AMC AIME USA(J)MO USAMO email geometry calculus

Hi guys,

After numerous interests in this forum, the registration for AMC-level contest is now [b]open[/b].

Benefits in registration include:

A) Updates about contest and reminder
B) Occasional documents with interesting problem
C) Automatic sending of problems on the contest

Things to include on registration are answering following questions: (You may simply copy and paste them with your answers)

Following are for student registration (if you're only one interested in the school)

1. Your name
2. Your grade (in school year 2008-2009)
3. Your school name and state
4. Your coach/proctor/parent if there is any
5. Participation of AMC and/or AIME and/or USAMO

Following are for coach registration (student may register first and get coach to register as well)

1. Your name
2. Your school name and state

Benefits of coach registration are that I might be able to create some special documents (sample problems maybe or something of that sort) for coaches to help their kids become interested.

Any questions may be addressed here or by emailing me directly at hdmatholympiad@gmail.com

Note that this is same as the second page's description. Once you registered, you'll be able to be notified of details as the contest nears.

[hide="Original Introduction"]
Hopefully everyone did well on their AMCs. For those who qualified, good luck on your AIME and USAMO. As for myself, this was my last year for AMC and unfortunately, due to snow day, I did not get to take the official AMC. I did take test afterwards but the room was very loud and constantly bothered that I scored very poorly. Not a good way to end AMCs in high school I suppose.

Anyway, later this year, I plan to hold a AMC-level contest in December. It'll be exactly same format as the actual AMC-12 and will have following AIME-level contest (again same with AIME but different name) and possibly USAMO as well (The problems are more than likely to be derived from other competitions and purely original ones though).

So, please take a time to cast your poll. The reason I'm doing this now is that I get most votes around this time than later. Thanks! [/hide]

[quote="cognos599"]I think this is a great idea and i am very interested in participating, but wouldn't it be more helpful to make an AIME test instead of an AMC 12 since AMC 12 has already passed. just a thought though.[/quote]

Keep in mind that the contest will start in December which is for [b]next[/b] AMC/AIME/USAMO preparation. There is no way that I can make a mock AIME for this year's.

[quote="CatalystOfNostalgia"]This sounds cool. Seems like alot of work for one person though...I would be willing to contribute problems/grade if you need me to (although I'm probably not qualified to grade a mock USAMO).[/quote]

It will be a lot of work and so far, I wanted to see how many are interested. I'll probably send problems to review to other users in this forum (probably bit older ones though; ones that are in college) and as for USAMO, I'll pick out problems but grading them will require getting help from other people.

[quote="SimonM"]I like this idea, although I would prefer it slightly earlier in the year (since the UK one is earlier (by then I will have taken SMC and BMO1)).

Other than that, it sounds fairly cool[/quote]

I apologize for this but judging from this year's UK Senior Challenge, next year's contest will be held around November? That's bit too early for me. I don't think it'll be possible to delay it.

Overall, I appreciate continued interest. Any question regarding this matter, the creation of problem, and etc, can be asked in here.  :)

[quote="Silverfalcon"]

[quote="SimonM"]I like this idea, although I would prefer it slightly earlier in the year (since the UK one is earlier (by then I will have taken SMC and BMO1)).

Other than that, it sounds fairly cool[/quote]

I apologize for this but judging from this year's UK Senior Challenge, next year's contest will be held around November? That's bit too early for me. I don't think it'll be possible to delay it.
[/quote]

Yeah, the 6th of November. Still... I may as well participate

[quote="miller4math"]since it is following amc 12 format, will it be amc 12 difficulty?

anyways, id like to participate[/quote]

Yes.

The rules will be 25 questions with multiple choice (A,B,C,D,E) with only [b]one[/b] as correct answer. The calculators will NOT be allowed.  Basically, anything that you are allowed to do in current AMC will permit in this exam.

As with calculator rules, I really do want this rule to be carried out so I'm thinking if I can actually contact the coaches of the math club or at least someone who can be a proctor for this. There is no prize or anything but since this is going to be bigger than mock AMC, I want to make it as official as possible. Plus, having ability to contact the coaches enable me to directly send the group of scores through one email (rather than sending one to every contestant) and a full solution packet online as well.

So, let me go onto little bit more details now that a lot of people are interested.

[hide="1. Difficulty of Problem Level and Area Coverage"]

[i]AMC-12 Level[/i]

Similar to actual AMC, this exam will contain several areas of high school math curriculum excluding pre-calculus and beyond but not limited to them. That is, the problems will draw ideas from number theory, combinatorics, basic counting and probability principles, and complex numbers/trig/advanced algebra.

[i]AIME Level[/i]

Similar to actual AIME, this exam will have integer answers (000-999) and 15 questions. The sources of problems will vary. The ideas will be taken from past AIMEs, local math competitions, and national competitions around the world. The actual level of difficulty of problems may vary but I'll try to pick out problems that come close.

[i]USAMO Level[/i]

Same rules as actual USAMO will apply and I'll try to get some advanced graders to actually grade these problems. Similarly, I will probably end up requesting/suggesting problem(s) on this level since it is practically impossible for me to write one up (the best thing I can do is take the actual problem and modify it bit ---- which might happen in one or two problems).

[i]For all three levels[/i]

Problems will have ideas on other competitions but problems themselves will be [b]original.[/b] [/hide]

[hide="2. Qualification and Awards"]

Qualification will be same as actual AMC. Now, there will be no awards since this is 100% Free, Just-For-Fun test. There is no way I can make awards or plaques, etc...  :P [/hide]

[hide="3. Exam Proctoring"]

With AMC/AIME, I prefer to have someone older than 18 years of age proctoring your test. It is best if this happens by your school teacher (not necessarily math coach since not everyone has math coach in the school) or your parent but again, that's not always the case.

In the case of homeschooled student, the policy is that a parent of the child will suffice.

The contest will go on for a week instead of one date.[/hide]

[hide="4. Sending out the Exams"]

The problems in PDF format (they are not necessarily $ \text{\LaTeX}$ed because I don't have it installed here ---- my computer is slow enough) but they will be typed in computer. They will be delivered to coach/parent. See the details in Section 5. [/hide]

[hide="5. Notification of Scores and Solutions"]

Case 1: Presence of Coach/Parent

The official scores and a copy of solutions in PDF format will be sent to the coach or parent. Each copy of solution will contain at least one way to approach the problem and solutions. But, they will be delivered [b]after[/b] the end of exam date (which will probably be a week instead of one day because that gives people more flexible time to take this and avoid situation of snow days) since there is a chance that student may "pretend" to be a coach.

Case 2: Absence of Coach/Parent - Individual-Participation

If this is the case, then the student himself/herself will receive problems first and solutions later directly to his/her email. The solutions will be same as the ones than the ones coaches would receive.

Optional: Student wants a copy of solution

If this is the case, then the student must email directly to me and request the solutions. But usually, this will not happen since coaches are allowed to produce multiple copies of solutions as long as it's not for sale. [/hide]

Any questions (I typed up this pretty fast although I had general idea in my head) can be addressed here.

@SimonM: That's good. It's sad that UK does their test too early. Just for suggestion to you, if you want to practice skills for UK (I'm saying this based on what I've seen from past three years of them), try Mu Alpha Theta problems. They are good for concentrating on the skills.

[quote="Silverfalcon"]Hi guys,

Hopefully everyone did well on their AMCs. For those who qualified, good luck on your AIME and USAMO. As for myself, this was my last year for AMC and unfortunately, due to snow day, I did not get to take the official AMC. I did take test afterwards but the room was very loud and constantly bothered that I scored very poorly. Not a good way to end AMCs in high school I suppose.

Anyway, later this year, I plan to hold a AMC-level contest in December. It'll be exactly same format as the actual AMC-12 and will have following AIME-level contest (again same with AIME but different name) and possibly USAMO as well (The problems are more than likely to be derived from other competitions and purely original ones though).

So, please take a time to cast your poll. The reason I'm doing this now is that I get most votes around this time than later. Thanks![/quote]
sorry about that dude. it stinks how some people who definitely deserve it, don't get it. im definitely interested. :-)

[quote="annoyingpi"]I think this is a great idea![/quote]

 :lol:

[quote="Silverfalcon"]

@SimonM: That's good. It's sad that UK does their test too early. Just for suggestion to you, if you want to practice skills for UK (I'm saying this based on what I've seen from past three years of them), try Mu Alpha Theta problems. They are good for concentrating on the skills.[/quote]

I will look at the Mu Alpha Theta problems. The main reason why the UK ones are so early is the transition between two operating bodies. It goes from the UKMT to BMOC between the SMC and BMO. Then there are two rounds until they select 20 people for the training weekend. My aim is to get a decent score in BMO 1 to in order to get a place in their IMO training group (I doubt I'd make the training weekend since its my last possible attempt :()

i would be very intersted but am not able to get a parent or coach to proctor.  What do you mean "deliver"  like mailed(hopefully not) i would like it to be emailed but what ever is most convinient for you its ok with me.  Ill be honest and wont cheat. Anyway it probably wont matter cause i probably wont make AIME!

I don't meant to be the antagonist in the crowd, but  --

-- For all your effort's worth, I think it would be better placed starting an "actual" contest in some area or by pumping up one that already exists and is already "official". I think that for a project of that size, it would be more beneficial overall if it were delivered to an audience that needs it than to an audience that's already satiated with competition mathematics.

-- From a more self-oriented perspective, it would probably be more resume worthy and you might be able to get some monetary compensation for your work. If you want, you could even do it like starting a business (though you likely won't make too much money at first); that's a very fun experience, if you haven't tried it already.

[quote="The Zuton Force "]-- From a more self-oriented perspective, it would probably be more resume worthy and you might be able to get some monetary compensation for your work. If you want, you could even do it like starting a business; that's a very fun experience, if you haven't tried it already.[/quote]

How would one go about doing this because i am very interested in trying!!!and have you done it cause you say its fun!

[quote="RussianRocket"]after the 2nd round will u post the questions and answers[/quote]

By "posting" as in this forum, no.

The contest problems and answers will only be sent from my email address. If you're registered and send in your problems, you will receive solutions to your problems for the first round. Now, that'll be automatically for you if you make it to the second round.

But, let's say you do not make to the second round. You can still receive problems and solutions even this is the case. However, you're going to have to wait until the second round is over. I do this for security reasons.

Here is a catch, however. Unless you're registered for my contest, you're not going to be able to receive problems no matter what. By this, I mean that if you send me PM here or leave a message to request problems without registering, you will not receive them.

[b]I also want to stress the importance that problems and solutions that you'll receive for your eyes only. You're NOT given permission to print them in mass production or email to your friends. In order to do so, you must send me an email requesting to do so. This is a free contest but without such rules, my documents and possibly, faked versions of them could fly in cyberspace. [/b]

I'm pretty sure I'm going to do the same for the last round. But that's a long way from today.

A HUGE HUGE ANNOUNCEMENT!

I just sent email with PDF for information covering the rules and structures of this contest. If you registered and did not receive this document, you MUST give me an email to hdmatholympiad@gmail.com

It is very crucial!!

And this is to a contestant named David Li: If you read this, I could not send the document to your email address. I'm not sure why so please email me with other address than what you registered with.

[quote="Xantos C. Guin"]I'm assuming that the time limits for the first two rounds are 75 minutes and 3 hours respectively? This is probably kinda obvious but it is not explicitly stated in the information PDF.[/quote]

Yes, that is correct. I meant to do it and I recall doing the time thing for last round but I must've forgotten to do so for the first two rounds.

[quote="hurdler"]I assume we can use graphing paper like on the amc's and aime, (the other things like protractor and compass were mentioned in the document, but not graphing paper, sorry for being picky). I find that most amc's/aime have at least one problem in which graphing paper seems helpful.

