this is in responce to a situation i'm going through

my parents wouldn't let me go to the duke math contest this weekend because:
a) there is a lot of traffic on the interstate (go figure)
b) the trip would consist of going through charlotte a little after rush hour. more (dangerous) traffic....
c) part of the trip would be a night

now remember that a responsible adult would be driving and that i'm 17 years old. my parents feel it is a great risk to my safety if i were to take the trip. i am very mad right now as i looked forward to going to this contest.  :mad:
so, how protective are your parents?

oh man you should elope (metaphorically speaking) DUKE IS PRETTY AWESOME (in math. in basketball, it is the devil)

my parents care too much about me and always need to know where i am. Which is good and bad.

My parents are really protective, at least compared to a lot of people I know.  I'm not allowed to go to the Homecoming dance, or any other school dance, not allowed to have a girlfriend untill I'm like 23, not allowed to go to any concerts, and theres a whole bunch of stuff.

But what's kind of funny is that I'd probably be allowed to go the Duke math contest.

My parents are selectively overprotective. They'd have no problem in putting me on a plane to some math competition, but I doubt they'd let me go to an out-of-state paintball tournament.

I don't wanna participate in your poll, because I am 21 (actually, I just became 21 TODAY  :D ), making me already three years adult in belgium, and now even in USA i can buy alcochol (which i prefer not to)

I was never much of a party goer, or someone who would get into trouble
My parents may be more protective than others, but not a lot was enforced, because I wasn't pushing the boundaries
When I was seventeen I could definitely go away at night or stuff.  I also participated in the IMO 2002, however we were overprotected  ;)  by the IMO people there  :D .

But do I get this straight?  Will you be guided by an adult (for god sake you are an adult minus epsilon :o ) ALL the time?  What I find strange is that you seem to say that the ride is the biggest problem.  Perhaps your parents don't trust the driving skills of the person that will take you?

I really hope they don't do this again, I sympathise with you, they should realise that this is exactly the kind of thing you need, both on a math level as a personal level, to get out and see some new stuff

I am grounded for the rest of my life without doing anything (or the equivalent of grounded).

My parents really don't mind what I do as long as I have a fairly decent reason and tell them (i.e I can skip school whenever I want and go wherever, whenever).

I like the system.

[quote="yif man12"]While you're there, might as well get laid and elope the real way too.[/quote]

oh man that'd be awesome I should do that at ARML and be all like whoaaa

Right...so let's stay off the stuff about getting laid (despite how Mystic doesn't know what it is). And also the partying bit...

This won't help the original poster but it certainly is not extreme as other propositions here

apparently when the IMO was held in cuba, one belgian participant connected with a cuban team member and flew to cuba each year for her!
when i went to IMO2002, my team leader told me they were married and lived in paris! :huh:

My parents are in between. It really depends compared to whom. They have really odd reasons for not letting me do specific things, but other things they are like whatever.

[quote="furious"]
now remember that a responsible adult would be driving and that i'm 17 years old. my parents feel it is a great risk to my safety if i were to take the trip. i am very mad right now as i looked forward to going to this contest.  :mad:
so, how protective are your parents?[/quote]

I went to the Duke Math Meet with Joe and Mr. R.
Are you seriously suggesting Mr. R is a responsible adult? :P

Let me give you an example.  Mr. Neal was following our car in his own, so Mr. R was giving him fake blinkers to try to get him to make a wrong turn.  Mr. R might be a genius, but he's still a kid.

[quote="mysmartmouth"]I went to the Duke Math Meet with Joe and Mr. R.
Are you seriously suggesting Mr. R is a responsible adult? :P

Let me give you an example.  Mr. Neal was following our car in his own, so Mr. R was giving him fake blinkers to try to get him to make a wrong turn.  Mr. R might be a genius, but he's still a kid.[/quote]
joking around and being responsible are two completely different things. and yes, i am suggesting that Mr. R is a responsible adult.

[quote="mysmartmouth"]I was just kidding.  But ask Joe about the comment card Mr. R filled out at Wendy's. :rotfl:  :rotfl:[/quote]

lol, that was great...but it was Hardee's :|

Ah, it's a shame you weren't able to go, David   :moose:

[quote="fredbel6"]the roughest thing I did  :o  on IMO was going to a cybercafe, 200 metres away from the designated mall!! :ninja:

The german team however went 'marching' throught town :|[/quote]

'marching'? Does this imply some select instances of nonmarching?

Er...back on topic please. You can talk about Duke somewhere else.

[quote="furious"]now remember that a responsible adult would be driving and that i'm 17 years old. my parents feel it is a great risk to my safety if i were to take the trip. i am very mad right now as i looked forward to going to this contest.  :mad:[/quote] honestly, I'd simply say 'see you tomorrow', and go. You'd be really stupid to listen to that.

If the danger is not real, just go, they cannot do you anything when you return, as nothing happened. As protective parents, they should be happy to see you back well.

Mine also used to be very protective... they indeed need a little push forward from time to time. ;)

About the marching, yes I do recall a moment when a german was not marching, curiously enough when playing mafia with him :ninja:

But back on topic : I don't know Peter, simply disobeying can strain relations? Although the poster is seventeen of course. It is my perception that European youngsters here can in general do more than americans?  Or am I wrong?

[quote="Peter VDD"]honestly, I'd simply say 'see you tomorrow', and go. You'd be really stupid to listen to that.[/quote]
obviously, you don't know my parents  :P
i asked them after the team left what would happen if i went anyways. they would have drove to duke and picked me up that night and not allowed me to compete the next day. then they would take my car, xbox, tv, and probably computer. they would have made my life totally horrible. sad, isn't it.  :help:

[quote="fredbel6"]
It is my perception that European youngsters here can in general do more than americans?  Or am I wrong?[/quote]

Interesting question. Well, I also get the impression that Asian parents in general are more protective, but it could just be a generalization.

My parents are okay. They just are willing to do anything to prevent me from going on AoPS and buying those AoPS books.

Yeah, it seems there is a culture clash here
Most parents here are extremely proud when their kid goes to olympiad training or to competitions, and help them doing that (my father and mother helped me out to reach certain locations several times)
Both my parents have studied , but there also parents who haven't studied that are happy to see their child doing wel
I actually know some parents that support olympiad, but only find it sad to see their child hanging out again in the quite "tame" world of mathematicians (you know what I mean, going to an olympiad training isn't the same as hanging out with some strangers in the middle of the night in a rough neighbourhood)

But furious, you said something about your car?
I do not understand?  You can't go with an adult but they do let you drive?  I know in USA minors can drive in some states, but still...

Belgium is so different, minors buying alcochol is extremely common, but a 16 year old driving is unimaginable

Minors can drive in every state, I believe, except New Jersey where you can't get a regular license until you're 18.  But the large majority of states set 16 as the regular license age.

furious' parents let him drive to school, around town, and to other places when he needs to.  I believe their major issue was not necessarily the drive per se, but the fact that he would have been in a car at night going through a big city, Charlotte.  The traffic through Charlotte wasn't at all bad...we went [i]by[/i] Charlotte...not through the downtown part of it or something.

Anyway, what I don't understand is how furious' parents can leave him home alone for a week but not let him go on a [b]math competition[/b] at Duke.  Also keep in mind that they let him go to Penn State for ARML over the summer, although I don't think there was much night driving...

my parents are very strict (not THAT overprotective)
when i got a cut from soccer, they scold me

[quote="fredbel6"]It is quite obvious I assume that the debated contest has passed and that the damage has been done now... :( ?[/quote]

Yes, the contest already took place.

He never said he was older than 65. You just kind of presumed that.
For all we know, he could've been an "accountant" when he was 8, in some game he played, then he retired or something.

So, maybe he wasn't lying.  :lol:

[quote="236factorial"][quote="MysticTerminator"][quote="fredbel6"]Forgive my poor English, scolding is just blaming right, or is it more agressive, also involving yelling

and why are YOU scolded for a cut from soccer, do i understand they blame you for a wound at football

seems i have the easiest parents on this forum? :?[/quote]

eh my parents would probably let me do whatever, I just don't WANT to do anything sketchy in particular[/quote]

I thought you were a retired accountant (age>65)?  :roll:  See, I knew you were lying.[/quote]

Um, maybe he still lives with his parents?

um, I don't see how I contradicted myself? AT ALL? your parents don't necessarily die when you hit 50 or anything; c'mon now be serious, people! :)

No, but it'd be very worse if they'd still be overprotective when you were 17...


I have seen a few trolls and spammers in GS and other places. Don't encourage them.

A few suggestions:
1) Post sources for every problem
2) latex everything

Also, the following article contains some good info:
[quote]
Netiquette (neologism, a portmanteau formed from "Internet etiquette") is a catch-all term for the conventions of politeness recognised on Usenet, in mailing lists, and on other electronic forums such as internet message boards. These conventions address group phenomena (such as flaming) with changes in personal behaviour, such as not posting in all uppercase, not (cross-)posting to inappropriate groups, refraining from commercial advertising outside the biz groups and not top-posting. RFC 1855 is a fairly lengthy and comprehensive set of such conventions.

The rules of netiquette are slightly different for newsgroups, web forums and IRC (Internet Relay Chat). For example, on Usenet it is conventional to write in standard English and not use abbreviations such as "u" for "you" or "ne1" for "anyone". These abbreviations are only slightly more likely to be tolerated on web forums, but are almost universal on IRC where, since discussion is real-time, they serve the practical purpose of speeding the flow of conversation. Many IRC users look down on this form of conversation, though. Issues such as the level of tolerance for off-topic discussion or spoilers may also vary from one newsgroup, forum, or channel to another. The rule of thumb in any of these discussion mediums is to "lurk before you leap"—get a feel for the local conventions before diving into conversation and inadvertently embarrassing oneself. Also, read the FAQ if there is one.

The following is a list of some of the more common rules associated with netiquette on Usenet.

    * The most important rule of netiquette is, "Think before you post". If what you intend to post will not make a positive contribution to the newsgroup and be of interest to several readers, do not post it! Personal messages to one or two individuals should not be posted to newsgroups—use private e-mail instead.
          o This is sometimes stated in other forms, such as "Remember the human."

    * When following up on an article, quote the minimum text necessary to give some context to your reply and be careful to attribute the quote to the correct person. If the article to which you are responding was cross-posted to several groups, edit the distribution ("Newsgroups:") header to contain only those groups which are appropriate to your reply, especially if the original message was posted to one or more inappropriate groups in the first place.

    * Re-read and edit your posting carefully before you post. Check the spelling, grammar, and capitalization. Typing in all capital letters usually denotes screaming or yelling. Keep your lines to fewer than 70 characters. Do not post test messages except in their appropriate groups; wait until you have something to say. When posting humorous or sarcastic comments, it is conventional to append an emoticon, but do not overuse them.

    * Before asking a question, read the messages already in the group and read the group's FAQ if it has one. When you do post a question, follow it with, "Please reply by mail and I will post a summary if requested," and make sure you do post a summary if requested; or, if only a few people were interested, send them a summary by mail. This avoids umpteen people posting the same answer to the group and umpteen others posting "me too"s.
          o However, many newsgroups follow the rule "Post here, read here" (that is, they explicitly discourage posters from requesting offline answers to topical questions). This is because there may be lurkers who would benefit from seeing the discussion unfold online, and because questioners often do forget to post summaries of their findings.

    * Be proud of your postings, but do not post just to see your name in pixels. Remember: your future employer may be reading.

There is also a netiquette rule for how to deal with someone who has violated one of the rules of netiquette:

    * If you believe someone has violated netiquette, send him or her a message by private e-mail; do not post a follow-up to the offending post. Be polite. The author may not have realized his or her mistake, may be a beginner, or may not even have been responsible for the "crime" — his or her account may have been used by someone else, or the address may have been forged. Furthermore, a person who breaks netiquette over and over may be doing it intentionally to disrupt the group (see Internet troll), in which case public flaming over the violation would amount to what is termed feeding the troll.

[/quote]

That article should be cited: http://en.wikipedia.org/wiki/Netiquette .

perhaps there should be something in there about not deviating from the topic and posting text that has no correlation to the topic at hand.

so as to high school basics, is it mostly just geometry and  algebra, excluding trigonometry and algebra II?

No, trig problems are welcome, as are most algebra II problems.

Don't worry-those Qs are appropriate to HSB. Also, welcome to aops!

[quote="mj93"]oh woops.  I'm brand new to the forum and I think the question I posted might go in the intermediate section... what do I do?[/quote]

If you do indeed post a problem in the wrong section by accident, it is best to notify a moderator of the forum in which you posted the problem (i.e. PM).  Most of the time, problems will be noticed relatively soon, but if you notice your mistake, it is always better to ask a moderator.

[quote="LevatorNasolabialis"]I did the same thing but I put to hard of a q in the middle school area[/quote]

Then notify the moderators of whatever forum you posted the problem in.

