Hi, I welcome you all to the Getting Started Forum if you're new here and even you're above the level, I just want you to know that there's a contest that is [b]open to everyone[/b] no matter where you live in AoPS site.

Informations from other thread that I wrote:

[quote="Silverfalcon"]COP, stands for Contest of Opportunity, is a contest that is open to EVERYONE in the Earth. This is a private messaging sending contest where contestants are given 1 week to do each round. Sometimes, it may little bit longer or lesser than 1 week but average is one week.

Rules for each round

Round 1 consists of very basic 5 questions from at least 5 different branches of math. Often times, the questions are coming from geometry, algebra, basic number theory, counting, sequence and series. The best way to study for this round is to just learn a lot of stufffs. The contestants take 1 week to solve the problems. The only way you can get to the next round is if and only if you get a perfect score. This is not hard since all you have to do is solve 5 basic problems in 1 week.

Round 2 consists of 7 questions that are in level of Getting Started and Intermediate. Some questions just require basic knowledge while some questions, you actually have to work out. This round is also about one week. The qualifying is only two top score. If there's a tie, we have a tiebreaker round, or round 3.

Round 3, if it's tiebreaker, consists of two proof for a week. The two proof questions are basic but I look for elegant solutions. There is no specific rules for how scoring works but simply, I grade them.

Round 4, is the Championship Round. In the First COP, there were 5 questions but from the Second COP and so on, there will be 15 questions, approximately in level of AMC-10 to AMC-12, perhaps easy AIME style questions. The question is not impossible but they require thinking and specific knoledge.

How to Prepare for This?

The best way to prepare is to study many math competitions. Especially for Round 4, it is highly recommend that one contestant look over AIME or old AHSMEs for practice because that helps a lot. For other rounds, look over easy AMC-8 or any other elementary math contests. I'm planning to write a math problem book, just for this competition, and only thing is I'm not sure how long it'll take. If I publish one, it would contain rougly 100 problems. But that's for later on.[/quote]

I'm going to add few links on this thread for the old tests and their answers. Hope this would clear many questions up.

Past 2 rounds, I deleted, so they are not up here.

[b]3rd COP Rounds[/b]

[url=http://www.artofproblemsolving.com/Forum/topic-27094.html]Round 1 Problems[/url]

[url=http://www.artofproblemsolving.com/Forum/topic-28890.html]Round 1 Solutions[/url]

[url=http://www.artofproblemsolving.com/Forum/topic-28904.html]Round 2 Problems[/url]

[url=http://www.artofproblemsolving.com/Forum/topic-29386.html]Round 2 Solutions[/url]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=29675]Round 3 Problems[/url]

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=31324]Round 3 Solutions[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31211]Round 4 Problems[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32157]Round 4 Solutions[/url]

[b]4th COP Rounds[/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35184]Round 1A Problems[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=36035]Round 1A Solutions[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=37208]Round 1B Problems[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38555]Round 1B Solutions[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39259]Round 2 Problems[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39597]Round 2 Solutions[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=252316#p252316]Round 3 Problems[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42692]Round 4 Problems[/url]

[b]COP 5th Rounds[/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50026]Round 1A Problems[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=322233#p322233]Round 1A Solutions[/url]

[url=http://www.artofproblemsolving.com/Forum/post-330164.html#p330164]Round 1B Problems[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=340926#p340926]Round 1B Solutions[/url]

And from COP 3rd, I made a special contest for middle schoolers. The rule is same for time but you will be given 10 questions and you have to send at least 7 solutions and get them right. If you get them right, you'll be sent to the round 2 with qualifiers from round 1 of COP.

[b]COPSD 1st[/b]

[url=http://www.artofproblemsolving.com/Forum/topic-27236.html]COPSD 1 Problems[/url]

[url=http://www.artofproblemsolving.com/Forum/topic-27847.html]COPSD 1 Solution[/url]

[b]COPSD 2nd[/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35186]COPSD 2A Problems[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=225843#p225843]COPSD 2A Solutions[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=COPSD&t=37209]COPSD 2B Problems[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=COPSD&t=39213]COPSD 2B Solutions[/url]

[b]COPSD 3rd[/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50030]COPSD 3rd A Problems[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=51090]COPSD 3rd A Solutions[/url]

Note that some links doesn't necessarily include solutions. In that case, please copy the problem and ask on this thread. This thread is going to stay as long as I'm moderator and I do not want old thread [b]revived[/b].

Thanks.

[quote="DarkKnight"]The competition is currently in progress. I think it's in the proof round (tiebreaker round before championship). Don't know when the next one will be though,[/quote]

Yes.

I'll post a notifying thread when the new one starts.

can we submit answers even if we did not participate in the earlier ones?

Hall of Fame

COP 1:

Winner: theone853
Second placer: RC-7th

COP 2:

Winner: alan
Second placer: RC-7th

COP 3:

Winner: Jimmy
Second placer: Farazy

COP 4:

Winner: The Original Pi
Second placer: BornForMath

Shortlist

[url=http://artofproblemsolving.com/Forum/viewtopic.php?p=194622#194622]1[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=205140#205140]2[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=244928#p244928]3[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=55290]4[/url]

Several additional rules regarding the competition.

[b]Two test dates[/b]

Starting from COP 4th, I decided to have A and B version for Round 1 and COPSD. Qualifiers from these four rounds (COP Round1A, COP Round1B, COPSD A, COPSD B), they will all send into same round, COP Round 2. I know this will make a lot of qualifiers to sort but this also means ----more competitive round.

Usually, B test will be posted a week or two after A. They would not contain same topics but their level will be generally same. But that is what [b]I[/b] think on problems so it may be different from one person to another. I'm going to try my best to keep their level same though. Each problem will be in either number theory, combinatorics, geometry, algebra, and other topics that are widely covered in math competitions. The chance of seeing the problem that you have no idea on COP Round1A, COP Round1B, COPSD A, or COPSD B are very rare.

For COP Round1A and B, you only need to send the answers. To qualify, you need to get all questions right.

For COPSD A and B, you have to send at least 7 SOLUTIONS (That is, answers and explanations on them) in order to get credit. If you get at least 7 right, you'll be qualified.

[b]Calculator Use[/b]

Unless the use of calculator is restricted on specific problem, the use of calculator is perfectly allowed. However, you are not allowed to put [b]equations[/b] on calculator and therefore, finding the answer. Those problem will either look for careless mistake or clever manipulation of variables or numbers. This means same for computer answers.

[b]Question contributions[/b]

This usually only happens in Round 4, the Championship Round. I specifically ask a person who is not attending the competition and are in level of Olympiad or beyond. I ask them to give me one question with explanation at level of last few AIME. This is done to make my contest bright and also, to prevent anyone from getting perfect.

[b]Eligiblity on Outside AoPS[/b]

This is topic I'm still studying on how to achieve. From 1st COP, the COPs had been taken by AoPS users. I'm glad that I'm getting good support from it and I thank AoPS users who attended the competition. But maybe little bit later or if there is any chance, I'll be glad to have this contest nationalwide or at least taken by people who are not AoPS users. I had one contestant asking his friend to take the test and I accepted but I don't think I ever got to receive the answer from him. So, I'm still working on this topic.

[b]Additional Resources[/b]

AoPS now provides students with great amount of Olympiad problems. Here is the link for USA part (only because this is the only place that has at least some easy problems)

http://www.artofproblemsolving.com/Forum/resources.php?c=182

These are pretty important rules. So, if you think anything applies to you, please read them. Thank you.

WAIT... Where are the questions for theis round

My bad. I forgot to sticky it back.

Here is the direct link. Answers are due today 11:59:59.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50026

MAybe this has been asked before, but what's a shortlist?

Shortlist is basically collection of problems from other contests that will help you practice on the real COP.

I stopped doing it after a while but who knows, I might be doing them again.  :)

Getting Started Forum: Problem of the Day is a very good place for anyone to practice for COP. I usually post the problems that are in this level but somewhat requires you to do work. In another word, problems are not hard but you have to work on it ---> which is what you have to do on COP.

After 5th COP, the 6th COP will appear much later. I want to take sometime and study more to produce better questions. Right now, I'm getting out of the questions.

This thread will be stickied as long as I moderate this forum. Thanks.  :)

[quote="joseluis"]hello i am new here can some body tell what can de news do in the contest or the maratons i mean if there are some thing that we would do te participate in the cops[/quote]

5th COP is on its 3rd round now and sadly, there will be no COP for about couple month (I'm guessing the 6th one will be around next year's March).

While then, you can hang here and solve problems in Problem Solving Marathon.

No.

The COP is officially done. However, I plan to write a new contest in future but it's not anytime soon.


function
limit
topology
calculus
calculus computations

Let $ f(x)$ be a real-valued function defined for all real numbers $ x$ such that $ |f(x)\minus{}f(y)|$ $ \leq$ $ \frac{1}{2}|x\minus{}y|$ for all $ x,y$ , then find the number of roots of $ f(x)\equal{}x$

[quote="Xaenir"]Let $ f(x)$ be a real-valued function defined for all real numbers $ x$ such that $ |f(x) \minus{} f(y)|$ $ \leq$ $ \frac {1}{2}|x \minus{} y|$ for all $ x,y$ , then find the number of roots of $ f(x) \equal{} x$[/quote]

Exactly one, by Banach fixed poin theorem.

@rafaelm: Can you elaborate , please.
What is "Banach fixed point theorem" ?

[quote="Xaenir"]What is "Banach fixed point theorem" ?[/quote]
Also known as the "contraction mapping lemma."

Suppose $ X$ is a complete metric space and $ f: X\to X$ satisfies the condition that there is a number $ K,\ 0\le K < 1,$ such that $ d(f(x),f(y))\le Kd(x,y)$ for all $ x,y\in X.$ Then $ f$ has a unique fixed point. That is, there exists exactly one point $ p\in X$ such that $ f(p) \equal{} p.$

Outline of the proof: Choose an arbitrary $ x_0.$ Let $ x_1 \equal{} f(x_0),x_2 \equal{} f(x_1),$ etc. Prove that $ d(x_n,x_{n \plus{} 1})\le K^nd(x_0,x_1).$ Use this to prove that $ x_n$ is a Cauchy sequence. Let $ p \equal{} \lim x_n.$ Use the continuity of $ f$ to show that $ f(p) \equal{} p.$ Use the contraction condition directly to show that if $ f(p) \equal{} p$ and $ f(q) \equal{} q,$ then $ p \equal{} q.$


Let n>2 and $ x_1\dots x_n$ be non negative reals. Prove the inequality

$ (x_1x_2x_3.......x_n)^\frac{1}{n}\plus{}\frac{1}{n}\sum_{i<j}|x_i\minus{}x_j|\ge\frac{x_1x_2x_3.......x_n}{n}.$


a parking lot has 5 spots remaining and 5 different cars in line waiting to be parked. if one of the cars is too big to fit in the two outermost spots, in how many different ways can the five cars be arranged?

is there a combination/permutation(or factorial) way to solve this?

[hide]yes; place the biggest car first (it has 3 possibilities since it cannot go on the 2 outer places)
then you are placing 4 cars in 4 spots. there are $ 4!\equal{}24$ ways to do this.
the answer is $ 3 \cdot 24\equal{}72$[/hide]

In general, with problems like these, it's easiest to first place items that have limits on where they can go. Then you can just multiply a couple combinations/permutations rather than going into some obscure, complex method.

Heuristic: Do the difficult decisions first then move to the simple ones


inequalities
algebra
polynomial
inequalities proposed

If $ u,v,w$ are nonnegative numbers, then

$ 25\plus{}30(u\plus{}v\plus{}w)\plus{}8\left(u^2\plus{}v^2\plus{}w^2\right)\plus{}125u^2v^2w^2\geq 8uvw\left[21\plus{}4(u\plus{}v\plus{}w)\right],$

with equality if and only if $ u\equal{}v\equal{}w\equal{}1.$

[quote="Ji Chen"]If $ u,v,w$ are nonnegative numbers, then

$ 25 \plus{} 30(u \plus{} v \plus{} w) \plus{} 8\left(u^2 \plus{} v^2 \plus{} w^2\right) \plus{} 125u^2v^2w^2\geq 8uvw\left[21 \plus{} 4(u \plus{} v \plus{} w)\right],$

with equality if and only if $ u \equal{} v \equal{} w \equal{} 1.$[/quote]
let $ u\plus{}v\plus{}w\equal{}p ,uv\plus{}vw\plus{}wu\equal{}\frac{p^2\minus{}q^2}{3},uvw\equal{}r$
it is equivalent to
 $ 25\plus{}30p\plus{}8(p^2\minus{}2\frac{p^2\minus{}q^2}{3})\plus{}125r^2\minus{}8r(21\plus{}4r)$
$ \equal{}(\frac{84}{125}\plus{}\frac{16p}{125}\minus{}\frac{\sqrt{35379\minus{}9558p\minus{}696p^2\minus{}6000q^2}}{375}\minus{}r)(\frac{84}{125}\plus{}\frac{16p}{125}\plus{}\frac{\sqrt{35379\minus{}9558p\minus{}696p^2\minus{}6000q^2}}{375}\minus{}r) \geq 0$
because $ r \leq \frac{p^3}{27}$
we just need to prove
$ \frac{84}{125}\plus{}\frac{16p}{125}\minus{}\frac{\sqrt{35379\minus{}9558p\minus{}696p^2\minus{}6000q^2}}{375}\minus{}\frac{p^3}{27}$
$ \geq \frac{84}{125}\plus{}\frac{16p}{125}\minus{}\frac{\sqrt{35379\minus{}9558p\minus{}696p^2}}{375}\minus{}\frac{p^3}{27} \geq 0$
and
$ (\frac{84}{125}\plus{}\frac{16p}{125}\minus{}\frac{p^3}{27})^2\minus{}(\frac{\sqrt{35379\minus{}9558p\minus{}696p^2\minus{}6000q^2}}{375})^2
\equal{}\frac{(125p^4\plus{}750p^3\plus{}2511p^2\plus{}3780p\plus{}2025)(p\minus{}3)^2}{91125} \geq 0$
the equality holds if and only if $ u\equal{}v\equal{}w\equal{}1$

we can prove it by only AM-GM,but i think it is better to try first before see my proof because the proof need a very important skill :)

[hide]
$\leftrightarrow$

$14(a+b+c+2a^2b^2c^2-5abc)+8(a^2+b^2+c^2-\frac{(a+b+c)^2}{3})+24(2abc-\frac{a+b+c}{3}-1)^2+(abc-1)^2\ge 0$[/hide]


Are we allowed to discuss the problems from round 3 now that the deadline has passed, or should we hold off in case there were problems with submissions, etc.?  If we shouldn't yet, when can we begin discussing them?

