Hey everyone,
When are you all starting ARML Practices?  How long do you practice each week at your team?

~lyra

I will start
We are starting tonight, we practice 2.5-3hrs a week.

We start in September, and have about one practice a month until March. We'll have about 5 more practices this year, each one about 3 hours long, on Saturdays.

Anyone in our area is welcome to come to these practices. We don't start worrying about who's on what team until April.

Minnesota will notify team members at the beginning of April, roughly.  Maybe a bit later.  Then they have 3 Saturday practices that run almost all day in May.

SFBA is getting started up the weekend after Easter.  We do practices every week from then until Memorial Day weekend, about 3 hours a week.

Alabama starts like three days before the competition.

[quote="Kent Merryfield"]Si - to what extent are people willing to come during AP Exam weeks?[/quote]

Hmm, I've never really been aware of that conflict.  For the MN practices, they're more or less mandatory (and held on weekends), so I don't think anyone skipped because they had AP tests.  Prom tended to be the most common reason for skipping.  I don't really know about Bay Area kids; I'll have to see if there's a relative drop in attendance if I'm at that week's practice.  Those are Sunday afternoon practices, so I would think there's not too much conflict unless people are that determined to cram the night before the test.

Florida is the same as Alabama. We start 3 days before the competition. Maybe that's why we do so poorly  :(

NC has two day-long practices, and we practice on the bus ride up. Moral: who needs practice? :wink:

Georgia usually does pretty well at ARML... we start about two months before and practice every other week or so.

Since October, Mr. Crawford has held ARML practices at the San Diego Math Circle on Saturdays at least once per month.

As for making it onto a team, the SDMC directors look at our AMC/AIME scores and our SDML scores (along with some other factors).

Last year on the 5 hour bus ride we cubed, played pokemon, magic cards, and other nerdy stuff. No studying there  :roll:.

Attendance is semi-mandatory for the Lehigh Team. It's understandable if you have a conflict, but it might affect your position on Fire. Hmm... although we take average performance. So like... last year some seniors came to 4 of the 8, but got a 8 on each of the individual rounds each time, for a total of 32, which was a decently high average. Of course they were on Fire.

I suppose AIME and AMC scores are important along with scores from the Lehigh Valley Math Contest.

PA's team is sweet. Join if you can. Free lemonade and cookies, and best of all, we're only three hours from State College. Yay! We are the only ones without a coach bus, it's funny.

In the parking lot: Greyhound, Greyhound, Greyhound, Yellow Bus!

Upstate New York doesn't start practicing until the Thursday before.

However, virtually our entire team goes to NYSML, which has similar format to ARML.  So that kinda counts as practice.

I think we're having four practices, we already had one at the end of march and the other three are later on, in late april and may. 3.5 hours long each approximately iirc.

we start on the 15th and do one every week.
go SFBA! :)

Hmm.  We'll probably practice the weekend before.  And maybe the weekend before that, too.   :oops:

We have 3 3-hour practices starting in March, plus the runoff test.

I started in the beginning of March, three-hour sessions, once a week.


Let triangle XOY be a right-angled triangle with angle XOY=90. Let M and N be midpoints of legs OX and OY, respectively. Given that XN=19 and YM=22, find XY.

(A)24      (B)26      (C)28      (D)30         (E)32

[quote="ckck"]I know you added both equations in the first step to the second, but how did you get $\frac{YO^{2}}{4}$ and $\frac{XO^{2}}{4}$?[/quote]
$M$ and $N$ are midpoints.

So, $ON=\frac{YO}{2}$.


2a
________________________
root of (a-b) + root of (a+b)

This is probably the easiest type of problem for math people like you guys at mathlinks, but please solve this in the next 10 minutes cuz i have math next.

How do you take care of the denominator in

$\frac{1}{\sqrt{2}+\sqrt{3}}$

?

Try the same thing here.

You multiply it by the denominator over itself so that what it is multiplied by equals one. Just you make it - not + Now what?

For future reference, if you have a sqrt(a)+sqrt(b) in the denominator, you want to multiply (sqrt(a)-sqrt(b))/(sqrt(a)-sqrt(b)) to the fraction to get rid of the square roots, leaving you with a-b.

-interesting_move


Hi every one

I have some questions about submitting a problem to the American Mathematical Monthly.

1) Should one provide an answer to ones problem, or the problem itself is the only thing that is required?

2) How long does it take, on average, for the problem to be processed and published?

3) Where should we send our problems? (I have an old address, and I'm not sure if it is still the same address.) And can we submit a problem via email?

Thanks beforehand

What is the American Mathematical Monthly?  :huh:

Hmm, that sounds like an interesting magazine.

Wickedestjr, please try to make your posts more useful so that when people read them they don't feel like they've just wasted their time.

And also Google is your friend. GOOGLE IS YOUR FRIEND!! Now don't forget it again.  :wink:

Perimorac, [url]http://www.maa.org/pubs/monthly.html#Sub[/url] will give you some information (Google is your friend as well.  :) ). The way it is worded there, it seems as if sending the solution to a problem is not necessary. But I don't see why you wouldn't send the solution if you had it, and why you send a problem if you didn't have the solution (journals are not solving your problems).

Of course you can send it to them by email, the email address is on that web page.

As for how long the processing will take, I don't know.

I submitted a problem about five months ago and yesterday I got their positive answer.

And today they announced me that it takes about two months for the problem to be published and then over one year for the solution to be published. So, 5 + 2 + 12 = minimum 19 months from the submission.


HELLP ME!!!!! I need major help on the AMC's!!! I've taken practice tests but I'm not ready!!! And they're only a week and a half away!!!!!!!!!!!!!!!!!!!!!! Plus I've got pneumonia so i can't study! HELLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLPP!!! :rt3: = ?????????????? AHHHHHHHHHHHHHHHHHHHhh

thanks, 1+1=3 yepthatzme

[i][Moderator note: Edited to remove caps and excessive exclamation points.  You got your point across.][/i]

Check out the old mock tests.

If it's a week and some left, you still got a lot of time. Read over your tests and try to look and find the information as quick as possible.

Plus, don't use too much expressions. It will get them disappeared plus it's a public forum so please think of others.

Regards,

Silverfalcon

i thought the AMC's are in middle feburary. I'm in GMP and thats what we were told unless you are in 1 or 2

The AMC 8 is given in November. The AMC 10 and AMC 12 are given on either of two dates in February. See the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11693]AMC FAQ[/url] on this forum for much more information.

so i guess he is taking the amc 8.  i was talking about the amc 10.  This site has some links to amc test problems so check it out for help


superior algebra
superior algebra unsolved

On p.23 of the lecture notes on representation theory of finite groups at http://www.maths.gla.ac.uk/~ajb/dvi-ps/groupreps.pdf, it defines a k-central division algebra to be a k-division algebra where k is in the center. But if $ a \in k$ and $ b \in D$, then $ ba \equal{} (b)(a1_D) \equal{} (ab)(1_D) \equal{} ab$ by the bilinearity of the product in an algebra, so all k-algebras should be k-central. What is the problem here?

Edit: Thanks! It's cleared up now.

You misread the definition a little bit.  $ D$ is called $ K$-central if $ K\equal{}Z(D)$, not if $ K\subset Z(D)$.

The definition is that $ k$ is exactly the center.


In a room with floor dimensions 24'*48' and each tile is 1" long, how many tiles does the interior diagonal go through?

Can you generaize for dimensions x and y? :ninja:

[hide="This?"]
Assuming that the tiles are square are $1''$ square, we have a floor of $288*576$ tiles.
Since this is in the ratio $1: 2$, the interior diagonal goes from one corner to the opposite corner of $2$ by $1$ rectangles. Therefore, if there $576$ columns of tiles, it goes through one tile in each column: $576$ tiles.[/hide]


[hide="i dunno if this is right....."](2^n+1)(3^n+1)+(2^n+2)(3n+1)=6^n
since the exponents are added when multiply u can rewrite like this:
(2)(3^2)(2^n)(3^n)+(2^2)(3)(2^n)(3^n)+6^n=
18(6^n)+12(6^n)+6^n
31(6^n)!!!!
*gasp* waddaya know. [/hide]

this problem was from a standard 8th grade romanian math book. should it really be in the high school forum?

Yes, just because it says 8th grade in Romania it is not the same as 8th grade in USA. Use your common sense. If you are too good in math, this does not mean the entire forum must post more harder problems like this one in MOEMS just to match up your level. Maybe you do not know that you advanced in level of problem solving. If you think too easy, go try harder problems in higher forums, not posting harder problems in lower forums like Middle School

[quote="h_s_potter2002"]Actually, he was the one who posted the problem.

But I'd keep it here.  I dunno if you could use whats taught in middle school for this problem.[/quote]
I know, but he is complaining why this problem cannot be in Middle School just because it says 8th grade Romania


search
linear algebra
linear algebra unsolved

If $A \in M_{n}(\mathbb R)$,and all entiries of $A$ is positive,Can we prove that $A$ has a positive eiganvalue?
 :?:

It's [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1734772401&t=34219]Perron-Frobenius[/url].
In a book that I have, there is a whole chapter dealing with this...

some proofs here:
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=975859863&t=34219

Thank you,too,but it is hard to read harazi's solution...


Let $ X$ be a topological space and $ Y$ a subset of $ X$. If $ Y$ is irreducible,

how to prove that $ \bar{Y}$ is irreducible?

Thanks!

Hm, just by using the definition of irreducibility of a subset of a topological space. ;) What have you found out so far?


analytic geometry
geometry
Princeton
college
calculus
calculus computations

Hi, I'm studying for the AP Calc BC exam which will be on this Wed.

I'm self-studying for this, so I don't have any AP teacher to ask this...

How can I know the Lower/Upper Limit when I am finding the Area beneath a Polar Curve?

I'm currently using the Princeton Review Book to study, but the book assumes that I finished all of Pre-Calc., and does not explain how to find these boundaries.

Can anyone help? thx.

David Lee

That's phrasing the question far too generally for anyone to be able to help you. Try posting a specific example of what you're talking about.

srry, here's an example.

> Find the area inside the smaller loop of the limacon $ r \equal{} 2cos{\theta}$.

In the Princeton Review book, after graphing the curve, they say

> Because in the inner loop, r sweeps out the region as $ \theta$ goes from $ \frac {2\pi}{3}$ to $ \frac {4\pi}{3}$, these are our limits of integration.

But, I don't get how they got the " $ \frac {2\pi}{3}$ to $ \frac {4\pi}{3}$ ". Is it just graphical analysis?

Those are angles for which $ r\equal{}0.$ (Or would be, if you didn't have a typo in what you wrote.)

So, for these type of questions, do I just set r = 0 and solve for theta?
(just because the AP exam tends not to use graphs that do not intersect the x-axis)

The $ x$ axis is completely irrelevant. Note the wording of the problem you quoted: the area [i]inside[/i] the curve, or one segment of the curve. No one said a thing about [i]beneath[/i] any curves.

ahhh

now i understand :)
Thx Kent, I think I won't have any problems with the polar coordinate problems on the AP exam now


A segment has endpoints P(1, 1) and Q($ x, y$). The coordinates of the midpoint of segment PQ are positive integers with a product of 36. What is the maximum possible value of $ x$?

Call the midpoint of $ PQ$ point $ M$.  $ M$ has coordinates $ (a,b)$.  For $ x$ to be at its maximum, $ a$ must be at its maximum, which occurs at $ (36,1)$.  Thus, the coordinates of Q are $ (71,1)$ and $ x \equal{} \boxed{71}$ at its maximum.

Didn't understand it correctly
please explain it i description


If $ x$ and $ r$ are directly proportional and $ x\equal{}5$ when $ r\equal{} 25$, then what is $ x$ when $ r \equal{} 40$?


Prove that the ecuation doesn't have any solutions in positive integer numbers:
 [quote]$ x^2\plus{}y^2\plus{}z^2\plus{}u^2\equal{}2xyzu$[/quote]

I think it's quite obvious applying Fermat's infinite descent method.