Also, please try to use correct grammar / spelling / punctuation.

This is just a reminder that if spam/commerical advertising is posted, please DO NOT reply. Replying will just bump the topic so more people will see it. Just send a PM to a mod and we'll take care of it  :wink: .

[quote="10000th User"]But, can also mods get the admins attention so that these abusers can be IP-banned permanently or other stronger measures?[/quote]

We generally attempt to do so, though the admins don't always respond immediately...

[quote="soltixnj"]Has anyone actually ever been banned from this website?[/quote]

Naturally.


trigonometry
geometry
area of a triangle
Heron\u0027s formula
inequalities unsolved
inequalities

[quote="gotztahbeazn"]Find, with proof, the minimum value of $(x+y)(y+z)$, where $x,y,z$ are positive real numbers satisfying the condition $xyz(x+y+z) = 1$.[/quote]
Easy. We have $(x+y)(y+z)=xz+y(x+y+z)\ge 2\sqrt{xyz(x+y+z)}$

[quote="gotztahbeazn"]Find, with proof, the minimum value of $(x+y)(y+z)$, where $x,y,z$ are positive real numbers satisfying the condition $xyz(x+y+z) = 1$.[/quote]
Or:
From: $xyz(x+y+z)=1$
$\longleftrightarrow xz(x+y)(y+z)=1+x^2z^2$
$\longleftrightarrow (x+y)(y+z)=\frac{1+x^2z^2}{xz} \geq 2.$
It's too easy. :(

yeah... easy problem... this is ussr '89 (10th grade) too...

a partner of mine find a nice proof to this one... consider $\triangle ABC$ with sides $a=x+y, b=x+z, c=y+z$, then by heron's formula and the condition we have $(ABC)=1$, but, $1=\frac{ac\sin B}{2}\leq \frac{(x+y)(y+z)}{2}$. the result follows.


induction
linear algebra
matrix
combinatorics proposed
combinatorics

Let $A$ be the set of $n$-digit integers whose digits are all from $\{ 1, 2, 3, 4, 5 \}$. $B$ is subset of $A$ such that it contains digit $5$, and there is no digit $3$ in front of digit $5$ (i.e. for $n = 2$, $35$ is not allowed, but $53$ is allowed). How many elements does set $B$ have?

maybe there is a smarter way, but this is a more straightforward way.
Let $x_n,y_n,z_n$ denote respectively the number of good tuples of length $n$, starting with $3,5$ and neither 3 nor 5 respectively. then we have the following relation(these may or may not contain 5):
 $\left(\begin{array}{ccc}{ccc} 1&0&1\1&1&1\1&1&1\end{array}\right)\left(\begin{array}{ccc} x_ny_nz_n\end{array}\right)=\left(\begin{array}{(ccc)}x_{n+1}y_{n+1}z_{n+1}\end{array}\right)$.
Also we have $\left(\begin{array}{c}{ccc}x_{1}y_{1}z_{1}\end{array}\right)=\left(\begin{array}{(ccc)}1\1\3\end{array}\right)$.
one can check by induction that if the above matrix is $A$, then $A^{n+1}=\left(\begin{array}{ccc}{ccc}F_{2n}&F_{2n-1}&F_{2n}\F_{2n+1}&F_{2n}&F_{2n+1}\F_{2n+1}&F_{2n}&F_{2n+1}\end{array}\right)$
 where $F_n$ is the $n$th Fibonacci number($F_0=F_1=1,F_2=2$ etc)
 what we need is of course $\left(\begin{array}{ccc}{(c)}1&1&1\end{array}\right)A^{n-1}\left(\begin{array}{(ccc)}1\1\3\end{array}\right)=F_{2n}+3F_{2n-1}$.
to include $5$'s we just throw away all the sequences without $5$'s from this, which is just $4^n$.
(hopefully everything is alright)

Isn't it like this?
Let $a_n$ be n-digit without $3$ in front of $5$, and $b_n$ be n-digit without $3$ in front of $5$ and ending with $3$
We have $a_{n+1} = 5(a_n-b_n)+4b_n$
and $b_{n+1} = a_n$
And we get $a_{n+1}-5a_n+a_{n-1} = 0$
So $a_n = A(\frac{5+\sqrt{21}}{2})^n+B(\frac{5-\sqrt{21}}{2})^n$ where $A,B$ are constants we have to solve.
And then we take out the $4^n$ without a $5$.


geometry
analytic geometry
trigonometry
quadratics
Pythagorean Theorem
geometric series
complex numbers

Here are some nice problems from this local competition.  Some very easy questions, although some very unique and challenging ones too.  Here's the cream of the crop:

individual round 1999
1
A list of consecutive positive integers beginning with 1 is written on the blackboard.  One number is erased.  The average of the remaining numbers is $35 \frac{7}{17}$.  What number was erased?

2
Triangle ABC has points D, E, and F on AB, BC, and CA respectively, such that AD = 2, and DB = 3 and the area ABC = 10.  If the area of ABE equals the area of DBEF, find the area of ABE.  (I don't know if this will help, but the figure given has the lines AE and DF drawn in.)

3
A boat has sprung a leak, and water is coming in at a uniform rate.  Some water has already accumulated before the leak is discovered.  At this point, 12 men can pump the boat dry in 3 hours, while 5 men would require 10 hours.  How many men are required to pump it dry in 2 hours?

4
All right... I'll try my best to explain this one, as it relies on the diagram.  An equilateral triangle ABC intersects a circle, such that all sides of the triangle go thru the circle (hardest part to explain...).  Points F, G, H, J, D, and E all lie on both the circle and the triangle.  G and F are on AC (G closer to A) such that AG = 2, GF = 16, FC = 3.  H and J are on AB (H closer to A) such that AH = 3.  D and E both lie on BC (D closer to B, although it doesn't matter).  Find DE.  Someone could make a diagram of that (hint).

5
A unit square has an equilateral triangle inscribed inside, with one vertex of the triangle shared by the square.  The other two vertices of the triangle are on sides of the square, not on a vertex.  Find the area of the triangle.

6
The salt concentration in a solution is measure in pounds of salt per gallons of water (lb/gal).  A 60 gallon tub is initally empty.  One solution enters the tub from pipe A at 2 gal/min with a salt concentration of 0.1 lb/gal.  Another solution enters the tub from pipe B with a salt concentration of 0.4 lb/gal.  After 5 minutes, the salt concentration in the tub of 0.3 lb/gal.  What was the flow rate of pipe B in gal/min?  (It doesn't specify, but I think you should assume the tub is infinitely big... no overflowing.)

team round 2002
7
Start at the origin in the x-y coordinate system.  First, go along the angle 0 degrees for a length of 1 unit.  Then move from that point along the angle 1 degree for a length of 1 unit.  Continue in this fashion, incrementing by 1 degree each time until you go to moving at a 90 degree angle for a length of 1 unit.  Find the x-coordinate of the final point reached.

8
Let $a_n$ be the nth positive integer k such that the greatest integer not exceeding $\sqrt k$ divides k.  For example, $a_1 = 1$, $a_2=2$, $a_3=3$, $a_4=4$, $a_5=6$, $a_6=8$, $a_7=9$, $a_8=12$, etc.  Find $a_{20}$, $a_{10000}$, and a formula for $a_n$.

9
Find all solutions to the system:

$a^2 + bc = -1$
$ab + bd = 0$
$ac + cd = 0$
$ac + d^2 = -1$

[hide="Problem 1"]  Since the average of any arithemetic sequence from 1 to an integer  $x$ is $\frac{x+1}{2}$, we can guess that the sequence ends at about 70.  Since the mean has a fraction with a 17 in the denominator, the sequence's number of terms is one more than a multiple of 17.  Than there are 69 original terms.  We simply find that $69\times 35=68\times (35+\frac{7}{17})+x=2145=2148+x$ and so $x=7$.[/hide]

[hide="Problem 5"]
We know that one side of the triangle is 1, as it shares the side with the unit square.  Let the altitude be $h$, then by the Pythagorean Theorem, we know that $1^2=h^2+\left(\frac{1}{2}\right)^2$ or $h=\sqrt{\frac{3}{4}}$.  Then the area is $\frac{\left(\frac{\sqrt{3}}{2}\right)\times 1}{2}=\frac{\sqrt{3}}{4}$.[/hide]

Samath, on 5, the triangle and square share a VERTEX, not side.  Perhaps I should have specified further... I'll edit the original post.

[hide="3"]
Suppose each man can pump $q$ water per hour .Let the rate of water flowing in be $k$ and the initial volume of water in the boat be $V$ . If $n$ is the number of man , $t$ is the time taken to pump out all the water , we have the following equation

$V+kt=nqt$ since the volume of water going into the boat must equal to volume of water pump out the boat .

first equation yields $V+3k=36q$
second equation $V+10k=50q$

Solve them and obtain $V=30q$ , $k=2q$

Hence when $t=2$ , $30q+(2q)(2)=nq(2)$ $\implies n=17$
[/hide]

I'm sorry I posted it with "hide", but it didn't hide.(Maybe some typo.
I'm posting solution to 4.

[hide]
We know that from an external point A,
$AT^2=AB.AC$,  where AT is the length of the tangent, ABC is a secant and B and C lie on the circle.

Thus on drawing the diagram(which I am unable to in this solution), we get
$AG.AF=AH.AJ$
$\Rightarrow AJ=12$
Since the triangle is equilateral, $BJ=6$
Similarily, $BJ.BH=BD.BE=108$ and $CE.CD=CF.CG=57$

I was unable to solve after this. Please if you can solve then post.
I am posting what I have tried. :P [/hide]

[hide="Problem 2"]
$\triangle ABE=\triangle BDE+\triangle ADE$
$BDFE=\triangle BDE+\triangle FDE$

So $\triangle ADE=\triangle FDE$

If we regard $DE$ as the base of $\triangle ADE$ and $\triangle FDE$, the triangles must have the same height, which implies that $AF\parallel DE$.

$AC\parallel DE\Longrightarrow\frac{AD}{DB}=\frac{CE}{EB}$

$\therefore BE=\frac{3}{5}BC$

$\therefore \triangle ABE=\frac{3}{5}\triangle ABC=\boxed{6}$[/hide]

Varun, there are no tangents.  The sides are all intersected.  Nice solutions guys.
My teacher found the problem solving packet (team round) from 2002... so I'll add the good ones to the list also.

Bump... I would like to see solutions to some of these problems.

OK I agree with all answers posted except samath's problem #5.
[hide="#5"] If the triangle intersects the side of the square $x$ from the vertex, the sidelength of the equilateral triangle is $\sqrt{1+x^2}$, but you will also see that the side length is also $\sqrt{2(1-x)^2}$.  Solving this yields $x=2-\sqrt{3}$, thus the area of the triangle is $28\sqrt{3}-48$.[/hide]
I loved number 9.
[hide="#9"] $b(a+d) = 0, c(a+d) = 0$ this means we can solve it for two cases $a+d=0$ and $b=c=0$.  In the case where $b=c=0$, $a^2 = -1 \rightarrow a = \pm i$ and $d^2=-1\rightarrow d = \pm i$, which yield the ordered quadruples $(i,0,0,\pm i)$ and $(-i,0,0,\pm i)$.  The other case is when $a = -d \rightarrow ac+a^2 = -1$ and $a^2+bc=-1$.  This yields $(b-a)c = 0$, which comes to two sub cases $a = b$ and $c = 0$, with $c = 0$ we end up with $(\pm i, x, 0, \mp i)$, so when $b=a, a^2+ac+1 = 0 \rightarrow c = -\frac{1}{a}-a$, which yields the final quadruple $(a,a,-\frac{1}{a}-a,-a)$.[/hide]

Unfortunately, I do not have the official solutions, only my own, and those only for some... 5 I do not think is right, I think you mean $\sqrt{x^2 + 1}$ at the beginning.  9 probably is right... you got a lot more solutions than me  :lol:

[quote="deej21"]

6
The salt concentration in a solution is measure in pounds of salt per gallons of water (lb/gal).  A 60 gallon tub is initally empty.  One solution enters the tub from pipe A at 2 gal/min with a salt concentration of 0.1 lb/gal.  Another solution enters the tub from pipe B with a salt concentration of 0.4 lb/gal.  After 5 minutes, the salt concentration in the tub of 0.3 lb/gal.  What was the flow rate of pipe B in gal/min?  (It doesn't specify, but I think you should assume the tub is infinitely big... no overflowing.)

team round 2002
7
Start at the origin in the x-y coordinate system.  First, go along the angle 0 degrees for a length of 1 unit.  Then move from that point along the angle 1 degree for a length of 1 unit.  Continue in this fashion, incrementing by 1 degree each time until you go to moving at a 90 degree angle for a length of 1 unit.  Find the x-coordinate of the final point reached.