It's usually one week after the submission deadline or so.

DO NOT DISCUSS YET

Indeed - please do not discuss the problems until next Monday.

Just wondering, why the one week delay? Have some people not sent their answers?

Mainly because we'd like to have nearly all the submissions in our hands before there's any discussion.  Sometimes there are problems with faxes and emails (missing pages, etc) that we'd like sorted out before public discussion.

YAY. Discussion begins tomorrow. YIPPEE.


For [tex]a,b,c>0[/tex], prove that

[tex]
\displaystyle\frac{a^2+b^2}{a^4+b^4}+\displaystyle\frac{b^2+c^2}{b^4+c^4}+\displaystyle\frac{c^2+a^2}{c^4+a^4} \leq \displaystyle\frac{a+b+c}{abc}
[/tex]

Easy one.We have:
[tex] \frac{a^2+b^2}{a^4+b^4} \leq \frac{1}{ab} [/tex]
Now,just add them up.


[hide="first question"]Well, ABC in base three is equivelent to 9A+3B+C in base 10, and n digit integer $a_{1}a_{2}\cdot\cdot\cdot a_{n}$ in base three = $a^{1}*3^{n-1}+a_{2}*3^{n-2}+\cdot\cdot\cdot+a_{n}$ in base ten. Also, the digits in base three have to be  either 0, 1 or 2. That tells us that there are 5 digits in the base three number. There is one 81, so the first digit is one.
We subtract that from 145  to get 64. There are 2 27's, so the second digit is 2. We subtract and get 10. There is one nine, no threes and one one. That makes the number 12101 in base three. If you wantbase x, then the base x number $a_{1}a_{2}\cdot\cdot\cdot a_{n}$ is equal to $a^{1}*x^{n-1}+a_{2}*x^{n-2}+\cdot\cdot\cdot+a_{n}$ in base 10.[/hide]

I'm not good at explaining mods, but here's a sample problem.

Find the remainder when 2004^2004 is divided by 7. Can you generalize for n^n?

[hide]First of all, the answer for 2004^2004 is 1.

Second of all, the answer for n^n depends on the remainder for n divided by 7, and then divided by 6, so we may have a problem.

Third, are we allowed to use mods on MathCounts?[/hide]

well these types of problems pop up at times at nationals, it's not really using mods but finding patterns though

and yeah, it's kinda caseworkish for $n^{n}$, like neil said, each value of $n^{x}$ mod 7 loops every 1, 2, 3, or 6 as x increases (kinda weird to explain), so yeah, you'd have to work in mod 7 and mod 6 cases. It's not a question with a nice simple answer.

It might be a little advanced, but doesn't Fermat's Theorem make short work of problems like those? (When you are dividing by a prime, that is)

Check out [url=http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_Totient_Theorem]Euler's Totient Theorem[/url], which helps when you're taking n^something mod x where n and x are coprime. It's a little more advanced than Mathcounts, but still enjoyable.[/url]

I don't know what the rest of the discussion is about as it doesn't really relate to this problem

Find the remainder when 2004^2004 is divided by 7. Can you generalize for n^n?
[hide]Let x = n mod 7. Then it's just x^n mod 7 which can be found using a pattern. 2004 mod 7 = 2 and we want 2^2004 mod 7. So let's look at some mods of 2

$2^{1}$ = 2 mod 7
$2^{2}$ = 4 mod 7
$2^{3}$ = 1 mod 7
$2^{4}$ = 2 mod 7

So it looks like we have a repeat. When you find the repeat, call the number of terms before it repeats y. In this case, it's 3. Find n mod y. In this case, n mod y is 0. This means we want the yth term or $\boxed{1}$.[/hide]


What is the number of inches in the perimeter of a square with an area of two square inches? Express  your answer in simplest radical form.

The area of a square: $ A \equal{} s^2$

Inputting the area: $ 2 \equal{} s^2$

Solving for the side: $ \sqrt{2} \equal{} s$

The length of the perimeter: $ P \equal{} 4s$

$ P \equal{} \boxed{4\sqrt2}$


probability
AMC
AIME
geometry
USA(J)MO
USAMO
number theory
\/closed

i remember reading somewhere that the cost to enroll in the C&P class is 150 dollars without the accompanying textbook and 180 with the textbook...now my question is...when they say 180 dollars with the accompanying C&P textbook, does that include the solutions manual as well?

Which would help more in remedying a gaping weakness in Counting and Probability for the AIME? WOOT or the Intermediate class? I have the basics, but not much more.

I looked at the WOOT schedule but the main topics are geometry, algebra, and Number Theory, as well as Problem Solving, but nothing much to do with combinatorics or counting. Would WOOT still be a wise choice if I just want to make it to USAMO?

Thanks!

If your goal is to improve with counting specifically, then the Intermediate Counting class is the way to go.

If your main goal is passing the AIME, and you don't care at all about the USAMO, then our intermediate classes (and upcoming books) would in general be the way to go.  That said, it's usually best to train a little beyond your goal!

Oh, no, I do care about the USAMO.

I want to get at least 21 points after I make USAMO, but if I'm not able to get past AIME in the first place because of counting ineptitudes, what's the use of that kind of goal? :)

Would WOOT help me with making the USAMO as much as it will in preparing for the USAMO? Otherwise, I'm afraid I'll have to shift my focus from WOOT to Intermediate Counting and Probability, which I don't want to do, because WOOT is a lot more fun!

Another option is to take WOOT, and get the Interm Counting & Probability book when it is out (September, hopefully) and work through it on the side (or during history class if your history classes are anything like mine were).

Good idea. Approximately how much will the book cost?

We haven't set the price yet, but the book and solutions (total a bit over 600 pages) should be around 50 dollars.


Hello i have this problem:

If a+b+c>=abc then proove that

a^2 + b^2 + c^2>= sqrt(3)*abc

a,b,c are real numbers greater then 0.

applying caushy shwartz inequalityto the reals (a,b,c) and (1,1,1)
$ 3(a^2\plus{}b^2\plus{}c^2)>\equal{}sqrt/(a\plus{}b\plus{}c)>\equal{}sqrt/3abc$

[quote="aadil"]applying caushy shwartz inequalityto the reals (a,b,c) and (1,1,1)
$ 3(a^2 \plus{} b^2 \plus{} c^2) > \equal{} sqrt/(a \plus{} b \plus{} c) > \equal{} sqrt/3abc$[/quote]
Can you post your solution written in LATEX.
I think you are wrong.


Find the greatest prime factor of : $ 330^2 \plus{} 330 \plus{} 1$

It should be a factoring problem, but I can't think of how to factor it.

Thanks for your help

Edit: Calculators aren't supposed to be used. In a competition, this problem is given out with 2 other problems in a 10 minute time limit. So there can't be too much guessing. I know that factoring normally won't work. But factoring backwards could work. Like I tried doing:

$ (330 \minus{} 1)(330^2 \plus{} 330 \plus{} 1) \equal{} 330^3 \minus{} 1^3$

So it would be the same thing as:
$ \dfrac {330^3 \minus{} 1^3}{7*47}$

(329=7*47)
And then I got stuck there, so it didn't really work.

And the person who gave me the problem says that he remembers that part of the solution involves first adding everything up and then manipulating it in some way (the problem is really really old so he can't find the solution anymore). It doesn't seem right so maybe he remembered it wrong, but someone could try to solving the problem this way.
It also definitely does not need anything outside of factoring (like mods).

I don't see a way to factor something of the form $ x^2\plus{}x\plus{}1$, but I know that the answer is
[hide]5749[/hide]

Well since this expression takes the form of the 3rd cyclotomic polynomial, $ \displaystyle\Phi_{3}$, we know it's irreducible over $ \mathbb{Z}$....so trying to factor this expression isn't a very promising approach i don't think.

Instead, examining this under various small moduli would probably do the trick. Namely, modulo 19 after testing the other first few primes.


parameterization
geometry
geometric transformation
rotation

Don't know if this is hard enough to be in the Advanced Physics forum, since we may have missed something obvious, but my friend and I were discussing how one could analyze the motion of, say, a ball rolling down a hill with a friction coefficient such that some slipping / sliding occurs (in other words, some of its motion is translational).

In essence, we want to be able to tell "how much sliding" occurs given parameters $\theta, \mu$, etc., so that we can calculate the work done by friction, which would allow us to analyze the motion.  But we couldn't come up with a convincing calculation.

Can anyone help us?

lets make a similar problem.
say we have a billiard ball of mass $m$, radius $r$.  someone hits the ball (along with its center), and the ball starts to slide first, we would like to know how much does the ball slides before it starts to roll.

solution:
the friction force $f$, slows the ball down, and eventually stops the translational motion of the ball; and after the translational motion of the ball stops, due to the torque caused by $f$, it will start to roll.
we have $a=-\mu g$
$\tau=fr=I\alpha\Rightarrow \mu mgr=\frac{2mr^{2}}{5}\alpha\Rightarrow \alpha=\frac{5\mu g}{2r}$
on the other hand we have:
$v=v_{0}+at=v_{0}-\mu gt$   (1)
and $\omega=\omega_{0}+\alpha t=\frac{5\mu g}{2r}t$    (2)

when translational motion stops, $t_{1}$ we have
$v=r\omega$
combining (1) and (2) gives:  $v_{0}-\mu gt_{1}=r(\frac{5\mu g}{2r}t_{1})$
$\Rightarrow t_{1}=\frac{2v_{0}}{7\mu g}$
now by using (1) the speed of the ball when rolling starts:
$v_{1}=\frac{5}{7}v_{0}$
so the distance the ball travels before it starts rolling is:
$x=\frac{v_{1}^{2}-v_{0}^{2}}{2a}=\frac{12v_{0}^{2}}{49\mu g}$

use the same argument for coming down the hill.

[quote="amirhtlusa"]the friction force $f$, slows the ball down, and eventually stops the translational motion of the ball; and after the translational motion of the ball stops, due to the torque caused by $f$, it will start to roll.[/quote]

But what stops the ball from rolling before its translational motion stops?

um...what exactly are you asking? :maybe:

well, you wont have both rolling and slipping at the same time, friction will first slow down the center of mass; which stops the translational motion.
after the translational motion stops, the same force will provide a torque which causes rotation. you wont have rotation and translation at the same time.

[quote="amirhtlusa"]you wont have rotation and translation at the same time.[/quote]

 :?:

Is this certain?  I find this hard to swallow... perhaps I am not thinking clearly right now.

You [i]will[/i] have both rotation and translation at all times. When you hit the billiard ball and give it a certain translational speed $v_{0}$, the friction force ($f$) will slow down the translational motion, but will speed up the rotational motion due to its torque. Assuming a constant friction force, and labeling the final translational and rotational velocities with $v$ and $\omega$, respectively, we get
\[v_{0}-v = \frac{f}{m}\cdot t,\]
\[\omega_{0}= \frac{fR}{2mR^{2}/5}\cdot t.\]
After the slipping ceases, the ball will roll with constant translational/angular velocity; these velocities are what I mean by 'final'. There won't be any acceleration because the work of the friction force on slowing down the ball will now exactly balance the work of its torque on accelerating its rotation. That's why the friction force of rolling without slipping doesn't exert [i]any[/i] work.

Back to the above equations: we can divide the first with the second and then use the non-slipping condition ($\omega R = v$), and we'll get
\[\frac{v_{0}}{\omega}-R = \frac{2R}{5},\] and from here we can find the final angular velocity
\[\omega = \frac{5v_{0}}{7R}.\]
We also have
\[v = \omega R = \frac{5v_{0}}{7},\]
and now we can find the total kinetic energy of the ball
\[T = \frac{mv^{2}}{2}+\frac{I\omega^{2}}{2}= \frac{25mv_{0}^{2}}{98}+\frac{5mv_{0}^{2}}{49}= \frac{5}{14}mv_{0}^{2},\]
if I haven't made a mistake while adding fractions.
The initial kinetic energy is
\[T_{0}= \frac{1}{2}mv_{0}^{2},\]
so the work of the friction force is
\[A = T_{0}-T = \frac{1}{7}mv_{0}^{2}.\]

Rolling down the hill is essentially the same.


Let $ d_n$ be the sum

$ d_n \equal{} \frac{1}{n\plus{}1} \plus{} \frac{1}{n\plus{}2} \plus{} \ldots \plus{} \frac{1}{2n}$

Prove that

$ d_n \equal{} 1 \minus{} \frac{1}{2} \plus{} \frac{1}{3} \minus{} \frac{1}{4} \plus{} \ldots \plus{} \frac{1}{2n\minus{}1} \minus{} \frac{1}{2n}$

[b]I have done this Qsn using the Principle of Mathematical Induction but i want to know if there are other methods.[/b]

Thank You  :)

Let $ H_n \equal{} \sum_{k\equal{}1}^{n} \frac{1}{k}$ and note that the first expression can be rewritten $ d_n \equal{} H_{2n} \minus{} H_n$.  Why is this equivalent to the second expression?


trigonometry
ratio
modular arithmetic
real analysis
integration
limit
real analysis unsolved

$D_n(x)=\frac{1}{2\pi}\sum_{k=-n}^n e^{ikx}$ denotes the Dirichtlet-Kernel.