Also, I find that my coordinate planes seem to be often poorly drawn........[/quote]

Yes, you can use them.

I didn't put them because I didn't know how many people actually used them. Basically, you can use everything you can on AMC on this contest (and remember, that means no calculator too!).

Important announcement!

Second round is DONE! I have to just post problem and solutions for #14 and #15 but other than that, the second round's problems and solutions are DONE and saved on my computer. They should be undergoing proofreading by some people soon (I just have to decide who I'll give to).

Now, with the last round, I'm planning to use 6 olympiad problems from other contests excluding USAMO because olympiad problems are practically impossible for me to make, and I want to have problems and solutions done before I go away to the college.

PLEASE let me know if there is any objection to this matter. What I'll do is to pick out 6 problems and cite them in both problems and solutions. I'll vary them as in real USAMO though.  :)

[quote="mathemonster"]when is round 1 silverfalcon? u have my email, and i think i registered did i? if i did will u just be sending the contest by email?[/quote]

If you received a reply after registration, you should've been replied. Check if you have received my problem solving strategy article. If you didn't and you emailed me, then you have not ben registered. But if you received the document, yes, you are registered.

Round 1 is to be around late December this year so you have time to relax and practice until this comes. The second round'll be a month before AIME and similarly for last round in respect to USAMO.

[quote="SigmaTheta"]Are you going to have this competition every year?[/quote]

Not sure why I just read this message now... Hmm, AoPS must've not given me an email notification? I don't know...

The competition is definite for this year. Rest years depend on how much time I have and how much interest arised from this (like how many will actually compete after registratin). Writing math problems AND solutions take tons of time so I can't really say one way or another at this point.

AND IF YOU DIDN'T REGISTER, YOU SHOULD! SEE THE FIRST POST IN THE FIRST THREAD FOR DETAILED INFORMATION!

I'll be reviving this thread every couple weeks to make sure new (and old returning visitors) can read the thread and register.

This contest is open to EVERYONE so Europeans and other non-United Statesers are more than welcome to participate.


Alright, so I get this problem, which says find all solutions $(n,m)$ to the equation

\[ n^4+n^3+n^2+n+1=m^2  \]

such that $n\in\mathbb{N}$, $m\in\mathbb{Z}$

So I did it and here's my solution, is it actually believable/conclusive, and does anyone have a better solution?

edit> oh my bad, someone just mentioned to me it's a problem off the PROMYS application, I didn't know...sorry, nobody go and post solution to this or anythin..

It's a common enough problem, really.  I've already written up my solution for this problem, but just in case, you should just delete this post.


function
combinatorics open
combinatorics

$ n$ points are placed on a straight line so that distance between any two points is an non zero integer and unique. Find the function $ n\mapsto f(n)$ where $ f(n)$ is the minimum possible distance between the farthest separated points.


Find the value of $\frac{\text{lcm}\{1,2,3,4,...,200\}}{\text{lcm}\{102,103,104,...,200\}}$.

EDIT: This should probably be moved to High School Basics?

[quote="neelnal"]They are one and the same. Any number below or equal to 100 has a multiple from 101 to 200.

Hence, the LCM is one and the same since 1 to 100 don't matter anyways.[/quote]

Could you elaborate a bit more on that, i dont really get what you are trying to say :huh:  :ninja:  :maybe:

[quote="robinhe"][quote="neelnal"]The answer is 1.

They are one and the same. Any number below or equal to 100 has a multiple from 101 to 200.

Hence, the LCM is one and the same since 1 to 100 don't matter anyways.[/quote]

Sorry you're answer isn't right.

[hide="Big hint"]$101$ is prime.[/hide][/quote]

shoot sorry I forgot that the bottom set starts from 102 not 101.

So, the answer is 101.

[hide="Formal solution"] We consider an arbitrary prime $p$, and consider the highest power of $p$ that divides the numbers on the top and the numbers on the bottom.

For $p < 101$ the result is the same since the highest multiple of the highest power of each of those primes is the same in both sets (specifically:  it is not $101$, and it is between $100$ and $200$).

For $p = 101$ we find that it only divides the top set.

And $p > 101$ only appear once on both the top and the bottom.

The only difference in the prime factorizations is the $101$ on top, so our answer is $\boxed{101}$. [/hide]

Edit:  [hide="Simpler solution"] Let $M = lcm(102, 103, ... 200)$.  Obviously $1, 2, 3, ... 100 | M$, which means

$M = lcm(1, 2, 3, ... 100, 102, ... 200)$

Our given expression is then $\frac{101M}{M}$. [/hide]

[quote="jgao"]I dont get your solution  :maybe:[/quote]

The answer is 101.

Why? Because:

Any number below or equal to 100 has a multiple from 101 to 200.

Hence, The LCM of 1 to 200 is the same as that from 101 to 200. However, since 101 is not a member of the denominator, we take that out and it is a prime, and hence, it is our answer.

[quote="jgao"]I dont get your solution  :maybe:[/quote]

[hide="What I was doing in the long solution"] Basically, if you have some numbers

$n_{1}, n_{2}, ... n_{k}$

And they have prime factorizations

$n_{i}= \prod_{j=1}^{m_{i}}p_{j}^{e_{i,j}}$

Then

$lcm(n_{1}, n_{2}, ... n_{k}) = \prod_{j=1}^{max(m_{i})}p_{j}^{max(e_{i,j})}$

In words, the lcm of a set of numbers has to have the property that for any prime that divides any of those numbers, the highest power of that prime that divides any of that set must also divide the lcm.

So I consider those powers for three different cases, and find that they are the same except for the case $p = 101$. [/hide]

[hide="What I was doing in the short solution"] $M = lcm(102, 103, ... 200)$ obviously has the property that $102, 103, ... 200 | M$.

[b]Lemma:[/b]  Given two numbers $a, b$ such that $|a-b|+1 \ge k$, there exists a multiple of $k$ such that $a \le mk \le b$.

For $k = 1, 2, 3, ... 99$ then there exists a multiple of $k$ between $102$ and $200$ which means that $k$ also divides $M$.

And for $k = 100$ obviously $200 | M \implies 100 | M$.

That means that $M = lcm(102, ... 200)$ is also a multiple of $1, 2, 3, ... 100$, and so

$M = lcm(1, 2, 3, ..., 100, 102, 103, ... ,200)$

Which means that

$lcm(1, 2, 3, ..., 100, 101, 102, 103, ... 200) = 101M$

Since we need a number divisible by $101$ and $101$ is prime (which means $gcd(M, 101) = 1$).  [/hide]


i think it has been already discussed here on the forum , it`s a problem given by mr . Laurentiu Panaitopol at OBJM 2003, if i`m corect

Posted some time before and unsolved at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=122522[/url].

[quote="mychrom"]U can write 1+x*x=1+x*x. We can observe that 1 & x*x are simultaneosly perfect squares, so i think the best solution is to use Bergstom ine :lol: q.[/quote]

i don`t think it can be solved that way, the original solution was with cauchy indeed but after we use the fact that $b\leq \frac{1+b^{2}}{2}$ so $\frac{1+a^{2}}{1+b+c^{2}}\geq \frac{1+a^{2}}{1+\frac{1+b^{2}}{2}+c^{2}}$, another solution i think is with Jensen after the same "trick".


AMC
USA(J)MO
USAMO
logarithms
probability
expected value

21. On the USAMO, one must write out solutions and proofs to six problems. Each problem is graded out of seven points. When, Carl takes the USAMO, there is a 1/12 chance that he will solve a given problem. Of the six problems, two have answers in the form of numbers or algebraic expressions, and Carl has a 1/1100 chance of guessing the answer right if he does not solve the problem. Define this type of problem as a “guessable” problem. If Carl solves a problem, there is a 1/2 chance he will solve it with a brute force solution, and a 1/8 chance that his solution will have a hole. It is impossible that he will have a hole in a brute force solution. If a brute force solution results in four points, a solution with a hole results in three points, a correctly solved problem that is neither brute force or containing a hole results in seven points, a correct guess to a guessable problem results in two points, anything other than this results in zero points, and there are no other factors, what is Carl’s expected score? Express your answer as a mixed number.

There is a 0% chance of this being right, but is it [hide]0[/hide]

[hide]
For the unguessable problems:
There is a $\frac{1}{12}\cdot\frac{1}{2}=\frac{1}{24}$ chance Carl will get a problem right with a brute force solution.  Multiply by 4 for the point value.  There is a $\frac{1}{12}\cdot\frac{1}{8}=\frac{1}{96}$ chance Carl will get a problem right with a solution with a hole.  Multiply by 3 for point value.  There is a $\frac{1}{12}\cdot\frac{3}{8}=\frac{1}{32}$ chance Carl will get a problem completely correct without brute force.  Multiply by 7 for point value.  His expected point value, then, for each of the 4 guessable problems is $\frac{5}{12}$.   Multiply by 4 to get $\frac{5}{3}$ for the guessable section.
For the guessable problems:
Add on $\frac{1}{13200}$ for the chance Carl gets the problem right by pure luck.  That brings the expected value for the 2 guessable problems to $\frac{5501}{6600}$ for the guessable problems.

Add it all to get $\frac{16501}{6600}$ as Carl's expected score.  In mixed form it's $2\frac{3301}{6600}$.[/hide]

That has a $\ln{e^{100}}\%$ chance of being wrong.


[quote="pascal_1623"]Given a 3 by 7 rectangular array of dots. how many triangles can be formed whose vertices are dots in the array?
I can never get these ones...[/quote]

[hide]There are 21 dots. $\binom{21}{3}=1330$ ways to choose 3 dots. There are $3\binom{7}{3}=105$ ways to choose a horizontal line and $7\binom{3}{3}=7$ ways to choose a vertical line. There are 18 more ways to form straight lines. 1330-(105+7+18)=$\boxed{1200}$[/hide]


AMC AIME MIT college probability USA(J)MO USAMO

1500 SAT I
700 writing
800 physics
800 bio
800 math IIC
800 chinese

5s on the following APs before the apps are sent in
physics C mechanics
physcis C E/M
bio
calc AB
calc BC
statistics

some Okay, but not outstanding, scores on competitions..
my last years amc=107.5 and aime=3

random clubs... lol i created the "trading card club" for when yugioh was popular..

small amounts of comunity service and extracurricular activities..

and finally.. a gpa of around 3.3 raw..

i will be a senior next year.. and i HATE college's emphasis on GPA...

so wut school u think thats good enough for and why would MIT or Cal tech reject me if u think that would happen...

my phylosophy is.. why would grade matter if you could do better than all those straight A students on standerdized tests.. as little respect as i have for these tests, i still think they do measure a degree of skill.. and why would grade matter if you learned the material? isnt knowledge the ultimate goal to learning? ya.. hardwork helps, but if its only gonna get u a 3 or 4 or AP, wut makes college think those with high gpa and mediocare test scores are more qualified?

why should colleges focus more on hard work than "you learned the material?" Because, in the real world, hard work is what matters. Not if you know small facts about the biology of a certain bug, or whatnot. If the extent of your application is scores, you aren't prepared for the real world. Why do grades matter more than standardized tests? I think it is because grades show how much you are willing to work for something. Grades span a long period of time and a large amount of homework, projects, and tests. Grades show effort more than one test on one day shows it. Granted, it is rather easy to make all As at some schools, but the point is that you are doing the best that your school has to offer. ALL schools (not just Ivys and top-tiers) want to know that you are taking the most challenging courses that your school offers and excelling at them. That's challenging yourself, and it shows hard work.

