FTW
MATHCOUNTS
geometry
3D geometry
pyramid
AMC 8

1) It varies from chapter to chapter and depends on the difficulty of the year. I would assume that a score in the forties would make it in most chapters.

2) Actually, I struggle with this too. Just practice a lot and be confident in your own ability? Also, don't think that's it's the most important thing in your life. It's not.

3) It really depends on how long the question is. Also, if your opponent has already answered incorrectly, take as long as you need and make sure you get the right answer. If you don't have an answer and there's almost no time left, guess. It can't hurt. Also, you might want to practice your "twitching" skills; aka, when you see a problem you know you can solve in less than 3 seconds, buzz in immediately. Very helpful. This is one of the things FTW doesn't help with. It's also very dangerous. If you do have people to practice with, consider yourself lucky. I don't.

4 and 5: There's several things that I think are useful.
1- Fast reading/skimming skills are very useful here. If you can skim right through the problem, get the information you need, and answer, it's better for you. I think I'm pretty good at this, since I read a lot. I don't necessarily read tough books, but they're long and have boring parts. Sometimes I just skim stuff.
2- Practice. Once again, if you have someone who you can actually practice with, you're very lucky. Some poor people like me don't have anybody to practice countdown with. Grr. The more you practice, the more used to it you will be.
3- Mental math. The more math you can do mentally in your head, the better. It takes forever to actually write stuff down on paper. However, this can be dangerous unless you've practiced a lot. If you make a mistake while doing it in your head, you might not notice it and you will get the question wrong. I suggest doing any problems that you can do without paper...well, without paper.
4- Get faster. I actually suggest FTW for this. Countdowns and 10-25 second games really make you go faster. It's also better to do time games, since you actually get points based on your time. Of course, FTW has its limits, such as repeating problems and no twitching practice, but it can certainly help you.

The most important thing, although it may not seem helpful for countdown, is to do lots and lots of problems. They introduce you to new techniques that you may not have noticed before, and you can use those when you see a problem that is related to a previous one. This is probably the most important thing in competitive math. Even if you see a problem you've already done, try to think of a faster way to solve it. This is also quite helpful for written. For example, #28 (?) on last year's National Sprint is almost exactly like #23 on the 1999 AHSME, I belive. I had seen that question, so I got the one on the sprint right, and relatively quickly too. It was also one of the two sprints in the twenties I solved :P

[quote="pinkpanther"]Well, Chapter is coming up and I think I have a good chance of getting into Count Down.  These are my questions below:

(1) What is a minimum score you think I must get to get into Chapter countdown?

(2) Since this is my first time, I am afraid of getting nervous and the pressure might make me do bad.  Is there anything I can do to stop this from happening?

(3) What is a nice time to aim for answering to questions?

(4) How should I practice?

(5) How should I improve my speed and accuracy together?

Thank you!![/quote]

1. it matters which state ur in. If ur state is easy at least above 30.

2. um no the first time i was in the countdown i was extremly nervous but i beat like 5 people.

3. the first problems should be less than 3-5 seconds.
after that it should be 8-20

4. aops for the win

5. see # 4

Since I'm pretty much a failure at CD (OK, I won my (unofficial) state countdown, but that's because everyone else failed on that day and I was extremely lucky), I will just emphasize a few points that I think are important, in no particular order.

a.) Do many problems from lots of random contests! You'll get familiar with lots of good techniques (i.e. what works when), and you'll also be surprised how many problems there are in MATHCOUNTS that repeat almost exactly from the old tests.
b.) Be very confident of yourself. One thing that closely relates to this is that you shouldn't ever doubt yourself anywhere. I have a lot of trouble with these things, and I think it really slows me down.
c.) Study mental math and arithmetic tricks. These can save lots of time, and are often more accurate than normal methods.
d.) Practice...? This is probably the most important part. The more experience one has in something, the better they will do in it. This can help you with your reading and speed and can get you much more used to the real countdown climate. It always helps to minimize the chance of the competitive jitters screwing you up.

[quote="pinkpanther"]Well, Chapter is coming up and I think I have a good chance of getting into Count Down.  These are my questions below:

(1) What is a minimum score you think I must get to get into Chapter countdown?
Depends where you live. It could be around 40 or around 29.

If you get 40+ you should make CD. If you don't that would be astounding to me.

(2) Since this is my first time, I am afraid of getting nervous and the pressure might make me do bad.  Is there anything I can do to stop this from happening?

A little nervous is good! Take practice tests, and FTW might help.

In my experience (And I've taken a LOT of competitions), usually you feel nervous before the competition, but during the actual exams, you tend to relax a little. That is, assuming you don't look at #1 on the School Sprint and have no idea how to do it. Then you might panic  :P

But seriously, you really won't be nervous during the tests. Trust me. I've done at least 50 competitions.

(3) What is a nice time to aim for answering to questions?

On the sprint:

First 10 questions: shoudln't use more then maybe 1 minute a question.
Next 5: Use about 1 minute 5 seconds.
Next 5: About a minute 15 seconds.
Last 10: Work down as you can.
Attempt to leave at least 5~10 minutes to check.

Target: no more then 3 minutes a question. Try to leave some time to check your work

(4) How should I practice?

Definately do old tests. You might regonize methods in actual competition problems similar to the ones you did.

Also, FTW helps, and if you can find a group of people (your team is good, if you have one) that can CD each other.

(5) How should I improve my speed and accuracy together?

Practice, Practice, and Practice.

Thank you!![/quote]

That's about all I can give you, except focus on geometry techniques. Many geometry problems that look hard are actually pretty easy.

And check your work a lot on counting problems. Its very easy to make stupid mistakes on these more than other subjects.

Hope this helped.

~BOGTRO.

Well, I'll answer No. 2. The BEST way not be anxious about the test and/or countdown is to NOT STUDY AT ALL the day before the test. Trust me.

It may seem ridiculous BUT IT WORKS. Besides, cramming stuff in before the test is not good and neither can you achieve much in just 1 day.

Ok, so I'm looking at the question that asks about how much time you should spend, and I say there is definitely no fixed time schedule for problems. Admittedly, the sprint should be in some order of difficulty and the first few should be done the most quickly, but depending on what you're good at, you might find an extremely difficult problem in 10-20, and find the last few problems relatively easy. I think that you should skip problems you think you can't solve immediately so you solve all the ones you can more quickly. Then you can go back.

Target, you should definitely at least look at both questions. Do the one that seems more easier to you. If there are three minutes left and you still haven't solved it, don't think you have to switch immediately. It depends on whether you think you can solve the other question more quickly and how close you are to finishing the current question. Also, double check.

Double checking is extremely important. However, you don't need to check [i]every single[/i] problem. What I do is if I get an answer to a problem that I'm not sure about, I mark that problem with a question mark so I remember to check it. If I am sure about my answer to a problem though, I don't mark it and I don't come back to it. I somewhat double check immediately after doing the question, so I don't need to spend time later getting used to the question again.

For countdowns, there is definitely no helpful answer for the timing. I don't think that the harder problems are necessarily at the end. This is where recognizing techniques is helpful, so you can finish a question faster.

[quote="pinkpanther"]Well, Chapter is coming up and I think I have a good chance of getting into Count Down.  These are my questions below:

(1) What is a minimum score you think I must get to get into Chapter countdown?

(2) Since this is my first time, I am afraid of getting nervous and the pressure might make me do bad.  Is there anything I can do to stop this from happening?

(3) What is a nice time to aim for answering to questions?

(4) How should I practice?

(5) How should I improve my speed and accuracy together?

Thank you!![/quote]

[hide="1"] okay, it depends on your chapter. my chapter, which is extremely beastly, is better than most STATES, so we require like a 42 to make countdown. on the other hand, some of our neighboring chapters with dumber kids only require in the thirties.
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[hide="2"] have confidence. i see that you are confident that you'll get into cd, which is good. also, if you get nervous, realize that you cannot drop more than one spot. in my opinion, countdown is the best part of mathcounts (it can actually be, yes-fun). i dont know if its allowed, but when they call you up for countdown, take a bottle of water. it'll help...trust me.
[/hide]

[hide="3"] some problems, which are solely computational (for example, evaluate $ 4\cdot2^4\plus{}5$), should be answered as soon as possible. if you see a problem, take a second to look at it and see how much time it'll take. for example, in the problem above, you should buzz in, and THEN do the computations. but, as xpmath said, DONT RUSH IF THE OTHER PERSON GETS IT WRONG. also, because it is so fast-paced, some people (including me) feel compelled to say their answer as soon  as they are called upon. take your time saying your answer (i mean, before we graduate, but...yeah)
[/hide]

[hide="4"] FTW can be amazingly helpful (like it was for me, the person who got pwned 3-0 at chapters AND states) and it can also be very annoying. if you are new, try countdowns by yourself on ftw, because you will not recognize many problems. also, you need to realize that most problems on ftw are not actually meant to be solved as cd questions (okay, well some of them are...), they are meant to be amc 8 or sprint/target questions. also, see if you can find countdown rounds from previous years and do those...PM me, i have some (few, but some.)
[/hide]

[hide="5"]i like how you phrased this question. speed and accuracy. as much as speed matters in cd, ACCURACY MATTERS more-clearly a slow, correct answer is far mor valuable than a lightning quick, yet utterly wrong answer. but, still work on speed, as you wont have a chance to answer otherwise. :P haha...i'd do the same things i said in number 4, along with working on mathcounts in general (chapter and school rounds are generally better practice for countdown). be confident, work hard, and great things will come! =D
[/hide]

i really hope that helped, i spent a while writing it.

[quote="pinkpanther"]Well, Chapter is coming up and I think I have a good chance of getting into Count Down.  These are my questions below:

(1) What is a minimum score you think I must get to get into Chapter countdown?

(2) Since this is my first time, I am afraid of getting nervous and the pressure might make me do bad.  Is there anything I can do to stop this from happening?

(3) What is a nice time to aim for answering to questions?

(4) How should I practice?

(5) How should I improve my speed and accuracy together?

Thank you!![/quote]

(1) Like Nikey said, a lot depends on how good your chapter is. Mine is really good, so I'll probably need a 40 or higher, while chapters that have kids that aren't as good will only need mid-20s.

(2) People react differently to adrenaline; I find adrenaline makes me think faster, but I'm jumpier and wouldn't be surprised if I slam the buzzer by accident before I get the answer. Other people can't concentrate and really freak out. Breathing deeply(4 counts in, 4 counts out) might help you calm down, and just relax. Wear comfortable clothes, like a sweatshirt and not-too-tight jeans and try to relax. Competitions are fun, but if you don't do as well as you want to, so what? It sucks if you don't make state, but the world won't collapse.

(3) Depends how good your opponent is. If the opponent's better than you, think fast and don't be afraid to go with your gut instinct. Like if you see a problem that says something like "Find the product of the 5th and 3205th term", chances are the answer's 0, so go with that. If the opponent's not as good as you, spend a little more time to make sure you don't make stupid mistakes. To check your answers, just give it a quick look to make sure it makes sense. If the question asks for the arithmetic mean of a say, 30, 32, 42, and 84, and you got 23, you must have done something wrong.

(4) FTW's pretty good, except the questions repeat and you can't practice the twitching thing xpmath described. If you become the kind of uber-memmer I am, it's probably not worth it to spend 6 hours a day on FTW(even though I do :P ), but there are other ways to practice too. If you can get someone willing to help you, a timer, and a bunch of old countdown problems, you can get that person to read the questions and you can use the timer to see if you can get faster. I personally use a chess clock, since you can hit it to simulate buzzing in. It's even better if you can practice with people.

(5) Practice.

1, 3.  As everyone else said, depends on your chapter.  If, for example, you have a countdown superstar in your chapter, you'll need to shoot for negative time (i.e. buzzing in before the moderator is done reading the question).

2. Confidence.  You own the room.  Imagine how awesome everyone watching will think you are when they see your prowess at mathematics.  I suggest speaking forcefully but evenly into the microphone (if there is one) as a start.  You'd be surprised how much that can help your frame of mind.

4, 5.  I've found that the most important part of countdown is having your brain "well-lubricated".  By that I mean you see the problem, you immediately recognize what's being asked and how to find it.  Then you quickly solve it without stopping or second-guessing yourself.  And on top of that, you learn to do this without sacrificing accuracy.

[quote="pinkpanther"]Well, Chapter is coming up and I think I have a good chance of getting into Count Down.  These are my questions below:

(1) What is a minimum score you think I must get to get into Chapter countdown?

(2) Since this is my first time, I am afraid of getting nervous and the pressure might make me do bad.  Is there anything I can do to stop this from happening?

