations + reflection another), and in three dimensions you add rotoinversion to the mix and possibly something else I've forgotten.  The important idea is that there are a small number of parameters, which can be exploited as follows:  every special orthogonal matrix is the [url=http://en.wikipedia.org/wiki/Matrix_exponential]matrix exponential[/url] of a skew-symmetric matrix, whose entries can be modified continuously.  (This corresponds to continuously modifying some parameters that can be thought of as angles but does not require writing down a homotopy.)  This idea looks particularly nice in two dimensions.


"It would be very discouraging," Graham remarks, "if somewhere down the line you could ask a computer if the Riemann hypothesis is correct and it said, 'Yes, it is true, but you won't be able to understand the proof.'"
http://www.fortunecity.com/emachines/e11/86/certain.html

I'm not sure if that's possible...could a computer be programmed like that?

This thread should be moved to the Round Table Forum. It seems sufficiently serious enough to be moved there.

That was really an interesting article, though maybe a little overly apocolypic. This debate over whether computer assisted proofs are valid has been going on since the Four Color Map Proof came out in the 1970s. And it is now universally accepted as a valid proof by everyone. And all the claims that computer assisted proofs would become the new "rage" in theorem proving has turned out to be way off. Deep theorems in mathematics are still overwhelmingly proven in the traditional way, without the aid of computers for anything more than supporting evidence. Nor is it true that deep theorems are being proven with less frequently. In fact the last 4 years has seen the proving of some of the most famous problems in mathematics.

It's not a new thing for proofs of difficult and complicated theorems to be misunderstood or left in doubt, by the mathematical community for a long period of time. It has happened with several great theorems throughout history; see Fourier or Cantor or Godol for example.

But it was still an interesting article. And William Thurston is a brain of the highest order, as well as a very nice guy.


Let $( a_n )^{\infty}_{n = 1}$ be a sequence taking values in the set $\{ 0, 1 \}$. Is is true that
\[ \sum_{n = 1}^{\infty} \frac{a_n}{2^n}  \]
is rational if and only if there is some $k_0$ and a positive integer $m$ such that for all $n \geqslant k_0$ we have $a_{n + m} = a_n$. In other words $a_n$ is periodic for $n$ sufficently large.
Also, can you give any additional conditions for rationality of any serie.
Thanks in advance.

In any base (not just two as given here), the rational numbers are exactly those whose sequence of digits gets periodical.
The same holds for $p$-adic numbers, too.


[quote="irantst"]hi find all f:R--->R that for any real x,y
((x+y)^2)-f(xy)=f(x)+f(y)[/quote]

Let $ P(x,y)$ be the assertion $ (x\plus{}y)^2\minus{}f(xy)\equal{}f(x)\plus{}f(y)$

$ P(0,0)$ $ \implies$ $ f(0)\equal{}0$

$ P(x,0)$ $ \implies$ $ f(x)\equal{}x^2$ which, obviously, is not a solution.

So no solution.

[quote="pco"][quote="irantst"]hi find all f:R--->R that for any real x,y
((x+y)^2)-f(xy)=f(x)+f(y)[/quote]

Let $ P(x,y)$ be the assertion $ (x \plus{} y)^2 \minus{} f(xy) \equal{} f(x) \plus{} f(y)$

$ P(0,0)$ $ \implies$ $ f(0) \equal{} 0$

$ P(x,0)$ $ \implies$ $ f(x) \equal{} x^2$ which, obviously, is not a solution.

So no solution.[/quote]

Maybe a typo... :(

[quote="pco"][quote="prester"] Maybe a typo... :([/quote]

Maybe just a set of invented problems[/quote]

I think you're right... :mad:

[quote="irantst"]this is invented because I produce equation and I try to solve it and when I got in trouble I ask.[/quote]

Front page :

  Algebra Unsolved Problems : Problems you couldn't solve [u][b]and to which you know that there is a solution [/b][/u](i.e. a problem from a contest, etc.) but you don't know it.

  Algebra Proposed & Own Problems : Problems[u][b] that you have already solved [/b][/u]and you are interested in second opinions or solutions.

  Algebra Open Questions : An open question is a question that has no known solution up to this moment, and it is not known wheter the problem has one or not.

[quote="irantst"]find all f:R--->R that
f(f(x)+y)=f(x+2y)+x+y[/quote]

Let $ P(x,y)$ be the assertion $ f(f(x)\plus{}y)\equal{}f(x\plus{}2y)\plus{}x\plus{}y$

$ P(x,f(x)\minus{}x)$ $ \implies$ $ f(x)\equal{}0$, which obviously is not a solution.

So no solution.


function
real analysis
real analysis solved

Let f:R+->R a continuous function such that f(x)<=f(nx) fro every x>0 and every natural n.Prove that there exist lim n->oo f(x)

Check this out:

[url]http://www.mathlinks.ro/viewtopic.php?t=3289[/url]

:lol:
This is one problem that is arising in the forum: it's hard to know which problems were posted and which ones were not.
 :)  :(  :?  :D  :lol:  :cool:  ;)  :blush:  :maybe:  :P
Sorry I got bored.


inequalities quadratics algebra polynomial trigonometry function

[quote="t0rajir0u"][hide="The two-variable case, one-liner"] $(a^{2}+b^{2})(c^{2}+d^{2}) = (ac+bd)^{2}+(ad-bc)^{2}\ge (ac+bd)^{2}$ [/hide] [hide="The general case, method 1"] The quadratic polynomial $P(x) = (a_{1}x-b_{1})^{2}+(a_{2}x-b_{2})^{2}+...+(a_{n}x-b_{n})^{2}$ has either one or no roots, so its discriminant is less than or equal to zero.  This is given by

$\left(-2a_{1}b_{1}-2a_{2}b_{2}-...-2a_{n}b_{n}\right)^{2}-4 \left( a_{1}^{2}+a_{2}^{2}+..+a_{n}^{2}\right) \left( b_{1}^{2}+b_{2}^{2}+...+b_{n}^{2}\right) \le 0$

Dividing by $4$ gives our desired inequality. [/hide] [hide="The general case, method 2"] Let $\vec{v}= \left< a_{1}, a_{2}, ... a_{n}\right>$ and let $\vec{u}= \left< b_{1}, b_{2}, ... b_{n}\right>$.  Then

$\vec{v}\cdot \vec{u}= \left(a_{1}b_{1}+a_{2}b_{2}+...+a_{n}b_{n}\right) = | \vec{v}\parallel \vec{u}| \cos \theta \le | \vec{v}| | \vec{u}| = \sqrt{ \left( a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}\right) \left( b_{1}^{2}+b_{2}^{2}+...+b_{n}^{2}\right) }$

As desired. [/hide][/quote]

ah.. nice..

[hide="The general case, method 3"] We can induct on the number of terms in each sequence.

That is, the inequality is easy to prove for $n = 2$.  Assume it holds for $n = k$.