[quote="kentliu"]small amounts of comunity service and extracurricular activities..

and finally.. a gpa of around 3.3 raw..[/quote]
Those are the two reasons I see. A lot of colleges seem to place a tremendous emphasis on community service, and that GPA isn't really that great.

I've been worrying about colleges not liking my gpa too-- that seems to be the only major flaw I have floating on my record-- that my GPA is iffy-- I think it's a 3.4 or 3.5 (without weighted grades for APs  versus regular classes) but I've been trying to improve it, and it has slowly improved... I just don't do too well gradewise in English and History classes, though I love the classes... *sigh*

I agree that colleges should look at your effort and such... but there has to be a way to really explain your problems to the college, that would be nice to be able to do.

My high school has turned me down as a tutor, when I know the material and teach it well, so that the kids that  do not know the material or want to tutor can tutor for their service hours. It's a major problem, but it also now shows on my record that I haven't done any community service since my middle school years-- because everything I do is off the records.

Not that it matters so much whether I get into a great college, I can't afford much and if I'm borderline getting in, I'm probably lower on the line for scholarships.

sorry for ranting in the thread, I went a little off the original topic...

[quote="kentliu"]how can you say that hard work is what matters in the real world if those hard workers cant learn something and do that well?
hard to comprehend such an idea...

also.. i am persuing maybe a PhD and maybe do research.. and i am sure you need more than hard work for such a career..[/quote]
Research is certainly a lot of hard work. Also, on your college applications, you don't mention that you're planning on something after undergrad - they don't care.

you do need more than hard work, but the thing is, there are a LOT of "above average" people in this world, all of which are capable of getting a PhD. see, you can be smart and talented and all of that naturally, but it's not that impressive unless you actually do something with it besides taking a lot of tests.

You're right, standardized tests are wonderful for measuring some abilities.  In particular, the ability to take standardized tests.  Grades are a far more interesting measurement, including as they do a mastery of material, an ability to perform unwanted tasks like going to class or doing homework, and a reasonably good attitude so as to not make teachers despise you.  If I were picking a college class, I would be far more interested in those attributes than the ability to correctly fill in little bubbles on a piece of paper.

By the way, my list of possible reasons follows.  The emphasis is mine.

[quote="kentliu"][b]so wut school u think thats good enough for[/b] and why would MIT or Cal tech reject me if [b]u[/b] think that would happen[b]...
[/b]
my [b]phylosophy[/b] is.. [b]why would grade matter[/b] if you could do better than all those straight A students on[b] standerdized[/b] tests.. as little respect as i have for these tests, i still think they do measure a degree of skill[b].. a[/b]nd why would [b]grade [/b]matter if you learned the material? [b]isnt[/b] knowledge the ultimate goal [b]to[/b] learning? [b]ya.. hardwork[/b] helps, but if i[b]ts[/b] only [b]gonna[/b] get [b]u[/b] a 3 or 4 or AP, [b]wut[/b] makes [b]college think[/b] those with high gpa and [b]mediocare[/b] test scores are more qualified?[/quote]

well... theres another thing grade measures, ur ability to cheat... and that is far from an overstatement... i dont know about ur school.. but i dont know a person in mine tat hasnt cheated... many who cheat on everything and thats how they succeed thru honor and ap classes...

anyways... lets get onto the other topic..  got any idea wut a resume like tat would qualify me for?
i dunno if this helps but i started this very cool extracurricular activity a little late.. but still cool.. http://www.jisan.org

also.. i dont know wut those emphasis mean..

It's far easier to cheat on one exam than it is to cheat ones way through an entire course -- you can't cheat showing up to class, except in rather extreme circumstances.

A serious question for you, though:  what does it matter where we say that you are qualified for?  How is that going to affect anything?  Getting into colleges is a lottery.  You can increase or decrease your chances of "winning," but it's all a probability game.  You seem to have a good resume, so you'll probably have relatively good chances of getting into "good" universities.  What else is there to say, and how would anything I say make any difference in whether or not you get into a college you want?

[quote="kentliu"]1500 SAT I
700 writing
800 physics
800 bio
800 math IIC
800 chinese
[/quote]

That's far from an unheard-of test score profile at the top schools, and it's a frequently encountered test profile for Asian-American boys intending to major in math or science, so it ends up not making you stand out so very much. I know eleven-year-olds who can get 800s in the SAT I math section, so we all know that that is an easy enough test. Giving any room at the top on the verbal side or on the writing test is dangerous if you are applying to the top of the top schools, especially these days.

[quote="kentliu"]5s on the following APs before the apps are sent in
physics C mechanics
physcis C E/M
bio
calc AB
calc BC
statistics
[/quote]

Again, this is quite typical of the applicant pool in which you are competing. There is no English, history, or even economics on the list of AP tests, and the calc BC is a very commonly taken AP test.

[quote="kentliu"]some Okay, but not outstanding, scores on competitions..
my last years amc=107.5 and aime=3[/quote]

That would be a whole lot more impressive if only you were three or four years younger.

[quote="kentliu"]random clubs... lol i created the "trading card club" for when yugioh was popular..

small amounts of comunity service and extracurricular activities..[/quote]

Has the Yu-Gi-Oh club fizzled out? I would believe that you were really into the club a lot more if you had spelled "Yu-Gi-Oh" in the usual manner that it always appears in in print.

The "small amounts" of community service and extracurricular activities is a killer. It makes you look socially isolated and less versatile, not to mention less likely to use your fancy degree for the good of society.

[quote="kentliu"]and finally.. a gpa of around 3.3 raw..[/quote]

That all depends on what kind of school you go to, and what kind of classes are offered there. It is not promising that you don't have any AP credits on the language and humanities side. The high school that I went to was so mediocre, in those days, that the class valedictorian, who had a grade average well above 4.0 (that's how weighting was done in our school, with "double A" grades), who was a really diligent girl, didn't get sufficient scholarships to be able to attend her first-choice school. If you go to a high school at which a LOT of kids go on to the very top schools, then the issue for you is what about the other kids in your class who outrank you? If you go to a typical United States high school, you're hardly even in the running with that kind of grade average, because the high test scores indicate "able" while the low grades indicate "lazy." That's how most admission officers will think about the issue. (I am very sympathetic to the idea that some kids are too smart for their high school classes, and know better than to waste their time with them, but then you have to show what else you are doing with your time, and you haven't shown that above.)

[quote="kentliu"]i will be a senior next year.. and i HATE college's emphasis on GPA...[/quote]

Grade inflation in my high school back in the 1970s was ridiculous, because that is how the school kept parents satisfied when the parents worried about the kids having good grades. We actually had FIVE "valedictorians," because grade inflation was so bad. The one I referred to above was the lone smart valedictorian--the other four were ditzy brown-nosers. One of the ditzy valedictorians didn't know, as a senior, what the word "apathy" meant, and as a classmate of mine acidly commented, she didn't care either. Colleges do regard test scores precisely because there is no standardization of high school grades, but high school grades are so high these days, on average, that one simply has to get high grades--or take really difficult classes--to look competitive.

I can't think of any positive reason for the top schools to accept any one person who is just like a hundred other applicants they get every year. Maybe you would get lucky, but more likely than not you wouldn't. An Asian-American male applying to the schools you desire is Tweedledum to Tweedledee in the applicant pool. The classic profile of Chinese-American boy strong in math but weak in English is so commonplace that the way to stand out is to become a lot better at English. Perhaps you have browsed around this forum enough by now to have seen my posts in which I advise college applicants to READ, READ, READ, and READ to build up their verbal skills. That will show in your writing.

[quote="kentliu"]my phylosophy is.. why would grade matter if you could do better than all those straight A students on standerdized tests.. as little respect as i have for these tests, i still think they do measure a degree of skill.. and why would grade matter if you learned the material? isnt knowledge the ultimate goal to learning? ya.. hardwork helps, but if its only gonna get u a 3 or 4 or AP, wut makes college think those with high gpa and mediocare test scores are more qualified?[/quote]

The crucial difference between a standardized test and a homework assignment is that if the homework assignment is any good (many high school homework assignments are not), then you have to generate the answers from a blank page, so you really have to write. In principle, a monkey could haphazardly mark the correct answers on a multiple-guess standardized test, but only a thoughtful human being can write a coherent paragraph of prose. JBL has already pointed out several errors of English expression in your post above. You seem to have made many misspellings into finger habits, which is a very dangerous thing to do. I cut people a lot of slack if they are native speakers of Chinese writing English, because I know (from my study of Chinese) that the two languages have very different structures, and it is particularly hard for NEW learners of English with a family background of speaking Chinese to write English well, but some people manage to learn from the English-speaking environment that surrounds them, and those who don't end up looking like they have less ability to learn in general.

The crucial point is that the very top schools mostly admit people who have BOTH good grades (or kind-of-good grades in hard courses) AND high test scores AND something else besides. You would be an automatic admit at most state universities, but you seem to desire to go to a more competitive kind of school. Improving a LOT on the AMC tests and (possibly) qualifying for USAMO will occur after the date by which you have to apply to college, right? What else can you do between now and your application date that will really help you stand out, NATIONALLY?

My 23 year old sister is visiting me, sitting right here, and wants to know if you have any friends your own age?

Let me "hazard a prediction here" that your social life is sorely lacking, and you have nothing better to do than rave about the ditzy brown-nosers that you wished you were (and make otherwise, fine young people wish they were the "smart 11 year olds, that [b] you [/b]know.")

how about diversifying your APs and SAT IIs? almost all of them are somehow math or science related. i would think that would hurt you, because math and science are usually related together. if you had english as one of those subjects, it would've helped a ton.
and please, don't think you're alone. i'm sure there's hundreds if not thousands of people like you in your position, except they don't go and post their anger on a forum.
btw, i don't mean any offense to you by my post. I'm sincerely sorry if you take it that way because it wasn't intentional.

Wow, this spiraled into incivility rather quickly, didn't it?

[quote="blahblahblah"]
In any event, I think that we are reaching the point where college admissions really need quantifiable data about who should get in (statistically at least) and who shouldn't.  Because the pressure to get into an elite college is so high and the availability of resources which can easily increase the performance of students on standardized tests (mainly through showing them how to avoid trick questions) is considerable, I think a significant portion of the applicant pool for elite universities is becoming more and more homogenous and thus the selection process is becoming more and more arbitrary.

This is, of course, a big problem, but I think that the system is too resistant to change to have something like the AMC administered to the average high school student.  Perhaps a two-tiered SAT is the solution, but this would similarly lead to a whole can of worms.[/quote]

I couldn't agree more with this first paragraph.

I also essentially agree with the second.  I don't think anyone was proposing an AMC for the average high school student -- just to those interested in expressing high levels of achievement.

You're right that a new can of worms would be opened by a higher level standardized test.  But the question is whether or not the consequences would justify the effort.

I think they would.  I think our whole system is currently so backward in that there is an incentive system behind the curtain which does not at all represent a meritocracy.  Nor does it represent the system that is sold to students publicly.

Hey Joel, insofar as I applied to Princeton last year and got in (it was the only US school I applied to), I have a reasonable amount of familiarity with the US system.

I'm not advocating a numbers only formulaic approach to university admissions (which, unfortunately, is basically how it is done in Canada), but I am saying that the standardized component of admissions does not provide any normative standard from which we can assess the relative (standardized test) merits of two 'excellent' applicants.