Let us have $ x,y,z,u$ as the numbers satisfying our equation. Because the number $ x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}u^{2}$ is even, that means we have an even number of odd numbers.
I) if all the numbers $ x,y,z,u$ are odd, than $ 4|LHS$ but $ 4\not |RHS$
II) if only 2 of the numbers are odd than $ 4\not |LHS$ and $ 4|RHS$
So it's obvious that all the numbers are even. Put $ x\equal{}2x_{1} \ y\equal{}2y_{1} \ z\equal{}2z_{1} \ u\equal{}2u_{1}$
We land in an equation $ x_{1}^{2}\plus{}y_{1}^{2}\plus{}z_{1}^{2}\plus{}u_{1}^{2}\equal{}8x_{1}y_{1}z_{1}u_{1}$
Analogically we have that all numbers $ x_{1},y_{1},z_{1},u_{1}$ are even.
By that method we have, that $ \forall_{k\in \mathbb{N}}x_{k}^{2}\plus{}y_{k}^{2}\plus{}z_{k}^{2}\plus{}u_{k}^{2}\equal{}2^{2k\plus{}1}x_{k}y_{k}z_{k}u_{k}$
Which gives us that $ \forall_{k\in \mathbb{N}}\frac{x}{2^{k}},\frac{y}{2^{k}},\frac{z}{2^{k}},\frac{u}{2^{k}}\in \mathbb{Z}$
which obviously can't be true :wink:


Several people have asked me for these so I am attaching my lecture notes for calculus, multivariable calculus, and linear algebra.

This set is calculus, based off of the Honors Single Variable course, which uses Spivak's "Calculus", 3E.

If you spot typos or errors, please tell me.

I'm teaching a Calculus II for first semester freshmen, most of whom have AP Calculus AB credit. A few of them have told me that they're worried that their high school course lacked depth or was insufficiently rigorous. I'm mostly going to tell them not to worry, and that I'll cover and emphasize what I need. But just to have something to keep and point to, I decided to slap together a one-page reference sheet.

eh do you have a typo (sorry, minor, but I just noticed) in the file Lecture Notes 5? Section: Rigorous Definition of Limits, first paragraph, you said $ \delta < 0$ instead of $ \delta > 0$. I'm trying to understand the delta-epsilon definition of a limit, and I'm reading these notes. They're very comprehensive. :)

[quote="JRav"]Anybody have feedback for me?[/quote]

Sure, I do...

The indices in your [b]signature[/b] are kind of misleading... :lol:

Best of Luck,

Well, I was referring to the one that appears in your Forum-profile.

See you!  :wink:

Oh, that one.  Yes, I probably should edit it but I'm lazy.


real analysis
integration
real analysis unsolved

Suppose A and B are two Lebesgue measurable subsets of R, each one of strictly positive but finite measure. Prove that there exits
a number c in R such that  m((A+c)^B)>0

Note: ^ stands for intersection

Let $ f: c\rightarrow{m}((A\plus{}c)\cap{B})\equal{}\int\chi_A(x\minus{}c)\chi_B(x)dx$. $ \int{f}(c)dc\equal{}\int\int\chi_A(x\minus{}c)\chi_B(x)dxdc\equal{}\int\chi_B(x)m(A)dx\equal{}m(A)m(B)>0$. Thus the measure of the set $ \{c;f(c)>0\}$ is positive.

Hello Sir,

Thanks a lot, could you please help me to solve this problem using the definition of Lebesgue outer measure.


Find all functions $ f: \mathbb N \rightarrow \mathbb N$ such that for each $ m,n\in\mathbb N$ then:
$ f(m^2\plus{}n^2)\plus{}f(m\plus{}n)\equal{}f(m)f(n)\plus{}1$

[quote="hien"]Find all functions $ f: \mathbb N \rightarrow \mathbb N$ such that for each $ m,n\in\mathbb N$ then:
$ f(m^2 + n^2) + f(m + n) = f(m)f(n) + 1$[/quote]
We find $ f(0)=1$ then $ f(x^2)=1,\all x\in \mathbb N$ Fix $ m$,and let $ n\in N$,hence $ f(n)=1$ for all $ n\in N$
Am I right?

your solution isn't right! :lol:
from $ f(x^2)=1,\all x\in\mathbb N$ not->$ f(n)=1,\all x\in\mathbb N$
can U detail it?I think not easy :wink:

The most diffcult of this problem is: from $ f(x^2)\equal{}1$ for all x in N we have to show that $ f(x)\equal{}1$ for all x in N

I didn't résole this probleme but I find ...

f(m²+1)+f(m+1)=f(m)*f(1)+1 ===> f(m+1)-f(m)=1-f(m²+1) (because f(1)=1)

so if you can prove that f(m)#0  (for all m)    you can conclur f(m+1)-f(m)=<0

now you can prove with recurence demonstration f(m)=1
 -f(0)=1
 - suppose f(m)=1  ; f(m+1)-f(m)<=0 ===> f(m+1)<=1 ==> f(m+1)=1 ( because f(m)#0)

(but the probleme i didn't prove f(m)#0)


Does anybody have any recommendations on [b]software for visualising, manipualting, and analysing 3D Mathematical Objects [/b] that is preferably, available free online! This software would be e.g. for drawing an arbitrary tetrahedron, drawing a plane (or part of it) say $\perp$ to a edge/face etc. and observing/analysing some properties....

Any suggestions!!!! ( This is sort of like manipulating Chemistry Molecules in 3D!!!)

Thanks!

No recommendations?!?!?!?!?!?!?!?!  :(  :(  :(  :?  :?  :?

Somebosy must have made such a software. It's almost indispenisble till we don't get R2D2's Hologram Projector!!!!

well, as far as 3D surfaces and curves goes, i use TI 89. it is not a software but ....

not sure, but i think Maple does this stuff, of course it is not free,


Question:There is a contest where someone has to guess how many marbles are in a box. The only information they are given is that there a three red identical marbles and ? white marbles.The number of permutations for all the marbles is 35. How many white marbles are there?

Since there are 3 red marbles I think there are 6 permutations of red marbles.So if you take that away from 35 your left with 29 permutations of white marbles.I think this question's answer has to do with the equation n!/p!q! where n is the number of items and p and q the number of repititions, but it is hard to tell with these type of problems. I can get confused with these type of problems(involving distinguish and indistingishability) and I wish there was an easier way to solve them without having to remember certain equations. Sometimes recognizing whether or not the order is involved is a problem to. Can anyone help me? Thank you.

35 is really not a big number, just list all the permutations on a paper.

You have to assume each marble of the same colour is identical [it says so for the red ones, and will be the same for white. If it wasn't whats the point of having colours anyway?].  So if there was 1 white marble, you get 4 permutations - WRRR, RWRR, RRWR, RRRW. Its easy enough to solve from here (just by checking 1, 2, 3..).

Here is a nice way to do it:
There are x blue and 3 red marbles.
So number of permutations=$(x+3)!/3!x!=35$
$(x+3)!/x!=210$
$(x+3)(x+2)(x+1)=7*6*5$
x=4

I'd rather do it this way than listing all the numbers on paper.
And when you multiply both sides by 6, you would be better off leaving it as 7*6*5 instead of 210, I just listed that so there wouldn't be as much confusion.

[quote="h_s_potter2002"]In a permutation order matters, right?

So if you had three red marbles in a row like this: r1 r2 r3, would that be different from this: r2 r3 r1, beucase there are still three red marbles in a row, it's just that are in different positions.[/quote]

Yes. The ones where order doesn't matter are combinations.

Identical  means, one can not  tell the difference between  one red ball and the other red ball.

If I use R (for red ) and B  (for Black

Then  RRBBR
and    BBRRR

are  diffent permutations,  ( they have the same combination - 3 res, 2 black -order here does not matter)

But RRR is one permutaion  (If you can distinguish  the balls by  numbring then r1,r2,r3 then, and only then  you can count  - as 6)

Hope this helps.

[quote="hatsup123"][quote="RC-7th"]When the numbers get bigger, I think it's better to use algebra rather than guessing.[/quote]
I would too.[/quote]

Ok, would you like to tell me how to solve the more complicated problem, say, 5 red marbles, x white marbles, 15504 permutations, with algebra?

That would be the same way that I did it before:
(x+5)!/(5!x!)=15504
Factoring the right, multiplying by 5!, and simplifying (x+5)!/(x!) gives you
(x+5)(x+4)(x+3)(x+2)(x+1)=19*17*5*3^2*2^7
You could, algebraically, solve for x, or get a computer to do it, or you could just find the factors that work.
20*19*18*17*16=19*17*5*3^2*2^7 so x=15.

Well, you can approximate (x+5)(x+4)(x+3)(x+2)(x+1) by (x+3)5.  Solving the resulting equation gives x approximately 14.944.  Rounding to an integer gives x = 15.  You can then check that x = 15 solves the original equation.

So I wouldn't say that you need to check many cases.

[quote="TripleM"]My point is, that there is no way you can algebraically solve that problem for anything higher than 3 red marbles. [even three is pushing it, unless you know the cubic formula off the top of your head.] It still comes down to checking various cases.[/quote]

Yes, you have to do some checking, but listing all of the cases would be hard if the number was something as big as 15504. I was trying to point out how paganinio's method wouldn't work for higher numbers.


how many sets {a,b,c} of 3 different, positive single-digit integers have the property that $a^{2}+b^{2}+c^{2}$ is  a perfect square?

[hide]Well...

There is {2,3,6},  {1,4,8}

Wonder if there are anyothers...[/hide]

[size=75][color=darkred]Hidden.
-nebula42[/color][/size]

[quote="frost13"][hide]is zero a "positive single-digit integer"??????[/hide][/quote]
Also note that the question asks for three different values.

EDIT: Never mind, I just realized something.  :oops:

By the way, this question looks like it doesn't really have a very good MathCounts solution (other than long casework, of course), so unless someone finds one, I may move this problem to one of the high school forums.

yeah, this will require lots of modular arithmetic, and even after that, it will turn into brute force.

[quote="frost13"][hide]is zero a "positive single-digit integer"??????[/hide][/quote]

i think he was talkin to jhredsox


function
inequalities
integration
calculus
real analysis
real analysis solved

Let $f:[0,1]\to\mathbb{R}$ be a continuous function s.t. $\int_x^1f(t)dt\ge\frac{1-x^2}2$ for all $x\in[0,1]$.
Prove that $\int_0^1f^2(x)dx\ge\frac13$.

Interesting. So far, I've got that the integral of the square is $\ge\frac8{27}.$ Not quite $\frac13$, but getting there.

Integrate the given inequality on [0;1]

$\int_{0}^{1}\left(\int_{x}^{1}f(t)dt\right )dx\geq \int_{0}^{1}\frac{1-x^2}{2}dx=1/3$

LHS integration by part gives

$\int_{0}^{1}xf(x)dx\geq 1/3$

To finish use Cauchy-Schwarz inequality

nice solution moubi...
I don't see where is the problem in what I am doing. :(
Let $F$ be a primitive of $f$.
Then $\int_x^1f(t)dt=F(1)-F(x)\ge \frac{1-x^2}2$
by derivating this we get that
$-f(x)\ge -x\Rightarrow f(x)\le x\Rightarrow f^2(x)\le x^2$  and now by integrating between 0 and 1 $\int_0^1 f^2(t)dt\le\frac13$....
Where is the mistake?

[quote="Kuba"]$F(1)-F(x)\ge \frac{1-x^2}2$
by derivating this we get that
$-f(x)\ge -x\$ [/quote]
Not correct.
Another thing in this problem $-f(x)\geq -x$ does not imply $f(x)^2 \leq x^2$


Let $x_0, x_1, x_2, \ldots$ be the sequence defined by
$x_i= 2^i$ if $0 \leq i \leq 2003$
$x_i=\sum_{j=1}^{2004} x_{i-j}$ if $i \geq 2004$

Find the greatest $k$ for which the sequence contains $k$ consecutive terms divisible by 2004.

consider the sequence mod 2004. it is equivalant to the following recurrence:
x_0=1, x_1=2^1, x_2=2^2 ... x_2003=2^2003, x_n=x_n-1 ... + x_n-2004.
so the recurrence is periodic. so the terms {1, 1, 2^1 ... 2^2003} occur again in that order somewhere on in the sequence. we can see that the 2004 terms before that are 1, followed by 2003 '0's. so we can have 2003 consecutive terms all divisible by 2004. We cannot have 2004 such terms, because if then all ensuing terms are divisible by 2004 - which then contradicts the periodicity of the recurrence mod 2004. And we are done.

[quote="vinoth_90_2004"]we can see that the 2004 terms before that are 1, followed by 2003 '0's.[/quote]

Sorry, I think you've meant:

We can see that the first term before that is 1, and the 2003 terms before this 1 are '0's.

