calculus
email
Princeton
college
graph theory

The American Mathematical Society and the Math Zoom Academy are sponsoring "Who Wants to Be a Mathematician" - a math contest with Southern California high school students competing for cash and prizes.  The event will take place on December 6, 2008 at California State University Fullerton in the Titan Student Union (directions and a map are at [url]http://asi.fullerton.edu/tsudir.htm[/url]). Questions will cover pre-calculus topics (including logic and history of mathematics). The top prize in the game is $ \$$2000.
[b][u]How to participate:[/u][/b]
[b]For schools:[/b]
  - Request the qualifying test from WWTBAM by email to myb@ams.org
  - Administer the test to students at your school who sign up and send us the test papers. We will score the papers and choose 8 students (at most one from each individual school) to participate in Who Wants to be a Mathematician, and the remaining students will be put in the selection pool for the Southern California Math Olympiad (see below for details).  The test is closed book ([i]no calculators[/i]) with a time limit of ten minutes.
  - Return the completed tests to the attention of Mike Breen (either fax 401-331-3842 or mail: Mike Breen c/o AMS; 201 Charles St.; Providence, RI 02904), so that they arrive by Wednesday, November 5, 2008.

[b]For individuals:[/b]
Fill and submit the online registration form at [url]http://www.mathzoom.org/scmoreg.php[/url]
and Math Zoom Academy will notify you the qualifying test date, time, and location.

We will choose eight students based on their scores on the test, with no more than one student from any one school. We will notify the teachers of the contestants who have qualified once the deadline has passed. Those students who do qualify to be contestants will be required to be at the game site at 9:00 on the morning of December 6 for practice and orientation.

There will be a math talk (aimed at high school students) at 10:00 that morning by Dr. Paul Dreyer of the RAND Corporation. Dr. Dreyer does a wonderful job of communicating to students his enthusiasm for mathematics. Dr. Dreyer is also a Math Zoom instructor.

In addition to Who Wants to Be a Mathematician, there will also be the [u][b]Southern California Math Olympiad - SCMO[/b][/u], which is a dynamic and entertaining contest. The top 40 scorers on the enclosed qualifying test (excluding the 8 contestants selected for the WWTBAM) are eligible to participate in this team contest. The names of the top 40 scorers and the team assignments will be announced on the [url=http://www.mathzoom.org]Math Zoom website[/url] and through email two weeks before the event.  See below for more details about the activities.

Admission is free to both the contests and talk. We hope to see many students from your school. Teachers, family and friends are all welcome.

If you would like more information, feel free to contact Mike Breen (email: myb@ams.org, phone: 800-321-4267) or Kevin Wang at the Math Zoom Academy kwang@mathzoom.com, phone: (714)389-2818. The contests and talk will make for a fun, exciting and educational event.

Best wishes,

Mike Breen, AMS Public Awareness Officer
Kevin Wang, Math Zoom Academy

[u][b]More information on activities for December 6:[/b][/u]
Format of the Southern California Math Olympiad:

A set of math problems will be given to all the teams.  Members in each team collaborate on the problem, and the first team that correctly answers a question will "own" that question.  Someone in a team raises his or her hand in order to answer a question, and they will be called on in a predefined order.  Each team has two chances to answer each question that has not been "owned".  At the end (when either all math questions are owned or the hour is up), the team that "owns" the largest number of questions wins the contest.

Note: The top 40 scorers on the enclosed qualifying test (excluding the 8 contestants selected for the WWTBAM) are eligible to participate in this team contest. The names of the top 40 scorers and the team assignments will be announced on the Math Zoom website and through email two weeks before the event.  The contestants are required to confirm their attendance one week before the event.  If anyone selected is not able to attend, the spot will be given to alternatives.

[u][b]Tentative schedule:[/b][/u]
9:00 Contestants for WWTBAM arrive for practice.
9:15-9:45 Contestants for SCMO arrive for practice; other students and parents arrive.
9:45-10:00 Announcement of the 40 top scorers and the 8 WWTBAM contestants.
10:00-10:30 Math Seminar: Dr. Paul Dreyer: Six Degrees of Graph Theory: Kevin Bacon, Paul Erd$ \"o$s, William McKinley, and Paul Dreyer.
10:30-10:45 Break
10:45-11:45 Southern California Mathematical Olympiad
11:45-12:00 Break
12:00-1:00 Who Wants to be a Mathematician
1:00-1:30 Award ceremony

For those who already emailed me to signed up: I have you down for the time slot you prefer.  See you then.

For other people:

We have scheduled the following dates, times, and locations for the qualifying test.  Please choose the time and location most convenient for you and reply this email to let us know.

(1) Irvine: Friday 10/31, 7pm, 4255 E. Campus Drive, Suite 290, Irvine, CA 92612.
(2) Arcadia: Saturday 11/1, 10am, 41 W Santa Clara St, Arcadia, CA 91007.
(3) Arcadia: Saturday 11/1, 3:30pm, 41 W Santa Clara St, Arcadia, CA 91007.
(4) Walnut: Saturday 11/1, 7pm, 19720 E. Walnut Dr. South, #100, Walnut, CA 91789. (Princeton Review site)

The test will be for 10 minutes.  Please be on time.  I am looking forward to seeing you at one of those time and locations.

If you registered online but will take / have taken the test with your school, please also let me know (PM or email).  Thanks!


Who like Super Smash Bros Brawl? (Wii)

How about Halo? (Xbox)

Ratchet and Clank? (PlayStations)

Dragonfable? (Online)

Second Life? (Online)

Post ur  fave games here!

I like SSBB. My favorites are Ike, Meta Knight, and Marth.

YES!!! ME too. i like mr game and watch too. also, charzard,(pokemon trainer)

I like contra, fifa, series Rocman... and My favourite game, maybe, is Rocman

meta knight, marth, pit spamping AKA spam-camping

SSBB - always with Pit
FFVII, FFX, FFXII are all amazing.
But Neverwinter Nights is the greatest game ever :)

[quote="matt276eagles"]Guitar Hero 1/2/80's/3 and Rock Band 2.

Here's a Youtube video of me playing Through the Fire and Flames, one of the hardest Guitar Hero songs ever:

http://www.youtube.com/watch?v=QrEY7uK5NaM&feature=channel

/shameless plug :D[/quote]

here is one you should all see:
http://www.youtube.com/watch?v=IIkySmW1YnM&feature=channel_page

Pit is like one of the worst characters in Brawl. People use him just for his spamminess.

I use Marth or Lucario.

i really think mewto would be betta that lucario.
but i like toon link

Me?
Didn't I already answer this on a thread..
well anyways,
Kirby.
THE POWER ABSORBER!
I love him :)

Ok this is really funny, because i don't wanna spoil it, so use kirby and suck up ROB then press b.
See what sound he makes. it's funny.

[quote="matt276eagles"]Guitar Hero 1/2/80's/3 and Rock Band 2.

Here's a Youtube video of me playing Through the Fire and Flames, one of the hardest Guitar Hero songs ever:

http://www.youtube.com/watch?v=QrEY7uK5NaM&feature=channel

/shameless plug :D[/quote]
omg superdock your a beast!!! although I have not played guitar hero, i can already tell that what you did was insane

I just woke up from a conjecture power :stretcher:  :wow:  nap that lasted three years avoiding the power of spam(the meat) and aops forum postings.  This subject was new so I will well... um.. play along....meh.  Man I feel old... Did anybody from this forum remember Chuck Norris facts as a favorite and Teh GAME as another classic.  SSMB RULES!!!!!
Die spam characters.

I am currently addicted to Plants vs Zombies. Favorite games of all time, though, include Pokemon and the Golden Sun series for the GBA. Golden Sun: The Lost Age may very well be the best GBA game of all time.

Portal FTW. Too bad once you're good its super-quick.

My favorite game is Tennis. Except I'm a little sore I lost a match 4-6 1-6 and Del Potro beat FEDERER in US Open

Yea, he did.

*sigh*, i guess either Federer is getting too old to play PRO tennis or he just sucked that day.

Federer's only 27 or 28. My guess is that he'll retire in about 5 years. :(

Hello Everyone,
Here are my favorite Top 10 Xbox Live Games.

1. Halo
2. Star Wars: Knights of the Old Republic
3. Ninja Gaiden
4. Project Gotham Racing 2
5. NBA Street Vol. 2
6. Grand Theft Auto Double Pack
7. Prince of Persia: The Sands of Time
8. Crimson Skies: The High Road to Revenge
9. Splinter Cell: Pandora Tomorrow
10. Rainbow Six 3

which are yours??

@Amherstclane: You shouldn't post on topics more than a couple months old.

Going along:

For the Win;
Browsing YouTube to find something to watch while playing FTW;
Mathcounts Practice.

Those are my favorite games. No more video games for a while, even though I have a DS and a Wii.

Scribblenauts is the best game ever. I mean, which game allows you to call on the Flying Spaghetti Monster and Cthulu?(If I spelled those right)

[quote="meewhee009"]I have this addiction to the Sims 3. SORRY IT'S JUST SO FUN!!!! DX xD ohhh... and yah :{D[/quote]
Another person who likes to killz their simz.

I've tried the demo for Halo, and I'm hoping I get it. Unfortunately my mouse acts weird sometimes. >.<
Other than that I have 5 computer games.

I can make you and kill you... over and over again :evilgrin:.
Anyways...
I'm still hooked on that game :)
And those Nancy Drew Computer Games... Adventure Role-Play FTW.

I like almost any Mario game.  There's this really addicting one called Super Mario 63 (yes, 63).

I also like FTW.  But nobody's ever there when I go in, so I never end up playing.  :(

Hmmm....
I played Animal Crossing: City Folk (Wii) for almost my whole Thanksgiving vacation.
SUPER ADDICTING!
Really!

And New Super Mario Bros. Wii isn't that bad, either.
I love toad in his propeller suit!  :oops:

SSBB is pretty fun.  My favorite characters are Kirby (Oh yeah) and Falco, fox, or wolf (all practically the same).  Picacu is a total button hacking character.  You can just press down B and u own.  Snake is slow but I like his homing missiles.

Halo 3 rocks so much.
I am beast with the sniper.

[quote="gogators"] the Flying Spaghetti Monster [/quote]

Really? What does it look like?

1.HALO
although I think it all went downhill after the 1st one

WITH SSBB i totally pwn with marth
hes way better than ike and is just for pros

Everyone who doesn't know how to use him thinks he sucks

1. Call of Duty 4:Modern Warfare
2. Final Fantasy IV (DS version)
3. Battlefield: Bad Company
4. Far Cry 2
5. Uncharted 2
6. Final Fantasy VII
7. Halo 3
8. Crysis
9. Call of Duty 6: Modern Warfare 2
10. Fallout 3

My favourite are:

Halo 2, 3 and ODST
Fable 2
Brütal Legend
99Nights

[quote="Joe10112"]Been so long. I have not been on here in like a long time, and my topics are still here!

I still like all the ones I had in my opening post.[/quote]

That's because the topics are posted in order people post them...

here is my list...

Prince of Persia 1,2,3 and 4
God of War 1 2
GTA: San Andreas
Call of Duty 1,2,3 and 4
Devil May Cry 1,2,3 and 4
Far Cry 1 and 2

[quote="andrewjjiang97"][b][i][color=red][size=200]PLEASE LOCK THIS[/size][/color][/i][/b][/quote]

This was written because people kept bumping 1/2 to 1 year threads. Please stop.


$3<p$ prime integer

prove $p^2$ divides  the numerator $\sum_{k=1}^{p-1}\frac{1}{k}$


I really want to take an online class this summer, but I didn't realize that Intermediate Complex/Trig wouldn't be offered.

Now I'm indecisive over these two classes (Number Theory or Counting).

I can easily do about 70-80% of the problems in the Intro to Number Theory "Do You Need This?" test but practically everything in Inter. Counting and Probability is new to me.

I'm confused over which class I should take...

Couldn't you take both? I've taken two classes at once a couple of times, during the school year even, and you should have more free time in the summer. Though if you already know 70-80% of the number theory course, I'm not sure if you need to take it.