I'm not confining my views to math, either - I think that, for example, the SAT II writing test is not in need of overhaul to the same degree as the math component, but it simply doesn't test anything but writing.  There should be some compulsory literature component to the SAT IIs (I think there is an English lit exam, but its not mandatory like writing is).
[quote]
What incentive could you possibly offer them to take a test that will clump them into a group at the bottom of scores?[/quote]

As for this, there's no 'incentive' to take SAT IIs, save for the simple fact that universities require it.  The universities have to take the lead in any hypothetical situation we may propose, because SATs have no worth outside of being a yardstick (and a mediocre one at that) for university entrance.  A new, harder standardized test (pretty much across the board) would not be an end-all or a be-all, but it would at least provide a more accurate assessment of the relative abilities of two students (as opposed to the current system which provides an accurate assessment of the students ability to take the SAT)

I think that the human factor is one of the most important factors in admissions.  But people can lie about their accomplishments, they can affect a persona in writing that does not reflect their accurate character, and they can go through an interview without acting in character.  However, given that standardized testing is administered in an environment free of outside influence (which is what we'd hope), it does provide  a basis of comparison between different students.  Now, what we really want is for it to provide a basis of comparison between different students that has a strongly positive correlation with 'ability', insofar as 'ability' relates with skills which universities expect students entering their universities to have.

[quote="JBL"] How [i]do[/i] you deal with students who have achievements that don't just read as a list of scores on math competitions?
[/quote]

I think a better question is "how do you deal with universities that value money more than merit?"

I think you are erecting a straw man as nobody has argued that the hypothetical test we are discussing should simply take over as the ultimate admissions criterion.  The student that has already run a succesful business can still be given admission.  The student with 5's on their x-teen AP exams can still be given admission.  The 1320 SAT student that raised their 7 younger siblings while working part time can still be granted admission.

But what would be the harm in using a test that distinguishes students with a broad array of problem solving knowledge from their peers?

You say most people would lose out -- but isn't that the point -- that there are many people who are unrightfully hoisted up as equals to those who are really capable of changing the world?  Isn't this partially because we set the bar too low?

[quote="blahblahblah"]
I think that the human factor is one of the most important factors in admissions.  But people can lie about their accomplishments, they can affect a persona in writing that does not reflect their accurate character, and they can go through an interview without acting in character.[/quote]

I'll go one step farther.  I think people do lie and play a game the entire way through the process.  This usually comes as a shock to those who rely on their merit, but unfortunately that's reality.

And one step farther...I think think that top universities have a directive to accept a large number of such sociopaths -- because sociopaths are successful to a degree way beyond their merit.  They are more likely to run companies.  They are more likely to get rich.  They are more likely to get elected to office.  They are more likely to have influence to lend to the universities that got them there.  And they don't even have to do anything for anyone to do it.  They simply learn how to act in the show.

Okay, so I'm convinced now -- scrap my idea of a better test.  Nobody's going to use it anyway.

At the end of the day, a private university is a business.  They answer to no one regarding who they admit, who they waitlist, and who they reject.

For any new standardized test to come along, it would almost have to be vetted by the admissions offices of these same private schools.

So if you believe there's some sort of agenda - which, of course, there is because its a business - then that same agenda will be present in the formation of any new test.

Otherwise, that test is completely doomed to failure once everyone catches on that private schools don't use it to make admissions decisions.

So I agree with you...it would take an incredible amount of inertia to go from the status quo to a world where there's a standardized test that even resembles in the slightest way any national math competition.

On the other hand, consider the downside to attempting this.  In creating a new standardized test for math, you are implicitly saying that there is something unsatisfactory with the SAT's math section.  For the vast majority of the world, the math part of the SAT gets the job done fairly accurately...describing, with a single number, the skill level of any given student at mathematics.

But when you introduce another standardized test, you introduce an entire new set of barriers for disadvantaged kids and lower income classes to traverse in order to enter the world of "elite education."  Even worse, we send the signal to people who already can't do well on the SAT that they REALLY are stupid at math, helping reinforce the sort of general hopelessness that so many have regarding the subject.

So the situation I see is one of little upside to the creation of such a test, and a lot of downside.  If you know how to fix it while avoiding the perception problem I describe above, you could make a whole lot of money in the process while doing a lot of people a lot of good.  I just don't see it happening, for a whole lot of reasons.

A private university is a business, but it's also a university.  If it's goals do not resemble those of a university, we should cease to acknowledge it as one.

I agree that it's elite universities that must push for these changes.  At times they do, so I disagree that it's out of the question that such an inertia change could happen.

Nobody is suggesting that we so away with the SATs for the bulk of students -- but those students are not competing seriously for spots at elite universities in the [i]status quo[/i].

I'm not sure I understand your downside arguments at all.  The SAT [i]is not satisfactory[/i] for distinguishing top students.  So what's the harm in acknowledging reality?

And I think the argument about disadvantaged and lower income students is garbage.  Growing up in a low income family didn't keep me from learning math.  A lack of a good school math team doesn't either.  We live in the internet age in which any kid can walk into a public library and download far more interesting material than any of us could growing up.

If the problem is that many kids feel really stupid in math because of the existence of high level tests, the AP exams should be done away with, right?  Where is the threshold at which these students feel that the existence of a level of testing makes them stupid?

Additionally, I think most kids who feel hopeless or stupid feel that way because of their environments.  I fail to see any real link between an SAT III and such feelings.  I think these feelings are addressed almost entirely by their parents, teachers, and peers.

Also, are you suggesting that elite university admissions should be based on the feelings of average or less than average students?

None of your downside seems to me to address the filtration process of allowing top students to distinguish themselves.  It seems to have much more to do with a general state of self esteem which, while an extremely important issue, is entirely orthogonal to the goal of such testing.

And one must also remember: "elite" universities? They are called that... FOR A REASON! Just because one cannot attend said elite university shouldn't invoke a feeling of "hopelessness" (or else 99.99999% of the United States would feel "generally hopeless"). In stating that, you are almost undermining the value of an education system outside of these "elitest colleges." I try and continue to remind you guys of this: it's not the opportunities you are handed. It's what you do with them that makes you a great person.

And on the SAT Math thing: My humble opinion is that the SAT Math section works fine for what it is supposedly used for: distinguishing the "good" from the "bad." Now, within the "good" section, I realize that there are "stellar" and "above average" but I don't think that the SAT can feasibly be designed to distinguish BOTH the "stellar" from the "above average" AND the "below average" from the (very respectable) "trade school" (I do not wish to be demeaning in this, I hope you understand). If one were to design a harder SAT, as many of you suggest, then we would have quite the opposite trouble: the people that are now getting 600 and below would then be getting 400 and below. This is unfair to THEM. So, what is one to do? I really don't think there is much one CAN do. I really believe that this particular glitch in the system is up to the universities themselves to fix, by educating themselves on such elite tests such as the AMC, AIME, Mathcounts, etc. If they want the best students (which they most surely claim to), then they should be able to actively pursue them.

[quote="requiem"][quote="MCrawford"]
I'll go one step farther.  I think people do lie and play a game the entire way through the process.  This usually comes as a shock to those who rely on their merit, but unfortunately that's reality.[/quote]

So instead of marking 50 noes on the ACT registration booklets I should mark the most impressive ones yes?  Will do![/quote]

I have no idea what's on the ACT registration booklet, but I don't believe I made a statement about what philosophy anyone [i]should[/i] follow.

although increasing the difficulty of the sat would distinguish the truly good students, it would make the average people more vague... two people who now scores a 650 and 700 might become like 650 and 660 after the change.. so it looks like there isnt much of a difference.

i think the current system is good, with both SAT and AMC.. so colleges could distinguish one 800 from another by the AMC score...

[quote="kentliu"]i think the current system is good, with both SAT and AMC.. so colleges could distinguish one 800 from another by the AMC score...[/quote]

If colleges looked at AMC scores and understood them, I would say this is true.  So far as I know, only MIT explicitly asks about them, though a few other colleges might.  Mid-level universities and below probably don't need to distinguish beyond the SAT, so in that regard, I think you're right.  I think most of this conversation is centered around a standard that only a few dozen universities at most might need to evaluate.

[quote="MCrawford"][quote="kentliu"]i think the current system is good, with both SAT and AMC.. so colleges could distinguish one 800 from another by the AMC score...[/quote]

If colleges looked at AMC scores and understood them, I would say this is true.  So far as I know, only MIT explicitly asks about them, though a few other colleges might.[/quote]

Caltech asks about AMC scores in its application form (not currently posted on the Web, as the application deadline has passed for this year). A few other colleges do. More to the point, Harvard, Stanford, and Princeton surely evaluate applications with an awareness of the AMC program. The top math schools are also aware of accelerated standard curriculum programs, like [url=http://www.itcep.umn.edu/annual/aluminfo.html]UMTYMP[/url]. I think the problem is rather more that the [b]students[/b] don't know what a good math preparation is until it's too late, rather than that the colleges don't know how to find good math program applicants. Now AoPS is helping students learn how to be better prepared to make strong applications, and I think the strong applicants will mostly find places in strong math programs, as they generally have.

Yale's app still asks for (optional, of course) ASHME scores. heh.

[quote]Caltech asks about AMC scores in its application.... A few other colleges do. More to the point, Harvard, Stanford, and Princeton surely evaluate applications with an awareness of the AMC program. The top math schools are also aware of accelerated standard curriculum programs.[/quote]

The top schools are certainly aware of the MAA (or is it all AMC now?) high school math competitions and their general meaning; saying you won the USAMO, an IMO medal, or (maybe) a statewide olympiad will not be lost on them.  But I doubt most admissions personnel are up to date on the year-to-year or even decade-to-decade evolution of the contests, or the addition of newer small contests.  (The AMC director wrote in a thread here, that there are now 19 (!) paths to the USAMO, as compared to just one in the old days.)

I guess the same is true of Westinghouse, RSI, ISEF and the other science competitions
and programs.  The colleges know the general facts about the major ones but not necessarily the details of what it means to be Westinghouse finalist versus ISEF winner.

currently:
accepted by: UCR, UCD, UCLA
Rejected: MIT

caltech coming bak this sat or monday, my expectations aint high..


Find a closed form for $S_{n}$ if

$S_{n}=F_{1}F_{2}+F_{2}F_{3}+F_{3}F_{4}+\dots+F_{n}F_{n+1}$ where $F_{n}$ is the $n^{th}$ Fibonacci number.

[hide="how i started"]
Obviously $S_{n}=S_{n-1}+F_{n}F_{n+1}$.
Also after a little work you should be able to find a way to express $F_{n}F_{n+1}$ in terms of either Lucas numbers of Fibonacci numbers.
[/hide]

[hide="hint"]Use Binet's formular $F_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta}$.[/hide]

[hide]${F(n)=\frac{2^{-n}\left(\left(1+\sqrt{5}\right)^{n}-\left(1-\sqrt{5}\right)^{n}\right)}{\sqrt{5}}}$

${F(n) F(n+1)=\frac{1}{5}2^{-2 n-1}\left(\left(1+\sqrt{5}\right)^{n}-\left(1-\sqrt{5}\right)^{n}\right) \left(\left(1+\sqrt{5}\right)^{n+1}-\left(1-\sqrt{5}\right)^{n+1}\right)}$
${F(n) F(n+1)=\frac{1}{5}2^{-2 n-1}\left(-\left(-1+\sqrt{5}\right)^{2 n+1}-(-1)^{n}2^{2 n+1}+\left(1+\sqrt{5}\right)^{2 n+1}\right)}$
[/hide]

to solve the other equation, convert it into a Sum

the great Sigma

he


OK, so the question is, how do I create (in the simplest way possible) a simple package that automatically loads a few definitions (that are active once \usepackage{mypackage} command appears in the document)?

And to make things complicated (really?) how about adding other random commands, like \setlength ?

From "the not so short introduction to latex2e"

[quote]
6.1.5 Your Own Package
If you define a lot of new environments and commands, the preamble of your
document will get quite long. In this situation, it is a good idea to create
a LATEX package containing all your command and environment definitions.
You can then use the \usepackage command to make the package available
in your document.