  Darij

i think i have seen that problem before in the IMO 2003 short listed


$ABC$ - acute triangle,
$H$ - orthocentre,
we know that $AB=CH$.
Find the measure of angle $ACB$.

It's not hard to calculate $AB=CH \tan \angle ACB$  :wink:

could you write something more...how did you achieve this ?

CH=2rcos(C)
AB=2rsin(C)
So 2rcos(C)=2rsin(C)   and C=45


I've currently finished an associates degree in computer networking and am going for a bachelors in electrical engineering come next fall. However, it's been two years since I've done any higher level math. I took both PreCal and Trig in high school, but I don't really remember anything. So I'm looking for something that can give me a refresher course in algebra, precal and trig, while getting my foot in the door when it comes to calculus. I've heard that calculus for dummies is good, but it's not really comprehensive. I don't have too much time either about 3 months and outside work I don't have too much free time. TIA

The answer depends on how calculus is taught at your target college. It makes sense to choose a precalculus book that is aligned to whatever calculus book they use. The webpage of your college may have a textbook list; if it does not, ask a professor who is teaching precalculus there.

I do a significant amount of academic advising, and my university admits a significant number of transfer students, mostly from community colleges. I encounter students in the general situation of johnec22 frequently: transfer students wanting to pursue a degree in math, science, or engineering who have so far taken very few of the required lower division math and science courses for that degree.

I should say I encounter such students with distressing frequency, as the prognosis often isn't a rosy one. If you haven't taken calculus, you probably also haven't taken the calculus-prerequisite physics course. Now look at the degree requirements for an EE degree, and start looking at the chains of prerequisites on courses. How many things have a circuit theory course - or two circuit theory courses - as prerequisites? And those circuit theory courses require what - differential equations? The physics E&M course? The chains of prerequisites are going to be long.

As an EE major, you're essentially a freshman, and it's unlikely that you'll be able to finish the degree in less than four additional years. The fact that you already have an associates degree won't speed the process very much at all. I would add that we would make you take and pass a placement test before we would let you enroll in our Calculus I course - else, we would have you take one of the courses that precede calculus. Other campuses might have similar systems. In fact, there's a lot to be said in your situation for actually going ahead and enrolling for a precalculus or trigonometry course rather than trusting that you can accomplish what you need to through self-study.

Did your community college offer a calculus sequence, a "physics for engineers" sequence, a general chemistry course? Because those would have been the pathway to transferring as an engineering major with a chance to finish in a reasonable amount of time.

I don't believe that it's totally for loss. Now I haven't had a physics class, but I probably can take one over the summer classes don't start till June 1st.  I did well in both classes and I know that I can refresh on what I have already learned once. If I walked into a calculus class with a good foundation PreCal and Trig would that be enough? Also their is a parallel program two years at the community college and two years at the university.

So say you did some independent studying with Calc 3, is there a way to test out of it at a school that generally doesn't allow these tests? It would be a waste of time to waste a semester while you could be taking more advanced math courses.

The short answer is, it depends on the school.  For example, at MIT, you can take the exam for any course during freshman orientation, and if you pass it, you don't have to take the course (If you fail, it doesn't show up on your transcript).


calculus
derivative
function
limit
topology
complex analysis
complex analysis unsolved

Let $ f$ be $ C^1$ function on open set $ U$ in complex plane

if $ f$ preserve angle, i.e. when $ w_1 , w_2$ be any complex number and $ P \in U$ then

angle between $ \lim _{t \rightarrow 0} \frac{f(P\plus{} tw_j ) \minus{} f(P)} {t}$ $ (j \equal{} 1 , 2)$ is equal to angle between $ w_1 , w_2$

then $ f'(P)$ exist.


trigonometry
inequalities
geometry
circumcircle
triangle inequality
trig identities
Law of Cosines

Given $\triangle ABC$, prove the following two inequalities:
[list]1. $\cos{A}+\cos{B}+\cos{C}\geq 1$.

2. $\cos{A}+\cos{B}+\cos{C}\leq \frac{3}{2}$, for an acute triangle $ABC$.
Is this inequality also true for a non-acute triangle as well?[/list]

[quote="nebula42"]
2. $\cos{A}+\cos{B}+\cos{C}\leq \frac{3}{2}$, for an acute triangle $ABC$.
Is this inequality also true for a non-acute triangle as well?[/list][/quote]

This is true by Jensen's inequality because $\cos x$ is concave in the interval $[0,\frac{\pi}{2}]$, and it's not true for non-acute triangles.

Actually, 2 [b]is[/b] true for non-acute triangles, it's just not trivial by Jensen...
[hide]
Let $x=\tan \frac{A}{2}$, $y=\tan \frac{B}{2}$, and $z=\tan \frac{C}{2}$. Then $x,y,z$ positive real numbers (since $x=0$ would imply $A=0$) and $xy+yz+zx=1$, which follows from
\[z=\cot \left( \frac{A}{2}+\frac{B}{2}\right)=\frac{1-xy}{x+y}\]
Then we have
\[\cos A=2\cos^{2}\frac{A}{2}-1=\frac{2}{x^{2}+1}-1=\frac{2(xy+yz+zx)}{(x+y)(x+z)}-1\]
and similar relations for the other angles. So the inequality is equivalent to
\[2(xy+yz+zx)\left( \frac{1}{(x+y)(x+z)}+\frac{1}{(y+x)(y+z)}+\frac{1}{(z+x)(z+y)}\right) \leq \frac{9}{2}\]
or, after clearing denominators,
\[8(x+y+z)(xy+yz+zx)\leq 9(x+y)(y+z)(z+x)\]
But we have $(x+y+z)(xy+yz+zx)=(x+y)(y+z)(z+x)+xyz$, so the inequality is equivalent to $(x+y)(y+z)(z+x)\geq 8xyz$, which follows immediately from AM-GM on each of the three factors:
\[(x+y)(y+z)(z+x)\geq 2\sqrt{xy}\cdot 2\sqrt{yz}\cdot 2\sqrt{zx}=8xyz\]
We can prove 1 as well. This will follow from the inequality
\[(xy+yz+zx)\left( \frac{1}{(x+y)(y+z)}+\frac{1}{(y+x)(y+z)}+\frac{1}{(z+x)(z+y)}\right) \geq 1\]
which, after clearing denominators, becomes
\[(x+y+z)(xy+yz+zx)\geq (x+y)(y+z)(z+x)\]
or $xyz\geq 0$, clearly true. In fact, the inequality in 1 is strict; we have
\[\cos A+\cos B+\cos C>1\]
[/hide]

I also know another way to prove #1, but it's not extremely pretty.  :P
[hide]Let $a,b,c$ be the side lengths of the triangle. Then by the Triangle Inequality we have $a+b-c>0$ (or $\geq$, depending on whether you allow degenerate triangles or not). Similarly, $b+c-a>0$ and $c+a-b>0$. Multiplying these all together, we obtain
$(a+b-c)(b+c-a)(c+a-b)>0$,
which by expanding out (a lot of computation) and using the Law of Cosines ($\cos{A}=\frac{b^{2}+c^{2}-a^{2}}{2ab}$, etc.) transforms into $\cos{A}+\cos{B}+\cos{C}>1$.
I think this is basically the same as the other proof, though...[/hide]

Also, what is the equality condition for part 2? It's an equilateral triangle, right? (Since from AM-GM equality occurs when $x+y=y+z=z+x$, or $x=y=z$.)

Would it be possible to prove part 2 for a non-acute triangle (and possibly part 1 too) using something like the following (assuming $\angle C \geq 90^\circ$):
$\cos{A}+\cos{B}+\cos{C}=\cos{A}+\cos{B}-\cos{(A+B)}$?

[quote="joml88"][quote="arqady"][quote="nebula42"]
2. $\cos{A}+\cos{B}+\cos{C}\leq \frac{3}{2}$, for an acute triangle $ABC$.
[/quote]
It's true for all real $A,$ $B$ and $C$ such that $A+B+C=\pi.$ :wink:[/quote]

scorpius already said this.[/quote]
No, joml88. scorpius  said:
[quote="scorpius119"]Actually, 2 [b]is[/b] true for non-acute triangles, it's just not trivial by Jensen...
[/quote]
But I mean that it's true for all $A,$ $B$ and $C$ such that $A+B+C=\pi.$
For example, for $A=-1,$ $B=-1$ and $C=\pi+2.$ :wink:

Nice solution, Virgil. But in your solution we have to prove that $\frac{r}{R}\le\frac{1}{2}$. Now, here is another solution for [b]2.[/b]
We have:
$\sum\cos A\le\frac{3}{2}\Longleftrightarrow 2\cos\frac{A+B}{2}.\cos\frac{A-B}{2}+1-2\sin^{2}\frac{C}{2}\le\frac{3}{2}$
$\Longleftrightarrow 4\sin^{2}\frac{C}{2}-4\sin\frac{C}{2}.\cos\frac{A-B}{2}+\cos^{2}\frac{A-B}{2}+1-\cos^{2}\frac{A-B}{2}\ge 0$
$\Longleftrightarrow\left(2\sin\frac{C}{2}-\cos\frac{A-B}{2}\right)^{2}+\sin^{2}\frac{A-B}{2}\ge 0$
[b]General:[/b] $\sum yz\cos{A}\le\frac{x^{2}}{2}\qquad\forall x,\,y,\,z$

[quote="April"]Nice solution, Virgil. But in your solution we have to prove that $\frac{r}{R}\le\frac{1}{2}$. [/quote]

that is trivial...the circumradius is twice the radius of the 9pt circle, and the incircle is trivally inside the 9pt circle, so we are done


Princeton college MIT algebra polynomial probability Harvard

So, Princeton's prospective freshman visiting weekend is April 12-14 this year, I believe. Prospective freshmen may request particular hosts. So if you're going and you'd like to be hosted by an AoPSer / Princeton Math Club member, I can arrange that (I'm the president of the Math Club). Just PM me.

I'd also be happy to help you organize an AoPSer gathering at some point in the Math Dept building, if enough of you are going here (instead of MIT) in order to warrant it. ;)

Why, oh WHY do they have to be at the same time?

[quote="quantum leap"]Woah, you're the president of the math club?  Were you the Nathan that gave the talk on cyclotomic polynomials that one Monday in February, or was that another Nathan?[/quote]
That's a different Nathan.

[quote="MithsApprentice"]If you're going to MIT CPW and would like to come along when I haul (insert word for posterior) to Princeton on Tuesday or something after CPW, let me know. :D[/quote]
I can arrange for someone (possibly myself if you come at a good time) from the Math Dept to meet you (pl?) if you'd like an informal math-oriented tour in addition to/instead of the Orange Key tour.

[quote="MithsApprentice"]Speaking of an orange key tour...Nathan, why is Princeton's orange so bright? I mean...how does that work with the t-shirts and sweaters and stuff.[/quote]
rofl.... uh.... you just get used to it. Or you don't, in which case you don't wear half of the free shirts you get. (But you get so many free shirts that it's not a great loss...)

I will be coming for Thursday night and Friday :D

I'll be there Thursday morning (my flight from LA is on Wednesday night).

What math classes do you think we can visit on Thursday or Friday (I only saw Analysis III open to visitors - I'd be quite lost listening in to that class, though I could see how some math classes are taught at Princeton)?

I don't know if you're "not supposed" to go to certain classes, but in my experience with the Math Department at Princeton, you can generally do whatever you want as long as you don't cause trouble. So just show up to whatever class you want, sit/stand in the back and be quiet, and all will be well.

It's possible that some classes are not marked "open to visitors" because there aren't many free seats. In that case you should not force any enrolled students to stand while you sit -- they're the ones who are paying $\$$35000 a year tuition for the opportunity to be in that class.

Here are my recommendations:

[hide="A typical upper level math class"] You should go to Math 330, Analysis I, from 1:30 - 2:20 on Thursday or Friday. The professor, Elias Stein, is one of the best profs at Princeton and the creator of the analysis sequence. The analysis sequence is one of the star features of the Princeton undergrad math curriculum.

The class will be starting Dirichlet's theorem soon. The material is probably not very accessible unless you have already seen a bit of the theory of Fourier analysis. However if they are already proving the theorem and you know something of number theory you should at least have some appreciation of what they are doing even if you can't follow the technical details. [/hide]

[hide="A financial math class"] I am taking ORF 527, a course on stochastic calculus. This is about as abstract as math gets in the finance department. If you're interested in both math and finance, you might want to visit this class to see how it is over in the finance department.