(3) What is a nice time to aim for answering to questions?

(4) How should I practice?

(5) How should I improve my speed and accuracy together?

Thank you!![/quote]

(1) It depends on how hard the chapter sprint/target rounds are and who you're up against. I actually got 33 at chapter, and I got 9th place, but somebody who did better than me was taken out of the countdown round for some  reason, so I ended up competing in it and got 4th. I would say that you need at least a score of 30 to make the cd round, but you'd be really lucky to make cd with a score of 30 anyway.

(2) Like Sli Si said, you need to feel confident about yourself and how amazing you would appear to the people watching if you won a round. I know how pressuring it is. One round I was asked to find the area of a pyramid, and I got the right answer, but I added pi to the end of it, as if it were the area of a cone, so I ended up missing the question. Once you get used to it, then you will be able to do it with out being very nervous.

(3) If it is a really easy question that is basically mental math, then I would say you would have to solve it in about three seconds. If it is one of those problems that looks hard but has a really easy way of solving it, then you would probably have to solve it in about five seconds. If it is easy, but requires a little bit of computation, then you would have to solve it in about ten seconds. If it was a really hard problem, then you would probably be safe getting it in 20-30 seconds.

(4) Solve lots of problems, and see how many solutions you can find to each problem, and see which solution is fastest/easiest and try to remember that solution for future reference. Once you have done this, then you should compete in solving problems with others in a countdown round. I don't think that countdown rounds are your greatest concern in mc, unless you want to earn a few trophies at chapter or state  :lol:. If you make it to nats, and you feel really confident, then that is when you really need to practice countdown.

(5) Pretty much doing what i said for the previous question. But you have to stay confident and don't get nervous.

what chapter were u guys in
and i'm in greater houston one so i need OVER 40 :wink:

I'm not sure if this has been mentioned, but FTW is extremely helpful; it helped me get to nats :D

[quote="akhil0422"]I was in greater Houston last year. I made two TOTALLY CARELESS mistakes. I would've gotten 46/46 but no... I got 44/46.

I came in 7th place; I FAILED MISERABLY IN COUNTDOWN. Even on FTW, my rating is 1199 after many games. I'm not fast at all :([/quote]

i'm kind of the opposite...i'm not that good at the written part but pwn at cd (except 4 bobby shen-but that's different)

Here's an answer to one question.
3) I find that in countdown round, you should buzz in if you have an answer that your 1/2 sure of. If you wait until you are totally sure that your answer is correct, it might be too late. DON'T use this technique on any other round.[/b]

Since you bring that up, it's very important not to second guess yourself, unless your opponent has already gotten it wrong. If you get an answer and your opponent still hasn't buzzed in, buzz in anyway.

oh and if u're facing a [b]really good [/b]person, as soon as u just find HOW to solve a problem, just hit that buzzer

my strategy against bobby shen: buzz in early, sneeze or something, then answer :D
i dont know if i'm in greater houston...last time they said it was bay area or something but i'm pretty sure sugar land is NOT bay area

Do you guys know Bobby Shen?

Whenever I am in the cd round I always rest my right hand on top of the buzzer, and then I solve the problem as quickly as possible, and then when I am really close to solving the problem I hit the buzzer and then solve the rest and answer the question.

I think that Darryl Wu did that at the countdown round because there was one question where he hit the buzzer but didn't give an answer.

According to 5849206328x, the finger is faster than the hand. Just keep your finger on the buzzer; no need to press it with your whole hand.

5849206328x is pretty dang beast at countdown, so pay attention to his (rarely given) advice.

it is also generally accepted not to take my advice very literally, unless it's hidden in some way :wink:

[color=white]basically, try not to make unnecessary motions when you get the answer, such as raising your hand off the buzzer to semi-slam it.  this reduces to "stay calm throughout the question," which is also important so that you don't forget your answer after hitting the buzzer. I would know[/color]

[quote="pinkmuskrat"]You are left-handed, right?

I'm pretty awful at countdown. Fortunately, for uys, it's just for fun, not official.[/quote]

They have MATHCOUNTS in Monaco?

thanks for the tips...fortunately, my countdown IS official. that should help me this year...we'll see.

[quote="mathluver2.718"]i suck at cd rounds.  :P  :wink:  lucky for me, mine's not official. :)[/quote]

haha same :)

i fail at cd :(


Dream of all glorious Greek or Roman childs, to free the world, achieve the glory beyond Achilles, Hercules, rival Prometheus?

the friendship between the liberators and the liberat-ees.

After seeing so many affecting scenes, if one's eyes is not filled with tears, he must be ironhearted.

Answer my question before or after you enjoying these photos, please.
P.S. Do not forget to save a copy of this post if you like it. I don't know when this post will be deleted because it is inappropriate in some subtle standard.

I don't really get what the point of these posts/pictures is, nor do I understand the question you are asking.

I second that opinion.

What the hell are you doing?

IMMORTALITY!!  TAKE IT!!  IT'S YOURS!!!

Take [i]that[/i] Patrick Stewart.   :lol:

Buffalo, there seems to be popular demand for an explanation of this thread.  I too am curious: aside from yet another place for you to air your grievances about my tyrranical rule, what does thread this mean?

Basically I think he is trying to point out how all tyrants are trying to make their invasion look like liberation, both in real life as in movies.
Those last two pics, what do they mean?

Basically Buffalo, I don't get you.  First you asked whether or not you could post some pictures.  You started a whole debate about it and moderators and stuff,...without the pictures.
Now that they are here, you post non offending pictures (those are not offending pictures at all! in my opinion) but with no explanation????

Aren't there any pictures of the Chinese incursion into Tibet?

AFTER EDIT:

An easy Google search finds these links:

http://news.bbc.co.uk/1/hi/world/asia-pacific/1803898.stm

http://www.studentsforafreetibet.org/article.php?id=422

and plenty more.

Yeah, I guess it is common for the invader to justify his invasion as liberation. At least in the case of the Second Sino-Japanese War, it is without doubt that many civilians were killed by the Imperial Japanese Army.

In response to the last post, I'm with Michael Parenti. The Shangri-La Myth of Tibet is just that, a myth. The only people lobbying for a removal of the "oppression" are the people who were in power before, no surprise.

I can't understand why so many people requested explaination, enough have those pictures said, I think.
In the movie Alexander, when Alexander entered Babylon, what did Ptolemy say? "He was loved by all". Yeah, Persians loved the one who had just killed tens of thousand of their friends, their family, grab whatever they had, that's really great.
What did president Bush say? Iraqi will greet USA army with roses. Equally great, but with suspect of plagiary.
What happened in Dec. 1937 In Nanjing? Japs slaughtered more than 300,000 unarmed Chinese people in the way that human beings can never imagine, compared with these, what Hitler did to the Jews looks just like pastime of a naughty boy. Wanna see some photos? I would like to show some, unfortunately, these are called improper topics according to the moderator.
One thing these Japanese enjoyed to do is, taking photos of all these enormities, as a proof of their braveness/glory(?) to show off to their friends, their families. Does this remind you something, happened recently? Those heroes in mideast are undoubtly much more civilized than their Japanese precursors, you have to admit.
What are Japanese doing now? They tell their children in the textbooks, the Japanese Imperial Army went to liberate Chinese at that time, soldiers behaved like angels, they saved the people from danger, they handed round candy to the kids, they loved the people and the people loved them.
What do you expect to see in USA's history books later? Which pictures are more likely to be cited to describe what happened in Iraq? The one showed USA soldier saving dying Iraqi kid or those maltreatments?
I wonder how Britain teach the opium wars, the East India Company to their children, anyone, willing to help me?
And do USA tell their children how did they transform millions of american natives into "good indians"? I saw a cartoon a long time ago, in which they says because Ameicans need elbow room, so there was the Westward Expansion. Do Americans need another elbow room now?

The last two pictures showed why soldiers go there, take everything that's not nailed down. This is the reward to their excellent services. Go and check your nearby museums, how many of those treasures did you "borrow" from the other countries?

For those who keep using words like Tibet and Turkestan, you are challenging all Chinese--or almost all, to be more precise.

JBL, if I'm not mistaken, you told me that you were doing something protesting the war(one year before?), how is that going? Have you made any progress?

[quote]For those who keep using words like Tibet and Turkestan, you are challenging all Chinese--or almost all, to be more precise. [/quote]
And your point is... :?

And in those two pics, what is the green stuff the soldier is holding up, and what are they going to 'grab' in band of brothers?

Buffalo, you talk about Native Americans?  You better check up on your United States education system.  In the late eighteenth century, I believe, there was a book written called [i]A Century of Dishonor[/i].  It was [u]all[/u] about the mistreatment of the Native Americans by American settlers and the Manifest Destiny.  While yes, the Dawes Act and the Indian Removal Act were harsh and pretty much oppressive, I think in the 1930's FDR was able to get the Indian Reorganization Act through Congress, which recognized the mistreatment and tried to amend.  Biased material such as that with the Japanese textbooks is not so common in America, where we complain about every little thing.

More importantly, the last time America had a war like Iraq was thirty years ago.  Do you know how Vietnam is portrayed in American classes?  Like the disaster it was.  I suspect, although cannot be absolutely positive, that my US History textbook included photos of Vietnamese civilians who were the victims of napalm attacks; it certainly included reference to the My Lai massacre.

Buffalo, it is going well.  Substantially fewer than half of Americans think it is right that we went to Iraq and roughly one third approve of the behavior of the president.  It is seeming increasingly likely that at least one house of the new congress, to be elected in November, will have a Democratic Party majority which will enable the investigation of the behavior of the current administration.  America isn't likely to leave Iraq right now.  But Americans have realized what a terrible idea it was to begin with, and many of the lies surrounding the waging of the war are now part of the public record.  Members of the administration, perhaps including the vice president, are under investigation by an independent prosecutor.  What empire in history can say something comparable?  Not the Japanese, sure, and not the Chinese, either.

@erik: Tibet didn't have to be shangri-la for the Chinese invasion to have been bad.  Iraq certainly wasn't; China in the thirties certainly wasn't.  What it did have to be is a place where civilians aren't routinely [url=http://hrw.org/english/docs/2005/04/27/china10549.htm]tortured[/url] or [url=http://hrw.org/english/docs/2005/09/17/china11754.htm]kidnapped[/url].

[quote="Buffalo"]For those who keep using words like Tibet and Turkestan, you are challenging all Chinese--or almost all, to be more precise.[/quote]
What does this mean?  That Chinese citizens don't like it when you point out the hypocrisy of their paens against oppression and tyrrany while they are defending the behaviors of their own government?  Shocking!

[quote="bleumoose"]I'm probably going to look like a total ignoramous (borrowed that from the Cracker Barrel game, yep) for asking this, but could someone please post a summary of what this thread is supposed to be about? I'm rather confused. What is the argument? What are the main positions?[/quote]
I second that request.

[quote="JBL"]
@erik: Tibet didn't have to be shangri-la for the Chinese invasion to have been bad.  Iraq certainly wasn't; China in the thirties certainly wasn't.  What it did have to be is a place where civilians aren't routinely [url=http://hrw.org/english/docs/2005/04/27/china10549.htm]tortured[/url] or [url=http://hrw.org/english/docs/2005/09/17/china11754.htm]kidnapped[/url].[/quote]

Once again, I'm with Michael Parenti. It may be true that there is torture and kidnapping performed by the government against citizens living in the T.A.R. It may have been even worse under the traditional rule of the Dalai Lama.

[quote="bleumoose"]I'm probably going to look like a total ignoramous (borrowed that from the Cracker Barrel game, yep) for asking this, but could someone please post a summary of what this thread is supposed to be about? I'm rather confused. What is the argument? What are the main positions?[/quote]

It's just a highly elitist way of stating that the winners of wars do their best to make themselves look good.

...  :huh:  but isn't that kind of a "duh" statement? Surely that can't be all there is to it.

[quote="blahblahblah"][quote="Iversonfan2005"]WHOA, I UNDERSTAND EVERYTHING NOW, IT IS A COMMON DREAM, OMG, IT ALL MAKES SENSE!!!!

POLITICAL RIOTS, WAR, OPPRESSIVE LEADERS, OMG!!!! IT IS HISTORY!

I SOLVED THE RIDDLE YAY![/quote]

You're not funny nor are you even witty, and that goes for all of you who begin your posts with 'WHOA' and proceed to write in capital letters.[/quote]

Your bash on someone's retarded post was also absolutely hilarious, ironically.  I guess one doesn't have to act like a prick to make someone feel shameful.


