algebra polynomial logarithms conics quadratics calculus blogs

[quote="7h3.D3m0n.117"]it's ok. not really hard. you learn and review factorization of polynomials, logarithms, conic stuff, etc.[/quote]

I think he's saying that Algebra II isn't covered middle school. But many people do take it in middle school.

well, i can assume that if you are in a middle school class for algebra II, then most of the people there are probably used to being at the top of their class. If they hit one bump in the road, and everyone else gets it, they might say "what's going on? you're not better than me? are you?" I don't think it's all on your shoulders about low class morale. If half of the class gets something, then the minority will obviously feel kind of down. Eventually, they'll all be part of the minority. Even though it's a high honors class, they're used to flying first class through the world of math and when they don't, they automatically think "ok, this is too hard I give up". It can be hard after riding high to accept the fact that they're not numero uno in the math class. I don't think they're having a problem with you, maybe just with themselves.

Classroom Math is for basic Algebra I. The quadratic formula is probably the upper bound of the allowed difficulty here.

That's not fair you get to learn Algebra II in your middle school. I'm stucked with stuff I already know, Algebra I.

[quote="Ihatepie"]what grade are you in?

Are you in a genius school? because I am in 7th grade and am doing algebra now with the 8th graders. There are only about 4 other 7th graders in my class. I get 100s all the time and am at the top of my class and people call me a "genius" and the other 7th graders "extremely smart" There is noone in my school who is double advanced all though I might do algebra 2 next year[/quote]

Hey, I did that last year! Weird coincidence.

[quote="kstan013"]I took Algebra II in 6th grade.[/quote]

really? you mean your second algebra class right?

you must be really smart... did you do good in that class and if you did, did they beat you up because they got beaten by a 4th grader? :P :starwars:

[quote="kstan013"]they didn't beat me up! I got an A in there[/quote]

ok I have an 97 average in my first year of algebra right now and am in 7th grade and am at the top of my class but I would be pissed if a 4th grader came and beat me in my grades all though you wouldn't  :P

[quote="kstan013"]that would be a rather weird situation if I was on both ends of the spectrum :huh:[/quote]

I wonder what grade are you in now and what math?

[quote="kstan013"]I am in 8th grade and I am only taking these AoPS courses. I would have taken Calculus I this year but I need to take at least one class in the high school while I'm actually "in high school" :D[/quote]

I'm thinking of doing algebra 2 next year(I'm doing algebra 1 now)by doing AoPs courses. I thought I was really ahead of everyone but compared to you I guess I'm not...

Finally I find people like me... Algebra I by 4th grade.

In my school (it rocks) the 8th graders (in the "gifted program") do Honors Geometry.  Only one 8th grader does pre-calc.

[quote="7h3.D3m0n.117"]I'm in Algebra II and we are reviewing the FOIL method for multiplying binomials in class right now. Ever since we started "complicated" stuff in class the morale on the class has been pretty low, probably because of me. I always get the correct answer before they can start the problem (which I try to stall answering the question but I wait instead and put my hand up later).
For example, people complained that $(9x+2)^{2}$ was equal to  $81x^{2}+4$. When the teacher finally corrected all of them they said that it didn't make sense. (Which of course I knew the answer was $81x^{2}+36x+4$.  :lol: )
Today, when my teacher was explaining what FOIL was again, I got bored and started my math homework. I finished before she completed her lecture...Of course the person next to me watched and told like her friends and they were shocked...Eventually within the next 10 minutes, news circulated around the class that I already completed the homework.
Besides that, I tutor people in my math class during class if they don't get something. And if they make a mistake they usually end up saying "I don't get it." My only problem is that I can't raise their morale and help them at the same time  :).

Is there a good way to raise morale to people who just don't get it?

P.S. I was even considering to pass Algebra II right then and there but I wasn't familiar with some of the stuff they covered (near the end).[/quote]
Now here is some advice:
What you are experiencing is not going to go away.
It will get worse as time goes on.
It is called "aopserism".
Here are some symptoms:
You have always had a talent for math
School math is seeming more trivial every day
During math class you could get an extra hour of sleep and be caught up
You think that real math lies in competitions, and you have a thirst to learn it
Other people wonder how on earth you are so good at math.
Like I said, it doesn't go away, but it gets worse...

yea i pretty much have those symptoms lol sept im afraid of sleeping because i worry that i will be in detention lol

[quote="bpms"]It is called "aopserism".
Here are some symptoms:
You have always had a talent for math
School math is seeming more trivial every day
During math class you could get an extra hour of sleep and be caught up
You think that real math lies in competitions, and you have a thirst to learn it
Other people wonder how on earth you are so good at math.
Like I said, it doesn't go away, but it gets worse...[/quote]

:o That's exactly how I feel most of the time in Precalc...

"You would have skipped your class (that is, advanced early, not committed truancy) if you only knew how". I would have wanted to skip Algebra II except that I had a great teacher. And now this year is all "review".

"The seniors in your class come to YOU for help"

Algebra I by fourth grade :rotfl:

[quote="nutz_for2.718281828"]suggestion:

The only reason for not getting it is because they think that they can't.

Teach them that they CAN do it. Don't just show the answer, guide them to the solution.[/quote]
That's not true, some people just lack the ability...

some lack the ability but most do try their hardest...its just that when they fail (whether by slight miscalculation or by serious error) they don't bother trying again

public schools don't teach them how to do math correctly

this is probably why americans are, in general, worse at math than foreigners -- because most foreign parents teach their kids math themselves when they are very young, and so they learn how to do it correctly -- not the american way

[url=http://sixthform.info/maths/index.php?m=200610#136]interesting blog entry[/url]

[quote]AP - Kids who are turned off by math often say they don't enjoy it, they aren't good at it and they see little point in it. Who knew that could be a formula for success?

The nations with the best scores have the least happy, least confident math students, says a study by the Brookings Institution's Brown Center on Education Policy.

Countries reporting higher levels of enjoyment and confidence among math students don't do as well in the subject, the study suggests. The results for the United States hover around the middle of the pack, both in terms of enjoyment and in test scores.

In essence, happiness is overrated, study author Tom Loveless says.

"We might want to focus on the math that kids are learning and just be a little less obsessed with the fact that they have to enjoy every minute of it," said Loveless, who directs the Brown Center and serves on a presidential advisory panel on math.

"The implication is not 'Let's go make kids unhappy,' " he said. "It's 'Let's give kids better signals as to how they're performing, relative to the rest of the world.' "

Other countries do better than the United States because they seem to expect more from students, he said. That could also explain why high performers in other nations express less confidence and enjoyment in math. They consider their peer group to be star achievers.

Even efforts to make math relevant may be irrelevant, says the study, released today.

Nations that try to teach math in terms of daily life have the lowest test scores.

All this is not easy to compute. Math teachers typically don't avoid enjoyment, confidence and relevance in their math lessons. They strive for those things.

Loveless says pleasing kids has come at the expense of mastering skills.

His findings come from the 2003 Trends in International Mathematics and Science Study, a test of fourth-graders and eighth-graders across the globe. Along with answering math questions, students were asked whether they enjoyed math and whether they usually did well in it.

The eighth-grade results reflected a common pattern: The 10 nations whose students enjoyed math the most all scored below average. The bottom 10 nations on the enjoyment scale - including Japan, Hong Kong and the Netherlands - all excelled.[/quote]

Entry in MichaelFaraday's blog, I believe, which I later quoted in mine. It kind of made me mad (at the current system)...

that's rather interesting. Perhaps your results might not be true? Less happy people and people who don't try won't beat people who work their behinees off, unless they are miraculously smart.

speeding up was never good for me.

I took AP Calculus BC when I was a sophmore, (that class barely had any juniors and 'smart' seniors, (if not.. u go to AB)

Anyways I got an A in the class, but I was somewhat too confident that I'm having a really hard time doing Linear Algebra right now with college students.  I rather would have slowed down because some of the concepts were not even close to what I learn in other classes.

If you want to speed up, speed up in everything , not just in math or i gets really hard later on and everything that seemed easy will get difficult.

By the way, I've seen plenty of genius Algebra students. That's because Algebra is extremely basic in its concept. Trust me that just because you are good in Algebra doesn't mean you will succeed in Calculus or other advanced topics because I've seen a plenty of 'ex-Geniuses' that fell behind.

"Ex-geniuses" don't fall behind because they don't understand the class. It seems to me that it is because they get lazier or something.

You know, frankly, I think this topic belongs in the Round Table. Any moderators agree?

Well, in the binomial class, show them that they are wrong by putting in a convinent number for x. In your example, 2.

Also, try to teach them how to evaluate multiplications of polynomials. Also, give them some practice problems. At least it works for me. :) I got past pre-calc in 7th grade!(I'm still in 7th grade)

Talk to the teacher about it.  Then do your own math during the class period.

I find the classroom math in my school to be trivial as well.  My teacher once told me, "I'll find something to make up."  LOL

But I am really respected in my math class because I am so good at it.  I have no problem helping other students.  Also, when my teacher says that there was one perfect score in the class, everyone turns to me.

[quote="h_s_potter2002"][quote="kstan013"]yep. I took Algebra I in 4th grade w/8th graders :help: It was kind of nerve-racking[/quote]

I wish I could have done this.

Actually, there was a guy who did this in my school district, who graduated last year.  But his mom was on the school board.  I think that had sometihng to do with it/[/quote]
Well, my mom is not on the school board, my dad is not on the school board, my grandpa's roommate's 3rd cousin is not on the school board either :D

Your time in school will pass quickly.

You don't feel that you are fitting in, because you are different.  Welcome to the world, everyone is different in some way.  You should find some way to learn more.  What if you can find extending problems in the same area as is being covered by your class?  Your teacher should be able to help.  If not, seek something else.  Get a textbook that is more advanced and don't just waste your time in the class.

As far as the other students are concerned, see if your teacher will allow you to help teach others in your class that are having difficulty.  I found throught my career (including college) that I learned things best when teaching them to others.  You need to be very careful to be gentle and not condensending when talking to others in your class who are not as strong as you are.  If you find the right college, you will find one where you are the one being streched and not the one leading the class.  Prepare for this emotionally, learn how to "dig yourself out of a hole" and how to "run faster than you are currently able" with concepts.  Don't worry you won't run out of material to study.

If this doesn't work for yourself, I suggest going to the head of the department or someone else who can help and get an independent study on something not covered in your classes or future classes.  Something like numerical methods of solving equations, by hand or computer.  This isn't really taught in school, but is the method used to solve real problems in the real world.

Find alternative ways to solve the problems in your classes.  What happens if the methods you currently use don't work for some reason?  Find fast ways to check work, this can be useful on college tests when it isn't baby math and you have a couple of minutes extra on a problem you can't solve.  If you are familiar with numerical methods and can eliminate three of four answers in the time left, you may be able to improve your score.  Think creatively and find the correct answer, not just an application of what I suggested off the top of my head.  Bottom line -- don't ever waste time you could be learning something.

Why are you raising your hand to answer the problem to discourage them anyway? If you can get the correct answer instantly, then it is not much of a challenge for you and it is for other people so when you do what you do, you basically take the point away for them to do the problem as you revealed the answer and you frustrate them which will frustrate them with math which will lead to them giving up which will lead to bad grades which will in turn influence other classes which will make them not go to a good college which will ruin their life.

Way to ruin lives.

[quote="7h3.D3m0n.117"]I'm in Algebra II and we are reviewing the FOIL method for multiplying binomials in class right now. Ever since we started "complicated" stuff in class the morale on the class has been pretty low, probably because of me. I always get the correct answer before they can start the problem (which I try to stall answering the question but I wait instead and put my hand up later).
For example, people complained that $(9x+2)^{2}$ was equal to  $81x^{2}+4$. When the teacher finally corrected all of them they said that it didn't make sense. (Which of course I knew the answer was $81x^{2}+36x+4$.  :lol: )
Today, when my teacher was explaining what FOIL was again, I got bored and started my math homework. I finished before she completed her lecture...Of course the person next to me watched and told like her friends and they were shocked...Eventually within the next 10 minutes, news circulated around the class that I already completed the homework.
Besides that, I tutor people in my math class during class if they don't get something. And if they make a mistake they usually end up saying "I don't get it." My only problem is that I can't raise their morale and help them at the same time  :).

Is there a good way to raise morale to people who just don't get it?

P.S. I was even considering to pass Algebra II right then and there but I wasn't familiar with some of the stuff they covered (near the end).[/quote]

Judging by the rather conceited tone of this post, I am not too surprised that your classmates' "morale" is low. Seriously, if you know the answer to every question why don't you just step aside and let others attempt them? It seems like this post is just another way of bragging about how much better you are than your classmates. So here's an idea: stop trying to show off and start trying to genuinely help.