Most of the other ORFE classes are going to be less mathematical and more computational. But I like this class. It is about at the level of the 300-400 level undergrad math classes and it is kinda equivalent to MAT 391.

We are currently doing martingales and we will soon start with stochastic integration. If you have had some abstract probability theory before you might understand the lecture but otherwise you will just be observing what the lectures are like. [/hide]

[hide="an intro math class"] You could attend any of Math 215, 217, or 218. These all meet 10-10:50 on Friday (and 217 also meets 11-11:50, there are 2 sections). If you're going to major in math, you'll probably do these three classes in your first 3 semesters. The professors are usually some of the more senior professors in the department (a strategy for attracting math majors). These are basically the equivalent of Math 55 at Harvard, but a little less intense.

They are all intro classes, so if you've seen the subject at all before you should know what's going on. 215 is real analysis (like convergence, continuity theorems, etc). 217 is linear algebra. 218 is multivariable calculus (proof-based, with some basic differential geometry). [/hide]

Also, the Math Club will have an event on Friday afternoon. I'll post details here once I have them. It will be a lecture, probably about number theory, and there will probably be food.

edit: for anyone who will be on campus Thursday morning and doesn't mind waking up early, MAT306 Graph Theory at 8:30 and MAT325 Topology at 9:00 are possibilities as well. Both of the professors are good, and you'd have a chance of understanding the lectures too. (Actually I don't know what's going on right now in 306.) I am in MAT325. I like that class too.

The Math Club will be hosting an event on Friday for the prospective freshmen (and Math Club members). Everyone is welcome! Here are the details:

We will have lunch followed by a lecture by Professor Ramin Takloo-Bighash. The talk will be on a variety of problems in number theory and geometry which are solved by techniques of (abstract) algebra. Ramin says he will talk about "Pythagoreans, irrational numbers, geometric constructions, squaring a circle, the number pi, equations, King Solomon, different kinds of infinity, St Augustine, algebraic numbers and transcendentals, and just how lucky we are."

The event will take place from 12:00 - 1:30 PM on Friday, April 13 in Fine Hall room 322 (on the third floor).

I hope to see you there!


For each positive integer $ n$, define the polynomial

$ P_n(z)\equal{}1^3z\plus{}2^3z^2\plus{}...\plus{}n^3z^n$

Do the zeros of $ P_n$ lie inside, outside, or on the unit circle $ |z|\equal{}1$?

Posted already:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=210984


Find the number of distinct integral solutions of  $ x^{4} \plus{}2x^{3} \plus{}3x^{2} \minus{}x\plus{}1\equiv 0\, \, \left(mod\, 30\right)$ where $ 0\le x<30$.

$\textbf{(A)}\ 0  \qquad\textbf{(B)}\ 1  \qquad\textbf{(C)}\ 2  \qquad\textbf{(D)}\ 3  \qquad\textbf{(E)}\ 4$

Let $ A$ define that polynomial. If $ A\equiv0\mod{30}$, then $ A \equiv 0 \mod2, A\equiv 0 \mod3....$and so on $ \mod{n}$, n being positive factors of 30 (greater than 1). Putting $ A$ into $ \mod2$ and you will find that $ x\equiv1\mod2$. Which means that x is odd. When putting $ A$ into $ \mod3$, $ 0 \mod3$, $ 1\mod3$ and $ 2 \mod3$ represent all odd numbers, but you find that x is congruent to $ 1\mod3$. So with what we have so far we have that x could be $ {1,7,13,19,25}$. We could plug this in and see which ones work, but we can still narrow it down a bit more. If you look at these numbers in $ \mod5$, you see that they are congruent to $ {1,2,3,4,0}$, but only one of these can be right, so the answer is $ \boxed{B}$.

[quote="modularmarc101"]When putting $ A$ into $ \mod3$, $ 0 \mod3$ and $ 2 \mod3$ represent all odd numbers[/quote]
Well, $ 1$ is odd...

Ah poop. Then i guess i don't have a solution... Il try again.

Ok, i fixed it, it should be right now.

[hide]Denote the given expression by $ E$. Because of the free term "$ \plus{}1$", we see that $ x\not\equiv 0\pmod{2,3,5}$

a) $ x\equiv 1\pmod{2}$

b) For $ x\equiv\pm 1\pmod{3}$ we get

$ E\equiv 1\pm 2\plus{}3\mp 1\plus{}1\equiv 2\pm(2\minus{}1)\equiv 2\pm 1\pmod{3}$, so only plus sign works, hence $ x\equiv 1\pmod{3}$

c) For $ x\equiv\pm 1\pmod{5}$ we get

$ E\equiv 1\pm 2\plus{}3\mp 1\plus{}1\equiv \pm(2\minus{}1)\equiv \pm 1\pmod{5}$, so no solution in this case.

For $ x\equiv\pm 2\pmod{5}$ we get

$ E\equiv 1\pm 1\plus{}2\mp 2\plus{}1\equiv 4\pm(1\minus{}2)\equiv 4\pm (\minus{}1)$, so only minus sign works, hence $ x\equiv 3\pmod{5}$

From $ x\equiv 1\pmod{2},x\equiv 1\pmod{3}, x\equiv 3\pmod{5}$ we conclude there's exactly one solution (namely $ 13$), hence the answer is $ B$.[/hide]


A movie place charges 7 dollars for an adult ticket and 4 dollars for a child ticket. On a recent night, the sales of children tickets were 3 times the sale of adult tickets. If the total amount collected for the tickte sales WAS NOT greater than 2000 dollars, what is the greatest number of adults who could have purchased the tickets.

[quote="indianamath"]couldn't you also use system of equations?[/quote]
too much work
[hide="another idea"]
modular congruences[/hide]

Who's Kramer????

What are modular congruences?

Is there a simple way to solve this question or not?

you can just have a equaiton with an inequality..

lmao jeez you're scaring him

well let's say that each family has one adult and three children. this means that each family pays 19 dollars, and only whole families come to see the movie. (see why each family has exactly one adult and three children?)

now, at most, 105 families can come and see the movie (19*106 > 2000, but 19*105 isn't), so at most, 105 adults could have purchased a ticket.


function
induction
number theory unsolved
number theory

[quote="kvmedya"]Find all functions [b]F:N*N--->N[/b] satisfying:(for all natural [b]a,b[/b])
[list][b]1.f(a,a)=a
2.f(a,b)=f(b,a)
3.(a+b)f(a,b)=bf(a,a+b)[/b][/list]

 :alien:[/quote]
The unique solution I think is clear, and in fact the function is the ${\rm lcd}(a,b)$ function, because satisfies the 1. and 2. condition, and moreover  $a\cdot b=\gcd(a,b)\cdot {\rm lcd}(a,b)$ then
\[(a+b){\rm lcd}(a,b)=b\, {\rm lcd}(a,a+b)  \iff \gcd(a,b)=\gcd(a,a+b)\]
which is clear.

Best Regards.

RX TCM

Yes, but you have to prove there aren't any other functions too, of course...

Call $c=a+b$, then condition 3 becomes

\[(c-a)f(a,c)=f(a, c-a)c.\]

Thus if we have defined $f(a, c)$ we can also define $f(a, c-a)$ and vice-versa.

Note that by applying $(a,c) \rightarrow (a,c-a), c>a$, since we are defining progressively smaller pairs, eventually the process must terminate (or otherwise we'd go into negative numbers which are undefined), which only happens when we get a pair of the form $n,n$ (because then by applying condition 3 the next pair is $n,0$ which is undefined). Also note that if $d \mid a, d \mid c \Rightarrow d \mid c-a$, so $d \mid n$, then also $(a, c) \mid n$. Besides, if for any pair $(a,c), d \mid a, d\nmid c$, then $d \nmid c-a$, so a number that isn't a common divisor of $a$ and $c$ can't be a common divisor of two numbers in the same pair, and then can't divide $n$. Then $n=(a, c)$.

Now using condition 3 repeatedly, starting from $f(n,n)=n$ and going back up the chain up to $f(a,c)$ it's easy to check that it would give us $\frac {ac}{n}$.

[quote="Severius"]Yes, but you have to prove there aren't any other functions too... [/quote]

By definition this is clear enough in my opinion. I mean,  you can use an induction process to probe the unicity of the value of $f(m,n)$ for every pair, try to do it, I don't want to write every detail in every topic, I don't  do new IMO's exams, I preffer that you try to complete the probes.

Sorry and good luck.
RX

[quote="Severius"]Yeah, I already did that, using the chain going from $(a,c) \rightarrow ... \rightarrow (n,n), n=(a,c)$ backwards it is trivial.

Anyhow, anything non-trivial should be proved, otherwise how do you know you're right? ;)[/quote]
Well, don't if I understood but ${\rm lcd}(a,c)={\rm lcd}({\rm lcd}(a,c),{\rm lcd}(a,c))$ but please, can you write an SPECIFIC example. I am sure the solution is unique.

Best Regards.

RX

[quote="Severius"]I never said the solution wasn't unique, I just said that in general you can't be sure something non-trivial and unproven is right... Anyhow, what do you mean by $rmlcd$?[/quote]
OOooopsss, I wrote $\LaTeXe$ too fast  :?   , but now fixed.


function
trigonometry
geometry
trig identities
Law of Sines

This is my last problem for my presentation:

A V-gauge is used to find the diameters of pipes.  The advantage of such a device is that it is accurate and rugged, and has no moving parts to break down.  The measure of angle AVB is 54°.  A pipe is placed in the V-shaped slot and the distance VP is used to predict the diameter.  The line VP is calibrated by listing the corresponding diameters as its units.  This, in effect, establishes a function between VP and d.

a.	Suppose that the diameter of a pipe is 2 cm.  What is the distance VP?
b.	Suppose that the distance VP is 3.93 cm.  What is the diameter of the pipe?
c.	Find the formula for d in terms of VP.
d.	Find a formula for VP in terms of d.

Please show all detailed work.  Thank you.

[quote="Jhonny Ruiz"]where is point P located[/quote]

Hi Jhonny,

I am providing an attachment that may help.  It is backward, but I think you can figure it out.

Thank you.

Jhonny,

P, in the attachment, is shaded, and is between B and V.

Thank you.

I am sending back to top, because I added the attachment.  Hope that helps.

Thank you.

Thanks for the attachment. It helped a lot
I am vey uncomfortable with trigonometry. Please correct me if I'm wrong. Here it goes:

We know that the angle made by AVP is 54° and based on the diagram we can say that the diameter of the pipe is in front of this angle. Also we know that triangle $QVP$ has to be isosceles since both lines $AV$ and $VB$ tangent to the circular face of the pipe.

 $54+2x=180$
$x=63$
$x$ being the measure of angle $VPQ$.

We use Law of sines to find the relation between $VP$ and $d$. We have:

$\frac{\sin 54}{d}=\frac{\sin 63}{VP}$
From here we can derive:

$VP=\frac{d\sin 63}{\sin 54}$ And

$d= \frac{VP\sin 54}{\sin 63}$

Jhonny, thanks for the effort.  I think I have figured it out.  The other board helped me verify.  See if you agree with the following:

a. Suppose that the diameter of a pipe is 2 cm.  What is the distance VP?

We have

1cm = VP * tan 27°

Thus,

VP = 1 / tan 27° = 1/0.5095 = 1.96(cm)

b. Suppose that the distance VP is 3.93 cm. What is the diameter of the pipe?

We have

CP = VP * tan 27°

Thus,

CP = 3.93 * tan 27° = 2.00 cm

Thus, the diameter of the pipe is

2CP = 4.00 cm

c. Find the formula for d in terms of VP.

We have

CP = VP * tan 27°

Thus,

d/2 = VP * tan 27°

Thus,

d = 2 * VP * tan 27° = 2 tan 27° * VP = 1.02VP

d. Find a formula for VP in terms of d.