Have you considered intermediat number theory?

surge,

It sounds like you know Intro Number Theory well enough to take the Intermediate Number Theory course.  If you can only take one course, decide which kinds of probems motivate you most: counting/probability or number theory.

I personally believe that going with high motivation classes first makes the educational process easiest.


geometry
rectangle
pigeonhole principle
search
function

Points in the plane are each colored with one of two colors: red or
blue. Prove that there is a rectangle whose vertices are of the same
color.

[hide]
Clearly on any line we choose there are either at least 3 red points or 3 blue points.
Let's assume WLOG that there are 3 red points, A, B and C.

Through A, B and C draw perpendiculars to the original line. Then draw 4 more lines intersecting those perpendiculars and parallel to the original line.

We have reached a pattern like this:

[code]
R - R - R
|   |   |
X - X - X
|   |   |
X - X - X
|   |   |
X - X - X
|   |   |
X - X - X
[/code]

On each row there must be no more than 1 red point, otherwise we have reached the rectangle. If on one row there are 3 blue points next line must have at least 2 blue points and those blue points together form a blue rectangle.

So we may assume each row has 1 red point and 2 blue points. The two blue points can be arranged in 3 ways: (B-B-R, B-R-B, R-B-B) and there are 4 rows to be filled in. By pigeon hole principle two rows must have the same arrangement and we have a blue rectangle.

[/hide]

Extension:

a) Show that for any finite number of colors you can find such a rectangle?

b) What about a countably infinite number of colors?

The first extension is easy with Pigeonhole; $ n\plus{}1$ by $ n^{n\plus{}1}$ lattice rectangle colored by $ n$ colors must contain a monochromatic rectangle.

The second I posted in Combinatorics Own & Proposed problems. Apparently, it is equivalent with the Continuum Hypothesis, which is independent of ZFC. Which is bad. So apparently the 2nd is unknown.

[quote="madshiet"]it's an easy one because i've solved it in 30 seconds. Let there be $ n$ points in the plane colored with either blue or red. We first prove it's true for quadratileral. For this it's enough to have only $ 7$ points in the plane which id possible.
So we've at least $ 7$ points, so we choose two points in the same colour and we begin to draw edge from this to another point under it, then go straight to another point and finally up to the second point we've. Just proven.[/quote]

But by an unlucky coincidence we could end up on a line where there is no other point in the color we seek :)

[quote="hipsishopsis"]
 WLOG [/quote]

With law of gravity?

[quote="JBL"]"Without loss of generality," as any decent search engine could have told you.[/quote]
oh god I always thought WLOG was some kind of computer language

[quote]Clearly on any line we choose there are either at least 3 red points or 3 blue points. [/quote]

why?.............

Because there are more than 4 points on each line?

Funny story: when I was at MOP, after a lesson, one of my fellow students came up to me and asked, "What was that omega-log function, anyhow?"

[quote]If on one row there are 3 blue points next line must have at least 2 blue points[/quote]

 :o  why is this true?

Read the whole argument: each row has at most 1 red point.  Thus, every row has at least 2 blue points.  If on one row there are three blue points and on another row there are two blue points, then there is a blue rectangle.

alright

how many rows do you need to fill total to make sure the configuration repeats- and is this number significant at all?

Again, please go back and reread the argument.  There's an application of the pigeonhole principle here, and hipsishopsis chose a grid of the appropriate size to make the pigeonhole argument work.


calculus
derivative
function
LaTeX
calculus computations

Can anyone help me with these problems? I suppose there are some formulas for them that I don't know:

1. A company has cost function C(x) = 1000 + 10x + x^2. Find the average cost of producing 100 units.

2. A company has cost function C(x) = 1000 + 10x + x^2. Find the marginal cost of producing 100 units.

3. A company has cost function C(x) = 2000 + 50x + x^2 and demand function p(x)= 100. How many units should it make to maximize its profit.

Plus, my problem with these questions are I don't know what marginal cost, demand, and profit means and what their relationship is. Can someone please explain?

1. I'm assuming "cost function" means: total cost to produce the $ x$-th item. In that case average cost would be $ \frac {\text{total cost}}{\text{number of items}} \equal{} \frac {C(100)}{100}$ in your case.

2. Marginal cost is the change in total cost that arises when the quantity produced changes by one unit. In other words the derivative. So $ C'(100)$ in your case.

3. I'll let someone else handle this. I haven't really encountered this type of maximizing.

P.S. You have perfectly typed LaTeX equations. Just put $ \$$ around them

[quote="sliver"]3. A company has cost function C(x) = 2000 + 50x + x^2 and demand function p(x)= 100. How many units should it make to maximize its profit.
[/quote]
I know how to do this with $ C(x)$ and $ S(x)$, where $ C(x)$ is the cost function and $ S(x)$ is the gross profit function. Maybe your $ p$ is my $ S$. What you do is take the derivative of both functions and set them equal to each other (i.e. $ C'(x)=p'(x)$). This yields local maximum profits and maximum losses. In your case this yields only one value, which should be your maximum profit.
Interestingly, in your case, the answer is $ -25\text{ units}$, so I'm guessing that $ S(x)\not=p(x)$.

I would guess that $ p(x)$ stands for the "unit price the company can get if $ x$ units are to be sold". Usually it is some non-increasing function of $ x$, so it seems to fit but, of course, I may be completely off track here. Anyway, if it is what I think, the objective is to maximize $ xp(x)\minus{}C(x)$.

[quote="fedja"]I would guess that $ p(x)$ stands for the "unit price the company can get if $ x$ units are to be sold". Usually it is some non-increasing function of $ x$, so it seems to fit but, of course, I may be completely off track here. Anyway, if it is what I think, the objective is to maximize $ xp(x) \minus{} C(x)$.[/quote]
That would work, $ C'(x)\equal{}\frac{d}{dx}[x\cdot p(x)]$ would give [i]positive[/i] 25 units as the outcome. Good thinking, fedja.


calculus
function
trigonometry
algebra
polynomial
calculus computations

Hi, could someone provide me with any hints or tips as to how to evaluate the following limit? I tried breaking it up, l'hopitals rule, and dividing by x^2 with no success...

Evaluate the limit as x -> 0 of (x-sin[x]) / ([x^2]*arctan[x])

Thanks very much for your assistance.

-Jeff

We are given $\frac{x - \sin(x)}{x^2\arctan(x)}$.

Recall that $\sin(x) = x - \frac{x^3}{3!} + O(x^5)$ and $\arctan(x) = x - O(x^3)$. So we have

$\frac{x - x + \frac{x^3}{3!} - O(x^5)}{x^3 - O(x^5)}$

which means that the limit should be $\frac{1}{6}$, as the higher order terms are negligible when $x$ is small enough.

L'Hôpital would work, but would require that you differentiate the numerator and denominator three times each. How do I know that? It's visible in Xevarion's work.

I don't recommend using L'Hôpital. In fact, I seldom recommend using L'Hôpital.

Power series are an indispensible tool of calculus. When I teach Calculus II, I insist that my students memorize - yes, truly, solidly, memorize - the power series centered at zero for each of the following functions:

$\frac1{1-x},\ \ln(1+x),\ \arctan x,\ e^x,\ \cos x,\ \sin x,\ (1+x)^{\alpha}.$

With the series thoroughly memorized, you look at this problem and say, "$x-\sin x$ near zero? That cancels the $x$ terms and we might as well have $\frac{x^3}6.$"

From the power series / Taylor polynomial point of view (in other words, exactly what Xevarion did), this is an [i]easy[/i] problem, one that an experienced person should be able to do in his or her head.


Hi,

Does anyone know of any kind of Math contests for people who have graduated from school & are currently working in the industry?

In my case, it is presently not possible for me to give up full-time employment but I would love to do Math nonetheless.

Thanks,
Ashok

[quote="a_ashokkumar"]I have no formal degree in Mathematics. I have a Masters in Electrical Engineering from IIT-Bombay.

Thanks,
Ashok[/quote]Hmm, I am not sure if you can participate or not in the IMC (as a contestant) but it's worth to try :) You can write an email to the organizing party of the IMC. Their contacts can be found on [url=http://www.ucl.ac.uk/~ucahjej/imc/imc2005/contact.html]this page[/url]. Good luck! :)


This is another attempt at getting a marathon going in intermidiate topics. Read the rules carefully since they're sightly different than other marathons. If you post a solution, please (really please) post a problem. Anyone can participate. NO SPAM PLEASE. Also, if a problem is not solved for two days, we'll just skip the problem. The difficulty rules of intermidiate topics apply.

Okay. I have just one last request, please number the problems :)

Problem 1: Prove that the sum of measures of perpendiculars from any point on a side of a rectangle to the diagonals is constant.

Source: CPIG

[hide]Let $ ABCD$ be the rectangle and let $ X$ be on $ AD$. Let $ O$ be the center of the rectangle. Let $ X_A$ be the feet of the perpendicular to $ AO$ and $ X_D$ the feet to $ DO$. Then $ 2[\triangle AOD] \equal{} AO\cdot XX_A \plus{} DO\cdot XX_D \equal{} (XX_A \plus{} XX_D)AO$ (AO=DO)[/hide]

Problem 2:
Find, with proof, all natural numbers $ n$ such that $ n$ is divisible by all natural numbers less than $ \sqrt {n}$


function
geometry
3D geometry
number theory
modular arithmetic

I think that the functions $6x^{2}+12x+8$ and $3x^{2}+3x+1$ canno't be a cube for $x>0$. prove it!

6x^2 + 12x + 8 = (x+2)^3 - x^3
AND
3x^2 + 3x + 1 = (x+1)^3 - x^3

Thus, when rearranged they would satisfy the following equations:

(x^3) + (3x^2 + 3x + 1) = (x+1)^3
AND
(x^3) + (6x^2 + 12x + 8) = (x+2)^3

But Fermat's last theorem claims that there are no solutions to the equation:
a^3 + b^3 = c^3
where a, b and c are integers (apart from if they are all 0 which is not possible with your problem).

Thus, your function is proved that it can never be a cube. Modular arithmetic may work, too. Use the fact that all cubes are 0, 1 or 8 mod 9 whilst all squares are 0 or 1 mod 4.

I actually derive it from Fermat's last theorem, by a geometric way, do you see it? and how does the modular arithmetic helps? thanks.
also, if you can prove that  $z(6x^{2}+12x+8)+y(3x^{2}+3x+1)$ for $x,z,y \in \mathbb{N}$ canno't be a cube, you actually prove FLT for the case $n=3$   :)


$ a$,$ b$ and $ c$ are  three positive real  numbers .
prove that
$ \frac{1}{3}(a\plus{}b\plus{}c)\minus{}\sqrt[3]{abc}\leq (\sqrt{a}\minus{}\sqrt{b})^2\plus{}(\sqrt{b}\minus{}\sqrt{c})^2\plus{}(\sqrt{c}\minus{}\sqrt{a})^2$