ΠΡΟΒΛΗΜΑ

 Αν  ΒΔ ,ΓΕ  είναι δύο ύψη τριγώνου και Ν είναι σημείο της πλευράς ΒΓ ώστε οι γωνίες ΒΝΕ και ΓΝΔ να είναι ίσες, τότε  η ΑΝ είναι κάθετη στη ΒΓ .

[color=blue][b] Σχόλια[/b][/color]

 Μόλις τώρα κοίταζα το ερώτημα αυτό και βρήκα μια έμμεση απάντηση. Μήπως βλέπετε καμιά ευθεία απόδειξη(όχι με άτοπο ) , διότι δεν βλέπω τίποτα αυτή τη στιγμή;
 Το ερώτημα αυτό έχει να κάνει με την  τελευταία άσκηση που έβαλα. Στο mathlinks υπάρχει και μια λύση από τον  Yetti, την οποία τώρα μελετώ(βάλτε στην αναζήτηση delegat και θα σας οδηγήσει στο πρόβλημα). Επό εκεί προέκυψε και το παρόν ερώτημα.
 Μπάμπης
ΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧ

Έστω ότι η ΕΝ τέμνει τον κύκλο του BEDC στο σημειο Q . Τότε $ \angle{ENB}\equal{}\angle{BNQ}$ άρα $ \angle{QNO}\equal{}\angle{ENO}$ και αφού OQ=OE οι περιγεγραμμένοι κύκλοι των ENO , NQO είναι ίσοι . Και επιπλέον ΝΟ κοινή πλευρά άρα $ \angle{NQO}\equal{}\angle{NEO}$  , το όποίο σημαίνει ότι τα τρίγωνα ENO , NQO είναι ίσα άρα το Β είναι μέσο του τόξου EQ , το οποίο δείχνει ότι η DN είναι διχοτόμος της $ \angle{EDN}$ άρα αν η ED τέμνει τη BC στο Η , τότε Η,Ν συζυγή αρμονικά προς τα Β,C άρα το Ν είναι το ίχνος του ύψους ...

[quote="silouan"]Έστω ότι η ΕΝ τέμνει τον κύκλο του BEDC στο σημειο Q . Τότε $ \angle{ENB} \equal{} \angle{BNQ}$ άρα $ \angle{QNO} \equal{} \angle{ENO}$ και αφού OQ=OE οι περιγεγραμμένοι κύκλοι των ENO , NQO είναι ίσοι . Και επιπλέον ΝΟ κοινή πλευρά άρα $ \angle{NQO} \equal{} \angle{NEO}$  , το όποίο σημαίνει ότι τα τρίγωνα ENO , NQO είναι ίσα άρα το Β είναι μέσο του τόξου EQ , το οποίο δείχνει ότι η DN είναι διχοτόμος της $ \angle{EDN}$ άρα αν η ED τέμνει τη BC στο Η , τότε Η,Ν συζυγή αρμονικά προς τα Β,C άρα το Ν είναι το ίχνος του ύψους ...[/quote]

Σιλουανέ,
 ωραία συλλογιστική ! Δεν φανταζομουν ότι θέλει τόσο πολύ ... θεωρία(μέχρι αρμονικότητα) για να αποδειχθεί. Και αυτο γιατί με άτοπο βγαίνει σε μια γραμμή.
 Σε ευχαριστώ πολύ και θα κρατήσω τη λύση σου στο αρχείο για χρήση!

 Καλό βράδυ - Μπάμπης


Να δείξετε ότι δεν υπάρχει γνησίως αύξουσα συνάρτηση $ f: N \to N$ έτσι ώστε:
$ a)$ $ f(2)\equal{}3$
$ b)$ $ f(mn)\equal{}f(m)f(n)$

Συζητήθηκε εδώ και δόθηκε μάλιστα και πολύ ωραία γενίκευση απ' τον Ηλία!
[url]http://mathematica.gr/forum/viewtopic.php?f=50&t=3066[/url]


Hi,
Did anybody here solved the Character Recognition problem in the USACO training pages?
If so, can anyone help me.. I'm stuck in this problem..

Thanks in advance.

can a moderator please move this thread to the Computer Science and Informatics forum..

thanks,


Not using calculators, find the exact measure of acute angle $x$ satisfying:
$\cos x=\frac{1}{\sqrt{1+(\sqrt{6}+\sqrt{2}-\sqrt{3}-2)^2}}$


number theory
greatest common divisor
number theory unsolved

Suppose $ x, y, z$ are positive integers satisfying the equation

$ \frac{1}{x} \minus{} \frac{1}{y} \equal{} \frac{1}{z}$,

and let $ h$ be the highest common factor of $ x, y, z$.

Prove that $ hxyz$ is a perfect square.

Prove also that $ h(y\minus{}x)$ is a perfect square.

Nice!
Let \[ x \equal{} ha\,\,,\,\,\,y \equal{} hb\,\,\,,\,z \equal{} hc\]
Your equality is equivalent to,
\[ (b \minus{} a)(b \plus{} c) \equal{} {b^2}\]
but \[ \gcd (b \minus{} a,b \plus{} c) \equal{} 1\]
so for a certain integer d we have,
\[ b \minus{} a \equal{} {d^2}\]
and so,
\[ h(y \minus{} x) \equal{} {h^2}(b \minus{} a) \equal{} {(hd)^2}\]
and,
\[ hxyz \equal{} {z^2}h(y \minus{} x) \equal{} {(zhd)^2}\]

[quote="Evariste-Galois"]
but
\[ \gcd (b \minus{} a,b \plus{} c) \equal{} 1\]

[/quote]
Not clearly! You only have (a,b,c)=1, you can not get gcd (b - a,b + c) = 1.

You can get it, in fact,
(b-a)(b+c)=b²
Let gcd(b-a,b+c)=d
d/b-a and d/b+c  so d/(b-a)(b+c)=b²
so d/a² d/b² d/c²
so d/(gcd(a,b,c)²=1  so d=1.
I thought that it was clear!
 :wink:


Prove that if, x,y and z are positive numbers with sum 1, then $ \frac{x^2\plus{}3xy}{x\plus{}y}\plus{}\frac{y^2\plus{}3yz}{y\plus{}z}\plus{} \frac{z^2\plus{}3zx}{z\plus{}x} \leq 2$    :huh:
 :rotfl:

[quote="Pirkuliyev Rovsen"]Prove that if, x,y and z are positive numbers with sum 1, then $ \frac {x^2 + 3xy}{x + y} + \frac {y^2 + 3yz}{y + z} + \frac {z^2 + 3zx}{z + x} \leq 2$    :huh:
 :rotfl:[/quote]

It's very easy with AM-GM and Cauchy.As
\[ \sum{\frac{xy}{x+y}}\leq\sum{\frac{\left(\frac{x+y}{2}\right)^2}{x+y}}=\frac{1}{4}\sum{(x+y)}=\frac{1}{2}(x+y+z)\leq \sum{\frac{y^2}{x+y}}\]
Thus $ \sum{\frac{y^2-xy}{x+y}\geq 0 \Longrightarrow \sum{\frac{y^2-xy}{x+y}=\sum{\left[(x+y)-\frac{x^2+3xy}{x+y}\right]}\geq 0}}$
which is equalent to $ \sum{\frac{x^2+3xy}{x+y}}\leq 2$

[quote="Pirkuliyev Rovsen"]Prove that if, x,y and z are positive numbers with sum 1, then $ \frac {x^2 \plus{} 3xy}{x \plus{} y} \plus{} \frac {y^2 \plus{} 3yz}{y \plus{} z} \plus{} \frac {z^2 \plus{} 3zx}{z \plus{} x} \leq 2$    :huh:
 :rotfl:[/quote]

use this identity:
$ \frac {x^2}{x \plus{} y} \plus{} \frac {y^2}{y \plus{} z} \plus{} \frac {z^2}{z \plus{} x} \equal{} \frac {y^2}{x \plus{} y} \plus{} \frac {z^2}{y \plus{} z} \plus{} \frac {x^2}{z \plus{} x}$
we get
$ \frac {x^2 \plus{} 3xy}{x \plus{} y} \plus{} \frac {y^2 \plus{} 3yz}{y \plus{} z} \plus{} \frac {z^2 \plus{} 3zx}{z \plus{} x} \equal{}$  $ \frac {x^2 \plus{} 6xy \plus{} y^2}{2(x \plus{} y)} \plus{} \frac {y^2 \plus{} 6yz \plus{} z^2}{2(y \plus{} z)} \plus{} \frac {z^2 \plus{} 6zx \plus{} x^2}{2(z \plus{} x)}$
so just need to prove
$ \sum{\frac {x^2 \plus{} 6xy \plus{} y^2}{2(x \plus{} y)}}\le 2(x \plus{} y \plus{} z)$
that's equivalent to
$ \sum{\frac {(x \minus{} y)^2}{2 (x \plus{} y)}}\ge 0$
obvious true.


One-fourth of one-third of a number is one-half. What is the
number?

$ \frac {1} {4}$ of $ \frac {1} {3}$ would be $ \frac {1} {12}$

so, $ \frac {1} {12}$ of $ x$ is $ \frac {1} {2}$

therefore, we get the equation

$ \frac {x} {12} \equal{} \frac {1} {2}$

$ 2x \equal{} 12$

$ x \equal{} 6$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $H$ be a subgroup of additive group of rational numbers with the property that $\frac{1}{x} \in H$ for every nonzero element $x$ of $H$.

Prove that $H=0$ or $Q$

uhm...
this could be given as a definition of rationals:
well, if $H =\{0\}$ it's just obvious.
now, say $H \neq \{0\}$. if $x \in H \Rightarrow x^{-1}\in H$ then $H$ is a field (since all operations are working in the right way), then it contains an isomorphic copy of $\mathbb{Q}$. but since $H \subset \mathbb{Q}$ you have your thesis.

otherwise, working "by hands", you can say: let $\frac{p}{q} \in H$, with $p,q \neq 0$ and coprime. now, since $H$ is additive subgroup, $\frac{kp}{q} \in H$ for all $k\in \mathbb{Z}$. but this implies $\frac{q}{kp} \in H \Rightarrow \frac{hq}{kp} \in H$, and you can "produce" all rationals this way.

Sorry perhaps silly question, but in the first argument, you have an additive group structure, and multiplicative inversion, but why multiplication? You immediately say it is a field?

uhm.. you're right...
you don't have multiplication, for elements of $H$, of course ;)
but you have "multiplication" for elements of $\mathbb{Z}$, even if it's "hidden" in iteration of sums and differences...
i don't know if i was able to explain, sorry :p

(you're totally right, anyway ;))

The key fact: An additive subgroup is a $\mathbb{Z}$-submodule.

If we have the reciprocals of the integers, we have everything, since we have $k\cdot\frac1n$ for each pair of integers $k$ and $n$.
If we have the integers, we have their reciprocals.
If we have 1, we have the integers.
If we have some integer $n$, we have $\frac1n$ and we have 1.
If we have some nonzero rational number $\frac nm$, we have $n=m\cdot\frac nm$, and we are done.

if $\frac pq\in H$ then $p=\frac{pq}q\in H$. So $\frac1p\in H$. So $1\in H$. Now for each $m,n\in\mathbb Z,m,n\in H$. So $m,\frac1n\in H$. Now if $m=kn+r$ then $\frac mn=k+\frac rn\in H$


i realy want to purchase AOPS vol1 but how since i'm not in the U.S.i live in africa  and i will like to take part and win contest but the books are not good and so i need directives cause i want to purchase the AOPS book

Go to the [url=http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php]bookstore[/url] and place an order.


ceiling function
geometry
perimeter
combinatorics unsolved
combinatorics

[quote="OHO"]What is the minimun number of dominoes which can cover the chessboard so that no more domino can be added?[/quote]

Haha I misread the question to be an $m \times n$ board :P, but found I couldn't do it, and could only show a lower bound which asymptotically tends to the answer, which is $\left\lceil{\frac{mn}{3}}\right\rceil$.