[quote="zhaobin"]for $a \ge b \ge c \ge 0$
Prove
\[ a^xb+b^xc+c^xa \ge ab^x+bc^x+ca^x  \]
where $x \ge 1$[/quote]
We rewrite the inequality in this form:
$(a^x-b^x)(b-c)-(b^x-c^x)(a-b)\geq 0$
$\Leftrightarrow (a-b)(b-c)(\sum_{i=0}^{x-1}a^{x-1-i}b^i-\sum_{i=0}^{x-1}b^{x-1-i}c^i)\geq 0$
which is clearly true. :)
P.S. I forgot $x$ is real...  :blush:


abstract algebra
Ring Theory
number theory
prime numbers
superior algebra
superior algebra solved

Can the property of being finitely generated be checked locally? More precise:
Let $R$ be a noetherian commutative ring and $M$ be an $R$-module. Denote by $M_p$ the localization of $M$ at the prime ideal $p$ of $R$. Suppose $M_p$ is finitely generated for every prime ideal $p$ of $R$. In that situation, can we conclude that $M$ is finitely generated?

I think the answer is yes because the unit ideal is generated by finitely many elements

Here's a simple example to show that the answer is negative (I hope I'm not missing anything):

Take $R=\mathbb Z$, so $R$-modules are just abelian groups. Now let $\mathcal P$ be the set of prime numbers, and take $M=\bigoplus_{p\in\mathcal P}(\mathbb Z/p\mathbb Z)$. Clearly, $M$ is not finitely generated. Its localizations, however, are cyclic modules over the corresponding localizations of $\mathbb Z$.

Hmm, I needed some time to recognize the author of this thread, but I'm beginning to remember - gradually  :wink:
Thanks for solving this 'oldie'  :D


Find the greatest possible product of a four digit number and a three digit number with 7 distinct digits.

I know that the thousands/hundreds digits are the most important, and the units is the least important.

The amount of the thousands you have is the most important.

so you have

abcd x 9ef. next would be the 'a' spot, since that controls the # of hundreds.
8bcd x 9ef. # of hundreds is more important then the tens spot.
87cdx9ef. now you have a simple 2x2, which is easier to fill out.
8753x964

I am pretty certain this is the highest, and how I would solve it. But someone else take a less, -guessing- approach?

The answer and reasoning is right, but here is a more logic-based method:
Think about the increase based on each digit. First write in form
 abcd x efg = p , where p is the product. It is obvious that 8 will go in for a, as the most important digit. The hard part if finding which is more important, b or e, c or f, d or g.

Note that the digits we will insert will have the following properties:

a > b > c > d
e > f > g
p = e x 100 x abcd + f x 10 x abcd + g x abcd.
p = a x 1000 x efg + b x 100 x efg + c x 10 x efg + d x efg
It is obvious that
 g x abcd > d x efg
 f x abcd > c x efg
 e x abcd > b x efg
Since an increase in e, f, or g will result in a higher increase in product [i] p [/i] than an increase in b, c, or d would, respectively, we can conclude that
e > b
f >  c
g > d
So, we can see that the product must be 8753 x 964.


Let $ c$ be a positive constant.
Find the maximum and minimum  value of the function $ f(x)=\sqrt{x+c}-\sqrt{c^{2}-x^{2}}+\sqrt{c-x}$.

We have $ f(x) = \sqrt{c+x}+\sqrt{(c+x)(c-x)}+\sqrt{c-x}$. So the function is only defined when $ -c \leq x \leq c$.

We observe that the value of $ \sqrt{(c+x)(c-x)}$ has a maximum value of $ 2c$ when $ x = 0$. Now plotting a graph of $ g(x) = \sqrt{c+x}+\sqrt{c-x}$, we see it also has a maximum value of $ 2\sqrt{c}$ when $ x = 0$. So the maximum value is $ \sqrt{c^{2}}+2\sqrt{c}= \boxed{c+2\sqrt{c}}$ when $ x = 0$.

From the graph of $ g(x)$, we see also that it is minimum when $ x = c$ or $ x =-c$. Incidentally, $ \sqrt{(c+x)(c-x)}= 0$ when $ x = x$ or $ x =-c$ and this must be minimum. So the required minimum value is $ \boxed{\sqrt{2c}}$ when $ x = c$ or $ x =-c$.

[hide]I got $ \sqrt{c+x}+\sqrt{c-x}=1$ but I can't figure out how to siplify that[/hide]

Maybe this helps?
[hide]$ \sqrt{x+c}+\sqrt{(x+c)(c-x)}+\sqrt{c-x}=(\sqrt{x+c}-1)(\sqrt{c-x}-1)-1$[/hide]

[quote="aidan"] .....We observe that the value of $ \sqrt{(c+x)(c-x)}$ has a maximum value of $ 2c$ when $ x = 0$. Now plotting a graph of $ g(x) = \sqrt{c+x}+\sqrt{c-x}$, we see it also has a maximum value of $ 2\sqrt{c}$ when $ x = 0$. So the maximum value is  .......
[/quote]

The function is $ f(x)=\sqrt{x+c}-\sqrt{c^{2}-x^{2}}+\sqrt{c-x}$.
(and not $ f(x)=\sqrt{x+c}+\sqrt{c^{2}-x^{2}}+\sqrt{c-x}$.)
Just be a little careful, you are not summing   "your two functions"  .. you are taking  their difference... The month when I made the most money and also spent the most money need not be the month when  my net savings increased the most .:)

[quote="aidan"]Which part of my reasoning is wrong? And the answer is in terms of c right?[/quote]

We need to classify some cases depeding on the values or range of $ c$.

This problem might have seemed to difficult for Intermidate Topics....

Ok :blush:  didn't see that minus sign. Now things have changed a little bit :P Maybe it is hard for intermediate haha

[quote]didn't see that minus sign. Now things have changed a little bit...[/quote]

May be ! x=0 could still be  a maximum (or a minimum or even a point of inflexion)  depending on the value of $ c$..
Your answer could still be correct with some clarifying statements ...:)

The problem is not really that hard, if one  is allowed to use calculus ...
[hide]For first derivative to be zero, You get something like:
${ \sqrt{c+x}-\sqrt{c-x}= 4x}$   (Eq 1)
One solution,  so, obviously is when x=0,
Obviously also: $ (\sqrt{c+x})^{2}-(\sqrt{c-x})^{2}= 2x$ (Eq 2)
Dividing eq 2  by eq 1 (when x is not zero) we get
${ \sqrt{c+x}+\sqrt{c-x}= 1/2}$    (Eq 3)
Adding eq 1 & 3, you get a very simple equation ($ x^{2}= \dots$ ) and rest is easy. [/hide]


1 Find such all (p,q), where p and q are prime numbers so that,
a. 3p-q=30
b. p^2-q=287
God bless you.
NB : May I ask you who is from Indonesia among all of the users here ?

[quote="Qruni"]1 Find such all (p,q), where p and q are prime numbers so that,
a. 3p-q=30
[/quote]
$ 3p-30 = q$
$ 3(p-10)=q$
So, $ 3|q$ and so $ q=3 \implies p=13$.

The second problem is easy. Note, $ q=2$. Why?


Let $\triangle ABC$ is a triangle with circumcentre $O$ and orthocentre $H$ Prove that with any $M \in$ circle diameter $OH$ then
$MG^2-OG^2=\frac{MO^2}{3}$

Let $G$ be the centroid of triangle.We know that $G \in OH$ and $HG|OG=2$.Let $GH=2OH=2x$.Then
$MO^22x+MH^2x=3x(MG^2+2x^2)$
$2MO^2+MH^2=3MG^2+6x^2$
$2MO^2+9x^2-OM^2=3MG^2+6x^2$
$\frac{MO^2}{3}=MG^2-x^2$
$\frac{MO^2}{3}=MG^2-OG^2$
That's all ;)


Hello. I just got back my score for the AMC 10A and I got a 124.5. But after I wrote the contest I calculated my score based on the answers and was almost sure that I had a 130.5. Even though 6 may not seem like a big difference, in my school 130.5 would have been the highest with two other people now beating me with a 126 and 129. Do you recommend anything, because this is my first time writing the AMC. Thanks.

Did you circle all your answers in your booklet?

You can check the AMC website to make sure that everything that you circled matches up with the answers.  Then you can calculate your score that way (6 points correct, 1.5 points blank, and 0 points wrong).

It seems like you
1. Miscalculated your score
2. Misbubbled an answer because losing 6 points means you missed a question that you thought you got right.
This does however mean that you didnt skip/omit a question you thought you had right because that would be only a 4.5 swing.

But I wouldnt worry too much about AMC school winner...what I think you should be more concerned about is your index score (6 points can make a big difference), so you should calculate it first by comparing answers with your booklet answers.

Then if it still doesnt match up and you are willing to spend some money (25 or so?), you can have the AMC regrade it.

Ask your teacher or contest manager for the record of what we actually read from your answer sheet.
It's in the email we sent to the teacher/contest manager, at the end in the CSV file.  Compare to what you wrote down.   Calculate your score form that record.   Remember however, the official answers are the blackened circles on the scantron sheet.

Steve Dunbar
MAA Director, American Mathematics Competitions

Thank you for your help. I will ask my teacher.


Plz help me solve this problem:
 A cake is sliced roughly in half, the largest piece being chosen each time, the other pieces discarded. We will assume that a random cut creates pieces of proportions
  P={(2/3,1/3) with prob. 3/4; and (2/5,3/5) with prob. 1/4}

Thus, for example, the first cut(and choice of largest piece) may result in a piece of size 3/5. Cutting and choosing from this piece might reduce it to size (3/5)(2/3) at time 2, and so on. How large, to first order in the exponent, is the piece of cake after n cuts?

I tried it using the law of large no.s but i am getting strange answers(the cake is getting bigger after a few cuts??). Plz help!

define a random variable $ X$ by: $ P(X\equal{}\frac{2}{3})\equal{}\frac{3}{4}\equal{}1\minus{}P(X\equal{}\frac{3}{5})$. If the inital volume of the cake is $ V_0$ and $ X_1, X_2, \ldots, X_k$ are iid random variables with same law as $ X$, then the size of the cake after $ n$ cuts is precisely: $ V_n \equal{} V_0 X_1 X_2 \ldots X_{n}$. As you said, you can use the Law of large numbers to show that $ \frac{\ln(X_1) \plus{} \ldots \plus{} \ln(X_n)}{n} \to E[\ln(X)] \equal{} \mu_X$ almost surely. Hence, almost surely:
$ V_n \equal{} V_0 \exp(\minus{}n (|\mu_X| \plus{} o(1))$
because $ \mu_X<0$. Where is your problem ?

Thank you.. i had done a similar thing.. only diff was i took logarithms to the base 2.. and i got my answer to be 2^(-nH(x)) where H(x) is defined to be the entropy of the rv. x = -E(logp(X)).{strange ans was bcoz of a silly mistake in the exponent sign}.

Just a question though.. How do u say that the answer so obtained is within the first order of the exponent of the actual cake size after n cuts?


x,y,z are positive integers.prove that:
8xyz>=(x+y+z-xy/z)(x+y+z-xz/y)(x+y+z-zy/x)

[quote="behdad.math.math"]x,y,z are positive integers.prove that:
$ 8xyz\geq(x\plus{}y\plus{}z\minus{}xy/z)(x\plus{}y\plus{}z\minus{}xz/y)(x\plus{}y\plus{}z\minus{}zy/x)$[/quote]
It's true even for all real positive numbers. :wink:

[quote="behdad.math.math"]x,y,z are positive integers.prove that:
8xyz \ge\ (x+y+z-\frac{xy}{z})(x+y+z-\frac{yz}{x})(x+y+z-\frac{xz}{y})[/quote]
Assum $ a\plus{}b\plus{}c\equal{}1$

$ Ineq \leftrightarrow 8x^2y^2z^2 \ge\ (1\minus{}xy)(1\minus{}yz)(1\minus{}zx)$

This is an old problem which have been post in ML . And I know it 's problem of tranvanluan


algebra
polynomial
abstract algebra
advanced fields
advanced fields unsolved

Let $ k$ be a field, $ k'$ any extension of $ k$, $ f(X)$ an irreducible polynomial over $ k$, and $ \alpha$ a root of $ f$. Show that $ k(\alpha)\otimes k'$ is isomorphic, as a $ k'$-algebra, to $ k'[X]/(f(X))$.