% Demo Package by Tobias Oetiker
\ProvidesPackage{demopack}
\newcommand{\tnss}{The not so Short Introduction to \LaTeXe}
\newcommand{\txsit}[1]{The \emph{#1} Short
Introduction to \LaTeXe}
\newenvironment{king}{\begin{quote}}{\end{quote}}

Figure 6.1: Example Package.

Writing a package basically consists of copying the contents of your document
preamble into a separate file with a name ending in .sty. There is
one special command,

\ProvidesPackage{package name}

for use at the very beginning of your package file. \ProvidesPackage tells
LATEX the name of the package and will allow it to issue a sensible error
message when you try to include a package twice. Figure 6.1 shows a small
example package that contains the commands defined in the examples above.[/quote]

I think it's pretty simple but I didn't try it yet

hmm, not to sure about that...

\renewcommand{\thesection}{\Roman{section}}

is the same as

\def\thesection{\Roman{section}}

maybe \def will define a new command if it doesn't exist and will replace a command if it does exist.  \newcommand and \renewcommand would be two more "precise" command.  The opposite of \def would be \providecommand which work like \newcommand but if the command is already define, latex will silently ignore that line.

If it's not clear, it's because it isn't clear in my head... :?

and about new packages, I tried it and it's as easy as it looks.  Just don't forget to put your package in the right directory in latex and to refresh.

And a personnal advice:

\usepackage{frrectoverso}
%C:\texmf\tex\latex\frrectoverso

this way, if you want to modify your package, you won't look for it for hours...

[quote="Valentin Vornicu"]I feel the need to ask then: what's the difference between \def and \newcommand actually?  :?[/quote]
[url=http://www.amazon.co.uk/exec/obidos/ASIN/0321173856/qid=1132140188/sr=8-1/ref=sr_8_xs_ap_i1_xgl/202-1512953-5013434]Kopka & Daly[/url] discuss this point. Paraphrasing what they say, \def is a primitive TeX command; such commands form the bedrock of any format and will always do exactly what the programmer expected. On the other hand equivalent LaTeX commands could do more. For example, \newcommand checks for name clashes with existing commands.

The book then discusses the fragility of using commands with \def (ie premature interpretation when used as an argument of other commands) which doesn't happen with LaTeX commands like \newcommand. It concludes with the observation that there may well be occasions where \def has to be used but this should not be done lightly as it could risk future incompatibility.


There was some problem with coloring hexagons, and one was labeled with an R, and the problem said something like &quotRquot instead of actual quotation marks.

I just saw another problem which asked for the area of a mirror given the width and dimensions of its frame, and the problem referred to a picture, but no picture was included.

there was also one about topsoil where there was no space between a word and the subsequent price.

and one about Sam driving a car where an apostrophe was bad and came out as a funny a (the problem had a graph and asked for average speed)

Thanks for your notes.

A good way to report these problems is to review the questions at the end of the game.  You can do this by clicking the review button in the lobby.  Then on the problem that is in error, click the report a problem button.  This lets us see exactly which problem you are finding an error on.

OK I wasn't aware that this option was available. I'll do that in the future.


If $x$ and $y$ are real numbers that verify the relation $x^{2}+y^{2}+168 = 24x+10y$ then $x \in [11,13]$ and $y \in [4,6]$.


geometry
trapezoid
parallelogram
geometry unsolved

In a trapezoid $ABCD$ with $AD \parallel BC , O_{1}, O_{2}, O_{3}, O_{4}$ denote the circles with diameters $AB, BC, CD, DA$, respectively. Show that there exists a circle with center inside the trapezoid which is tangent to all the four circles $O_{1},..., O_{4}$ if and only if $ABCD$ is a parallelogram.

[quote=Nevergiveupbtw]Assume the existion of circle O, we have O will external tagent with (O2),(O4) or internal tagent with them.
If ABCD is not a parallelogram, AC intersect BD at K, AB intersect CD at H.
The case O external tagent with (O2),(O4) at M and N : we have M,N, K are collinear.
So that O2M and O4N are parallel but O is the intersection of them, contradiction.
Similarly for 2nd case.
If ABCD is a parallelogram we are easy to show that the circle with center K and radius is a half of AD-AB( or AB-AD)[/quote]

Your solution is not clear, remaining a few questions.

BTW if anyone would give a clear or an official solution?

I simply cannot believe that there were no full solutions posted for this absolutely wonderful geometry problem for 12 years.

(Nevergiveupbtw's solution is wrong, because by Monge Theorem we have $M,N,H$ colinear, not $M,N,K$ colinear.)

I will now post the official solution for this problem.

We will denote the midpoints of $AB, BC, CD, DA$ as $O_1, O_2,O_3,O_4$.
The proposition that there exists such a circle is equivalent to the proposition that for some $O$ inside $ABCD$,
$$OO_{1}+O_{1}A=OO_{2}+O_{2}B=OO_{3}+O_{3}C=OO_{4}+O_{4}D$$
Firstly, if $ABCD$ is a parallelogram, we can easily see that for $O=AC\cap BD$ satisfies the condition.

Now we suppose $ABCD$ is not a parallelogram, and we prove that there does not exist such $O$.
Suppose there exists such $O$ satisfying the undesired condition by contradiction.
Without loss of generality we assume $AD<BC$ and $AB\leq CD$.

Denote the midpoints of $AC$ and $BD$ as $P$ and $Q$, respectively.
Since $AD\parallel BC$, we easily see that $\square AO_{4}QO_{1}$, $\square BO_{1}PO_{2}$, $\square CO_{2}QO_{3}$, $\square DO_{3}PO_{4}$ are parallelograms.
We notice that since $ABCD$ is not a parallelogram, $O=P$ and $O=Q$ does not satisfy the undesired condition.

[b]Case 1.[/b] $O$ is on the interior or boundary of $\square DO_{3}QO_{4}$.

Let $H=OO_{1}\cap QO_{4}$. We see that
$$OO_{1}+O_{1}A=OH+HO_{1}+QO_{4}=OH+HO_{1}+QH+HO_{4}\geq OO_{4}+QO_{1}=OO_{4}+O_{4}D$$
and equality holds if and only if $O=Q$, which is a case we have already ruled out, so contradiction.

Now if $O$ is on the interior or boundary of $\square AO_{4}PO_{1}$, $\square BO_{1}QO_{2}$ or $\square CO_{2}PO_{3}$, by the same way we have a contradiction.
Therefore the only case left is...

[b]Case 2.[/b] $O$ is on the interior of $\triangle PO_{2}Q$.

Without loss of generality we assume that $O$ is on the interior or boundary of $\triangle QO_{2}O_{4}$.
We have $QO_{4}+QO_{2}\geq OO_{2}+OO_{4}$, which means $O_{1}A+O_{3}C\geq OO_{2}+OO_{4}$.
Also, $OO_{1}+OO_{3}>O_{1}O_{3}=O_{1}P+PO_{3}=O_{2}B+O_{4}D$.
Add up both sides, and we have
$$OO_{1}+O_{1}A+OO_{3}+O_{3}C>OO_{2}+O_{2}B+OO_{4}+O_{4}D$$
which leads to a contradiction. $\blacksquare$

[quote]
We will denote the midpoints of $AB, BC, CD, DA$ as $O_1, O_2,O_3,O_4$.
The proposition that there exists such a circle is equivalent to the proposition that for some $O$ inside $ABCD$,
$$OO_{1}+O_{1}A=OO_{2}+O_{2}B=OO_{3}+O_{3}C=OO_{4}+O_{4}D$$
[/quote]
Could you explain this part?


floor function
graph theory
combinatorics open
combinatorics

Given [ne!] +1 points  in plane .Color it by one in n colors .Proved that there are a triangle with three sides in the same color

I do not see how you could obtain colored sides by colouring only a finite number of points. Maybe are you talking about triangles with three vertices with the same color? In that case, the given bound can easily be reduced to $2n+1$ from the pigeon-hole principle.

Pierre.

The problem probably concerns coloring all edges of a complete graph with $\lfloor e \cdot n! \rfloor+1$ vertices (at least it's correct and not so unsharp).
Then it's also posted before.

@nkht-tk14: If you want help, answer to what pbornsztein said instead of crying for help.


calculus
integration
function
trigonometry
limit
calculus computations

Show that a function $ f(x)\equal{}\int_{\minus{}1}^1 (1\minus{}|\ t\ |)\cos (xt)\ dt$ is continuous at $ x\equal{}0$.

Three things must be satisfied:

1) f(0) must be defined, which it is because f(0) = 1

2) $ \lim_{x \to 0}f(x)$ must exist, which is 1 in this case.

3) $ \lim_{x \to 0}f(x) \equal{} f(0) \equal{} 1$

Thus f(x) is continuous


i TOOK THE cOMPUTER SCIENCE aB EXAM I WAS wondering if someone can post the A  multiple- choice questions to see if any of them are the same.!

I didnt take the comp sci test this year, but isn't discussion of the multiple choice questions illegal?

plus, im pretty sure that we all took the same test.

[quote="Mariag"]isn't discussion of the multiple choice questions illegal?[/quote]

Yes. Some MCQs are reused to maintain a somewhat stable level of difficulty so  the College Board does not allow test takers to discuss them.


LaTeX
inequalities unsolved
inequalities

[quote="Master"]let a,b,c>0
and ab+bc+ca=1
prove that
(a^3+b^3+c^3)/abc>=6sum(a(a^2/(1+a^2))[/quote]

I don't understand the right hand side part, can you explain?

I think that you made a mistake in typing,master.If not it is easy and the equality doesn't occur :? .
We have:
[tex] 6\sum a\frac{a^2}{a^2+1} \leq 3(a^2+b^2+c^2) [/tex] (since $a^2+1 \geq 2a$).
[tex] \frac{a^3+b^3+c^3}{abc}\geq \frac{(a^2+b^2+c^2)(a+b+c)}{3abc} \geq 3(a^2+b^2+c^2) [/tex] (since $a+b+c=(a+b+c)(ab+ac+bc) \geq 9abc $)
 :)

i will write it again, thank a lot Anh Cuong

and ab+bc+ca=1
prove that
(a^3+b^3+c^3)(1/a^2+1/b^2+1/c^2)>=6{[(a[a^2/(1+a^2)]^1/2+b[(b^2/(1+b^2)]^1/2+c[(c^2/(1+c^2))]^1/2]}

 :blush: if you have time AnhCuong, please "translate " it into LaTex for me, please :blush:

I think you should change the title:"Easy...yes,nice...maybe".
By Cauchy,we have:
[tex] (a\sqrt{a} \sqrt{\frac{a}{a^2+1}}+b\sqrt{b} \sqrt{\frac{b}{b^2+1}}+c\sqrt{c} \sqrt{\frac{c}{c^2+1}})^2 \leq (a^3+b^3+c^3) (\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{a}{a^2+1}) \leq \frac{3 \sqrt{3} (a^3+b^3+c^3)}{4} [/tex]
By Cauchy too,we have:
[tex] (a^3+b^3+c^3)^2(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})^2 \geq 27 \sqrt{3} ( a^3+b^3+c^3) [/tex]


it used to be admin birch, navier, yangs, remain, poincare, hodge, p vs np, and new memeber.
What's the status now?

Where?  All I see are the normal ones.  Except for that Alcumus developer who has "Developer"

The titles for normal users are unchanged. Administrators can apparently change their titles; rrusczyk chose "Boss", for example.


there's http://www.number2.com which is a free online study prep thing. also, check out whatever review books are available at your local or school library. it doesn't really matter what book you get.