Proof that

$D_n(x)=\{
\begin{array}{cc}
\frac{1}{2\pi}\frac{\sin (\frac{(2n+1)x}{2})}{\sin (\frac{x}{2})} & \textnormal{if     }  x\not \in 2\pi \mathbb{Z}\\
\frac{2n+1}{2\pi}  & \textnormal{if     }  x \in 2\pi \mathbb{Z}
\end{array}$

You can prove it by direct computation using complex numbers. Have a look at Harmonic Analysis (Katznelson) for instance.

This is a basic calculation, so it's worth reproducing here. The key thing is that it is a finite geometric series. For such a sum, if the ratio is not equal to 1, then

$\text{sum}=\frac{(\text{first term})-(\text{next term after the last})}{1-(\text{ratio})}.$

Assume that $e^{ix}\ne1$ - that is, $x\not\equiv0\pmod{2\pi}.$

$2\pi D_n(x)=\sum_{k=-n}^ne^{ikx}=\frac{e^{-inx}-e^{i(n+1)x}}{1-e^{ix}}$

Now multiply by $\frac{e^{-ix/2}}{e^{-ix/2}},$ getting $\frac{e^{-i(n+\frac12)x}- e^{i(n+\frac12)x}}{e^{-ix/2}-e^{ix/2}}$

$=\frac{e^{i(n+\frac12)x}- e^{-i(n+\frac12)x}}{e^{ix/2}-e^{-ix/2}}$

$=\frac{e^{i(n+\frac12)x}- e^{-i(n+\frac12)x}}{2i}\cdot\frac{2i}{e^{ix/2}-e^{-ix/2}} = \frac{\sin\left(\left(n+\frac12\right)x\right)}{\sin\left(\frac{x}2\right)}.$

Divide that by $2\pi$ to get the quoted formula for $D_n(x).$ (Actually, I wouldn't divide by $2\pi$: I would have that in the definition of the convolution integral. But that's just a matter of convention, not substance.)

For $x\equiv0\pmod{2\pi},$ use the fact that $\sum_{k=-n}^n1=2n+1.$

Followup questions:

Compute $K_n(x)=\frac1{n+1}\sum_{j=0}^nD_j(x)=\frac1{2\pi(n+1)}\sum_{k=-n}^n \left(1-\frac{|k|}{n+1}\right)e^{ikx}.$

Show that $K_n(x)\ge0$ for all $x$ and that $\int_0^{2\pi}|K_n(x)|\,dx=1.$

Show that $\lim_{n\to\infty}\int_0^{2\pi}|D_n(x)|\,dx=\infty.$

Thank you Kent Merryfield.   I've seen my gap


Let a,b be real number such that for any integer p,q then $ap+bq\in I$.Prove that for all$0<s_1<t_1<1,0<s_2<t_2<1$there exists an positive integer n such that$s_1<\{na\}<t_1,s_2<\{nb\}<t_2$


I look at it and the last updated for some people is like 2013! Does anyone else have this problem?


function
logarithms
induction
inequalities proposed
inequalities

If $a_i>0$, $i=\overline{1,n}$ then prove that
$\left(1+\frac1{a_1(1+a_1)}\right)\cdot\ldots\cdot\left(1+\frac1{a_n(1+a_n)}\right)\ge \left(1+\frac1{p(1+p)}\right)^n$ where $p=\sqrt[n]{a_1\cdot\ldots\cdot a_n}$

We just need to make sure that the function $f(x)=\ln{(1+\frac{1}{e^{x}\cdot(e^{x}+1)})}$ is convex. ;)

[quote="billzhao"]Is there a more elementary way?[/quote]
yes, induction

Cauchy induction.
But I think that with that convex function is the easiest way to do it .

:D yes, but,induction is method elementary :D
funny pro


algorithm
linear algebra
matrix
algebra
function
domain

$\mbox{Whats the inverse of } \left(\begin{array}{cc}12&7\\-7&-4\end{array}\right) \mbox{ ?}$

for this problem:
$\left(\begin{array}{cc}12&7\\-7&-4\end{array}\right) . \left(\begin{array}{cc}a&b\\c&d\end{array}\right)= \left(\begin{array}{cc}1&0\\0&1\end{array}\right)$
and you solve the system... There some better algorithms for large matrix...

There's a very curious strategy to find matrix inverse [it is essencially solving the system proposed by igor]. Let me see if I have the domain of TeX so that I can do this by making no mistake

$\left(\begin{array}{cccc}12&7&1&0\\-7&-4&0&1\end{array}\right)$

Now you perform some matrix operations to transform the initial matrix into the identity one. The matrix that remains on the right, after the end of the steps is the inverse matrix [why?]

Example (beggining of the step)

$\left(\begin{array}{cccc}1&\frac{7}{12}&\frac{1}{12}&0\\-7&-4&0&1\end{array}\right)$
$\left(\begin{array}{cccc}1&\frac{7}{12}&\frac{1}{12}&0\\0&\frac{1}{12}&\frac{7}{12}&1\end{array}\right)$

and so on...;-)

Because is a 2x2 matrix you can also try
if $A = \left(\begin{array}{cc}a&b\\c&d\end{array}\right)$

Then $A^{-1}$ be found with formula

$A^{-1} = \frac 1{ac-bd} \left(\begin{array}{cc}d&-b\\-c&a\end{array}\right)$


A jar contains two red marbles, three green marbles, ten white marbles and no other marbles. Two marbles are randomly drawn from this jar without replacement. What is the probability that these two marbles drawn will both be red? Express your answer as a common fraction.

There are $ 15$ marbles total. So we have $ \frac{2}{15}*\frac{1}{14}\equal{}\frac{1}{105}$. So the answer is $ \boxed{\frac{1}{105}}$.


inequalities
function
inequalities proposed

This problem has many solutions. Mine is the following.
It is easy to show
$ 1) abc\leq 1$ and $ (\sum a)\leq 3$
$ 2) (\sum a^4)\geq 1/3* (\sum a^3)*( \sum a)$
$ 3) (\sum a^3) +6abc\geq  (\sum a^2b+b^2a)+3abc =  (\sum a) (\sum ab)$
Opening parenthesis gives us to demonstrate:

$ 8(\sum ab)+a^2b^2c^2\leq 16 +3abc(\sum a)$

We see $1/3abc(\sum a)\geq a^2b^2c^2$
So it remains to prove:

$ 8(\sum ab)\leq 16 +8/3abc(\sum a)$ or

$ (\sum ab)\leq 2 +1/3abc(\sum a)$

From 2) and 3) we have:

$1/2  = 9/18 =\frac{3(\sum a^4)}{18}\geq \frac{(\sum a^3)*(\sum a)}{18}$

$1/2 +1/3abc(\sum a)\geq \frac{(\sum a)*((\sum a^3)+6abc)}{18}\geq \frac{(\sum ab)*(\sum a)^2}{18}$

It remains to prove $3/2 + \frac{(\sum ab)*(\sum a)^2}{18}\geq (\sum ab)$

$3/2\geq (\sum ab)(1- \frac{(\sum a)^2}{18})$

We see that $\frac{(\sum a)^2}{18}\leq 1/2$

So $(\sum ab)(1- \frac{(\sum a)^2}{18})\leq1/3(\sum a)^2(1- \frac{(\sum a)^2}{18})\leq 3/2$

Which gives us multiplying by 3:

$  9/2 + \frac{(\sum a)^4}{18}\geq (\sum a)^2$  - AM-GM inequality for 2 numbers. c.t.d.

By applying AM-GM inequality, we obtain
     $LHS \leq \sum \frac{1}{4-\sqrt{\frac{a^4+b^4}{2}}} = \sum \frac{1}{4-\sqrt{\frac{3-c^4}{2}}}$
  Denote $u = \frac{3-c^4}{2}, v = \frac{3-c^4}{2}, w = \frac{3-c^4}{2}$. Then $0< u, v, w<\frac{3}{2}$ and $u+v+w = 3$
  Let $f(u) = \frac{1}{4-\sqrt{u}}$
  Then $f'(u) = \frac{1}{2\sqrt{u}(4-\sqrt{u})^2}$
          $f''(u) = \frac{-u^{\frac{-3}{4}}(1-\frac{3}{4}u^{\frac{1}{2}})}{(4u^{\frac{1}{4}} - u^ \frac{3}{4})^3}$
      Hence $f''(u) < 0$ for every $ 0<u<\frac{3}{2}$
      By apply Karamata 's inequality for the function that is concave down, we obtain the result.  :)

Is it true that $x^2+y^2+z^2=3$ and $x+y+z>23/9$ implies
\[x+y+z\geq2+xyz?\]

[quote="prowler"]
Opening parenthesis gives us to demonstrate:

$ 8(\sum ab)+a^2b^2c^2\leq 16 +3abc(\sum a) (\sum ab)$
[/quote]

why?

Sorry, of course without $( \sum ab)$.
Mechanical mistake posting solution.

Great proof, Prowler. :)

Only this is useless

[quote]We see that $\frac{(\sum a)^2}{18}\leq 1/2$[/quote]

as you use only

[quote]$(\sum a)^2 \geq 3\sum ab$[/quote]

I think this one is more elegant.

It is easy to prove that : $( \sum \frac{1}{(4-ab)}) \leq  ( \sum \frac{1}{(4-a^2)}) \leq ( \sum (a^4+5)/18) .$

The rest follows.

How to prove

[quote]$\displaystyle ( \sum \frac{1}{(4-ab)}) \leq ( \sum \frac{1}{(4-a^2)})$[/quote]

It remembers me $\sum \frac {1}{a^2} \geq \sum \frac{1}{ab}$ but how to relate this one at our first inequality ?

and

[quote]$ ( \sum \frac{1}{(4-a^2)}) \leq ( \sum \frac{a^4+5}{18} $ [/quote]

This last $\frac{1}{(4-a^2)}) \leq \frac{a^4+5}{18} $ becomes

$(a-1)^2(-a^4-2a^3+a^2+4a+2) \geq 0$

As $a^4 \leq 3$ it becomes

$a^2+4a \geq 1 +2a^3$

Is it true for $a^4 \leq 3$  ?????
Thanks.

Thank you very much , Anto.

[quote]I always remember your super solution for the very hard Harazi "sweet inequality" :) [/quote]

[quote="Ghang Hwan"][quote="Anto"]Well, the first inequality follows from : $ \frac {2}{4-ab} \leq \frac {1}{4-a^2} + \frac{1}{4-b^2}$

The second : $ \frac {1}{4-a^2} \leq \frac {a^4+5}{18}$  is equivalent to :
$0 \leq 4a^4+2-a^6-5a^2 $  or  $ 0 \leq (a^2-1)^2(2-a^2) $  which is true since $ a^4 \leq 3 $ and as a result $ a^2 \leq 2 .$

Sorry that i didn't explain it before, but it would lose its elegance.  :P[/quote]

I think your solution is  :!:  :!:  :!:  :!:  :!:  :!:  :!:  :!:  :!:[/quote]

what do you mean? :huh:

Solution:
We have:
$ \frac {1}{4 \minus{} b^2} \plus{} \frac {1}{4 \minus{} c^2}\ge\frac {4}{8 \minus{} b^2 \minus{} c^2}\ge\frac {2}{4 \minus{} bc}\Rightarrow \sum\frac {1}{4 \minus{} bc}\le\sum\frac {1}{4 \minus{} a^2}$
But $ \frac {1}{4 \minus{} a^2}\le\frac {a^4 \plus{} 5}{18}\Leftrightarrow (a \minus{} 1)^2(2 \minus{} a)\ge 0$

The following solution, based on Cauchy-Schwartz inequality, is dedicated to [b]arqady[/b], a person who helped me very much.

[url=http://www.artofproblemsolving.com/community/c6h1178544p5696767]NHN inequality[/url]
Let $a,b,c>0$ and $a^4+b^4+c^4=3$ .For $k \geq 4$, prove that
$\frac{1}{k-ab}+\frac{1}{k-bc}+\frac{1}{k-ac} \leq \frac{3}{k-1}.$

[quote=Sasha] For any positive reals $ a$, $ b$, $ c$ satisfying $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 3$, prove the inequality

$ \frac{1}{4\minus{}bc}\plus{}\frac{1}{4\minus{}ca}\plus{}\frac{1}{4\minus{}ab}\leq 1$.[/quote]
[url=http://www.artofproblemsolving.com/community/c6h1138683p5344973]Strengthening:[/url]
For all positive real numbers $x,y,z$ such that $x^2+y^2+z^2=3$ . Prove that$$\frac{1}{4-x}+\frac{1}{4-y}+\frac{1}{4-z} \leq 1.$$

[quote=Sasha]Let $ a$, $ b$, $ c$ be positive reals such that $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 3$. Prove that $ \sum\frac1{4 \minus{} ab}\leq1$, where the $ \sum$ sign stands for cyclic summation.