$\left( a_{1}b_{1}+...+a_{k}b_{k}+a_{k+1}b_{k+1}\right)^{2}= \left( a_{1}b_{1}+..+a_{k}b_{k}\right)^{2}+2 \left( a_{1}b_{1}+...+a_{k}b_{k}\right) \left( a_{k+1}b_{k+1}\right)+( a_{k+1}b_{k+1})^{2}$

$\left( a_{1}^{2}+...+a_{k}^{2}+a_{k+1}^{2}\right) \left( b_{1}^{2}+...+b_{k}^{2}+b_{k+1}^{2}\right) = \left( a_{1}^{2}+...+a_{k}^{2}\right) \left( b_{1}^{2}+...+b_{k}^{2}\right)+\left( a_{1}^{2}+...+a_{k}^{2}\right) b_{k+1}^{2}+a_{k+1}^{2}\left( b_{1}^{2}+...+b_{k}^{2}\right)+( a_{k+1}b_{k+1})^{2}$

Then we just have to prove

$2 \left( a_{1}b_{1}+...+a_{k}b_{k}\right) \left( a_{k+1}b_{k+1}\right) \le \left( a_{1}^{2}+...+a_{k}^{2}\right) b_{k+1}^{2}+a_{k+1}^{2}\left( b_{1}^{2}+...+b_{k}^{2}\right)$

Which follows from AM-GM and our inductive step.

This is a bad proof, since it gives you a very poor idea of what's really going on. [/hide] [hide="The general case, method 4"] Rearrangement.

This is possibly the worst proof, since it gives you [b]no[/b] idea what's really going on  :rotfl: [/hide]

[color=darkblue][b]The Lagrange's identity :[/b] $\sum_{k=1}^{n}a^{2}_{k}\cdot\sum^{n}_{j=1}b^{2}_{j}=\left(\sum_{i=1}^{n}a_{i}b_{i}\right)^{2}+\sum_{1\le k<j\le n}\left|\begin{array}{cc}a_{k}\ & a_{j}\\ b_{k}& b_{j}\end{array}\right|^{2}\ .$[/color]

Proof for generalisation for $k$ sequences is [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=579428#579428]here[/url].

[quote="t0rajir0u"]Correct me if I'm wrong, but isn't that just a special case of Jensen's Extension of Holder's Inequality?[/quote]
 Yes, you are right, it is a special case.
[quote="t0rajir0u"]Ah, sorry, it's not.

Let $a_{i}, b_{i}, ..., l_{i}, \alpha, \beta, ... \lambda > 0$ and $\alpha+\beta+...+\lambda = 1$.  Then

$( \sum_{i=1}^{n}a_{i})^{\alpha}) ( \sum_{i=1}^{n}b_{i})^{\beta}... ( \sum_{i=1}^{n}l_{i})^{\lambda}\ge \sum_{i=1}^{n}( a_{i}^{\alpha}b_{i}^{\beta}... \l_{i}^{\lambda})$[/quote]
(the second quote from [url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=102983+php+viewtopic+forum+ro+mathlinks+http&t=102923[/url]).
But thus you mistoke here, because $\text{Jensen's Extension of Holder's Inequality}\supset\text{my inequality}\supset\mathbf{M4RIO}\text{'s inequality}$. I thought you were right and the first $\supset$ isn't correct.  :blush:

Suppose $v, w \in V$ an inner product space equipped with a norm $\|\cdot\|$ and inner product $\langle \cdot, \cdot \rangle$. If you are not familiar with this notation, you could imagine $V$ is $R^{n}$ and the inner product is the usual dot product.

The usual assumptions on the inner product are:
(1) $\langle u, u \rangle = \|u\|^{2}\ge 0$ and $\langle u, u \rangle = 0 \Leftrightarrow u=0$
(2) $\langle u, v \rangle = \overline{\langle v, u \rangle}$
(3) $\langle \lambda u, v \rangle = \lambda \langle u, v \rangle$ for $\lambda \in \mathbb{C}$

I think that's it. Now, Cauchy is
$\langle u, v \rangle \le \|u\|\|v\|$
From (1),
$0 \le \|u-\lambda v\|^{2}= \|u\|^{2}-\lambda\langle u, v\rangle-\overline{\lambda}\langle v, u \rangle+|\lambda|^{2}\|v\|$
Assume that $u, v \ne 0$ and take $\lambda = \frac{\langle v, u\rangle}{\|v\|^{2}}$
Plug that in and we get
$0 \le \|u\|^{2}-\frac{|\langle u, v\rangle|^{2}}{\|v\|^{2}}$
whence the desired result.

Now you can use Cauchy on infinite sequences, on (appropriate classes of) functions, etc. ;) Of course, letting $V = R^{n}$ (as I mentioned before) gives you the Cauchy you're familiar with. But do not chain such a marvellous beast into the crushing confines of finite dimensions!! :D :D :D

PS Wikipedia has this apparently (their notation is slightly different)
http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequality
Note their comment about the triangle inequality.  :lol:  :lol:

[quote="Xevarion"] But do not chain such a marvellous beast into the crushing confines of finite dimensions![/quote]

Well played, sir.


For positive integers $ n$, define $ A(n)$ to be $ \frac {(2n)!}{(n!)^{2}}$. Determine the sets of positive integers $ n$ for which

(a) $ A(n)$ is an even number,

(b) $ A(n)$ is a multiple of $ 4$.

$ A(n)$ is even is equivalent to $ v_2((2n)!)-{v_2((n!)^2)} \ge 1\Leftrightarrow v_2((2n)!)-2v_2((n!)) \ge 1$ and so we have

$ \sum_{i=1}^\infty{[\frac{2n}{2^i}}]-2\sum_{i=1}^\infty{[\frac{n}{2^i}}]\ge 1 \Leftrightarrow n \ge 1+[\frac{n}{2}]+[\frac{n}{4}]+[\frac{n}{8}]+.....$

it's well-know that $ [a]+[b] \le [a+b]$  Using this lemma we get

$ [\frac{n}{2}]+[\frac{n}{4}]+[\frac{n}{8}]+.....\le [n(1-\frac{1}{2^a})]\leq n-1$ where $ a$ is the last power that ends Legendre sequence
we conclude that all natural number satisfy our condition

Now if $ A(n)$ is a multiple of $ 4$ it means that $ n \ge 2+[\frac{n}{2}]+[\frac{n}{4}]+[\frac{n}{8}]+.....$

if $ n=2^a$ is a power of two we have $ n \ge 2+[\frac{n}{2}]+[\frac{n}{4}]+[\frac{n}{8}]+.....=2+n(1-\frac{1}{2^a})$ which is equivalent to $ n\ge2^{a+1}$ which is a contradiction. Hence we conclude that n is not a power of $ 2$
 we have already found that $ [\frac{n}{2}]+[\frac{n}{4}]+[\frac{n}{8}]+.....\le [n(1-\frac{1}{2^a})]$ where a is the latest power .
Since $ n$ is not a power of two we get $ \frac{n}{2^a}>1$ which implies that $ [\frac{n}{2}]+[\frac{n}{4}]+[\frac{n}{8}]+.....\le [n(1-\frac{1}{2^a})] \leq n-2$
Hence all natural number excluding the power of two are solution .