For the vocab. it doesn't really matter what book, but for the practice tests Kaplan's the best. Princeton Review is the worst.

Going to College board. (no pun intended)

The books I recommend for building up SAT verbal scores are the genuine College Board book about the SAT test, and then reading any and every book that you like to read. Read about something you like. After you read one book, read about what you are curious about after reading that book. And so on. Subscribe to some magazines or to the New York Times national edition too. (AT LEAST read your town's local newspaper, even if it is a rag like my town's local newspaper.)

READ, READ, READ, and READ. That will help.

I just learned classical Latin and Greek roots, suffixes, and such.

Helps a lot for those sentence completions.

i donit know, english is my second language and i got 650 and that's great for me..
i just practice with an SAT book for  3 weeks.

Knowing alot of latin and greek really helps in my opinion.  I've never studied any vocabulary before and I can derive a large portion of the words.  It never hurts to study though.
Latin: 52% of English
Greek: 10% of English
Other: the rest
Those are approximations, but it just shows how much latin can help you if you have a true mastery of the subject.

Most high scorers I know don't recommend devoting large amounts of time to pure vocab build-age. I mean you can pwn the vocab and still not score high. It's far more important to read a lot and get familiar with the English language itself, including analysis of literature.

But if you must do vocab, then find yourself a good 200-400 word (don't do something crazy like 1000 words) list and take it slowly. Try to do 10-20 per day.

Get yourself some Thoreau, some 18th-19th century poets, and a subscription to the Economist. Once you can read all those without trouble, I say you're more than set for vocab/reading/writing. (It's really not too bad on the SAT - though I remember there was a time when "effervescent" seemed like a really hard word to me) :)

[quote="star99"]I would like to know, which books or websites do you recommend for the SAT, especially in the vocabulary area?[/quote]

Maybe [url=http://www.sparknotes.com/]www.sparknotes.com[/url]


[quote="boukharfane"]x,y and z are reals>0.
prove that (x+y)^z+(y+z)^x+(z+x)^y>2[/quote]

i don`t think it`s true. check x=y=z and chose a convenient x

[quote="boukharfane"]x,y and z are reals>0.
prove that (x+y)^z+(y+z)^x+(z+x)^y>2[/quote]
The hint see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42452


algorithm function algebra polynomial calculus integration logarithms

Is there an Algorithm for integrating elementary functions? Elementary functions here refers to the class of funtions containing y=x ,y=exp(x), y=const and closed under multi, addi, iterating, contrafunctioning. Some mathematicians have studied on part of this question including Liouville, and Matlab also has a program for this. But the "elementary func" in their research seems incomplete, usually dosn't allowing iter and contra and in Matlab don't seems containing contrafunctioning.

Is it true?

Practically all elementary functions cannot be integrated in elementary terms (they way to prove this is similar to they way one proves that some polynomials cannot be solved by radicals).
In general, there is the Risch algorithm that decides if such an elementary integral exists and what it is.

I searched for this Risch algorithm. And find " The Risch algorithm is used to integrate elementary functions. These are functions obtained by composing exponentials, logarithms, radicals, and the four operations (+ − × ÷).". Which means the "elementary" that the Algo solves donot contain contrafunction, for example $ (exp^{2}(x)\plus{}x^{2})^{\minus{}1}$, does it?

Indeed, the most common definition of elementary only allows $ \plus{},\minus{},\cdot,/,\exp,\log$, solving polynomial equations (so only contrafunctions for polynomials) and composition. So Risch only works for those.
The proofs for existence of nonelementary integrable elementary functions of your kind still hold after some adaptions. I'm not really into the way Risch works, but as far as I know about it, there is some chance of adaption (but this will get very difficult). Theoretically, it's just about differential Galois theory on some differential field, where the original field is that of elementary functions and yours is just some extension of it.


[quote="Nexmus"]Just wondering, i've seen people with pictures in their signature. How do i get one in my sig?

Just an example, the user "myth" has an opera browser image in his sig.[/quote]

Find the website of the picture, then use this code:

[code][img](your url)[/img][/code]

If you want to create your own picture, you can either use ImageShack or make your own personal gallery in AoPS.

When I add it it says:

You have 1 image, when only 0 images is allowed.

how do i fix this?


Yes, I didn't know you could have blinking tex either :D. I'm using Firefox 1.5.0.4 on Gentoo Linux that is all ~x86. \[\left(p+1\right)\,\left \lceil{{\left(2\,p+1\right)\,\left \lceil{{\left(2\,p+1\right)\,\left \lceil{{\left(2\,p+1\right)\, \left \lceil{{\left(2\,p+1\right)\,\left \lceil{{\left(2\,p+1 \right)\,\left \lceil{{\left(2\,p+1\right)\,\left \lceil{{\left(2 \,p+1\right)\,\left \lceil{{\left(2\,p+1\right)\,x}\over{\left(p+1 \right)^{2}}}\right \rceil}\over{p+1}}\right \rceil}\over{p+1}}\right \rceil}\over{p+1}}\right \rceil}\over{p+1}}\right \rceil }\over{p+1}}\right \rceil}\over{p+1}}\right \rceil}\over{p+1}}\right \rceil\] [code]\left(p+1\right)\,\left \lceil {{\left(2\,p+1\right)\,\left \lceil
 {{\left(2\,p+1\right)\,\left \lceil {{\left(2\,p+1\right)\,
 \left \lceil {{\left(2\,p+1\right)\,\left \lceil {{\left(2\,p+1
 \right)\,\left \lceil {{\left(2\,p+1\right)\,\left \lceil {{\left(2
 \,p+1\right)\,\left \lceil {{\left(2\,p+1\right)\,x}\over{\left(p+1
 \right)^2}} \right \rceil}\over{p+1}} \right \rceil}\over{p+1}}
  \right \rceil}\over{p+1}} \right \rceil}\over{p+1}} \right \rceil
 }\over{p+1}} \right \rceil}\over{p+1}} \right \rceil}\over{p+1}}
  \right \rceil[/code]

I use firefox 1.5.0.4 (win 2000) and my $TeX$ doesn't blink. :(
This is what your code looks like for me solafidefarms:
$\left(p+1\right)\,\left \lceil{{\left(2\,p+1\right)\,\left \lceil{{\left(2\,p+1\right)\,\left \lceil{{\left(2\,p+1\right)\, \left \lceil{{\left(2\,p+1\right)\,\left \lceil{{\left(2\,p+1 \right)\,\left \lceil{{\left(2\,p+1\right)\,\left \lceil{{\left(2 \,p+1\right)\,\left \lceil{{\left(2\,p+1\right)\,x}\over{\left(p+1 \right)^{2}}}\right \rceil}\over{p+1}}\right \rceil}\over{p+1}}\right \rceil}\over{p+1}}\right \rceil}\over{p+1}}\right \rceil }\over{p+1}}\right \rceil}\over{p+1}}\right \rceil}\over{p+1}}\right \rceil$

[quote="ZetaX"]I second that (same FireFox, Windows XP). Also it happens to do the same when I post it myself.
But don't fix it, it's nice :D
@solafidefarms: how have you found that¿[/quote]

How did I find that equation? I was using xmaxima, kind of like mathematica or maple, and did the following commands:

f(x):=(p+1)*ceiling(x*(2*p+1)/((p+1)^2));
tex(f(f(f(f(f(f(f(f(x)))))))));

and it gave me that formula. :lol: Maybe the code is TeX and the LaTeX renderer can't handle TeX?

It blinks in everything I've tried: (Firefox on Mac, Safari on Mac, Mozilla on Windows). I don't think this has anything to do with browsers.

[quote="b-flat"]It could also be because the image is so large that it takes a long time to generate/refresh, etc.  This might not be it, but you could try it out by posting incredibly large blocks of $\LaTeX$ in the test forum.[/quote]Or you might not try that  :!:

It blinks here too, but I have absolutely no idea why.

Seems like it just exceeds the maximum vertical size allowed for the picture. I edited it so that it should end with division by $p+2$ now  but this part is clearly not there.
${\left(p+1\right)\,\left \lceil{{\left(2\,p+1\right)\,\left \lceil{{\left(2\,p+1\right)\,\left \lceil{{\left(2\,p+1\right)\, \left \lceil{{\left(2\,p+1\right)\,\left \lceil{{\left(2\,p+1 \right)\,\left \lceil{{\left(2\,p+1\right)\,\left \lceil{{\left(2 \,p+1\right)\,\left \lceil{{\left(2\,p+1\right)\,x}\over{\left(p+1 \right)^{2}}}\right \rceil}\over{p+1}}\right \rceil}\over{p+1}}\right \rceil}\over{p+1}}\right \rceil}\over{p+1}}\right \rceil }\over{p+1}}\right \rceil}\over{p+1}}\right \rceil}\over{p+1}}\right \rceil}\over{p+2}$
the end

It blinks because the $LaTeX$ extends onto 2 pages.  ImageMagick converts the 2 pages into an animated gif. :roll: Looks like one of treething's nauseating animations.


Good luck to all taking the AMC this Tuesday.

[b]Do not discuss the test until noon Thursday ET. [/b]  This includes no talk like 'That was hard' or 'I almost ran out of time', etc.

We will be hosting a Math Jam Thursday evening at 7:30 PM ET (4:30 PM PT) to discuss the test.  See you there!

[quote="DPopov"]Have there been any announcements about how this year's test will compare to tests in the past (easier/hard)?[/quote]

Oh, if only they were that kind.

Thanks!

Thursday, February 3 12:00 noon ET is a good time to open discussion.  By then the contest will be over in all time zones world-wide, and there will even be a few students taking the contest unofficially on Wednesday.

Steve Dunbar
AMC Director

I'll be done with the AMC 12 almost exactly 12 hours from now (it's 1:10 here). I'm too excited to go to bed  :D.

Treething has taken the AMCs today :) , 8 hours ago.

usually higher than 100 or 120....you should read their website for any statistics you want to know.

[quote="probability1.01"]Does anyone know what the threshold for AMC 12 and AMC 10 normally are? I know there's the 100 and 120, but they also take a percentile--is that usually any lower than the solid thresholds?[/quote]

Percentile on 12 is, as far as I know, always higher than the 100 number. As for 10, last year was the first year the 120 points "requirement" was given. On 10A, it was 10 points below the mark, and on 10B, percentile was higher than the 120 mark.

When AMC 12 says top 5%, does it mean top 5% in the state, school, nation...?

Top 5% of all test takers period. Meaning generally the US, and anyone else who might take it outside the US

Does it really make a difference?(assuming most schools/states are equal)

No, but not all states/schools are equal.


Find the max value of $ k$ for which is true that

$ a^3\plus{}b^3\plus{}c^3 \geq 3abc \plus{}k(a\minus{}b)(b\minus{}c)(c\minus{}a)$

where $ a,b,c$ are positive real numbers.

[quote="manlio"]Find the max value of $ k$ for which is true that

$ a^3 \plus{} b^3 \plus{} c^3 \geq 3abc \plus{} k(a \minus{} b)(b \minus{} c)(c \minus{} a)$

where $ a,b,c$ are positive real numbers.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101916

Thank you very much, Arqady. :)

I didn't remember it was already posted. Sorry.


geometry
analytic geometry
graphing lines
slope
probability
algebra
polynomial

Put solutions for the USC test here.  [url=http://www.math.sc.edu/contest/problems.html]Here's[/url] the link to the problems.