[i]Alternative formulation:[/i] For any positive reals $ a$, $ b$, $ c$ satisfying $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 3$, prove the inequality

$ \frac{1}{4\minus{}bc}\plus{}\frac{1}{4\minus{}ca}\plus{}\frac{1}{4\minus{}ab}\leq 1$.[/quote]

Hello this is mine solution : we proceed by chebyshev's associate technique : start: note that the ineqn is equivalent to proving $$\sum \frac{ab-1}{4-ab} \ge {0}$$ then let $ab=x,ac=y,bc=z$ also by the $\sum {a^{4}} \ge \sum {{(ab)}^{2}}$ which is obtained by AM,GM on $a^{4}+b^{4}$ and its family $\cdots$  which means $ 3 \ge{\sum x^{2}}$ and lets WLOG : ${x}\ge{y}\ge{z}$  we write the ineqn to prove as : $$\sum \frac{1-x^{2}}{-x^{2}+3x+4} \ge {0}$$ also note that ${-x^{2} + 3x + 4}\ge{-y^{2}+3y+4}$ and now we finish by chebychev's : $${\sum \frac{1-x^{2}}{-x^{2}+3x+4}} \ge {{\frac{1}{3}}(\sum 1-x^{2}){(\sum \frac{1}{-x^{2} + 3x + 4})}} \ge {0} $$

For all positive real numbers $x,y,z$ such that $x^2+y^2+z^2=3$ . Prove that$$  \frac{3-x^2}{4-x}+\frac{3-y^2}{4-y}+\frac{3-z^2}{4-z} \leq 2 $$
[hide=Thanks.][quote=PSS_73939133][quote=sqing]For all positive real numbers $x,y,z$ such that $x^2+y^2+z^2=3$ . Prove that$$ \frac{yz}{4-x}+\frac{zx}{4-y}+\frac{xy}{4-z}  \leq 1 $$[/quote]

[hide=hint] note that $\frac{1}{2}(\frac{y^{2}+z^{2}}{4-x}) \ge {\frac{yz}{4-x}}$ and simply fractions[/quote]

[/hide]


Tami had 1800 baseball cards. She sold $ \frac{1}{2}$ of them and then
gave away $ \frac{4}
{5}$ of her remaining ones. How many cards does
Tami have left?


let positive numbers $a;b;c$ satisfies $ab+bc+ca=1$
Find a minium value of  $5a^2+16b^2+27c^2$

I think it is wrong.
Equality case does not exist.

AM-GM implies:
$\frac{3a^2+12b^2}{2}\ge\sqrt{3a^2\cdot 12b^2}\Rightarrow 3a^2+12b^2\ge 12ab$
$\frac{4b^2+9c^2}{2}\ge\sqrt{4b^2\cdot 9c^2}\Rightarrow 4b^2+9c^2\ge 12bc$
$\frac{18c^2+2a^2}{2}\ge\sqrt{18c^2\cdot 2a^2}\Rightarrow 18c^2+2a^2\ge 12ca$
Adding these inequalities we get:
$5a^2+16b^2+27c^2\ge 12(ab+bc+ca)=12$
Equality occurs for $a=1$, $b=\frac{1}{2}$, $c=\frac{1}{3}$.


Prove that

$\displaystyle \sqrt[2n]{n!}\leq \displaystyle \sqrt[2n+2]{(n+1)!}$

$\forall x \in\ \mathbb{N}$

Enjoy it! :D  :)  :)

[quote="Diogene"]That's right because of $\displaystyle \sqrt[n]{n!}\leq \displaystyle \sqrt[n+1]{\frac{(n+1)!}2}$   :P
 :cool:[/quote]

Yes. Whta's your proof??

$\displaystyle \sqrt[n]{n!}\leq \sqrt[n+1]{\frac{(n+1)!}2} \Longleftrightarrow (n!)^{\frac 1n - \frac 1{n+1}} \leq (\frac{n+1}2)^{\frac 1{n+1}} \Longleftrightarrow \sqrt[n]{n!} \leq (\frac{n+1}2)$, and the last inequality is trivial (think AM_GM)
:cool:


LaTeX
geometry
geometric transformation
number theory unsolved
number theory

SOLVE IN POSITIVE INTEGERS:X^2+Y^2+W^2=(Y^2-Y+1)^2

Please use $\LaTeX$ :) (see http://www.mathlinks.ro/Forum/viewtopic.php?t=5261 ).
Also don't write in capitals please.

In translation: find all solutions to $x^2+y^2+z^2=(y^2-y+1)^2$


[quote="Dimitris X"][quote="KapioPulsar"]i[i])[b]ndo yparxei 8etikos N tetios wste h dekadikh anaparastash tou 2000^N  arxizei me thn akolou8ia twn yhfiwn 200120012001[/b][/i]i)[b]ndo yparxei 8etikos N tetios wste h dekadikh anaparastash tou 2000^N  arxizei me thn akolou8ia twn yhfiwn 200120012001[/b][b][/b][/quote]
Τι μου θυμίζει,τι μου θυμίζει????χεχε
Έχω την εντύπωση ότι τέτοια θέματα δουλεύονται με το θεώρημα για τη πυκνότητα του kronocker.....
Σωστά???[/quote]   Swsta  :wink:  :lol:

[quote="Dimitris X"][quote="KapioPulsar"]i[i])[b]ndo yparxei 8etikos N tetios wste h dekadikh anaparastash tou 2000^N  arxizei me thn akolou8ia twn yhfiwn 200120012001[/b][/i]i)[b]ndo yparxei 8etikos N tetios wste h dekadikh anaparastash tou 2000^N  arxizei me thn akolou8ia twn yhfiwn 200120012001[/b][b][/b][/quote]
Τι μου θυμίζει,τι μου θυμίζει????χεχε
[/quote]

Πολύ ωραίο και γνωστό πρόβλημα που έχει πέσει σε Μεσογειάδα (του 2001). Εχει αναρτηθεί επίσης (π.χ.) εδώ:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1150782252&t=50019[/url]

Lhmma  :[hide]estw $ d0,d1,d2,..,d(n\minus{}1)$ mia akolou8ia n yhfiwn tote yparxei N>0 akereos tetios wste h dekadikh anaparastash tou 2000^N arxizei me afthn thn akolou8ia
 An z enas arhtos ari8mos tote ka8w diastima (z,b) periexei toulaxiston ena stoixio tou sunolou A={nz+m|n,m anoikoun Z(akereoi)) ara ws amesh sunepia exoume oti an z arhtos tote to sunolo twn pragmatikwn ari8mwn ths morfhs A einai pukno panw ston pragmatiko a3wna dhladh ka8e diastima  (z,b) periexei enan apeiro ari8mo stoixion aftou tou sunolou[/hide]

 :!:

[quote="Dimitris X"]
Τι μου θυμίζει,τι μου θυμίζει????χεχε
Έχω την εντύπωση ότι τέτοια θέματα δουλεύονται με το θεώρημα για τη πυκνότητα του kronocker.....
Σωστά???[/quote]

Στο μάθημα προετοιμασίας το περασμένο καλοκαίρι έγιναν και άλλες παρόμοιες ασκήσεις που λύνονται με το παραπάνω πολύ χρήσιμο θεώρημα. Αν θέλει κάποιος ας τις αναρτήσει


Given $ a,b,c>1$ and $ ab\plus{}bc\plus{}ca\equal{}2abc$.Prove that:
$ 5(a\plus{}b\plus{}c)\minus{}4abc\geq\ 9$ :)

[quote="Allnames"][quote="quykhtn-qa1"]Given $ a,b,c > 1$ and $ ab \plus{} bc \plus{} ca \equal{} 2abc$.Prove that:
$ 5(a \plus{} b \plus{} c) \minus{} 4abc\geq\ 9$ :)[/quote]
Set $ a \equal{} \frac {1}{x}.etc$
The inequality becomes
\[ 5(xy \plus{} yz \plus{} zx)\ge 4 \plus{} 9abc
\]
with $ x \plus{} y \plus{} z \equal{} 2,x,y,z < 1$

Now $ f(x,y,z) \equal{} LHS \minus{} RHS$.Assume $ z\ge y\ge x$ so $ z\ge \frac {2}{3}$
\[ A \equal{} f(x,y,z) \minus{} f(\frac {x \plus{} y}{2},\frac {x \plus{} y}{2},z) \equal{} \frac {(x \minus{} y)^2}{4}(9z \minus{} 5)
\]
It is trivial to see $ A\ge 0$.So if your inequality is true,the last step will be very easy  :lol:[/quote]
 the solution of you very long :wink:
this ine is SCHUR 'inequality :oops:

[quote="quykhtn-qa1"]Given $ a,b,c > 1$ and $ ab \plus{} bc \plus{} ca \equal{} 2abc$.Prove that:
$ 5(a \plus{} b \plus{} c) \minus{} 4abc\geq\ 9$ :)[/quote]
Let $ x,y,z \ge 0$ and $ a \equal{} 1 \plus{} x , b \equal{} 1 \plus{} y,c \equal{} 1 \plus{} z$

$ p \equal{} x \plus{} y \plus{} z$ , $ q \equal{} xy \plus{} yz \plus{} zx$ , $ r \equal{} xyz$

So we must prove : $ p \ge 2q$ then $ q \plus{} 2r \equal{} 1$
     $ \frac {3}{4} \le q \le 1$
By SHUR : $ f(p) \equal{} p^3 \minus{} 4pq \plus{} \frac {9}{2}(1 \minus{} q) \ge 0$ , $ f(2q) \equal{} 8(1 \minus{} q)(\frac {9}{16} \minus{} q^2) \le 0 \equal{} > p\ge 2q$

equality will be for ( $ a \equal{} 1$ and $ b \equal{} c \equal{} 2$ ) or ( $ a \equal{} b \equal{} c \equal{} \frac {3}{2}$) :)


What is the largest value of [tex]k[/tex] such that

[tex]a^2+b^2+c^2+\sqrt{kabc} \leq 2[/tex]

holds for any non-negative [tex]a,b,c,[/tex] satisfying [tex]a+b+c=1[/tex]?

Using the AM-GM inequality we get:
[tex] (\frac{a+b+c+k}{4})^2 = (\frac{1+k}{4})^2 \geq \sqrt{kabc}[/tex]

By the requirement [tex] a+b+c = 1 [/tex] and the symmetry we know that [tex] a\leq b\leq c\leq 1[/tex] and thus [tex] a+b+c = 1 \geq a^2+b^2+c^2 [/tex]

Then we have: [tex] 1+(\frac{1+k}{4})^2 \leq 2 \Leftrightarrow k^2+2k-15 \leq 0 [/tex] for which the maximum k is [tex] 3 [/tex].

Is this correct? This is my first post here and I'm somewhat a beginner when it comes to problem solving.

toub wrote:[tex] a\leq b\leq c\leq 1[/tex] and thus [tex] a+b+c = 1 \geq a^2+b^2+c^2 [/tex]

You're on the right track, but can you improve this bound?

Hint (spoiler): [hide]a^2+b^2+c^2 <= (a^2+b^2+c^2+2ab+2bc+2ca)/3[/hide]

Toub, I don't understand the "Then we have:" ?

I found this problem very difficult !

I think that k=75 is the answer.

First setting a=b=c=1/3 gives k <= 75.

Homogeneizing, we have to prove :

 :Sigma: a + 5 :sqrt: (3abc (:Sigma: a)) <= 2(:Sigma: a)

Rearranging, squaring, ... this leads to the ugly :

8 (:Sigma: a:^3:b) + 18 (:Sigma: ab) + (:Sigma: a^4) >= 35 (:Sigma: a:^2:bc)

which is true by rearrangement.

Yep, [tex]k=75[/tex] is right... another way to prove it:

After noting that [tex]a=b=c=\frac{1}{3}[/tex] implies that [tex]k\leq 75[/tex], it suffices to prove that the inequality

[tex]a^2+b^2+c^2+\sqrt{75abc} \leq 2[/tex]

holds for all [tex]a,b,c,[/tex] satisfying [tex]a+b+c=1[/tex].

We have [tex]a^2+b^2+c^2=\displaystyle\frac{3(a^2+b^2+c^2)}{3}[/tex]

[tex]\leq \displaystyle\frac{a^2+b^2+c^2+2(ab+bc+ca)}{3}=\displaystyle\frac{(a+b+c)^2}{3}=\displaystyle\frac{1}{3}[/tex], from rearrangement.

Now it suffices to prove that [tex]\sqrt{75abc} \leq \displaystyle\frac{5}{3}[/tex]

or after squaring and dividing by 75,

[tex]75abc \leq \displaystyle\frac{25}{9}[/tex]

[tex]abc \leq \displaystyle\frac{1}{27}[/tex]

This is true from AM-GM,

[tex]a+b+c \geq 3\sqrt[3]{abc}[/tex]

[tex]\displaystyle\frac{1}{27} \geq abc[/tex].    [tex]\blacksquare[/tex]

a+b+c <= 1/3 ??

Either I'm missing something really obvious  :oops: , either a=1, b=c=0 is a major problem  :?:

[quote="ThAzN1"]We have [tex]a^2+b^2+c^2=\displaystyle\frac{3(a^2+b^2+c^2)}{3}[/tex]

[tex]\leq \displaystyle\frac{a^2+b^2+c^2+2(ab+bc+ca)}{3}=\displaystyle\frac{(a+b+c)^2}{3}=\displaystyle\frac{1}{3}[/tex], from rearrangement.
[/quote]

Is this reversed?


Given a vector $ (x_1, \dots, x_{10})$ of independent random variables that are all drawn from a uniform distribution on $ (0,1)$, perform the following operation 999 times:

1. Find the maximum of the $ x_i$. Let's say it's $ x_k$.
2. For each $ j \ne k$, replace $ x_j$ with $ x_j/x_k$.
3. Replace $ x_k$ with a new independently drawn $ U(0,1)$ random variable.

At the end, you have a new random vector $ (y_1, \dots, y_{10})$. What is the distribution of $ y_1$?