Our problem is proven ! :)

[quote="makar"]does anyone has any alternate solution?[/quote]

For (a) I have found another proof

we have
$ \frac {(2n)!}{(n!)^2} \equal{} \frac {(2n)!}{n!(2n \minus{} n)!} \equal{} \frac {2n(2n \minus{} 1)!}{n!(2n \minus{} 1 \minus{} (n \minus{} 1))!} \equal{} 2\frac {(2n \minus{} 1)!}{(n \minus{} 1))!((2n \minus{} 1) \minus{} (n \minus{} 1))!}$ $ \equal{} 2C_{2n \minus{} 1}^{n \minus{} 1}$
Hence $ A(n)$ is even for all natural number  :D


what are some of the best english grammar books available ?

It really depends on if you're a native English speaker.

Hard to answer if you don't say what kind of grammar you are looking for.  There are "prescriptive" grammars which tell you how to write and speak 'proper English,' but there are also "descriptive" grammars, which tell you how the English language functions in practice, usually on a deep linguistic level.  The former will teach you how to write "good English" for your English teachers, but are often wrong about what correct English means, and which rules are important for grammatical English.  A descriptive grammar will tell you how the English language works, but it (probably) won't help you make better grades on essays.

Hi Nick31. Have you seen the following?:

http://www.amazon.com/Schaums-Outline-English-Grammar-Ehrlich/dp/0071359850

There are many good grammar books. I suggest the SAT writing Barron's book. It is quite tough, and helps you through a lot of grammar.

Wren and Martin is a classic. It is still used today in Britain and other commonwealth countries, namely India.


function
topology
real analysis
real analysis unsolved

(a) Let $ X$ be a set and $ R$ a ring of continuous real-valued functions on $ X$. Let $ \bar{R}$ be the ring of all real-valued functions that can be uniformly approximated by functions in $ R$. If $ f \in \bar{R}$ and $ \sup_{X}|f(x)|<1$, show that $ cf \in \bar{R}$ for all $ c \in \mathbb{R}$.

(b) Let $ X$ be a compact Hausdorff space and $ R$ a ring of continuous real-valued functions on $ X$ such that $ 1 \in R$, and for each pair of distinct points $ x$ and $ y$ there is a function $ f$ in $ R$ such that $ f(x) \neq f(y)$, and $ |f(z)|<1$ for all $ z \in X$. Show that every continuous real-valued function on $ X$ can be uniformly approximated by a function in $ R$.

[quote] (a) Let $ X$ be a set and $ R$ a ring of continuous functions [/quote]

So sorry. $ R$ is a ring of functions. The word 'continuous' should not be there.

[quote="chungyc"](a - corrected version) Let $ X$ be a set and $ R$ a ring of real-valued functions on $ X$. Let $ \bar{R}$ be the ring of all real-valued functions that can be uniformly approximated by functions in $ R$. If $ f \in \bar{R}$ and $ \sup_{X}|f(x)| < 1$, show that $ cf \in \bar{R}$ for all $ c \in \mathbb{R}$.[/quote]This is just a blown up version of the fact that, given some real number $ |q| < 1$, then the (not necessarily unitary) ring $ R[q] \equal{} \{mq^n|m\in \mathbb Z, n\in\mathbb N_{ > 0}\}$ is either the zero-ring (iff $ q \equal{} 0$) or is dense in $ \mathbb R$.

Reduce to the case that $ f$ does not vanish on $ X$ by splitting off the zero set $ f^{ \minus{} 1}(0)$ of $ f$. Take $ f^2$ if $ f$ has negative values. Thus, assume that $ f > 0$.

Then mimic the proof that $ \mathbb Q$ is dense in $ \mathbb R$ to show that every constant $ c\in\mathbb R$ can be uniformly approximated by some element in the ring $ R[f] \equal{} \{mf^n|m\in \mathbb Z, n\in\mathbb N_{ > 0}\}$ generated by $ f$. Then $ \mathbb R\subset \overline{R[f]}$, in particular the ring $ \overline{R[f]}$ is a unitary $ \mathbb R$-algebra.

For general $ f$ which possibly has zeros in $ X$ we have $ cf\in\overline{R}$ ($ \forall c\in\mathbb R$).

[quote="chungyc"](b) Let $ X$ be a compact Hausdorff space and $ R$ a ring of continuous real-valued functions on $ X$ such that $ 1 \in R$, and for each pair of distinct points $ x$ and $ y$ there is a function $ f$ in $ R$ such that $ f(x) \neq f(y)$, and $ |f(z)| < 1$ for all $ z \in X$. Show that every continuous real-valued function on $ X$ can be uniformly approximated by a function in $ R$.[/quote]

This is implied by the version of the Stone-Weierstrass theorem given in the first posting of [url=http://www.mathlinks.ro/viewtopic.php?p=1246803#1246803]this thread[/url]: $ \overline{R}$ is either $ C(X)$ or it is the ideal of functions vanishing in a common point, say $ x_0\in X$.

If $ \overline{R}\equal{}\{f\in C(X)|f(x_0)\equal{}0\}$ for a point $ x_0\in X$, then $ X$ cannot be a  singleton, since $ 1\in R$. Pick another point $ x_1$ different from $ x_0$. Then there is a function $ f\in R$ with $ 0\equal{}f(x_0)\neq f(x_1)$ and $ \|f\|_0<1$. Then $ g: \equal{}\sum_{n\ge 0}f^n\in\overline{R}$. $ g(x_0)\equal{}0$ implies $ 1\equal{}1(x_0)\equal{}0$. Contradiction.


[quote="spinos"]Can you explain me, how is it something like that imposible

If a>0 proved that the series (alternating series)
a-a^1/2+a^1/3-a^1/4+... devergent

but!!! the series below converge!
(a-a^1/2)+(a^1/3-a^1/4)+...

a-(a^1/2-a^1/3)-(a^1/4-a^1/5)-...

PLEASE HELP ME![/quote]
because your serie divergent!.
if serie divergent and any next member of serie change the sign then you can put $ ($ and $ )$ such as that serie will converge and will be equal to any $ A$ number

as extremal said above, know that a [b]conditionally [/b]convergent series can always be rearranged (by selective regrouping) in such a manner that the resulting series can be shown to diverge, or even converge to practically any sum that you desire! [this is due to riemann]

in contrast, an [b]absolutely [/b]convergent series will [i]always [/i]diverge if any rearrangement diverges and [i]always [/i]converge (to the same sum) if a certain rearrangement is shown to converge to a particular sum.

the series you have, i.e. $ \sum\;u_n$ converges but $ \bigg|\sum\;u_n\bigg|$ diverges... and so, $ \sum\;u_n$ is said to be conditionally convergent.

The first series has a much more basic problem: the terms don't go to zero. As $ n\to\infty$, the partial sums from $ 1$ to $ n$ oscillate between two values- and these two values are the sums of your second and third series.

Thanks all of you for your help!
Can you give me the proof, why the socond and the third series converge?