Proof:

Define an exterior length/width as a length/width of a domino that touches a side of the board
Define an interior length/width as a length/width of a domino that does not touch the sides of the board

Consider any solution with $x$ dominoes
There are $m(n-1)+n(m-1)$ possible domino positions
Each domino eliminates at most $7$ possible positions (inclusive of its own)
However each interior length must be adjacent to at least $1$ other domino, otherwise an extra domino can be placed beside the original

Consider the possible positions eliminated adjacent to each side of the dominoes
For each interior length at least $1$ of the possible positions are double-counted
For each interior width at least $0$ of the possible positions are double-counted
For each exterior length/width there are $0$ possible positions

Let $i$ be the number of exterior lengths and $j$ be the number of exterior widths
We have $3i+2j \geq 2(m+n)$ otherwise a domino can be placed somewhere along the perimeter of the board

Thus the total number of positions eliminated by all the dominoes is at most
$x+\left(2(2x-i)+(2x-j)\right)-\left\lceil{\frac{1}{2}(2x-i)}\right\rceil$
$= 6x-2i-j+\left\lceil{\frac{i}{2}}\right\rceil$
$\leq 6x-\frac{3}{2}i-j+\frac{1}{2}$
$\leq 6x-(m+n)+\frac{1}{2}$

By our original assumption,
$6x-(m+n)+\frac{1}{2}\geq m(n-1)+n(m-1)$
$\Rightarrow x \geq \left\lceil{\frac{2mn-\frac{1}{2}}{6}}\right\rceil = \left\lceil{\frac{mn}{3}}\right\rceil$

In the case of an $8 \times 8$ board, $x \geq \left\lceil{\frac{64}{3}}\right\rceil = 22$, so we need at least 22 dominoes. A solution using 22 dominoes is not hard to find:

[code]
 -- -- |
-- -- ||
 -- --|
-- -- --
 -- -- |
-- -- ||
 -- --|
-- -- --
[/code]

[quote="weiquan"]Haha I misread the question to be an $m \times n$ board :P, but found I couldn't do it, and could only show a lower bound which asymptotically tends to the answer, which is $\left\lceil{\frac{mn}{3}}\right\rceil$.[/quote]

Just a small note: $mn>1$ otherwise the statement $3i+2j>2(m+n)$ is incorrect because there's not enough space for even $1$ domino!

But I haven't been able to find a general construction to satisfy my lower bound, except for $m,n$ where $3|(mn)$. The smallest example where I cannot even find one construction for is a $10 \times 10$ board. $\left\lceil{\frac{100}{3}}\right\rceil = 34$ but the best constructions I can find all require $35$ dominoes. Does anyone have any ideas? Maybe my lower bound is not sharp for the general case? :maybe:

Edit: I got a construction for $10 \times 10$, but it seems hard to generalize...

[quote="weiquan"][quote="weiquan"]Haha I misread the question to be an $m \times n$ board :P, but found I couldn't do it, and could only show a lower bound which asymptotically tends to the answer, which is $\left\lceil{\frac{mn}{3}}\right\rceil$.[/quote]

But I haven't been able to find a general construction to satisfy my lower bound, except for $m,n$ where $3|(mn)$.

Edit: I got a construction for $10 \times 10$, but it seems hard to generalize...[/quote]

Hmm... I've been trying $(m,n)=(10,22)$ and I could not find a construction that uses $\left\lceil{\frac{220}{3}}\right\rceil = 74$ dominoes, while there are many using $75$ dominoes. Can anyone find one, or prove that it is impossible? If it may help, the only construction I have for $10 \times 10$ is:

[code]
 -- | | |
-- |||||||
 --| | | |
-- -- |--
 -- --| --
-- |-- --
 --| -- --
| | | |--
||||||| --
 | | | --
[/code]

Oh no I made a small mistake, albeit one with important consequences.

The lower bound is almost always correct :P , but the proof is wrong. In particular, the lower bound is correct for all $m,n>1$, and should be $\left\lceil{\frac{mn-1}{3}}\right\rceil$ for $m=1$ or $n=1$.
The corrected proof for $m,n>1$ is as follows:

Define an exterior length/width as a length/width of a domino that touches a side of the board
Define an interior length/width as a length/width of a domino that does not touch the sides of the board

Consider any solution with $x$ dominoes
There are $m(n-1)+n(m-1)$ possible domino positions
Each domino eliminates at most $7$ possible positions (inclusive of its own)
However each interior length must be adjacent to at least $1$ other domino, otherwise an extra domino can be placed beside the original

Consider the possible positions eliminated adjacent to each side of the dominoes
For each interior length at least $1$ of the possible positions are double-counted
For each interior width at least $0$ of the possible positions are double-counted
For each exterior length/width there are $0$ possible positions

Let $i$ be the number of exterior lengths and $j$ be the number of exterior widths
We have $3i+2j+4 \geq 2(m+n)$ otherwise a domino can be placed somewhere along the perimeter of the board, with the exceptions are the $4$ corners

Thus the total number of positions eliminated by all the dominoes is at most
$x+\left(2(2x-i)+(2x-j)\right)-\frac{1}{2}(2x-i)$
$= 6x-\frac{3}{2}i-j$
$\leq 6x-(m+n-2)$

By our original assumption,
$6x-(m+n-2) \geq m(n-1)+n(m-1)$
$\Rightarrow x \geq \left\lceil{\frac{2mn-2}{6}}\right\rceil = \left\lceil{\frac{mn-1}{3}}\right\rceil$

But notice that for equality to hold,
1)	the $4$ corners of the board have to be empty
2)	the interior lengths are adjacent to exactly one other domino
3)	the interior widths are not adjacent to any other domino

If equality can hold,
Consider the configuration near the top-left corner
Case 1:
[code]
 --
--
[/code]
	This forces
[code]
 -- --     --
-- -- ... --
 -- --     --
[/code]
	and the top-right corner is not empty $\Rightarrow\Leftarrow$
Case 2:
[code]
 --
|
|
[/code]
	This forces
[code]
 -- --     --
| -- -- ... --
|| -- --     --
 |
[/code]
	which is simply impossible

Therefore equality cannot hold, thus $x \geq \left\lceil{\frac{mn}{3}}\right\rceil$
For $m=1$ or $n=1$ it is obvious that the lower bound on $x$ is attained in the following construction
[code]
 -- -- ... --
[/code]

But I still don't know of a construction for $(m,n)=(10,22)$ with $74$ dominoes, and I suspect that it doesn't exist. Can anyone prove or disprove that the lower bound I obtained is attainable for the general case?


Let $d(n)$ be the number of divisors n has. find all n such that $d(n)=\frac{n}{4}$

How can $n$ be smaller than its number of divisors?  :?


calculus
algebra
polynomial
derivative
function
integration
calculus computations

Can anyone help with these problems? Any help would be welcomed.

A particle moves along the x-axis so that its velocity is given by [i]v(t) = 3t^2 - 2t - 1[/i] at any time [i]t is less than or equal to 0[/i]. The position [i]x(t) is 5 for t =2[/i].

A. Write a polynomial expression for the position of the particle at any time [i]t is less than or equal to 0[/i].

B. For what values of [i]t, 0 is less than or equal to t less than or equal to 3[/i], is the particle's instantaneous velocity the same as its average velocity, on the interval [0,3]?

C. Find the total distance traveled by the particle from time [i]t = 0[/i] until time [i]t = 3[/i].

Thanks in advance

I'll give you some hints to do these

A.  Remember the velocity is the derivative of the position, and you have been given the velocity funciton and the position at ond point.

So, $x(t) = \int{v(t) dt} + C$, and you need to solve for C using the fact that $x(2) = 5$.

B. Remember that the average velocity is $\frac{x_f - x_0}{t_f - t_0} = \frac{x_f - x_0}3$.  You  now need to find the final and starting position for $0<t<3$.  Then set this equal to the equation for velocity and solve for $t$.

C. Since the total distance traveled is the integral of the velocity, all you need to calculate is,

$D = \int_{t_0}^{t_f}{v(t) dt}$

You can IM me if you still need help (the links are below).  I'll probably be online, but I might not be.


abstract algebra
search
Ring Theory
superior algebra
superior algebra unsolved

$ A$ is a commutative Noetherian ring.　$ X$ is a indeterminate element on $ A$.

Does such an $ A$ exist that satisfies $ A \cong A[X]$?

Sorry.

I wanted to say "Does A that is a commutative Noetherian ring and fills A ≅ A[X] exits ?".

Language notes on the original post:
- "commutative" has a typo in the first line
- It's "indeterminate". "Indeterminant" is not a word.
- That last sentence is nonsense. A sentence that probably means what you want: "Does such an $ A$ exist that satisfies $ A\cong A[X]$?" That's still awkward, so I would have instead rewritten the whole post as "Does there exist a commutative Noetherian ring $ A$ such that $ A$ is isomorphic to $ A[X]$, where $ X$ is an indeterminate?" The last phrase might not even be necessary, since it's so widely accepted as a standard.

Suppose $ A$ is a commutative ring such that $ A[X]$ is isomorphic to $ A$; let $ f$ be this isomorphism, mapping $ A[X]$ to $ A$. We construct an infinite ascending chain of ideals in $ A[X]$ as follows:
Let $ I\equal{}I_1$ be the ideal generated by $ X$. The quotient $ A[X]/I_1$ is naturally isomorphic to $ A$, so we can tweak $ f$ to an isomorphism $ f_1$ mapping $ A[X]$ to $ A[X]/I_1$. Now, let $ I_2\equal{}f_1(I)\plus{}I_1$; this is an ideal in $ A/I_1$, and again $ (A[X]/I_1)/I_2$ is isomorphic to $ A$. Repeat this process indefinitely: $ f_n$ is an isomorphism from $ A[X]$ to $ A[X]/I_n$ and $ I_{n\plus{}1}\equal{}f_n(I)\plus{}I_n$. These $ I_n$ are an infinite ascending chain.

By the contrapositive, if $ A$ is Noetherian, $ A[X]$ is not isomorphic to $ A$.

I know you love correcting people jmerry, but "indeterminant" is certainly a word. :)  It is usually used as an adjective but can also mean "indeterminate".

Thank you for answering my poor English. I corrected the article.

[quote="LydianRain"] but "indeterminant" is certainly a word.[/quote]
Based on the top few results of a search, it exists solely as a very common mistaken version of "indeterminate", with no other meaning. I'll stick with the one that the Firefox spellchecker recognizes.

[quote="jmerry"]Now, let $ I_2 \equal{} f_1(I) \plus{} I_1$; this is an ideal in $ A/I_1$[/quote]
However, $ I_1$ is an ideal in $ A[X]$. Therefore, $ A/I_1$ is not defined.

He clearly means it is an ideal in $ A[x]/I_1$, but here it doesn't really matter because we are presupposing an isomorphism between $ A$ and $ A[x]$.

Yes, that's what I meant. I used $ A[x]$ in the next phrase, after all.


Denote by $Q(n)$ the sum of the digits of $n.$ Determine all $n$ such that \[n + Q(n) + Q(Q(n)) = 2004.\]

there is not a lot of possibilities isn't it (because n<2004 Q(n)<=1+9+9+9 ...)

[hide="solution"]if n=9a+x, then q(n)=9b+x and q(q(n))=9c+x. a>b>c. 9(a+b+c)+3x=2004. 3(a+b+c)+x=668. x is a positive integer less than 9. since 668=2(mod 3), x must be 2 (mod 3). Guess and check from there[/hide]

[hide]
Let n = 1000a+100b+10c+d where a,b,c,d are from 0 to 9 inclusive.

1000a+100b+10c+d+a+b+c+d+Q(Q(n))=2004

Since a+b+c+d = 10e+f (it will be a two digit number)
1001a+101b+11c+2d+e+f=2004 where a,b,c,d,e,f are digits. There, we have a system of equations.
To eliminate d:
999a+99b+9c+2(a+b+c+d)+e+f=2004
999a+99b+9c+20e+2f+e+f=2004
999a+99b+9c+21e+3f=2004
or
333a+33b+3c+7e+f=668
but 7e+f=a+b+c+d-3e
334a+34b+4c+d-3e=668
but if a=2, we get 2000 and it works... 2000+2+2=2004 (b=c=d=e=0)
Letting a = 0
34b+4c+d-3e=668
The maximum value of b,c,d=9 which would mean 9+9+9=27 so e=2
34(9)+4(9)+9-3(2)=345... therefore, a cannot equal 0.
So if a = 1 we get
34b+4c+d-3e=334
If c = d = 9, we get
4(9)+9-3e=334-34b
45-3e=334-34b
-3e=289-34b
34b-3e=289 ... if b>2, e=2 (if b<2, e<0 so it cannot work)
34b-6=289
34b=295
b=8.676470588... what does this number mean? It's the smallest b can get assuming c and d are as large as possible. Since b is a digit, b > 8.. therefore b=9

34(9)+4c+d-3e=334
306+4c+d-3e=334
4c+d-3e=28
4c+d=28+3e
e must be at least 1 since the digits currently total 10 (a=1,b=9)
if c+d>9, then e=2

Case 1:
e=2
we get 4c+d=34 where c+d>9
Solutions in this case would be {8,2}, {7,6}

Case 2:
e=1
we get 4c+d=31 where c+d<10
There are no solutions.

Therefore, the only solutions to this problem are 2000, 1982 and 1976.[/hide]

Hope I did that right..


The hardest problem from the Polish Olympiad this year - 3rd stage (no-one has done in regular competition time):

Prove that in every convex ploygon with area $1$ we can find a convex hexagon with area at least $\frac{3}{4}$.

Enjoy.

I believe this is from ISL 2004. It's somewhere on the forum, in a thread started by Valentin and where the problem was solved by Myth, I think.