[quote="alekk"]just a precision:
let's call $ x_k(n)$ the maximum of $ (x_1(n), x_2(n), \ldots, x_{10}(n))$. Then one defines $ x_k(n \plus{} 1) \equal{} U(0;1)$. The question is: do you define, for $ j \neq k$:
$ x_j(n \plus{} 1) \equal{} \frac {x_j(n)}{x_k(n)}$ or $ x_j(n \plus{} 1) \equal{} \frac {x_j(n)}{x_k(n \plus{} 1)}$
I would be inclined to think that it is the first variant (to keep things inside $ [0;1]$). Is it ?

[edit]: no, if this is the first variance (everything stays inside [0;1]), then one sees that $ y_1$ is also uniform on $ [0;1]$[/quote]

It's the first version - I have now corrected the question.

induction seems to show that for all $ n$, the random vector $ (x_1(n), x_2(n), \ldots, x_{10}(n))$ has same law as $ (U_1, \ldots, U_{10})$ where $ U_i$ are independant random variable uniformly distributed on $ [0;1]$. Isn't it ?

Yes. Each step leaves everything the same in distribution.

So what about the following modification:

Given a vector $ (x_1(0), \dots, x_{10}(0))$ of independent random variables that are all drawn from a uniform distribution on $ (0,1)$, perform the following operations:

1. Find the maximum of the $ x_i(n)$. Let's say it's $ x_k(n)$.
2. Put $ x_k(n\plus{}1)\equal{}U_{n\plus{}1}$, where $ \{ U_i \}_{i \geq 1}$ is a sequence of iid random variables uniformly distributed on $ (0;1)$.
3. For each $ j \ne k$, put $ x_j(n\plus{}1)\equal{}\frac{x_j(n)}{x_k(n\plus{}1)}$.

Does the law of the random vector $ (x_1(n), x_2(n), \ldots, x_{10}(n))$ converges in distribution as $ n \to \infty$


give the exact value of

$\cos \frac{\pi}{9}\cos \frac{2\pi}{9}\cos \frac{4\pi}{9}$

[quote="M4RI0"]

$8E sen\frac{\pi}{9}= sen\frac{8\pi}{9}$
$8E=1 \Rightarrow E=\frac{1}{8}$[/quote]

How did you do that?  I'm bad at radians, so I converted to degrees:

$8E\sin 20^\circ = \sin 160^\circ$

$8E= \frac{\sin 160^\circ}{\sin 20^\circ}$

$E= \frac{\frac{\sin 160^\circ}{\sin 20^\circ}}{8}$

$sin160^\circ=sin20^\circ$, more generally $sin\alpha=sin(180^\circ-\alpha)$

[quote="shobber"]http://www.mathlinks.ro/Forum/viewtopic.php?t=103990 They are the same. Why you posted two times?[/quote]

good question.  i saw them in two different places when i was studying similar trig ideas.  It wasn't until after i solved the second one that i realized how similar they were.

[quote="M4RI0"]$E=cos\frac{\pi}{9}cos\frac{2\pi}{9}cos\frac{4\pi}{9}$

Multiplying by $8sen\frac{\pi}{9}$ and by doube angle formula:

$8E sen\frac{\pi}{9}= sen\frac{8\pi}{9}$
$8E=1 \Rightarrow E=\frac{1}{8}$[/quote]

nvm got it.

$2 \cdot 4 \cdot \sin\frac{\pi}{9}\cdot \cos\frac{\pi}{9}\cdot \cos\frac{2\pi}{9}\cdot \cos\frac{4\pi}{9}= 2 \cdot 2 \cdot \sin\frac{2\pi}{9}\cdot \cos\frac{2\pi}{9}\cdot \cos\frac{4\pi}{9}= 2 \cdot \sin\frac{4\pi}{9}\cdot \cos\frac{4\pi}{9}= \sin\frac{8\pi}{9}=8E\sin\frac{\pi}{9}E=\frac{1}{8}$

[quote="M4RI0"]$E=cos\frac{\pi}{9}cos\frac{2\pi}{9}cos\frac{4\pi}{9}$

Multiplying by $8sen\frac{\pi}{9}$ and by doube angle formula:

$8E sen\frac{\pi}{9}= sen\frac{8\pi}{9}$
$8E=1 \Rightarrow E=\frac{1}{8}$[/quote]

Srry I'm not that familiar with trig also, what is $sen(A)$?

Oh, wow.  I knew that.  I don't know what I was thinking....

[quote="lotrgreengrapes7926"][quote="computer_nuke"]So why did M4RI0 write $sen$ instead of $sin$? What's special about it?[/quote]That's just how they write it in Huacho-Lima (maybe Peru). It's the same thing, nothing special. :D[/quote]

Hmmm, awsome, I didn't know they had different $sin$ notations in different countries  :D .

What do you use to express tangent function? $\tan$ or $\text{tg}$?

I think the onlyone that is different is $sin$.

English: sin, cos, tan.
Spanish: sen, cos, tg.

sen=sin, cos=cos, tan=tg


John sold a fridge at a discount of $100 and made a 25% profit. He sold another identical fridge at a discount of
$300 and made a loss of $30. What is the usual selling price of the fridge?

[hide="Answer"]Discount of $270$ gives the price of the fridge. Discount of $100$ gives a 25% profit, so $170$ is 25%. Therefore, the normal price of the fridge is $\boxed{680}$ dollars.[/hide]

That is right for what John bought it at, but I meant what price does John usually sell it. :)

my answer[hide]x-100=5/4y                     x=original price
                          x-300=y-30                     y=the price that the shop bought it at.
             therefore x=y+270 and x=5/4y+100
             y+270=5/4y+100, so 270=1/4y+100 and therefore 1/4y=170
             therefore y=$\$680, insert that into\"x=y+270\" and x=$\$950.
[b]THE ANSWER THEREFORE IS $950.[/b][/hide]

flamesofdestiny is correct

[hide]Let x be the price of the normal selling price, y be the price of the price John bought it at.

A profit of 25% is 1.25y, a loss of 30 dollars is y-30.

So,

$x-100=1.25y$
$x-300=y-30$

Solving that system, $x=950$[/hide]


The number of real solutions to the equation $|x-2|+|x-3|=1$ is

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ \text{more than 3}$

[hide]
$(C)$ ?
I got 2 and 3 as the only working ones.[/hide]

[hide="Neither is correct"]The answer is $E$, but not because there are four solutions, but because there are infinitely many solutions. Namely, any $x$ such that $2\leqslant x\leqslant 3$ satisfies the equation. Try to solve it for a change, and leave the guesswork aside :P[/hide]

[quote="4everwise"][hide]The solutions to the equation $|x|+|y|=c$ can be found by creating 4 equations:

$x+y=c$, $x-y=c$, $-x+y=c$, and $-x-y=c$. Similar situation here, so there should be 4 solutions. (Unless two of them yield the same solution. :P)[/hide][/quote]

[hide="The second case..."]The second case $x - y = c$ yields $1 = 1$, which has infinite solutions.[/hide]

[hide="I do not like to solve equations"]If $x<2$, -2x+5=1, x=-1 which is less than or equal to -2.

$2\lex\le3$, 1=1, which is true. Thus any x with a value between 2 and three (including 2 and 3) works, and there are infinite solutions. [/hide]


calculus
integration
trigonometry
function
algebra
rational function
real analysis

Can someone show me how to finish this by hand?

\[\int^{2\pi}_{0}\int^{s}_{0}\sqrt{s^2-r^2sin^2\theta}\textsf{ }r\textsf{ }dr\textsf{ }d\theta\]
\[u = s^2-r^2sin^2\theta\]
\[du = -2r\ast sin^2\theta\textsf{ }dr\]
\[=\int^{2\pi}_{0}\int^{s^2-s^2sin^2\theta}_{s2}\frac{\sqrt{u}}{-2sin^2\theta}du\textsf{ }d\theta\]
\[=\int^{2\pi}_{0}\frac{u^{\frac{3}{2}}}{-3sin^2\theta}|^{s^2-s^2sin^2\theta}_{s2}\textsf{ }d\theta\]
\[=\int^{2\pi}_{0}\frac{cos^3\theta \ast s^3 + s^3 }{-3sin^2\theta } \textsf{ }
d\theta\]

The sign in your last integral is wrong.
$=\int^{2\pi}_{0}\frac{s^3 -s^3cos^3\theta}{3sin^2\theta }d\theta=\frac{s^3}{3}\cdot\frac{2-\cos^2t-\cos t}{\sin t}\Big|_0^{2\pi}$

AHH!  You're right!  Thank you, but can you show me what steps you took to get to the right hand side of your equation there?

These are standart methods:
\[\int\frac{\cos^3t}{\sin^2 t}dt=\int\frac{1-\sin^2t\,d\sin(t)}{\sin^2 t}\]
(+integral of rational function)

\[\int\frac{dt}{\sin^2t}=-\cot(t)\]
(well known fact)


A circle of radius 1 has area π (pi), which clearly exceeds the area of two squares each with side length 1. What is the fewest number of pieces that the two squares need to be cut into so that the pieces will all fit, none overlapping, into the circle?

Construct a square of area $ 2$ and sizes $ \sqrt{2}$ by dividing one square in $ 4$ equal triangles.

Then it has to be fewer or equal to $ 5$


abc  atriangle  if $3A+2B=180$deg
prove that
$a^{2}+c b=c^{2}$

[hide="Hint"]
OK, let's write angle at A $2\alpha$. Using the condition and the fact that A+B+C=180, angles B and C are equals to $90-3\alpha$ and $90+\alpha$.
Now use Sine law to get that $b=\frac{\cos{3\alpha}}{\sin{2\alpha}}$ and $c=\frac{-\cos{\alpha}}{\sin{2\alpha}}$. Now the equality $a^{2}+bc=c^{2}$ turns into (after dividing both sides by $a^{2}$ and multiplying by $\sin^{2}{2\alpha}$)
$\sin^{2}{2\alpha}-\cos{\alpha}\cos{3\alpha}=\cos^{2}{\alpha}$. The rest is easy
[/hide]

I was looking old questions see that
ı did a synthetic solution
[url=http://imageshack.us][img]http://img441.imageshack.us/img441/4229/bisectorbt4.gif[/img][/url]
Shot at 2007-08-09


trigonometry
logarithms
email
integration
real analysis
real analysis unsolved

say, what would the [b]complete [/b]solution to the following differential equation
look like (i.e. $ \text{C. F.\plus{}P. I.}$) ? :maybe:

$ \textbf f\;''(\theta)\plus{}\textbf f\;(\theta)\;\equal{}\;A\plus{}B\cdot\theta\plus{}C\sin\theta\plus{}D\cos\theta\plus{}E\cdot\theta\cdot\cos\theta\plus{}F\cdot\sin\theta\cdot\ln\sin\left(\frac{\theta}{2}\right)$

( where $ A,B,C,\ldots,F$ are [i]arbitrary [/i]constants )

hello, the solution of your problem looks terrible, if you are interested to see it, send me please a mail with your email adress and you will get the solution.
Sonnhard.

[quote="misan"]say, what would the [b]complete [/b]solution to the following differential equation
look like (i.e. $ \text{C. F. \plus{} P. I.}$) ? :maybe:

$ \textbf f\;''(\theta) \plus{} \textbf f\;(\theta)\; \equal{} \;A \plus{} B\cdot\theta \plus{} C\sin\theta \plus{} D\cos\theta \plus{} E\cdot\theta\cdot\cos\theta \plus{} F\cdot\sin\theta\cdot\ln\sin\left(\frac {\theta}{2}\right)$

( where $ A,B,C,\ldots,F$ are [i]arbitrary [/i]constants )[/quote]

$ f(\theta)\equal{}C_1\cos\theta\plus{}C_2\sin\theta\plus{}\int\limits_0^\theta \sin(\theta\minus{}s)\left[A \plus{} B\cdot s \plus{} C\sin s \plus{} D\cos s \plus{} E\cdot s\cdot\cos s\plus{} F\cdot\sin s\cdot\ln\sin\left(\frac {s}{2}\right)\right]\,ds$


function integration inequalities algebra polynomial calculus derivative

The following is known about the function $ f$

($ i$)  $ f$ is analytic in $ \mathbb{C}\setminus \{0\}$, f has a pole of order 2 with residue 1 at $ z=0$;
($ ii$)  There exists constants $ A>0$ and $ R>0$, s.t. $ |f(z)|\leq A|z|$ if $ |z|\geq R$;
($ iii$)  $ \int_\gamma zf(z)dz=4\pi i$, ($ \gamma$ is the positively oriented unit circle)
($ iv$)  $ f(1)=-2$ and $ f(-1)=2$

Determine $ f$ !

Let $ g(z)=z^{2}f(z)$.  Then $ g(z)$ is entire by (i).  Now multiplying the inequality in condition (ii) on both sides by $ |z|^{2}$ gives $ |g(z)| \leq A |z|^{3}$ for $ |z| \geq R$.  Since $ g$ is polynomially bounded, it is, here, a polynomial of degree $ 3$ which can be proven separately.

So write $ g(z)=c_{0}+c_{1}z+c_{2}z^{2}+c_{3}z^{3}$.  This implies, $ f(z)=\frac{c_{0}}{z^{2}}+\frac{c_{1}}{z}+c_{2}+c_{3}z$.

Condition (i) says $ c_{1}=1$.  Condition (iii) says $ c_{0}=2$.  And condition (iv) yields a $ 2\times2$ system of equations which yields:  $ c_{2}=-2$ and $ c_{3}=-3$.

So barring any mistakes, I have:  $ f(z)=\frac{2}{z^{2}}+\frac{1}{z}-2-3z$.