We have

CP = VP * tan 27°

Thus,

d/2 = VP * tan 27°

Thus,

VP = d/2 tan 27° = 0.981d


Let a,b,c be sides of a triangle.Prove that
 $ a^3b\plus{}b^3c\plus{}c^3a \geq a^2b^2\plus{}b^2c^2\plus{}c^2a^2$

[quote="quykhtn-qa1"][quote="8826"]Let a,b,c be sides of a triangle.Prove that
 $ a^3b \plus{} b^3c \plus{} c^3a \geq a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2$[/quote]
It 's IMO 1983 :wink:[/quote]
Sorry, I don't know

Nice problem. the inequality is:
$ a^3b\plus{}b^3c\plus{}c^3a \geq a^2b^2\plus{}b^2c^2\plus{}c^2a^2$
$ \Leftrightarrow \sum a^2b(a\minus{}b) \geq 0$
we have $ c\plus{}b\geq a \Leftrightarrow b\geq a\minus{}c$
then it's suffice to prove that, $ \sum a^2(a\minus{}b)(a\minus{}c)\geq 0$ wich is Schur's inquality.

[quote="8826"]Let a,b,c be sides of a triangle.Prove that
 $ a^3b \plus{} b^3c \plus{} c^3a \geq a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2$[/quote]
Can we use Muirihead here???

[quote="ElChapin"]No, we can't, since the left side of the inequality is only cyclic and not symmetric.

:)[/quote]
That was my argument too....


What is the smallest integer greater than $ \sqrt{300}$?


As $\angle B = \angle D = \angle F$, so $\angle BCA = \angle BAC = \angle DEC = \angle DCE = \angle FAE = \angle FEA = \theta$.
Consider $\triangle ACE$. By the Law of Sine, the given $BA=BC$, $DC=DE$ and $FE=FA$,
\[\frac{\sin \angle AEB}{\sin \angle CEB}\frac{\sin \angle CAD}{\sin \angle EAD}\frac{\sin \angle ECF}{\sin \angle ACF} = \frac{\frac{BA}{\sin(\theta+\angle EAC)}}{\frac{BC}{\sin(\theta+\angle ECA)}}\frac{\frac{DC}{\sin(\theta+\angle ACE)}}{\frac{DE}{\sin(\theta+\angle AEC)}}\frac{\frac{FE}{\sin(\theta+\angle CEA)}}{\frac{FA}{\sin(\theta+\angle CAE)}}=1.\]
By the Ceva's theorem, $AD,BE,CF$ are concurrent.

[quote="ALGEBRA"]in the haxagonal $ABCDEF$ the angles of $B,F,D$ are equal so $AB=BC$ and $CD=DE$ and $EF=FA$ prove that the lines of $AD,BE,CF$ are concurrent[/quote]

This follows from the Kiepert theorem ( http://www.mathlinks.ro/Forum/viewtopic.php?t=30114 post #2): Since AB = BC, CD = DE and EF = FA, the triangles ABC, CDE and EFA are isosceles (with bases AC, CE and EA, respectively), and since the angles at B, F, D are equal, i. e. we have < ABC = < CDE = < EFA, these isosceles triangles are directly similar, so we can apply the Kiepert theorem to triangle ACE and obtain that the lines AD, CF, EB concur; in other words, the lines AD, BE, CF concur. That's all.

  darij


[hide]
When we make the coefficient of $ x^{88}$ we select $ 88$ $ x$'s from the expression $ (x \plus{} 1)(x \plus{} 2)(x \minus{} 3)...(x \minus{} 90)$ as well as two other numbers.
So effectively the coefficient of $ x^{88}$ is when we multiply all pairs of numbers from the set $ 1,2, \minus{} 3,4,5, \minus{} 6... \minus{} 90$ and add them all together.

Now consider the expansion of $ (1 \plus{} 2 \minus{} 3 \plus{} 4 \plus{} 5 \minus{} 6... \minus{} 90)^2$
This counts all pairs of numbers that we need (twice) as well as the sum of all squares from $ 1\rightarrow 90$

So we need to evaluate

$ \frac {1}{2} \left[(1 \plus{} 2 \minus{} 3 \plus{} 4 \plus{} 5 \minus{} 6... \minus{} 90)^2 \minus{} (1^2 \plus{} 2^2 \plus{} 3^2...90^2)\right]$

we can simplify by spliting $ (1,2, \minus{} 3,4...)$ into three arithmetic sequences of $ 30$ terms
The first is $ 1,4,7... \Rightarrow S_{30} \equal{} \frac {30}{2}(2(1) \plus{} (30 \minus{} 1)3) \equal{} 1335$
Every term from the next two sequences have a difference of $ 1$
$ \Rightarrow (1 \plus{} 2 \minus{} 3 \plus{} 4... \minus{} 90) \equal{} 1335 \minus{} 30 \equal{} 1305$

Now the sum of the squares to $ 90$ can be simplified to $ \dfrac{90(90 \plus{} 1)(2(90) \plus{} 1)}{6} \equal{} 247065$

So $ \frac {1}{2}[1305^2 \minus{} 247065] \equal{} 727980$

Answer: $ 727980x^{88}$
[/hide]


linear algebra
matrix
inequalities
ratio
complex numbers
superior algebra
superior algebra solved

Let M_n(C) with n>=2 so that:
|det(In+zA)|<=|1+det(A)| for any z in C with |z|=1.
Prove that:
a)det(A) is a real and pozitive number;
b)there is a real number a so that A^n=aIn.

Take this det(A+B)+det(A+ \varepsilonB)+...+det(A+ \varepsilon  n-1B)=n(det(A)+det(B)). Where  \varepsilon is a primitive root of x n+1=0.   All you have to use are basic properties of determinants and of  \varepsilon.

In the problem
n|det(A)+1|=|det(A+I)+det(A+ \varepsilonI)+...+det(A+ \varepsilon  n-1I)| \leq |det(A+I)|+|det(A+ \varepsilonI)|+...+|det(A+ \varepsilon  n-1I)| \leq {for z=1/ \varepsilon  i} n|1+det(A)|.
This means that all the inequalities above are equalities so
|det(A+ \varepsilon  iI)|=|det(A)+1| for all i and the complex numbers det(A+ \varepsilon  kI) are colinear.
Can't remember now how to continue. Get back to it later. Maybe somebody else could prove it.

I put P(x)=det(In+xA).
Let P(x)=anx^n+..+a1x+1. It's obvious that an=det(A). Let  \varepsilon be the primitive root of the eqn x^n-1=0. Then I did the same thing as amfulger. You get
|P( \varepsilon ^k)|=|an+1| for all k from 0 to n-1 and, also,
P( \varepsilon ^k) are all equal because the sum of their moduli (is this the plural? :D ) = modulus of their sum, so they are all collinear and the ratio of each 2 must be positive, but since they have equal moduli, they must all be equal. Let P(1)=P( \varepsilon)=..=P( \varepsilon^(n-1))=z with z a complex number with modulus |an+1|. Then P(x)=an(x-1)*..*(x- \varepsilon ^(n-1)) + z = an(x^n-1)+z = an*x^n+z-an, but the last coefficient of P is 1, so P(x)=an*x^n+1, so |1+an*x^n|<=|1+an| for all x with |x|=1. It should be really easy to derive an positive real from here. This is just for (a). I haven't even thought abt (b) yet.

Sorry  \varepsilon is a primitve root of x n-1, not x n+1.


[color=darkred]Pentru o dreapta mobila $d$ care trece prin varful $A$ al triunghiului $ABC$ definim punctele $\{\begin{array}{c}U\in d\ ,\ BU\perp d\\\ V\in d\ ,\ CV\perp d\\\ MU\perp AB\ ,\ MV\perp AC\end{array}$.
Sa se determine maximul ariei triunghiului $UMV$ si pozitia punctului $M$ in care se atnge maximul cautat.

[b]Observatie.[/b] Va recomand sa tratati mai intai cazul particular $A=90^{\circ}$ pentru care $M\in BC$ (dovediti !).[/color]

Hello my friends  :lol: I saw the exercise which was proposed from Virgil Nicula and seems to be a very interest exercise so I wonder if anybody could translate it for me in English :maybe:

For a mobile straight line $d$ that always crosses through vertex $A$ of the triangle $ABC$ we choose points so that $\{\begin{array}{c}U\in d\ ,\ BU\perp d\\\ V\in d\ ,\ CV\perp d\\\ MU\perp AB\ ,\ MV\perp AC\end{array}$.
Determine the maximum area for the triangle $UMV$ and the position of the point $M$ when the maximum occurs.

[b]Note.[/b] I suggest that you first analyse the particular case for $A=90^{\circ}$ in which $M\in BC$ (prove !). :)

Thanks a lot my friend Topor for your time you spent to translate the exercise for and for your note (your idea I think is correct)  :)


[quote="konstantine"]Γεια σας και πάλι παιδιά.Λοιπόν την ανισότητα μου την ξεφτιλίσατε αλλά θα βάλω και εγώ μιά λύση αρκετά στοιχειώδη(με bcs και AM-GM).Κάτι άλλο όμως.Πρόσφατα κατα τύχην παρατήρησα ότι στην αναλυτική γεωμετρία αν αφαιρέσουμε κατά μέλη τις εξισώσεις δύο κύκλων βρίσκουμε το ριζικό τους άξονα.Υπάρχει κάτι τέτοιο ως γνωστή πρόταση η αν οχι μπορεί κανείς να δώσει κμια απόδειξη?[/quote]

auto ego toulaxiston den to gnoriza...alla sigoura tha uparxei os gnosti protasi h opoia einai profanis dioti afou afaireseis kata meli exeis to eksis:
$ (x\minus{}x_0)^2\plus{}(y\minus{}y_0)^2\minus{}R_0^2\equal{}(x\minus{}x_1)^2\plus{}(y\minus{}y1)^2\minus{}R_1^2$ diladi $ d_0^2\minus{}R_0^2\equal{}d_1^2\minus{}R_2^2$ htoi oi dunameis na einai ises......----------> rizikos axonas.

good observation though.


calculus
integration
trigonometry
derivative
calculus computations

Find the maximum and minimum values of $ \int_0^{\pi} (a\sin x \plus{} b\cos x)^3dx$ for $ |a|\leq 1,\ |b|\leq 1$.

Note that you are not allowed to solve in using partial differentiation here.

$ \int_{0}^{\pi}(a\sin x\plus{}b\cos x)dx\equal{}\frac{4}{3}a^3\plus{}2ab^2$.

So the maximum value is $ \frac{10}{3}$ when $ (a,b)\equal{}(1,1)$, and the minimum value is $ \minus{}\frac{10}{3}$ when $ (a,b)\equal{}(\minus{}1,1)$.


Okay this is a pretty easy warm-up problem for anybody that is having trouble with the problems in this section. This is for their level. Please don't solve it if it is a type of problem that yuo have mastered.

Problem;

At a restaurant, you can get 1 main dish, 2 side dishes (that can be the same or different), or 1 dessert. There are 12 main dishes, 6 side dishes, and 4 desserts. How many possible meals can you get.

Please don't answer if you are good at these kinds of problems only if you are still learning them.

[quote="Wickedestjr"]No, that is incorrect. I'll wait for somebody else can get it right. You're getting the right idea, but that is not the answer. [color=orange]It isn't 6 choose 2, because the dishes can be the same. It is only 6 choose 2 if the two side dishes had to be different. [/color][/quote]
6 choose...1?

Okay, let me tell you about permutations and combinations.

Combinations are when you can't use the same thing twice.
Permutations are when you can use the same thing twice.
Those are two very important words that you need to know.
You only use the choosing formulas when it is a combination.

This problem is a permutation.
Myyellowducky82's idea is wrong, because #1; it is using the choosing formula, and #2; there are [color=orange]2[/color] side dishes. Not 1.

[quote="nick42"]what?? isn't that idea similar to last year's state target round problem #2?

ok, well if you insist on a permutation, the answer if 21*30*4=$ \boxed{2520}$.

right?  :maybe:[/quote]
that sounds right...[this problem is really familiar]

[quote="Wickedestjr"]that can be the same or different[/quote]

It's a permutation hint hint.

[hide="I will hide the answer..."]There are $ 12$ different ways to get $ 1$ main dish.  There are $ 6$ different ways to get $ 1$ side dish, so you can get $ 36$ different groupings of side dishes.  However, you can get either $ 2$ side dishes or $ 1$ dessert, and there are $ 4$ dessert possibilities, so in total, there are $ 40$ possibilities for the second part of the problem.

Multiply them: $ 12 \times 40 \equal{} \boxed{480}$.

Hmmm, was that "or" supposed to be an "and"?[/hide]

Would you like me to tell you the real answer?

And if that really is an "or", what two things does it apply to:

2 sides OR a dessert

or

1 main dish and 2 sides OR a dessert

But the question of the "or" (and that's a pretty big one) has not yet been clarified.