And Rafal, like you, I am not too big of an inequality fan either, but I think a lot of Mathlinkers would disagree (because the inequality section has the most number of threads and posts!  :idea: ) :D

I don't want to make a silly off-topic here, I just think that inequalities are less interesting to other problems because of the fact that a great minority of solutions include applying inequalities which you know and those without you couldn't handle the problem (from inequality XXX + inequality YYY we have the given one...). The fact that the inequality section has the biggest number of posts is rather a consequence of the fact that many olympiad begginers start their training from inequalities (I did so), because it's the most simple thing (in the begginig).

[Edit] Anyway...thanks for the link ;)


algebra polynomial induction calculus derivative function inequalities

Show that every polynomial with leading coefficient 1, where all coefficients except for the leading one are negative or zero (all cannot be zero), has exactly one positive root of multiplicity one.

I've thought some about it but Im not satisfied (I was thinking something like induction over the degree, which implies that the derivative also has exactly one postive root and so on). My thoughts where something like:

Assume it holds for every polynomial of degree $ n$. Now consider a polynomial of degree $ n\plus{}1$. If the polynomial has no constant coefficient, i.e. a root in $ x\equal{}0$, then everything follows from the induction assumption since multiplying with $ x$ doesn't change the positive roots. Otherwise, $ f(0)$ must be negative. From the induction assumption it follows that the derivative only has one positive root. We know that $ f(0)$ is negative, and that $ f(x)$ goes to infinity as $ x$ goes to infinity, so we have at least one postive root. Since the derivative has only one positive zero, we can not have two roots. The only thing is, if the derivative is positive in $ x\equal{}0$, how do I show that the derivative is not 0 in the root of $ f(x)$ (inflection points), which would voilate the multiplicity of the root. I dont really like this since it's pretty geometric and heuristic, things you can see by "looking" at the graph and so on.. would like a nicer proof.

This is immediate from Descartes' Rule of Signs - [url]http://en.wikipedia.org/wiki/Descartes%27_rule_of_signs[/url]

Note that the derivative doesn't necessarily have leading coefficient one, but a multiple of it will.

Thanks

However I got this problem from a friend, and Descartes' Sign Rule shouldn't be used. Haven't looked at it so much, perhaps there is a simple proof for this special case?

Why Descartes' Sign Rule shouldn't be used? You can see this for proof http://www.cut-the-knot.org/fta/ROS2.shtml

Here's an argument without Descartes's Rule.  For concreteness, I'll suppose the polynomial is $ x^9 \minus{} x^7 \minus{} 1$, but it doesn't matter much.  We are looking for solutions of $ x^9 \equal{} x^7 \plus{} 1$.  Let $ y$ be the reciprocal of $ x$.  Dividing both sides by $ x^9$, we get the equation $ 1 \equal{} y^2 \plus{} y^9$.  The right side is an increasing function of $ y$ (for positive $ y$), so there can be only one positive solution.

Thanks Ravi.. but how do we handle the simplicity of the root?

That's a straightforward inequality, which I'll also illustrate with Ravi's example.  If $ r$ is the unique positive root of $ r^9 \equal{} r^7 \plus{} 1$, then

$ 9r^9 \equal{} 9(r^7 \plus{} 1) > 7r^7 \implies 9r^8 > 7r^6$,

so $ r$ cannot also be a root of the derivative.


(from AHSME 1999 #27):

In triangle ABC, 3*sin A + 4*cos B = 6 and 4*sin B + 3*cos A = 1.  Find the measure of angle C.

(A) 30
(B) 60
(C) 90
(D) 120
(E) 150
(F) Just kidding!  Only 5 answer choices.

[hide]Something strange happened here, but I think with a bit of numerical guesswork we can find the answer. Square the two equations and then add them together. Then we get

9 sin^2 A + 24 sin A cos B + 16 cos^2 B +

9 cos^2 A + 24 cos A sin B + 16 sin^2 B =

37

or

9 + 24 (sin A cos B + cos A sin B) + 16 = 37

24 sin(A+B)=12

sin(A+B)=1/2.

The problem is that now we don't know whether A+B is 30 or 150 degrees. Now the best thing to do is to notice that in the first equation, if 3 sin A + 4 cos B = 6, then sin A and cos B must each be fairly close to 1. Thus A must be quite close to 90. Thus A+B can't possibly be 30 since B is positive. Thus A+B = 150, or C=30.[/hide]


linear algebra
matrix
function
linear algebra unsolved

$ S \subset \mathbb{R}^{n \times n}$ is said to be totally regular if $ S \minus{} \{0\} \subset GL(n.\mathbb{R})$, it is,  every non zero matrix of $ S$ is invertible. ¿Can you show a totally regular subspace of dimension $ > 1$ in $ \mathbb{R}^{2 \times 2}$, and in $ \mathbb{R}^{3 \times 3}$, and in $ \mathbb{R}^{4 \times 4}$.

$ n \equal{} 2$: Let $ S \equal{} \mathbb RE_2 \plus{} \mathbb RA$ where $ A$ is a matrix with no real eigenvalue and positive determinant.

$ n \equal{} 3$: Assume there are $ A,B\in M_3(\mathbb R)$ such that the function $ f: \mathbb R^2\setminus\{0,0\}\to\mathbb R\setminus\{0\}$, $ (x,y)\mapsto\det(xA \plus{} yB)$ is well-defined. Since $ f$ is continuous, we have either $ f > 0$ or $ f < 0$ (use that connectedness is preserved by continuous mappings). The relation $ x^3f(1,y/x) \equal{} f(x,y)$ for $ x\neq 0$ then implies that $ x^3 > 0$ for all $ x\neq 0$ which is adsurd.

$ n \equal{} 4$: Patch something together using case $ n \equal{} 2$.

can you write down a full detailed solution, please?


If
$loga=6logb$
$2logb=logc$
$logc=45$

Find $y$
$y = \log \sqrt[5]{{\frac{{a^3 b^4 }}{{c^2 }}}}$

[hide="solution"]Rewriting a, b and c, we have
$\log a=6\log b=3\log c=3\times 45=135$
$\log b=\frac16\log a=135/6=\frac{45}2$
$\log c=45$

Therefore,
$y=\log\frac{a^{\frac 35}b^{\frac 45}}{c^{\frac 25}}$
$=\frac35\log a+\frac45\log b-\frac25\log c$
$=\frac35\times135+\frac45\times\frac{45}2-\frac25\times45=81$[/hide]


trigonometry
trig identities
cosine half angle

[hide="Solution to A"]$ \cos^4\theta\equal{}(\cos^2\theta)(\cos^2\theta)\equal{}(1\minus{}\sin^2\theta)(\cos^2\theta)\equal{}\cos^2\theta\minus{}\cos^2\theta\sin^2\theta$
$ \sin^4\theta\equal{}(\sin^2\theta)(\sin^2\theta)\equal{}(1\minus{}\cos^2\theta)(\sin^2\theta)\equal{}\sin^2\theta\minus{}\cos^2\theta\sin^2\theta$
$ \therefore \cos^4\theta\minus{}\sin^4\theta\equal{}\cos^2\theta\minus{}\cos^2\theta\sin^2\theta\minus{}\sin^2\theta\plus{}\cos^2\theta\sin^2\theta\equal{}\cos^2\theta\minus{}\sin^2\theta\equal{}\cos(2\theta)$

Applying this identity, $ \cos^4\left(\frac{\pi}{24}\right)\minus{}\sin^4\left(\frac{\pi}{24}\right)\equal{}\cos\left(2\cdot\frac{\pi}{24}\right)\equal{}\cos\left(\frac{\pi}{12}\right)\equal{}\cos\left(\frac{1}{2}\cdot\frac{\pi}{6}\right)$.

By the cosine half angle identity we have $ \pm\sqrt{\frac{1\plus{}\cos\frac{\pi}{6}}{2}}\equal{}\pm\sqrt{\frac{1\plus{}\frac{\sqrt{3}}{2}}{2}}$, which simplifies to $ \boxed{\frac{\sqrt{2\plus{}\sqrt{3}}}{2}}\approx{0.966}$.[/hide]

[hide="solution A"]

$ \cos ^4 \left( \dfrac{\pi}{24}\right) \minus{} \sin^4 \left( \dfrac{\pi}{24}\right)$

$ \equal{} \left(\cos ^2 \left( \dfrac{\pi}{24}\right) \minus{} \sin^2 \left( \dfrac{\pi}{24}\right)\right).\left(\cos ^2 \left( \dfrac{\pi}{24}\right) \plus{} \sin^2 \left( \dfrac{\pi}{24}\right)\right)$

$ \equal{}\cos \dfrac{\pi}{12}$

$ \equal{}\dfrac{\sqrt {2\plus{} \sqrt{3}}}{2}$

$ \equal{}\dfrac{\sqrt{6} \plus{} \sqrt{2}}{2}$
[/hide]

[hide="solution B"]

$ 1 \minus{} \cot (23^o) \equal{} \dfrac{(\sin (23) \minus{} \cos (23))}{\sin (23)}$

Now
 $ \sin (23) \minus{} \cos (23) \equal{} \sqrt {2} \times \left(\dfrac{1}{\sqrt {2}}\sin{23} \minus{} \dfrac{1}{\sqrt {2}}\cos{23}\right) \equal{} \sqrt {2}\sin (23 \minus{} 45) \equal{} \minus{} \sqrt {2} \sin (22)$

$ \Rightarrow 1 \minus{} \cot (23) \equal{} \dfrac{ \minus{} \sqrt {2} \sin(22)}{\sin(23)}$

Applying the same concepts to $ 1 \minus{} \cot (22)$ we get

$ (1 \minus{} \cot (23))(1 \minus{} \cot (22)) \equal{} \left( \dfrac{ \minus{} \sqrt {2} \sin(22)}{\sin(23)} \right).\left( \dfrac{ \minus{} \sqrt {2} \sin(23)}{\sin(22)} \right) \equal{} 2$
:) [/hide]


limit
calculus
derivative
calculus computations

$ \lim_{x\to0}\frac{(1\plus{}x)^\frac{1}{3}\minus{}(1\minus{}x)^\frac{1}{3}}{x}$

i posted this is the highschool forum cause thats what im in, and figured the question was way too easy for the college forum..

anyway someone mentioned using l'hopitals rule, but I was never taught that in class yet and I dont think thats the way the teacher wanted us to go about the problem, so if there is any other way to go about calculating the limit i'd like to know please :)

thanks

Use $ A^3 \minus{} B^3 \equal{} (A \minus{} B)(A^2 \plus{} AB \plus{} B^2)$ to rationalize the numerator.

well ive tried doing that, but then I end up with an even harder function..

unless Im not doing it how you figured I might

I end up with

$ ((1 \plus{} x)^\frac {1}{9} \minus{} (1 \minus{} x)^\frac {1}{9})((1 \plus{} x)^\frac {2}{9} \plus{} ((1 \plus{} x)(1 \minus{} x))^\frac {1}{9} \plus{} (1 \minus{} x)^\frac {2}{9})$

which doesnt seem to make things any easier :o

Set $ A\equal{}\sqrt[3]{1\plus{}x},\ B\equal{}\sqrt[3]{1\minus{}x}$.

by doing that and letting $ x \equal{} a^3 \minus{} 1$ I end up with

$ \frac {(a^\frac {1}{3} \minus{} b^\frac {1}{3})(a^\frac {2}{3} \plus{} (ab)^\frac {1}{3} \plus{} b^\frac {2}{3})}{(a \minus{} 1)(a^2 \plus{} a \plus{} 1)}$

but if x approaches zero, a and b both approach 1, leaving what I have left with 0/0 still  :|

EDIT::

[quote="Croata"]Your limit equals,

$ \underset{x\to 0}{\mathop{\lim }}\,\frac {{{(1 \plus{} x)}^{\frac {1}{3}}} \minus{} 1 \minus{} {{(1 \minus{} x)}^{\frac {1}{3}}} \plus{} 1}{x} \equal{} \underset{x\to 0}{\mathop{\lim }}\,\left\{\left( \frac {{{(1 \plus{} x)}^{\frac {1}{3}}} \minus{} 1}{x} \right) \minus{} \left( \frac {{{(1 \minus{} x)}^{\frac {1}{3}}} \minus{} 1}{x} \right) \right\}.$

Study $ \underset{x\to 0}{\mathop{\lim }}\,\frac {{{(1 \plus{} x)}^{\frac {1}{3}}} \minus{} 1}{x}.$ By making the substitution $ u \equal{} \sqrt [3]{1 \plus{} x}$ we have $ u\to1$ and $ x \equal{} u^3 \minus{} 1$ so the limit becomes $ \underset{u\to 1}{\mathop{\lim }}\,\frac {u \minus{} 1}{{{u}^{3}} \minus{} 1},$ which is not hard to compute.