As for that provable fact regarding the growth estimate:

Cauchy's Integral Formula for derivatives gives:

$ |g^{ (k) }(0)| \leq \frac{k!}{2 \pi}\int_{|z|=R}\left|\frac{g(z)}{z^{k+1}}\right| dz \leq \frac{k!}{2 \pi R^{k+1}}\int_{|z|=R}A|z|^{3}dz = \frac{k!}{R^{k-3}}\rightarrow 0$ as $ R\rightarrow \infty$ for $ k\geq 4$


Pythagorean Theorem
geometry
calculus
calculus computations

1. The problem statement, all variables and given/known data

Two cars start moving from teh same point. One travels south at 60 mi/h and the other travels west at 25 mi/h. At what rate is the distance between the cars increasing two hours later?"

2. Relevant equations

So far I have the equation..
(z^2) = (x^2) + (y^2)

I know dy/dt = 60 mi/h and dx/dt = 25 mi/h.

3. The attempt at a solution

I implicity differentiated the first equation and ended up with:
dz/dt = (50x +120y)*(1/2z)

Since I do not know what x or y is I am unable to find the rate the distance between the two cars is increasing.

Thank you so much for your help!

Well, you know that the cars were traveling at given constant speeds and that they were traveling for 2 hours... so find $ x$ and $ y$ by multiplying the corresponding speeds by the 2 hours that passes... $ z$ can then be found from the Pythagorean theorem (which you have stated).


The [i]subset-sum[/i] of a set of positive integers is the sum of all sums of all non-empty subsets of the set. Let $ N$ be the [i]subset-sum[/i] of the set consisting of the first $ 7$ powers of $ 10$:
\[ \left\{10^0,\,10^1,\,10^2,\,10^3,\,10^4,\,10^5,\,10^6\right\}\]What is the remainder when $ N$ is divided by $ 10$?

$ \mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ \text{None of these}$

[hide="solution"]There is a 1-1 correspondence between the subsets with and without $ 10^k$, namely removing/adding this value, so each $ 10^k$ is included in half the subsets.  (We will keep the empty set in, because it has a sum of 0 and makes the number of subsets nice.)  There are 7 independent choices that define each subset, which are whether each of the $ 10^k$ are included or not, so there are $ 2^7 \equal{} 128$ different subsets.  Then each $ 10^k$ is counted half of this, or 64 times.  We have the subset-sum is 64(1111111) which, when taken mod 10, gives 4(1)=4 mod 10.  Thus the answer is $ \boxed{C}$[/hide]

[hide]
The only units digit that a sum of the elements in a subset of $ N$ can be is 1. That is when $ 10^0$ is included in the subset.
There are
$ \binom{6}{0} \plus{} \binom{6}{1} \plus{} \binom{6}{2} \plus{} \binom{6}{3} \plus{} \binom{6}{4} \plus{} \binom{6}{5} \plus{} \binom{6}{6}$
or $ 64$
subsets that include $ 10^0$.
These subsets all have a sum ending in 1.
Therefore, the last two digits of the subset-sum of $ N$ is 64.
$ 64\equiv 4 \bmod{10}, \boxed{(C)}$
[/hide]

[hide="Solution"]Take the set $ \mod 10$ to get
\[ \{1,0,0,0,0,0,0\},\]
so only sets containing $ 10^0$ will have a sum of $ 1$; the rest sum to $ 0$.

If we fix the first element as $ 1$, we could have a 1 element set, 2 element set, etc; so the number of ways to do this will be
\[ \sum_{k\equal{}0}^{6}\binom{6}{k}\equal{}2^6\equiv \boxed{\textbf{(C)}\ 4}\pmod{10}.\][/hide]

I failed the sprint so badly...  :(


Let $a,b \in (1,\infty)$, $a$ not equal to $b$.
Solve $2^x \cdot (a^x + b^x) + x \cdot (a^x + b^x) = (a+b)^x + x \cdot (a+b)$

$a^x+b^x \geq 2(\frac{a+b}{2})^x \Longleftrightarrow (\frac{2a}{a+b})^x+(\frac{2b}{a+b})^x \geq 2 \Longleftrightarrow y^x+(2-y)^x \geq 2$ which is true because $y^x$ is convex. so we have that :
$2^x2(\frac{a+b}{2})^x+x(a^x+b^x) \leq (a+b)^x+x(a+b)$ or
$(a+b)^x+x(a^x+b^x) \leq x(a+b)$.
Now let $x \geq 0$ we put $a=1+t$ so
$(a+b)^x+x(a^x+b^x) = (1+t+b)^x+x((1+t)^x+b^x) \geq 1+x(t+b)+x > x(1+t+b)=x(a+b)$ which is tru only for $x=0$ which is the solution to the equation.
Now suppose that $x<0$ and let $x=-z$ we will get that
$\frac{1}{(a+b)^z} -z(\frac{1}{a^z}+\frac{1}{b^z}) \leq -z(a+b) \Longleftrightarrow \frac{1}{(a+b)^z} \leq z(\frac{1}{a^z}-a+\frac{1}{b^z}-b) < 0$ and which is not possible. So the only solution is $x=0$

[quote="RaduSorici"]So the only solution is $x=0$[/quote]

2=1 ????  ;)

I really don't get your solution.

RaduSorici, could you please explain what you did here?


let $ x,y,z$ nonnegative numbers . and $ xyz \equal{} 1$
prove that :

$ \frac {1}{\sqrt{1 \plus{} 3x^3}} \plus{} \frac {1}{\sqrt{1 \plus{} 3y^3}} \plus{} \frac {1}{\sqrt{1 \plus{} 3z^3}}\,\geq\,\frac {3}{2}$

[quote="mr.inequalities"]let $ x,y,z$ nonnegative numbers . and $ xyz \equal{} 1$
prove that :

$ \frac {1}{\sqrt {1 \plus{} 3x^3}} \plus{} \frac {1}{\sqrt {1 \plus{} 3y^3}} \plus{} \frac {1}{\sqrt {1 \plus{} 3z^3}}\,\geq\,\frac {3}{2}$[/quote]
Try $ x\equal{}y\equal{}2$ and $ z\equal{}\frac{1}{4}.$

The inequality of that form:
$ k\in Z \ \prod_{i \equal{} 1}^{n}a_{i} \equal{} 1$
$ \sum_{cyc}\frac {1}{\sqrt {1 \plus{} 3a_{1}^{k}}}\geq \frac {n}{2}$
holds only for 2 variables :wink:

Well, we can always take $ x_{i} \equal{} a_{i}^{k}$ and we will have to prove
$ \sum\frac {1}{\sqrt {1\plus{}3x_{i}}}\geq \frac {n}{2}$ (we still have that the product of new variables is 1)
For $ n\geq 3$ we can build a counterexample (in a similar way argady did).
For n=2 look for easy proof here:
http://www.mathlinks.ro/viewtopic.php?t=185707 :wink:


number theory
relatively prime
number theory unsolved

Let $m>1$ an integer. We denote $PF(m)$ the set of prime factors of $m$.
Let $m,p$ rispectively an integer $>1$ and a prime such that $PF(m)\subseteq PF(p-1)$. To characterize primes q, relatively prime with m, such that

$p^m\equiv1(q)$

solutions are found for
$m=\varphi(q)$
with $(\varphi(q),q)=1$ but this is true if primes appear in m only at power 1.
so to charactirize q just decompose p-1 in prime factors,construct PF(p-1).
now take all subsets of PF(p-1) multiply all primes in any different subset
previously constructed and you get a q for each such subset.
i guess this is good


Put 2007 balls into 200 distinct boxes. There are always $ n$ boxes that have the the same number of balls inside. Find the maximum of $ n$.

Consider the following arrangement:

$ 11$ boxes have $ 1$ ball each,
$ 11$ boxes have $ 2$ balls each,
and so on, upto:
$ 11$ boxes have $ 18$ balls each,
and $ 2$ boxes have $ 63$ balls each.

From this arrangement it is clear that $ n_{max}\leq 11$

Now suppose $ n_{max}$ is $ 10$ (or less)
Then minimum total number of balls is for the following arrangement:

$ 10$ boxes have $ 1$ ball each,
$ 10$ boxes have $ 2$ balls each,
and so on, upto:
$ 10$ boxes have $ 20$ balls each.

Therefore total number of balls is at least $ 2100$ which is a contradiction.
Therefore, $ n_{max}>10$

The two conditions together imply that $ n_{max}\equal{}11$

[quote="patham"]Consider the following arrangement:
From this arrangement it is clear that $ n_{max}\leq 11$
[/quote]
I wonder how do you get it?

Suppose the answer is $ 12$
[quote]There are always n boxes that have the the same number of balls inside[/quote]
This implies that "There are always 12 boxes that have the the same number of balls inside."
which is clearly false, as can be seen from the given arrangement.


real analysis
advanced fields
advanced fields unsolved

Prove that $ \ds \bigcap_{m = 1}^{\infty} \bigcup_{n = 1}^{\infty} \left ( r_n - \frac {1}{2^{n + m}}, r_n + \frac {1}{2^{n + m}} \right )$ is a second Baire category set in $ \mathbb{R}$, with Lebesgue measure 0.

If $ \{r_n|n\in\mathbb{N}\} \subset \mathbb{R}$ is dense in $ \mathbb{R}$ (and that's obviously what you meant to write) the statement about category is a direct consequence of Baire's theorem since $ \bigcup^\infty_{n \equal{} 1} \left( r_n \minus{} \frac {1}{2^{n \plus{} m}},\,r_n \plus{} \frac {1}{2^{n \plus{} m}}\right)$ is open and dense in $ \mathbb{R}$ for all $ m$.

Furthermore $ \lambda\left(\bigcup^\infty_{n \equal{} 1} \left( r_n \minus{} \frac {1}{2^{n \plus{} m}},\,r_n \plus{} \frac {1}{2^{n \plus{} m}}\right)\right) \leq \sum^{\infty}_{n \equal{} 1} 2^{ \minus{} (n \plus{} m \minus{} 1)} \equal{} 2^{ \minus{} m \plus{} 1} \to 0,\,m\to\infty$, thus $ \lambda\left(\bigcap^{\infty}_{m \equal{} 1} \bigcup^\infty_{n \equal{} 1} \left( r_n \minus{} \frac {1}{2^{n \plus{} m}},\,r_n \plus{} \frac {1}{2^{n \plus{} m}}\right)\right) \equal{} 0$ by monotonicity.


So I was the dude who got extremely lucky and managed to go home with \$120 solely because I sat in the front row for the team round (luckily, I am just taking calculus this year). This enabled me to place 2nd overall, which is cool, but rather useless.

Therefore, I will post it here that I acknowledge the fact of the obvious flawed method of grading. Hopefully that will anyone feel better that they didn't win any cash. Personally, I think of it like winning a raffle (which never happens to me, so yay).

Yes, I thought that part of the competition was very strange too. I sat in the second row, and missed only question as a result. That was pretty randomly unfair and they should probably go back to the other team round they used to have.


[quote="Quevvy"][quote="ckck"]How many three-digit numbers between 100 and 30 are there for which one of the digits equals the sum of the other two?[/quote]
Um none? do you mean 300?[/quote]
Oops. Sorry. Yeah thats what I meant. Edited. :blush:

[quote="ckck"]How many three-digit numbers between 100 and 300 are there for which one of the digits equals the sum of the other two?[/quote]
[hide]for the 100s:
the other 2 digits must differ by 1 so then there are 18 of those.

for the 200s:
the other digits must differ by 2 or be 1 1 so then you have 17 of those

so then there are 35 numbers[/hide]

Yes, I think you are correct, thers 17 in the 200s, counted 19 but there are only 8, not 9 combinations that differ by 2


absolute value
linear algebra
linear algebra unsolved

Let $A\in\mathcal M_n(R)$ with the property that at least $n^2-n+1$ of it`s elements have their absolute values smaller than $\frac{1}{n}$.
 Prove that $A$ has a minor of order $n-1$ such that it`s absolute value is greater than $|\det A|$.

you just have to find a column with all its elements that are smaller than $\frac{1}{n}$ wich is true because otherwise you'll have $n$ elements greater than $\frac{1}{n}$  thus at most $n^2-n$ elements smaller than $\frac{1}{n}$ :).

Yes the problem is solved once you have noticed that!


MATHCOUNTS FTW LaTeX geometry perimeter Support

Please no discussing on this thread. PLEASE read the rules:
[hide="Rules"]
[list][*] The competition will consist of 4 levels: School, Chapter, State, and Nationals. Each level will have two rounds: Sprint and Target.

[*] Your score for each level will be calculated by adding the number of questions correct on the Sprint Round to twice the number of questions correct on the Target Round. There is no penalty for incorrect answers or guessing.

[*] You must qualify to get to each level. All participants who sign up qualify for the School level. The top 20 scores from the School level will qualify for the Chapter level. The top 10 scores from the Chapter level will qualify for the State level. The top 5 scores from the State level will qualify for the National level. Based on National level scores, first, second, and third places will be awarded to the top three scorers. Please note that if you do not qualify, you may still take the test, and though it won't be graded by the coordinators, you may check your answers with the answers posted after the test.

[*] If there are any ties in the School, Chapter, or State levels, everyone who receives a qualifying score will qualify. Any ties for the top 3 positions after the National level will be broken through FTW tournaments.

[*] You will have 40 minutes for the Sprint test and 6 minutes for each pair of Target questions. This will be based on the honor system. Please, no cheating.

[*] The Target and Sprint Rounds will be posted on a Thursday. You will be given until Monday to finish the test. Anything submitted after 11:59:59 PM EST on Monday will be considered late. You will lose 2 points per day late after Monday. Anything submitted 3 or more days late will not be accepted and the participant is disqualified.

[*] Submit your answers by PMing any of the contest coordinators: ernie, PowerOfPi, pytheagle, or vallon22. It does not matter if they are in $ \text{\LaTeX}$ or not, as long as they are legible. Please clearly specify which answer belongs to which question number and round.