"At a restaurant, you can get 1 main dish, 2 side dishes (that can be the same or different), or 1 dessert. There are 12 main dishes, 6 side dishes, and 4 desserts. How many possible meals can you get?"

Literally, this means that you have a choice between:

1. One of 12 main dishes
2. 2 side dishes
3. One of 4 desserts

If you choose one of 12 main dishes, you have 12 choices. If you choose 2 side dishes, you have 6^2 = 36 choices. If you choose a dessert, you have 4 choices. Total, you have 12 + 36 + 4 = 52 different choices.

This is only is you interpret the problem literally.

However, what if the problem was:

"At a restaurant, you can get 1 main dish, 2 side dishes (that can be the same or different), [u][b]AND[/b][/u] 1 dessert. There are 12 main dishes, 6 side dishes, and 4 desserts. How many possible meals can you get?"

We would have 12 choices for the main dish, 6^2 = 36 choices for the side dishes, and 4 choices for the dessert. (12)(36)(4) = 1728 so we would have 1728 different meals to choose from.

[quote="myyellowducky82"][quote="ernie"]Is it supposed to be 2 side dishes and a dessert?

Because the word "or" changes the scenario...[/quote]

The original question:
At a restaurant, you can get 1 main dish, 2 side dishes (that can be the same or different), or 1 dessert. There are 12 main dishes, 6 side dishes, and 4 desserts. How many possible meals can you get.

So you can get only a main dish if you want.
Or you can get a main dish and 2 side dishes...etc.[/quote]

And 1 desert. Sorry about that.

12markkram34, that's exactly why we're wondering what that "or" is supposed to mean. It changes the problem completely. Also, if that really is an "or", then does the comma between "1 main dish" and "2 side dishes" imply another "or"?

That means that the question should be interpreted as:

"At a restaurant, you can get 1 main dish, 2 side dishes (that can be the same or different), [u][b]AND[/b][/u] 1 dessert. There are 12 main dishes, 6 side dishes, and 4 desserts. How many possible meals can you get?"

Yeah, sorry about that guys. I guess the only person that learned from this problem was the author.  :rotfl:  :rotfl:

[quote="mathgeniusxx"]Um.. would it be 12*6*6*4?[/quote]

Yes that is correct!!!

huh isnt it $ 12*15*4$
because you can get 12 main courses 6 pick 2 sides and 4 different desserts?
Im really confused

No, you can pick the same side dish twice.

[quote="Wickedestjr"]Here is the answer;

1728 = 12 x 6 x 6 x 4[/quote]
So getting side dish A and side dish B is different when they are switched? (Getting side dish B and side dish A.)
[hide="My Answer"]There are 12 choices for the main dish, 36 choices for the side dish, and 4 for the dessert. But, for the side, there are 21, because 6*5/2=15 are repeated, so it would be 12*21*4, which equals 1008.[/hide]

Yes ttocs. The way the problem is interpreted Side dish A and Side dish B is different from Side dish B and Side Dish A.

Wait, does that mean we have to divide by 2?

[quote="Wickedestjr"]But I think that it would be (6 x 6)/2 , because the dishes can be the same. I do not see why it would be 6 x 5.[/quote]

There are $ 6$ possibilities for the first main dish.  There would be $ 6$ possibilities for the second one as well.  However, there would actually be only $ 5$ possibilities because $ 1$ side dish is already used.


[quote="bpms"][quote="jhredsox"]what is $(x+y+z)^{3}$ again?

for some random reason, it popped out of my head :blush:

-jorian[/quote]
The proof of this is kind of above mc, but:
$\sum_{cyclic}x^{3}+\sum_{cyclic}3 \cdot x^{2}y+6xyz$[/quote]

i don't need the proof, i just need it simplified (i forgot it :oops: )

what's the cylic thing?

jorian

so $a^{2}+b^{2}+c^{2}+3abc^{2}+3acb^{2}+3cba^{2}+6abc$?

that looks messed up

jorian

that's what i was thinking, but then i thought you were to lazy to put the three variables in that section

thanks

jorian

Or you could be 1337 and use the multinomial theorem:
\[\left( \sum_{i=1}^{m}x_{i}\right)^{n}= \sum_{k_{1}+k_{2}+\ldots+k_{m}=n}\dbinom{n}{k_{1},k_{2}, \ldots ,k_{m}}x_{1}^{k_{1}}\cdot x_{2}^{k_{2}}\cdots x_{m}^{k_{m}}\]
where
\[\dbinom{n}{k_{1},k_{2}, \ldots ,k_{m}}= \frac{n!}{k_{1}!k_{2}!\cdots k_{m}!}\]

[quote="bpms"]cyclic means swap all variables of x y and z because I was too lazy to type up the whole thing.[/quote]

That is the [u]symmetric[/u] sum, I believe, and not the cyclic sum. Like, \[\sum_{cyc.}f(a,b,c) = f(a,b,c)+f(b,c,a)+f(c,a,b) \] \[\neq f(a,b,c)+f(a,c,b)+f(b,a,c)+f(b,c,a)+f(c,a,b)+f(c,b,a) = \sum_{symm.}f(a,b,c) \]

for some function $f(a,b,c)$ on three variables. In a cyclic sum, you [i]cycle[/i] through the variables in order; in a symmetric sum you take [i]all possible orderings[/i] of the variables.

[quote="kevenmee"]its not really that hard to figure it out by foiling[/quote]

foiling only applies when multiplying something of the form $(a+b)(c+d)$ because it stands for "first, outer, inner, last."  In $(x+y+z)^{3}$, there are more terms.

[quote="ragnarok23"][quote="bpms"]A suggestion.
For ugly problems where it seems messy because of large numbers, try to use variables to represent repeated numbers.

This method nearly trivialized a problem on ragnarok+eli's mock, and a problem from last year's HMMT power round (B)[/quote]
wait which one was that[/quote]
The "value of your name"
You should have said, "Sukhdeep, Hardik, Aneet, or Venkatramaanvaas is trying to find the value of his name" or something.
The second I read it I knew that I could just let the value be a variable.

[quote="cincodemayo5590"]

That is the [u]symmetric[/u] sum, I believe, and not the cyclic sum. Like,
\[\sum_{cyc.}f(a,b,c) = f(a,b,c)+f(b,c,a)+f(c,a,b) \]

\[\neq f(a,b,c)+f(a,c,b)+f(b,a,c)+f(b,c,a)+f(c,a,b)+f(c,b,a) = \sum_{symm.}f(a,b,c) \]
for some function $f(a,b,c)$ on three variables. In a cyclic sum, you [i]cycle[/i] through the variables in order; in a symmetric sum you take [i]all possible orderings[/i] of the variables.[/quote]

ohhh wow I never knew of that distinction. So Vieta is symmetric right? I always wrote everything as $\sum_{cyc}$, I guess not anymore.

Yeah, there was a big rant at camp about how you can't WLOG a<b<c<d<... if it's a cylic sum, you can WLOG a<b<c<d<... only if it's symmetric...


Hi everyone! I'm new to mathlinks, but I'd like some help on some problems that I've known for a while but I never was able to solve..

1) Construction: given the lengths of the three bisectors of a triangle, construct the triangle (with line and compass).

2) Oviously you are always able to divide a square in an even number of identical triangles. Prove that it is impossible to divide it in an odd number of identical triangles.

Thanks  :lol:

[quote="Mat_88"]1) Construction: given the lengths of the three bisectors of a triangle, construct the triangle (with line and compass).[/quote]

You will never be able to solve this. In fact, it is unsolvable: In general, one cannot construct the triangle from the lengths of its angle bisectors using compass and ruler only.

See American Mathemathical Monthly, January 1994, pp. 58-60. (I have not seen this article, I just saw a reference to it.)

  Darij

[quote="Mat_88"]2) Oviously you are always able to divide a square in an even number of identical triangles. Prove that it is impossible to divide it in an odd number of identical triangles.
[/quote]

In fact, there is a stronger result, it is impossible to divide a square into an odd number of  triangles with equal areas.

It was proved in the article by Paul Monsky ( American Mathematical Monthly 77 :2 (February 1970), 161-164).

oh,ok.. i will look for those articles.
thank you very much to both of you!
Matt


calculus
integration
calculus computations

find  the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

y=x^4, y=1; about y=2

i drew the graphs and shaded the regions,

here is what i did V k= pie {2-(x^4 k)^2} change of x k

then i factored it out (2-x^4) (2-x^4)

4-2x^4-2x^4+x^8

4-4x^4+x^8

then i antidifferentiated it

16 integral sign, x^5/5 + x^9/9

i plugged in 1 into the x signs and i got 14/45 and i multiplied it by 16 and i got 224pie/45

and the answer is 208 pie/45, so what did i do wrong, can you show me the right way to do it

So we have the region between $y=x^{4}$ and $y=1$.  We're revolving it around $y=2$.

[hide]So our equation will be (disc method):

$V= \pi \int_{-1}^{1}\left[ (2-x^{4})^{2}-(2-1)^{2}\right] ~dx$; And it is even so we can $\cdot 2$ and go from $0$ to $1$.
$V=2 \pi \int_{0}^{1}\left[ (2-x^{4})^{2}-1 \right] ~dx$
$V=2 \pi \int_{0}^{1}\left[ x^{8}-4x^{4}+3 \right] ~dx$
$V=2 \pi \left[ \frac{x^{9}}{9}-\frac{4x^{5}}{5}+3x \right]_{0}^{1}~dx$
$V=2 \pi \left( \frac{1}{9}-\frac{4}{5}+3 \right)$
$\boxed{V= \frac{208 \pi}{45}}$[/hide]

hey, why is 2 near the pie sign, shouldn't it be 4

Think of it as first we're swinging the main thing around which is from $2$ to $x^{4}$... So that is the region of $2-x^{4}$.

Next, we're gonna swing around the region between $2$ and $1$ which we're going to subtract out because it was accounted for in the last swing but we don't want that, its not part of the region specified.  So that it $2-1=1$.

how did you get 2 pie instead of just pie

You can just keep integrating as normal and then plug $-1$ in later on if you want in place of $0$ and get rid of the $\cdot 2$ if you want.

For example..  Say I were to tell you to swing the region between $y=1$ and $y=x^{2}$ around the x-axis.  Because $y=x^{2}$ is the same from both $-1$ to $0$ and $0$ to $1$...  We can just swing around the region from $0$ to $1$ and then multiply by $2$.  It's a property of even functions.

We couldn't do this with, for example, $y=x^{3}$ or something along those lines, an odd function.

As long as the upper and lower bounds are the same, yes.

oh, is that a theorem, i need to pay attention to those things


The problem that I am having trouble with is

(4a-2b-1)(-ab)

Would I write this 4a(squared)b+2b(squared)a+1ab

Any help would be greatly appreciated. Thanks.[/code]

Remember, the negative on the ab term carries through, so the 4a would also be negative, resulting in $ \minus{}4a^{2}b\plus{}2ab^{2}\plus{}ab$


summer program
Mathcamp
HCSSiM
PROMYS
SuMAC
Ross Mathematics Program
search

I have not been to Mathcamp so I cant tell you much about it, but I highly reccomend Hampshire College Summer Studies ([url]http://www.hcssim.org/[/url]) in Mathematics, which I attended last summer.  There are many other summer programs in mathematics that you should also consider (PROMYS, SUMaC, Ross, etc.) but alumni of those programs would be able to tell you more about those.

Thanks for your reply.

The problem is that I live in Canada and I'm not so sure if the camps in USA for Americans only or for Canada/USA.

Most of those camps will probably take anyone from Canada, but I'm not 100%.  You could certainly e-mial them and ask.

Also, tehre are several other threads like this in forums here.  You might do a search for the words "math camp" or "mathcamp."

My son has been to USA-Canada Mathcamp for the past 3 summers. It's always the highlight of his year! The USA-Canada part is that they usually hold it somewhere near the border to make it easier for both US and Canadians to go. A lot of the staff seems to come from Waterloo or MIT. But internationals go also. One year my son had a roommate from Greece.

Mathcamp is different from the other camps like PROMYS, Ross Young, etc in that there is a lot of variety in the offerings. The others are strictly number theory. At Mathcamp you get a huge smorgasbord of math related stuff to do, and you get to pick and choose whatever you are interested in each day.  Some of the courses run the entire length of the camp, others might be just a week or just a single lecture.