[quote="greatestmaths"]$ a$,$ b$ and $ c$ are  three positive real  numbers .
prove that
$ \frac {1}{3}(a \plus{} b \plus{} c) \minus{} \sqrt [3]{abc}\leq (\sqrt {a} \minus{} \sqrt {b})^2 \plus{} (\sqrt {b} \minus{} \sqrt {c})^2 \plus{} (\sqrt {c} \minus{} \sqrt {a})^2$[/quote]
Let $ a \equal{} x^6,$ $ b \equal{} y^6$ and $ c \equal{} z^6,$ where $ x,$ $ y$ and $ z$ are non-negative numbers. Then
$ \frac {1}{3}(a \plus{} b \plus{} c) \minus{} \sqrt [3]{abc}\leq (\sqrt {a} \minus{} \sqrt {b})^2 \plus{} (\sqrt {b} \minus{} \sqrt {c})^2 \plus{} (\sqrt {c} \minus{} \sqrt {a})^2\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(5x^6 \minus{} 6x^3y^3 \plus{} a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^6 \minus{} x^4y^2 \minus{} x^4z^2 \plus{} x^2y^2z^2) \plus{} \sum_{cyc}(4x^6 \plus{} x^4y^2 \plus{} x^4z^2 \minus{} 6x^3y^3)\geq0,$ which true by Schur and Muirhead.
The following stronger inequality is true too:
For non-negative numbers $ a,$ $ b$ and $ c$ prove that
\[ \frac {a \plus{} b \plus{} c}{3} \minus{} \sqrt [3]{abc}\leq \frac{(\sqrt {a} \minus{} \sqrt {b})^2 \plus{} (\sqrt {b} \minus{} \sqrt {c})^2 \plus{} (\sqrt {c} \minus{} \sqrt {a})^2}{3}\]

i think we can prove this inequality by another way

A interesting problem: $ \frac {a\plus{}b\plus{}c}{3}\minus{}\sqrt[3]{abc}\geq\frac{1}{9}\sum{(\sqrt a\minus{}\sqrt b)^2}$
First, we prove that
$ \frac {a\plus{}b\plus{}c}{3}\minus{}\sqrt[3]{abc}\geq\frac{(\sqrt a\minus{}\sqrt b)^2}{3}\equal{}\frac{a\plus{}b}{3}\minus{}\frac{2}{3}\sqrt{ab}$
or $ c\plus{}2\sqrt{ab}\geq 3\sqrt[3]{abc}$
By AM_GM inequality we have $ c\plus{}\sqrt{ab}\plus{}\sqrt{ab}\geq 3\sqrt[3]{abc}$.So, this inequality is true.
Similarly, we have $ \frac {a\plus{}b\plus{}c}{3}\minus{}\sqrt[3]{abc}\geq\frac{(\sqrt b\minus{}\sqrt c)^2}{3},\frac{(\sqrt c\minus{}\sqrt a)^2}{3}$
Then that inequality is true! :D

Oh, we have a Stronger-AM_GM inequality, it is:Let $ a_{1},a_{2},...a_{n}$ are non_negative numbers. Prove that
$ a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}\minus{}\sqrt[n]{a_{1}a_{2}a_{3}...a_{n}}\geq\frac{1}{n}((a_1\minus{}a_2)^2\plus{}(a_2\minus{}a_3)^2\plus{}..\plus{}(a_{n\minus{}1}\minus{}a_{n}^2))$ :D
It is nice?

[quote="thegod277"]A interesting problem: $ \frac {a \plus{} b \plus{} c}{3} \minus{} \sqrt [3]{abc}\geq\frac {1}{9}\sum{(\sqrt a \minus{} \sqrt b)^2}$
[/quote]
The following stronger inequaly is true too:
For non-negative $ a,$ $ b$ and $ c$ prove that
\[ \frac {a \plus{} b \plus{} c}{3} \minus{} \sqrt [3]{abc}\geq\frac {\left(\sqrt a \minus{} \sqrt b\right)^2 \plus{} \left(\sqrt a \minus{} \sqrt c\right)^2 \plus{} \left(\sqrt b \minus{} \sqrt c\right)^2}{6}
\]
[quote="thegod277"]Oh, we have a Stronger-AM_GM inequality, it is:Let $ a_{1},a_{2},...a_{n}$ are non_negative numbers. Prove that
$ a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n} \minus{} \sqrt [n]{a_{1}a_{2}a_{3}...a_{n}}\geq\frac {1}{n}((a_1 \minus{} a_2)^2 \plus{} (a_2 \minus{} a_3)^2 \plus{} .. \plus{} (a_{n \minus{} 1} \minus{} a_{n}^2))$ :D
It is nice?[/quote]
I think, for $ n\geq2$ the following inequality is nicer:
$ a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n} \minus{} n\sqrt [n]{a_{1}a_{2}a_{3}...a_{n}}\geq\frac {1}{n \minus{} 1}((a_1 \minus{} a_2)^2 \plus{} (a_2 \minus{} a_3)^2 \plus{} .. \plus{} (a_{n \minus{} 1} \minus{} a_{n}^2))$

[quote="arqady"]
The following stronger inequality is true too:
For non-negative numbers $ a,$ $ b$ and $ c$ prove that
\[ \frac {a \plus{} b \plus{} c}{3} \minus{} \sqrt [3]{abc}\leq \frac {(\sqrt {a} \minus{} \sqrt {b})^2 \plus{} (\sqrt {b} \minus{} \sqrt {c})^2 \plus{} (\sqrt {c} \minus{} \sqrt {a})^2}{3}
\]
[/quote]
In other words, we have
\[ \frac{a\plus{}b\plus{}c}3\minus{}\sqrt[3]{abc}\leq\frac23\max\{(\sqrt a\minus{}\sqrt b)^2, (\sqrt b\minus{}\sqrt c)^2, (\sqrt c\minus{}\sqrt a)^2\},\]
which is problem 6.1 in the Olympiad Inequalities (Pham Van Thuan, Le Vi).

Can you prove these inequalities, argady and pvthuan :maybe:

[quote="greatestmaths"]$ a$,$ b$ and $ c$ are  three positive real  numbers .
prove that
$ \frac {1}{3}(a \plus{} b \plus{} c) \minus{} \sqrt [3]{abc}\leq (\sqrt {a} \minus{} \sqrt {b})^2 \plus{} (\sqrt {b} \minus{} \sqrt {c})^2 \plus{} (\sqrt {c} \minus{} \sqrt {a})^2$[/quote]

$ \leftrightarrow 2(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}) \le \frac{5}{3} (a\plus{}b\plus{}c) \plus{} \sqrt[3]{abc}$
But I don't know where to go from here.

[quote="FOURRIER"][quote="greatestmaths"]$ a$,$ b$ and $ c$ are  three positive real  numbers .
prove that
$ \frac {1}{3}(a \plus{} b \plus{} c) \minus{} \sqrt [3]{abc}\leq (\sqrt {a} \minus{} \sqrt {b})^2 \plus{} (\sqrt {b} \minus{} \sqrt {c})^2 \plus{} (\sqrt {c} \minus{} \sqrt {a})^2$[/quote]

$ \leftrightarrow 2(\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}) \le \frac {5}{3} (a \plus{} b \plus{} c) \plus{} \sqrt [3]{abc}$
But I don't know where to go from here.[/quote]
Your inequality follows from AM-GM and Schur.
$ \frac{5}{3}(a\plus{}b\plus{}c)\plus{}\sqrt[3]{abc}\geq a\plus{}b\plus{}c\plus{}3\sqrt[3]{abc}\geq 2(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca})$

[quote="leepakhin"]
Your inequality follows from AM-GM and Schur.
$ \frac {5}{3}(a \plus{} b \plus{} c) \plus{} \sqrt [3]{abc}\geq a \plus{} b \plus{} c \plus{} 3\sqrt [3]{abc}\geq 2(\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca})$[/quote]

Thanks,

Its AM-GM, Schur, and AM GM again.

[quote="arqady"]
$ \Leftrightarrow\sum_{cyc}(x^6 \minus{} x^4y^2 \minus{} x^4z^2 \plus{} x^2y^2z^2) \plus{} \sum_{cyc}(4x^6 \plus{} x^4y^2 \plus{} x^4z^2 \minus{} 6x^3y^3)\geq0,$ which true by Schur and Muirhead.
[/quote]

how
$ \sum_{cyc}(4x^6 \plus{} x^4y^2 \plus{} x^4z^2 \minus{} 6x^3y^3)\geq0$ is true by Murhead?

[quote="FOURRIER"]
Why
$ \sum_{cyc}(4x^6 \plus{} x^4y^2 \plus{} x^4z^2 \minus{} 6x^3y^3)\geq0$
Is true by Murhead?[/quote]
Yes, for example!

I have a proof of it by AM-GM and rearangement,
But I still can't see [u]how[/u] it is true by murhead?


inequalities
geometry
function
inequalities proposed

if a,b,c>0 and a+b+c=6 then

\[ (3-{a})(3-{b})(3-{c})\leq {1}  \]

                                                                                Luca Victor Iliesiu

I think it's something wrong at this inequality. :?:

the new one is:
a+b+c=6
a,b,c>0
\[ (3-{a})(3-{b})(3-{c})\leq {1}  \]
sorry but the last one had just a little mistake :blush:

solution will be given soon(very soon)(i just want to know how many do it)

It comes from this:

$(b+c-a)(c+a-b)(a+b-c) \leq abc \leq \biggl(\frac{a+b+c}{3}\biggr)^3=8$

What is AM-GM???I'm only 7th grade :blush:

So am I.The inequality $\frac{a_1+a_2+...+a_n}{n}$>= $\ (a_1a_2a_3...a_n)^\frac{1}{n}$ is called AM-GM.(aritmetic-geometric)

We can assume that $a$,$b$,and $c$ are sides of a triangle.Then $(3-a)(3-b)(3-c)=(p-a)(p-b)(p-c)$,where $p=\frac{a+b+c}{2}$.We know that in any triangle,$p(p-a)(p-b)(p-c)=S^2$,where $S$=the area of the triangle.Now in our case we know $p=3$ therefore $(p-a)(p-b)(p-c)=\frac{S^2}{3}=(3-a)(3-b)(3-c)$ therefore all we have to do is prove that $\frac{S^2}{3}\leq 1 \equiv S^2\leq 3$.
To prove that we express $S$ in two different ways:
$S=r\cdot p$,where $r$=the radius of the inscribed circle
$S=\frac{abc}{4R}$,where $R$=the radius of the circumscribed circle.
Now by multiplying the last two relations we get that $S^2=\frac{r}{4R} \cdot p \cdot abc$.Now by Euler`s ineq we have that $2r\leq R$ therefore
$\frac{r}{4R} \leq \frac 18 (*)$
and it`s a known fact that $abc\leq (\frac{a+b+c}3)^3$ and in our case we get that
$abc\leq 8 (**)$.Now by multiplying the relations $(*)$ and $(**)$ we get the inequality that we want.Thus the problem is solved!Luca, nice problem.
Andrei Ciupan

[quote="silouan"]Sorry but Stanciou your solution is wrong. You cannot use AM-GM since you don't know the sign of each term .  ;) since  the expression $3-x$  where $0<x<6$  is not always positive[/quote]
   Please Silouan, call me Sorin, no Stancioiu.I understood where I've had mistakes.Sorry! :blush:

Nice solution Andrei...
___________________
And Sorin you just needed to make a discution id x>3 cause the first part of the inequality will be negative and the second part positive...Q.E.D

Yes, you're right.But because I know that Silouan is better than me, I said that my solution is completly wrong.

It was quite an easy one...Just that I didn't realise that

Note:this problem I've seen it in M.O.Drimbe's book.So, it isn't yours. ;)

I didn't know it was in a book...I first thought of it in a triangle like andrei did...I
P.S.:I don't have that book...But I think it's a good one if this kind of problems are in it :P ..Joke

sorry but I don't know functions:blush:...I'm only the 7th grade...

Functions are really easy. ;) I think this inequality can be done by Popoviciu.I'll try.

You must verify whether we have $\sum \sqrt[3]{3-a} \geq 0$ or not. But I guess it's not hard to prove.

good solution andrei...but it was already given...But don't worry be happpy :lol: ...You have your one solution(that with the triangle)

[quote="andyciup"]$xyz\leq (\frac{x+y+z}{3})^3$,for all $x,y,z \in \mathbb{R}$[/quote]

You are wrong . It is only holds for .[u] positive  [/u]  . If you do not belive me take
$x=y=-1$  and  $z=2$

[quote="silouan"][quote="andyciup"]$xyz\leq (\frac{x+y+z}{3})^3$,for all $x,y,z \in \mathbb{R}$[/quote]

You are wrong . It is [b][i]only[/i][/b] holds for .[u] positive  [/u]  .[/quote]

Not really. It can hold if some of the numbers are negative.

You can do it this way: $a^3+b^3+c^3 - 3abc = \left( a+b+c \right) \left( a^2+b^2+c^2 - bc - ca - ab \right)$. The second paranthesis is always non-negative. Thus, AM-GM for 3 numbers holds iff $a=b=c$ or $a+b+c \geq 0$. That's what I meant with
[quote]You must verify whether we have $\sum \sqrt[3]{3-a} \geq 0$ or not.[/quote]

[quote="luke999"]What does this have to do with
\[ (3-{a})(3-{b})(3-{c})\leq {1}  \]??
How do you prove it???[/quote]what do you mean ?this is the sorthest and nicest solution to this problem and it is also quite clear....