8
Let $a_n$ be the nth positive integer k such that the greatest integer not exceeding $\sqrt k$ divides k.  For example, $a_1 = 1$, $a_2=2$, $a_3=3$, $a_4=4$, $a_5=6$, $a_6=8$, $a_7=9$, $a_8=12$, etc.  Find $a_{20}$, $a_{10000}$, and a formula for $a_n$.
[/quote]
[hide="#6"] rate of salt (lb/min) = 2 gal/min * 0.1 lb/gal = 0.2lb/min, following in this fashion we know that $\frac{5min*0.2lb/min + 0.4lb/gal*B*5min}{5min*2gal/min+5min*B} = 0.3 lb/gal$ lookee here it doesn't matter that it took 5 minutes, by some algebra B = 4 gal/min.[/hide]
ok here's a cute one.
[hide="#7"]The problem breaks down into the sum $\cos(0^\circ)+\cos(1^\circ)+...+\cos(89^\circ)+\cos(90^\circ) = \sum_{i = 0}^{90}\cos(i)$  It is important to know that $\frac{\sin(a+b)-\sin(b-a)}{2} = \sin(a)\cos(b)$.  Therefore $\frac{\sin(0.5^\circ)}{\sin(0.5^\circ)}( \sum_{i = 0}^{90}\cos(i)) = \frac{1}{2\sin(0.5^\circ)}(\sin(0.5^\circ)-\sin(-0.5^\circ)+\sin(1.5^\circ)-\sin(0.5^\circ)+...+\sin(90.5^\circ)-\sin(89.5^\circ)) = \frac{1}{2\sin(0.5^\circ)}(\sin(90.5^\circ)-\sin(-0.5^\circ)) = \frac{\sin(0.5^\circ)+\cos(0.5^\circ)}{2\sin(0.5^\circ)} = 0.5(1+\cot(0.5^\circ))$.[/hide]

[quote="deej21"]8
Let $a_n$ be the nth positive integer k such that the greatest integer not exceeding $\sqrt k$ divides k.  For example, $a_1 = 1$, $a_2=2$, $a_3=3$, $a_4=4$, $a_5=6$, $a_6=8$, $a_7=9$, $a_8=12$, etc.  Find $a_{20}$, $a_{10000}$, and a formula for $a_n$.
[/quote]
After this I will need to do 2 and 4, and I will be contented. Mwahaha.
[hide="#8"]Ok this one was pretty sweet.  I will write only a small amount but you can prove it to yourself if you find it necessary.
$a_0 = 0, a_{3k+1} = a_{3k}+1, a_{3k+2} = a_{3k+1}+\sqrt{a_{3k+1}}, a_{3k+3} = a_{3k+1}+2\sqrt{a_{3k+1}}$. From this behavior, you know that: $a_{3k+1} = (k+1)^2, a_{3k+2} = (k+1)(k+2), a_{3k+3} = (k+1)(k+3)$, which can be simplified to one expression: $a_{3k+x} = (k+1)(k+x)$ for $x,k$ integers and $1\le x \le3$. Now that means $a_{20} = a_{3*6+2} = (7)(8) = 56, a_{10000} = a_{3*3333+1} = (3334)(3334) = 11115556$.[/hide]

I'm tired and I'm not sure how to do #2, so I'm just going to post my answer to #4, [hide="answer to #4"] $4\sqrt{7}$[/hide]

Hm, I'd like to see how you got 4.  Anyway, interesting solution to 7.  What I did was add $i \sin \theta$ to each term, then you have powers of one complex number, and I found the sum of that geometric series, and took the real part of it.  I get a slightly different answer (same value though) of $\frac{1 + \sin 1 - \cos 1}{2 - 2\cos 1}$

[quote="deej21"]
2
Triangle ABC has points D, E, and F on AB, BC, and CA respectively, such that AD = 2, and DB = 3 and the area ABC = 10.  If the area of ABE equals the area of DBEF, find the area of ABE.  (I don't know if this will help, but the figure given has the lines AE and DF drawn in.)

4
All right... I'll try my best to explain this one, as it relies on the diagram.  An equilateral triangle ABC intersects a circle, such that all sides of the triangle go thru the circle (hardest part to explain...).  Points F, G, H, J, D, and E all lie on both the circle and the triangle.  G and F are on AC (G closer to A) such that AG = 2, GF = 16, FC = 3.  H and J are on AB (H closer to A) such that AH = 3.  D and E both lie on BC (D closer to B, although it doesn't matter).  Find DE.  Someone could make a diagram of that (hint).
[/quote]
I like the $e^{i\theta}$ approach to #7, although I think telescoping is very satisfying.
Alrighty, now that I'm not so tired, I will show my solution to #4.
[hide="#4"] This problem uses only three ideas: power of a point, definition of an equilateral triangle, and how to solve seemingly ugly algebra.  Power of a point: $AG*AF = AH*AJ \rightarrow 2*18 = 3*(HJ+3) \rightarrow HJ = 9$. Equilateral: $AG+GF+FC = AH+HJ+JB \rightarrow JB = 9$. For the sake of simplicity let $BD = a, DE = x, EC = b$.  Power of a point: $BJ*BH = BD*BE \rightarrow 162 = a(a+x)$ and $CF*CG = CE*CD \rightarrow 57=(b)(b+x)$. Equilateral: $a+x+b = 21$.
\[ a(a+x) = 162 \] \[ b(b+x) = 57 \] \[ a+x+b = 21 \] Therefore, $a(21-b) = 162$ and $b(21-a) = 57$, now we have 2 equations-2 unknowns.  Subtraction yields, $21a-21b = 105 \rightarrow a-b = 5 \rightarrow a(26-a) = 162 \rightarrow a=13 \pm \sqrt{7} \rightarrow b = 8 \pm \sqrt{7} \rightarrow x = 21-21 \mp 2\sqrt{7} \rightarrow x= 2\sqrt{7}$.
This doesn't agree with the answer I initially wrote, because last night I couldn't solve a quadratic very well.[/hide]
Well...I'm still working on #2, has anyone solved it?

[quote="FMako"]Well...I'm still working on #2, has anyone solved it?[/quote]
Yeah, I did- See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=355687#p355687]post #7[/url] ;)

Ok good I was at least approaching it the right way.  I did it that way, but with a lot of unnecessary steps and other triangles.


Observations:

Most of the lifeforms on this planet are six-legged.

A smaller group of lifeforms are four-legged and are more advanced.

A few billion two-legged lifeforms are even more highly developed,
both intellectually and technologically.

Conclusion:

The dominant species on this planet is a dozen snakes.

Bacterium don't have legs either, do they? Unless flagella can be counted as legs...

Plants? Fungi? Protozoa? Bacteria? Archibacteria?

A human with his arms and legs chopped off?

come on guyz...it was just a joke...let's not overanalyze


For how many positive integers $ n$ does $ 1 \plus{} 2 \plus{} \cdots \plus{} n$ evenly divide from $ 6n$?

$ \textbf{(A)}\ 3\qquad
\textbf{(B)}\ 5\qquad
\textbf{(C)}\ 7\qquad
\textbf{(D)}\ 9\qquad
\textbf{(E)}\ 11$

The sum of the first $ n$ integers is $ \frac{n(n\plus{}1)}{2}$ so we want to find the cases when $ \frac{6n}{\frac{n(n\plus{}1)}{2}} \equal{} \frac{12}{n\plus{}1}$ is an integer. The answer is $ \textbf{(B)}\ 5$

(adding onto above note that $n$ needs to be positive so that is why the answer is $5$ not $6$)

It'd be really funny if MAA had made an option for $6$


It's similar:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59758
but easier. :wink:

[quote="Vasc"]I don't think it is easier.
[/quote]
I meant it. Certainly your inequality is harder. :lol:

It's not hard by Lagrange multiplier or even EV-method. But I hope there's more elementary solution. Do you have a elementary solution, VASC?

[quote="Vasc"]I don't think it is easier.
Show that the inequality is not true for $n=82$. :wink:[/quote]
Setting $x_{1}= ... = x_{n-1}$ and $x_{n}$ as large as possible doesn't give counterexample until $n = 223$. Even when I have computer [b]Vasc[/b] defeats me. :P

[quote="Vasc"]In the homogeneous inequality for $n=81$, equality occurs again when $x_{1}=x_{2}=...=x_{n-1}=1$ and $x_{n}=2$.
For this reason I called it "very special inequality".
For $n=82$, the homogeneous inequality is false when $x_{1}=x_{2}=...=x_{n-1}=1$ and $x_{n}=2$.[/quote]
When $n=81$, $x_{1}=x_{2}=...=x_{n-1}=1$ , $x_{n}=2$.
Then $x^{6}_{1}+x_{2}^{6}+\cdots+x_{n}^{6}=80+64=144$???
Am I miss  something :|

PS:but your inequality is equal to:
$x_{1}^{6}+x_{2}^{6}+\cdots+x_{n}^{6}=n$(n \le 81)
then:
$x_{1}^{5}+x_{2}^{5}+\cdots+x_{n}^{5}\ge x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}$

Then it can apply the half convex and half concave theorem:)
And it will reduce it to n-1 case like your n-1 EV-method.

[quote="Vasc"][quote="zhaobin"][quote="Vasc"]In the homogeneous inequality for $n=81$, equality occurs again when $x_{1}=x_{2}=...=x_{n-1}=1$ and $x_{n}=2$.
For this reason I called it "very special inequality".
For $n=82$, the homogeneous inequality is false when $x_{1}=x_{2}=...=x_{n-1}=1$ and $x_{n}=2$.[/quote]
When $n=81$, $x_{1}=x_{2}=...=x_{n-1}=1$ , $x_{n}=2$.
Then $x^{6}_{1}+x_{2}^{6}+\cdots+x_{n}^{6}=80+64=144$???
Am I miss  something ?[/quote]
Yes.
I wrote the "homogeneous" inequality, that is
$x_{1}^{6}+x_{2}^{6}+...+x_{n}^{6}\leq n(\frac{x_{1}^{5}+x_{2}^{5}+...+x_{n}^{5}}{x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}})^{2}$.[/quote]
Thanks :)


geometry
trapezoid
circumcircle
geometry proposed

Let $C_1$ and $C_2$ be two circles with respective centers $O_1$ and $O_2$.
These circles intersect in two distinct points A and B.
The line $O_1 B$  and resp. $O_2 B$ cuts again $C_2$ and(resp. $C_1$) at the
point F and
(resp. E).
The parallel to (EF) passing through B meets $C_1$ and $C_2$ respectively
in M
and N.
If B is between M and N, prove that :
MN = AE + AF.

The triangles $\triangle BO_1O_2 \sim \triangle AMN$ are similar, because $\angle AMN \equiv \angle AMB = \frac{AO_1B}{2} = \angle Q_2O_1B$ and $\angle ANM \equiv \angle ANB = \frac{AO_2B}{2} = \angle Q_1O_2B.$ The quadrilateral $O_1O_2FE$ is cyclic, because the isosceles triangles $\triangle O_1BE \sim \triangle O_2BF$ having the same base angles at the common vertex B are similar. Consequently, the triangles $\triangle AMN \sim \triangle BEF$ are also similar. Since their corresponding sides $MN \parallel EF$ are parallel, they are centrally similar, so that the corresponding sides $BE \parallel AM,\ BF \parallel AN$ are also parallel. Thus ABEM, ABFN are both cyclic trapezoids with the circumcircles $(O_1), (O_2)$, i.e., both are isosceles and with equal diagonals AE = BM and AF = BN, respectively. It follows that $MN = BM + BN = AE + AF.$


Well, I make games, and for one I need the graph equation that includes these folowing points:
(2.5,4)
(5,2)
(10,1)
(20,0.5)
(40,0.25)

You get the patern (when x=10, I want y to equal 1) I just can't figure out the equation for it, Please help me out.

Thanks,
-Puffs

Notice: $xy=10$.
So, a possible graph equation would be the hyperbola $y=\frac{10}{x}$.


AMC USA(J)MO USAMO search geometry incenter trigonometry

I've noticed that a lot of Olympiad problems use theorems or formulas that I've never heard of, but that are quite useful for solving the problems. I've never learned most (if not all) of these, but I'm hoping that I can do well on the 2006 USAMO, assuming I get that far. What are the most commonly used theorems/formulas? Links to Mathworld are fine, I'm just not sure what to search for. :)

Most Powerful Geometry Theorem:
Pythagorean Theorem :D

Otherwise its really important to know the all the properties of incenters, othercent, circumcenter. Also cyclic quadralaterals. Make sure you can do some trig manipulation

Look at the "Theorems and Formulas" forums in the Olympiad Section.

Just for kicks, here is a list of recent USAMO problems that require no (common) theorems whatsoever to solve. That is, you never really use anything that you just cite off the top of your head. (aka Cauchy, Ceva, Fermat, etc.)