At SUMaC in 2001, most of the people were from the US, but there was one person from Canada, one from England, one from Taiwan, and one from Singapore, so it's definitely possible to go there even if you aren't from the US (provided that you are a sophomore or junior right now). SUMaC material seems to be rather different from that of most of the other camps (Ross, PROMYS, etc.). You have two options for what to study: group theory or topology. I found it very enjoyable.

i havent been to mathcamp but ive been to mathpath, which is run by the same person.  There were many different branches of math that we explored, inlcuding geometry (euclidean, non linear, quadratic, i think thats what they were called), number theory, and history of math (which were actually pretty interesting.  In the afternoon, we had problem solving for different levels of students (mathcounts, amc, aime, olympiad, etc).  These problem solving classes are taught by the counselors, who have been through these competitions as students (kinda like MCrawford and ComplexZeta).  I enjoyed it.

Im aware that many math camps (including Ross and PROMYS) are focused primarily on number theory.  That was probably what made me decide to go to HCSSiM.  I can tell you from my experiences with HCSSiM that it is not focused on one branch of mathematics.  I ended up taking topological graph theory as my main course, which was a refreshing break from the standard topics generally found in math contests and which I also found extremely interesting.

The best part about HCSSiM though, in my opinion, was the relaxed atmosphere.  In the time that we werent doing math or other scheduled activities we were free to do pretty much anything (HCSSiM is not legally considered a "camp", but rather a "Summer Studies" which seems like a petty distinction but actally frees them of some liability constraints that other summer math camps have to deal with).  For example, on several equations I took a cab to a mall nearby with a group of friends.  I cant really compare HCSSiM to any other programs because I have not attended any of the others, but I can certainly highly reccomend it.

Concerning your worries about your Canadian residency -- though there were no students from out of the US last year, im sure that they would not have a problem with accepting a student living in Canada.

Actually.. There was 1 student from Quebec... I lived with him Ian!  HCSSiM is split into two three week sessions.  In the first session, students are split into two classes and taught basically the same things.  In the second session, students get to choose one major class and two mini classes (each one and a half weeks long) HCSSiM is lots of fun, and I would also highly suggest everyone to go there to learn how to prove things and famous theorems.

[quote="kool_dudy"]I found a MathCamp that is called [b]Canada/USA Mathcamp[/b]. I'm probably going to apply to it this summer althought I'm still working on the qualifying quiz.  I was wondering if you guys have ever been there and if you can give me any opinions and suggestions.

Here's the website...

http://www.mathcamp.org

Thanks.[/quote]

MathCamp is so awesome!!  (i went in '02)  there are quite a few people from Canada, actually, around 20?  i think.  they offer lots of really nice classes (Titu Andreescu comes for a week as does John Conway and they have really nice camp activites.  awesome place)  but it really depends.  how old are you?  and what maths have you done?  i was in 8th grade, it some of the stuff was beyond me, so yeah.

I'd like to plug the Canada/USA Mathcamp as well.  I attended in 2001 and 2002, and I am intending to attend this summer as well.  They have classes for all levels, and the classes vary from year to year, so that people can attend several years in a row and get more out of it each time.  I think I can safely say that I learned more mathematics in my two summers at Mathcamp than in any other way during my time so far in high school (I'm a senior).  My advice is, if you really like math - go.  If you put forth the effort, you can learn a lot and have a good deal of fun interacting with other math people.

Thanks for all your comments.

I'm really interested in MathCamp but I'm struggling with the problems and I doubt I can get as many as 5 or so.  First of all, I can't really get a start to any of them.

So...meh...I'll probably have to wait for next year, just to wait to improve my math skills before going although I really do want to go this year however...like I said, I can't solve those problems.

Once again, thanks for your comments.

[quote="kool_dudy"]I'm really interested in MathCamp but I'm struggling with the problems and I doubt I can get as many as 5 or so.[/quote]

Try submitting an application with however far you get on the problems, and see what they do.

I'm 14...I assume I'm really good at arithmetics but horrible at problem solving.

I'll try...thx TA

Hmm...is this individual project thing tetrahedr0n mentioned research? Or is it just some sort of silly project thing =)...? I just decided about a week ago (after I finished all the Ross problems...) that I didn't really want to go to one of these camps (MathCamp, Promys, Ross) and preferred to just learn some math on my own at home...But I think a research project would be really fun to do, even though I don't know what math research really is like--I have to start somewhere! I'd really like more information about it, too. How interesting and fun is it? Also, do you know if Promys and Ross have projects like this?

The first serious research project I did was essentially copying a proof of a difficult theorem from a book and writing it up in my own way. If by research you mean that you want to do your own math, that's really tough. I could never have come up with that proof that I learned. It is truly brilliant.

JBL did something and was semifinalist in the STS.

NEW YORK
Joel Brewster Lewis, 18, of Brooklyn, analyzed a vector game of his own invention for his Intel Science Talent Search project in mathematics. Based on a previously studied but unsolved game called Half-Liar, Joel created his own version of two-dimensional vector manipulations. After analyzing his two-person game of integer-valued vectors, he found all starting positions which allow the second player to win in a prescribed number of moves. Such games are useful in handling false positives in medical tests and corrupt signals in communications. The son of David Lewis and Ellen Winner, he earned perfect scores on his SATs. As math team captain at Stuyvesant High School in New York City, Joel spends every morning teaching fellow students. Last summer he volunteered as a teaching assistant at CCNY and eventually would like to join academia as a mathematics professor and researcher. Winner of numerous math honors, Joel hopes to contribute to the field by advancing mathematical research and by continuing to teach interested students. His hobbies include hiking, juggling, writing stories and debating political issues.

[quote="tetrahedr0n"]If mathcamp admits me, I will definitely go this summer. I have a few questions for alums though.

I have 7 problems with full solutions on the qualifying quiz. Does this sound reasonable for an average person that is accepted? Furthermore, I wrote out formal solutions, like I would send to USAMTS or another mail contest. Should I include extra things, like my comments on the problem, my early (unsuccesful) work, and how I came to my solution?

At the camp, how does the individual project thing run? Is one necessary? Should we come with a prepared idea of what kind of project we would like to work on? And what kind of time do you have to do this?

Classes at the camp: Generally, which classes are more interesting? The really long classes, like Linear Algebra, or Topology, which are several weeks, or shorter (1 week or less) sections? Are the problem solving classes very helpful, as opposed to reading/practicing from a problem book like engel? Also, how much material can be expected in a theoretical class, especially one of the longer ones?

And most importantly, how is the food?

Thanks in advance.[/quote]

um, the food...  eh, well the college changes each year, so the food varies, but then, if you bring enough monies, you can go and eat out.  seven problems ought to make it, i think.  and yeah, like usamts.

I haven't been to MathCamp, but I would recommend writing down lots of ideas about problems. Sometimes on such things I would write down something along the lines of ``This problem reminds me of this other problem because we can do this and get something nicer, and that's how we solve the other problem, but then I don't know how to finish off the solution after having done that.''

The qualifying quiz looks a bit more difficult than it has been in previous years, so I imagine that 7 correct solutions will hold you in good stead.  Of course, it would be best to keep working and also include progress on the other ones.  As far as format, I would suggest formal solutions as the way to go, but definitely put down any enlightening comments that you have.  They probably don't want the story behind each of your solutions, but they do want to know how you think mathematically.

The individual projects (at least in '01 and '02, when I attended) were run a bit haphazardly, and not everyone did them.  If you have some good ideas before you go, great, but don't sweat it if you don't.  The staff would be more than happy to get you acquainted with an interesting problem.  And of course, the learning is far more important than the "officialness" of the project.

Classes are very interesting in general, but the ones that are most interesting depend on your preferences.  Generally you will want to take some longer classes and some shorter ones - and of course, don't miss John Conway!  The problem solving classes, which are run by Dorin Andrica, are very good; I learned a lot from his Olympiad problem solving class.   The material covered in classes varies quite a bit, but tends to be more than you expect.  If you want a [i]real[/i] challenge, you can take the long "Moore-method" class, where students have to provide proofs of all the results; these are a lot of fun, but can be very taxing.

As Mathcamp is being held at Colby College this year, I can assure you that the food will be very good.  I think that Colby has been ranked in the top few colleges in the "best food" category by one of the college reviews.  The key to dining at Mathcamp involves two things in addition to food, however:
(1) The Round Table Theorem.  I think most of us know what this is; if not, I'll let someone else explain it.
(2) Pens, to write math on napkins.  Yes, this happens a lot.

I'm also planning to apply to Mathcamp this year.  I read on their website that international students could apply, so I thought I'll give it a shot since I'm going to the US this summer anyway.  I am an US citizen who's Taiwanese but grew up in Taiwan all of my life.  It'll be an interesting experience I think if I am accepted.  I came up with solutions for 8 questions, but I'm still not very sure how to write my solutions formally.  Is it necessary to prove your solution formally or is it enough to justify it clearly using reasoning and logic?

I'm also applying to PROMYS this year.  Wrote a letter to them concerning whether US citizens who live abroad could participate, but never got a reply back, but anyway, their quiz is much easier than the Mathcamp quiz, so I'll give it a shot.

I'll plug Canada/USA MathCamp -
I attended MathCamp this past summer when I was 14 years old.  It was an awesome experience - I was exposed to a ton of new math that I had never begun to imagine existed.

But to get a little more concrete... At MathCamp, there are basically no rules, and it's a ton of fun.  It's assumed that if you want to attend, you are mature and can take care of yourself.  So, you can attend whatever classes you want, and theoretically you don't have to go to any (but who would want to do that?).  Most people attend class whenever possible, and the classes are arranged so that during each time slot, classes of all different difficulty levels are available.  There are no "bed times" or "lights out," and no set time to wake up, either.  In effect, everything is optional.  All the people are really nice, fun, and smart.

As for the qualifying quiz, try to explain and prove your answers as much as possible.  Seven seems pretty solid... but I guess I don't have the authority to say that.

Oh yeah, the food at University of Puget Sound last year was pretty bad.. but they usually had cereal, cookies, fruit, and other generic stuff.  It's at a different location this year, though, so it'll vary.

Good luck!  =)

Oh yes, the round table theorem - if n people can fit around a round table, then n+1 people can fit around a round table =)

1. Hello Caroline!

2. As Ari commented, "[vegetable] is like drinking water, but chewier". You should know that, you were the yearbook girl. ;D

3. Colby College is rumoured to have the best food.

Hope to see you all there! I definitely will be reapplying.

[quote="Caroline"] In effect, everything is optional. [/quote]

Except the monday meetings of course. ;D And also the three rules:

1. Sign in between 8 and 10 PM every day.
2. Don't play with fire.
[b]3. Don't divide by zero without staff supervision.[/b]

The third directive takes precendence above all else.

As Brenda said: "You need way more staff supervision to divide by that many zeroes."

Colby is much nicer than UPS. It is one of those sterotypical NE liberal arts colleges. However, the weather in Seattle was a whole lot better. No grantees in Maine. The best thing is that you will be in Maine. The worst part is that you will be in Maine. Make your choice.

BTW. MC was awesome. I sucked at math and I still enjoyed it. (But don't assume that everyone else is like me).

Seattle weather was glorious. I just wish the food there backed it up...

I don't suppose there'll be as many extracurricular activities there though, what with Waterville being in the middle of nowhere. Is it close enough to the sea that we can go kayaking again? :D

(Yeah, a clumsy I tripped David Roe over in his kayak... that was funny ;) )

[quote="JP"]Comparing MathCamp, Promys, Ross, and other math summer programs, which one is the best?  Are those programs going to help me doing better in AIME and USAMO?[/quote]

There's another thread about this somewhere in this forum...

I don't think the goal of the programs is to improve your performance on the AMC tests, but I think that if you learn methods and approaches and stuff, you certainly won't get any worse..

These camps will help you in the sense that you'll have other people to talk to, and they will also want to do well. Andrew Dudzik (who went to MOP 2000) was a counselor at SUMaC 2001, and he gave me a lot of good advice for math contests. It was also nice that the best bookstore within a 300 mile radius (at least) of my house was half a mile away or so from where I was staying, and I went there almost every morning before class (which was in complete violation of one of the camp rules... I'm such a rebel).