I'm having trouble constructing a homomorphism $ k'[X]\rightarrow k(\alpha)\otimes k'$ that is surjective.

send $ x$ to $ \alpha \otimes 1$ and send $ k'$ to $ 1 \otimes k'$.  This makes a surjective map $ k'[x] \rightarrow k(\alpha) \otimes k'$  You can verify that $ f(x)$ is in the kernel of this map.

more generally: $ A/I \otimes_R B/J \cong (A \otimes_R B)/ \langle I \otimes B , A \otimes J \rangle$ for (commutative, unital) $ R$-algebras $ A,B$ and ideals $ I;J$ of $ A$ resp. $ B$.

in your problem, put $ R \equal{} k, A \equal{} k[x], I \equal{} (f), B \equal{} k', J \equal{} 0$. we don't have to assume that $ f$ is irreducible.

Why do we need not assume that $ f$ is irreducible?

well you don't need it in the proofs. this is especially clear when looking at the generalization above.

The irreducibility is important for the isomorphism in the end; without it, we merely have a surjective map from $ k'[X]/(f(X))$ to $ k(\alpha)\otimes k'$. The kernel of the map is generated by $ \frac{f}{g}$, where $ g$ is the minimal polynomial of $ \alpha$.

ah, I immediately replaced $ k(\alpha)$ with $ k[x]/f$ ...
but I mean, the essence of this isomorphism does not have to do anything with irreducibility (again, look at the generalization).


ratio
geometry
MATHCOUNTS
geometric series

Don't do this if you already know how!  :)

What does $ 1 \plus{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \frac {1}{16} \plus{} \frac {1}{32} \plus{} \frac {1}{64} \cdots$ equal? How is it that you can add an infinite number of numbers and come up with a finite result?

What does $ \frac {3}{4} \plus{} \frac {3}{16} \plus{} \frac {3}{64} \plus{} \frac {3}{256} \plus{} \frac {3}{1024}\cdots$equal?

hint: What happens if you multiply that whole thing through by 4?

Try come up for a general formula for these types of sums.

These are called [b]geometric series[/b].  :)

Just wanted to say, that this probly doesn't belong in classroom math....

And i know how to do these

Just trying to get people to open their mind :)

And it is fun to think about, even when you don't understand it  :) .

[hide="This will give it away"]Well, for the first one, each time you add the next term, you halve the distance to two, so if this goes on infinitely, you will get 2. Note: Because you cannot actually add infinitely, this will never reach 2, but merely approach it. This gets into limits, which basically says, in this case, that as the number of terms approaches infinity, the sum of the terms will approach 2. (For those of you who are geometrical learners, graph this to see that 2 is an asymptote.)

The second one: Same principle. Each time, you get closer and closer to 1 (You quarter the remaining distance), so as we add more and more terms, we will approach 1.

Before you look at the formula below, see [url=http://en.wikipedia.org/wiki/Geometric_series#Example]this[/url] to understand why the formula works (or at least what it's based on), and what the hint means.
The general formula for this is $ \displaystyle\sum_{k\equal{}0}^{\infty}ar^k\equal{}\frac{a}{1\minus{}r}$
Here, $ a$ will be the first term of the series, and $ r$ will be the geometric ratio (the number you multiply by each time).

The first series can be solved like this: $ \displaystyle\sum_{k\equal{}0}^{\infty}1\cdot\frac{1}{2}^k\equal{}\frac{1}{1\minus{}\frac{1}{2}}\equal{}\frac{1}{\frac{1}{2}}\equal{}2$

And the second: $ \displaystyle\sum_{k\equal{}0}^{\infty}\frac{3}{4}\left(\frac{1}{4}^k\right)\equal{}\frac{\frac{3}{4}}{1\minus{}\frac{1}{4}}\equal{}\frac{\frac{3}{4}}{\frac{3}{4}}\equal{}1$[/hide]

[quote="Textangle"]How is it that you can add an infinite number of numbers and come up with a finite result?[/quote]
This is a much more subtle point than it, at first, appears.  The Greeks, in particular, believed that it was not possible to add an infinite number of numbers and come up with a finite result; think [url=http://en.wikipedia.org/wiki/Zeno%27s_paradoxes#The_dichotomy_paradox]the dichotomy paradox[/url].  This is something that, before you understand what it really means to sum an infinite series, you have to take on faith.

[quote="Textangle"]What happens if you multiply that whole thing through by 4?[/quote]
The underlying assumption here is that you can multiply an infinite number of numbers by $ 4$ and the "sum" (whatever that means) will increase by a factor of $ 4$.  Again, before you understand what it means to sum an infinite series, you have to take this on faith.

As long as we're taking the Socratic approach, here's a question for you:  what is $ 1 \minus{} 1 \plus{} 1 \minus{} 1 \plus{} 1 \minus{} 1 \pm ...$?  ;)

[hide="Several answers"] If you manage to come up with the formula $ 1 \plus{} r \plus{} r^2 \plus{} ... \equal{} \frac{1}{1 \minus{} r}$, the answer you get is $ \frac{1}{2}$.  But that's ridiculous, isn't it?

On the other hand, you can write $ (1 \minus{} 1) \plus{} (1 \minus{} 1) \plus{} ... \equal{} 0 \plus{} 0 \plus{} ... \equal{} 0$.  That doesn't agree with our previous answer!

On the third hand, you can write $ 1 \plus{} (\minus{}1 \plus{} 1) \plus{} (\minus{}1 \plus{} 1) \plus{} ... \equal{} 1 \plus{} 0 \plus{} 0 \plus{} ... \equal{} 1$.  That doesn't agree with either of our previous answers!

Clearly [i]something is wrong with our intuition about the way infinite sums behave.[/i]  If you're curious about the resolution to [url=http://en.wikipedia.org/wiki/1_%E2%88%92_1_%2B_1_%E2%88%92_1_%2B_%C2%B7_%C2%B7_%C2%B7]this paradox[/url]... [/hide]
[quote="Textangle"]These are called [b]geometric series[/b].  :)[/quote]
Why?  This doesn't look like geometry at all.  :)

What is the answer to the second one? That must have been a real poser, because the value fluctuates...
Let's try series: $ \displaystyle\sum_{k\equal{}0}^{\infty}1\cdot\minus{}1^k\equal{}\frac{1}{1\plus{}(\minus{}1)}\equal{}\frac{1}{0}$
Division by 0 is undefined, which would explain why this problem has no solution (or at least, not a universally accepted one, because different solutions will give you different answers).

As for the name, I'm not quite sure, but I think it's because you could take a square, for example, and with a segment, divide it into halves, divide a half into halves, etc., and go on infinitely, but still have the whole square. This matches the solution to the first problem, if you let the largest triangle be of area 1.

That's what you get if you try to sum $ \sum_{k\equal{}0}^{\infty} 1^k$.  What you get if you sum $ \sum_{k\equal{}0}^{\infty} (\minus{}1)^k$ is, as I have stated, $ \frac{1}{1 \minus{} (\minus{}1)} \equal{} \frac{1}{2}$.  For the resolution, see the link.

[quote="t0rajir0u"]
Why?  This doesn't look like geometry at all.  :)[/quote]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=173294[/url].

Heh, for some reason this is amusing. It seems t0rajir0u is in every single thread I seem to look up :)

I still need to press this point further (it is never too early to start thinking like a mathematician!).  What if I let $ S \equal{} 1 \plus{} 2 \plus{} 4 \plus{} 8 \plus{} ...$?  Then

$ 2S \equal{} 2 \plus{} 4 \plus{} 8 \plus{} ... \equal{} S \minus{} 1 \Leftrightarrow$
$ S \equal{} \minus{} 1$

but how can that possibly make any sense?  Again, [i]something is wrong with our intuition of the way infinite sums behave.[/i]  (Please don't read either of the first two responses if you are one of those people who insists that the Mathcounts forum is for Mathcounts-level material.)

[hide="Answer 1"] Of course, the reason it doesn't make sense is exactly because it violates the definition of convergence:  the partial sums $ \sum_{k \equal{} 0}^{n} 2^k$ are obviously not bounded and hence this series does not [url=http://en.wikipedia.org/wiki/Convergent_series]converge[/url]. [/hide]
[hide="Answer 2"] There is a sense in which this identity can be made to work; it requires working with a different notion of convergence called the [url=http://en.wikipedia.org/wiki/P-adic_number]$ 2$-adic metric[/url].  [/hide]
[hide="The mathcounts-level response I can accept"] It is not hard to figure out the [i]finite[/i] geometric series formula

$ 1 \plus{} r \plus{} r^2 \plus{} ... \plus{} r^n \equal{} \frac {1 \minus{} r^{n \plus{} 1}}{1 \minus{} r}$.

If $ |r| < 1$ then we expect that as $ n$ becomes large $ r^{n \plus{} 1}$ becomes very, very small.  Hence if $ n$ goes to infinity we should expect that that term disappears and the resulting sum should be equal to $ \frac{1}{1 \minus{} r}$.

This logic clearly does not hold for the cases $ r \equal{} \minus{} 1, r \equal{} 2$ that I have been trying to confuse all of you with; the term $ r^{n \plus{} 1}$ sticks around and can't be ignored. :) [/hide]

I woke up in the middle of the night last night, and realized that I screwed up. My previous post should say $ \frac{1}{1\minus{}(\minus{}1)}\equal{}\frac{1}{2}$, as t0rajir0u said.  :blush:


General: $n^n | (n-1)^{n^{n+1}}+(n+1)^{n^{n-1}}$ if $n$ is odd and squarefree.

Proof:
If $p$ is an odd prime and $p^k|a-b$ ($k \geq 1$), we get that $p^{k+m}|a^{s p^m}-b^{s p^m}$ for all $m,s$.

Set $a=(n-1)^{n^2}$ and $b=-(n+1)$, then $a^{n^{n-1}}+(-b)^{n^{n-1}}$ is the desired number. Since $a-b = (n-1)^{n^2} + (n+1) \equiv (-1)^\text{odd} +1 = 0 \mod n$, thus $p|a-b$ for all primes $p|n$.
This now gives:
$p^n|a^{s p^{n-1}}-b^{s p^{n-1}}=(n-1)^{n^{n+1}}+(n+1)^{n^{n-1}}$
with $s=\left(\frac{n}{p}\right)^{n-1}$.
the result follows.

If $ n$ is even or not square-free, is there contradiction?

If $ n$ is even, then $ 4|n^n$, but $ (n\pm1)^{n^{n\mp1}}$ are both 1 mod 4, so their sum is not divisible by $ 4$.

So $ n$ must be odd, but the square-free condition is unnecessary.  By binomial expansion it suffices to show that for $ 1\leq k < n$, $ \left(n^{n\pm1}\atop k\right)$ are divisible by $ n^{n \minus{} k}$.

Let $ \alpha \equal{} \nu_p(n)$, i.e., $ p^\alpha \parallel{} n$.  Then $ \left(n^{n\pm1}\atop k\right) \equal{} \frac {n^{n\pm1}}{k}\left(n^{n\pm1} \minus{} 1\atop k \minus{} 1\right)$, so it suffices to show

$ \alpha(n \minus{} k) \leq \alpha(n \minus{} 1) \minus{} \nu_p(k)$,

and for this it suffices to observe

$ \nu_p(k) \leq \log_p(k) \leq \log_2(k) \leq k \minus{} 1 \leq \alpha(k \minus{} 1)$.

So the result is true for all odd $ n$ (Mathematica 5's buggy PowerMod notwithstanding).

Isn't this trivial, after some simple algebra to make the exponents equal, by LTE?

@Wolstenholme That's essentially what ZetaX's solution is doing.


geometry
perpendicular bisector
geometry solved

Given a triangle $ABC$, and let $M$ be the midpoint of its side $AB$. The perpendicular bisectors of the segments $AM$ and $BC$ intersect at $P$. The perpendicular bisectors of the segments $AC$ and $BM$ intersect at $Q$. Prove that the lines $PQ$ and $CM$ are perpendicular.

This one is nice and easy. In fact, remember the following theorem:

[i][b]Theorem 1.[/b] Given four distinct points A, B, C, D in the plane, the lines AB and CD are perpendicular if and only if $CA^2 - DA^2 = CB^2 - DB^2$.[/i]

Now to the solution of your problem: By Theorem 1, in order to show that the lines PQ and CM are perpendicular, it is enough to prove $CP^2 - MP^2 = CQ^2 - MQ^2$.

Now, since the point P lies on the perpendicular bisectors of the segments AM and BC, we have AP = MP and BP = CP. Hence, $CP^2 - MP^2 = BP^2 - AP^2$. Now, let V be the midpoint of the segment AM. Then, $\displaystyle AV = \frac12 \cdot AM = \frac12 \cdot \frac12 \cdot AB = \frac14 AB = \frac14 c$. Also, $\displaystyle BV = AB - AV = c - \frac14 c = \frac34 c$. But the point P lies on the perpendicular bisector of the segment AM, i. e. on the perpendicular to the line AM through the midpoint V of the segment AM. Hence, the lines PV and BA are perpendicular. Thus, after Theorem 1, we have $BP^2 - AP^2 = BV^2 - AV^2$.