[hide="16"]After calling the radius of $C_{1}$ as $x$ and drawing a tangent line from the center of $C$ to the intersection of the two bigger circle, we can draw a right triangle with legs $x$ and $x-3$ and hypotenuse $x+3$. Therefore $x^{2}+(x-3)^{2}=(x+3)^{2}\Longrightarrow x=12$.
[/hide]
[hide="17"]$\frac{1}{\frac{1}{60}+\frac{1}{90}}=36$
[/hide]
[hide="18"]Because the two triangles are similar and right triangles, $b$ is the diameter of the circle. $b=\frac{2}{3}a$ and the area of the triangle is $\frac{a^{2}}{9}\pi$.[/hide]

[hide="19"]All the possible lattice points lie on the line from $(2,0)$ and $(16,203)$. The lines slope is $\frac{29}{2}$, which means that for every two spaces along the positive $x$ axis, we have a lattice point, thus there are $8$ total lattice points within the interval. [/hide]

[hide="20"]Look at the two equations graphically. $x^{2}=y^{2}\Longrightarrow y=\pm x$. Drawing this, we see that it’s simply two mirrored lines intersecting at the origin. The second equation gives us a unit circle which can be shifter only along the $x$ axis. If the unit circle is placed at the origin, there are four intersections with the lines (four solutions). However, if the unit circle is shifted one unit either to the left or right, the lines intersect the circle at three points (one point at the origin). Therefore our solutions are $a=\pm 1$.[/hide]

 [hide="21"]We have to take into account that the coins are chosen without replacement. Either the first coin is counterfeit or the first coin is real. Our probability is thus: $\frac{9}{14}\cdot \frac{5}{13}+\frac{5}{14}\cdot \frac{9}{13}=\frac{45}{91}$.
[/hide]
[hide="22"]Eh..I just multiplied it out quickly…the digit is $3$.[/hide]

[hide="23"]The flower is the intersection of four equal circles with radius $a$. Choose any one of the circles. The area of the shaded circular region is the area of the semi circle minus the area of the iscoceles right triangle with it’s hypotenuse as the diameter of the circle. Thus, the two circular shaded region area is $\frac{a^{2}}{2}\pi-a^{2}$ However, there are four of these circular sections of two. Therefore, the total area is $2\pi a^{2}-4a^{2}$.[/hide]

[hide="24"] $3^{2}+x^{2}=6^{2}+10^{2}\Longrightarrow x=\sqrt{127}$. This is known as the British Flag Theorem and can be proven by dropping perpendiculars and calculating the area that way.[/hide]

[hide="25"] Not only does the question have to do with digits (divisibility by $9$), but one can easily use trial and error on this. The answer is $9$.[/hide]

 [hide="27"]There are $1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}=91$ total squares[/hide].

 [hide="28"]By adding the equations, we have $x_{1}+x_{2}+x_{3}=3-(x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})$. Consider a cubic polynomial with roots $x_{1}, x_{2}, x_{3}$, there are a finite number of sets $>4$ of a polynomial that satisfies this condition (through vieta) and contains real roots.[/hide]

Are numbers 29 and 30 calculus? I don’t know limits…

EDIT: mistake on #16

Yeah, they look like calculus to me.  They're not too hard calculus problems though so I'll post solutions tomorrow.

There is a chapter in AoPS about limits.  Are limits really calculus, or can I read that chapter and be able to do those problems?

[hide="29"](b) The slope of the tangent is $2u$, so the slope of the normal is $\frac{-1}{2u}$. Then $y=\frac{-1}{2u}x+u^{2}+\frac{1}{2}$. The answer is the limit as u goes to 0 of $u^{2}+\frac{1}{2}$, which is $\frac{1}{2}$. [/hide]

[hide="30"](a) The given quantity equals $\int_{0}^{1}e^{x}dx=e-1$.[/hide]

Calculus is pretty easy. It just has a bad reputation for being hard. Limits are even easier. They are not calculus, though they are analysis. You should probably read up on them if they cause trouble.

Note: if there were a chapter on calculus, you could read that chapter and be able to do those problems, too :wink: .

I read the AoPS book, so with a basic knowledge of limits, I attempted #30:

$\lim_{n\to\infty}{\frac{1}{n}\sum_{k=1}^{n}e^{k/n}}$
Using properties of limits, this can be rewritten as:

$\lim_{n\to\infty}\frac{1}{n}\cdot \lim_{n\to\infty}{\sum_{k=1}^{n}e^{k/n}}$

We can evaluate the letf-hand term as $0$, and get:
$0 \cdot \lim_{n\to\infty}{\sum_{k=1}^{n}e^{k/n}}$

Therefore, the limit is equal to $0$.  What did I do wrong?  Is this an indeterminate form?

[hide="EDIT"]
Maybe we can also evaluate the right-hand side:

$\lim_{n\to\infty}{\sum_{k=1}^{n}e^{k/n}}$
This is continuous, so we can just "plug in" the infinity for n..?
$\lim_{n\to\infty}{\sum_{k=1}^{\infty}e^{k/n}}$
Then we can use properties of limits, as well as definition of a summation, to rewrite this as:
$e^{\lim_{n\to\infty}{1/n}}+e^{\lim_{n\to\infty}{2/n}}...$
This will equal $e^{0}+e^{0}+e^{0}....=1+1+1+1+1...= \infty$.

Because we had the LHS equal to 0, and the RHS equal to $\infty$, is this an indeterminate form?
[/hide]

I know limits, but not integrals/derivitives.  Can problem 30 be explained without calculus (other than properties of limits)?

[quote="joml88"]Well, first of all, according to your limit properties we have $\lim_{n\to \infty}n\cdot \frac{1}{n}= \lim_{n\to \infty}n\cdot \lim_{n\to \infty}\frac{1}{n}= \lim_{n\to \infty}n\cdot 0 = 0$.

However, I tend to think that $\lim_{n\to \infty}n\cdot \frac{1}{n}=\lim_{n\to \infty}1 = 1$.

edit: $0\cdot \infty$ is an indeterminate form.  Also, that multiplicative identity [i]does[/i] work but both of the limits must be finite.

2nd edit: I'm pretty sure any method you use for this one will require calculus.  The limit in question is basically the definition of that integral jb has there.  You won't really need any of this stuff until you get into calculus anyway, so there is no reason to worry about it.[/quote]

OK, thanks, that makes sense.  I was trying to learn limits, but they seem too tied to other calculus concepts, which is weird, because I think, in calculus, you learn limits first?

Limits are the foundation of all of calculus.  Once limits were defined, differentiation, antidifferentiation/integration followed.  If you're talking about epsilon-delta stuff, that's not calculus and you can get around it if you just want to learn about limits.  Epsilon delta is for rigorously proving limits.  The only time I can think of when you will see calculus with the limits is l'Hopital's rule.

Can you explain this "epsilon-delta" stuff, or at least describe it and point to some resources?

I think [url=http://www.ocf.berkeley.edu/~yosenl/math/epsilon-delta.pdf]this[/url] does a good job of explaining them.  Ask if you have any questions.

"Ask if you [b]have[/b] any questions."  Don't you know me well enough Joe?  Anyway.... this is basically a technique used for proofs?

[quote="joml88"]Limits are the foundation of all of calculus.  Once limits were defined, differentiation, antidifferentiation/integration followed.  If you're talking about epsilon-delta stuff, that's not calculus and you can get around it if you just want to learn about limits.  [b]Epsilon delta is for rigorously proving limits.[/b]  The only time I can think of when you will see calculus with the limits is l'Hopital's rule.[/quote]

You also quoted me wrong.  I certainly did not say "Ask if you [i]need[/i] any questions."

And, no, I don't know you well enough Shawn.

-Jow

[quote="joml88"]OH DARN.  WHY DID MY PARENTS GIVE ME SUCH A HARD NAME TO SPELL :(

At least I pelled you're name write Sen.[/quote]

This is no joke.

u won't find limits on uscs anymore and you won't find them on the amc or aime. wait till calculus

How about solutions for numbers 7, 13, and 15?  I got 7, but I didn't do it hardcorely, I used common sense.

Also, I'm getting a different answer on problem number 16.  I did it the same way but I got down to the equation:
$x^{2}=12x$
$\boxed{x=12}$

Can you edit your solution (so when we get around to Latexing these, we'll have them ready to go)?

yea like for 7, "divide" the exponent...like literally, you know what i'm saying?

for 13, arc BC is $30^\circ$, so $\angle{O}=30^\circ$, and $BOC$ is iscoceles

for 15, largest value of $x=\pi/2$, and $\sin$ can't be bigger than $1$..

OK, I got 7 and 13 (stupid mistakes...) but 15 is still stumping me.  Solution?

um...i already pretty much gave the solution....


geometry
3D geometry
tetrahedron
function
floor function
algebra

1. A cube has volume 144 cubic cm.  Find the number of cubic cm in the volume of a tetrahedron whose vertices are one vertex of the cube and the midpoints of the three edges of the cube that are connected to the vertex.
[hide="ANSWER"]3[/hide]

2. How many zeroes are there to the right of the last non-zero digit of (5!)! ?
[hide="ANSWER"]28[/hide]

3. Market Town Department Store has a 25% discount on all goods at all times.  They advertised that they will increase the discount by 40% for the coming weekend.  If a radio was listed at 120 dollars before any discounts were applied, how many dollars would it cost after the weekend discount was applied?
[hide="ANSWER"]78 dollars[/hide]

[quote="LevatorNasolabialis"][hide]3 28 and 78[/hide] :)[/quote]

i think he wants to knw how u got it :D

[hide="1"]Area will be $ \frac{1}{3}Bh$The length of each side is $ \sqrt[3]{144}\equal{}2\sqrt[3]{18}$. The area is then $ \frac{1}{2}\cdot\frac{1}{3}\cdot\sqrt[3]{18^{3}}\equal{}3$[/hide]
[hide="2"]It's basically just asking how many times powers of 5 go into 120.
$ \frac{120}{5}\equal{} 24$
$ \frac{120}{25}\equal{} 4$. (I don't know how to put the floor function)
24+4=28.[/hide]
[hide="3"]Originally the discount is $ \$30$. After the weekend, the discount is $ 1.4\cdot{\$30}\equal{}\$42$.120-42=78[/hide]

[hide="#s 2 and 3"]

2: just find the powers of 5 in 5!

120/5=24
120/25=4.8, but we round down. 24+4=28

BTW, floor=[code]$\lfloor or \rfloor$[/code] whichever way u want it

3: before, u had a 25% discount. now, u have a (1.4)25% discount(35)

120*.65=78

[/hide]


The triangle $ABC$ is isosceles with $AB=AC$, and $\angle{BAC}<60^{\circ}$. The points $D$ and $E$ are chosen on the side $AC$ such that, $EB=ED$, and $\angle{ABD}\equiv\angle{CBE}$. Denote by $O$ the intersection point between the internal bisectors of the angles $\angle{BDC}$ and $\angle{ACB}$. Compute $\angle{COD}$.

what a bad question havent they got any questions its like a pre-middle school olympiad problem

Found this today [i](apologies for the necromancy)[/i] and, just like Jessen1111, I was a little confused:[quote=Jessen1111]can you explain the angle chasing ??[/quote]
[hide=so here's a slightly odd solution, for storage]Perversely writing $\measuredangle XY$ to mean the angle mod $180^\circ$ through which line $XY$ must be rotated anticlockwise to become parallel with $BC$, we're told that\begin{align*}\measuredangle BC&=0&&\text{by choice of angle baseline} \\
\measuredangle AB + \measuredangle AC &= 0&&\text{because }\triangle BAC\text{ is isosceles}\\
\measuredangle AB-\measuredangle BD &= \measuredangle BE-\measuredangle BC&&\text{because }\angle ABD = \angle CBE\\
\measuredangle AC-\measuredangle BD&=\measuredangle BD-\measuredangle BE&&\text{because }\triangle DEB\text{ is isosceles}\end{align*}
Substituting the first two into the sum of the second two gives $3\measuredangle BD = 0\pmod{180^\circ}$, so $\measuredangle BD=60^\circ$.