In 2005
#1
#4
#5
#6

In 2004
#2
#4

In 2003
#1
#3
#6

In 2002
#1
#2
#3
#4
#5
#6

In 2001
#1
#5

In 2000
#1
#3
#4

Feel free to argue with my logic :)

This year, #3 could also be solved without any real theorems past the really basic geo ones.  #1 in 2004 used only law of cosines and trivial inequality... I seem to recall that 2004's #3 was also fairly elementary.  So yeah knowing theorems isn't that important.  I think the only ones often used that they don't usually teach you at school are Fermat's Little Theorem, Ceva/Menelaus, and the various inequalities.

[quote="JesusFreak197"]Alright, I'll stick with what I already know for this year, and hope I do well. I already know some things, and I can prove others; for example, I didn't even realize that I knew how to prove Vieta's Formulas, because I just found out what they were last night when searching for a Quartic Formula on Mathworld. :P[/quote]

Yes, another thing is that many "theorems" you hear people cite are really more like lemmas and aren't too hard to prove.  I find this kind of thing particularly common in geometry.

I should note that I left out every geometry problem since they actually still do require you to learn something, whereas say one of those "boxes and balls" problems such as USAMO 2001 #1 can be solved by someone in fourth grade, if he/she possesses enough ingenuity and strong enough logic.


In right $ \triangle ABC$ with legs $ 5$ and $ 12$, arcs of circles are drawn, one with center $ A$ and radius $ 12$, the other with center $ B$ and radius $ 5$.  They intersect the hypotenuse at $ M$ and $ N$.  Then, $ MN$ has length:

[asy]size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair A=origin, B=(12,7), C=(12,0), M=12*dir(A--B), N=B+B.y*dir(B--A);
real r=degrees(B);
draw(A--B--C--cycle^^Arc(A,12,0,r)^^Arc(B,B.y,180+r,270));
pair point=incenter(A,B,C);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$M$", M, dir(point--M));
label("$N$", N, dir(point--N));

label("$12$", (6,0), S);
label("$5$", (12,3.5), E);[/asy]

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ \frac {13}{5} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ \frac {24}{5}$

$ MA \equal{} 12$ and $ MB \equal{} 5$. $ AB \equal{} MA \plus{} MB \minus{} NM \equal{} 13\implies{NM \equal{} 12 \plus{} 5 \minus{} 13 \equal{} 4} \boxed{(D)}$


Compute: $ 4\plus{}5\minus{}7\times9\div 3$.

you have to do order of operations.

so 7*9=63
63/3=21

4+5=9
9-21= -12


$ f(x)$ is defined for all integers $ x \geq\ 0$

$ f(a\plus{}b) \equal{} f(a) \plus{} f(b) \minus{} 2f(ab)$

and $ f(1) \equal{} 1$

Find $ f(2009)$ and $ f(2008)$

$ f(a\plus{}1)\equal{}f(a)\plus{}1\minus{}2f(a)\equal{}1\minus{}f(a)$
$ f(2)\equal{}0$
$ f(3)\equal{}1$
$ f(4)\equal{}0$
...
$ f(2008)\equal{}0,f(2009)\equal{}1$

It can easily be proved that zeroes and ones will be constantly changing (because if f(n)=1, f(n+1)=1-1=0, and if f(m)=0, f(m+1)=1-0=1)


integration
calculus
Putnam
calculus computations

Let $ f(z)\equal{}\int_{0}^{z}e^{\frac{\minus{}t^2}{2}}dt$

and $ f(z)\equal{}P_n(z)\plus{}R_n(z)$ where $ P_n(z)$ is the Taylor poly of degree 2n+1 about 0.

Show that $ P_n(z)\equal{}\sum_{k\equal{}0}^{n}\frac{(\minus{}1)^k z^{2k\plus{}1}}{k! 2^k (2k\plus{}1)}$

and $ R_n(z)\equal{}\frac{(\minus{}1)^{n\plus{}1}}{(n\plus{}1)! 2^{n\plus{}1}} \int_{0}^{z} e^c t^{2n\plus{}2}$, where for each $ t$, $ c$ lies between $ 0$ and $ \frac{\minus{}t^2}{2}$

:roll:  Try using first Fundamental theorem of calculus and then try finding the Taylor series and then integrate the the resulting series to get P(z). I think so.. I am not sure.. :blush:

Yes, this all just an application of Taylor's theorem with remainder.

In practice, this is an excellent way of approximating $ f(z)$ for $ z$ that aren't too large - say $ |z|\le 3$ or $ |z|\le 4.$

But for larger $ z,$ the convergence of the series is slower, and in a floating point arithmetic system, the accuracy of the alternating sums begins to degrade as we have differences of large numbers giving a small result.

In practice, what else could you do for approximation for large $ z?$

Slightly off topic but still relevant, this problem is reminiscent of  A4(2) of the 1950 Putnam [url]http://www.mat.itu.edu.tr/gungor/IMO/www.kalva.demon.co.uk/putnam/putn50.html[/url]


Hi all --

I'm going to be an undergraduate next year with a major in mathematics. I really think that I want to become a research mathematician, so I know that I must eventually get a Ph.D in mathematics. My question is: how do I know if I have what it takes to get a Ph.D? The very thought of coping with an unsolved problem is intimidating -- I have trouble enough with problems from the USAMO and Putnam!

Does a Ph.D in mathematics require someone to have otherworldly ingenuity to solve an unsolved problem? How will I know if I can do it?

Thanks,

-Arachnotron

I highly recommend [url=http://terrytao.wordpress.com/career-advice/]Terence Tao's career advice[/url], especially [url=http://terrytao.wordpress.com/career-advice/does-one-have-to-be-a-genius-to-do-maths/]this article[/url].

Also, the NYTimes put out an article within the past week on what it means to be a genius. See it [url=http://www.nytimes.com/2009/05/01/opinion/01brooks.html]here[/url]. It is rare for there to be people who have some innate ingenuity. It is usually just lots and lots of hard work.


A real-valued function f which is not identially zero has the property that for each pair of real number a and b,

            f(a+b) + f(a-b) = 2f(a) + 2f(b)

which of the following must be true?

a)f(0) = 1
b)f(-x) = -f(x)
c)f(-x) = f(x)
d)f(x+y) = f(x) + f(y)
e)A number T>0 exists with f(x+T) = f(x)

sorry for the horrible looks. im not good with LaTex yet.

wow... :surrender:  :omighty:  impressive
how did you know that f(x) was in the form of au^b though?

there is 2 hints for my think

but i dont know how to make a good proof of this

${f}$ can be write as a taylor series

From http://www.artofproblemsolving.com/Forum/posting.php?mode=quote&p=739253

"By a basic proof, the sum of two solution is a solution too.
And a solution times constant is a solution too.
${f_{1}}$ and ${f_{2}}$ are solution.

${f(u)=f_{1}(u)+f_{2}(u)}$

${f(x)=f(P+x)}$
${f_{1}(x)+f_{2}(x)=f_{1}(P+x)+f_{2}(P+x)}$
${f_{1}(x)-f_{1}(P+x)=-\left(f_{2}(x)-f_{2}(P+x)\right)}$

${f_{1}(x)-f_{1}(P+x)=0}$
${f_{2}(x)-f_{2}(P+x)=0}$ "

So  a x^b can be a solution

its not a proof, just a thougth

try use this in your problem to see that a sum of a solution is a solution

reversily,  a term of a solution can be a solution

it's not true for all function, but u need to show that...

hum..


Support
trigonometry
number theory
linear algebra

Has anyone here taken an EPGY course before? If so,
-How did you like it?
-Do you feel you learnt a lot?
-Was it challenging to get an A?
-Do you feel the instructors gave you enough support?
-Was the distance-learning aspect of the program a serious problem to overcome?

The reason I ask is that I'm considering the EPGY Light & Heat course.

Any information would be helpful.

Thanks.

Search. This has been asked before recently.  :)

I'm doing EPGY Number Theory and Computer Programming right now.

Number Theory is challenging.  I like it.

Computer programming is easy.  Right now they don't have any higher levels.

I developed a lot of the curriculum for JHU - CTY distance learning so I'm slightly biased.

That being said, I think EPGY is very good on the lower levels, as you get towards anything 6th grade and up, JHU -CTY's curriculum definitely is far superior ;)

EPGY's precalculus course was very good, although a bit light on trig, compared to other topics.  I'm taking EPGY's linear algebra class now, and while the course itself seems very easy, the tutors that students are assigned are very helpful and responsive.

I have a question for those who have taken epgy and cty and are in high school-- Does it count for credit and gpa? For example, if i take ODE or Light and Heat which is not in normal high school curriculum and i get A, can it count for graduation credits/ weighted gpa for my local high school? i think my high school requires 32 credits to graduate so if i independent study, am i able to graduate? thx

You need to speak to your school, and sometimes it can be a long hard battle.


Let $ a,b,c > 0$. Prove that

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq\frac {1}{2}(\frac {b}{a \plus{} c} \plus{} \frac {b \plus{} 2a}{b \plus{} c} \plus{} \frac {b \plus{} 2c}{a \plus{} b} \plus{} \frac {5}{2})$

[quote="Ligouras"]Let $ a,b,c > 0$. Prove that

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq\frac {b}{a \plus{} c} \plus{} \frac {b \plus{} 2a}{b \plus{} c} \plus{} \frac {b \plus{} 2c}{a \plus{} b} \plus{} \frac {5}{2}$[/quote]

er...try $ a\equal{}b\equal{}c$ ...  :o

[quote="kuing"][quote="Ligouras"]Let $ a,b,c > 0$. Prove that

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq\frac {b}{a \plus{} c} \plus{} \frac {b \plus{} 2a}{b \plus{} c} \plus{} \frac {b \plus{} 2c}{a \plus{} b} \plus{} \frac {5}{2}$[/quote]

er...try $ a \equal{} b \equal{} c$ ...  :o[/quote]

sorry my friends.... mystake of beating.
The expression correta is:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq\frac {1}{2}(\frac {b}{a \plus{} c} \plus{} \frac {b \plus{} 2a}{b \plus{} c} \plus{} \frac {b \plus{} 2c}{a \plus{} b} \plus{} \frac {5}{2})$


symmetry
group theory
algebra open
algebra

does anyone know any resources that present nicely the formulation of physical symmetry in mathematical (group theory) language? many thanks!

The only place i could think to look is http://www.mathworld.com


if 50x^2+25x-3= (Ax+B)(Cx+D) and D=-1 where A,B, and C are integers, find the value of (C/A - B) * D^5

Factor the quadratic to:

$ (10x \minus{} 1)(5x \plus{} 3)$, which implies that $ A\equal{}5$, $ B\equal{}3$, $ C\equal{}10$, and $ D\equal{}\minus{}1$.  Then $ \left (\frac{C}{A}\minus{}B \right ) \times D^5\equal{}\left (\frac{10}{5}\minus{}3 \right ) \times (\minus{}1)^5\equal{}(2\minus{}3) \times (\minus{}1)\equal{}\boxed{1}$.


Haven't done much algebra since highschool and now I'm getting ready to take the math placement test for community college. Is there any material i.e. books, online tutorials etc. anyone can recommend?

Here is a link to the sample questions like the ones that will be on the test. Just open the pdf and scroll down to page 3.

http://www.lacitycollege.edu/services/matriculation/docs/questions2006-1.pdf

I guess this post belongs to College subforum rather than here. Anyway, googling "elementary algebra online" gives a lot links. Here are just a few that seem most relevant

[url]http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/[/url]
[url]http://frank.mtsu.edu/~smcdanie/CSS_Site/ElemAlgOnline/ElemOnline_Main.htm[/url]
[url]http://www.mathpower.com/tutorial.htm[/url]
[url]http://www.homeschoolmath.net/online/algebra.php[/url]
[url]http://www.purplemath.com/[/url]

All of those are free resources that you can view without registration or anything (not all pages are properly maintained though). So, you have quite a lot of choices. If you want a paper textbook, just take some high school text, or go to some program like KUMON and see if they can give you their material related to algebra. As far as I understand, what you need is just developing basic routine skills, and those are acquired by solving as many exercises as you have time for following a given pattern. There is no really deep theory to learn there, so what you need is rather a good set of practice problem than a detailed tutorial. Some of the links above provide interactive tests but if you need more, Google is your friend. If you have some questions about elementary algebra, feel free to ask (probably, High School Basic or Intermediate subfora are appropriate for that).


This is like the conversation, but funny. We post witty things that tie up into a story.

MAJOR RULES:
No spam!
No inappropriateness!
No empty posts!
Absolutely no posts like these:

[quote]Robot: HAHA ILL STOMP YOU!!! *STOMP*[/quote]

[quote]and then something happened...[/quote]

And one more thing:

Please narrate the story in italics, and the statements in bold.