How many three-digit numbers are multiples of neither 5 nor 7?

There are $ 999\minus{}100\plus{}1\equal{}900$ three-digit numbers. Notice that if we simply subtract the multiples of $ 5$ and $ 7$ from this total, we would have subtracted the numbers that are multiples of both twice. However, we could eliminate this problem by noticing that multiples are both are all of the multiples of $ 5\times7\equal{}35$. Thus, our answer would be $ 900$ minus multiples of $ 5$ mines multiples of $ 7$ plus multiples of $ 35$. There are $ 180$ multiples of $ 5$, $ 128$ multiples of $ 7$, and $ 26$ multiples of $ 35$. Plugging these values into our expression we defined above, we find that there are $ 900\minus{}180\minus{}128\plus{}26\equal{}\fbox{618}$ numbers that are neither a multiple of $ 5$ nor a multiple of $ 7$.

This is only a little more detail on the previous solution:

There are \[ \left\lfloor \frac{999}{5}\right \rfloor \equal{}199\] multiples of $ 5$ from $ 1$ to $ 999$, where $ \lfloor x\rfloor$ denotes the greatest integer less than or equal to $ x$.  There are \[ \left\lfloor \frac{99}{5}\right\rfloor \equal{}19\] multiples of $ 5$ from $ 1$ to $ 99$ inclusive.  Therefore, there are $ 199\minus{}19\equal{}180$ multiples of $ 5$ between $ 100$ to $ 999$, inclusive.  It should be clear why this is the number of three-digit multiples of $ 5$.

Similarly, there are $ \left\lfloor \frac{999}{7}\right\rfloor \minus{}\left\lfloor \frac{99}{7}\right\rfloor\equal{}128$ three-digit multiples of $ 7$ and $ \left\lfloor \frac{999}{35} \right\rfloor \minus{} \left\lfloor \frac{99}{35} \right\rfloor \equal{}26$ three-digit multiples of $ 35$.

For the other details see the above solution.


What happened to the Chinese guy who got (if I remember correctly) a silver? Last I heard of him was when he was sitting in the back of our bus, one guy on our NZ team was secretly translating the chinese for us, apparently he wasn't eating or drinking anything. Wonder how his country would react..

[quote="sexynsmartjenny"]lol, maybe he was sent to a labor camp in china for bringing shame to his country.[/quote]
Your   guess   is    absolutely preposterous.
You   don't   know   China.

These   contestants   were    selected   by   very   very  cruel   and   fierce   competitio[color=#FF0000]n[b]s[/b][/color]，
so   they  are    very  very  very   spectacularly good  and   are  respected  by many  Chinese.

They       are [color=#FF0000] [b]all[/b][/color]  guaranteed    acceptance   into  the  best   universities  of  China.
So, in   China, olympiad   math     is  very     very   very   hot,  and    is   much  hotter  than  " TOKYOHOT".


geometry
circumcircle
geometric transformation
reflection
perpendicular bisector

Points O and H are the circumcenter and orthocenter of acute triangle ABC, respectively. The perpendicular bisector of segment AH meets sides AB and AC at D and E, respectively. Prove that $\angle DOA=\angle EOA$.

This was posted in the intermediate forums and no one responded with a solution...

$\Delta EHD$ is the reflection of $\Delta EAD$, so $\angle ADH = 2 \angle B = AOC$, and since $O$ and $H$ are isogonal cojugates, we know that $\angle DAH = \angle OAC$, so we have that $\Delta HDA$ and $\Delta COA$ are similar. Now we can easyly prove that $\Delta ADO$ and $\Delta AHC$ are similar.

Analogously, we prove that $\Delta AEO$ and $\Delta AHB$ are similar, so \[\angle DOA = \angle HCA = \angle HBA = \angle EOA\]


geometry
3D geometry
tetrahedron
AMC
USA(J)MO
USAMO
inequalities

Given a tetrahedron $VABC$ with edges $VA$, $VB$ and $VC$ perpendicular any two of them. The sum of the lengths of the tetrahedron's edges is $3p$. Find the maximal volume of $VABC$.

Denote $a=VA,b=VB,c=VC$, then $3p=a+b+c+\sqrt{a^{2}+b^{2}}+\sqrt{b^{2}+c^{2}}+\sqrt{c^{2}+a^{2}}$. Now, use AM-GM (note: volume of tetrahedron equal $\frac{abc}{6}$).

The same problem appeared already in USAMO 1976. As N.T.Tuan did not give the result, the maximal volume is $ \dfrac{p^3}{6\cdot(\sqrt {2} \plus{} 1)^2}$ with inequality if and only if $ a \equal{} b \equal{} c$.


Does interm counting and probability use alcumus in the course?

Probably not.

Maybe the instructors will pull out some problems, but I don't think they will "formally" use Alcumus.

No; we do not have many intermediate-level problems in Alcumus yet.

Yes, I have taken the intro C & P class and Alcumus is integrated with it.


Two and a half work crews take two and a half days to clear two and a half hectares of forest. How many work crews are required to clear four hectares in two days??

Notice that $ \frac{cd}h$ is constant.

Thus, we have $ \frac{2.5\cdot2.5}{2.5}\equal{}\frac{2c}4\implies2.5\equal{}\frac c2\implies c\equal{}\boxed5$.

Or, I just thought of it this way:

two and a half work crews can clear one hectare in one day
we need to clear 4 hectares in 2 days.
if we had two and a half work crews, we could only clear 2 hectares in 2 days
we need to double the work rate so we double the number of work crews which gives you 5 work crews

Yes, notice that that is the reason why $ \frac{cd}h$ is constant. :)


There is a board, and on it are three keyholes, one is a triangle, one is a square, and one is a circle.  If the shapes are regular polygons, the triangle has a side the same side length as the square, and the circle has the same diameter, and all of the measurements are length n, find the formula for the volume of a plug of minimum volume which can fit smugly into all three of the holes.

You mean maximum volume, not minimum, right?


[color=darkblue]Given are a fixed circle $w$ and two fixed points $B,C\in w$. Denote the middlepoint $D$ of the segment $BC$. For a mobile point $L\in w$ define the intersection $M\in LL\cap BC$ and the second intersection $N$ between the circle $w$ with the circumcircle of the triangle $LMD$. Prove that the line $LN$ passes through a fixed point (denote $XX$ - the tangent in the point $X\in w$ to the circle $w$).[/color]

$L$ moving on $\omega$ is $%Error. "Longrighleftarrow" is a bad command.
$, $M$ is moving on line $BC$ except segment $BC$. Denote by $H$ foot of altitude from $O$, (center of $\omega$) to $BC$. Let point $H \to H^{*}$ with inversion w.r.t. circle $\omega$ and radius $R$,(radius of $\omega$ ) . Now it is easy to see that line $BC$ is pole of  $H^{*}$ w.r.t. $\omega$, hence pole of $M$ passes through $H^{*}$ by the polar theorem. This is $\Longleftrightarrow$ to statemnt of the problem because line $LN$ is polar of the point $M$ w.r.t circle $\omega$

 q.e.d.


1.Find the equation of the hyperbola satisfying the given condition: Vertices at (9.0) and (-9,0); foci at (11,0) and (-11,0)

2.Find the focus and directrix of the parabola x2 = 12y

refrain from posting homework problems, these are obviously just using formulas that are in your book

I second Altheman's post. Moreover, stop posting problems over again. I saw you post the exact same questions a day or two ago. The topic is still in this forum.


it's quite easy, but how to use telescoping with their hint?

http://www.openmathtext.org/downloads.html

Junior problem seminar, page 49 :)

I didn't loook at the link, but do you multiply the LHS by 2/3, 4/5, 6/7, ... and the RHS by 1/1000?

yes  :)  :)  :) thanks quite nice! elegant

Wow you are smart.

I remember doing something like this at Mathpath.


Find the first number of the Fibonacci's sequence which ends in four 0.

The smallest $f_n$ such that $10000|f_n$ is $f_{7500}$.
One can easily check by hand that $16|f_n$ iff $12|n$. Similarly, $5|f_n$ iff $5|n$. Using this we can derive that $f_{5n}=f_{5(n-1)}+f_{5(n-2)} mod 25$, although this is tedious (just using Fibonacci recurrence). So taking every fifth term of the sequence behaves like the original sequence when it comes to divisibility by $5$, i.e. $25|f_n$ iff $25|n$. Reapplying this idea we get that $625|n$ for our required $n$; hence $n=12(625)=7500$.

...the answer, really isn't 7500 but 7501 ;)

[quote="ARTI"][quote="ZetaX"]To avoid such things like $7500 \leftrightarrow 7501$ I think you should give us the first terms of your fibonacci sequence.[/quote]
Fibonacci's first terms are the same for all people, I guess :D .They don't change: 0, 1, 1, 2, 3, 5, 8, 13, 21, ...
 :)[/quote]

Some people don't have a starting 0...

[quote="kueh"]

Some people don't have a starting 0...[/quote]

...should they?

I meant that you should tell us if you have $f_0=f_1=1$ or $(f_0=0,) f_1=f_2=1$, in other words what your starting index is and which value this fibonacci number has, because its really not the same standart for all people.


what is the sum of all positive integers less than 100 that are squares of perfect squares.

Two different answers...so would like to verify. 98 or 97?

Our goal is to find the sum of all powers of $ 4$ less than $ 100$. There are three; $ 1$, $ 16$, and $ 81$. Their sum is $ \boxed{98}$, thus our answer.

@KingSmasher, you got the right answer but you typoed in your solution...we are trying to find all fourth powers, not all powers of four

Thanks guys! Our school coach said the answer was 97, hence the doubt.


geometry
rectangle
modular arithmetic
combinatorics proposed
combinatorics

On a $2004 \times 2004$ chess table there are 2004 queens such that no two are attacking each other\footnote[1]{two queens attack each other if they lie on the same row, column or direction parallel with on of the main diagonals of the table}.

Prove that there exist two queens such that in the rectangle in which the center of the squares on which the queens lie are two opposite corners, has a semiperimeter of 2004.

Remember that ISL problem which said that if we have an even number of people around a table and they change places, there will be two who will have the same number oif people between them before and after the rearrangement?

Basically, that says that if $n$ is even, and $\sigma\in S_n$, then we can either find $i,j$ s.t. $|i-j|=|\sigma(i)-\sigma(j)|$ or we can find $i,j$ s.t. $|i-j|+|\sigma(i)-\sigma(j)|=n$. We apply it to $n=2004$, together with the observation that the conditions imply $|i-j|\ne|\sigma(i)-\sigma(j)|,\ \forall i\ne j\ (*)$, where $\sigma(i)$ is the row on which the queen standing on the column $i$ is placed ($(*)$ means that no two queens lie on the same line parallel to one of the diagonals).

I think we need more detailed references here, as it was obvious that we need to prove fact mentioned above.

I can't find the problem on Kalva anymore, because there are lots of problems to look through, and I don't remember the yewr, but I was pretty sure it had been discussed before.

Let's look at that problem I mentioned earlier:

For $i\in\overline{0,n-1}$, let $\sigma(i)-i$ denote the distance modulo $n$, in a counter-clockwise direction, between $i$ and $\sigma(i)$. If we assume that all $\sigma(i)-i$ are distinct, it means that they are $0,1,\ldots,n-1\pmod n$. If we add all these quantities, we get something which is not divisible by $n$ (it's $\frac n2\cdot(n-1)\pmod n$), but, on the other hand, $\sum(\sigma(i)-i)=0\pmod n$ because the residues $\{\sigma(i)\}$ are the same as the residues $\{i\}$, so we have a contradiction. This means that there are $i,j$ s.t. $\sigma(i)-i=\sigma(j)-j\pmod n$. This proves the assertion, I believe.


Prove that there are an infinite number of primes of the form $7n+4$ where $n$ is a positive integer.

Use this: if (a,b)=1 then there are an infinite number of positive integers n such that an+b is prime.

I think it difficult, but this would help http://www.mathhelpforum.com/math-help/showthread.php?t=3142

Do you mean that serious¿
This is a part of Dirichlet's theorem on primes in arithmetic progressions, which I would not call Intermediate.