[*] You are NOT allowed to use calculators of any type on the Sprint Round. You may use a calculator without a QWERTY keyboard on the Target Round. Calculations may also be done on scratch paper. Communication of any type, including internet, are not allowed. You may not talk with any other person during the test.

[*] Answers must be in the units specified in the problem. Including the unit in your answer will not affect the correctness of your answer. However, if the incorrect unit is included in the answer, the answer will be considered incorrect.

[*] All fractions must be simplified to lowest terms. Fractions not in this form will be considered incorrect.

[*] If you have any questions or concerns, please PM one of the coordinators: ernie, PowerOfPi, pytheagle, or vallon22.[/list]
[/hide]

The tests are attached.

Remember, you can submit answers to ernie, PowerOfPi, pytheagle, or vallon22 on or before Monday. Check the rules for more detail.

Ugh.  I got 38.   Judging from AIME15's mock MC where the cutoff was 41/46(and that was chapter, not school), I don't think I have much of a chance unless a bunch of ppl forget to turn in their test.  :(

[b]SORRY. THERE WAS A SLIGHT BUT IMPORTANT TYPO IN THE SPRINT ROUND! PLEASE RESUBMIT YOUR ANSWER TO NUMBER 28 IF YOU HAVE SUBMITTED YOUR ANSWERS. THANK YOU.[/b]

The typo is that the point should be (-2, 3) on #28, not (-1,3) as the problem originally said. An updated version of the Sprint Round is attached.

All scores sent out are subject to change by 1 point.

Could we get a table to see our scores and rankings at this moment in comparison to other people?

A reminder: We want to keep this thread for announcements only. Thanks.

I am compiling a list of every person's (who submitted answers) Sprint, Target, and total score, which [i]might[/i] be released. Also, I have had only 3 submissions (I don't know about the other graders).

There appears to have been an error in the answer key.

[b]The score you get is tentative. It will either go up two, go down two, or stay the same, but most likely up two.[/b]

You're final scores will all be posted when the round is over, as well as the answers. Thank you.

[size=134][b]On #6 in the target, express your answer as a decimal to the nearest tenth.[/b][/size]

Ladies and gentlemen, we have our first perfect, jxl28.

darn, now the message is to small and I have to type random words to fill it in... \

Hoogly boogly rabbits are us

On the bottom right of your post it will say "edit". Edit your post to TAKE OUT THE ANSWERS! Private Message Powerofpi, dont post your answers on the forum WHERE EVERYONE CAN SEE THEM.

I have PMed rrusczyk and The Hobbit about it.

EDIT: @The Hobbit: I sent you another PM telling you exactly what to do.

[quote="PowerOfPi"][b]Anything submitted now on is considered late. You have until or on Thurday to submit your answers. You will lose 2 points per day late.[/b][/quote]

I just saw this. If we submitted but resubmitted #6 target and #28 sprint, will we still lose points?

No. You're good, as I also said in a PM.

During the weekend, when all submissions should be sent in by.

I will probably do the Master's Round, [b]so anyone who is interested please PM me[/b].

There was a mistake in the answers for #5 on the target. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=320914]here[/url]. Updates for the scores will be posted soon.

EDIT: Here are the fixed results:

[hide="Results"]
[b]15!!!	-	41 [color=green](+2)[/color]
2lnx	-	41 [color=]green(+2)[/color]
bbgun34	-	36
TheMan1998	-	36 [color=green](+2)[/color]
pi31415926	-	35
jxl28	-	34 [color=green](+2)[/color][/b]
The Hobbit	-	30
also7	-	27

In bold are those who qualified for State through Chapter and School. Changes in scores are in color. I am not sure about 2lnx, because (s)he submitted answers to vallon22. If vallon22 or 2lnx still have the answers, PM me or post if the score should change.
[/hide]
[hide="Answers"]
Sprint Round
1. 45/512
2. 100
3. 2^12*3^4*5^4
4. 50
5. 100
6. 1/120
7. 30
8. 16/45
9. 10
10. 9
11. 81
12. 2sqrt2
13. 1,111,110
14. 50:3
15. 20*26^6
16. 48
17. 70
18. 120
19. 3/5
20. 18
21. 125
22. 18
23. 503/60
24. 7:38
25. 1
26. 64pi
27. (75sqrt3)/2
28. 3sqrt10
29. 57
30. 1875pi
Target Round
1. 3600-900pi
2. Sunday
3. 90
4. ---typo in problem---
5. 23/72
6. 165
7. 24
8. 1/(2pi)[/hide]

The qualifiers for Nationals don't change.

2nd EDIT: 2lnx's score fixed.

wow. I FAIL.
[color=red][size=25]But I still qualified! YESSSSS![/size][/color]

In addition to the two supporters of the CD round from this thread, I received two pms in support of a CD round.
Unfortunately, 4 people out of around 50 who signed up for the MMC wanting a CD round does not motivate me enough to write the problems.
There will not be a CD round.

Only two requests for Master's round. I don't think there will be one of these either.

Powerofpi hasn't been active on this forum.

I am not sure if the national round will be posted or not. I will check with pytheagle.

National round will probably be killer!  :what?:

[quote=intelligence_20]\signup$        $[/quote]

What's the need for this bump

oh wait its the wrong mathcounts test.... idk why this came up when i searched mock Mathcounts

1 wrong in the first 15   :wallbash:  :wallbash:   :wallbash:    :wallbash_red:

Wait i just realized that this is from 2009 lol


What's the best looking way to write delta y over delta x?
Using \delta ($\delta$) gives the wrong letter (looking for the small upright triangle) but  $\frac{\triangle{y}}{\triangle{x}}$ is awkward too.  Any suggestions?

try the following:

m = \tfrac{\Delta y}{\Delta x}


Kirez randomly selects a positive divisor of 120.

Lisa randomly selects a positive divisor of 84.

(a) What is the probability that Kirez and Lisa select the same number?

(Much Harder) (b) What is the probability that Lisa selects a number that is a divisor of the number Kirez selects?

Do these divisors include 1 and the number itself?

I think it should... but I'm not 100% sure.

err, i got 53/192 fer #2, not 54/192 as Chinaboy got...

I just took cases:

the only numbers that Lisa can choose that are factors of what the first guy chose are 12,6,4,3,2,1

Now, if Lisa chose 12, then Kirez must choose 120,60,24,12

So tha'ts 1/12*4/16.

If lisa chose 6, then kirez must choose 120,60,30,24,12,6
so thats 1/12*6/16.

And keep going

SO the answer is....o shoot. Chinaboy's rite.

My bad.

it is 54/192

Just a small word: I personally considered it easier to take Kirez's numbers in groups and count the number of possible divisors that Lisa could have chosen.  For instance, If Kirez picked a 12, 24, 60, or 120, then Lisa could pick any divisor of 12.  If Kirez picked a 6 or 30 then Lisa could pick any divisor of 6.  So on.  But it's purely a matter of what is easiest for you.


FIRST DAY - APRIL 12, 1997
TIME ALLOWED : 4 1/2 HOURS.

Problem 1.
In a triangle ABC which has a right angle at A, let H denote the foot of the altitude belonging to the hypotenuse. Show that the sum of the radii of the incircles of the triangles ABC, ABH, and AHC is equal to AH.

Problem 2.
The sequence $\{a_n\}, \{b_n\}$ are defined by $a_1 = \alpha, b_1= \beta$, and $a_{n+1} = \alpha a_n - \beta b_n, b_{n+1} = \beta a_n + \alpha b_n$ for all $n > 0$. How many pairs $(\alpha,\beta)$ of real numbers are there such that
$a_{1997} = b_1$ and $b_{1997} = a_1$ ?

Problem 3.
In a soccer league, when a player is transferred from a team X with x players to a team Y with y players, the federation is paid y-x billion liras by Y if y$\geq $x, while the federation pays x-y billion liras to X if x>y . A player is allowed to change as many team as he wishes during a season. In a league consisting of 18 leagues, each team starts the season with 20 players. At the end of the season, 12 of these turn out to again have 20 players, while the remaining 6 teams end up to have 16,16,21,22,22 and 23 players, respectively.
What is the maximal amount the federation may have won during this season ?

SECOND DAY - APRIL 13, 1997
TIME ALLOWED : 4 1/2 HOURS.

Problem 4.
The edge AE of a convex pentagon ABCDE whose vertices lie on the unit circle passes through the center of the circle. If AB = a, BC = b, CD = c, DE = d and ab = cd = 1/4, compute AC + CE in terms of a,b,c,d.

Problem 5.
Prove that for each prime p>=7, there exist a positive integer n and integers $x_1, x_2.., x_n, y_1, y_2.., y_n$ not divisible by p such that
$x_i^2 + y_i^2 = x_{i+1}^2 (mod p)$ for $i =1,2,..n$ (it is understood that $x_{n+1} = x_1$)

Problem 6.
Given an integer $n \geq 2$, find the minimal value of
$\sum_{1 \leq i \leq n} \frac {x_i^5}{x_1 + x_2 +..+ x_n - x_i}$
where $x_1,x_2..,x_n$ are positive real numbers subject to the condition $x_1^2 + x_2^2 +..+ x_n^2 = 1$.


Let $T$ be a real number satisfying the property : For any non-negative real numbers $a,b,c,d,e$ with their sum equal to $1$, it is possible to arrange them around a circle such that the products of any two neighboring numbers are no greater than $T$. Determine the minimum value of $T$.

Well, I've posted this a long time ago.. Anybody has a solution ?  :maybe:


conics
parabola
linear algebra
matrix
calculus
geometry
email

Wow that was a tough test because the time limit was killer

i made it through 63 problems.

when will we be able to see a list of the winners?

Well anyway I answered like 46 problems... hopefully i got them right =P   I am an 8th grader.... and I answered 1 for like 54 problems... maybe just by any chance i got one of those right? i guessed 1 on all of em when there were like 30 seconds left and I had done all i could do. Well kudos...... and remember let light and life shine through!

[quote="eryaman"]I think I answered 75 questions...not sure how many I got right though.[/quote]

that's impressive. how did you get through so fast? lots of tricks to simplify problems or what?

I don't think any of hte problems answers were 1

they probably figured people would just do the line through the column at the last second

Well... I am only in geometry... so I could only do like 46 of them after memorizing the sequence formulas and stuff.. Yeah... well i just felt like answering 1 anyway...

So how do you do all those vertex parabola focus stuff? anyone know?

WEll... I haven't learned them or heard of that matrix stuff well I saw my sister doing it in linear algebra a few years back but back then I was just a 3rd grader

OH well, i gotta prepare for vestavia hills tournament grrr i heard the tournament is evil.. it will be my 3rd math tournament i have ever been in

Just google for some basic pre calculus conics summary. It's a whole list of some pretty simple formulas you need to know to quickly do those problems

What division are you gonna compete in? I'm going to the Vestavia tournament too

I am most likely gonna enter as Geometry since im in 8th grade.

I think math is really fun and I've memorized lots of junk lately since I haven't really been on a "real" math team before now.

We have ASMA coming up whatever that is. WE had some practices called ASMA it was like just a few problems tho

So When do we know how we did on fall startup compared to other people?

Any tips for doing well on Vestavia?

ASMA?

oh yea
the math league

ugh
I hate math leagues

anyways, doing well at vestavia? Know like, every formula related to geometry, or be able to figure stuff out really quickly with all the basic formulas

i'm going to try to get my school to sign up for vestavia's tournament. should be interesting

[quote="now a ranger"]YAY! two months to prepare and memorize formulas for Vestavia lol

Yeah im gonna try to memorize not mesmerize all the formulas that i find because people at our school say they write their problems from going through formula sheets and drawing out.

yeah math leagues sound kind of boring because they are every month and don't end quickly enough =P[/quote]
MEMORIZING FORMULAS IS BAD.  Learn their proofs instead; derive them.  Try actually to [i]understand[/i] them.  Fomula sheets are [b][i][u]very very bad[/u][/i][/b].  Read [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]this[/url] AoPS article.  I know of what I speak.

If you don't have the prerequisite knowledge for something, simply compete in a lower division if you can—that's why lower divisions are there in the first place.

By the way, I think I'll be coming to Vestavia as well. :)   I'm looking forward to it.

[quote="Boy Soprano II"]
MEMORIZING FORMULAS IS BAD.  Learn their proofs instead; derive them.  Try actually to [i]understand[/i] them.  Fomula sheets are [b][i][u]very very bad[/u][/i][/b].  Read [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Prob_Solv.php]this[/url] AoPS article.  I know of what I speak.[/quote]

uh, no. Memorizing formulas without knowing their origin or how they are derived is bad. I agree, you should do what you can to understand where everything comes from, and it helps a ton in terms of trying to remember waht the formula actually is (cause you can quickly see how it comes about and everything), but if you don't memorize any formulas you're just putting yourself at a disadvantage. I wouldn't be anywhere near as fast as I am on math tests if I wasn't able to recall everything I can right now and I had to derive everything instead.

[quote]By the way, I think I'll be coming to Vestavia as well. :)   I'm looking forward to it.[/quote]

Comprehensive, I'm guessing? I'm going to Vestavia as well.

[quote="Elemennop"]Memorizing formulas without knowing their origin or how they are derived is bad.[/quote]
Hmm, this was what I meant to say.  I guess I was having a kind of allergic reaction to the term "formula sheet" or something. :P   Of course if you apply theorems enough, you will become familiar enough with them that you won't have to rederive them every time.  But nonetheless, it is good to be able to prove everything you use (up to a point—being able to derive geometry from Hilbert's axioms is going a bit far).