We're asked to find $\angle COD$.  Mod $180^\circ$, that's equal to $\measuredangle DO-\measuredangle CO$.  We're given one angle bisector ($2\measuredangle CO=\measuredangle AC+\measuredangle BC=\measuredangle AC$) and from the existence of the incentre we know another ($2\measuredangle DO=\measuredangle BD+\measuredangle AC$), so subtracting we get $2\angle COD=\measuredangle BD=60^\circ\pmod{180^\circ}$, so $\angle COD=120^\circ$.[/hide]

Got it! (most of the time I don't post because the solutions are already clear but this time I will clarify this one)

[quote=silouan]Let D,E inside AC
Let  $\widehat{ABD}=\widehat{CBE}=x$ . By Angle chasing we find that $\widehat{DOC}=180-B+x$ .(1)
But $O$ is the incenter in the triangle $DBC$  so
$\widehat{DOC}=90+\frac{\hat{B}}{2}-\frac{x}{2}$ (2)
(1)=(2)  so  $B-x=60$  and $\widehat{DOC}=120$ .
When $D\equiv C$  then $\widehat{DOC}=90$[/quote]

We have <ODC=1/2BDE=1/2DBE=1/2(B-2x), OCD=1/2B, so <DOC=180-1/2B-1/2B+x=180-B+x. On the other hand, B-2x=<DBE=<BDC=180-(B-x)-B
=> 180=3B-3x => 60=B-x. Substituting, we have <DOC=180-60=$\boxed{120}$.


A car has an engine which delivers a constant power. It accelerates from rest at time t = 0, and at t = t_0 its acceleration is a_0. What is its acceleration at t = 2t0? Ignore energy loss due to friction.
$ \frac{1}{2}a_0$
$ \frac{1}{\sqrt{2}}a_0$
$ a_0$
$ 2a_0$
$ \sqrt{2}a_0$

A mass is attached to an ideal spring. At time t = 0 the spring is at its natural length and the mass is given an initial velocity; the period of the ensuing (one-dimensional) simple harmonic motion is T. At what time is the power delivered to the mass by the spring first a maximum?
t=0
t=T/8
t=T/4
t=3T/8
t=T/2


Mr. Brennan has 7 boys and 4 girls in his statistics class. In how many ways can he pick 3 boys and 2 girls to give a group presentation tomorrow? (The order in which the boys and girls are chosen does not matter.)

Just to elaborate on what was previously said, 7choose3 is the same as 7*6*5/3!
which is 35, you do the same with the others one, 4 choose 2=6 so 6*35=210

To put in $LaTeX: \dbinom{7}{3} \cdot \dbinom{4}{2}= 35 \cdot 6 =210$.


inequalities
AoPSwiki
inequalities proposed

Suppose that

$ v_{m} \equal{} \frac {\sum_{1\le{k_{1}} < {k_{2}} < ... < {k_{m}}\le{n}}{x_{k_{1}}x_{k_{2}}...x_{k_{m}}}}{\binom{n}{m}}, v_0 \equal{} 1.$

Prove that

$ v_{m}^2 \minus{} v_{m \minus{} 1}v_{m \plus{} 1}\geq{0}$

holds $ \forall {x_1,x_2,...,x_{n}}\in{R}(n>m).$

If you cannot prove that general inequality , please just for m = 2 or m=3 to prove it.

[quote="Dumel"]it's Newton's inequality[/quote]

Yes,Dumel!Newton's inequality

    $ v_{m}^{2}-v_{m-1}v_{m+1}\geq{0}$

holds ${ \forall {x_1,x_2,...,x_n}\in{R}(n>m)}.$(G.H. HARDY:INEQUALITIES).

What's the point of this topic???   :roll:
You wanted to see if anybody know Newton's inequality?  :?:

There is a nice proof on the AoPSWiki here:
http://www.mathlinks.ro/Wiki/index.php/Newton%27s_Inequality


Consider all triangles $ ABC$ satisfying in the following conditions: $ AB\equal{}AC$, $ D$ is a point on $ \overline{AC}$ for which $ \overline{BD} \perp \overline{AC}$, $ AC$ and $ CD$ are integers, and $ BD^{2}\equal{}57$.  Among all such triangles, the smallest possible value of $ AC$ is

$ \textbf{(A)} \ 9 \qquad \textbf{(B)} \ 10 \qquad \textbf{(C)} \ 11 \qquad \textbf{(D)} \ 12 \qquad \textbf{(E)} \ 13$

[quote="azjps"][hide]Then $ AD \equal{} AC \minus{} CD$ and $ AB \equal{} AC$ are integers. By the Pythagorean Theorem, $ AD^2 \plus{} 57 \equal{} AB^2 \Longrightarrow 1 \cdot 57 \equal{} 3 \cdot 19 \equal{} (AB \minus{} AD)(AB \plus{} AD)$. Thus $ AC \equal{} AB \equal{} \frac {1 \plus{} 57}{2} \equal{} 29$ or $ \frac {3 \plus{} 19}{2} \equal{} 11 \Longrightarrow \mathrm{(C)}$.[/hide][/quote]

Why did you do this: $ 1 \cdot 57 \equal{} 3 \cdot 19 \equal{} (AB \minus{} AD)(AB \plus{} AD)$, and I don't get how it helped?

The key here is that the given side lengths are integers. Generally when this is the case, especially under a geometrical context, we want to find some way to get something to factor somewhere, so we can limit our answers to a certain number of cases, and then test to see which of the cases is (are) the correct one(s).

With the limited information given in the problem statement, it's pretty obvious the Pythagorean Theorem will be used, and as we are not told (nor can we find any information about) $ BC$, we must use $ \triangle ABD$. We know two sides are integers and the third is $ \sqrt{57}$, so we end up with $ 57 \equal{} (AB\minus{}AD)(AB\plus{}AD)$ by difference of squares. But $ AB\minus{}AD$ and $ AB\plus{}AD$ are also integers, so we just need to see what pair of numbers multiply to $ 57 \equal{} 3 \cdot 19$. These pairs would be $ (3,19),(1,57)$. Obviously $ AB \plus{} AD > AB \minus{} AD$, so either $ AB \plus{} AD \equal{} 57, AB \minus{} AD \equal{} 1$, or $ AB \plus{} AD \equal{} 19, AB \minus{} AD \equal{} 3$. These can easily be solved to find $ AB \equal{} 11, 29$, and we can confirm that $ 11$ yields a valid solution.


Find all reals $x$ satisfy $[x^2 - 2x] + 2[x] = [x]^2$

[quote="paradox102"]is $\big[ {x} \big]$ the greatest integer function?[/quote]
Yes.

Yes, $\frac 1 2$ doesn't work.  But the statement he made isn't true for all reals.  We're solving for x...

write x = k+y where k is the integer part and y is the mantissa

[x^2 - 2x] + 2[x] = [x]^2

$[y^2 + 2ky + k^2 - 2y -2k] = k^2 - 2k$

$k^2 - 2k + [y^2 + 2ky -2y] = k^2 -2k$

$[y^2 + 2ky -2y] = 0$

Let $f(y) = y^2 + 2(k-1)y$.  We need $f(y) \ge 0$ and $f(y) < 1$, and we want to determine the values of y that give this based on the parameter k.

The roots of f(y) are 0 and 2(k-1).  So if $k \le 0$ the roots are to the left of the y axis and we guarantee $f(y) \ge 0$.  If k = 1 then all values of y work.  If k=2 or higher then the curve is completely under the x axis.  So $k \le 1$.

For $f(y) < 1$, we require $y^2 + 2(k-1)y - 1 < 0$ in the interval $y \ in [0,1)$.  The quadratic opens up, and the roots of this quadratic are $-(k-1) \pm \sqrt{(k-1)^2 +1}$.  We want the positive root, so we see that y in $[0, \sqrt{(k-1)^2 + 1} - (k-1))$ is where y should be.

We combine these two conditions, so that $k \le 1$ and $y \in [0, \sqrt{(k-1)^2 + 1} - (k-1))$ describes the full nature of $x$.


AMC
AIME
function
AIME II
Pascal\u0027s Triangle

If we have a triangular array like the one shown [url=http://www.artofproblemsolving.com/Wiki/images/3/39/2007_AIME_II-13.png]here[/url]
and if any square above the bottom row is the sum of the two squares below
it, and if the bottom squares are labled $ a_0, a_1, ..., a_{10}$, what would the
topmost be equal to in terms of $ a_i$ for $ i \equal{} 0, 1, ..., 10$ (proof, please)?

Also, how do we find the total number of zigzag routes if we want to go from the top square to any of the squares at the bottom row (I know it's $ \binom{m}n$, where $ m$ is the number of rows and $ n$ is for the nth cell at that row, but not sure how to derive this)?

Merry Christmas!

[hide="Hint for your second problem"]Enumerate the boxes on a row with n boxes as 1..n, with the left-most box being 1.

On any move from one row to the row below, we can either move to the box to the left of the current box (+0 to the box number) or move to the box to the right (+1 to the box number).

Given that we make m moves, and we end on a cell number n, how can we recast this problem in terms of the box numbers?[/hide]

Got it; thanks.
(So for m rows, there must be m boxes traveled, and for the nth box in the (m-1)th row - with the first one labeled with the 0th row - there must be m-n boxes to the left and n to the right for a total of m boxes.)

Any idea for the first one?

[hide="The first problem"] By far the easiest way to do this problem is generating functions.  Letting $ A(x) \equal{} a_0 \plus{} a_1 x \plus{} ... \plus{} a_{10} x^{10}$, the entries in the $ n^{th}$ row of the triangle (starting from the bottom) are given by the middle coefficients of

$ (1 \plus{} x)^n A(x)$;

in particular, the topmost square is the central coefficient in $ (1 \plus{} x)^{10} A(x)$, that is, the coefficient of $ x^{10}$, which is just

$ {10 \choose 10} a_0 \plus{} {10 \choose 9} a_1 \plus{} ... \plus{} {10 \choose 0} a_{10}$.

You can also prove that the pattern above holds by induction.  Note also the special cases that $ a_i \equal{} 1$ or $ a_i \equal{} ( \minus{} 1)^i$. [/hide]
Edit:  Upon further consideration, the counting argument is just as simple.  Note that the result is linear in each of the bottom numbers, so it suffices to figure out the contribution from each number individually, so then all you have to do is count paths and the coefficient of each number is the number of paths from that number to the top, so the second question solves the first.

[quote="t0rajir0u"]Letting $ A(x) \equal{} a_0 \plus{} a_1 x \plus{} ... \plus{} a_{10} x^{10}$, the entries in the $ n^{th}$ row of the triangle (starting from the bottom) are given by the middle coefficients of

$ (1 \plus{} x)^n A(x)$[/quote]

How do we know it's $ (1\plus{}x)^n A(x)$, and what do you mean by the middle coefficients?  I'm not very good at generating functions.

For example, $ (1 \plus{} x) A(x) \equal{} a_0 \plus{} (a_1 \plus{} a_0) x \plus{} (a_2 \plus{} a_1) x^2 \plus{} ... \plus{} (a_{10} \plus{} a_9) x^{10} \plus{} a_{10} x^{11}$.  Discarding the constant and leading coefficients, we arrive at precisely the entries in the first row from the bottom.  In other words, multiplying by $ 1 \plus{} x$ has the same effect as the addition rule.  This, by the way, is precisely the same generating function that explains Pascal's triangle; this is actually the same phenomenon.