There's an easier way to prove it for a specific case like this. You don't need to prove the general thm. to prove this case  ;)

I know.
What I can prove elementary at the moment is: at least one of the sequences $7n+2$ and $7n+4$ contains $\infty$ many primes (and with some ugly work using Chebyshef's theorem on primes this could suffice to prove it).
The worst is: there is no proof purely using the idea of Euklid's proof of the existence of infinitely many primes (this idea works for example for all residues $\mod 24$).
I will try doing it by polynomial constructions, but till now they refuse to give a proof...
I just never heard of not to hard and elementary proofs for that cases.

I just checked again. I was mistaken. There is an elementary proof but it is fairly difficult. Most proofs will just cite Dirichlet's Theorem. My apologies

OK. Can some moderator please move it to the Number Theory forum¿

PS: before trying $\equiv 4 \mod 7$, try $\equiv 1 \mod 7$ or $\equiv -1 \mod 7$ (they should be easier than this one).


For any three positive reals $a,b,c$, we have

${ \frac{a^3+2abc}{b+c}+\frac{b^3+2abc}{a+c}+\frac{c^3+2abc}{b+a}\ge\frac{\sqrt{3}}{2}} \left(a \sqrt{a^2+2bc}+b \sqrt{b^2+2ca}+c \sqrt{c^2+2ab}\right)$

Apply Cauchy's ineq :
 $ [a(b+c)+b(c+a)+c(a+b)] . (\frac{a^3+2abc}{b+c}+\frac{b^3+2abc}{a+c}+\frac{c^3+2abc}{b+a}) \geq (a \sqrt{a^2+2bc}+b \sqrt{b^2+2ca}+c \sqrt{c^2+2ab})^2 $
  It 's enough to show :
 $ \frac{\sum{a \sqrt{a^2+2bc}}}{\sum{a(b+c)}} \geq \frac{\sqrt3}{2} $ , which is discussed in inequality 10
 :)


[quote="mf321"]Find the number of distinct integers among these numbers:
\[ floor(\frac{1^{2}}{2007})\]

\[ floor(\frac{2^{2}}{2007})\]
...
\[ floor(\frac{2007^{2}}{2007})\]
[/quote]
The first one equals 0, the second is also 0, and the third is 2007

There is an ellipsis there. You're supposed to consider 2007 values.

An alternate statement of the problem: find the cardinality of the following set.

\[ \{\left\lfloor\frac{k^{2}}{2007}\right\rfloor\mid k\in\mathbb{N}, k\leq 2007\}\]

I think the answer to this question is $ 1505$, but I am not absolutely sure. :maybe: Have any of you attended to solve this question?

I think the answer should be 1506 because I use C++ to brute force it  :lol: .
Does anyone have any mathematical solution to this one?

I think I have got the idea, I have miscounted 0 as one of the integers.
I just want to confirm with mf321, did you include 0 as one of the integers?

Yes. Firstly I also miscounted the number 0 like you  :blush:

This was Problem 12 from Singapore Mathematical Olympiads 2007, Open Section.

Let $ a_{i}\equal{}\text{ floor }\frac{i^{2}}{2007}$. Note that $ \frac{(n\plus{}1)^{2}}{2007}\minus{}\frac{n^{2}}{2007}\equal{}\frac{2n\plus{}1}{2007}\leq 1$ if and only iff $ n\leq\frac{2006}{2}\equal{} 1003$. Since $ a_{1}\equal{} 0$, we conclude that $ a_{1},\ldots,a_{1003}$ assume all the values from $ 0$ to $ 501$. We also conclude that $ a_{1004},\ldots,a_{2007}$ are mutually distinct integers. Therefore, the answer is $ 502\plus{}1004 \equal{}\boxed{1506}$.

yup Aidan, Thats the only way to go about solving this problem. I want to see people brute force this during the actual competition!! haha

Through brute force, they may get just this problem correct on the contest  :D


[quote="kunny"]This is very easy!

The answer is $2$ when $x=1$[/quote]
Yeeh, But how did you find it ?  :lol:

[quote="erdos"]Find The minimum of
$\root{3}\of{4-3x+\sqrt{16-24x+9x^{2}-x^{3}}}+\root{3}\of{4-3x-\sqrt{16-24x+9x^{2}-x^{3}}}$
with $-1\leq x\leq 1$[/quote]

Sorry, I was mistaken. The answer is $-1$ when $x=1$.

Here is my solution.

Set $a=\root{3}\of{4-3x+\sqrt{16-24x+9x^{2}-x^{3}}},b=\root{3}\of{4-3x-\sqrt{16-24x+9x^{2}-x^{3}}}$,

Let $y=\root{3}\of{4-3x+\sqrt{16-24x+9x^{2}-x^{3}}}+\root{3}\of{4-3x-\sqrt{16-24x+9x^{2}-x^{3}}}$
with $-1\leq x\leq 1$, we have $y=a+b$.Using $(a+b)^3=a^3+b^3+3ab(a+b)$, we get $y^3-3y=8-6x$.
Since $-1\leqq x\leqq 1$, we have $2\leqq y^3-3y \leqq 14$.The desired minimum value of $y$ can be obtained the inequality $2\leqq y^3-3y\Longleftrightarrow (y+1)^2(y-2)\geqq 0\Longleftrightarrow y=-1\ or\ y\geqq 2$
Therefore  the desired minimum value is $-1$ when $8-6x=2\Longleftrightarrow x=1$


Υπάρχουν επαρκή στοιχεία για την λύση της $ f'(x^2)\equal{}\minus{}f^4(x)$ στο $ (0,\plus{}00)$;
κι αν ναι πως λυνεται ;
Βεβαια με ενδιαφέρει με λυκειακές γνωσεις αλλα και με οποιες αλλες δεν με βλάπτει.
Ευχαριστώ.

Η συναρτησιακή σχέση που δίνεις δεν είναι διαφορική εξίσωση με τον κλασικό ορισμό τέτοιων εξισώσεων. Βλέπεις, στις διαφορικές εξισώσεις έχεις μια σχέση μεταξύ μιας συνάρτησης και των παραγώγων της αλλά χωρίς να συμμετέχει μια σύνθεση της άγνωστης συνάρτησης με άλλη συνάρτηση (το $ y'(x^2)$ στην προκειμένη περίπτωση). Για τη συναρτησιακή σχέση που δίνεις είναι εύκολο να βρει κανείς λύση ψάχνοντας για συναρτήσεις της μορφής $ x^a$ που να την ικανοποιούν. Προκύπτει ότι έχουμε λύση για $ a\equal{}\minus{}1$. Συνεπώς η $ f(x)\equal{}x^{\minus{}1}$ είναι λύση της συναρτησιακής σου εξίσωσης. Δεν ξέρω αν η λύση αυτή είναι μοναδική αλλά ενδεχομένως κάποιος άλλος φίλος της διαδικτυακής παρέας να βοηθήσει με αυτό.


Mrs. Read can knit one pair of children's mittens with a ball of yarn six inches in diameter. How many pairs of identical mittens can she knit with a ball of yarn twelve inches in diameter? Assume that the balls of yarn are rolled consistently.

The diameter is twice as large in the second case. Since this is a volume problem, it is 3D, so 2^3=8


Really nice, but not difficult. We have to prove the inequality

$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\geq\frac32$

for any positive reals a, b, c.

In fact,

$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)$
$=\left(\frac{a}{b}-\frac{a}{b+c}\right)+\left(\frac{b}{c}-\frac{b}{c+a}\right)+\left(\frac{c}{a}-\frac{c}{a+b}\right)$
$=\frac{\left(ca\right)^2}{abc\left(b+c\right)}+\frac{\left(ab\right)^2}{abc\left(c+a\right)}+\frac{\left(bc\right)^2}{abc\left(a+b\right)}$.

By [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15125]the Cauchy-Schwarz inequality in the Engel form[/url],

$\frac{\left(ca\right)^2}{abc\left(b+c\right)}+\frac{\left(ab\right)^2}{abc\left(c+a\right)}+\frac{\left(bc\right)^2}{abc\left(a+b\right)}$
$\geq\frac{\left(ca+ab+bc\right)^2}{abc\left(b+c\right)+abc\left(c+a\right)+abc\left(a+b\right)}=\frac{\left(bc+ca+ab\right)^2}{2abc\left(a+b+c\right)}$.

But $\frac{\left(bc+ca+ab\right)^2}{2abc\left(a+b+c\right)}\geq\frac32$, since $\left(bc+ca+ab\right)^2\geq 3abc\left(a+b+c\right)$, as

$\left(bc+ca+ab\right)^2-3abc\left(a+b+c\right)=\frac12\left(a^2\left(b-c\right)^2+b^2\left(c-a\right)^2+c^2\left(a-b\right)^2\right)$
$\geq 0$.

Thus, $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\geq\frac32$, and the proof is complete.

  darij

[quote="tmbtw"]Prove that for all a,b,c >0,following ineq is always right
        $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})\geq{\frac{3}{2}}$
:)[/quote]

Expand all, it is truely Murhead.
$[3,3,0]+2(a^4c^2+a^2b^4+c^4b^2) \geq [2,2,2] + [3,2,1]$
anyway, Darij solution is so good!  :)


[quote="qweretyq"][quote="joml88"]Evaluate \[(\sin3^\circ+i\cos3^\circ)^{60}.\][/quote]

its been a long time since i did complex numbers but if i remember correctly
[hide] demoivres  says $(cisx)^y = (cisy*x)$
so this would just be $(\cos60*3 + \ isin60*3) = -1$
[/hide][/quote]

Your logic does not quite follow.  You gave $(\cos 3^\circ+i\sin 3^\circ)^{60}$ but I asked for $(\sin 3^\circ +i\cos 3^\circ)^{60}$.

[hide]Multiply by $1=i^{60}$ to get $(-\cos 3^\circ +i\sin 3^\circ)^{60}$, which is $(-\cos (-3^\circ) -i\sin (-3^\circ))^{60}$, since $\cos x$ is even and $\sin x$ is odd.
That, in turn, is $(-\text{cis}(-3^\circ))^{60}=(-1)^{60}\text{cis}(-180^\circ)=-1$.[/hide]

oops read wrong.. i always do this in contests too  :blush:

ok someone help me out
[hide] $(sin 0 +icos 0)^x$ will give you i^x so then i cheated a little and used my ti-89 to find out that that equals $(cis(90x))$
and i found out that $(sin(3) + icos(3))^x$ gives you $(cis(87x))$ so then i realized that $(sin x +icos x)^y$ will equal $(cis((90-x)*y))$
so the answer to this problem is $cis(5220) = -1$ and i sort of understand how that works but i cheated :P so i need someone to explain this[/hide]
Thanks!
-edit-
o nvm i see that bictor has explained it

[hide]Is it just $\text{cis}(87^\circ)^{60}$, which makes it $\text{cis}((90^\circ-3^\circ)(60))$
$\text{cis}(5400^\circ-180^\circ)$ is obviously a multiple of $180\circ$ thus making the answer -1.[/hide]


Enumerate all the $ n\times n$ matrices with the following property: Every row  and every column is a permutation of $ \{1,2,\dots,n\}$. Two matrices may be considered different if one is obtained by another by a set of row exchanges.

Is there any algebraic structure for the set of these matrices ?

PS: I don't know whether I am posting in the right section.

The problem of enumerating [url=http://en.wikipedia.org/wiki/Latin_square]Latin squares[/url] is open.  This probably isn't what you meant by "algebraic structure," but any given Latin square can be thought of as giving the multiplication table of a [url=http://en.wikipedia.org/wiki/Quasigroup]quasigroup[/url], hence enumerating Latin squares is equivalent to enumerating finite quasigroups.  If you have an idea of how hard enumerating finite groups are, you should begin to see the difficulty here.


I have 10 distinguishable socks in my drawer: 4 white, 4 brown, and 2 blue.  In how many ways can I choose a pair of socks, provided that I get two socks of different colors?

There are 4*4 ways of getting white/brown, 4*2 of getting white/blue, and 4*2 ways of getting brown/blue. 16+8+8=32.