Therefore, $\displaystyle CP^2 - MP^2 = BP^2 - AP^2 = BV^2 - AV^2 = \left(\frac34 c\right)^2 - \left(\frac14 c\right)^2$. Similarly $\displaystyle CQ^2 - MQ^2 = \left(\frac34 c\right)^2 - \left(\frac14 c\right)^2$. Hence, $CP^2 - MP^2 = CQ^2 - MQ^2$, and we are done.

  Darij

Based on the same idea, but slightly shorter: let $T$ be the midpoint of $PQ$. We have $CM\perp PQ\iff PC^2+QM^2=QC^2+PM^2\iff PB^2+QB^2=PA^2+QA^2$, and this is obvious because in the triangles $APQ,BPQ$ the medians $AT,BT$ are equal. That's because $T$ is on the perpendicular bisector of $P'Q'$, where $P',Q'$ are the projections of $P,Q$ on $AB$.


induction
superior algebra
superior algebra unsolved

find the roots of polynom f

$f=1+\sum_{k=1}^{n}\frac{x(x+1)...(x+k-1)}{k!}$


Whenever n is a positive integer, for all nonnegative reals a_1, a_2, ……, a_n, find the minimum of

\sum_{i = 1}^{n}{\frac {a_i}{a_(i+1)+a_(i+2)}}

we suppose that a_(n+1)=a_1 and a_(n+2)=a_2


A sack contains blue and red marbles*. Consider the following game: marbles are taken out of the sack, one-by-one, until there is an equal number of blue and red marbles; once the number of blue marbles equals the number of red marbles, the game is over. In an instance of this game, it is observed that, at the end, $ 10$ marbles were taken out of the bag, and no $ 3$ consecutive marbles were all of the same color. Prove that, in said instance of the game, the fifth and sixth marbles were of different color.

*The original problem involved "stones."

[hide="rules"]
The rules are...
total of 10 marbles, red or blue
no 3 consecutive marbles taken out are the same color
the last x marbles cannot have $ x/2$ red and $ x/2$ blue marbles
my spot notation..._ _ _ _ _ _ _ _ _ _   1 is the first underscore, 10 is the last marble chosen
[hide="Case 1"]We are assuming that we will have enough marbles because the Cases will cover all cases (in other words, I was going to write a completely rigorous proof but I decided not to)
Case 1:  The middle two are blue.
It now looks like _ _ _ R B B R _ _ _
Case 1a:  B is the last one taken out.
The second to last taken out has to be blue or the game would have ended sooner.  Spot 8 has to be red to avoid the 3 consecutive rule.  However, because the last four marbles have an equal number of marbles, we have a contradiction.
Case 1b:  R is the last one taken out.  We need an a red in spot 9.  We need a blue in spot 8.  However, the last 6 marbles (spots 5-10) have an equal number of blue and red marbles, a contradiction.  Case 1b done.
Case 1 done.
[hide="Case 2"]Case 2:  The middle two are red.
_ _ _ B R R B _ _ _
Case 2a:  B is the last one taken out. Therefore, a blue goes in spot 9, a red goes in spot 8.  However, the last six marbles have an equal number of blue and red marbles again, a contradiction.  Case 2a done.
Case 2b:  R is the last one taken out.  Therefore, a red goes in spot 9, a blue goes in spot 8.  However, the last four marbles have an equal number of blue and red marbles, a contradiction.  Case 2b done.
Case 2 done.
We have covered all the cases, regardless of how many of each marble we have.  Therefore, because we have proved that the middle two cannot be the same color, they must be different colors.
QED  :)
 [/hide] [/hide][/hide]
Just out of curiosity, what's PRMO?

[url=http://ifem.math.uprm.edu/olimpiadas/index.php?modulo=home.php]Puerto Rican Mathematical Olympiad[/url] (It's actually OMPR, when you abbreviate its Spanish name, but 'PRMO' sounds cooler). See [url=http://www.mathlinks.ro/Forum/weblog_entry.php?t=216541]here[/url], if you don't "speaky the Spanish."

EDIT: [hide="Visual Proof"][img]http://img514.imageshack.us/img514/4376/marblesza2.gif[/img][/hide]

Is it necessary to prove both cases, or can you just assume WLOG that the two middle marbles are red?

Not really — in the "solution" I wrote for the exam, I just drew black and white dots, and generalized (I didn't say "WLOG," but I did implement the concept it implies). So, sure, generalize.


Woah. This was actually very well written. How did only a very little amount of people see this? From a first glance, Problem 23 looks very "cool", but at a deeper glance, it's way too easy IMO.

The last time this user was online was in 2013. :?[quote=AdiK]do you have the answer key?[/quote]

[quote=Possible]I would like to point out an error for 1. The question should write expected value.[/quote]

No, the question said that there is an equal number of correct answers for each answer choice.

[quote=AdiK]do you have the answer key?[/quote]

Here are the answers: [url]https://artofproblemsolving.com/community/c5h213727[/url]

[quote=Possible]Also, 21 never specifies if the dominos are a certain size or not.[/quote]

Yeah, that's true though.


Mafia
probability
calculus
Putnam
email
blogs
linear algebra

Soooo.  This thread is dedicated to randomness.  Or discussion of any topic.

42 days till I go to WATERLOO!!!!

SSSSSOOOOOOOOOOOOOOOO EXCITED!!!!!
The twins will rule again.  Mwuhahahaha.

Random as in this thread??  :maybe:  :D

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=103022[/url]

Small fonts, dedicated to last year's Seminar adventures!
Umm.. that's random!  :wink:

[size=59]So......how much wood can a woodchuck chuck if a woodchuck could chuck wood?  Well, anyone seen the latest Pirates of the Caribbean movie?  Whatcha think?

Hey who went to the Student life 101 other than Jenn?  Anything maybe of interest?[/size]

[quote="jpark"][size=34]David is the mafia!! No wait, maybe it's Victoria! Actually, it's Ian![/size][/quote]

[size=34]I'd like to see you try and wear 4 different seminar t-shirts   :wink:

BTW, Ian is like...the Godfather, we just have to figure out who his minions are.  Or....does he work alone?   :huh: [/size]

[size=34]Why don't you try first?  :wink: If you can wear 5 different shirts, I can wear 4![/size]

[size=59]I actually think I lost or left behind my t-shirt from 2004. lol...I dunno, I don't remember every seeing it at home.  Soo...it'd only be 4.   :lol: [/size]

[size=34]Darn. Oh well. Another Random post.
In the wonderful world of Waterloo, there are two primary life forms: geese and a Moose. The Moose speaks the truth all the time, and geese have 1/3 chance of speaking the truth. If a Wildlife Terrorist encounters 3 creatures, exactly one of which is the Moose, what is the probability that the terrorist actually finds the Moose, by asking the question "which is the Moose?"

Assume that the Terrorist is not very bright and cannot tell the difference between a Moose and a goose just by looking at it.[/size]

[size=59]WHOA hey anyone remember the Brazilian duck!!!!  oh wait...that might have at a seminar too long ago....hmm.  Oh well.   Hey, if you could have any pet in the world, what would it be? [/size]

First of all, I can't wait until you guys come to waterloo too!! :lol:  But you do have to open your mind a little... when you get to Waterloo, there's no curfew anymore, and there's more to life than mao and mafia.  In fact, the fun times really never end, if you know the right people ;).  The academic part is like 1/4 of the experience.

And like... this thread... OMGWTFBBQ?!?!?!?

But GAHHH... I HAVE FINALS!!!  And co-op in the fall, which sucks cuz I won't see you guys all the time... however, I'll probably be coming to waterloo at least a couple times a week for club-related stuff... and we should plan some fun times.

You should all join the juggling club; it's the BEST club ever, and they let you use all their stuff (including poi, clubs, devil sticks, cigar boxes, unicycles, diablos, staffs, contact balls... you name it!).  Sarah, there IS a fencing club here... lol.

And 42 is 101010 in binary!  WOOT!

[quote="Elyot"]
You should all join the juggling club; it's the BEST club ever, and they let you use all their stuff (including poi, clubs, devil sticks, cigar boxes, unicycles, diablos, staffs, contact balls... you name it!).  Sarah, there IS a fencing club here... lol.

And 42 is 101010 in binary!  WOOT![/quote]
LOL no worries, I definitely didn't plan on playing mafia and mao all day.
Oh I saw the fencing club.  Hmm, I'm not sure if i'm gonna do it, I'll check it out and see.  My coach wants me too, lol so I'll take a look

What happened to the Cheese club? I definitely remember you referring to it as the "BEST CLUB EVER" several time..  :D

[quote="Elyot"].

You should all join the juggling club; it's the BEST club ever, and they let you use all their stuff (including poi, clubs, devil sticks, cigar boxes, unicycles, diablos, staffs, contact balls... you name it!).  Sarah, there IS a fencing club here... lol.

[/quote]

do u know Paul? he's in both cheese club and juggling club. Joe West taught me calculus in 1B too. he's the BEST lecturer i've had so far!

:D PAUL yes!!!  Couple years ago he let me play with his juggling stuff, lol.

[quote="Elyot"]Dun worry Jenn... when Sarah has to take this in 2A, she'll tell you how cool it is.[/quote]

I don't mind listening to people talking about classes as long as I don't actually have to take it  :D

Maybe I should start talking about all the ugly, boring courses you'll have to take if you go for pure math, or all the exciting, interesting courses you'll get to take if you go for applied math!

Seeing as how you haven't taken any pure math courses, I don't think I'll even listen to that   :P

[quote="Sunny"]AH HAHAHAHAHAHA
GUESS WHO PASSED HER FRENCH IB EXAM

MUWHAHAHAHAHAHAHA
Wow, amazing.  like....wow.  :D[/quote]

Heheh... come on... as if you'd fail it.  ;)  Congrats!

Now you've just got a few years of painful courses at Waterloo to look forward to!

[quote="Elyot"]
Now you've just got a few years of [b]fun[/b] at Waterloo to look forward to![/quote]

[b]FYP[/b]

of course it'll be fun! You'll be there, Alex!!  :D

Hehe, I didn't know AoPS had blogs until now!!

A random comment for the random thread..  :blush:

[quote="Elyot"]

Heheh... come on... as if you'd fail it.  ;)  Congrats!

Now you've just got a few years of painful courses at Waterloo to look forward to![/quote]

Lol....funny...actually I was in danger of failing that one.  Umm...my french teacher is....umm...well.  :P  Some interesting stories.  Let's say she ends up using the dictionary more than her students!

LOL even if i tried to I can't fit it into my schedule.  I do want to go to France though...oh boy, I definitely want to.

RANDOM PICTURE:

[img]http://www.rit.edu/~andpph/photofile-c/exploding-egg_9893.jpg[/img]

Cool, eh?  It's an exploding egg.

Everyone should go see Snakes on a Plane. I caught an early showing and it's a real spectacle

I can think of a scene in the movie that corresponds to each of those pictures :D


inequalities
LaTeX
inequalities unsolved

I think you missed something when coding in latex, because from what you wrote it's somehting like that:
$ xyz\equal{}8^{3}$
Prove:
$ (\frac{1}{1\plus{}x})^{\minus{}1}\plus{}(\frac{1}{1\plus{}y})^{\minus{}1}\plus{}(\frac{1}{1\plus{}z})^{\minus{}1}\geq 1$
Which of course is equivalent to
$ 1\plus{}x\plus{}1\plus{}y\plus{}1\plus{}z\geq 1$
$ x\plus{}y\plus{}z\geq \minus{}2$ Which of course is not necessarily true, for ezample $ x\equal{}y\equal{}\minus{}8 \wedge z\equal{}8$ satisfy the condition but the inequality is not true for those numbers.