Estimate the values of $ a^{\frac1n}\minus{}a^{\frac1{n\plus{}1}}$ using the mean value theorem, and compare to series you already know.


probability
number theory
prime numbers
complementary counting

If two distinct numbers are selected at random from the first seven prime numbers, what is the probability that their sum is an even number? Express your answer as a common fraction.

The solution given was a complementary probability method. Alternatively, we can see that the only way we can get an even sum is if we add two odds. This is (5*6)/(6*7)=5/7. The answer is the same if your use combinations.

Here is that complementary counting method. There are 7C2= 21 ways total. You can get an odd number only by using the 2, and one of the other 6 numbers. The probability of this is 6/21=2/7. All the other sums are even. So, the answer is $ 1\minus{}2/7\equal{}5/7$

[hide = Sol]Because $2$ is the only even prime number, the rest are all odd. Thus, $\text{odd} + \text{even} = \text{odd}$. In other words, the only way for two primes to add up to an even number is if neither of the primes are $2$. Thus, we can choose $\dfrac{6}{7}$ numbers for the first choice, and $\dfrac{5}{6}$ numbers for the second. Multiplying these together, we get $\boxed{\dfrac{5}{7}}$.


How many different six-digit numbers can be formed using the digits
1, 2, 3, 4, 5 and 6 given that no digit is repeated in a number?

There are 6 choices for the first digit.  After picking this one, the second digit can be any of the 5 digits not used, so there are 5 choices.  Similarly, there are 4 choices for the third, 3 for the fourth, etc.  We get \[ 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\equal{}\boxed{720}\]


limit
number theory proposed
number theory

$ n\in\mathbb{N}$ and $ n\equal{}a_1\plus{}10a_2\plus{}100a_3\plus{}\ldots$ for $ a_i\in\mathbb{N}$.
Let
$ f(n)\equal{}a_1^3\plus{}a_2^3\plus{}a_3^3\plus{}\ldots$

For example:
\[ f(979)\equal{}9^3\plus{}7^3\plus{}9^3\equal{}1801\]

Let $ x\in\mathbb{N}$ and $ X\equal{}\{f(x),f(f(x)),f(f(f(x))),\ldots\}$. Show that $ |X|<\infty$.

[quote="LideR_M"]$ n\in\mathbb{N}$ and $ n \equal{} a_1 \plus{} 10a_2 \plus{} 100a_3 \plus{} \ldots$ for $ a_i\in\mathbb{N}$.
Let
$ f(n) \equal{} a_1^3 \plus{} a_2^3 \plus{} a_3^3 \plus{} \ldots$

For example:
\[ f(979) \equal{} 9^3 \plus{} 7^3 \plus{} 9^3 \equal{} 1801
\]
Let $ x\in\mathbb{N}$ and $ X \equal{} \{f(x),f(f(x)),f(f(f(x))),\ldots\}$. Show that $ |X| < \infty$.[/quote]
I think it is not hard problem . Because $ \lim_{n\to\infty}\frac{10^{n\minus{}1}}{n}\equal{}\plus{}\infty$ so exist an positive integer M such that for all $ n\geq M$ then $ 10^{n\minus{}1}>9^3.n$ . It shows that this sequence must be bounded or $ X$ has only finite elements .


Compute $ 29\cdot 103 \minus{} 4\cdot 29 \plus{} 29$.


number theory
prime numbers
number theory solved

n is a fixed even positive integer.we know that the only number a : 1<a<n that n|a^(a-1)-1 is a=n-1. Prove that n=2p (for some prime p)

Suppose that $n \neq 2p$, then there is an integer 1<a<n-1 such that n|(a^{a-1}-1)

1) n is of the form kp^2, k is positive integer, p is a prime number
let a=kp+1

2) n is of the form 2p_1...p_t, t>1, 2<p_1<...<p_t are prime numbers
let p be any odd prime number that divides n
let n=kp
consider the numbers (k+1)^k-1,(2k+1)^{2k}-1,...,((p-1)k+1)^{(p-1)k}-1
all of them are divisible by k
one of them is divisible by p (use complete residue system, there is j such that jk+1==-1(mod p))
therefore one of them is divisible by kp=n


does anyone know like the average SAT and ACT scores for people accepted into stanford?
and do they prefer the SAT or ACT?

any other information about getting into stanford is much appreciated!

thanks in advance!

http://www.ucan-network.org/ is a really helpful site, stanford:
[url]http://members.ucan-network.org/stanford_university2[/url]

The most important thing to know is that standardized test scores alone are not going to get you into Stanford (or any other top-tier school). On the other hand, to keep your chances good I would say you want 700+ in everything.


what is the probability that if you multiply two randomly selected two digit whole numbers, the result is greater than 100 and less than 200? express your answer as a common fraction.

how would u do this problem?

[hide="Hint"]

One of the numbers MUST be 14 or less.  Do case-by-case.[/hide]

[hide="Solution."]

$x_{1}$ = random 2-digit number

$x_{2}$ = random 2-digit number

total number of 2-digit numbers that can be chosen: 90

if

$100 < (x_{1})(x_{2}) < 200$

then

$10 \leq x_{1}\leq 14$

$10 \leq x_{2}\leq 19$

probability of $x_{1}$ being a viable number: $\frac{5}{90}$

probability of $x_{2}$ being a viable number: $\frac{10}{90}$

$P = \frac{5}{90}* \frac{10}{90}$

$= \frac{50}{8100}$

$= \frac{1}{162}$

[/hide]

Wow!! that's amazing for your very first post, nice job!!
Btw, $\preceq$ means maximizes not $\leq $ but its okay...

Have you been around these forums for a while? I am still amazed at how well you know LaTeX and everything...

Yeah, you have to use casework.

Afterwards's solution doesn't work because he includes cases like $14\times19$ which is larger than 200


[quote="stergiu"]Nice and simple  geometry  ![/quote]

 [b]question b) [/b]: Triangle DEZ is equilateral.

[quote="mathgeniuse^ln(x)"]
Now, we will show that ACDZ is cyclic.

[/hide][/quote]

  :) But this is very easy(two of the oposite angles are right !).

[quote="hell_ever"][quote="mathgeniuse^ln(x)"][hide]First, notice that DZ=DE=DC, so C,E,Z, are on the circle with center D.
[/quote]

How is DZ=DE?? :?:[/quote]

The rhombus and equilateral triangle.

And, yeah, I was being stupid, I didn't realize that the angles opposite were 90.

[quote="mathgeniuse^ln(x)"][hide]First, notice that DZ=DE=DC, so C,E,Z, are on the circle with center D.

Therfore, we see that angle CEZ has a measure of 30 degrees.

Then, if we show that ACZ is 60 degrees we are done.

Now, we will show that ACDZ is cyclic.