This question is a bit unclear.
Pretty much how many pairs of socks can you make, including each different sock, and excluding pairs of the same color sock.
So you first have to figure out what two colour socks can make a pair, and then you need to figure out how many pairs it can make with the seperate socks, so for determining the pairs with blue socks in them, there would be 8 pairs of socks for each blue sock, because 4 of the brown socks can each make a pair with each blue sock, same with the white socks.
Blue -
4 Blue-Brown pairs
4 Blue-white pairs

Other blue sock -
4 Blue-brown pairs
4 Blue-white pairs

And for the brown and white socks, each can make a pair with each other, so

Brown sock -
4 brown-white pairs

2nd brown sock -
another 4 brown-white pairs

3rd brown sock -
another 4 brown-white pairs

4th brown sock -
another 4 brown-white pairs

And then, the white socks make pairs with the brown socks, but those are already counted from the brown-white sock pairs.

So 4+4+4+4+4+4+4+4 = 4*8 = 32

Isn't this a much easier way to think?

[color=blue]Ways to choose two socks:[/color] $ \binom{10}{2}\equal{}45$
[color=red]Ways to choose two white socks:[/color] $ \binom{4}{2}\equal{}6$
[color=red]Ways to choose two brown socks:[/color] $ \binom{4}{2}\equal{}6$
[color=red]Ways to choose two blue socks:[/color] $ \binom{2}{2}\equal{}1$
[color=green]Ways to choose two different socks:[/color] $ 45\minus{}(6\plus{}6\plus{}1)\equal{}\fbox{32}$

I don't like this question because it is possible to get the right answer for the wrong reasons. 4*4*2=32. Maybe their should be 5 white socks so that the correct answer would be 5*4+5*2+4*2 = 38, whereas 5*4*2=40 is different.  For diagnostic purposes its good to have questions where it is unlikely that one will get the right answer for the wrong reasons. Apart from that I like the problem.

@deux:

I did that.  I multiplied 4x4x2 and that was 32.

Why don't we count the socks of the same color as identical? so that white1 and blue1 is the same pair as white2 and blue one? (like the 2person commitee problems)

If I were doing this problem, I'd use "How many are the same?" complementary counting. It's a logical choice.

If you want to see how many pairs you can make with all $4$ white and all $4$ brown socks, you multiply them to get a total of $16$ pairs.  Multiplying the $2$ blue and $4$ white, we get $8$.  And multiplying the $2$ blue with the $4$ brown gets us $8$.  [hide=So adding all three numbers together gets us]$32$ choices[/hide]

Rocket said 32. Please find a good reason to revive a thread next time.

[hide]The socks must be either both white, both brown, or both blue. If the socks are white, there are $\binom{4}{2} = 6$ choices. If the socks are brown, there are $\binom{4}{2} = 6$ choices. If the socks are blue, there is $\binom{2}{2} = 1$ choice. So the total number of choices for socks is $6 + 6 + 1 = \boxed{13}$.[/hide]

Subtract that from 45 and the you get the answer. That's the part we [b][i]don't[/i][/b] want.

[quote=PranavKammili]Solution:
The socks must be either both white, both brown, or both blue. If the socks are white, there are $\binom{4}{2} = 6$ choices. If the socks are brown, there are $\binom{4}{2} = 6$ choices. If the socks are blue, there is $\binom{2}{2} = 1$ choice. So the total number of choices for socks is $6 + 6 + 1 = \boxed{13}$.[/quote]

That is the number of socks you need to subtract from 45 because those are the combinations you don't want.

Can someone explain to me the issue with this way of solving the problem:

The first choice doesn't matter what you get, if you get a white sock, then you have $6$ choices for the second sock. If you get a brown sock, then you have $6$ choices for the second sock. If you get a blue sock, then you have $8$ choices for the second sock. So $6+6+8 = 20$

Well, first of all, it [I]does[/I] matter what you get on the first choice.

Imagine using Complementary Counting for Basic Combinations -_-

[hide=Complementary Counting]The number of total cases is C(10, 2) which is 45. The number of cases that they are the same are C(4, 2) + C(4, 2) + C(2, 2) which is 13. 45 - 13 = 32[/hide]

I entered in 13 at first :(

[hide=Solution]
For white and brown, we get $4\times(4+2)$, but that counts the combinations twice, so it is actually $\frac{4\times6}{2}$. Doing it similarly for each of the others (brown and blue, white and blue), it is $\frac{4\times6}{2}+\frac{4\times6}{2}+\frac{2\times8}{2}$. That's $12+12+8=\boxed{32}$.
[/hide]


[quote="Pirkuliyev Rovsen"]Solve  the equation (x^2+100)²+(x^3+100)³=0[/quote]

$ x\equal{}\minus{}5$ is a trivial solution. The problem is to show that this is the unique one.

[size=200]He  is where but solution? [/size]

This might help.
$ (x^2\plus{}100)\equal{}([x^2\minus{}25]\plus{}125)\\ \ (x^3\plus{}100)\equal{}([x^3\plus{}125]\minus{}25)$
Let $ (x\plus{}5)$ be $ k$
$ (x\minus{}5)\equal{}(k\minus{}10)$
Now see if this simplifies


Given that $ a,b,c \geq 0$, Prove that:

\[ \frac{a^3\plus{}b^3\plus{}c^3}{ab^2\plus{}bc^2\plus{}ca^2}\plus{}2 \geq \frac{a\plus{}b}{b\plus{}c}\plus{}\frac{b\plus{}c}{c\plus{}a}\plus{}\frac{c\plus{}a}{a\plus{}b}
\]

 :)

[quote="shaaam"]Given that $ a,b,c \geq 0$, Prove that:
\[ \frac {a^3 \plus{} b^3 \plus{} c^3}{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} 2 \geq \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b}
\]
:)[/quote]
After expansion we need to prove that
$ \sum_{cyc}(a^5b\plus{}a^4c^2\minus{}a^3b^3\plus{}a^4bc\plus{}a^3b^2c\minus{}2a^3c^2b\minus{}a^2b^2c^2)\geq0,$ which is obviously true.

[quote="shaaam"]Given that $ a,b,c \geq 0$, Prove that:
\[ \frac {a^3 \plus{} b^3 \plus{} c^3}{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} 2 \geq \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b}
\]
:)[/quote]

We just need to rewrite the inequality in the true form. :)
\[ \frac{{a^3  \plus{} b^3  \plus{} c^3 }}{{\sum\limits_{cyc} {ab^2 } }} \minus{} 1 \ge \sum\limits_{cyc} {\frac{{a \plus{} b}}{{b \plus{} c}}}  \minus{} 3
\]

\[ \frac{{\sum\limits_{cyc} {a^3 }  \minus{} \sum\limits_{cyc} {ab^2 } }}{{\sum\limits_{cyc} {ab^2 } }} \ge \frac{{\sum\limits_{cyc} {a^3 }  \minus{} \sum\limits_{cyc} {ab^2 } }}{{\sum {ab} \left( {a \plus{} b} \right) \plus{} 2abc}}
\]

which is obvious. :)

Note that if $ p\ge q>0$ we have $ \dfrac{p}{q}\ge\dfrac{p\plus{}r}{q\plus{}r}$ when $ r\ge 0$.

By rearrengment inequality $ a^3\plus{}b^3\plus{}c^3\ge ab^2\plus{}bc^2\plus{}ca^2$ then

$ 2\plus{}\dfrac{\sum{a^3}}{\sum{bc^2}}\ge$

$ 2\plus{}\dfrac{\sum{(a^3\plus{}b^2c)}\plus{}2abc}{\sum{(bc^2\plus{}b^2c)}\plus{}2abc}\equal{}$

$ \dfrac{\sum{(a^3\plus{}3b^2c\plus{}2bc^2\plus{}2abc)}}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\equal{}$

$ \sum{\dfrac{a^3\plus{}2a^2b\plus{}ab^2\plus{}ca^2\plus{}2abc\plus{}b^2c}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}}\equal{}$

$ \sum{\dfrac{(a\plus{}b)^2(c\plus{}a)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}}\equal{}$

$ \sum{\dfrac{a\plus{}b}{b\plus{}c}}$

:)

Thanks for nice solutions.  :)

Anyways, we have the following:

$ \frac {a^3 \plus{} b^3 \plus{} c^3}{ab^2 \plus{} bc^2 \plus{} ca^2} \plus{} 2 \minus{} \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {c \plus{} a}{a \plus{} b}$

$ \equal{} \frac {(a^2b \plus{} b^2c \plus{} c^2a \plus{} 2abc)(a^3 \plus{} b^3 \plus{} c^3 \minus{} a^2c \minus{} b^2a \minus{} c^2a)}{(ab^2 \plus{} bc^2 \plus{} ca^2)(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq 0$

 :)


For all prime p > 3, prove that [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/d575b3a849c56f6b1c1ee4b3fcb106c6.gif[/img]
EDIT: So sorry.. I forgot part of it.

It is true for p>3. My proof works, because $\sum_{k=1}^{p-1}\frac{1}{k}=0(mod p^{2})$ for p>3.


hello,

  Prove that $ \forall n \in \mathbb{N}$ , there's no number $ k\in \mathbb{N}$ and $ 3n^2 \plus{} 3n \plus{}7\equal{} k^3$

-=FAILED PROOF ALERT=-

This is equivalent to saying that there is no $ k$ for which the expression $ \frac { \minus{} 3 \pm \sqrt {9 \minus{} 4(3)(7 \minus{} k^3)}}{6}$ is an integer. (Which implies $ \sqrt {9 \minus{} 4(3)(7 \minus{} k^3)} \equiv 3 \pmod 6$).

That, in turn, implies

$ 9 \minus{} 12(7 \minus{} k^3) \equiv 3 \pmod 6$

$ \minus{} 75 \plus{} 12k^3 \equiv 3 \pmod 6$

$ 12k^3 \equiv 0 \pmod 6$

But that is true for all k... What did I do wrong now...

failedProofs++;

ARGH.

You didn't really prove anything above; many of your steps were not reversible. You merely showed that if there does exist such a $ k$, then it would satisfy that.

[hide="Solution"]
Take $ \bmod9$. The LHS (starting at $ 0$) cycles with period $ 3$ ($ 7,4,7,\ldots\pmod9$) but never hits $ 0,\pm1\pmod9$ which is required for the RHS to be a perfect cube.
[/hide]


Let $ a,b,c$ be nonnegative real numbers. Suppose that $ a\plus{}b\plus{}c\ge abc$. Prove that:

$ a^2\plus{}b^2\plus{}c^2 \ge abc.$

[quote="moldovan"]Let $ a,b,c$ be nonnegative real numbers. Suppose that $ a \plus{} b \plus{} c\ge abc$. Prove that:

$ a^2 \plus{} b^2 \plus{} c^2 \ge abc.$[/quote]
$ (a^2\plus{}b^2\plus{}c^2)^2 \ge (ab\plus{}bc\plus{}ca)^2 \ge 3abc(a\plus{}b\plus{}c) \ge 3a^2b^2c^2$.
So $ a^2\plus{}b^2\plus{}c^2 \ge \sqrt{3}abc.$ Done.

[quote=sqing]Let $ a,b,c$ be nonnegative real numbers. Suppose that $ a\plus{}b\plus{}c\ge abc$. Prove that:

$$ a^2\plus{}b^2\plus{}c^2 \ge \sqrt{3}abc.$$
$$(a^2+b^2+c^2)^2\geq3(a^2b^2+b^2c^2+c^2a^2) \geq 3abc(a+b+c)\geq 3a^2b^2c^2.$$[/quote]
Let $x=a^2$,$y=b^2$,$z=c^2$ and we prove that $(a^2+b^2+c^2)^2\geq3(a^2b^2+b^2c^2+c^2a^2)$, as it is known that $(x+y+z)^2\geq3(xy+yz+xz)$
Now, $(a^2b^2+b^2c^2+c^2a^2) \geq 3abc(a+b+c)$ can be solved by the inequality $x^2+y^2+z^2\geq xy+xz+yz$. The last inequality is obvious.