Yes:D i did make a mistake.
The right formulation is:

(1/(1+x))^(1/2) + (1/(1+y))^(1/2) + (1/(1+z))^(1/2),    where x,y,z are positive reals and   xyz=8^3

I think Jensen works

Let $ f(m) = \sqrt {\frac {1}{1 + x}}$ (which is convex)

Hence

\begin{align*} f(x) + f(y) + f(z) & \ge 3 f(\frac {x + y + z}{3}) & = 3\sqrt {\frac {3}{3 + x + y + z}} & = \sqrt {\frac {27}{3 + x + y + z}} \end{align*}

By AM-GM

$ x + y + z \ge 3 \cdot \sqrt [3]{8^3} = 24$

Hence $ \sqrt {\frac {27}{3 + x + y + z}} \ge 1$

And we are done

Hehehe, I tried very similarly and at one moment I made the same mistake as you did. (check the "obvious" inequality at the end. A similar "pseudo" proof ends in the same moment as your's:
$ \sqrt {\frac {1}{1 \plus{} x}} \plus{} \sqrt {\frac {1}{1 \plus{} y}} \plus{} \sqrt {\frac {1}{1 \plus{} z}}\geq 3\sqrt {\frac {1}{\sqrt [3]{(1 \plus{} x)(1 \plus{} y)(1 \plus{} z)}}}$
because $ \sqrt [3]{(1 \plus{} x)(1 \plus{} y)(1 \plus{} z)}\leq \frac {x \plus{} y \plus{} z \plus{} 3}{3}$ thus
$ 3\sqrt {\frac {1}{\sqrt [3]{(1 \plus{} x)(1 \plus{} y)(1 \plus{} z)}}}\geq 3\sqrt {\frac {3}{x \plus{} y \plus{} z \plus{} 3}}$
Unfortunately we are both wrong in the end :wink:

For now I'm giving up. The only thing I thought of that makes some sense is that our inequality will be proved if to the following inequality is proved:
$ \sum_{cyc}\sqrt {\frac {1}{x \plus{} y \plus{} 2}}\geq \frac {\sqrt {2}}{2}$
If this helps I can post a short proof for that.

If we look at x=e^a and such, we get a problem that we can attack with jensen (there is one inflection point so it is not trivial).

$ \because xyz\equal{}8^3$ $ \therefore (\frac{x}{8})(\frac{y}{8})(\frac{z}{8}) \equal{} 1$
$ \frac{x}{8}\equal{}a,\frac{y}{8}\equal{}b,\frac{z}{8}\equal{}c$
$ \therefore abc\equal{}1$
\[ L.H.S. \geq 1 \leftrightarrow \sum_{cyc}\sqrt{\frac{1}{1\plus{}8a}} \geq 1\]
substitution $ a\equal{}\frac{lm}{n^2},b\equal{}\frac{mn}{l^2},c\equal{}\frac{ln}{m^2}$
\[ \therefore \sum_{cyc}\sqrt{\frac{1}{1\plus{}8a}} \geq 1\]
\[ \leftrightarrow  \sum_{cyc}\frac{n}{\sqrt{n^2\plus{}8ml}} \geq 1\]
but \[ \frac{n}{\sqrt{n^2\plus{}8ml}} \geq  \frac{n^{\frac{4}{3}}}{l^{\frac{4}{3}}\plus{}m^{\frac{4}{3}}\plus{}n^{\frac{4}{3}}}\]
\[ \therefore \sum_{cyc}\frac{n}{\sqrt{n^2\plus{}8ml}} \geq 1\]
\[ \therefore L.H.S. \geq 1\]

Ok this last post is corect, bu i knew that solution and i would like to see another one, without using the inequality appolo has used.
Arquady  help!!

I think, the apollo's way ( Hoojoo Lee  :wink: ) is very nice and sharp:
Let $ \frac {1}{\sqrt {1 \plus{} 8a}}\geq\frac {a^{\alpha}}{a^{\alpha} \plus{} b^{\alpha} \plus{} c^{\alpha}}$ is true. We obtain
$ \frac {1}{\sqrt {1 \plus{} 8a}}\geq\frac {a^{\alpha}}{a^{\alpha} \plus{} b^{\alpha} \plus{} c^{\alpha}}\Leftrightarrow b^{2\alpha} \plus{} c^{2\alpha} \plus{} 2(ab)^{\alpha} \plus{} 2(ac)^{\alpha} \plus{} 2(bc)^{\alpha}\geq8a^{2\alpha \plus{} 1}.$
But by AM-GM we obtain:
$ b^{2\alpha} \plus{} c^{2\alpha} \plus{} 2(ab)^{\alpha} \plus{} 2(ac)^{\alpha} \plus{} 2(bc)^{\alpha}\geq8\sqrt [8]{a^{4\alpha}b^{6\alpha}c^{6\alpha}} \equal{} 8a^{ \minus{} \frac {\alpha}{4}}.$
$ \minus{} \frac {\alpha}{4} \equal{} 2\alpha \plus{} 1\Leftrightarrow\alpha \equal{} \minus{} \frac {4}{9}.$
Thus, $ \frac {1}{\sqrt {1 \plus{} 8a}}\geq\frac {a^{ \minus{} \frac {4}{9}}}{a^{ \minus{} \frac {4}{9}} \plus{} b^{ \minus{} \frac {4}{9}} \plus{} c^{ \minus{} \frac {4}{9}}}$ and done.

[quote="primoz2"]I have nothing against this solution (it is also the original solution from the 2002 shortlist), but i thought there might be a nicer solution, without using this method.[/quote]
The Jensen's inequality kills it easy enough. But it's very known too.


Four pieces of paper, one of each of the colors red, yellow, blue, and green, are randomly selected without replacement. What is the probability that the yellow piece is drawn first and the blue piece is drawn last? Express your answer as a common fraction.

whole outcome is 4!=24.

and if yellow and blue is drawn before, there's only 2 possibilities.

so its 2/24, which simplifies to 1/12.

answer : 1/12

[hide=Alternate Solution]
We find the probability of picking yellow on the first turn, then multiply this by the probability of not picking blue on the second turn, and multiply the product by the probability of not picking blue on the third turn. Thus, our answer is $\frac{1}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \boxed{\frac{1}{12}}$.  [/hide]


function
integration
topology
advanced fields
advanced fields unsolved

$C[a,b]$ is set of differentiable and continuous functions on $[a,b],A\subset C[a,b]$ and$A=\{x(t)|\int_{a}^{b}(|x(t)|^{2}+|x'(t)|^{2})dt \le c\},c=const>0$
prove that $A$ is compact set.
 :)

What topology (what metric) do you intend for the set $C[a,b],$ so that we know what "compact" means in this context?

ok here we use the metric
$d(x,y)=\max_{a\le t\le b}|x(t)-y(t)|$


Let arbitrary natural numbers label the vertices of a square. The number of any vertex can be replaced by the remainder of the division when the product of the numbers labeling any two other vertices is divided by a given prime p. (The three numbers are freely chosen out of the three.)
Prove that by applying the above rule for changing the numbers, using any prime p, in infinitely many steps we can reach the situation where the four numbers at the vertices are the same.

1) If one of these numbers is divisible by $p$ then it is clear we are able make four zeros. :)
2) Suppose all numbers are not divisible by $p$. Then we can perform following operations:
\[[a,b,c,d]\to[a,b,ab,ab]\to[ab,ab,(ab)^2,(ab)^2]\to[ab,ab,(ab)^2,(ab)^3]\to\]
\[\to[(ab,ab,(ab)^3,(ab)^4)]\to...\to[ab,ab,(ab)^{p-2},(ab)^{p-1}]=\]
\[=[ab,ab,(ab)^{p-2},1]\to[1,1,ab,(ab)^{p-2}]\to[1,1,1,1]\]
(we have used Fermat theorem).

That's all.


algebra
polynomial
search
number theory unsolved
number theory

let W(x) is a polynomial that W(x)=x^4 _3x^3 + 5x^2 _9x
find all different and integer numbers like a,b that W(a)=W(b)
thanks...

It will be helpful : [url]http://www.mathlinks.ro/viewtopic.php?search_id=1298744259&t=150626[/url]

oh dear im sorry but it wasnt useful at all


do u guys have any memorable/funny school stories?
i dont have any good one right now
there was an old thread about this, but the last post was over a month ago.

Quote from some 8th grade class president:

"Let's call ourselves the aminos, 'cause that's Spanish for friends!"  :D

[quote="ZhangPeijin"]Uhh...
This kid basically choked on his juice, and spat it all out.
It went EVERYWHERE.
So yeah.
It was disgusting...[/quote]

On a related note, a person in my 6th grade class ate strawberries in class, claiming that the teacher would not notice. He didn't......until my classmate threw up the strawberries somehow.

[hide="Gross"]it looked like blood and it smelled horrible.[/hide]

[quote="ZhangPeijin"]Umm.
Our health ed teacher decided to test our maturity,
So he just goes like.
<CENSORED>!
<CENSORED>!
<CENSORED>!
And the entire class was laughing like mad.
So he said we're all immature and gave us a F for the day.
:([/quote]

I wish that I had a teacher like that.  :rotfl:

[quote="ZhangPeijin"]Umm.
Our health ed teacher decided to test our maturity,
So he just goes like.
<CENSORED>!
<CENSORED>!
<CENSORED>!
And the entire class was laughing like mad.
So he said we're all immature and gave us a F for the day.
:([/quote]

wowww.... that's pretty sad...

wellllllll... i don't know if it's appropriate, but here it goes...

in English class, my friends and i were writing sentences on the board, and my friend was in the midst of writing "You look..." and so i said, "****". and she's like "okay, i'll write that"
and so she wrote that on the board, and my teacher asks, "Who wrote number 8?"
that isn't that funny...

the next day, we're talking about direct objects.. my friend who sits in front of me reads to the teacher, "The runner wore nothing." and i said, "Hey courtney, you should say something else.. something that relates to yesterday!" and she said, "**** shorts" we all started cracking up.

then our teacher said, "if someone walked by without knowing ANY context of this whatsoever, they will think we're perverts."

so Teddy (he's a bit of a pervert) says, "sweet! i'm shouting '**** shorts' to anyone who walks by!"
so this poor little Chinese kid walks by and teddy shouts it out the door... he walks by with the weirdest expression on his face. :D

Teacher: This sodium doesn't look too pure. I think I will add a bit more into the water.

Me: I don't think you should.

Teacher: Don't worry I know what I am doing.

[size=150]BANG!!!!!!!!!!!!!!![/size]

 :wink:  Told you so.

Haha... he probably isn't joking...
Hm this was pretty funny when it happened
I'm hugging a friend of mine, who has braces, in the hallway and...his mouth must have been opened or something because my hair got stuck in it  :huh:  we were both like wtf?
one of the pieces broke off heh
disgusting...

Well this isn't memorable but just on thursday, a coupld of kids I knew from school basically attacked me as I was going to the library.

Once there was this kid who threw up in the hallway. All it said was 'wet floor caution.' I say to my friend..."It may as well be a 'hazardous waste' sign. :rotfl:

Okay, so one day our physics teacher puts up a rocket engine in space problem on the board (you know, with the constant momentum and decreasing mass due to burning fuel) except he forgets to say at what rate the mass is going down. He says "okay I need to go out for a while" and then leaves, so we're all confused as to why the problem is not doable. Though we all forget about it five minutes later when he, (along with another physics teacher) is in a wagon, with a fire extinguisher (one that has CO2), and about to propel himself down the hallway with said extinguisher.

Oh yeah, he was wearing a superman cape also.

[quote="hm29168"]:rotfl: Was that a true story or what you wished would happen?[/quote]

It was true. Happened not long ago. I was the only one who wasn't surprised seeing that I saw it coming. LOL Teachers, can't teach them new tricks..........

It was also funny how she thought we are idiots despite the fact that we have two people who went to the Chemistry Olympiad Summer School, lol. Ironically the class is CH4 (Chemistry 4).

Ah, haha.
This was a bit embarrassing for me at the time... I'm sitting in a group of four in careers class, and I get up from my chair and quickly turn around. I trip on my tennis bag and quickly grab and  hold on to my rolling book bag for support.. it starts rolling and I fall..lol. TOO clumsy. :roll:  I get up laughing and say," I'm alright. I'm alright."

[quote="kimmystar94"][quote="illusiat"]all the chinese at my school call me a nerd and say i have no life, but the americans are like, "ur so smart i wanna be just like you please be my partner blahblahblah"

i think the chinese are jealous :lol:[/quote]

WELLL... mines sorta the opposite. its just some of the chinese don't care. and half the americans say you're amazing (as my friends) and the other half says, "freaking nerd, no lifer" (as my not so good relationship people). and when we have projects, everyone wants to be my partner. same for math games...[/quote]
Eh, same here.
Except they're both jealous.
:huh:

[quote="Nerd_of_the_Ages"]No one is jealous of me.
I'm glad.
Hurrah for the general sanity of society!

In 4th grade, my pants, ripped. 'Nuf said.