Notice that AD is parallel to BC so CAD=45 degrees, and ZAD=45 degrees.
Now, angle AEZ=120 degrees, so angle AZE=15 degrees.
But, ABZ is isosceles, so ABZ=15 degrees.  Therefore, EBC=30 degrees, and EDC=30 degrees, so ZDC=90 degrees.
But, we have ZAC=90 degrees, so AZDC is cyclic, and this implies that ACZ is 60 degrees, so we are done.[/hide][/quote]

 There are probably some ... holes in the above solution  :

  a) First we must prove that  $ DZ\equal{} DE$ , since we do not know that triangle $ DEZ$ is equilateral.To prove this ,  we see that $ ACDZ$ is cyclic, so $ \angle DZC \equal{} \angle DAC \equal{} 45$.

Hence , $ \triangle DCZ$ is isosceles, and thus  $ DZ\equal{}DC\equal{}DE$.

 The other steps are well done.

[b]Comment[/b]

[i] Trying to solve the problem , I made the following remarks :[/i]

 If we put $ AB\equal{}AC\equal{} a$ , then $ BC \equal{} a \sqrt 2 ,DZ \equal{} DC \equal{} a \sqrt 2$ , so $ CZ\equal{} 2a \equal{} 2 CA$. this means that $ \angle AZC \equal{} 30$ ect

[u]Babis[/u]

To sum-up and little bit more:
Take $F$ the foot of perpendicular from $C$ to $AD$; obviously,$CF$ is equal to the altitude of the right-angled and isosceles $\triangle ABC$, hence $2CF=CD$ and $\angle ADC=30^\circ$, so $\angle CBE=30^\circ$ and $m(\widehat{EBD})=m(\widehat{ABE})=15^\circ$, consequently $\triangle BDZ$ is isosceles ($30^\circ-75^\circ-75^\circ$) and, with $\triangle BED$ isosceles with $m(\widehat{BED})=150^\circ=2m(\widehat{BZD})$ implies $E$ the circumcenter of $\triangle BDZ$, hence $\Delta DEZ$ is equilateral, implying $CD=DE=DZ\ (\ 1 \ )$.
As shown, $ACDZ$ is cyclic, $AD$ the bisector of $\angle CAZ$; with $(1)$ we get $E$ is the incenter of $\Delta ACZ$, done.

Best regards,
sunken rock

$\left. \begin{gathered}
  EBCD\mathop {}\limits^{} romdus \hfill \\
  B\hat CA = D\hat CZ = 45^o  \hfill \\
\end{gathered}  \right\}\mathop {}\limits^{}  \Rightarrow \mathop {}\limits^{} B\hat CE = D\hat CE\mathop {}\limits_{}  \Rightarrow \mathop {}\limits_{} A\hat CE = Z\hat CE$


The equation $ 3y^2 \plus{} y \plus{} 4 \equal{} 2(6x^2 \plus{} y \plus{} 2)$ where $ y \equal{} 2x$ is satisfied by:

$ \textbf{(A)}\ \text{no value of }x \qquad\textbf{(B)}\ \text{all values of }x \qquad\textbf{(C)}\ x \equal{} 0\text{ only}$
$ \textbf{(D)}\ \text{all integral values of }x\text{ only} \qquad\textbf{(E)}\ \text{all rational values of }x\text{ only}$

[hide="Solution"]$ 6x^2\plus{}x\plus{}2\equal{}6x^2\plus{}2x\plus{}2\implies x\equal{}0\implies\boxed{\textbf{(C)}}$[/hide]

That was way too easy, so I must be doing something wrong.

hello, from $ 3y^2\plus{}y\plus{}4\equal{}12x^2\plus{}2y\plus{}4$ and $ y\equal{}2x$ we have $ 12x^2\plus{}2x\plus{}4\equal{}12x^2\plus{}4x\plus{}4 \Leftrightarrow x\equal{}0$. So the answer is $ \text{C}$.
Sonnhard.


demji wrote:[hide]
(xe4)+((xe2)(ye2))+(ye4)=((xe2)+(ye2))e2
[/hide]

Careful: the x 2y 2 term  is  2x 2y 2 if what you say holds true.

Hum...the answer is $\frac{x^2+xy+y^2}{x+y}$.

kunny

[hide]No, I don't think that's quite right.  x4 + 2x2y2 + y4 equals (x 2 + y2)2; the middle coefficient in the numerator of the given fraction, however, is 1 instead of 2, and so does not fit.  In addition, (x+y)3 does NOT equal x3+y3.  It equals (x+y)(x2+xy+y2).[/hide]


In right triangle $ABC$ with hypotenuse $\overline{AB}$, the legs are $AC = 8$ and $BC = 6$.  Point $D$ is the midpoint of $\overline{AB}$, and the perpendicular bisector of $\overline{AB}$ intersects $\overline{AC}$ at $E$.  Compute $DE$.
\[ \text{(A)}\ \frac{4}{3} \qquad \text{(B)}\ \frac{15}{4} \qquad \text{(C)}\ \frac{20}{3} \qquad \text{(D)}\ 10 \qquad \text{(E) None of these answers} \]

[hide]
Once drawn, it can be seen that triangle $ADE$ is similar to triangle $ACB$ by $AA$

We can label $\overline{AE}$ as $8-x$ and $\overline{EC}$ as $x$

Therefore we have $\frac{8-x}{5} = \frac{10}{8}$

Solve for $x$ and get $x = \frac{7}{4}$

Therefore $E.$ $None$ $of$ $these$ $answers$
[/hide]

windslicer, i think we were supposed to find DE, not EC

my solution requires a little more work...
[hide]because AC=8 and BC = 6, AB=10. Since D is the midpoint, AD=BD=5. We are told to draw a perpendicular bisector from D to E. Lets call side EC = X. AE then = 8-x, and so does BE because ADE is similar to BDE. We can find out X using triangle ECB and the Pythagorean Theorem. X = 7/4. Now, plug X into triangle ADE or BDE and use the Pythagorean Theorem. (6.25)^2 - 25 = 14.0625. Take the square root of that and it becomes 3.75, or 15/4. (B)[/hide]

Ahh ok, that's why my answer didn't fit with the others. Thanks for pointing it out! :P :)

[quote="The Math Mage"]The tangent of BC/AC is = tan(x/5) so $\frac {6}8$ = $\frac {x}5$ so x=to $\frac {15}{4}$[/quote]

Yeah, that's how I did it.


geometry
ratio
analytic geometry
graphing lines
slope
perpendicular bisector
Pythagorean Theorem

[hide="wrong answer"]the large square's side is $\frac{d\sqrt2}{2}$.  Its area would be $\frac{2d^2}{4}$, the side of the smaller square is $\frac{d-\frac{d\sqrt2}{2}}{2}$. Its area would be $\frac{2d^2-2d\sqrt{2d^2}+d^2}{8}$. This is looking very ugly. The ratio would be, after simplifying the denominators of the areas,

$\frac{2d^2}{\frac{2d^2-2d\sqrt{2d^2}+d^2}{2}}$

Rewrite $2d\sqrt{2d^2}$ as $d^2\sqrt{8}$.