Do the similar for $ \underset{x\to 0}{\mathop{\lim }}\,\frac {{{(1 \minus{} x)}^{\frac {1}{3}}} \minus{} 1}{x},$ and then we can say that $ \underset{x\to 0}{\mathop{\lim }}\,\frac {{{(1 \plus{} x)}^{\frac {1}{3}}} \minus{} {{(1 \minus{} x)}^{\frac {1}{3}}}}{x} \equal{} \underset{x\to 0}{\mathop{\lim }}\,\frac {{{(1 \plus{} x)}^{\frac {1}{3}}} \minus{} 1}{x} \minus{} \underset{x\to 0}{\mathop{\lim }}\,\frac {{{(1 \minus{} x)}^{\frac {1}{3}}} \minus{} 1}{x}.$[/quote]

genius.

thank you very much.. sometimes it takes only the simplest things. wow.

so thats what you wanted me to do kunny..

very interesting.

thanks for the help and methods guys, this class is a lot more thinking intensive than my previous ones  :blush:

[url=http://www.math.hmc.edu/calculus/tutorials/lhopital/]L'Hopital's Rule[/url]

Hopefully that helps.

The basic idea is that when you get a limit of the form
$ \frac{0}{0}$ or $ \frac{\infty}{\infty}$ you can look at the derivatives of the numerator and the denominator, and take the limit of the derivatives on the top and the bottom.

I think the link does a pretty good job explaining it.

When the denominator is just $ x$ or some linear expressionthat goes to zero, I wouldn't call it L'Hopital's rule. That limit [b]is[/b] a derivative; if we already know the function's derivative, we can apply that knowledge, but beware circular reasoning.


Let p a prime and let f(p)=(p^2-p)!/(p!)^(p-1). It is trivial to prove that f(p) is an integer. What can you say about f(p) mod p?
  Source: Created by Harazi

Well, it's not 0 mod p using method similar to that used to prove f(p) is integer :D

The answer is -1. We apply Wilson's theorem a couple of times :D

Yes, this is the correct answer. My solution is different:
        "Observe"  that the number is f(2p,p)*f(3p,p)*...*f(p(p-1),p), where f(x,y)=x!/(y!*(x-y)!). One can show that f(xp,p) mod p=x and thus the number is 2*3*...*(p-1) mod p, which is -1.


The integers a,b,c,d are such that ac,bc+ad,bd are divisible by a natural number n- Prove that ab and cd are also divisible by the number n-

The problem is incorrect. Let n=2, a=b=2, c=d=1: Then n divides ac, bd, bc+ad, but n does not divide cd=1.

However, there was a similar problem on the Hungary MO 1910:

[color=blue][i]Let $ a,b,c,d,n$ be integers such that $ n$ divides $ ac, bc\plus{}ad, bd$.
Prove that $ bc, ad$ are also divisible by $ n$[/i][/color]

Proof) Due to the conditions, $ n^2$ divides $ (bc\plus{}ad)^2$ and $ abcd$.
Therefore, $ n^2$ divides $ (bc\plus{}ad)^2 \minus{} 4abcd \equal{} (bc\minus{}ad)^2$.
$ \therefore n$ divides $ bc\minus{}ad$.

If we check that $ \frac{bc\plus{}ad}{n}$ and $ \frac{bc\minus{}ad}{n}$ have the same parity, we're done;
that implies $ \frac{\frac{bc\plus{}ad}{n} \plus{} \frac{bc\minus{}ad}{n}}{2} \equal{} \frac{bc}{n}$
$ \frac{\frac{bc\plus{}ad}{n} \minus{} \frac{bc\minus{}ad}{n}}{2} \equal{} \frac{ad}{n}$ are integers.

To check that, just observe that
$ (\frac{bc\plus{}ad}{n})^2 \equal{} (\frac{bc\minus{}ad}{n})^2 \plus{} 4 \frac{abcd}{n^2} \equiv (\frac{bc\minus{}ad}{n})^2$ (mod 4)


I just got my new copy of [i]Mental Floss[/i] today (a magazine) and I opened it and, to my joy, found an article on Paul Erdos. They mentioned Erdos numbers, so I decided to ask my math teacher what his was. Imagine my mouth dropping open when he quickly responded "oh, I'm a two." So wow, huh? My math teacher's a two. Which kind-of-sort-of-but-not-really makes me an Erdos 3. So, I was wondering if any of you guys knew or could find out what your Erdos number was. We're going to be a little lax on this poll, because we're all students here. Rules are as follows:

1. Paul Erdos has an Erdos number of 0. Everyone who co-wrote a paper with Erdos has an Erdos number of 1. Everyone who co-wrote a paper with someone who has an Erdos number of 1: they have an Erdos number of 2. And so on. It's like "Six Degrees of Seperation."
2. Since many of you probably haven't written a mathematical paper with a "real mathematician" yet (most of you being middle-schoolers), we're going to count the last string of our Erdos line a little differently: You must have taken a class with someone with a true Erdos number to join the chain. A "class" is defined as more than one session in which there is some sort of take-home problem set involved. Thus, a lecture does not count as a "class." The person who taught your class [i]must have[/i] co-wrote a paper with Erdos, someone who co-wrote a paper with Erdos, or someone who co-wrote a paper with someone who co-wrote a paper with Erdos (etc.).
3. You're Erdos number is defined as "infinity" if you have not taken a class with an Erdos finite number.

So what's your number?

I'm at 2 (the real way), and I got there with my very first publication. That paper even included both graph theory and metric spaces.

A link for this: The [url=http://www.oakland.edu/enp/]Erdos Number Project[/url]. The site includes lists of everyone with Erdos number 1 or 2.

I'm a 3; the link is through one of the coauthors on the paper jmerry was referring to. My Ph.D. advisor (Alberto Calderón) appears to have been a 3; he's not on the 2 list, but Antoni Zygmund is.

I think I've been in a class taken by a person who's been in a class taken by someone who co-published a paper with Erdos. Does that make me a 3 or 4?

I've taken three classes from people with Erdos numbers of two; Peter Borwein, Jason Bell, and Jonathan Jedwab at Simon Fraser University.

I have conducted (unpublished) research under the guidance of an Erdos 1 (Joel Spencer) and taken classes with at least 2 other Erdos 1s (Andras Gyarfas, Antal Balog).

[quote="curl"]Erdos number 0 here.[/quote]
There's internet service from beyond the grave? You do know that Erdös is dead, right?

Who of you voted Erdös number 1? Or was it just a joke?

I'm almost a two: my dad published with Erdos and is therefore a 1 (on the list at the site) plus I know at least 3 other 1s personaly. :)

I've published a paper, but that was all my own research.  It was my senior project when I was in high school, and a friend looked it over and agreed to post it on his site.

For now, I still probably lack one.

--KC

[quote="puuhikki"]I have wondered the definition of Erdős number. Assume that my Erdős number is $n$ and I have wrote the paper I got my Erdős number with mathematician $A$. After writing the paper I can't improve my Erdős number but $A$ decreases his Erdős number to $n-2$. Now will my Erdős number be $n$ or $n-1$?[/quote]

$n-1$.  Your Erdos number is the shortest chain connected you to Erdos, even if links in that chain were made after you wrote your paper.  So, if you write a paper with someone who later becomes Paul Erdos, your Erdos number would be 1, even if he wasn't Erdos at the time you co-authored with him.  ;)

I'm a three, although the chain is a little weak. I took a couple weeks of classes with George Thomas at Math Path a couple summers ago, and I think he's a two.

I'm at 3(the real way).

I've (co-)written an actual paper (with a bunch of people around my age and a mentor), which is on the arXiv.

The mentor of my paper, whose name I shall not mention, knows John Conway(personally).

John Conway has co-written a paper with Paul Erdös.

So that's it.

mine is infinite i have never written a mathematical paper as I am only in elementary school XD

I'm technically a 2 but it's kind of cheating because it's via a paper with Erdos listed posthumously as a coauthor. I don't know the details, but I guess they did at least use some comments Erdos had made, so he did contribute to the paper.

If posthumous doesn't count then I'm a 3.

mine is  [quote="TI-84"]"could not find value"[/quote]


function
group theory
superior algebra
superior algebra unsolved

By considering the action of the symmetric group $ S_n$ on $ \mathbb{C}^n$ by permutation matrices and using the orthogonality relations, one can prove the identity

$ \sum_{\pi \in S_n} \nu(\pi)^2 \equal{} 2n!, n \ge 2$

where $ \nu(\pi)$ denotes the number of fixed points of $ \pi$ (precisely the character of the above representation).  An alternate method using generating functions gives the following generalization:

$ \sum_{\pi \in S_n} \nu(\pi)^k \equal{} B_k n!, n \ge k$

where $ B_k$ are the [url=http://en.wikipedia.org/wiki/Bell_number]Bell numbers[/url].  Can this identity be proven directly, i.e. by a combinatorial argument?  Is there a natural interpretation in terms of, say, the $ L^k$ norm to the space of characters?  What about taking repeated tensor powers of the above representation?

http://www.mathlinks.ro/viewtopic.php?t=63678
(you may want to add your proof :D )


probability
inequalities
integration
calculus
analytic geometry
trigonometry
geometry

Let $ S: {(x,y,z)| x^2\plus{}y^2\plus{}z^2<\equal{}1}$ , $ S$ is a closed unit ball in $ xyz$ space.  Let's put two arbitrary points $ P$,$ Q$; $ P$ on boundary of $ S$ and $ Q$ in $ S$ except the boundary. When we think a rectangular parallelepiped that its diagonal is the segment $ PQ$ and each six faces are all parallel to $ xy$,$ yz$,$ zx$ plane, what is the probability that $ S$ completely includes this parallelepiped?

  I think 5 dimensional space is needed;;;;;

Actually, we don't. It's simpler than it looks.

First, we clearly want $ P$ to be distributed uniformly on the surface, and $ Q$ uniformly in the ball.
Let's think conditionally: if $ P\equal{}(x_0,y_0,z_0)$, what values $ Q\equal{}(x,y,z)$ work? If $ |x|>|x_0|$, then the corner $ (x,y_0,z_0)$ is outside the ball, which is bad. Similarly, we must have $ |y|\le |y_0|$ and $ |z|\le |z_0|$. If all three of those inequalities are satisfied, the parallelepiped fits entirely inside the one with corners $ (x_0,y_0,z_0)$ and $ (\minus{}x_0,\minus{}y_0,\minus{}z_0)$, which is inside the ball.
So, the volume of possible $ Q$-values that work is the volume $ 8x_0y_0z_0$ of that parallelepiped. The probability we seek is the average $ \frac{1}{4\pi}\int_{\partial S}\frac{8|xyz|}{4\pi/3}\,dA$ of this over the surface.
For that integral, I'll use cylindrical coordinates: $ x\equal{}r\cos\theta\equal{}\sqrt{1\minus{}z^2}\cos\theta, y\equal{}r\sin\theta\equal{}\sqrt{1\minus{}z^2}\sin\theta$ on the sphere $ r\equal{}\sqrt{1\minus{}z^2}$. We have that $ dA\equal{}d\theta\, dz$ (which is why I chose this parametrization rather than the usual spherical coordinates).
$ p\equal{}\frac1{4\pi}\int_{\minus{}1}^1\int_0^{2\pi}\frac{|8z(1\minus{}z^2)\cos\theta\sin\theta|}{4\pi/3}\,d\theta\,dz$
$ p\equal{}\frac{3}{2\pi^2}\left(\int_{\minus{}1}^1 |z\minus{}z^3|\,dz\right)\cdot \left(\int_0^{2\pi}|\cos\theta\sin\theta|\,d\theta\right)$
$ p\equal{}\frac{3}{2\pi^2}\cdot \frac12\cdot 1$
$ p\equal{}\frac{3}{4\pi^2}\approx 0.076$

Can we use the cylinderical coordinate system to do the integral $ \frac{1}{4\pi}\int_{\partial S}\frac{8|xyz|}{4\pi/3}\,dA$?
 We can't cover all the surface of $ S$ by cylinderical coordinates. Am I right?

No, you're not right. What does it take to specify a point on the sphere of radius $ R$? Two coordinates. The usual choice is longitude $ \theta$ and colatitude $ \phi$; the area element is then $ R^2\sin\phi\,d\theta\,d\phi$. For this problem, I chose longitude $ \theta$ and $ z$, for which the area element is $ R\,d\theta\,dz$. How are the two related? It's a substitution, and in this case it's really a one-variable substitution. $ \theta$ is the same, and $ z$ and $ \phi$ are related by $ z\equal{}R\cos\phi$.

I'll show what the integral looks like in the usual spherical coordinates:
$ p\equal{}\frac1{4\pi}\int_0^{\pi}\int_0^{2\pi}\frac{|8(\sin\phi\cos\theta)(\sin\phi\sin\theta)(\cos\phi)|}{4\pi/3}\cdot \sin\phi\,d\theta\,d\phi$

It's more complicated, and the substitution $ z\equal{}\cos\phi$ is probably the best way to do the integral anyway.