In 5th grade, I had to do a group writing assignment with lifeisacircle and another guy who refuses to have an AoPS account. We had to write about a story where your parents accidentally leave you in the supermarket. However, you must take turns writing. For example, I write one sentence, lifeisacircle writes one sentence, and so on We got a B for "too bizarre plot line." General plot line:

1. First we get stuck in supermarket
2. When everybody has left, we accidentally get an entire rack of 7-up bottles to fall on us, knocking us unconscious.
3. We wake up to find it is late night.
4. For no apparent reason, there is some kind of tiny robot security thing inside the building, and the door is locked.
5. We attempt to hijack it, but cause it to explode
6. Luckily, the explosion shatters the door
7. We escape
8. We find the police is chasing us for screwing around with stuff
9. We steal a Ferrari.
10. We drive away.
11. We find that we drove so fast that we are now in Earth's atmosphere
12. We accidentally keep on accelerating
13. For no apparent reason, we time travel and are now in ancient egypt.
14. The car is dead.
15. Random stuff occurs
16. For no apparent reason, there are 3 black holes in the sky, with ropes dangling out of them.
17 We climbed in
18. For no apparent reason, we are home

Really fun to do when the teachers claim that you're allowed to do whatever plot you want as long as you start out doing something, in this case being stuck in a super market. Just steal a car, make it have time travel, and have the bulk of your story in Egypt.[/quote]

Have you watched Monty Python?

Yes, but only one episode (on youtube). It was the one about the funniest joke in the world.

Why?

During a Geography Bee:

Question:
What continent has the least population?

Student's Answer:
Canada.

Yes, she is blonde.

Wow... just wow
haha last year whenever anaxerzia asked our science teacher a question, she'd always turn bright red.  :P
and the day i forgot to do all my homework was made into a HUGE deal... everyone knew about it lol
good times  :P

Um. In case anyone hasn't figured it out, I am a boring person. But my sister told me this (assumably true) story:

My sister has a male friend that has several friends she (my sister) doesn't know. Some other guy insulted one of my sister's friend's friends, and my sister's friend said, "You get between me and my friends and I get between you and your face!"

Un huh?
Uhh, after Math Practice one day, it was during winter and it was pretty dark, around 5.
These kids tackled me as soon as I stepped out of the front door.

[quote="Cyrazeno"]the 10 days I spent in Hawaii were simply awesome  :lol:[/quote]

For school?

I recall a friendly math competition in my math class and I complete crushed everyone. Won a lot of supporters that day  :P  and people still call me "Buddha."

I'm known for doing embarassing things and so are my friends. Once we were bored in class and one of my friends started clipping his fingernails. The teacher calls on him and tells him to stop reading his book and to pay attention, my friend says "im not reading a book, im clipping my fingernails". At first we all thought he was kidding but then he showed us the clippers everyone was laughing. Another time two years ago we were cleaning out are lockers on the last day of school. I found a rotten egg in my locker ( I think it was from an easter project, dont ask) and I tried throwing it in the trash can from about 20 feet away. I over shot it and it exploded over one of the cheerleaders heads. She smelled horrible for the rest of the day and so did the hall way since egg went everywhere.  :rotfl:  :rotfl:  :rotfl:


If $ secB=sinA+cosA$ show that

(i)$ tan^{2}B=sin2A$
(ii)$ cos2B=tan^{2}(\frac{\pi}{4}-A)$

[hide="Solution"]
i) $  \frac{1}{\cos A}=\sin A+\cos A\Leftrightarrow \frac{1}{\cos^{2}A}=(\sin A+\cos A)^{2}\Leftrightarrow$
$  \Leftrightarrow\frac{\sin^{2}A+\cos^{2}A}{\cos^{2}A}=1+\sin 2A\Leftrightarrow\tan^{2}A+1=1+\sin 2A\Leftrightarrow\tan^{2}A=\sin 2A$

ii) $ \cos 2A=2\cos^{2}A-1=\frac{2}{1+\sin 2A}-1=\frac{1-\sin 2A}{1+\sin 2A}$

$ \tan^{2}\left(\frac{\pi}{4}-A\right)=\left(\frac{1-\tan^{2}A}{1+\tan^{2}A}\right)^{2}=\left(\frac{\sin A-\cos A}{\sin A+\cos A}\right)^{2}=\frac{1-\sin 2A}{1+\sin 2A}$
[/hide]

Here's an even easier solution to number 1.

[hide]

$ \tan^{2}(b) = \sec^{2}(b)-1$ (pythagorean identity)

now substitute $ \sec^{2}(b)$ by $ (\sin(a)+\cos(a))^{2}$ and factor to get

$ \sin^{2}(a)+2\sin(a)\cos(a)+\cos^{2}(a)-1$ now use $ \sin^{2}(a)+\cos^{2}(a) = 1$ (again, pythagorean identity) to get

$ 2\sin(a)\cos(a)$

which equals the sine double angle formula $ \sin(2a)$.

[/hide]


A sum of 5000 dollars is invested at an interest rate of 9% per year. Find the time required for the money to double if the interest is compounded semi-annually?

Apparently we want to solve for $ x$ in \[ 5000\left(1\plus{}\frac{.09}{2}\right)^{2x}\equal{}10000\] where $ x$ is the time in years.  From here we use a calculator or something.

[quote="5849206328x"]Apparently we want to solve for $ x$ in
\[ 5000\left(1 \plus{} \frac {.09}{2}\right)^{2x} \equal{} 10000
\]
where $ x$ is the time in years.  From here we use a calculator or something.[/quote]

[hide="WARNING!  NOT CM LEVEL!"]Divide both sides by $ 5000$.

$ (1 \plus{} 0.045)^{2x} \equal{} 2$

$ (1.045)^{2x} \equal{} 2$

$ \log (1.045)^{2x} \equal{} 2$

$ 2x \log (1.045) \equal{} 2$

$ x \log (1.045) \equal{} 1$

$ x \equal{} \frac{1}{\log (1.045)}$[/hide]

Don't you need to take a logarithm of the other side too?

Oh oops, you are right, which means that the answer is totally different.


who knows the whole rank of the olympia?


geometry
3D geometry
rectangle
parallelogram
analytic geometry
geometry unsolved

Problem 1

Show that a cube can be dissected into n cubes for n>=55
-----------------------------------------------------------------------
Problem 2
Find with proof all possible values of the largest angle of a triangle
that can be decomposed into five disjoints triangles similar to it.
----------------------------------------------------------------------
Problem 3
A regular 4n-gone is decomposed into parallelograms.
Prove that there exist at least one retangle in the decomposition.
Find the sum of the areas of all rectangles from the decomposition.
(From Kvant (Quantum))

Hey moubi, when I read the first problem I remembered I nice problem from the romanian olympics this year for the 10th grade..although the problem is trivial not many people solved this(3 or 4).. at least in my county(judet)

In the interior of a cube we consider 2003 points. Show that we can divide in more than 2003^3 cubes so that any points from the given ones are inside one of the smallest cubes(not on its faces)..

cheers! :D

I think I have a soln for the first one, but it?s kind of hard to explain. Since any cube can be divided in 8 equal cubes, this means that if we divide it in n cubes, we can also divide it in n+7 cubes (we divide one of the 8 smaller cubes in 8 cubes). This means that it?s enough to prove that we can divide a cube in n cubes for n in the set {55, 56, 57, 58, 59, 60, 61}.

We can observe that a cube (we assume it has a unit side) can be divided in f(x)=3x^2+3x+2 cubes for any x>=1. For this we place a cube of side x/(x+1) in one of the corners of the initial cube and fill the rest with cubes of side 1/(x+1). If we count them we get 3x^2+3x+2. Then we can make the observation that if we can divide it in k cubes and in n cubes then we also can divide it in n+k-1 cubes (just like above for k=8). 59=f(3)+3*7, so we can divide it in 59 cubes. 55=f(2)+5*7. 57=1+8*7. Since f(2)=20 we can divide it in 20 cubes, so if we can divide it in n cubes we can also divide it in n+19 cubes. 60=1+2*19+3*7. 58=1+3*19. This only leaves 56 and 61. For 56 we do this: we place 4 cubes of side  which completely cover a face of the initial cube. Then, over these, we place 9 cubes of side 1/3. These completely cover the first 4 cubes. Then, over those we place 36 cubes of side 1/6 (the cube is filled since +1/3+1/6=1). We have placed 4+9+36=13+36=49 cubes. 49+7=56, so we got 56. I think I have sth that works for 61 too, but it seems kind of long, and I?ll only post it if smb REALLY wants me to (can't spend all day logged to the internet, you know)
 :D

Bye!

About the third one:

I don't have a complete soln, but this idea might help. I don't it's that hard to prove that every parallelogram has both sides parallel to 2 sides of the 4n-gon (this must be true for any polygon partitioned in parallelograms). Then we can form some "chains" of parallelograms (P_n) s.t. P_i and P_i+1 touch on a side (have a common line for a side) and P_1 has a side on a side of the 4n-gon (side 1) and P_n (the last in the "chain") has a side on the side 2 of the 4n-gon, which is parallel to side 1. For each pair of parallel sides 1 and 2 of the 4n-gon there exist sides 3 and 4 of the 4n-gon such that side 3 || side 4 and side 3 perpendicular to side 2. At the intersection of 2 chains of parallelograms corresponding to the 2 pairs (side 1,side 2) and (side 3, side 4) there must be a rectangle, since 1 perpendicular to 3 (and 1 to 4, and 2 to 3, and 2 to 4).

I think the area is n*l^2 where l is the side of the 4n-gon, but I'll have to work on that.

Indeed, it is possible that the whole side of $4n$-gon cuts into some smaller parts, which will be some sides of parallelograms. These part also cuts further and so on. Finally we can choose many small parts on the initial side of parallelogram, which are never divided further. Each this small part define its own chain. And this chain meets with all corresponding perpendicular parts once. It gives us a set of rectangles, those common area is $l^2$.

[quote="Lagrangia"]In the interior of a cube we consider 2003 points. Show that we can divide in more than 2003^3 cubes so that any points from the given ones are inside one of the smallest cubes(not on its faces)..[/quote]
We can choose such large $p>2003$ that all coordinates of the points are not of the form $k/p$. Then divide the cube into $p^3$ small cubes.

Maybe I misunderstood something?


geometry
ratio
cyclic quadrilateral
geometry proposed

$ ABC$ is a non-isosceles triangle. The incircle tangents to $ BC,CA,AB$ at$ D,E,F$ respectively. $ M,N$ are on lines $ EF$ such that $ BMNC$ is a cyclic quadrilateral. Denote $ K$ be the midpoint of side $ BC$. Prove that $ M,N,D,K$ lie on a circle.

Assume that $ AB<AC$. Let $ EF$ intersect $ BC$ in point $ G$. Let $ BC\equal{}a$, $ AE\equal{}AF\equal{}x$, $ BF\equal{}BD\equal{}y$ and $ CD\equal{}CE\equal{}z$.
Based on Menelaus theorem, we have $ \left(\frac{CE}{EA}\right)\left(\frac{AF}{FB}\right)\left(\frac{BG}{GC}\right)\equal{}1$ or
$ \left(\frac{z}{x}\right)\left(\frac{x}{y}\right)\left(\frac{BG}{BG\plus{}a}\right)\equal{}1 \Rightarrow \frac{BG\plus{}a}{BG}\equal{}\frac{z}{y} \Rightarrow 1\plus{}\frac{a}{BG}\equal{}\frac{z}{y} \Rightarrow BG\equal{}\frac{ay}{z\minus{}y}$  $ (1)$.

$ BMNC$ is a cyclic quadrilateral $ \Rightarrow GM.GN\equal{}GB.GC$   $ (2)$.
To show that $ MDKN$ is a cyclic quadrilateral, we have to prove that $ GD.GK\equal{}GM.GN$. By using $ (1)$ and $ (2)$, we have

$ MDKN$ is a cyclic quad     $ \Longleftrightarrow GD.GK\equal{}GB.GC \Longleftrightarrow (GB\plus{}y)(GB\plus{}\frac{a}{2})\equal{}GB(GB\plus{}a) \Longleftrightarrow$
$ GB(\frac{a}{2}\minus{}y)\equal{}\frac{ay}{2} \Longleftrightarrow \left(\frac{ay}{z\minus{}y}\right)\left(\frac{a}{2}\minus{}y\right)\equal{}\frac{ay}{2} \Longleftrightarrow a\equal{}y\plus{}z$.

The proof is complete.