After factoring out $d^2$, we have $\frac{2}{3-\sqrt{8}}$ which I'm sure is incorrect, but where is the error  :huh: [/hide]

[quote="riddler"]Find the area of the large square divided by the area of the small square.[/quote]

[hide]Draw in the triangle formed by the top corners of the square and the center of the circle. Split it vertically in half (with the radius). The hypotenuse is the radius, which is $\frac{s\sqrt{2}}{2}$ and the legs are each $s/2$. The area of the big square is obviously $s^2$. The radius minus half a side of the square is the side of the small square, or $\frac{s\sqrt{2}}{2}-\frac{s}{2}=\frac{s(\sqrt{2}-1)}{2}$. The ratio is $\frac{s}{\frac{s(\sqrt{2}-1)}{2}}=\frac{2}{\sqrt{2}-1}=2\sqrt{2}+2$,

So our answer is $4(\sqrt{2}+1)^2=4(2+1+2\sqrt{2})=\boxed {12+8\sqrt{2}}.$
[/hide]

Oh, I caught my mistake, but still I can't solve it now.

chess64, the answer is an integer.

Suppose the circle has radius 1 and the small square has side s.
Drawing a line from the center of the circle to a vertex of the small square on the circle, we have a right triangle with the hypotenuse the length of the radius.
$\left(\frac{s}{2}\right)^2+\left(s+\frac{\sqrt{2}}{2}\right)^2=1$
$\frac{s^2}{4}+s^2+\sqrt{2}s+\frac{1}{2}=1$.
$5s^2+4\sqrt{2}s-2=0$
$s=\frac{-4\sqrt{2}\pm\sqrt{32+40}}{10}$
$s=\frac{\sqrt{2}}{5}$ since $s$ is positive.
$s^2=\frac{2}{25}$.
The area of the large square is $\left(2*\frac{\sqrt{2}}{2}\right)^2=2$.
Thus, the ratio is $\frac{2}{\frac{2}{25}}=25$.

yes, chess63 you're answer is incorrect. the answer is an integer.

Draw the radius of the circle which is the perpendicular bisector of two sides of the small square.  Let the midpoint of the closer side be $D$.  Now let the point on the top right of the small square which is on the circle be $A$.  Let the bottom-left corner of the large square be $B$.  If we draw$AB$, it must pass through $M$.  Thus, we can put this in the coordinate plane, with the circle of radius 1, and tangent to the x and y axes.  Then this line has slope 2 and passes through the points $(1-\frac{1}{\sqrt{2}}, 1-\frac{1}{\sqrt{2}})$ and $(1, 1+\frac{1}{\sqrt{2}})$.  We'll use the second point.  Express this in point-slope form...
$y-1-\frac{1}{\sqrt{2}}=2x-2$
$y=2x-1+\frac{1}{\sqrt{2}}$
Find the solutions for this and $(x-1)^2+(y-1)^2=1$
$5x^2-10x+2x\sqrt{2}-2\sqrt{2}+\frac{9}{2}=0$
$x=\frac{10-2\sqrt{2}\pm\sqrt{18}}{10}$
$x=\frac{2-\sqrt{2}}{2}$
which is the bottom left corner of the square, or
$x=\frac{10+\sqrt{2}}{10}=1+\frac{\sqrt{2}}{10}$
This is the point we're looking for.  It means that half a side of the small square is $\frac{\sqrt{2}}{10}$.  This means that one side is $\frac{\sqrt{2}}{5}$, while one side of the larger square is $\sqrt{2}$.  This makes the ratio of their area $5^2=\boxed{25}$

Now that I see towersfreak2006's solution it's a lot easier and simpler than mine  :D Knew i was missing something, but didn't think it would be the pythagorean theorem!

I think gunes used the Power theorem (i.e., power of points).

[quote="236factorial"]I'm sort of confused how x=1.[/quote]
he took it into consideration that the large square had a side of $5$ ;)

After a while of pondering the chord bisection product theorem (the one gunes was mentioning), I finally got what gunes was doing. Very creative answer!

By the way, what is the power of a point theorem?

[quote="236factorial"]By the way, what is the power of a point theorem? [/quote]
I think he has explained in the previous posts ;)

[quote="frt"][quote="236factorial"]By the way, what is the power of a point theorem?[/quote]
That's the theorem gunes posted :)[/quote]

[i]That's[/i] it? I thought it was something more complicated than that!

[quote="236factorial"]

[i]That's[/i] it? I thought it was something more complicated than that![/quote]

Well usually the most useful theorems are extremely simple.  This one happens to be extremely simple, and extremely useful :)

Actually, 236factorial, Power of a Point is slightly more complicated than that.
I think there are two more versions involving tangents and secants.
For more information, look here:
http://www.cut-the-knot.org/pythagoras/PPower.shtml

oh, I knew those too, but I never knew that it was power of a point.


calculus
integration
geometry
conics
parabola
probability
calculus computations

ello

I originally posted this question on a programming forum, but haven't had a solution yet, so I'm throwing open to you math guys. Ignore the spod-talk bit unless you're familiar with computer programming yourself it might not make much sense:

[quote]I'm trying to write a function that'll take a set of results, assign a probability to each result, and then pick one result at random, according to its probability.

Soooo... I thought I'd calculate the area under sections of a basic parabolic function (y = x²) to generate the skewed probabilities.

Because of the nature of this curve, the amount of skew is determined by the range of the parabola used, as x tends to zero, the gradient tends to zero, so using 0 - 0.001 would be only slightly skewed. As x tends to infinity, the gradient tends to infinity, so using values 0 - 100 the probabilities would be heavily skewed.

to recap then, my function would be passed a skew factor, essentially the upper limit of the parabolic graph to use. All I need do is divide the whole range into equal sections according to the number of values and determine the definite integral for each range of the parabola assigned to each of the possible values. Make sense? Good.

[b]integrating y = x² gave me x³ / 3[/b]

from this I developed the following code:
vals = ["dog", "cat", "monkey", "horse", "marmotte", "hampster", "yetti", "ant", "mouse", "duck"];
skewfactor = 9;
trace(getProbabilities(vals, skewfactor));

function getProbabilities(vals: Array, skewfactor: Number): Array {

    var tot: Number = Math.pow(skewfactor, 3) / 3;
    var range: Number = skewfactor / vals.length;

    var probs: Array = new Array();
    var lastbound: Number = 0;
    var count: Number = 0;
    while(count < vals.length) {
        var ubound: Number = (count + 1) * range;
        probs[count] = ((Math.pow(ubound, 3) / 3) - (Math.pow(lastbound, 3) / 3)) / tot;
        lastbound = ubound;
        ++count;
    }

    return probs;
}

and it kinda worked. It produced a set of probabilities which looked like they fell in line with a simple parabola.

My problem is that no matter what skew value (i.e. max range) I pass in, it produces exactly the same results and I'm not sure if my maths is wrong, my theory on the nature of the curve is wrong or if my code is wrong.

Any thoughts, comments or pointers welcomed[/quote]

so have I got the correct thinking with regards to the nature of a parabolic curve? have I integrated the equation correctly?