S-a anuntat un lot largit de 20 de elevi pentru Balcaniada de juniori de matematica care urmeaza sa faca o pregatire centralizata la Bucuresti, cred ca la Liceul Nichita Stanescu in perioada de 14-28 mai a.c. ,insa de atunci nu a mai aparut nicaieri (Ori n-am gasit eu...) nici o informatie in plus, nici la liceu la mine, nici la Inspectoratul Scolar Judetean, si practic mai sunt doar cateva zile, ar trebui sa ne planificam plecarea,sa neluam bilete la tren etc..
  Ar fi bine daca am putea obtine niste informatii mai concrete(perioada exacta, ziua sosirii, locul-adresa de cazare, nr. de telefon etc.) in timp util de la cei abilitati, respectiv de la oricine este mai bine informat decat subsemnatul...
                 Felixman

[quote="Valentin Vornicu"]Eu trebuia sa postez un anunt, dar am uitat  :blush: luat fiind cu alte cele. Pregatirea se va desfasura intre datile 11-21 mai la Bucuresti. Intr-adevar elevii (si seniori si juniori) din provincie vor fi cazati la Nichita Stanescu (cei care au fost in lot in anii trecuti stiu unde este, cei care nu ... o sa incerc sa pun o harta mai incolo).

Barajele vor fi date marti 16 mai (si juniori, si seniori - cu mentiunea ca la seniori o sa fie un baraj de geometrie, ca si anul trecut), si inca 2 bucati pe 19-20 mai (si juniori si seniori). Dupa aceste baraje vor fi selectionate direct loturile pentru OBMJ si OIM, respectiv Yakutia. Componentii acestor loturi vor mai participa la o pregatire suplimentara in luna iunie.[/quote]
 Multumesc de raspuns, intr-adevar o harta ar fi buna, sau in orice caz adresa exacta a liceului Nichita Stanescu, pe care nu am reusit s-o gasesc pe internet nici cu Google, un nr. de telefon de la liceul respectiv, si daca s-ar putea sa stim exact cand trebuie sa fim acolo:  in seara de miercuri  10 mai sau de joi  11 mai, respectiv cand incepe pregatirea concret-in dimineata/dupa-masa de 11 sau in 12 mai? Si apropo, de curiozitate, unde (Si cand?...)se organizeaza anul acesta OBMJ? (N-am gasit nici despre asta nicaieri vreo informatie...)

Din cate stiu eu OBMJ e anul acesta la Chisinau

Am aflat intre timp cate ceva, respectiv urmatoarele: Liceul este mai cunoscut sub numele Grupul Scolar "Nichita Stanescu", este situat in sectorul 3, cartierul Titan, str. Lucretiu Patrascanu nr.3, telefon 021/3402330. Este situat in sud-estul Bucurestiului, intre bulevardele "Nicolae Grigorescu" si "1 Decembrie 1918", in apropierea unei salbe de strazi care incep cu "Barajul..."(...). Sper sa fiu de folos si altor interesati...

n-ar fi trebuit sa fim anuntati sau ceva...
k eu d ex nu stiam nik despre asta...k n-o venit nik d la inspectorat sau ceva d genu asta...

Iata si o "schita" preliminara a programului de pregatire (pentru seniori):
[code]VINERI	9 – 11	V. Vornicu 	Geometrie
	11 – 13	R. Gologan	Geometrie: rigiditate cu inegalitati

	16 –18	D. Schwarz, C. Popescu	Seminar

SAMBATA	9 –11	M. Becehanu	Geometrie: probleme deosebite
	11 – 13 	V. Vornicu 	Geometrie: tehnici de rezolvare

LUNI	9 – 11 	M. Andronache	Geometrie metrica
	11 – 13 	C. Popescu	Geometrie combinatorica

	16 – 18		Seminar

MARTI 	10 – 14		BARAJ GEOMETRIE: 3 probleme

MIERCURI	9 – 11 	B. Enescu
	11 – 13 	Alexandru Gica	Teoria numerelor

	16 – 18	D. Schwarz, C. Popescu,R. Gologan	 Combinatorica; seminar

JOI	9 – 11 	D. Schwarz	Combinatorica
		M. Becheanu	Teoria numerelor

VINERI	9 – 13:30		BARAJ TIP OIM, 4 probleme

SAMBATA	10 – 14:30		BARAJ TIP OIM 4 probleme

DUMINICA	13-14		REZULTATE
[/code]

PS Din cate stiu eu Ministerul a trimis acum ceva timp hartiile oficiale catre inspectoratele scolare din teritoriu anuntand datele exacte. O sa mai intreb sa vedem cand anume trebuie sa fiti la Nichita Stanescu exact, dar presupun ca ar fi bine sa fiti joi seara acolo (avand in vedere ca vineri la 9 avem prima lectie :) ).

Tinand cont de titlul forumului si implicit de cei care il viziteaza din acest motiv, am putea vedea si o schita de program de pregatire pentru juniori?
 P.S. As dori sa atasez si o harta pentru cine are nevoie sa gaseasca mai usor Grupul Scolar Nichita Stanescu...
  (Insa trebuie sa va autentificati ca sa devina vizibila...)

Gata, am facut rost de programele complete (atasate mai jos ca imagini).

Multumesc pentru programul de pregatire de la juniori, si nu numai in numele meu. Sper ca informatiile din acest forum au fost de folos si celorlalti componenti ai lotului largit pentru JBMO. Sa ne vedem cu bine joi/vineri! :)

[quote="andyciup"]Si aveti cumva o lista cu elevii participanti la aceste ore de pregatire?Nu de alta,dar si eu m-am calificat (juniori) si n-am primit nicio instiintare.[/quote]Cum adica o lista? Daca esti in lot trebuie sa participi :) Si ar cam trebui sa fii deja la Bucuresti (daca nu esti din Bucuresti).

[quote="andyciup"]Sunt din Bucuresti..sunt cel de pe locul 7 :|.. :idea:[/quote]Maine la ora 09:00 la Nichita Stanescu. Nu stiu de ce nu te-a anuntat nimeni dar ai harta mai sus cum sa ajungi.

Eu sunt ala batranu :) ..de 15 ani jumate..tot am dreptul sa vin la pregatirea lotului,nu?

Pana la urma a aflat cineva cand are loc Olimpiada si unde?


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

[hide="ANSWER"]$ \frac{4}{675}$[/hide]

Cylindrical drums measuring $ 3$ feet high and $ 2\frac{1}{2}$ feet in diameter are to placed in a storage area.  The floor of the storage area is a rectangle measuring $ 7$ feet by $ 7\frac{1}{2}$, and the drums cannot be stacked.  What is the greatest number of drums that can be placed in the storage area?

[hide="WHAT I DID"]
I treated the circular bases of the cylindrical drums as their circumscribing squares.  Then, knowing that the diameter of each of the circular bases was $ 2\frac{1}{2}$ feet, I squared $ 2\frac{1}{2}$ to get the area of the circumscribed square.  Following this, I took the total area of the storage floor, $ 7%Error. "feet" is a bad command.
\cdot7\frac{1}{2}$ and divided it by $ (2\frac{1}{2})^2$, getting $ 8.4$.  To get my final answer, I rounded down to $ 8$.  Is this the correct way to do it?  And why?[/hide]

First question:

Total possible:$ 90\times90\equal{}8100$

Then we count:
$ 10\rightarrow11\minus{}19\equal{}9$
$ 11\rightarrow10\minus{}18\equal{}9$
$ 12\rightarrow10\minus{}16\equal{}7$
$ 13\rightarrow10\minus{}15\equal{}6$
$ 14\rightarrow10\minus{}14\equal{}5$
$ 15\rightarrow10\minus{}13\equal{}4$
$ 16\rightarrow10\minus{}12\equal{}3$
$ 17\rightarrow10\minus{}11\equal{}2$
$ 18\rightarrow10\minus{}11\equal{}2$
$ 19\rightarrow10\equal{}1$

What we want:9+9+7+6+5+4+3+2+2+1=48

$ \frac{48}{8100}\equal{}\boxed{\frac{4}{675}}$

For the second question,
The number of drums that can fit in the room can be bigger.
Consider this scenario. You have two balls with a diameter of $ 10$ feet and you're trying to fit them in a rectangular box $ 25$ by $ 20$ feet without stacking. Conclusively, if we assumed the balls as a square, we would know that each ball would fill a 2-dimensional area of $ 100$ feet. Dividing it out, we would say that only $ 5$ balls can fit without stacking.

But wait! There's some extra space in a corner. We can arrange the balls, or circle, in a way that $ 1$ circle touches $ 4$ other circles instead of one circle touching $ 3$ other circles (notice in this scenario we can only have two rows of circles) (another way, try picturing the circles arranged like corners of a triangle instead of corners of a square). Ultimately, [hide]we can fit an extra circle.
$ \frac {500}{100} \equal{} 5$, $ \frac {500}{25\pi}\approx6$[/hide]

Try finding the area of the circle, then divide that by the area of the rectangle.
If you compare the numbers
[hide] $ \frac {15}{2} \times 7 \equal{} \frac {105}{2}$

$ (\frac {5}{4})^2\pi \equal{} \frac {25\pi}{16}$                ($ \pi r^2 \equal{} \text{area}$)

$ \frac {\frac {105}{2}}{\frac {25\pi}{16}} \equal{} \frac {840}{25\pi}\approx 10$ (we must round down)

$ \frac {\frac {105}{2}}{\frac {25}{4}} \equal{} \frac {210}{25}\approx8$
This result tells us that we can fit $ 2$ more drums in the box than if we assumed them as squares. [/hide]

[quote="Ihatepie"]srry to revive the topic, but I got a different answer for 1.

Since there are 90 2 digit numbers,

Wouldn't the  total possible be

89+88+87....+1 because order of the numbers doesn't matter?

that equals

(89)(90)/2 or 4005

Then, you find total possible,

10*11-19 or 9 possible
11*11-18 or 8 possible
12*12-16 or 5 possible
13*13-15 or 3 possible
14*14-14 or 1 possible

So 26 possible over 4005

26/4005[/quote]

no because you cna have something like 89*89, which you didnt include in your count of 90C2 becuase 90C2 implies two different nubmers, this problem doesnt imply that, and that means that if you have 89*89, theres a better chance that a 90*71 will occur because it can occur frontwords or backwords, so even if you add a 90 to that denominator it is still wrong.


LaTeX
inequalities
calculus
calculus computations

This forum is enabled for LaTeX.  Please use it.  See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 for more.

Also, I think your problem has a typo: shouldn't the exponent in the sum in the denominator of the left side be $ 2n$?

Ok , You are right the exponent is :2n ,not :2m but ,My Latex is damaged please forgive me
for this mistake and answer this problem :huh:

I think that the problem should read:

$ \frac{\sum_{i=1}^{m}(2i-1)^{2n}}{\sum_{i=1}^{m}(2i)^{2n}}\leq \left(\frac{2m}{1+2m}\right)^{2n}$

Namely the exponent in the right hand side of your inequality should be $ 2n$ rather than $ 2n+1$. Check your inequality for the case when

$ n=1$, I think it does not work in this case.

To prove it use the following trick:

$ \min \frac{a_{i}}{b_{i}}\leq \frac{\sum a_{i}}{\sum b_{i}}\leq max \frac{a_{i}}{b_{i}}$

Now in your case since

$ \frac{2k-1}{2k}\leq \frac{2m-1}{2m}$ for all $ k=1,\cdots, m$, so you get that your fraction is majored by

$ \left(\frac{2m-1}{2m}\right)^{2n}\leq \left(\frac{2m}{1+2m}\right)^{2n}$.

Didlica : Your proof is right but  if the exponent in the right side be:  (2n+1)
then it is better Approximation  try to prove it.


3.00 mol of gaseous carbon dioxide expands isothermally (in thermal contact with the surroundings: temperature = 15.0ºC) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are 10.0 L and 30.0 L, respectively.

1. Choose the correct option for the change in entropy of the system ($ \Delta S_{sys}$) and of the surroundings ($ \Delta S_{sur}$):

(a) $ \Delta S_{sys}> 0$ and $ \Delta S_{sur}\equal{} 0$.

(b) $ \Delta S_{sys}< 0$ and $ \Delta S_{sur}> 0$.

(c) $ \Delta S_{sys}> 0$ and $ \Delta S_{sur}< 0$.

(d) $ \Delta S_{sys}\equal{} 0$ and $ \Delta S_{sur}\equal{} 0$.
2. Calculate $ \Delta S_{sys}$, assuming $ CO_{2}$ to be an ideal gas.