Now, would I recommend this for a full 3-dimensional coordinate system? No, the hybrid coordinates $ \rho,z,\theta$ have problems. They cover everything cleanly (up to the same issues with the $ z$ axis that both cylindrical and spherical systems have), but they suffer by not being locally orthogonal.

Thanks jmerry, I understand what you say.  :) (Now i see your approach goes through the five dimensional space..... )


limit
logarithms
calculus
calculus computations

[hide="Answer"]$ \lim \sqrt[n]{a_n} \equal{} \lim \sqrt[n]{a_1 \cdot \frac{a_2}{a_1} \cdot \frac{a_3}{a_2} \cdot ... \cdot \frac{a_{n\minus{}1}}{a_{n\minus{}2}} \cdot \frac{a_n}{a_{n\minus{}1}}} \equal{} \lim \frac{a_n}{a_{n\minus{}1}} \equal{} \lim \frac{a_{n\plus{}1}}{a_n}$,

because we can show that $ \lim \sqrt[n]{a_1 \cdot a_2 \cdot a_3 \cdot ... \cdot a_n} \equal{} \lim a_n$, for $ a_n > 0$.[/hide]

I'm not sure that's really an answer since either statement is proved by Cesaro Stolz lemma

Read here for proof of CS lemma: http://www.math.ksu.edu/~nagy/snippets/stolz-cesaro.pdf

[quote="Carcul"]

because we can show that $ \lim \sqrt [n]{a_1 \cdot a_2 \cdot a_3 \cdot ... \cdot a_n} \equal{} \lim a_n$, for $ a_n > 0$.[/quote]

why?

[hide="Because"]$ L \equal{} \lim \sqrt[n]{a_1 \cdot a_2 \cdot a_3 \cdot ... \cdot a_n}$. Then

$ \ln L \equal{} \lim \ln \left(\sqrt[n]{a_1 \cdot a_2 \cdot a_3 \cdot ... \cdot a_n}\right) \equal{} \lim \frac{\ln a_1 \plus{} \ln a_2 \plus{} \ln a_3 \plus{} ... \plus{} \ln a_n}{n} \equal{} \lim \ln a_n$

Therefore, as $ \ln L \equal{} \lim \ln a_n$ then $ L \equal{} \lim a_n$.[/hide]

[quote="wizard"]Suppose that $ a_n > 0$ is a sequence of positive numbers and suppose that the limit $ \lim_{n\rightarrow\infty}\frac {a_{n \plus{} 1}}{a_n}$ exists.  Then prove that $ \lim_{n\rightarrow\infty}\sqrt [n]{a_n}$ exists and
\[ \lim_{n\rightarrow\infty}\frac {a_{n \plus{} 1}}{a_n} \equal{} \lim_{n\rightarrow\infty}\sqrt [n]{a_n}
\]
[/quote]

If $ L\equal{}\lim_{n\to\infty}\sqrt[n]{a_n}$ then
$ ln(L) \equal{}\lim_{n\to \infty} \frac{ln(a_n)}{n}\equal{}^{S.C.} \lim_{n\to\infty}\frac{ln(a_{n\plus{}1})\minus{}ln(a_n)}{(n\plus{}1)\minus{}n}\equal{}\lim_{n\to\infty} \ln \left( \frac{a_{n\plus{}1}}{a_n}\right)$ and the conclusion follows.


Let $ \{a_0,\ldots,a_9\}\equal{}\{0,\ldots,9\}$. Consider the sum $ n\equal{}\sum_{i\equal{}0}^9a_i\cdot 10^i$. What is the largest prime factor that $ n$ can have and what permutation gives that number?

I don't know if you expect that someone could do this without a computer...

Anyway, we have $ 9876541023 \equal{} 3^2 \times 1097393447$. We'll prove this is the best possible. (Just don't ask how we know that thing is a prime)

Let's say a natural number is special if it has exactly $ 10$ digits, all of them distinct. The problem asks for the largest prime divisor a special number can have. Notice all special numbers are divisible by $ 9$, since the sum of their digits is $ 1\plus{}2\plus{}...\plus{}9\equal{}9\times5$. So, if $ n$ is a special number less than $ 9876541023$, then $ n$ has a prime factor less than $ 1097393447$, otherwise it would be greater than $ 9 \times 1097393447 \equal{} 9876541023$, a contradiction.

So all we have to do now is check that no special number greater than $ 9876541023$ has a prime factor greater than $ 1097393447$. Suppose $ n$ is a special number such that $ \frac{n}{9}$ is not a prime. Then, the largest prime divisor of $ n$ is less than $ \frac{\frac{n}{9}}{2} \leq \frac{9876543210}{18} \equal{} 548696845 < 1097393447$. So, if there is any special number such that its largest prime divisor is greater than $ 1097393447$, it has to be of the form $ 3^2 \times p$, where $ p$ is a big prime. We'll see no special number grater than $ 9876541023$ can be written this way.

*$ 9876543210, 9876543120, 9876543102, 9876543012, 9876542310, 9876542130, 9876541320, 9876541302, 9876541230, 9876541032$ are discarded since they are divisible by $ 2$.
*$ 9876543201$ is divisible by $ 7$.
*$ 9876543021$ is divisible by $ 29$.
*$ 9876542301$ is divisible by $ 27$.
*$ 9876542103$ is divisible by $ 53$.
*$ 9876542031$ is divisible by $ 27$.
*$ 9876542013$ is divisible by $ 13$.
*$ 9876541203$ is divisible by $ 1181$  :!: .

This completes the solution   :huh:


probability
geometry
FTW
calculus
analytic geometry
number theory

I know this thing was posted before ([url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=211123&search_id=518175723]here[/url]), but I wanted to see if there would be any significant changes in people's opinions from then.

BTW: Mine is number theory, since algebra is too boring and counting + probability is too...hard. :( Geometry is OK, because you can rebuild anything that is on the diagram, so it's probably my second favorite.

[quote="akhil0422"]For me, geometry is favorite because it is unique. There are nice shapes and angles unlike other branches which are mainly boring, tedious algebra.[/quote]

Bleh and many geometry problems aren't boring and tedious?

Heh, actually, everything can be boring and tedious, but also can be elegant.

Number theory for me.

Ummm.... They're all amazing. Each is different from the others, and beautiful in its own way.

geo = favorite

Most of the time, it's not tedious or anything. It's just that you have to see something and you write a couple of equations and solve. Number Theorey is too weird, algebra is too plain, and C&P is probably 2nd favorite.

[quote="FantasyLover"]Algebra  :)[/quote]

same, cuz its ezpz

[quote="Poincare"]Number theory is really really fun, but not at the MC level. So, I put in Geo.[/quote]
I really meant for this to be general, not just in MathCounts. I probably shouldn't have put it in the MC forum then.

Counting and Probability PWNS

and then algebra cuz most of its epic brute force and calc(ulator) bashing

and then nt and finally geo

im surprised fanfan did not say calculus

1. ALGEBRA
2. NUMBER THEORY
3. COUNTING
4. GEOMETRY

I like all the subjects a lot? :P  It really depends what day it is, I usually like geometry the best, though.

[quote="funtwo"]i suppose coordinate bash falls under geometry?[/quote]

Heh, favorite thing in geo.

[quote="funtwo"]i suppose coordinate bash falls under geometry?[/quote]
I suppose it is. :D

Except that in coordinate geometry there's waaaaay too many numbers.

Coordinate bashing falls under Geometry because it's Analytical Geometry, I believe. All branches of math have problems that can be considered tedious and problems that can be considered fun. I'm not very good at any branch of math, but Counting & Probability is probably my favorite. I also like Algebra and Number Theory, and Geometry's okay.

Definitely geometry! Its so fun to manipulate shapes!

on a 5 point scale, here goes:

1. C&P-4.2
2. Geometry-3.8
2. Algebra-3.8
4. Number Theory-3.7

yeah...not much difference

I'd hate to have to choose! They're all so interesting/cool. :wink:


In volume 1 of the AOPS textbooks, i don't understand the solution to exercise 5-16, it doesn't seem like a clear cut solution.  Could someone elaborate on this solution as it would greatly clear this up.  Thx

the problem...i think thats wat he means....

[quote="demji"]EXERCISE 5-16  Show that the square of any integer is congruent to either 0, 1, or 4 mod 8.[/quote]
Very easy:
What are different forms for integers in mod 8? They are $0\mod 8$, $\pm 1\mod 8$, $\pm 2\mod 8$, $\pm 3\mod 8$, $4\mod 8$

Square them to give $0\mod 8$, $1\mod 8$, $4\mod 8$, $9\equiv 1\mod 8$, $16\equiv 0\mod 8$. Finished

Here's a method:

Every integer is either even or odd.  If even then it can be expressed as $2m$, if odd it can be expressed as $2n+1$.

If even:
Then the square of the integer is $4m^2$.  If $m^2$ is even then $m^2=2a$, so $4m^2\equiv4\times2a\equiv8a\equiv0\pmod8$.  If $m^2$ is odd then $m^2=2a+1$, so $4(2a+1)\equiv4\pmod8$.

If odd:
Then the square of the integer is $4n^2+4n+1=4n(n+1)+1$.  We see that either $n$ is even, or $n$ is odd.  If $n$ is even, then $4n(n+1)+1\equiv0+1\equiv1\pmod8$.  If $n$ is odd, then $n+1$ is even, then it is also $\equiv1\pmod8$.

So the only possibilities are $x^2\equiv0,1,4\pmod8$.

thx 10000, i forgot about the negatives... thats one thing i just miss in mods, the negatives, i like to think of everything as positive since when i learned mods in 7th grade we did the clock arithmetic and not the regular way... I understand it now


In a large collection of coins, the probability X that a head will be obtained when a coin is tossed varies from one coin to another and the distribution of X in the collection is specified by the following p.d.f:

f1(x)= 6x(1-x) for 0<x<1 and 0 otherwise.

Suppose a coin is selected at random from the collection and tossed once, and that a head is obtained. Determine the conditional pdf of X for this coin.

-Prob & Stats by Degroot

I thought I'd posted an answer to this - maybe I never clicked "submit"? Well, it's lost now - but I do remember the result, which is the that conditional density asked for is $ 12x^2(1\minus{}x)$ on $ [0,1].$ Maybe if I have the energy, I'll rewrite the argument, but not tonight.

Hate it when that happens. Well, whenever you get the chance.
The suspense is killing me.

Thanks in advance.

Say the event that happens is Y.

f(y|x) is Bernoulli with outcome X.

you have f(x) and f(y|x).  Their product gives you f(x,y). Integrate to find the marginal distribution of y etc...


We consider a polyhedra which has exactly two vertices adjacent with an odd number of edges, and these two vertices are lying on the same edge.

Prove that for all integers $n\geq 3$ there exists a face of the polyhedra with a number of sides not divisible by $n$.

I have not a complete proof yet :(

Anyway, we can easily prove that there is a triangular face (here we use that there are two odd vertices, but don't use that they are connected), so we are only to prove that it is impossible that each face has a number of sides, which is divisible by 3.

how does that finish the problem?  the problem says "for all $ n \geq 3 $... "

If we have a triangular face it means that number of its sides is not divisible by any $n\geq 4$ !

I realized that even don't need what I wrote above.

Draw this planar graph on the plane. Suppose all faces are divisible by $n$. Take an arbitrary face and remove all its edges! We will obtain a new planar graph and all its faces will be divisible by $n$ again (note that we need properly count number of sides in each face; see remark 2). We also see that new graph contains only two odd vertices.
Perform it step by step. Finaly we will obtain a planar graph consisting the only edge, which connects two odd vertices, but in this case we have the only face with two sides.

Remark 1. We need that two odd vertices are connected for the last step. Otherwise we will obtain a chain, so number of sides of the only face could be divisible by some number $n\geq 3$.

Remark 2. The number of sides on the yellow face in the planar graph below is 9.

I hope it is correct.

I think I can proe that thing about divisibility with $3$, Myth:

First of all, it suffices to assume that all our faces are triangular (I'll say why at the end). Assume we have managed to orient the edges so that all the triangles, except, maybe, for one of them, is oriented, either in a clockwise or an anticlockwise direction. Then the last one must also be oriented, because we can write the clockwise cycle on the dges of that triangle as a sum of cycles of the other triangles, and those are all $0$ (by a cycle I mean the value we get when we go around the triangle in the mentioned direction and we put a $1$ for edges we traverse directly, and a $-1$ for edges we traverse inversely).

Now, let's look at the polyhedron obtained from our polyhedron by deleting the common side of the two odd vertices. We have a polyhedron with even vertices only. We can color it chessboard-style (this need further justification, but I think that's not hard to give), and then orient each black face in a clockwise direction. The white faces will automatically be oriented in a counterclockwise direction. We go back to our original polyhedron, and orient the "bad" edge however we want. We have a polyhedron in which exactly one trianlge is not oriented, and that's a contradiction.