I'm applying too, but as they sent me the invitation only 3 days ago...  I'm gonna have to try hard.  Letsee, I'm 15 and I got 6 problems solved and pretty sure about the solutions to 4 and and 2/3 of the 6.  #7 is an interesting problem, I am pretty sure I encountered something like it before, now I'll just have to go dig through a mountain of stuff I saved since 7th grade...  So is there like a cut off?  And I saw only 100 people can get in, so around how many usually apply?  And is admittance by age&quiz score or is it just quiz score?  Thx in advance.

[quote="Morukumo"]I'm applying too, but as they sent me the invitation only 3 days ago... .[/quote]

Hi.  I'm one of the organizers of Mathcamp.

I'm sorry you didn't get the invitation until just a few days ago: they were mailed in March (!) but bulk mail seems to have been particularly slow this year.  Anyway, you're far from the only person in this situation.  If you can get your application done by Thursday, that would be best; but if you want to take the weekend to finish it, that should be fine.

We will take 110 students this year,  which will mean that there are around 70 spots for new students (alumni of the program can return automatically, if they wish).   We don't take your age into account when we're looking at the quiz, but admittance is definitely not by quiz alone: we take into account everything that we ask for.  (More than once, someone has gotten into Mathcamp because their personal statement has been so interesting that we just had to meet them!)  Typically, if you correctly solve 7 or 8 problems, it's a very safe bet that you'll get in.   5 or 6 is starting to take your chances -- lots of people who solved 5 or 6 problems have gotten in, but definitely not everyone.  But no guarantees that these numbers will hold true again this year.

I'm glad you're enjoying the quiz!

Application sent!  Darn, been doing this the whole week, lol, and a math contest that I'm supposed to represent my school for is tommorrow, and I haven't looked at a single problem yet.

My earlier posting about the summer math camp did not come out the way I intended it.  Could any one suggest a summer math program, which will help me most in doing well in AIME/USAMO?  I know we are supposed to do math for the love of it etc  but right now my goal to go to a summer math camp is to do well in AIME/ USAMO.  I am hoping for a serious input, not your camp is better than mine type of discussion. I went to SUMaC last summer. Thanks

[quote="beta"]I have full solutions to #1,2,3,4,6,8,9. I didn't try hard on #7 but I have 1/3 of solution. #5 is so poorly worded that I am not going to even try to understand. #10 sounds too advanced for me I am not going to try. I just left all my solutions at school today, I guess I am not going this year  :cry:

Also do they even accept 8th-graders?[/quote]

First of all, the deadline is not totally solid - you still have a chance if you send it in a few days after April 29 (or so I've heard).

Also, are you currently an eighth grader?  I went last summer, and I am currently a ninth grader, so I'd say yes :)

[quote="beta"]yes, I am currently an 8th-grader.[/quote]

Then by all means, apply, and good luck :)

I am actually wondering how hard is it to these math camps/....I wanted to apply this summer..but i ended up chickening it out becuase i was afriad of a rejection letter....so yea...can anyone sort of tell me like how many quations they asnwered for like the Canada/USA math cmap or SuMAC or any other summer math camps and just what you think these math camps are looking for so I could get a feeling of wther to chicken out next summer or not...Thanks!!

I was told at camp that about 40% of people who applied got a place this year (Canada/USA Mathcamp) - and on the website it said most people answer about 5 questions...i think....

I was at SUMaC 2004 this summer, and I tried to answer all of the entrance problems, but when we went over the problems at camp, I discovered that I got a problem or two messed up (out of 8). It's less about what you get as your answer, and more about what your thought processes are. The program itself is as hard as you want to make it. 'Twas a really fun and exciting experience for me. i definitely recommend it.  :lol:


Prove the 3 succesive terms of a G.P. would form the sides of a Triangle if the common ratio 'r' satisfies the inequality:

1/2(root5-1) < r < 1/2(root5 +1)

WLOG we can assume the sides are 1,r, and $ r^2$.
So we must have $ r^2\plus{}r>1$, where the intercepts are $ (\minus{}1\pm\sqrt{5})/2$
Also, we must have $ r\plus{}1>r^2$,  where the intercepts are $ (1\pm\sqrt{5})/2$
So we see these two are always true in the range you gave.
$ r^2\plus{}1>r$ is true for any r, as it is $ (r\minus{}1/2)^2\plus{}3/4$>0


The equation $ x^3\plus{}6x^2\plus{}11x\plus{}6\equal{}0$ has:

$ \textbf{(A)}\ \text{no negative real roots} \qquad
\textbf{(B)}\ \text{no positive real roots} \qquad
\textbf{(C)}\ \text{no real roots}  \\
\textbf{(D)}\ \text{1 positive and 2 negative roots} \qquad
\textbf{(E)}\  \text{2 positive and 1 negative root}$

[hide]
Factor it into
$ (x \plus{} 3)(x \plus{} 1)(x \plus{} 2)$
Therefore, the roots are $ x \equal{} \minus{} 1, \minus{} 2,$ and $ \minus{} 3$
Thus, the answer is $ \boxed{B}$, no positive real roots.
[/hide]

[hide="Click for solution"]
By Descartes' Rule of Signs, the answer is $ \boxed{\textbf{(B)}}$.
[/hide]


algebra
polynomial
real analysis
real analysis unsolved

Find the coefficient of $x^{50}$ in $(\sum_{n=1}^{\infty}x^{n})^{3}$

$(\sum_{n=1}^{\infty}x^{n})^{3}=(\frac{1}{1-x}-x)^{3}$, then expand and use the following general power series expansion:

$\frac{1}{(1-x)^{\alpha}}=\sum_{n=0}^{\infty}\frac{\Gamma(n+\alpha)}{n!\Gamma(\alpha)}x^{n}$, for $|x|<1$, $\alpha >0$ and you should get it.

Or a combinatorial explanation: this coefficient is the number of ordered triples of positive integers $(a,b,c)$ such that $a+b+c=50.$ The first thing we do with that is note that it is the same as the number or ordered triples of nonnegative integers such that $a+b+c=47.$ Imagine a row of dots and slashes, with $47$ dots and $2$ slashes. Let $a$ be the number of dots before the first slash, $b$ be the number of dots between the two slashes, and $c$ the number of dots after the second slash. We can arrange this in $\binom{49}2=\frac{49\cdot48}2=1176$ ways. So the coefficient is $1176.$


The vertices and midpoints of the sides of a regular $2007$-gon are to be numbered with numbers $1,2,...,4014$, so that the sum of three numbers on every side is the same. Show that such a numbering is possible


AMC
USA(J)MO
USAMO
search
Diophantine equation

What is the smallest n > 1 for which the average of the first n squares is a square?

A quick search using Mathematica shows the answer is 337.
I believe that this problem is from IMO 1990 Longlist.

The problem is equivlent to find (m,n) such that (n+1)(2n+1)=6m 2 .
Multiply with 8 we get (4n+3) 2 -3(4m) 2 =1.
substitute an+3 with x and 4m with y we have a Pell equation: x 2 -3y 2 =1.
check the first solutions for whether x=3 and y=0(mod 4) we got the smallest solutions are (x,y)=(7,4) and (1351,780).
(x,y)=(7,4) gives n=1, so we look at (x,y)=(1351,780).
That gives n=337 and m=195.

Yeah.

[quote]I believe that this problem is from IMO 1990 Longlist.[/quote]

kalva shows it as a usamo though, so i don't know

i think kalva has some of the problems switched. some of the usamo problems in math olympiad challenges dont appear on the site.


inequalities
calculus
derivative
inequalities proposed

Hi!

$x,y$ and $z$ are positive real numbers. What is the least value of
\[{{(x^{3}+3)(y^{3}+6)(z^{3}+12)}\over{(x+y+z)^{3}}}\]

[quote="zhaobin"]Sorry,you are right. :blush:[/quote]

I have a suspicion that the minimum value is when $(x^{3},y^{3},z^{3})=(3,6,12)$ I also think that it is solvable using derivatives :wink:


How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?

$ \textbf{(A)}\ 41\qquad
\textbf{(B)}\ 42\qquad
\textbf{(C)}\ 43\qquad
\textbf{(D)}\ 44\qquad
\textbf{(E)}\ 45$

[hide="Solution"]The tens digit must be integral, so the first and last digit must both be either odd or even.

Odd case: 111 -> 999. There are 25 odd three digit numbers such that the tens digit is the average of the hundreds and ones digit.

Even case: 210 -> 888. There are 20 even three digit numbers such that the tens digit is the average of the hundreds and ones digit.

Therefore, there are 45 total. $\boxed{E}$[/hide]

Do you even need casework?
[hide]
9 x 1 x 5 = 45
E
[/hide]

[hide="Solution"]Let the digits be $A, B, C$ so that $B=\frac{A+C}{2}$. In order for this to be an integer, $A$ and $C$ have to have the same parity. There are $9$ possibilities for $A$, and $5$ for $C$. $B$ depends on the value of both $A$ and $C$ and is unique for each $(A,C)$. Thus our answer is $9 \cdot 5 \cdot 1 = 45 \implies E$.[/hide]

[hide]Think of $xyz$

$y$ must be a whole number, the average of $x + z$ is only a whole number when $x$ and $z$ are both odd or both even. There are $10$ integers $x$ or $z$ can be.

Treating $0$ as even, number of odd/odd combos $+$ even/even combos $= 5^2 + 5^2 = 50$
Total combos $- 5 xy0s = 50 - 5 = \boxed{45}$[/hide]

Amctrivial grind: If the middle digit is $1$, there are $2$ ways. If the middle digit is $2$, there are $4$ ways. So on until $5$, where there are $9$ ways. For $6$, there are $7$ ways. So on, until $1$ way for $9$. The answer is $2+4+6+8+9+7+5+3+1 = \boxed{45}$.

[hide = sol]

A unique middle digit can be generated if the two digits beside them are both odd or even.

If they are both odd, there are $5^2 = 25$ ways to choose them.

If they are both even, and nonzero, there are $4^2 = 16$ ways to choose them.

If the last digit is 0, then there are an extra $4$ cases.

So the answer is $25 + 16 + 4 = 45$


linear algebra
matrix
function
logarithms
algebra
polynomial
linear algebra unsolved

Let   $A= \left(\begin{array}{cc} a & b\\ &\\ b & 2d-a \end{array}\right)$  with $a,b,d \in {\mathbb R}$. [b]Find all real numbers $a,b,d$ for which   $\ln(A)$  is a real matrix .[/b]

A necessary condition:

If $e^B=A,$ then $\det A = \det e^B=e^{\text{tr}B}.$

If $B$ is real, then $e^{\text{tr}B}>0.$

So the necessary condition is that $\det A>0$; in other words, $a^2+b^2<2ad.$

I haven't yet checked whether this is sufficient.

The necessary and sufficient conditions are $\det A>0,\,a>0$ - which is to say, $A$ is positive definite.

In general, a real symmetric $n\times n$ matrix $A$ has a real symmetric logarithm if and only if $A$ is positive definite.

This can be proved simply by diagonalizing $A.$

[quote="Kent Merryfield"]The necessary and sufficient conditions are $\det A>0,\,a>0$ - which is to say, $A$ is positive definite. .......[/quote]
Indeed  ! Thanks for interest !
 It may be observed that when $ A $ is a   $2\times 2 $ real matrix having sectrum  $ \sigma_A:=\{\lambda_1,\lambda_2\} $ is its spectrum , then for
  $ f$$: D\to {\mathbb C}, \sigma_A\subseteq D , $ we have
\[
f(A):=\left\{\begin{array}{lcl}\displaystyle \frac{f(\lambda_2)-f(\lambda_1)}{\lambda_2-\lambda_1} A
+\displaystyle \frac{\lambda_2 f(\lambda_1)-\lambda_1 f(\lambda_2)}{\lambda_2-\lambda_1} I    & , &  \lambda_2\ne \lambda_1 \\
&&\\
\displaystyle f'(\lambda_1) A+
\left(f(\lambda_1)-\lambda_1 f'(\lambda_1)\right) I    & , &  \lambda_2=\lambda_1 ,\; \mbox{[when}\;   f'(\lambda_1)  \; \mbox{exists ] }
\end{array}\right. ,
\]
$ I $ being the unit $ 2\times 2 $ matrix.

[b]Remark:[/b] [i] In fact, according to theorem of Hamilton-Cayley it follows that the more "sophisticated" matrix functions  $  f(A)  $,  $ A  $ being   $ n\times n $, [/i] [b] are the polynomial functions  of degree $n-1 .$ [/b]