(this is not an example)
Story: [i]At the city hall, a ruckus is imminent. The residents of Mayville are arguing about taxes.[/i]
[b]Mayor: QUIET DOWN!!! Thank you. I understand that you have trouble with taxes-[/b]
[b]Young man: That's for darn sure!![/b]
[b]Mayor: and I think that I have got a solution.[/b]
[i]Everyone instantly inquired what it was.[/i]
[b]Mayor: You keep quiet and the town will cut down in the school quality.[/b]
[b]School addict: NOOOO!!!!!!![/b]
[i]Everyone starts up again.[/i]
[b]Mayor: OK!! I'll have a meeting with the town council about this.[/b]
[i]That appeases the crowd right then.

30 minutes later, the council is at a meeting.[/i]
[b]Sheriff: I say that we tell them to have less income. With all that income, the'll be having a-plenty of taxes in a quarter![/b]
[b]Mayor: Well, the previous suggestions are put to vote. All in favor of deducting some quality in school, say aye.[/b]
[i]No one says aye.[/i]
[b]Mayor: All in favor of cutting down the oak tree?[/b]
[b]Sheriff: That is a historic landmark! We-[/b]
[i]The other members of the council interrupt him and contradict him. Then 2 out of 5 people say aye.[/i]
[b]Sheriff: I-[/b]
[b]Mayor: The ayes have it. Council ajourned.[/b]

[i]Meanwhile, Rutt and Tuke happened to be passing by.[/i]
[b]Rutt: Gee, eh, they have a funny way of voting?[/b]
[b]Tuke: Yes, sure, eh! Umm ... shall we go and meet that guy?[/b]
[b]Rutt: Trample off, I am digesting now![/b]
[i]Tuke moves forward alone.[/i]
[b]Tuke: Hello there, smallish mayor! How d'ya do eh?[/b]

OOOHHH! I forgot one thing:

No bother bear parts! I'll let it go just this time. Also, try to keep it funny.

[b]Mayor: Sheriff, how would you ike some moose for supper?
Sheriff: Much abliged!
Tuke: What's that thing with two holes in it?[/b]
[i]The sheriff levelled the gun, and pulled the trigger. All that came out was a "BANG!" sign.[/i]
[b]Tuke: I better get outta here!![/b]
[b]Rutt: I keep tellin' ya, never meet a human![/b]
[i]Then Rutt and Tuke trample off.[/i]

[i]story: while tuke and rutt ran off, they met a crazy old moose-eating mathematician.[/i]
[b]MA: me like moose very much.[/b]
[i]story: having seen tuke and rutt, the mathematcian wanted them very much. so he devised a plan to capture them, a proof to show that his plan will work, and another proof to show that he will never eat the moose because he would never carry out his plan (being faithful to the public's and arsemesomething's ideas of a mathematician).[/i]

[i]Story:The crazed mathmatician showed his devios plan to his fiendesh helper, Quosie.[/i]
[b]MA:  This is the most brilliant idea in the world![/b]
[i]Story: He boomed.[/i]
[b]Quosie:I see.[/b]
[b]MA:   Literaly or figuretively [/b][i]Story: his master asked .[/i]
[b]QU:Exactly[/b][i]Story: came the reply.[/i]


[quote="Steve Adrian"]Date: Thu, 1 Dec 2005 10:00:47 EST
To: (list of ARML coaches)
Subject: A.R.M.L. Reurns to UNLV

The ARML Executive Board is please to announce that for 2006 we will move
our Western Site from San Jose to UNLV in Las Vegas, Nevada. The move was
precipitated to hopefully spur growth at the western site. Las Vegas is
accessible by air from practically anywhere and the rates are usually
quite good as compared to San Jose. The move also allows us to keep the
rates charged to attendees of the western site at a comparable rate to
the other sites. Remember - ARML 2006 is on June 2-3. [/quote]

charge to attendees go down? hmm..... of all places, Las Vegas I thought would cost more.

[quote="peter"]So inconvenient for SFBA. :mad:[/quote]
But the rest of us all have to travel anyway. Las Vegas will be a slightly shorter bus trip for us from Southern California. For teams like Washington which have to fly, it's probably cheaper and more convenient to fly to Las Vegas than to San Jose.

And one other thing: in each of the last two years, SFBA did not even know how many teams you were going to field until late Friday afternoon. Maybe if you have to make travel plans, you'll get a firmer count on how many people will attend.

xxreddevilzxx: We'll be staying on the campus of UNLV. Part of the cost is determined by what the university charges for its dorm space. Since I wasn't part of those negotiations, I don't know what the relative costs there are. As I suggested above (and Mr. Adrian suggests in his message), your travel costs are likely to be lower.

(Of course, if travel convenience/travel costs are an issue ... State College, PA ???)

Expect to see ALOT LESS people from SFBA this year, due to the fact that we now have to fly.


probability
MATHCOUNTS
AMC
AIME
function
number theory
prime numbers

You go to the school BOY'S PE locker room with lockers numbered from 1-1000. In every 5 lockers (randomlY) there is one cockroach. If you have all the lockers open when you enter and close every otherl ocker (2) and then open or close (depends if hte locker is open or closed) every third locker and then continue wiht every 5th locker (as you see the pattern is prime numbers) What is the probablity (after going to the largest prime number less than 1000) that you will bump into an open locker with a cockroach in it?

IF THIS IS TOO COMPLICATED ASK ME

[quote="uglyman621"]IF THIS IS TOO COMPLICATED ASK ME[/quote]

It's too complicated.

I worked out an answer but it was a weird number so I'm not sure that it is correct.

hold on,
FIRST:
which locker do you start at?
SECOND:
[quote="uglyman621"]What is the probablity (after going to the largest prime number less than 1000) that you will bump into an open locker with a cockroach in it? [/quote]What?! do you mean? going to the locker (locker #s)? or counting off lockers (every _ lockers)?
THIRD:
When do you start skipping a new number of lockers?

I just need this clarified, thanks

It'll be nice if there weren't all those typos, but it's pretty clear to me.
Yes, and which locker do you start at each time?

[quote="chess64"][quote="nat mc"]For a locker to be open, it must have an odd number of prime factors. It cannot have five, so it either has 1 or 3.
Answer:$\frac{\text{(number of numbers with 1 prime factor)+(number of numbers with 3 prime factors)}}{5000}$[/quote]

In other words, $\frac{\text{(number of primes)+(number of numbers with 3 prime factors)}}{5000}$[/quote]

Actually no, because 81 only has one prime factor

How do you figure out how many numbers less than 1000 have 1 or 3 prime factors?

Well, the long way would be to go through all the cases, but I don't know a shorter way

Well, for 1 prime, it could either be a prime $p$ or $p^n$. There's 168 primes less than 1000.

Then, for $p=2$, there's also $2^n$, until $n=9$.
$p=3$, $3^n$ until $n=6$.
$(p,n)=(2,9),(3,6),(5,4),(7,3),(11,2),(13,2)...(31,2),(32,1)...(997,1)$

So $n_1-1+n_2-1+\cdots=25$. $25+168=193$. So that's how many numbers less than 1000 have only 1 prime factor, I guess .

But how many $p^2$ and $p^3$ and $p^$s and so on are there.  Then we have 3 to work with too

[quote="superman"]i like nat's answer[/quote]

 :huh: what is it?

[quote="superman"][quote="nat mc"]For a locker to be open, it must have an odd number of prime factors. It cannot have five, so it either has 1 or 3.
Answer:$\frac{\text{(number of numbers with 1 prime factor)+(number of numbers with 3 prime factors)}}{5000}$[/quote][/quote]

That's not a final answer  :roll: We're in the process of trying to simplify it, lol  ;)

but it is hard...so just leave it like that!

its the easy way

[quote="chess64"]For 3 prime factors, there are 168 possibilities for the first one. Then there are 167 for the 2nd, and 166 for the third, as long as $p_1 \cdot p_2 \cdot p_3 \leq 1000$. Then we have to consider $p^n$ where $p>1$. This is going to be hard  :([/quote]

WE also have
${p_1}^{n_1}\cdot{p_2}^{n_2}\cdot{p_3}^{n_3}\leq{1000}$

[quote="chess64"]So how are we going to do this?  :? Mods, isn't this too hard for mathcounts? If not, can you give us a hint or something...[/quote]
Since uglyman hasn't clarified the problem, maybe it's a "trick" question.  Cockroaches always flee from open lockers to closed lockers.  Therefore, the probability that a randomly chosen open locker contains a cockroach is $\hbox{zero}$.

Assuming you really want to find the number of open lockers, this problem would require a [b]ton[/b] of casework.  But, the casework shouldn't be much harder for a top MATHCOUNTS student than for an AIME student.

[I guess I should also mention that there are some [b]very advanced[/b] functions, related to this problem, but I don't think they actually get to the solution any easier than the casework advanced MATHCOUNTS students can handle.  Such methods are also way beyond the level of this forum.]

You are on the right track by categorizing locker numbers, depending on how many distinct prime factors they have.

You are correct that there are exactly 193 lockers with exactly 1 prime factor.  There are no lockers with 5 or more distinct prime factors.  And there is exactly 1 locker with zero distinct prime factors (Locker #1).

You can also enumerate all of the lockers with exactly 4 distinct prime factors.  (There are 23 such lockers.)

That still leaves 783 lockers to be distinguished as having 2 or 3 distinct prime factors, and that's where the casework gets very nasty.  If this were an AIME question, I would wager there would be [b]no[/b] correct answers.

If you are interested in pursuing this, let's redefine the problem to be more reasonable.

The locker room contains locker numbers 1 through 100.  Start with all lockers open.  Following the pattern of closing and opening lockers given by uglyman, how many lockers will be open when he is finished.  [Additional information:  There are exactly 25 prime numbers less than 100.]

I get 43.
For 1 prime factor,
start with powers of two then three and so on
2^ 1 through 6
3^ 1 through 4
5^ 1 through 2
7^ 1 through 2
The other 21 just have 1
So it is 21+the 14 we counted out=35

Now for three,
start with all powers as 1
2,3,5
2,3,7
2,3,11
2,3,13
2,5,7
all work.
Now with one of the primes squared
2^2,3,5
2^2,3,7
2,3^2,5

thats 8

35+8=43

[quote="rcv"]
You are correct that there are exactly 193 lockers with exactly 1 prime factor.[/quote]

Yay!  :D

I also wrote a reliable prgm on the calc and it worked, now I'm running it for 1000.  It might take a while

I tried writing one, but it got too confusing  :? and I gave up, lol.

Mines really not too confusing.  I used lists.  put the numbers in list one and how many times it would be opened and closed in list two, and found prime numbers, added one to list 2 that were divisible, and then found how many odd numbers are in list 2

I GOT IT!!!
The answer to the original question is 117/1250.
There are 468 numbers less than or equal to 1000, that have an odd number of prime factors!!

[quote="nat mc"]I get 43.[/quote]
Correct.

[quote="nat mc"]There are 468 numbers less than or equal to 1000, that have an odd number of prime factors!![/quote]
Correct.

But if a locker number has an odd number of distinct prime factors, wouldn't the locker door be *closed* when you are done?

Oh, woops I thought they would all be closed.  So, out of 100, 57 would be open.  Out of 1000, 532 would be open.

(if they started open)

rcv, how did you do it?

[quote="nat mc"]rcv, how did you do it?[/quote]
About the same as you and chess.  Some manual casework, along with a spreadsheet for the hard cases.


function
analytic geometry
calculus
real analysis

You can graph coordinates of imaginary numbers.  By increasing the # of coordinates, technically, you should be able to make an imaginary function shouldn't you?  And correspondingly, if you can make a function out of it, you can do calculus on it.

my question : Is imaginary functiosn feasible?

Yes.  The study of complex functions of complex variables is known as complex analysis.

http://mathworld.wolfram.com/ComplexAnalysis.html
http://en.wikipedia.org/wiki/Complex_analysis

Real-variable calculus is often known as "Real analysis," hence this is in some sense an extension of real calculus.

Cool. That's going to be fun.

EDIT: not joking.


I know that logarithms aren't allowed for mathcounts, but consider this problem from this year's hand book. It can be clearly solved with guess and check in under 3 minutes, but logs can get a solution in under 1 min. Are you guys sure that logs are never necessary for mc?

I also saw a problem in saab that said something like this: After every friend of yours visits, $\frac{1}{128}$ of your current level of sickness goes away. If you are completely sick then how many visits does it take for you to be under 16% sick?

This problem is super tedious without logs, and the only other way is to graph it and find the intersections.

I have always thought that logs were unnecessary for mathcounts, but can someone clarify?

if you can use them go ahead
but they are unnecessary
can you write up your solution to number 6(don't do all the steps and all)
that wording confused me so much that when i read how to do it i didn't understand any of it(and they didn't say much about logarithms)

lol that was the exact same problem...i didn't bother reading it it was too long  :lol:

There was a question like this on a previous team round. The way I would do this is, assuming that calculators are allowed, create the equation and then do guess and check for $x$. But then of course, that may take longer than just using logarithms.