[quote="mr.danh"]$ ABC$ is a non-isosceles triangle. The incircle tangents to $ BC,CA,AB$ at $ D,E,F$ respectively. $ M,N$ are on lines $ EF$ such that $ BMNC$ is a cyclic quadrilateral. Denote $ K$ be the midpoint of side $ BC.$ Prove that $ M,N,D,K$ lie on a circle.[/quote]
Let  line $EF$ cut $BC$ at $P.$ Then the cross ratio $(B,C,D,P)$ is harmonic $\Longrightarrow$ $\overline{PD} \cdot \overline{PK}= \overline{PB} \cdot \overline{PC}=\overline{PN} \cdot \overline{PM}$ $\Longrightarrow$ Points $M,N,D,K$ are concyclic.


[quote="12345678900"]A regular pentagon has a perimeter of 4855.  Bill is exactly three thousand metres from the center of the pentagon.  What is the probability that three sides are visible to him?

Thanks in advance![/quote]

I calculated 98.6%, if this is the correct answer i will share my solution =)


trigonometry
induction
real analysis
real analysis unsolved

Prove that the series $1+\sin (1)+\sin (\sin (1))+...$ diverges.

For all $x\in (0,1)$, $\sin x>x-\frac16x^{3}$. We prove inductively that $\sin^{n}(x)\ge \frac1{n+1}$ (interpreted with composition). For each $n\ge 1$, $\frac1n-\frac1{n+1}=\frac1{n(n+1)}\ge frac1{2n^{2}}>\frac1{6n^{3}}$. By the induction hypothesis, $\sin^{n-1}(x)\ge \frac1n$, so $\sin(\sin^{n-1}(x))\ge \sin(\frac1n)>\frac1{n+1}$.

By the above, the series diverges by comparison to the harmonic series. A tighter estimate would give that $\sin^{n}x$ is comparable to $\sqrt{3}n^{-\frac12}$.


An ant is crawling on the edges of a cube, starting from one vertex. How many edges can it go through the most if it can go on every edge only once?

Nine. We can map it in 2-d as a square within a square with segments connecting their corresponding vertices to each other. As every vertex has three edges coming out of it, one edge coming out of it must not be traversed (unless it's the starting vertex), meaning at most nine vertices edges can be traveled upon.


I took number theory this summer and found it's difficultly being 2.5 out 5 (1 being easy) should I take mathounts basics or advanced.

Take a practice chapter test (available on mathcounts.org, you can find the 2009 chapter.) If you have difficulty getting around a 40 or higher, then the basics class is probably for you. If you easily get 40+, and want to prepare for states/nationals, then try the advanced.

What other math classes have you taken?  Have you ever participated in MATHCOUNTS before?

I have not taken any math classes bescide school and I have never particed in mathcounts

[quote="Blazer123"]I think you should talk the pre/post tests for the courses and see how well you do with them.  If they are generally easy, go with advanced mc.[/quote]

That seems like good advice.  Also, note that if whichever class you choose doesn't fit well after the first class or two, you can switch to the other.

I'm taking Advanced Mathcounts/AMC8 in a few weeks no matter what(I am NOT dropping the class), but there was no pretest in the class introduction.

I qualified for State Mathcounts but I typically can only get in the 23-32/46 in state tests(with the exception of anything before 2004 where I do a bit better)   If I were to do it over again, would Mathcounts/AMC8 Basics have been a better fit?

My Dad talked with DPatrick a couple of months ago and he wasn't sure which class I should take and I didn't want this to be an extremely discouraging experience.

Also, is there a "pretest" of sorts for the class I could take to size up what I need to know so I could review it before the class?

Also, as far as Mathcounts Chapter, I usually do around 23-28 on sprint when I practice and around 6-8 on target.

Just trying to size up what I need to review.

If you've made States, Advanced MATHCOUNTS will be a better fit. (I wasn't qualified for State Competition as a 7th grader, but after taking this class, I was able to make it to Nationals.) If you really work hard in the class, you'll also improve a lot. It seems like you're doing pretty good on the Chapter problems, so you don't have to worry about taking the Basics course.

I don't really think there's a pretest for this course, but the pretests for the Introduction courses are roughly the same difficulty, so I would recommend you trying that.

There are no pre/post tests for the Problem Series classes. You'll just have to follow the scoring recommendation that is on the page. Basically, for Advanced you should be pretty confident you can make State Mathcounts. Also, you can't just take the class for an hour and a half and come out knowing everything. You might want to study the AoPS Vol.1

Good luck:)


algebra
polynomial
function
LaTeX
geometry
geometric transformation
reflection

How do I graph the polynomial function below?

f(x) = 1 - (x + 1)(^4)/3

This is is all good but HOW do I graph the given polynomial function?

which one is it?

as i said in many other posts, first plot the x intercepts. then mark signs in between. then plot points in between. then connect them with a smooth curve, ending each end with an arrow pointing up or down.

Sorry...it is the first one with the degree of 4.

[hide="using transformation geometry"]
$f(x) = 1-\frac{(x+1)^{4}}{3}$
$f(x) = \frac{1}{3}(3-(x+1)^{4})$

From here, let a new function $g(x) = x^{4}$.

Since, instead of an x being inside the parentheses, there is an $x+1$ term being raised to the fourth we move the function $g(x) = x^{4}$ over to the left one.  Then, since the $(x+1)^{4}$ is being negated, the y-values of $(x+1)^{4}$ are negated to become $-(x+1)^{4}$.

So far, writing this in transformation geometry, we have $r_{x-axis}(T_{-1,0}(g(x)))$

Now, from here, we increase the y-values by three.  And then, lastly, the function is amplified by 1/3.  If you know what the function $g(x) = x^{4}$ looks like, then using reflections and translations, we have:

$f(x) = \frac{1}{3}T_{0,3}(r_{x-axis}(T_{-1,0}(g(x))))$

The toughest part is to imagine each transformation.[/hide]


[quote="{x}"]Nie znalazlem lepszego okreslenia problemu od tytulowego "Hardkoru". Sadzcie sami:
$n$ jest dodatnia liczba calkowita, $d ( n )$ liczba dzielnikow $n$ a $d_1 < d_2 < \ldots < d_{d ( n )}$, wszystkimi dzielnikami $n$.
Udowodnij ze:
\[ \lim_{n \rightarrow \infty} \frac{\sum_{j = 1}^{d ( n ) - 1} \left( \frac{d_{j + 1}}{d_j} - 1 \right)}{\log n} = 1  \][/quote]
Can u translate this into english please :) ?

Here you go:
Let $d ( n )$ be the number of divisors of $n$ and $d_1 < d_2 < \ldots < d_{d ( n )}$ all divisors of $n$.
Prove that
\[ \lim_{n \rightarrow \infty} \frac{\sum_{j = 1}^{d ( n ) - 1} \left( \frac{d_{j + 1}}{d_j} - 1 \right)}{\log n} = 1  \]
I'd love to see someone solve it (tough, I've got my little idea too ;)).

Zadanie jest bardzo proste. Zastosuj kryterium Raabego.

mi sie wydaje ze cos tu jest nie tak, przeciez jesli n bedzie pierwsze to na gorze dostaniemy $n-1$, czyli $\limsup= \infty$


algebra
function
domain
Ring Theory
number theory unsolved
number theory

[quote="baz"]Working in $\mathbb{Z}[\sqrt{-2}]$, it is not too difficult to show that the only solutions to $y^{2}+2=x^{3}$ in integers are given by $(x,y)=(3,\pm 5)$.  Is there maybe an easier proof that is elementary?[/quote]

I read in E.T.Bell's book that the elementary demonstration of this result was proven by Euler. However, he spend 7 years on the proof. So I dont think it so nice to follow.

[quote="baz"]Thanks sylow theory!  I am now more satisfied with the algebraic number theory proof.[/quote]

There is actually another way to show this. Maybe you heard of "elliptic curves". But that is more complicated then the unique factorization domain proof.


function
topology
algebra proposed
algebra

If $ f(0)>0$, and $ f(x)<0$ for a particular $ x> 0$ is the existence of a minimal (positive) zero of the function $ f$ guaranteed?

$ \{0\}$ is a closed set in $ \mathbb{R}$, so $ f^{\minus{}1}(\{0\})$ is also a closed set. Then $ A\equal{}f^{\minus{}1}(\{0\})\cap [0,x]$ is also closed, so if $ y\equal{}\min A$, then $ f(y)\equal{}0$ and $ f(z)\neq 0$ for all $ 0<z<y$.

I have used several propositions.
1. For a continuous function $ f$ and a closed set $ C$, $ f^{\minus{}1} (C)$ is also a closed set.
2. $ \{0\}$ is a closed set.
3. A bounded closed set contains the minimal element.


The numbers $ \minus{}2$, 3 and 4 are each used once and substituted at random for $ a, b$, and $ c$ in the equation $ ax \plus{} b \equal{} c$. What is the probability that $ x$ is negative? Express your answer as a common fraction.

[quote="GameBot"]The numbers $ \minus{} 2$, 3 and 4 are each used once and substituted at random for $ a, b$, and $ c$ in the equation $ ax \plus{} b \equal{} c$. What is the probability that $ x$ is negative? Express your answer as a common fraction.[/quote]

$ x\equal{}\frac{c\minus{}b}{a}$.

There are two cases: c-b is positive, a is negative, or vice versa.

The first case has 1 way, while the second has 2 ways. Thus, $ \frac{1}{2}$.


The wall of a room is made up of 20 square panels. Each panel is one of four possible colors. There are 6 panels of color A, 2 panels of color B, 4 panels of color C, and 8 panels of color D. A ladybug randomly lands on the wall but never on panels of the same color on consecutive lands. What is the probability that the ladybug's first and third landings will both be on color A? Express your answer as a common fraction.

We can have ABA, ACA, or ADA, with probabilities $ \frac6{20}\cdot\frac2{14}\cdot\frac6{18} \equal{} \frac1{70}$, $ \frac6{20}\cdot\frac4{14}\cdot\frac6{16} \equal{} \frac9{280}$, and $ \frac6{20}\cdot\frac8{14}\cdot\frac6{12} \equal{} \frac3{35}$, respectively, so the total probability is $ \frac1{70} \plus{} \frac9{280} \plus{} \frac3{35} \equal{} \boxed{\frac {37}{280}}$.

The answer is 9/100 because it doesn't matter what is in between. It's just 6/20 * 1 * 6/20. The question never said WHAT IS THE PROBABILITY THAT IT WILL LAND ON A THE FIRST AND THIRD TIMES AND WILL NEVER LAND CONSECUTIVELY ON SAME COLORED PANELS the question said THE LADYBUG WILL NOT LAND ON THE SAME COLOR PANELS CONSECUTIVELY.

math154 is right, I think. It's second landing can't be on A so it's on B, C or D.

[quote="kobebryant24"]The wall of a room is made up of 20 square panels. Each panel is one of four possible colors. There are 6 panels of color A, 2 panels of color B, 4 panels of color C, and 8 panels of color D.[b][size=200] A ladybug randomly lands on the wall but never on panels of the same color on consecutive lands.[/][/size] What is the probability that the ladybug's first and third landings will both be on color A? Express your answer as a common fraction.[/quote]

 :huh:

That means it doesn't land on blue, then again on blue.

No, I understand that, that's why I made it really big. I don't udnerstand why people answer questions without reading them first.


The radius of a particular circle inscribed in an equilateral triangle is 2 units.  What is the perimeter of the triangle?  Express your answer in simplest radical form.

Answer: $ 12\sqrt3$

My answer: $ 6\sqrt3$; Why:

If the radius of the inscribed triangle is 2, then half the side is $ 2\sqrt{3}$, so the perimeter is $ 12\sqrt{3}$.

The sides of the triangle will be 2*sqrt 3 * 2.since it will form a 30-60-90 triabgle if you connect the atitudes or medians of the triangle. so the perimeter is 12 sqrt 3.

If a circle is inscribed in an equilateral triangle, its radius is one third of the triangles height. Our radius is $ 2$, so the height of the triangle is $ 6$. Here we have an $ x$, $ 6$, $ 2x$ right triangle, where x = one half side. Using the Pythagorean Theorem...

$ x^{2} \plus{} 6^{2} \equal{} (2x)^{2}$
Thus
$ x^{2} \plus{} 6^{2} \equal{} 4x^{2}$
Subtract $ x^{2}$ from each side to get
$ 6^2 \equal{} 3x^2$
$ 36 \equal{} 3x^2$
$ 12 \equal{} x^2$
$ \sqrt{12} \equal{} 2\sqrt{3} \equal{} x$
$ (x\cdot2)\cdot3 \equal{}$ Perimeter
$ ((2\sqrt{3})\cdot2)\cdot3$ = Perimeter
$ \boxed {\boxed {\boxed {12\sqrt{3}}}}$ is the perimeter.