Now, to see why we can restrict our attention to triangles, given a convex polygon with $3k$ vertices, it's easy to triangulate it so that each vertex lies on an even number of edges in the triangulation, so we do this with each face of our polyhedron.

Edit:  :blush: oops.. I didn't see your last post.

Another solution: Take the planar graph, remove the troubling edge, and colour this in red and blue, with no 2 neighboring faces coloured the same. Only one face won't have the no. of edges divisible by $n$, i.e. the one obtained by joining the 2 faces that had a common edge the one removed. Now let's say this face is colured blue. Count the edges after all red faces. We will have all the edges counted once this way, so $m-1$ is divisible by $n$. On the other hand counting these on the polyhedra we obtain $2m=M_n$, imposible.


advanced fields
advanced fields unsolved

Hi all,

1.  What are the minimum Hamming distance between any 2 code words for a single bit parity check?

My answer is 2 since if you have 1 bit difference, the parity bit will also be different, that makes it a minimum of 2.  However, I can say it but I do not know how to proof it? any help would be appreciated!

Thanks.

To be able to detect a one-bit error, you must have a code in which any two words have distance at least $2$. Indeed, if there are two words at distance $1$, then a one-bit error can change one word into the other, and this will not be detected. I don't know if there is anything else to add here.

Of course, in all the working single-error correcting codes I know, the distance between code words is 3. If two code words are distant by 2, there there would be two points "in between" that are each a distance of 1 from the code words. If the received message is one of those in-between words, then which way should we resolve it?

With a minimum distance of 3, we can have the radius-1 closed balls around each code word be disjoint, leading to no such ambiguities.

[quote]To be able to detect a one-bit error, you must have a code in which any two words have distance at least 2. Indeed, if there are two words at distance 1, then a one-bit error can change one word into the other, and this will not be detected. I don't know if there is anything else to add here.[/quote]

is there any systematic way to prove it? It seem like i know that there should be at least a distance of 2, but then I don't know how to go about proving it.  any hints?

Let me make what I said above stronger. A Hamming distance of 2 between code words is not far enough. The distance must be at least 3.

Suppose $x$ and $y$ are code words and $d(x,y)=2.$ Then there must exist $z$ such that $d(x,z)=d(z,y)=1.$ (In fact, there are two such words - just pick one of them for this argument.) Then a single-bit error could turn $x$ into $z$ and a different single-bit error could turn $y$ into $z.$ If $z$ is received as a message, it is impossible to determine whether the intended message was $x$ or $y.$

To illustrate: suppose we use 2-bit words, and that we try to make $11$ and $00$ code words. You receive $10.$ What was sent?

In fact, to make a single-error correcting code with two code words, you need at least three bits. Using $111$ and $000$ as code words works. Now if you receive, say, $101,$ we'll know how to decode it. But $d(111,000)=3.$

I thought it should be 2. When you're comparing 2 words
for example, using even parity:

$10010$  $0$

$10000$  $1$

The last digit is a parity bit, so in this case, these 2 only have a distance of $2$, right?  Since the first one has even # of 1's so the parity bit is 0.  The bottom one has odd # of one so the parity bit is 1.  So isn't 2 a minimum when comparing 2 words with single-bit parity?

You receive:

$10000$ $0$

An error has been [i]detected[/i] - you know that message is incorrect. But the error cannot be [i]corrected[/i]. You cannot tell me what the correct message should have been. I thnk you're stumbling over the difference between error-detecting codes and error-correcting codes.


geometry
rectangle
AMC
AMC 10
AMC 10 A
symmetry
analytic geometry

In rectangle $ ADEH$, points $ B$ and $ C$ trisect $ \overline{AD}$, and points $ G$ and $ F$ trisect $ \overline{HE}$.  In addition, $ AH \equal{} AC \equal{} 2.$  What is the area of quadrilateral $ WXYZ$ shown in the figure?

[img]http://data.artofproblemsolving.com/images/2006_AMC10A-17.png[/img]

$ \textbf{(A) } \frac 12 \qquad \textbf{(B) } \frac {\sqrt {2}}2\qquad \textbf{(C) } \frac {\sqrt {3}}2 \qquad \textbf{(D) } \frac {2\sqrt {2}}3 \qquad \textbf{(E) } \frac {2\sqrt {3}}3$

[hide]

notice that HXA and BYC are both isosceles right angle triangle .Since AH=2 , BC=1and HC=$2\sqrt{2}$

using phytagoras to find HX=$\sqrt{2}$ and CY=$\frac{1}{\sqrt{2}}$

hence XY=HC-HX-CY=$\frac{1}{\sqrt{2}}$

Since it is symmetry , $WXYZ$ is a square and hence its area is $\frac{1}{2}$

$\boxed{A}$
[/hide]

hehe, the diagram on the test was actually elongated, so the square looked like a diamond

[hide="Answer"]$WXYZ$ is a square, first of all. Then, we see that every triangle is 45-45-90. Take $CH=2\sqrt{2}$. $XH=\sqrt{2}$, and $CY=\frac{\sqrt{2}}{2}$. Therefore, $XY=2\sqrt{2}-\sqrt{2}-\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$, and the area of the square is $\frac{1}{2}\Rightarrow \boxed{A}$.[/hide]

how come i can't see the image in the first post? It just says "Image" and there's no link

I don't believe that any of the old images show up unfortunately. But images for this particular problem, among many others, can be found on the [url=http://www.artofproblemsolving.com/Wiki/index.php/2006_AMC_10A_Problems/Problem_17]wiki[/url].  :)  *advertises*

[hide=" My somewhat wordy solution"]If we drop a perpendicular from $ \mbox{B}$ to the line $ \mbox{AB}$ and we call the pt. of intersection $ \mbox{P}$. We know that $ {\mbox{BP}}$ is $ \frac{1}{\sqrt{2}}$,(for the sake of convenience, we will not rationalize the denominator yet). If we repeat the process dropping a perpendicular from $ C$, and call the point of intersection Q, then we get that $ \mbox{CQ}$, is also $ \frac{1}{\sqrt{2}}$ and since $ \mbox{BP}$ = $ \mbox{WZ}$ and $ \mbox{CQ}$ = $ \mbox{WX}$, the area of WXYZ is simply:

$ (\frac{1}{\sqrt{2}})^2$  $ \Longrightarrow$ $ \frac{1}{2}$, $ \fbox{\mbox{A}}$[/hide]

give xy coordinates for each point (after representing the lines with xy lines on the cartesian plane and solving for the intersections)......then use shoelace.


Suppose there are $n$ vertices in a simple connected graph, and each vertex has degree of no more than $3$. If any two vertices are connected by a path of length $1$ and $2$, what is the maximum value of $n$?

[hide]n=4.
Just draw a square where every vertex connects.
Since the distance between any 2 vertices is 1 and 2 and there's only a maximum of 3 connections per vertex, non higher polygons will suffice.  However, since the connections are between any 2 vertices, one must have a polygonal configuration.[/hide]

what do you mean by "If any two vertices are connected by a path of length 1 and 2"?


I had a problem solving this problem...

I converted the word problem into equations....

solve X Y and Z

X + Y + Z = 564
X - 25 = (Y*2)/5
Z + 25 = X * 1.2

NEED HELP DESPERATLEY!!!

[hide]okay, naming you the equations you gave a, b, and c.
From b, you can get that $x = 	\frac{2y}{5}+25  $
so if you sub that into a, then you get $ \frac{2y}{5}+25 +y+z=564$ and we call this a'
c tells you that $1.2x=z+25$... so if you multiply everything in b by $ 1.2$, then you get $1.2x-30=\frac{2.4y}{5}$ or $1.2x=\frac{2.4y}{5}+30=z+25$, so $z=\frac{2.4y}{5}+30-25 = \frac{2.4y}{5}+5$
If you sub that in to a', then you get $ \frac{2y}{5} + 25+y+\frac{2.4y}{5}+5=564$, and if you solve for y, then you get $\frac{9.4y}{5}= 534$ and $y =284.0425516.... $wow, weird answer, i'm not gonna find x and z, but this is kinda weird... does the problem tell you to round?[/hide]

[hide]Z = 1.2X - 25 so by substitution X + Y + Z = 564 becomes X + Y + 1.2X - 25 = 564 or 2.2X + Y = 619. Solving for x and y with the equations 2.2X + Y = 619 and X - (2/5)Y = 25 we get X = 145 and Y = 300. Pug those into the first equation and get 145 + 300 + Z = 564, so Z must be 119. X = 145, Y = 300, Z = 119.[/hide]

tarquin, your solution is incorrect. Check your answers once again.

wait, after I looked at tarquin's solution, i check the last few numbers, if z = 119, then x would equal 144.... but I didn't check the rest of the stuff.... is mine right?

Rewrite the equation system:
                               x + y + z  = 564
                             5x - 2y + 0.z =125
                             6x +0y -  5z  =125
Now, using Cramer's Method, we obtain
                                                        x=6515/47 quickly.
Locating this value in the system,        y=6643/47 and z=13350/47

                                                                   fatih sahin

Woah yeah I think I messed up adding 564 and 25 haha.

Geeze...

The reason why I asked AoPS users was because I was keep getting decimals...

this is like number of books problems, which means I shouldn't get the decimals...

 so I think I did my translating process wrong...ARG!!!

( though I had the exact same solution and answer as Lucy's )

Yeppy...

I must ';ve done the translating wrong...


Let $p$ be an odd prime and $a$ be any integer which is not congruent to 0 modulo p. Prove that the congruence $x^2=-a^2 mod p$ has solutions if and only if $p$ is congruent to $1 mod 4$.

$a$ is not null modulo $p$ so it's the same as showing that $x^2 = -1\ mod\ p$ has solution.

By Fermat "little" theorem we have $x^{p-1} = 1 \ mod\ p$

Suppose we have $x^2 = -1\ mod\ p$ and $p = 3\ mod\ 4$, then $(p-1)/2$ is odd and then ${x^2}^{(p-1)/2} = -1$ which contradicts FLT.

And if $p = 1\ mod\ 4$, by Wilson's theorem $\prod_{x=1}^{(p-1)/2} x(p-x) = -1\mod p$
Since $x(p-x) = -x^2$, $\prod_{x=1}^{(p-1)/2} x$ is a solution to our equation.


If a point is randomly selected inside a 1'$ \times$1' square, what is the probability that it will be less than 6'' from the center of the square? Express your answer as a common fraction in terms of $ \pi$.

We can use relative units to see that the square is a 2 by 2 and the "successful region" is a circle with it's center at the center of the square and a radius of 1. Thus, our answer is $ \boxed{\frac\pi4}$.

A $ 1'\times 1'$ square is a  $ 12"\times 12"$ square. Drawing this out we see that the point must be in a circle of radius $ 6"$. Thus we get

\[ \frac{6^2\pi}{12^2}\equal{}\boxed{\frac{\pi}{4}}
\]


calculus integration inequalities function derivative logarithms trigonometry

Let $\mathcal{F}$ be the set of functions $f: [0,1]\to\mathbb{R}$ that are differentiable, with continuous derivative, and $f(0)=0$, $f(1)=1$. Find the minimum of $\int_{0}^{1}\sqrt{1+x^{2}}\cdot \big(f'(x)\big)^{2}\ dx$ (where $f\in\mathcal{F}$) and find all functions $f\in\mathcal{F}$ for which this minimum is attained.

[hide="Comment"]
In the contest, this was the b) point of the problem. The a) point was simply ``Prove the Cauchy inequality in integral form''.
[/hide]

You can just use $\int_{0}^{1}\sqrt{1+x^{2}}f'(x)^{2}dx \int_{0}^{1}\frac{1}{\sqrt{1+x^{2}}}dx \ge \left( \int_{0}^{1}f'(x) dx \right)^{2}$. Basically, we have $f(x) = c \ln (x+\sqrt{1+x^{2}})$ for an appropriate constant c.

This doesn't seem quite right but don't you have to use the fact that
\[\int_{0}^{1}\frac{1}{\sqrt{1+x^{2}}}dx=\left[\sinh^{-1}x\right]_{0}^{1}\]
Where ${\sinh(x)=\frac{e^{x}-e^{-x}}2}$ so $\sinh(0)=0$. Thus the above integral is just $\sinh^{-1}1$ which seems difficult... it appears to be \[\int_{0}^{1}\frac{1}{\sqrt{1+x^{2}}}dx=\left[\sinh^{-1}x\right]_{0}^{1}=\ln(1+\sqrt{2}) \]
Oh I think I get it now... by considering the equality case below you don't have to actually deal with solving for $\sinh^{-1}x$ because it follows from $x=1,k=1,c=0$ in the end...
\\I guess skip the following paragraph
Because that would be harsh on a national olympiad, don't you think? Or is it really so important to have the derivative of the inverse hyperbolic sine function memorized...? (I pulled this one out of the back of my calc book)
And also I'm getting that the minimum occurs when
\[f'(x)=\frac{k}{\sqrt{1+x^{2}}}\implies f(x)=k\int (1+x^{2})^{-1/2}=k\ln(x+\sqrt{1+x^{2}})+C \]
oops I guess that's exactly what you had...