[quote="vishalarul"][hide="No one solved the problem yet!"]That type motor is $1,440,000,000$ Newton-Seconds.
A type $3$ motor goes from $5$ to $10$.
A type $4$ motor goes from $10$ to $20$.
A type $5$ motor goes from $20$ to $40$.
In general,
A type $n$ motor goes from $10(2^{n-4})$ to $10(2^{n-3})$
Guess a really large multiple of $2$:
$2^{30}<1440000000<2^{31}$. So the answer is just $n=34$. Someone check this.[/hide][/quote]
no that's not the answer

[quote="archimedes1"]It doesn't take that long:
[hide] The maximum impulse for each type can be modeled by 2.5*2^(t-1) where t is the Type.
So, 2.5*2^(t-1) is greater than 1440000000.
2.5 * 2^(t-1) > 1440000000
2^(t-1) > 576000000

Now, we know 2^10 = 1024.  So, 2^10 * 2^10 = 2^20 = approx. 1000000+. We are trying to get past 576000000, and we know that 2^9 is 512.  Multiplying gives 2^29 = about 512000000+, which still isn't enough. Therefore, our number is 2^30, the closest power of 2 that's higher than 576000000.

So, since 2^30 = 2^(t-1),
t = 31.[/hide][/quote]
haha that is what my coach told me to do instead of logarithms  :wink:


Does anyone have an suggestions on any books to read to help prepare for chem 13 and chem olympiad.

There are innumerable threads about this in the forum already.


trigonometry
quadratics
number theory
algebra proposed
algebra

Assume that $ \theta$ is a rational multiple of $ \pi$
i) If $ cos \theta \in Q$ ,determine with proof all possible value of $ cos \theta$
ii) If $ tan \theta \in Q$ ,determine with proof all possible value of $ tan \theta$

[hide="algebraic number theory solution"] Let $ \phi \equal{} \frac {\theta}{\pi}$. Since $ \phi$ is rational, there is some integer $ k$ such that $ k\phi$ is an even integer, and therefore $ e^{ik\pi \phi} \equal{} e^{ \minus{} ik\pi \phi} \equal{} 1$.  So the numbers $ e^{i\pi \phi},e^{ \minus{} i\pi \phi}$ are $ k$th roots of unity, and hence are algebraic integers.  The sum of algebraic integers is also an algebraic integer, so $ e^{i\pi \phi} \plus{} e^{ \minus{} i\pi \phi} \equal{} 2\cos \pi \phi$ is an algebraic integer.  The only rational numbers that are also algebraic integers are the integers themselves, so $ 2\cos \pi \phi\in \mathbb{Z}$, so we must have $ \cos \theta \in \left \{ 0, \pm \frac {1}{2}, \pm 1\right \}$.

If $ \tan \theta$ is rational then we must have that for $ \theta \neq \frac{\pi}{2}$
\[ \cos 2\theta \equal{} \frac {1 \minus{} \tan^2 \theta }{1 \plus{} \tan^2 \theta}
\]
is also rational.  But we just established that this can only be if $ \cos 2\theta \in \left \{ 0, \pm \frac {1}{2}, \pm 1\right \}$, so we must have that $ \frac {1 \minus{} \tan^2 \theta }{1 \plus{} \tan^2 \theta}\in \left \{ 0, \pm \frac {1}{2}, \pm 1\right \}$.  By solving each of the corresponding five quadratic equations, it's easy to check that the only rationals solutions are $ \tan \theta \in \{ 0, \pm 1\}$. [/hide]


Let triangle $ ABC$. Prove that
$ 3a^2\plus{}2b^2\geq c^2\plus{}4S$  :(

A double post.  Please continue your discussion here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193704[/url]


There are 100 people at a party. Prove that there are two people such that, of the remaining 98 people, there are at least 48 of them, each of whom either knows both or else knows neither of the two. Assume that 'knowing' is a symmetric relation.

I don't think the question statement is correct.  You could, for example, have all of the people not know the first person but know the second person.  So you would have that everyone knows exactly one of the two people...


algebra
polynomial
modular arithmetic
algebra unsolved

Let $A$ be the set $\{2,7,11,13\}$. A polynomial $f$ with integer coefficients possesses the following property: for each integer $n$ there exists $p \in A$ such that $p|f(n)$. Prove that there exists $p \in A$ such that $p|f(n)$ for all integers $n$.

If for each $p\in A$ exists $x_{p}$, such that $p\nmid f(x_{p})$, then there exists $x$, such that $x\equiv x_{p} \pmod{p}$ for all $p$, by CRT. It is then easy to chek that $\gcd\left (\prod_{p\in A}p ,f(x)\right )=1$.


geometry
HMMT
geometric transformation
homothety

Let a triangle $ABC$  and  $E,F,D$ the points that the incircle touches $AC,\ AB,\ BC$ respectively . If $ID$ intersects $EF$ at $K$ then prove that $AK$ passes through the midpoint of $BC$ .

This problem has shown up many times on this forum. The first time I remember seeing this problem was at HMMT 2007. I believe there is a homothety solution.

See here: http://www.mathlinks.ro/Forum/viewtopic.php?t=147114  :P

I have a proof which I didn't see on the links mentioned by April and Darij, and I think it is very nice so here it goes.

Denote $M$ the midpoint of $(BC)$ and consider the line $\rho$ the parallel through $A$ to the line $BC$.

Now because $A(B,M,C,\infty)=-1$, denote the intersection $X \in \rho \cap EF$, so we have that $(XENF)=-1$, where $N \in AM \cap EF$.

Therefore $X$ lies on the polar of $N$, but so does $A$, therefore $XA=\rho$ is the polar of $N$, therefore $IN \perp BC$.

So $ID, EF, AM$ are concurrent in $N$.


MATHCOUNTS
USAMTS
AMC
AIME
summer program
Mathcamp
USA(J)MO
\/closed

[url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Our current Math Jam schedule[/url] includes a number of exciting events.

In addition to the MATHCOUNTS prep, USAMTS round 4, and Alternate AIME Math Jams, Janet DeCarlo of the College Board will be hosting a Math Jam on April 13 discussing the Siemens-Westinghouse Scholarship Program with interested students, parents, and teachers and Mathcamp Deputy Director David Savitt will discuss Mathcamp on April 20.

In order to attend a Math Jam you must first log on the Art of Problem Solving Forum at [url=http://www.artofproblemsolving.com/Forum/index.php]www.artofproblemsolving.com/Forum/index.php[/url], then click the Classroom button on the left panel of the site up to 15 minutes before the Math Jam begins.  The Virtual Classroom should then open automatically.

[quote="beta"]Will there be a mathjam on USAMO strategies, or solutions to USAMO problems?[/quote]

We were originally planning on holding one, but it does not look like we have the time unfortunately.

I'm sure there will be plenty of discussion of the problems in the forums.

Other than the one on Aug 10th, when will MATHCOUNTS prep math jams be for the 2005-06 year? Please :)

The schedule only mentioned olympiad stuff (which isn't bad or anything...)

I have not yet scheduled MATHCOUNTS Math Jams.  We typically schedule 3 or more at a time for upcoming months 2-3 times per year.  I'll probably schedule at least a couple after we are done setting up WOOT.


There are 16 tablespoons in a cup. A recipe for 4 people uses 1 tablespoon of garlic powder. How many cups of garlic powder does it take to make the same recipe for 80 people? Express your answer as a mixed number.

We have $ \frac{80}{4}\equal{}20$ tablespoons of garlic powder. Dividing by $ 16$, we get $ \frac{20}{16}\equal{}1\frac{1}{4}$ cups of garlic powder needed for 80 people.


inequalities
search
inequalities proposed

Let $ a,b,c$ be positive . Prove that
$ \frac {a^4}{(b \plus{} c)^4} \plus{} \frac {b^4}{(a \plus{} c)^4} \plus{} \frac {c^4}{(a \plus{} b)^4} \plus{} 2\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq \frac {7}{16}$

This problem is inspired from http://www.mathlinks.ro/viewtopic.php?t=252965 :)

A second inequality is
$ a,b,c$ are positive . Prove

$ \sum_{cyc}\frac{a^2}{(b\plus{}c)^2}\plus{}\sqrt{8}\sqrt{\frac{abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}} \geq \frac{7}{4}$

Both of them weaker than this
\[ \left( {\frac{a}{{b \plus{} c}}} \right)^2  \plus{} \left( {\frac{b}{{c \plus{} a}}} \right)^2  \plus{} \left( {\frac{c}{{a \plus{} b}}} \right)^2  \plus{} \frac{{10abc}}{{\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}} \ge 2
\]
And here is the solution for the above inequality

[quote="Honey_S"]Both of them weaker than this
\[ \left( {\frac {a}{{b \plus{} c}}} \right)^2 \plus{} \left( {\frac {b}{{c \plus{} a}}} \right)^2 \plus{} \left( {\frac {c}{{a \plus{} b}}} \right)^2 \plus{} \frac {{10abc}}{{\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}} \ge 2
\]
And here is the solution for the above inequality[/quote]

Nice. Actually, there is a similar problem in which I have been studied some time ago. http://www.mathlinks.ro/viewtopic.php?search_id=619390218&t=115485

But personally, I think the solution is not greatest because it cannot solve the general problem.  :)

[quote="nguoivn"][quote="Gibbenergy"]
But personally, I think the solution is not greatest because it cannot solve the general problem.  :)[/quote]
Did you meant your general problem is:
Find Min: $ P \equal{} \sum\ \frac {a}{b \plus{} c} \plus{} k (\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)})^n$
With $ 1 \geq\ n \geq\ 0$  :maybe:[/quote]

No, it is not. I think Albanian Eagle in this post [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=112118[/url] proved it.
I am interested in terms $ \sum_{cyc} \ \frac{a^m}{(b\plus{}c)^m}$ where $ m$ is integer greater than zero.

But thanks for interest in my topic.  :)

[quote="nguoivn"]
Find Min: $ P \equal{} \sum\ \frac {a}{b \plus{} c} \plus{} k (\frac {abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)})^n$
With $ 1 \geq\ n \geq\ 0$  :maybe:[/quote]
Yes, its' very easy  :)
For the harder general problem:
Find Min: $ P \equal{}\sum\ (\frac{a}{b\plus{}c})^m \plus{} \frac{kabc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
satisfy equality occurs when $ a\equal{}b\equal{}c$ and $ a\equal{}b; c\equal{}0$
I think it's not trues with $ m \geq\ 7$  :wink:

[quote="Honey_S"]Both of them weaker than this
\[ \left( {\frac {a}{{b \plus{} c}}} \right)^2 \plus{} \left( {\frac {b}{{c \plus{} a}}} \right)^2 \plus{} \left( {\frac {c}{{a \plus{} b}}} \right)^2 \plus{} \frac {{10abc}}{{\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}} \ge 2
\]
And here is the solution for the above inequality[/quote]

It's also found out by me. See here http://www.mathlinks.ro/viewtopic.php?search_id=333697881&t=195311


limit
real analysis
real analysis theorems

Recently I received a problem, which reduced to the following on an exam:

Problem: Show $\displaystyle \sum_{n=0}^\infty (\sqrt{n+1}-\sqrt{n}) z^n$ converges around the unit circle except at $z = 1$.

My solution: (please note I did not realize there is a theorem which is a consequence of Abel Summation that shows this)

Assume $z \ne 1$ and $|z| = 1$. Then let us multiply the series by $(1-z)$, which is a nonzero constant. Noticing $\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n} }$, we get

$(1-z) \sum_{n=0}^\infty \frac{z^n}{\sqrt{n+1}+\sqrt{n}} = \sum_{n=0}^\infty \frac{z^n - z^{n+1}}{\sqrt{n+1}+\sqrt{n}}$.[color=red] It is not hard to see that this is equivalent (with absolutely no rearranging of any terms in the sequence) that this is equal to[/color]

$1 + \sum_{n=1}^{\infty} (\frac{1}{\sqrt{n+1}+\sqrt{n}} - \frac{1}{\sqrt{n}+\sqrt{n-1}})z^n.$

Now we intend to show this series is absolutely convergent. Recalling $|z| = 1$, we examine
$\lim_{N \rightarrow \infty} \sum_{n=1}^N \left|\frac{1}{\sqrt{n+1}+\sqrt{n}} - \frac{1}{\sqrt{n}+\sqrt{n-1}}\right| = \lim_{N\rightarrow \infty} \sum_{n=1}^N \frac{1}{\sqrt{n}+\sqrt{n-1}} - \frac{1}{\sqrt{n+1}+\sqrt{n}} = \lim_{N\rightarrow \infty} 1 - \frac{1}{\sqrt{N+1}+\sqrt{N}} = 1$.

Therefore this series is absolutely convergent, implying it is convergent, which implies the original series is convergent as well.

I received no credit for this solution, but I think it is possible my solution is valid? My professor's argument was that you can't possibly rearrange a series that is not absolutely convergent into one that is. At the time, I did not want to argue without a concrete argument. However, while I agree that the original series was not absolutely convergent, after multiplying by $(1-z)$, it is absolutely convergent. And since multiplying by $(1-z)$ is a valid operation for all $z \ne 1$, my proof should still be valid.