I'm getting pretty desperate now! :P

Any two parabolas are similar, and you're scaling to normalize them- you have to normalize the area of your slice to 1 to get a probability, and you're scaling horizontally to normalize that dimension as well (by looking at equal slices scaled to that width).
After this scaling, there is only one parabola. You will always get exactly one eighth of the area in the first half.


geometry
rectangle
3D geometry
prism
perimeter
analytic geometry
LaTeX

[quote="falconwing64"]I looked through the "What we really need to know" thread, hoping for some help.  I noticed it got a little sidetracked.  I want to start a new thread with [b]only[/b] formulas - nothing else.  Hopefully, this will help everybody, esp. those who are going to nats.  We could even sticky it.

[b]One last note: put all formulas in red.[/b]

Number of diagonals in an n-sided polygon: [color=red]n*(n-3)/2[/color][/quote]

Why don't you just ignore all the other conversation there,  and only look at the formulas (there were a LOT there and I'm sure the diagonals one was mentioned there already)

i tried doing that...however, i realized it was kind of strenuous searching for the formulas...copying...pasting...finding more i'd missed the first time...etc.

The number of squares a diagonal goes through in a m*n rectangle: [color=red]m+n-GCF(m,n)[/color]

Sorry about that...i fixed it.

sum of an infinite geometric sequence: [color=red]A/(1-r) where A is first term, r is the rate, and r<1[/color]

For a rectangular prism (see the rectangle one)...

[color=red]a + b + c - gcf(a, b) - gcf(a, c) - gcf(b, c) + gcf(a, b, c)[/color]

Based on the I-EP

i graduated from mc 2 years ago, but how about these for any given triangle:

$R = abc/4A$

$A = rs$

$A = \sqrt{(s)(s-a)(s-b)(s-c)}$

$a,b,c$ = sides
$R$ = radius of circumscribed circle
$r$ = radius or inscribed circle
$A$ = area of triangle
$s$ = semiperimeter = half the perimeter; $\displaystyle\frac{a+b+c}{2}$

[quote="pkothari13"]i graduated from mc 2 years ago, but how about these for any given triangle:

$R = abc/4A$

$A = rs$

$A = \sqrt{(s)(s-a)(s-b)(s-c)}$

a,b,c = sides
R = radius of circumscribed circle
r = radius or inscribed circle
A = area of triangle
s = semiperimeter[/quote]

(Semiperimeter is half the perimeter)

[quote="xingster14"]I think the one for figuring out the area of a polygon on a x-y coordinate grid when given the positions of the vertices is:

[color=red]                                -------(x1, y1)
          y1 *x2   (x2, y2)  x1 * y2
          y2 *x3    (x3, y3)  x2 *y3
          y3 *x1    (x1, y1)  x3 *x1
       +_______      add       +_______
             =  N            &             M

| N-M |  /   2   = Area of polygon!

So first you list the coordinates as shown where you go from one vertex to another adjacent/connecting vertex and then list the vertex you started with again at the end. You multiply the two numbers who are diagonal to each other and then multiply them. After this, you add each of the two columns as shown. Then you find the difference of numerical value between the two sums and then divide it by 2. [/color]

Don't know the name of this formula, but it is useful! ..And i need to learn LaTex...
If it is no trouble, can someone make this formula clearer to see?[/quote]

I don't understand it... can someone who does put it in LaTeX please?

[quote="xingster14"]I think the one for figuring out the area of a polygon on a x-y coordinate grid when given the positions of the vertices is:

[color=red]                                -------(x1, y1)
          y1 *x2   (x2, y2)  x1 * y2
          y2 *x3    (x3, y3)  x2 *y3
          y3 *x1    (x1, y1)  x3 *x1
       +_______      add       +_______
             =  N            &             M

| N-M |  /   2   = Area of polygon!

So first you list the coordinates as shown where you go from one vertex to another adjacent/connecting vertex and then list the vertex you started with again at the end. You multiply the two numbers who are diagonal to each other and then multiply them. After this, you add each of the two columns as shown. Then you find the difference of numerical value between the two sums and then divide it by 2. [/color]

Don't know the name of this formula, but it is useful! ..And i need to learn LaTex...
If it is no trouble, can someone make this formula clearer to see?[/quote]

Can you demonstrate with an example?

[quote="pkothari13"]Think of it like this.....

If you have three points $(x_1,y_1), (x_2,y_2),$ and $(x_3,y_3)$, the area of the triangle with those vertices is:
$\displaystyle\frac{1}{2}
\left|
\begin{array}{ccc}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1\end{array}
\right|.
$

Obviously, any negative answers should be made positive. If you don't know, then that function is called the "determinant" of a matrix.[/quote]

Ohhh! That's what xingster meant. I thought you could only do this with triangles, though? Xingster said polygon...

[quote="mysmartmouth"][quote="pkothari13"]Think of it like this.....

If you have three points $(x_1,y_1), (x_2,y_2),$ and $(x_3,y_3)$, the area of the triangle with those vertices is:
$\displaystyle\frac{1}{2}
\left|
\begin{array}{ccc}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1\end{array}
\right|.
$

Obviously, any negative answers should be made positive. If you don't know, then that function is called the "determinant" of a matrix.[/quote]

I am having trouble with matrices (I get an image of Agent Smith [b]every time[/b]).  Can someone please do an example with clear steps?  This looks like a formula worth learning.[/quote]

Yeah, me too, except I get an image of Neo owning Agent Smith :)... Ok, here it is explained:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32380&postorder=asc&start=15]Look at the bottom of the page, the post by natmc[/url]

[color=red]"McNugget" Formula: The greatest positive integer that cannot be obtained by $Am+Bn$, where $m$ and $n$ are given integers and $A$ and $B$ can be any integer, can be determined by $\bold{mn-m-n}$.[/color]

What's the formula for the sum of the first n triangular numbers?

[quote="chess64"]The $n$th tetrahedral number, which is the sum of the first $n$ triangular numbers, is $\frac{n(n+1)(n+2)}{6}$.[/quote]

If you watch the flashbars, over that <----- way, you will see this formula occasionally pop up, but for a different use.  I forget what it was though...

[quote="pkothari13"]Think of it like this.....

If you have three points $(x_1,y_1), (x_2,y_2),$ and $(x_3,y_3)$, the area of the triangle with those vertices is:
$\displaystyle\frac{1}{2}
\left|
\begin{array}{ccc}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1\end{array}
\right|.
$

Obviously, any negative answers should be made positive. If you don't know, then that function is called the "determinant" of a matrix.[/quote]

I think you can do it with any polygon too, you just have to list the verticies in order:
   $(x_1,y_1)$
           :downto:
  $(x_2,y_2)$
             :downto:
$(x_3,y_3)$
             :downto:
$(x_4,y_4)$
          :downto:
$(x_1,y_1)$

Multiply $x_1$ to  $y_2$ , $x_2$ to $x_3$ and so on..... by the arrows.
So you find the sum of these: $(x_1y_2)+(x_2y_3)+(x_3y_4)+(x_4y_1)$

And then you multiply the other side, which $y_1 * x_2$ , $y_2 * x_3$ , .....
So you find the sum of these: $(y_1x_2)+(y_2x_3)+(y_3x_4)+(y_4x_1)$

Finally it is $[(x_1y_2)+(x_2y_3)+(x_3y_4)+(x_4y_1)]$ $-$ $[(y_1x_2)+(y_2x_3)+(y_3x_4)+(y_4x_1)] = N$

If $N$ is negative, then make it positive, because it is the difference between the two sums. $Divide$ $N$ $by$ $2$ and get the area of the polygon

Whew glad that's over!