This is a Calculus of Variations problem.

I suggest you use the http://en.wikipedia.org/wiki/Euler-Lagrange_equations


In an analysis book i found this proof for the uniqueness of a limit of a function.

"It is easy to show that when a Limit of a function f(z) exists at a point a,it is unique.To do this ,we suppose that

lim f(z) =l and lim f(z)=m as z----> a (z goes to a).

Then,for any positive number ε,there are positive numbers r,δ such that

           |f(z)-l |< ε   whenever 0< |z-a |<r

                       and

             |f(z)-m |<ε whenever 0< |z-a |<δ

     So if   0< |z-a |<θ ,where θ denotes the smaller of the two Nos r and δ,we find that

 |l-m |= |(f(z)-m)-(f(z)-l) |=< |f(z)-l | + |f(z)-m | < ε +ε=2ε

          But   |l-m | is a nonnegative constant, and ε can be chosen arbitrarily small.
Hence

l-m =0 , or l=m."

is that proof correct??

[quote="JRav"]Why was this posted 3 times?[/quote]

NOT only three times,to plenty other forums.

The right question is if somebody can tell me whether the above proof is correct or not and WHY.

In another forum i wrote my version w.r.s to the correctness of the proof.

But that was MY IDEA of the proof .

I am very interested to know what others think about the proof and why>

Thanks

Posting in multiple forums is not a good idea, as the others will probably just get deleted.

[quote="τρικλινο"]The right question is if somebody can tell me whether the above proof is correct or not and WHY.[/quote]
I don't think anyone's said anything other than "the above proof is definitely correct; what step are you having trouble with?" and you've [b]continually[/b] refused to answer this simple question.


Let $ x_1,\ x_2,\ \cdots ,\ x_n$ be real numbers satisfying the conditions:

$ \left\{ \begin{array}{ll} |x_1| \equal{} 999 & \quad \\
|x_n| \equal{} |x_{n \minus{} 1} \plus{} 1|\ \ for\ n \equal{} 2,\ 3,\ \cdots \ , 2008 & \quad \end{array} \right.$

Find the least value of $ x_1 \plus{} x_2 \plus{} \cdots \plus{} x_{2008}.$

[hide="Hint"] Use the fact that $ x_n^2\equal{}x_{n\minus{}1}^2\plus{}2x_{n\minus{}1}\plus{}1$.[/hide]


algebra
polynomial
linear algebra
matrix
function
domain
calculus

So we did this example in class, and just need some help understanding it.

T: P2 --> P2            //T(transformation) where P2 is polynomial of degree 2
Basis for P2: {1, t, t^2}
T( f(t) = f '' (t) + 4f ' (t) )

Solution:
//We just plugged in...
T(1) = f '' (1) + 4f ' (1)           = 0 + 0 //first column
T(t) = f '' (t) + 4f ' (t)             = 0 + 4 //second column
T(t^2) = f '' (t^2) + 4f ' (t^2) = 2 + 8t //third column

//This is the matrix we got...
0 4 2
0 0 8
0 0 0

I just want to make sure that I understand that the matrix is 3x3 because the dimension of the basis is 3, correct? Now I just need help understanding the matrix. What confuses me is how the 4 (gotten from 4f ' (t) ) is on the same row as the 2 (gotten from f '' (t) ).  Shouldn't the matrix be like this?:

0 0 2
0 4 8
0 0 0

Since both the 4 and the 8 are gotten from f ' (t). Just one last question which is bugging me also. Suppose the basis is switched up a bit, from: {1, t, t^2} to {t^2, t, 1}. The matrix would then be:

0 0 0
8 0 0
2 4 0

 :wacko:  How did that happen?
T(t^2) = f '' (t^2) + 4f ' (t^2) = 2 + 8t //first column
How did the 2 go on the bottom this time? If I was following the previous example, I would have thought that the first column of the matrix would be:

2
8
0

Can anyone shed light on these questions? Thanks a lot in advance!

Both the rows and columns of your matrix are labeled with the basis elements $ 1,t,t^2$. The columns are labeled with the version in the domain, which is why $ 2\plus{}8t$ is a column. The rows are labeled with the version in the codomain, which is why $ 4$ and $ 2$ are in the same row. Nothing is labeled with derivatives, so what derivative things come from doesn't really matter.

Your last step applies a change of basis; since $ T$ maps a space to itself, we apply that change of basis to both sides. I'll add an extra row and column for labeling:
$ \begin{pmatrix} &1&t&t^2\\1&0&4&2\\t&0&0&8\\t^2&0&0&0\end{pmatrix}$ becomes $ \begin{pmatrix} &t^2&t&1\\t^2&?&?&?\\t&?&?&?\\1&?&?&?\end{pmatrix}$ How do we fill that in? Since all we did was permute, we'll just permute the matrix entries; the $ (t,t^2)$ entry will be the $ (t,t^2)$ entry in whatever new position that is in, and so on. That $ (t,t^2)$ entry is the $ 8$; it was (second row, third column), but the new labels make it (second row, first column). The $ (1,t^2)$ entry is the $ 2$; (first row, third column) becomes (third row, first column). The $ (1,t)$ entry is the $ 4$; (first row, second column) becomes (third row, second column). The rest of the entries are zero, and the matrix becomes $ \begin{pmatrix} &t^2&t&1\\t^2&0&0&0\\t&8&0&0\\1&2&4&0\end{pmatrix}$.


geometry
integration
real analysis
real analysis unsolved

derive a formula for the volume of intersection of n cylinders having the same radii?  what about different radii for each?
2 and 3 cylinders is easy what about higher n?

You left out a lot of details, Tr, about such things as angles of intersection, or whether the axes of the cylinders are concurrent.  But here's a formula that takes care of a number of cases. Suppose that the cylinders all have the same radius, and that the axes of the cylinders all lie in the same plane. Consider for each line that is the axis of a cylinder the set of points in that plane that line within 1 unit of the line. The intersection of these sets is a convex polygon $P$. (If you have the lines concurrent and the angles equally spaced, then it will be a regular polygon; but that special case isn't necessary for what follows.) Let $A$ be the area of polygon $P$. Then the volume of the intersection of cylinders of radius $r$ and with these axes is

$2\int_0^r A\cdot (r^2-y^2)dy = \frac43 Ar^3$.


I had a really hard time dealing with word problems in school, so I'm approaching math a little differently with my son.  I taught him how to multiply without handing him a set of multiplication tables -- I want him to know that 3 x 4 means three sets of four objects, so he can multiply in his head, but he hasn't memorized the chart (yet).  Am I screwing him up for life or helping him understand the theory instead of simple memorization (which is my intent)?  What do you think is the best way to learn things like this?

I'm no expert, being only a kid, but when I learned to multiply, I first learned the concept of having a number of groups containing a certain number of objects each.  After understanding this, I memorized the multiplication tables.  This seems to have been effective, because I still understand what I am doing when I multiply, and I can also compute equations quickly.  =)

Yay!  You're teaching him concepts and not rote memorization!  Hooray, Cool and AWESOME! All my best math teachers did this. The bad ones were the ones who made us memorize stuff.

You are definitely not screwing him up for life. You've got a wicked smart little munchkin who most likely needs to understand the underlying principle of things  - in fact, I'm guessing that you'd have a terrible time stopping him from figuring it out on his own.  :D

You might also want to look into any local resources for gifted kids, or any internet based ones, too.  I'd recommend a search for gifted mailing lists. These can be wonderful places to get support for when you are tearing your hair out.

Lots of luck!

Elise

whoa...a teacher that teaches concepts....and their not even a teacher...KUDOS TO YOU!!!


[quote="drunner2007"]why do people memorize thousands of digits of pi and why do some professional mathematicians spend most of their time calculating digits of pi?

in either case, aren't there other theoretical problems that are more deserving of their attention? just wondering what everyone thinks[/quote]

You're right, but maybe they do it for fan and love, althought there are more important things that calculating $\pi$.........

Well I know only one student who had the $\pi$ bug and he unfortunately didn't make it through first year

I think 3.14159 is just fine

I know e=2.7182818284 by heart...because of this French mnemonic : 'Tu aideras à rappeler ta quantité à beaucoup de docteurs amis'='You will help by telling your quantity to many doctor friends' :roll:

The people that memorize: often crazy or have too much spare time...
The ones that calculate it: this is often not to really find out the digits but to test computers and algorithms. But well, most times you could do better things, right.


any hint on this problem?

If a and b > 2 are any positive integers, prove that 2^a + 1 is not divisible by 2^b - 1.

[hide="Solution"] This is clearly impossible if $a \le b$, so $a > b$.  Then $2^{b}| 2^{a}$, so

$2^{a}+1 \equiv (1)^{b-a}+1 \equiv 2 \bmod (2^{b}-1)$

QED. [/hide]
 In fact, this allows us to show that for $a > b$ we have $gcd(2^{a}+1, 2^{b}-1) = 1$.  This result is also true if $b$ is odd, even if $a \le b$.  Can you prove it?  (The only case where this doesn't hold, $a \le b, 2 | b$ is trivial, especially the case $b = 2a$.)

[hide="Solution"] [b]Lemma:[/b]  $gcd(a^{n}-1, a^{m}-1) = a^{gcd(n,m)}-1$

Then

$gcd(2^{2a}-1, 2^{b}-1) = 2^{gcd(2a, b)}-1$

Since $b$ is odd, $gcd(2a, b) = gcd(a, b)$, but since

$gcd(2^{a}-1, 2^{b}-1) = 2^{gcd(a, b)-1}$

We divide out and conclude that

$gcd(2^{a}+1, 2^{b}-1) = 1$ [/hide]


notice the equivalent form $n(pq+qr+rp)=pqr$ which allows us to notice $n<p,q,r$ so $p,q,r\not|n$ so $n=1$ and the only slution is $(p,q,r)=(3,3,3)$

Sorry, I didn't see the word "prime"  :blush:

[quote="eLeM"][quote="gollywog"]Find all $p,q,r$(primes) and $n$ (natural numbers) satisfying equation:

$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}= \frac{1}{n}$
I'm curious how will ya solve it[/quote]

Multiply equation by $p, q, r$.

$pq+qr+rp=\frac{pqr}{n}$

Since $p, q, r$ are prime, we have 4 cases:

1) $n=pqr$
2) $n=pq$
3) $n=p$
4) $n=1$

Case 1: $pq+qr+rp=1$ gives no solutions
Case 2: $pq+qr+rp=r$ gives no solutions, since $pq=r(1-p-q)$
Case 3: $pq+qr+rp=qr$ gives no sulutions, since $p(q+r)=0$
Case 4: $pq+qr+rp=pqr$

$qr=p(qr-q-r)$

We have that, $q=p$

It follows that:

$r=pr-p-r$
$p=r(p-2)$

From there $p=r$

So we have $p=q=r$ and $3p^{2}=p^{3}$. From there $p=3$

So we have only solution $(p,q,r,n)=(3,3,3,1)$[/quote]

Exactly my solution

btw. lol @ zetax, ur a bit tired today ain't ye?


[rove that, given two points, and a line not containing both, that there is a point on the line such that, when it and the other 2 points are connected to form an acute triangle.

[quote="nateharman1234"]I think I have a counter example, consider a line perpndicular to the line containing the two points, but where the intersection with that line is outside the two points.[/quote]

that's a very good counterexample, then the triangle drawn must be right, so the statement is proven false by counterexample.


Any idea as to how to calculate permutations using Google's calculator ??? I mean , you can say 10 choose 5 , but I dont see any simple option to calculate permutations...BTW , do any of you know any downloadable calculator that calculates permautations ??

http://www.google.com/help/calculator.html

That's all the functions in google calc... if it's not there then it's not on the calc...

People don't bother adding permutations because they are only trivially different from combinations.  If you understand the mathematics behind them, you can translate from one to the other without any difficulty.

or, u can just use the factorial signs... i find that much easier than looking for permutation and conbination symbols

Hey JBL ,
I do know what permutations and combinations are.Only looking for a feaure that can calculate them easily.....
BTW Mariag ,
Saying 10 choose 4 is much much easier than saying $10! / 4! * (10 - 4)!$

As far as permutations are concerned , of course, 10! / 4 ! is easy enough , but I somehow hate using the "Shift " key .


I want to attend the online classes which could help me through state competition.. But do you [b]have [/b]to pay for the money for classroom?

Math Jams are free...check the Math James section under "Community" for a list of upcoming ones...

However, all online classes cost money...yes...