Whether I am right or wrong, can someone please explain to me the legitimacy/fallacy of my argument?

Thank you

I went back in and edited your post to put the key phrase in color.

It's not quite true that this involves no rearrangement. It's subtle, but there is a rearrangment. Look at it this way:

$\lim_{N\to\infty}\sum_{n=0}^N\frac{z^n-z^{n+1}}{\sqrt{n+1}+\sqrt{n}}$

$=\lim_{N\to\infty} \left[1+\sum_{n=1}^N\left(\frac1{\sqrt{n+1}+\sqrt{n}} -\frac1{\sqrt{n}+\sqrt{n-1}}\right) z^n-\frac{z^{N+1}}{\sqrt{N+1}+\sqrt{N}}\right].$

You must now make the separate claim that $\lim_{N\to\infty}\frac{z^{N+1}}{\sqrt{N+1}+\sqrt{N}}=0.$

That is true, and you can make that claim. Once you do that, the rest of your argument follows - and actually covers the same ground as the summation-by-parts argument. Note that you did not prove convergence for $z=1,$ which is good, because that would be false. Note also that multiplying by $(1-z)$ did not by itself create an absolutely convergent series - it couldn't have. You had to make the rearrangement to do that, which does leave a leftover term. Only by controlling that leftover term can you fully justify the argument.

Ah, thank you very much for your response. I will talk to my professor about this. It's not that I needed points; I made this post out of mathematical curiosity, and it's good to know that such a technique is valid.

One more question.

So for instance, multiplying by $1 - z$ did not create an absolutely convergent series because for instance plugging in $z = -1$ is a direct contradiction of this, but the leftover term is what makes this work out?

I just want to make sure, so that if I use something like this in the future I know exactly what I need to do. Thanks.


[img]http://www3.sympatico.ca/yuethomas/math/261.gif[/img]

I have a solution, but I don't think it's either elegant or efficient.

Both of those are substantially different than mine... I'll post mine in a couple of days' time.

I have two solutions.  like osiris said, I'll post when the due date to the set is up (http://cms.math.ca/Competitions/MOCP/2003/prob_sep.html)
oct 15th it seems
today is too early

The equilateral triangle question is my favorite though, have you managed to solve it Tom?

Aye, no... I know it's the first isogonal centre. I just don't know what the heck to do with it.

I only solved two questions this time around. I've been tinkering with the tetrahedra question forever with little success...

Speaking of which, I need to print out September solutions. ^_^

speaking of which, do you have any clue when is the TLE?

I personally haven't had any experience with TLE. I asked someone who has written it - apparently it was used to select people to go to winter camp. But now it doesn't matter (much) any more, as winter camp people are selected by their performances on COMC.

Mid-November, he said.

thanx for the info.  btw John Sun got one of the two perfect COMC papers (Ralph was the other) last year but he did not get invited to Winter Camp.  who's your source?  I dont recall anyone from AYJ last year at winter camp ;)

From what I heard, John didn't perform very well on TLE and that was why he wasn't selected. So it matters somewhat... eh. It's unlikely that I'll go there anyway.  :wink:

AYJ has been in a slump the past few years. Chen Shen, who went to London camp with me and has been doing MOCP for a while, wrote the TLE last year and told me all that.

Do you have any information regarding APMO 2003? Chen and I wrote it this spring as additional contestants, and we haven't heard anything about the results. It's been a LONG time.

Ahh Chen...  okay well for APMO, the results were emailed to me by professor Bill Sands.  but as far as i know a lot of ppl who wrote it did not get their results back.  technically in order to get recognition you must at least get a bronze medal, which meant a score of 17 or above i believe, which is the top x (i dont remember the number is either 10 or 15) in which case you receive a certificate.  our school was under supervision when writing the TLE.  but everybody else... well lets just say the rules were a little less strict...  hope they change that this year...  the questions in the TLE were brutal!!  especially to those who had never done olympiad questions before!!  (they wont even put an easy one up)

I wrote the USAMO and i also did not get any results back :(  i REALLY REALLY wanted to know how i did on that just as a comparison to the competitors in the states :(

Bleh, my school wouldn't let us write any American competitions this year... says there aren't enough teachers to supervise.  :evil:

my solution #1 (#2 is too long and tedius so i wont post it here)

[hide]

LHS

=x(x-sqrt(x+y)(x+z))/(xy+yz+zx)+y(y-sqrt((y+z)(y+x))/(xy+yz+zx)+z(z-sqrt(z+x)(z+y))/(xy+yz+zx)

<=x(x-(x+y+x+z)/2+y(y-((y+z)+(y+x))/2)+z(z-((z+x)+z+y)))/2)/(xy+yz+zx) (AM-GM)

=(xy/2+xz/2+yz/2+yx/2+zx/2+zy/2)/(xy+yz+zx)

=1

[/hide]

[quote="bugzpodder"]my friend and I suspect this is used as a selection process for ppl writing the Time Limited Exam as well as Winter Camp.  Barbeau said he had been trying to find a balance in these sets so that everybody could get one or two in the least.[/quote]

Wait, Barbeau as in Edward J Barbeau?

Yup.

(I don't know how long it'll take for me to post my solution for that question... I just formatted my computer.  :cry: )


calculus
integration
geometry
geometry open

Find the surface and the volume of the circlar ring (the form of the ring is exactly as inner car tube) with internal diameter d and external diameter D.

There is only one not solved problem sent by me - it is this one. If no one don't want to solve this problem I'll stop spamming mathlinks. I posted it because i think it seems easy and maybe it is easy. I'm wondering if the intersection of the car tube with a radial plane is a circle, and if the car tube is a cylinder. The ideas that came to me for solving this problem - were by using calculus - integrals and limits. What is your opinion about this problem and what ideas you have about the solution?

Borislav, please be patient. Maybe some people don't think this problem is attractive, but that doesn't mean it won't eventually be solved by someone, and it also doesn't mean that people regard your post(s) as spam.

As for me, I don't even understand this problem. You want surface and volume of a right cylinder over a ring (difference of two concentric circles), cut by planes parallel to the base plane? Then it's rather trivial... Else, it can quickly turn out pretty ugly.

Please state the problem in precise mathematical terms so that we have a basis for an objective discussion.

  darij

Please, look at the attached file.
All intersections with "vertical" planes through the center are circles (I suppose).
The intersections with "horizontal" plane through the center includes all points from bigger circle that don't belongs to smaller circle of two concentric circles widh radii = D/2 and d/2.
The "center" is something very interesting...i dont know how to define the therm - center of "life-belt".
Inner car tubes and life-belts are very interesting figures. A friend of mine said me that in university courses they had learn such bodies. If you still don't understand me - please ask more questions.

P.S. I suppose something like integrals or limits may be used here...maybe some engineer already had solved the same problem.

Ah! This body is called a ring torus.
See http://mathworld.wolfram.com/Torus.html , but it's a different question how to prove those formulae - though both of them are obvious by integration.

  Darij

It is very interesting! I'm very happy now! It will be very good problem for math olympiad in my opinion - when I was 11-th grade it was shown how to calculate volumes and surfaces by using limits. I think if it is proposed in math competition there not will be many people that will solve this problem. Thorus is "mistuqie" body and indeed it will be very good excercise in geometry and algebra. It will be even good excercise for college students in integrating. Do you share my opinion?

P.S. I sent my solution to my other topic you responded.


Im using the following:
\documentclass[letterpaper,fleqn,twocolumn,twoside,9pt]{report}
\usepackage[top=1.3cm,left=1.2cm,bottom=2.64cm,right=1.2cm]{geometry}

By this it can be seen that im using a two column setting and giving the margins around the text. Im wondering how I can set the width (distanse) between the two columns of text?

Also wondering if Latex accept 9pt and if the scalings like \Large work the normal way.

Sincerely,
Kristian

The distance between the columns is given by \columnsep so for example you could use \setlength{\columnsep}{1em}.

For 9pt use the [url=http://www.tex.ac.uk/tex-archive/help/Catalogue/entries/extsizes.html#Docs]extsizes package[/url] where you would use \documentclass[letterpaper,fleqn,twocolumn,twoside,9pt]{extreport} instead of \documentclass[letterpaper,fleqn,twocolumn,twoside,9pt]{report}

Thanks again for the help. Im impressed by how dedicated you are to answering the questions (answers come quickly).


Already discussed. Only brute force works here. At least this is what I think.

This is some Bulgarian problem... 1997 maybe.


function
calculus
derivative
algebra unsolved
algebra

Solve this system
$ (1\plus{}4^{2x\minus{}y})(5^{1\minus{}2x\plus{}y})\equal{}1\plus{}2^{2x\minus{}y\plus{}1}$
$ y^3\plus{}4x\plus{}ln(y^2\plus{}2x)\plus{}1\equal{}0$

Any solution
I have just guessed it has only one root  $ (0,\minus{}1)$

hello, by substituting $ 2x - y = t$ we have the first equation in the form
$ 5^{1 - t} + 4^t\cdot5^{1 - t} - 2^{t + 1} - 1 = 0$. By defining a function
$ f(t) = 5^{1 - t} + 4^t\cdot5^{1 - t} - 2^{t + 1} - 1$ we get from the first derivative
$ f'(t) = - 5^{1 - t}\ln(5) + 4^t\cdot5^{1 - t}(\ln(4) - \ln(5)) - 2^{1 + t}\ln(2)$ that
$ f(t)$ is strictly monotonously decreasing, so we have $ t = 1$ asthe uniquely solution of $ f(t) = 0$, thus we have $ 2x - y = 1$. By substituting $ 2x = y + 1$ in the
second equation and defining $ g(y) = y^3 + 2y + 3 + \ln(y^2 + y + 1)$ we get:
$ g'(y) = \frac {3y^4 + 3y^3 + 5y^2 + 4y + 3}{y^2 + y + 1}$. The numerator is always positive, because $ h(y) = 3y^4 + 3y^3 + 5y^2 + 4y + 3$ has at the point
${ y_{min} = - \frac {1}{12}(135 + 4\sqrt {3001})^{\frac {1}{3}} + \frac {31}{12(135 + 4\sqrt {3001})^{\frac {1}{3}}} - \frac {1}{4}}$ a global minimum
with $ g(y_{min}) \geq2 > 0$ thus the solution of our system is $ x = 0,y = 1$.
Sonnhard.

Thank Dr Sonnhard Graubner,but your proof which follows $ (hy)>0$ is quite long
Ai all thanks for your help
Have funs :)


number theory
greatest common divisor
induction
linear algebra
matrix
function
strong induction

One solution was given by Peter in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=54800 post #4 (thanks ZetaX for the link), but since it is probably not the proof you want to hear, here is the classical one:

First we show the "addition formula" of the Fibonacci numbers:

[color=blue][b]Theorem 1.[/b] Let $F_0$, $F_1$, $F_2$, ... be the Fibonacci sequence, defined by $F_0=0$, $F_1=1$ and the recursive equation $F_{n+2}=F_{n+1}+F_n$ for all nonnegative integers n. Then, for any two positive integers n and m, we have $F_{n+m}=F_nF_{m+1}+F_{n-1}F_m$.[/color]

You can prove Theorem 1 by double induction (it's boring but it works, of course), but a more elegant [i]proof of Theorem 1[/i] uses matrix multiplication:

We will prove that $\left(\begin{array}{cc}F_{n+1}&F_n\\F_n&F_{n-1}\end{array}\right)=\left(\begin{array}{cc}1&1\\1&0\end{array}\right)^n$ for every positive integer n.

In fact, for n = 1, this is evident, since $\left(\begin{array}{cc}F_{n+1}&F_n\\F_n&F_{n-1}\end{array}\right)=\left(\begin{array}{cc}F_2&F_1\\F_1&F_0\end{array}\right)=\left(\begin{array}{cc}1&1\\1&0\end{array}\right)=\left(\begin{array}{cc}1&1\\1&0\end{array}\right)^n$.

Now, assume that the identity $\left(\begin{array}{cc}F_{n+1}&F_n\\F_n&F_{n-1}\end{array}\right)=\left(\begin{array}{cc}1&1\\1&0\end{array}\right)^n$ holds for some n; we have to show that the same identity holds for n + 1, i. e. that we have $\left(\begin{array}{cc}F_{n+2}&F_{n+1}\\F_{n+1}&F_n\end{array}\right)=\left(\begin{array}{cc}1&1\\1&0\end{array}\right)^{n+1}$.