[quote="kchande"]The popup thing says the formula is n(n+1)(n+2)
                                                         -----------------
                                                                 6
for 1^2 + 2^2 + 3^2 + ........................................n^2
if that is what you mean[/quote]

No that's the sum of squares...the formula looks similar but is not the same.

[color=red]Volume of a frustrum with radii p and r and height h[/color]
$V=\displaystyle \frac{(p^2+pr+r^2)h\pi}{3}$
[color=red]
Area of intersection of two circles with radii p and r and distance between the centers d[/color]

$\displaystyle \frac{p+r+d}{2} \rightarrow s$

$\displaystyle \frac{2\sqrt{s(s-p)(s-r)(s-d)}}{d} \rightarrow h$

$2\sin^{-1}{{\displaystyle \frac{h}{p}}} \rightarrow \alpha$

$2\sin^{-1}{{\displaystyle \frac{h}{r}}} \rightarrow \beta$

$A=\displaystyle \frac{(\alpha{p^2}+\beta{r^2})\pi}{360}-\displaystyle \frac{p^2\sin{\alpha}+r^2\sin{\beta}}{2}$

The thing Xingster is talking about is called the "Surveyor's Formula." It can be written this way:

If the vertices are $(x_1,y_1), (x_2,y_2), ..., (x_n,y_n)$, then:

$2A = \displaystyle \left| \begin{array}{cc}x_1 & y_1  \\x_2 & y_2\end{array} \right| + \displaystyle \left| \begin{array}{cc}x_2 & y_2  \\x_3 & y_3\end{array} \right| +...\displaystyle \left| \begin{array}{cc}x_n & y_n  \\x_1 & y_1\end{array} \right|
$

Where $\displaystyle \left| \begin{array}{cc}a & b  \\c & d\end{array} \right| = ad-bc$.

Tell me if this doesn't make sense, and I'll try to clarify.

[quote="mysmartmouth"]
I am having trouble with matrices (I get an image of Agent Smith [b]every time[/b]).  Can someone please do an example with clear steps?  This looks like a formula worth learning.[/quote]

[img]http://img.photobucket.com/albums/v259/drspyder/idiot.jpg[/img]

[color=red]
xCy, where x > y

[(x)(x - 1)(x - 2)...(x - y + 1)]/[(1)(2)(3)...(y)]

xPy, where x>y
(x)(x - 1)(x - 2)...(x - y + 1)
[/color]

In other news, Comfort Eagle > all.

[quote="mysmartmouth"][quote="pkothari13"]Think of it like this.....

If you have three points $(x_1,y_1), (x_2,y_2),$ and $(x_3,y_3)$, the area of the triangle with those vertices is:
$\displaystyle\frac{1}{2}
\left|
\begin{array}{ccc}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1\end{array}
\right|.
$

Obviously, any negative answers should be made positive. If you don't know, then that function is called the "determinant" of a matrix.[/quote]

I am having trouble with matrices (I get an image of Agent Smith [b]every time[/b]).  Can someone please do an example with clear steps?  This looks like a formula worth learning.[/quote]if you guys really want to learn that polygon problem, i have it on a pdf, btw, it is called the surveyor's method, and sometimes the shoestring method

Finite differences:

This method is a little advanced but can be used to find a term of a sequence.

This method can be used on any sequence of the form 2,2,2,2,2 ; 1,2,3,4,5 ; 1,3,6,10,15 : basically any arithmetic-type sequence.

Ex. For a sequence 5,11,19,29:

Write out the first few terms: 5,11,19,29

Find the difference of consecutive terms. If the differences arent the same, continue doing finding differences until they are all the same.

[code]
5   11   19   29
  6     8   10
    2     2
[/code]

Then use this formula: [img]http://mathworld.wolfram.com/fimg1599.gif[/img]=[img]http://mathworld.wolfram.com/fimg1600.gif[/img], where $n$ is the number of the term you want to find, $a\sub{0}$ is the first term of the first row of the above sequence, $b\sub{0}$ is the first term of the second row, etc...

NOTE: Based on my experimentation, I have concluded that in most cases you would have to substitute $n-1$ for $n$ in the equation above. Don't know why.

This may look like a lot of work, but it's actually quite easy and can be very helpful when you have complex sequences.

[quote="NoSoupForYou"]Finite differences:...
NOTE: Based on my experimentation, I have concluded that in most cases you would have to substitute $n-1$ for $n$ in the equation above. Don't know why. ...[/quote]
I think finite differences are a very powerful tool for finding patterns.  Sometimes applicable to MATHCOUNTS, but frequently applicable in competitions just above MATHCOUNTS level.

The RHS of your formula looks correct.  You must count the terms of the sequence starting with 0.  In the sequence {5, 11, 19, 29, 41, 55}, 5 is the zeroeth term and 29 is the third term.  [$a_0$ is the zeroeth term of the sequence.  $b_0$ is the zeroeth term of the first difference.  $c_0$ is the zeroeth term of the second difference.  Etc.]

I don't teach that formula to my MATHCOUNTS students.  I think it is rote memorization and it might push something more important out of their heads.  Instead, I tell them to find a difference pattern.  If they are convinced they have found a pattern, they should assume the pattern continues.


The functions $f(n)$, $g(n)$ of natural number $n$ are difined  as follows.
$f(n)$ =the residue when n is divided by $7$.

$g(n)=3f\left(\sum_{k=1}^7 k^n\right)$

Choose a number which you would like,and find the value of $g(n)$.
This value of $g(n)$ is your point in this problem.

Heh, thats probably the coolest way I've seen an easy question asked before..

FLT gives n=6 pretty quickly. Or if you didn't notice that you could just simplify it for even n as odd gives 0, that works out quickly also..

After all,how many points could you get? :D

It is clear $f(n)\leq 6$, so max points $\leq 18$, and we can get $18$ if $n=6$, since FLT give $a^6=1\pmod{7}$ for all $a$, which is not divisible by 7, so $\sum_{k=1}^{7} k^6=1+1+1+1+1+1+0=6$ modulo 7.

Um, you did notice that I had solved it up the top :P


If $n$ is a positive integer which can be expressed in the form $n=a^{2}+b^{2}+c^{2}$, where $a, b, c$ are positive integers, prove that for each positive integer $k$, $n^{2k}$ can be expressed in the form $A^2 +B^2 +C^2$, where $A, B, C$ are positive integers.

That easily follows from Legendre theorem that only integers of the form $ 4^{\ell}(8k\plus{}7)$ cannot be represented as the sum of three squares.

I still requires a bit of work then.
But the problem is about showing it in a simpler way, I assume.