3. Calculate $ \Delta S_{sur}$.

4. Calculate the change in entropy of the universe. Does your answer agree with the Second law of Thermodynamics?

1.temp. constant so $ \Delta U$ is 0 since work done is negative,so heta is gained by the gas so heat is gained by the gas.so entropy of teh system increases at the expense of surrounding so i think it should be C

What is the entropy of the universe??? And how is it found? (I'm a thermodynamic noob)

In thermodynamics "universe" is the system under study plus the surroundings with which the system can interact by heat, work, or matter transfer. So, $ \Delta S_{univ} \equal{} \Delta S_{syst} \plus{} \Delta S_{surr}$.

How would you find the change of the surroundings ($ \Delta S_{sur}$)in this problem?

The heat and work transfers that occur with surroundings can be considered reversible, since during the process the surroundings are in thermodynamic equilibrium. Therefore,

$ \Delta S_{surr} \equal{} \frac{Q}{T}$,

where Q is the heat transferred between system and surroundings.

doesnt the entropy of the universe always increase for a spontaneous process?
but for processes such as crystallization( which is spontaneous), doesnt the entropy of the universe actually decrease?

In this problem, since it is carried out reversibly, the entropy of the universe will remain constant.

Consider this. During free expansion, the entropy of the universe should increase according to second law, but in free expansion, work done=heat exchanged=change in internal energy=0.
If there is no heat exchanged, how, can there be any increase in entropy. Yes, the disorderedness of the system increases, but how do you explain this?

[quote="VARUNARASU"]doesnt the entropy of the universe always increase for a spontaneous process?[/quote]

Yes, because a spontaneous process is irreversible.

[quote="VARUNARASU"]but for processes such as crystallization( which is spontaneous), doesnt the entropy of the universe actually decrease?[/quote]

According to the second law of thermodynamics, that cannot happen. We must also take into account $ \Delta S_{surr}$.

[quote="Zimrock"]Consider this. During free expansion, the entropy of the universe should increase according to second law, but in free expansion, work done=heat exchanged=change in internal energy=0.
If there is no heat exchanged, how, can there be any increase in entropy. Yes, the disorderedness of the system increases, but how do you explain this?[/quote]

The entropy of the universe and of the system must increase because the free expansion is an irreversible process. However, since S is a state function, $ \Delta S$ for this irreversible process is the same for a reversible process between the same initial and final states. Using the first law we can write for a reversible process on an ideal gas

$ dU \equal{} \delta Q \plus{} \delta W \equal{}> \delta Q \equal{} dU \minus{} \delta W \equal{}> \delta Q \equal{} C_v dT \plus{} PdV \equal{} C_v dT \plus{} \frac{nRT}{V} dV$

because for an ideal gas $ C_v \equal{} \frac{dU}{dT}$. Now divide by T and using the second law we get

$ dS \equal{} C_v \frac{dT}{T} \plus{} nR \frac{dV}{V}$

and integrating between states 1 and 2 we have

$ \Delta S \equal{} C_v \ln \frac{T_2}{T_1} \plus{} nR \ln \frac{V_2}{V_1}$

considering that Cv doesn't vary significantly with temperature. In a free expansion the temperature of the gas remains constant, because dU = 0, and so

$ \Delta S_{free\,\,expansion} \equal{} nR \ln \frac{V_2}{V_1}$.

As we can see, because the volume of the gas increases its entropy must also increase.


How many even, three-digit positive integers have the property
that exactly two of the integer's digits are equal?

Case 1: $ \overline{xyx}$

$ 9 \cdot 9 \cdot 1 \equal{} 81$

Case 2: $ \overline{xxy}$

$ 9 \cdot 1 \cdot 9 \equal{} 81$

Case 3: $ \overline{yxx}$

$ 9 \cdot 9 \cdot 1 \equal{} 81$

In total $ 3 \cdot 81 \equal{} \boxed{243}$.

There are 118 ... split into 2 groups:those that end in 2,4,6,8 and those that end in 0

Enjoyed this. A really nice one.

[hide=Solution]We complementary count. The total amount of three-digit even positive integers is $450$. Now we find the amount with no repeating digits. We have two subcases: If the last digit is zero, there are $9 \cdot 8$ ways. If the last digit is not zero, there are $4 \cdot 8 \cdot 8$ ways. Summing these gives $328$. The only ways to have all three digits the same are $222, 444, 666, 888$ for a total of $4$. Adding to $328$ gives $332$. Our answer is hence $450-332=\boxed{118}$.[/hide]

really? this isn't really that nice, in my opinion; seen a lot better in cp :P

I did casework, not sure where I got wrong though.
Case 1: 0 is unit digit
9 choices for if it's _00, and 9 for if _ _ 0
Case 2: [2,4.6.8] unit digit
_ 22 [9 choices for this as 1-9 all work] 2_2 [10 choices] 22_ [5 choices]
adding them up but minus 2 as you count 222 three times
9+10+5-2 = 19+3 = 22 but it says 25.
Where did I go wrong? Thank you


Given acute triangle $ABC$, altitude $AD$ othorcenter $H$, $M\in DH,BM\cap AC=B',CN\cap AB=C'$ prove that $ABC$ is isosceles triangle iff $BB'=CC'$

[quote="skywalker"]Given acute triangle $ABC,$ altitude $AD,$ othorcenter $H,$ $M\in DH(M\neq D),\,BM\cap AC=B',\,CN\cap AB=C'.$
Prove that $ABC$ is isosceles triangle iff $BB'=CC'$[/quote]
We will prove the problem:
[quote][b]Problem.[/b] [color=darkred]Given acute triangle $ABC,\,AB>AC,$ altitude $AD,$ othorcenter $H,$ $M\in DH\,(M\neq D),\,BM\cap AC=B',\,CN\cap AB=C'.$
 Prove that: $BB'>CC'$[/color][/quote]

[b]Solution. (not of me)[/b]  :D
Contruct the ray $[Bx,\,[Cy$ so that: $\widehat{xBC}=\widehat{ACB},\,\widehat{yCB}=\widehat{MBC}.$
We have:
$\blacktriangleright\;AB>AC\Longrightarrow \widehat{ABC}<\widehat{ACB}\Longrightarrow\widehat{ABC}<\widehat{xBC}\Longrightarrow$ ray $[BA\in \widehat{xBC}$
$\blacktriangleright\;AB>AC\Longrightarrow DB>DC\Longrightarrow MB>MC\Longrightarrow\widehat{MBC}<\widehat{MCB}$
$\Longrightarrow \widehat{yCB}<\widehat{MCB}\Longrightarrow$ ray $[Cy\in\widehat{MCB}$
Denote $L=Cy\cap AB$, we have: $\widehat{CC'L}>90^\circ\Longrightarrow CL<CC'\quad (1)$
Denote $K=Bx\cap Cy\Longrightarrow\begin{cases}L\in [CK]\;\text{because}\; [BA\in\widehat{xBC}\\ \triangle BKC=\triangle CB'B\end{cases}$
$\Longrightarrow CL<CK=BB'\quad (2)$
From $(1)$ and $(2)$ we have: $BB'>CC'$

And what is your solution, [b]skywalker[/b]? :wink:


[quote="K81o7"][hide="hint"] $ \frac{A^{2}}{A^{2}\plus{}B^{2}}\equal{}\frac{1}{1\plus{}\frac{B^{2}}{A^{2}}}$[/hide][/quote]

This one is making me feel old and out of practice.   I'm still lost.

[quote="cbehan"]This one is driving me nuts...I think I'm overcomplicating it.

$ (\frac{A^{2}}{A^{2}\plus{}B^{2}})C \equal{} D$[/quote]
[hide="Solution"]
$ \frac{A^{2}}{A^{2}\plus{}B^{2}}\equal{}\frac{D}{C}$

$ A^{2}\equal{}A^{2}\frac{D}{C}\plus{}\frac{B^{2}D}{C}$

$ A^{2}\left(1\minus{}\frac{D}{C}\right)\equal{}\frac{B^{2}D}{C}$

$ A^{2}\frac{C\minus{}D}{C}\equal{}\frac{B^{2}D}{C}$

$ A^{2}\equal{}\frac{B^{2}D}{C\minus{}D}$

For $ \frac{B^{2}D}{C\minus{}D}\geq 0$,we have:

$ A\equal{}\pm\sqrt{\frac{B^{2}D}{C\minus{}D}}$

ofc with suitable assumption in solvin'...  :wink: [/hide]


Please, is in your books defined this sequence of operations ?

   $ 8 : 4 \cdot 2$

or we must put $ (, )$  ?

Babis

Hi stregiu. I spotted your question this morning but I'm not sure how to answer as I don't think I know what the question means. However, as no-one else has answered I'll give it a go. Do you mean how would "8 divided by 4.2" be written in the AoPS books? I've never seen any of the books but I imagine it would be 8 followed by the division sign (a horizontal line with a dot over the middle and one below it - I can't reproduce this as my computer is on its last legs) followed by 4.2. I guess your comma refers to the fact that a comma is used for the decimal point in Continental Europe. Perhaps if you gave some examples I could help a bit more.

Nope, stergiou asked whether "8 divided by 4, multiplied by 2" is defined in the books, or the brackets are needed to define the order of operations.

Interesting enough, I failed a similar problem in 5th grade (i.e. 10 years ago)... and I'm still not sure who was right, my teacher or \me :)

Now it makes sense and we proceed from left to right.

Yes, the expression would be $ \frac{8}{4} \cdot 2$.

Many thanks !

The question was reffering on the fact if we need brackets or not . In the greek school books this was not cleared enough and when a teacher gave a problem , many students had problem. As teachers , we all discuss this situation in a greek forum. So , I asked you to see what you have met in your text  books , in order to have a most wide opinion.

 Thanks again - Babis


Zan tosses one fair 6 sided die with faces labelled 1 through 6 and one fair 4 sided die with faces labeled 1-4. If at least one die shows a three what is the probability that the sum is 5. Express your answer as a common fraction.

I think that the answer is 1/12 but the answer key says it is 2/9.

The reason your answer is off is because of conditional probablility.  you are saying there are 2out of 24 ways to get a 3 and a sum of 5 (2,3) and (3,2) which is right, but there are not 24 possible outcomes.   Since we only want to look at the roles where a 3 showed up, you only want to look at these : (3, 1); (3, 2); (3, 3); (3, 4) ; (1, 3); (2, 3); (4, 3); (5, 3); (6,3) (don't count (3,3) twice!!)

So of the rolls that include a 3, only 2 had sums of 5.

Hope that helped.

Frost

oh i understand now thanks for your help  :D


trigonometry
trig identities
Law of Cosines
AMC

I know it's a bit late for this, but... from Kalva's solution (http://www.kalva.demon.co.uk/usa/usoln/usol911.html):

[quote]We have $b\cdot sin(C) = c\cdot sin(2C)$ and $c^2 = a^2 + b^2 - 2ab\cdot sin(C)$[/quote]
Can someone explain why the Law of Cosines suddenly has a sine in it?  I am confused.

[quote="zanttrang"]I think it's a typo, as in the next step they assume $\cos C$ rather than $\sin C$.  However, the solution still doesn't make sense to me, so I may be wrong.[/quote]
It IS a typo. Since his next step is: $\frac{\sin{2C}}{\sin{C}}=\frac{2\sin{C}\cos{C}}{\sin{C}}=2\cos{C}=...$
So it should've been $\cos{C}$.


Each edge of a cube is colored either red or black. Every face of the cube has at least one black edge. The smallest possible number of black edges is

$\text{(A)} \ 2 \qquad \text{(B)} \ 3 \qquad \text{(C)} \ 4 \qquad \text{(D)} \ 5 \qquad \text{(E)} \ 6$

[hide]I guess-and-checked with pictures and I got 3?[/hide]

[hide]There are 6 faces.  Each edge belongs to 2 faces.  So we need 3 black edges. B.[/hide]

[hide="Answer"]Each two faces share one edge. We have six faces, so our answer is $\boxed{B}$.[/hide]

[hide]Each edge will take care of two sides. With 6 sides, the answer is 3. [/hide]