Yeah, I'll be doing comprehensive.

dude I don't even know what hilbert's axioms are wtf

what is vestavia I interest myself

yeah boysoprano what are hilbert's axioms? the furthest i've gone into higher math is group theory

WEll... I was sick from school today so I never found out what I got. I think i will go online and learn orgiins of stuff..

I will be entering GEometry section of Vestavia as well.. First Vestavia tournamet i have been to

[quote="MysticTerminator"]
what is vestavia I interest myself[/quote]

Vestavia high school

like google it to find out what it is (woah)

I answered 79 and got 69 right. Lost by 1 point to the highest scorer in my school  :(

Okay, so it was known for some time that Euclid's axiomatic system of geometry was riddled with holes for some time, so in the early twentieth century, David Hilbert set out to axiomatize geometry rigorously, from about twenty very basic axioms.  You can see the result in Hilbert's book [i]Foundations of Geometry[/i], which is fairly inexpensive.  It is rather amazing.  I don't really think it counts as "higher math", apart from the fact that he proves that the geometric axioms are consistant as long as the real axioms are consistant (by setting up analytic geometry, I think).

But anyhow, the point is that there are over twenty axioms, and it was two thousand years or so before anybody was really able to fix Euclid's unsound work.

(Oh yeah, here is the Vestavia tournament website :  [url]http://math.gosuapm.com/tourn[/url].  And this year, unless I'm mistaken, it's on the ninth, not the tenth, which is a Sunday.)

Yea, that site's a year old. You should probably email the address on that site (kay tipton) and ask about registering for the tournament if you want to attend

[quote="Phelpedo"]What's wrong with Euclid?

And why is the memorizer a "he" but the problem solver a "she" in that article?  :wink:[/quote]
Lots of things are wrong with Euclid.  In the first place, in his many definitions, he makes several bizarre assumptions, which, by modern standards, is just sleazy (for example, he says that a circle is symmetric about its diameter in its definition).  Second, many of his proofs are not logical.  For example, in merely beginning of the first book, we can find fault with propositions 1 (how do we know that the circles intersect at all?), 4 (how is the principle of superposition implied by the axioms and definitions Euclid listed anywhere?), and 5 (where does Euclid say that we can select an arbitrary point on a line?).  As Hilbert said, you should be able to substitute "tables, chairs, and coffee mugs" for "points, lines, and planes", and your theorems should still work if you assume the axioms you listed.

Furthermore, Euclid lived before people realized that it was necessary to have undefined terms to avoid circularity, and his definitions are accordingly convoluted.  But in his defense, it was two thousand years before anybody could improve on [i]The Elements[/i].

As for he vs. she, I think that's just Mr. Rusczyk trying to maintain a gender balance by switching the pronoun every now and then.

grrr... I am in 8th grade and I got a 31. Poo....

Oh well, speed isn't really my thing. Next up Vestavia.

[quote][quote]Where can I find scores of Fall STartup that happened last thursday?[/quote]

They probably won't be up for a while; it took em almost a month last year to get it out[/quote]

     Official scores will be announced by Monday, October 16th.  Yes, we realize that waiting is annoying, but we want to be sure we get it right.  Because we dislike multiple choice tests, grading is somewhat tedious.  Your coaches do the first grading, but each of them grade slightly differently or make a few mistakes, so we double-score all of the answer sheets that are in contention for awards to catch any errors and make sure that grading was consistent across the board.  However,to double-score the answer sheets we have to wait for them to be mailed back to us, which also takes time.

Thank you,

Well. I wanted to refresh this thread, and also say.

CAn anyone help me with geometry problems? like on geocities.com/grissom_math

like the 2nd half of the geometry test of 2003

maybe?

AND cant wait till scores come up.. not really

Soooo....
Can someone send me a link for this competition  :D

[url]http://www.natassessment.com/[/url]

sadly my school isn't taking this stuff this year... i wanted more plaques :P


geometry
incenter
parallelogram
circumcircle
geometric transformation
homothety
ratio

Given a triangle $ ABC$ with $ \angle A = 60$ degrees,  prove that $ AP= AT$ where $ P = ID %Error. "meet" is a bad command.
AA'$ where $ I$ is the incenter, $ D$ midpoint of $ BC$ and $ AA'$ is the altitude from $ A$ to $ BC$, $ A' \in BC$ and $ T$ is the point of contact of the Nine point circle and incircle.

Let $ X,Y$ be the tangency points of the incircle $ (I,r)$ and the A-excircle $ (I_a)$ with $ BC.$ Let $ X'$ denote the antipode of $ X$ WRT $ (I).$ $ A,Y,X'$ are clearly collinear $\Longrightarrow$ $ DI$ is the X-midline of $ \triangle AXY$ $ \Longrightarrow$ $ APIX'$ is a parallelogram, hence $ AP \equal{} IX' \equal{} r.$ On the other hand, let $ N$ be the 9-point center of $ \triangle ABC$ and $ M,L$ the midpoints of $ AC,AB.$ Simple angle chase gives

$\angle NLB \equal{} 180^{\circ} \minus{} \angle B \minus{} 30^{\circ} \equal{} 60^{\circ} \plus{} \angle B \plus{} \angle C \minus{} \angle B \minus{} 30^{\circ} \equal{} \angle NMA$

Since $ NM \equal{} NL \equal{} \frac {_R}{^2},$  we deduce that $ N$ is equidistant from $ AC,AB$ $ \Longrightarrow$ $ N \in IA.$ So the Feuerbach poin $ F_e$ of $ \triangle ABC$ lies on $ AI$ as well. Further, $F_e$ is midpoint of $ AI,$ since $ AI \equal{} 2r$ $ \Longleftrightarrow$ $ \angle A \equal{} 60^{\circ},$ hence $ AP \equal{} AF_e \equal{} r.$

Well Luis, my friend, actually I wanted to prove $ AT = r$ using another approach.

Anyway, I have two beautiful ways for proving $ AT = r$ and $ AP = r$. (Yours was extremely good as well)

[b]Lemma 1:[/b] $ AT = r$

Proof:
                Consider the nine point circle. We know by definition that the nine point circle is the circumcircle of triangle $ ABC$.  So assume $ D,E,F$ are midpoints of $ BC, CA, AB$ respectively. Therefore, circumcircle of triangle $ DEF$ is the nine point circle.

                Consider a point $ K$ such that $ Ak = r$, on $ AI$.  First, since $ AI = r %Error. "cosec" is a bad command.
{\frac {A}{2} = r%Error. "cosec" is a bad command.
{\frac {\pi}{6}} = 2r}$ where $ r$ is the inradius.
So, $ k$ will be the midpoint of $ AI$.

                   By a homothety of center $ G$ and ratio $ \frac { - 1}{2}$, triangle $ ABC$ maps to triangle $ DEF$. ($ G$ is the centroid)
Therefore, $ \angle EDF = \frac {\pi}{3}$ .

                   By a homothety of center $ A$ and ratio $ \frac {1}{2}$, triangle $ BIC$ maps to triangle $ FKE$ . This implies $ \angle FKE = \angle BIC = \frac {2\pi}{3}$.
This is because $ \angle BIC = \frac {\pi}{2} + \frac {\pi}{6} = \frac {2\pi}{3}$.

                   Therefore, $ \angle FKE + \angle FDE = \frac {2\pi + \pi}{3} = \pi$. This implies $ FKED$ is a cyclic quadrilateral.

                   This implies $ K$ lies on the nine point circle .

                   But $ IK = r$ (Since $ K$ is midpoint of $ AI$) . Therefore, $ K$ lies on the incircle as well.

                   Thus, we have that $ K$ lies on both the incircle and nine point circle .

                   By Feurbach Theorem, we know that the incircle is internally tangent to the nine point circle.

                   Since $ K$ lies on both incircle and nine point circle, $ K$ must be the internal point of tangency of the incircle and nine point circle.

                   This implies $ K = T$. Since $ AK = r$, it implies that $ AT = r$.

[b]Lemma 2: [/b]  $ AP = r$.

      In a triangle $ ABC$, let $ D$ be the midpoint of $ BC$, $ I$ as incenter, $ P$ is the intersection of $ DI$ with the altitude $ AX$ of triangle $ ABC$.

Consider the circumcircle of triangle  $ ABC$ with circumcenter $ O$. Now, if $ AI$ meets the circle at some point $ X$, $ X$ is the midpoint of arc $ BC$. This implies that $ OX$ is the perpendicular bisector of $ BC$.

        Let the point of tangency of the incircle with side $ AC$ be $ Q$. So, $ IQ$ is perpendicular to $ AC$.

       Now, $ \angle IAQ = \angle BAX = \angle BCX = \frac {1}{2} \angle A$ .

       This implies that the right triangles $ AIQ$ and $ CXD$ are similar. Also, since $ \angle IAP = \angle IXD$ and $ \angle AIP = \angle XID$ , triangles $ AI$ and $ XID$ are also similar.

Hence, $ \frac {AP}{XD} = \frac {AI}{IX}$  and    $ \frac {AI}{XC} = \frac {IQ}{XD}$.

 Since $ IX = XC$, we have $ AP = AQ$.

Well, actually wanted to prove $ AT = r$ using another approach and so I thought of using $ AP = r$ and trying to prove triangle $ ATR$ is isoceles.

(Should have put it in proposed problems section).

Can someone please post a solution for $ AP = AT$ by angle chase using the fact that $ AP = r$ and not using $ AT=r$?Like an angle chase etc?

[quote]Given a triangle $ \triangle ABC$ with $ \angle A \equal{} 60^{\circ}$. Let $ (I)$ be the incircle of $ \triangle ABC$ and $ D$ be the midpoint of $ BC$. Denote $ A'$ by the projection of $ A$ onto $ BC$ and $ P$ by the intersection of $ ID$ and $ AA'$. Let $ T$ be the [i]Feurebach point[/i] wrt $ \triangle ABC$. Prove that $ AP \equal{} AT$.[/quote]
Let $ D,$ $ E,$ $ F$ respectively be the midpoints of $ BC,$ $ CA,$ $ AB$ and $ A'',$ $ B'',$ $ C''$ respectively be the tangency points of $ (I)$ with $ BC,$ $ CA,$ $ AB$. Let $ T'$ be the midpoint of $ AI$, then it is followed that $ T'$ is the circumcenter of $ \triangle AB''C''$. But since $ \angle A \equal{} 60^{\circ}$, hence $ \angle B''T'C'' \equal{} 120^{\circ}$. But $ \angle B''A''C'' \equal{} 120^{\circ}$, thus $ \angle (T'B'',T'C'')$ $ \equal{} \angle (A''B'',A''C'')$, which implies $ T'\in (A''B''C'')$ $ \equiv (I)$. Further, since $ T'F\|IB$ and $ T'E\|IC$, which yields that $ \angle (T'F,T'E)$ $ \equal{} \angle (IB,IC)$. In the other hand, due to the fact that $ IB,$ $ IC,$ $ A''C''$ and $ A''B''$ form a concyclic quadrilateral, then $ \angle (IB,IC)$ $ \equal{} \angle (A''C'',A''B'')$. Also from the beginning, we have known that $ \angle (A''C'',A''B'')$ $ \equal{} \angle (AB,AC)$ and with a notice that $ AFDE$ is parallelogram. As the result, $ \angle (T'F,T'E)$ $ \equal{} \angle (DF,DE)$. Therefore, $ T'$ also lies on $ (DEF)$, which is also the $ 9 \minus{}$ point circle wrt $ \triangle ABC$. As the result, $ T'$ is the [i]Feurebach point[/i] wrt $ \triangle ABC$ $ \Longrightarrow$ $ T'\equiv T$. Now, by the [i]Newton theorem[/i] for the "quadrilateral" $ ABA''C$ with $ (I)$ its inscribed circle, we obtain that $ ID$ also passes through the midpoint $ M$ of $ AA''$. Then it is followed that $ TM\|IA''\|AP$, but $ T$ has already been the midpoint of $ IA$, thus $ M$ is also the midpoint of $ IP$. The diagonals of quadrilateral $ APA''I$ intersect each other at their midpoints $ \Longrightarrow$ $ APA''I$ is parallelogram. Therefore, $ AP \equal{} IA'' \equal{} IT \equal{} AT$. Our proof is completed then. $ \square$


Prove that a linear operator T:$ C^{n}$->$ C^{n}$
 is diagonalizable if for all  $ \lambda  \in C$,$ Ker(T\minus{} \lambda I)^{n}\equal{}Ker(T\minus{} \lambda I)$
 where I is the nxn indentity matrix

[quote="nkht-tk14"]Prove that a linear operator T:$ C^{n}$->$ C^{n}$
 is diagonalizable if for all  $ \lambda \in C$,$ Ker(T \minus{} \lambda I)^{n} \equal{} Ker(T \minus{} \lambda I)$
 where I is the nxn indentity matrix[/quote]If it is diagonizable, that means that there is a basis of eigenvectors.  Consider such a basis, and take a look at what $ T\minus{}\lambda$ does to linear combinations of these basis vectors.

Actually, he wants it the other way around.

WLOG assume that $ \lambda \equal{} 0$ is an eigenvalue for $ A$, so that we're trying to show that equality actually holds in $ a \geq g$ (multiplicities).

We actually have that $ \ker \, A \equal{} \ker \, A^2$. Thus, by the rank-nullity theorem, we have that $ \text{rank} \, A \equal{} \text{rank} \, A^2$. In particular, the restriction of $ A$ to $ \text{Im} \, A$ is actually a bijection. Therefore, $ N(A) \cup R(A)$ spans everything, and wrt the basis $ \mathcal B_{N(A)} \cup \mathcal B_{R(A)}$, our matrix looks like $ \begin{pmatrix} O & O \\
O & A \end{pmatrix}$, where $ A$ is invertible.