But, in fact,

$\left(\begin{array}{cc}1&1\\1&0\end{array}\right)^{n+1}=\underbrace{\left(\begin{array}{cc}1&1\\1&0\end{array}\right)^n}_{=\left(\begin{array}{cc}F_{n+1}&F_n\\F_n&F_{n-1}\end{array}\right)}\cdot\left(\begin{array}{cc}1&1\\1&0\end{array}\right)$
$=\left(\begin{array}{cc}F_{n+1}&F_n\\F_n&F_{n-1}\end{array}\right)\cdot\left(\begin{array}{cc}1&1\\1&0\end{array}\right)$
$=\left(\begin{array}{cc}F_{n+1}+F_n&F_{n+1}\\F_n+F_{n-1}&F_n\end{array}\right)=\left(\begin{array}{cc}F_{n+2}&F_{n+1}\\F_{n+1}&F_n\end{array}\right)$ (since the recursion of the Fibonacci numbers yields $F_{n+1}+F_n=F_{n+2}$ and $F_n+F_{n-1}=F_{n+1}$),

and the induction is complete. Hence, the formula $\left(\begin{array}{cc}F_{n+1}&F_n\\F_n&F_{n-1}\end{array}\right)=\left(\begin{array}{cc}1&1\\1&0\end{array}\right)^n$ is proven.

Thus,

$\left(\begin{array}{cc}F_{\left(n+m\right)+1}&F_{n+m}\\F_{n+m}&F_{\left(n+m\right)-1}\end{array}\right)=\left(\begin{array}{cc}1&1\\1&0\end{array}\right)^{n+m}$
$=\underbrace{\left(\begin{array}{cc}1&1\\1&0\end{array}\right)^n}_{=\left(\begin{array}{cc}F_{n+1}&F_n\\F_n&F_{n-1}\end{array}\right)}\cdot\underbrace{\left(\begin{array}{cc}1&1\\1&0\end{array}\right)^m}_{=\left(\begin{array}{cc}F_{m+1}&F_m\\F_m&F_{m-1}\end{array}\right)}$
$=\left(\begin{array}{cc}F_{n+1}&F_n\\F_n&F_{n-1}\end{array}\right)\cdot\left(\begin{array}{cc}F_{m+1}&F_m\\F_m&F_{m-1}\end{array}\right)$
$=\left(\begin{array}{cc}F_{n+1}F_{m+1}+F_nF_m&F_{n+1}F_m+F_nF_{m-1}\\F_nF_{m+1}+F_{n-1}F_m&F_nF_m+F_{n-1}F_{m-1}\end{array}\right)$.

Hence, particularly, $F_{n+m}=F_nF_{m+1}+F_{n-1}F_m$, and Theorem 1 is proven.

An easy property of Fibonacci numbers is:

[color=blue][b]Theorem 2.[/b] For any positive integer n, the numbers $F_n$ and $F_{n+1}$ are coprime.[/color]

[i]Proof of Theorem 2.[/i] We will prove Theorem 2 by strong induction over n:

For n = 1, we have to show that the numbers $F_1$ and $F_2$ are coprime, what is clear, since $F_1=F_2=1$.

Now, assume that, for some integer n, the numbers $F_n$ and $F_{n+1}$ are coprime. Then, the numbers $F_{n+1}$ and $F_{n+2}$ must also be coprime (in fact, if these numbers would have a common divisor > 1, then, since $F_{n+2}=F_{n+1}+F_n$, this divisor would also divide the number $F_n$, and thus be a common divisor of the numbers $F_n$ and $F_{n+1}$, but this contradicts the fact that the numbers $F_n$ and $F_{n+1}$ are coprime). Hence, the induction is complete, and Theorem 2 is proven.

Now, we can show the required fact:

[color=blue][b]Theorem 3.[/b] For any two positive integers n and m, we have $\gcd\left(F_n;\;F_m\right)=F_{\gcd\left(n;\;m\right)}$.[/color]

[i]Proof of Theorem 3.[/i] We will prove Theorem 3 by induction over n + m.

If n + m = 2, then, since n and m are positive integers, we must have n = m = 1, and it remains to prove that $\gcd\left(F_1;\;F_1\right)=F_{\gcd\left(1;\;1\right)}$, what is trivial.

Now, let k be a positive integer, and assume that the assertion of Theorem 3, namely the equation $\gcd\left(F_n;\;F_m\right)=F_{\gcd\left(n;\;m\right)}$, holds for any two positive integers n and m with n + m < k. We will then show that it also holds for any two positive integers n and m with n + m = k.

In fact, let n and m be two such integers. If n = m, then the assertion of Theorem 3 is trivial, as it becomes $\gcd\left(F_n;\;F_n\right)=F_{\gcd\left(n;\;n\right)}$. So we only have to consider the case when $n\neq m$. We can WLOG assume that n > m. Then, n - m is a positive integer. Now, applying Theorem 1 to the positive integers n - m and m, we get $F_{\left(n-m\right)+m}=F_{n-m}F_{m+1}+F_{\left(n-m\right)-1}F_m$. In other words, $F_n=F_{n-m}F_{m+1}+F_{\left(n-m\right)-1}F_m$.

According to this equation, every common divisor of the numbers $F_n$ and $F_m$ must also divide $F_{n-m}F_{m+1}$, and thus divide $F_{n-m}$ (since it is coprime with $F_{m+1}$ (since it is a divisor of $F_m$, and by Theorem 2, we know that $F_m$ is coprime with $F_{m+1}$)). Hence, every common divisor of the numbers $F_n$ and $F_m$ must also be a common divisor of the numbers $F_m$ and $F_{n-m}$. The converse is even more obvious: Since $F_n=F_{n-m}F_{m+1}+F_{\left(n-m\right)-1}F_m$, every common divisor of the numbers $F_m$ and $F_{n-m}$ must divide the number $F_n$, and thus it must be a common divisor of the numbers $F_n$ and $F_m$.

Hence, the set of all common divisors of the numbers $F_n$ and $F_m$ coincides with the set of all common divisors of the numbers $F_m$ and $F_{n-m}$. Particularly, the greatest common divisor of the numbers $F_n$ and $F_m$ coincides with the greatest common divisor of the numbers $F_m$ and $F_{n-m}$. In other words, $\gcd\left(F_n;\;F_m\right)=\gcd\left(F_m;\;F_{n-m}\right)$. Now, since m + (n - m) = n < n + m = k, we know that the assertion of Theorem 3 is true for the integers m and n - m, so we have $\gcd\left(F_m;\;F_{n-m}\right)=F_{\gcd\left(m;\;n-m\right)}$. Finally, $\gcd\left(m;\;n-m\right)=\gcd\left(n;\;m\right)$ (in fact, every common divisor of the numbers m and n - m must also be a common divisor of the numbers n and m, and conversely). Combining all of these equations, we get $\gcd\left(F_n;\;F_m\right)=\gcd\left(F_m;\;F_{n-m}\right)=F_{\gcd\left(m;\;n-m\right)}=F_{\gcd\left(n;\;m\right)}$;
thus, the assertion of Theorem 3 is proven for our numbers n and m.

Thus, the proof of Theorem 3 is complete.

  Darij

To simplify things, for your theorem 2, you can prove that $F_{n+1}F_{n-1}-F_n^2=(-1)^n$ by the matrix you gave (just take determinants!), which immediately  establishes that $F_{n+1}$ and $F_n$ are coprime.

thanks enescu, darij .I have to addmit that from all the solutions in this forum, most i like darij`s solutions

Well, I didn't produced a solution. I only showed a link. Anyways, something more can be said about this problem.
A sequence $(a_n)_{n\ge1}$ of positive integers is called a Mersenne-Fibonacci sequence if the following holds:
\[ \gcd{(a_m,a_n)}=a_{\gcd{(m,n)}} \]
A sequence $(b_n)_{n\ge1}$ of positive integers is called a Dedekind sequence if $\gcd{(b_m,b_n)}=1$ whenever $n$ does not divide $m$ and $m$ does not divide $n$.
There is a theorem stating that $(a_n)$ is a Mersenne-Fibonacci sequence iff there exists a Dedekind sequence $(b_n)$ such that
\[ a_n=\prod_{d|n}{b_d} \]
In the case of the sequence $(F_n)$, its Dedekind sequence is
\[ b_n=\prod_{d|n}{F_d^{\mu (\frac{n}{d})}} \]
where $\mu$ is the Mobius function, that is, $\mu (1)=1$, $\mu (p_1p_2\ldots p_k)=(-1)^k$ if $p_1,\ldots,p_k$ are distinct primes and $\mu (n)=0$ if the square of a prime divides $n$. For instance, the first terms of $(b_n)_{n\ge1}$ are  $1,1,2,3,5,4,13,7\ldots.$

Now that is something interesting.. Thank you enescu


Let $ a,b,c,d\geq0$ s.t. $ a^2\minus{}ab\plus{}b^2\equal{}c^2\minus{}cd\plus{}d^2$.Prove that:
$ (a\plus{}b)(c\plus{}d)\geq 2(ab\plus{}cd)$.

Assume that $ ab\leq cd$.
Since both side of the inequality is non-negative, it is equivalent to
$ (a+b)^2(c+d)^2\geq 4(ab+cd)^2$
Rewrite the codition as $ (c+d)^2=(a+b)^2-3ab+3cd$ and substitute to the above inequality:
$ (a+b)^2((a+b)^2-3ab+3cd)\geq 4(ab+cd)^2$
Rewrite, and expand this, we obtain:
$ ((a+b)^2)^2+(-3ab+3cd)((a+b)^2)-4(ab)^2-4(cd)^2-8abcd\ge 0$
The codition also gives us $ (a+b)^2=c^2+d^2-cd+3ab\geq cd+3ab$, so
$ ((a+b)^2)^2+(-3ab+3cd)((a+b)^2)-4(ab)^2-4(cd)^2-8abcd \\
=((a+b)^2+4cd)((a+b)^2-3ab-cd)+4ab(cd-ab)\geq 0$
Which is always true for any a,b,c,d satified the condition and our assumption.
Equation holds when $ \{\begin{array}{l}c=d\\ab=cd\\a^2-ab+b^2=c^2-cd+d^2\end{array}\Leftrightarrow a=b=c=d$.


real analysis
topology
real analysis theorems

If $ D \subseteq R^m$is an open set and $ \varphi  \in C^1 (D,R^n )$, then $ \varphi (D)$ is a set of measure zero in $ R^n$when m < n.

Proof:

We have the theorem that if $ \psi : G \to F$is a smooth mapping of open sets in $ R^n$ and $ A \subseteq G$ has Lebesgue measure zero then $ \psi (A)$ has measure zero also.

Let $ \pi : R^n  \to R^m$be the projection onto the first m coordinates. $ \varphi  \circ \pi : R^n  \to R^n$is smooth and $ D \times \{ 0\}  \subseteq R^n$ has measure zero in $ R^n$ since $ R^m  \times \{ 0\}  \subseteq R^n$has measure zero.

Hence $ \varphi (D) \equal{} \varphi  \circ \pi (D \times \{ 0\} )$ has measure zero.

---
I am not sure if I have correctly applied the theorem. I am also worried that I did not use the assumption that D is open.

The open hypothesis is irrelevant; the statement is easier if we make things smaller. The only reason to assume it is so that $ C^1$ will make sense.
Your proof is correct.


Dear Math Enthusiast,

     Our first competition of the year, the 2005 Fall Startup Event, took place on Thursday, September 29th.  Because students are asked to solve 100 problems in just 30 minutes, the problems were designed to be fast and easy for those familiar with them, requiring students to quickly recall common equations and relations, and to skip problems that they did not immediately know how to do.  The first-place winners in each division are:

School:  Saratoga High School, Saratoga, CA
Middle School Student:  Alan Chou, Grissom High School, Huntsville, AL
9th-grade Student:  Jack Chen, Ladue Horton Watkins High School, St. Louis, MO
10th-grade Student:  Yolanda Ma, Mark Keppel High School, Alhambra, CA
11th-grade Student:  Evan Zhao, Vestavia Hills High School, Vestavia Hills, AL
12th-grade Student:  Jessica Wang, A & M Consolidated High School, College Station, TX

The recognized scores in each division were:

Schools:  151-254
Middle School Students:  17-54
9th-grade Students:  36-64
10th-grade Students:  51-80
11th-grade Students:  53-81
12th-grade Students:  53-78

     Complete results for the competition (the top 25 scores in each division) are available on our website; take a look to see who in your area you can congratulate!

Thank you,

Tom Clymer
Director of Academic Competitions
National Assessment & Testing
http://www.natassessment.com

P.S.  Our second contest, the 2005 Team Scramble, is a team test in which your school assembles a single team consisting of as many students as you can recruit.  That team has 30 minutes to answer 100 problems covering a broad range of topics and difficulties, offering confidence and challenges to students of all abilities.  The 2005 Team Scramble will occur on Thursday, November 3rd, 2005, and has a registration fee of fifty dollars.

[quote="Tom Clymer"]10th-grade Student: Yolanda Ma, Mark Keppel High School, Alhambra, CA[/quote]
Hooray!

i'm an idiot :D

didn't fill my Grade on the answer sheet, got 59.

should have been 14th in the 12th Graders.. :)