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Announcements and invitations will be going out across the country over the next month regarding America's First Internet Mathematics Competition.

AHSIMC's inaugural mathematics competition will take place the week of September 13-17, 2004. Teams of 5 high school students will compete against the nation's best students nationwide, with the top 5% of participating teams winning awards.

This event is open to public and private high school students in the United States, and is FREE.

For further information, go to http://www.ahsimc.com or http://www.americanmathcompetition.com.

"4. Calculators and computers ARE allowed. However, to receive credit for any correct answer, full work must be shown. Failure to accompany correct answers with correct mathematical logic will result in no credit given."

Does full work need to be shown for the short answer problems?

[quote="Rep123max"]Do the team members have to be at the same school?[/quote]

If not, could we have a bunch of AoPS teams?

I want to be on an AoPS team for this contest.

Q.  Do team members have to be from the same school?

A.  No.  See below in this thread for a detailed explanation to this question.

Q.  Do you need to show work for short-answer questions?

A.  Yes.  The only exception to this are questions pertaining to history of math.  In this case, you may simply write an answer.

By the way, I'll mention that we are going to update our page to reflect that test submissions may be made in Tex, in addition to MS Word.

I will continue to check this board daily, so feel free to ask additional questions.

Simon, i think it's just high school kids...

'9. All participating students must be enrolled in a public or private high school located within the United States of America. Failure to meet this criterion will result in automatic disqualification and revocation of any prizes awarded. '

Maybe you could run a team though, but i dont know if that's very interesting.

Sorry, Simon.

"All participating students must be enrolled in a public or private high school located within the United States of America. Failure to meet this criterion will result in automatic disqualification and revocation of any prizes awarded."

Looks like no.

And TeX would be nice.  I plan to use that.

I know I'm not eligible. That doesn't mean I'm not allowed to want to participate.

To follow up, AHSIMC has met regarding team members from different schools, and decided the following:

"Each school represented within a team must also be represented by a faculty sponsor."

So if you have 2 students from School A, 2 from School B, and 1 from School C, you must have 1 faculty sponsor from School A, B, and C.

Please format the Word registration file to reflect this (with all faculty sponsors listed) and provide mailing addresses for each, so we can get you the necessary registration material.

Feel free to email AHSIMC if you need further clarification on this or have any other questions, and I'll check back here and post daily.

In order to facilitate participation as much as we can, AHSIMC is allowing people to register all the way up to right before the test starts (and even during the testing period, although obviously you'll have less time to complete it)...

so you can register once you know who your faculty sponsor will be.

good question.

No plans to post additional practice questions, and no special rules for homeschoolers.  Clearly, homeschoolers need to be able to demonstrate (upon request) that they actually are high school students being taught at home, but other than that, you're good!

To the first question, bear in mind that the practice questions are more to familiarize potential test-takers with the format of the test (ie short answer, chain-reaction, and long answer questions) than the content of those questions.  For more details about test content, visit our FAQ on http://www.ahsimc.com or http://www.americanmathcompetition.com.

The website layout and design are very nice. : )

Thank you!  We hope the site is easy to navigate and understand.

We think you'll like the AHSIMC test experience itself just as much.

do you need 5 members to register a team?  can you have less?

assuming you don't need 5, how would this affect prizes?  if three people on a team won, could they each get a gaming system and split the cash value of the 2 other systems?

i guess you could just sell the other systems.  regardless, i love that video game consoles are being awarded as prizes.

[quote]I don't believe this will effectively remove the incentive to cheat.  First of all, the fact that the prize IS a video game console means that people who otherwise would not try their hand at a math contest will, if only because they like video games.  I believe that most people who do math contests are honest, but I can't say the same of hardcore video gamers, who live in a world of Gamesharks and cheat codes.

I do agree that the test format and prizes will encourage people to not include more than five people.  But we're not just talking about cheating by including more than five people - there are numerous other ways to cheat.  What if a student posts a question from the test on Dr. Math, or asks his teacher how to do a question, without them knowing that the question is from the AHSIMC?  I don't think any incentive (or absence of it) will prevent a student from doing that.[/quote]

First off, the reason behind the prizes being video game consoles in the first place is that we feel like its a widely appealing prize to students who also excel at math.  Many, many smart US high school students enjoy video games as a hobby, and this hobby in no way correlates with honesty in test taking, as was insinuated by an earlier post.

I tend to agree with the line of thinking that some cheating - or at least attempts to cheat, as it will be fairly difficult to cheat - will not affect the outcome.

As far as this cheating occurring by students posting on Dr. Math, etc...we feel like the sheer quantity of students nationwide taking this test will serve as a large deterrent to posting online, as someone out there reading the same board you are is probably aware that the question you're asking is on this year's AHSIMC.

We will also be monitoring this board and others for cheating violations on this year's AHSIMC, but we aren't extremely worried because simply the prizes are not divisible by more than 5 people per team, and also because of the difficulty of the questions.

We know very well that the AHSIMC will be treated with some skepticism prior to some successful years of operation, so there's not much more to say than we feel we have the situation under control.  Should we be wrong about this, we will certainly adapt our test to better fit the constraints of internet security.

[quote="Me"]Another idea would be people who are ineligable to take the test, either because they are in another country or in college.  I know a couple of college kids who would gladly help me out on this test (though again, I really won't cheat).[/quote]

Quote is for emphasis.

I was wondering what your policy for looking up how to solve these problems is?  Is it considered immoral to look for methods to solve them during the testing time?

[quote="Magnara"]I was wondering what your policy for looking up how to solve these problems is?  Is it considered immoral to look for methods to solve them during the testing time?[/quote]

Do you mean illegal?  Some people might consider it immoral in the sense that they're not solving the problem entirely by themselves, but it may still be within the rules.

I have a question: are students allowed to consult books, mathworld, etc?

Don't worry about cheating. Video game consoles are not worth the risk. Think about it, winning a national prize would definitely gain publicity... if someone unknown student suddenly was winning a national math contest, hm...

BTW, where's the discussion regarding USACO cheating? I haven't done the USACO for awhile, but I was always very curious regarding the number of people cheating.

[quote="webhappy"]BTW, where's the discussion regarding USACO cheating? I haven't done the USACO for awhile, but I was always very curious regarding the number of people cheating.[/quote]

[url=http://mathcounts.org/meeting/Upper_Forum_DB/Queries/Active_Forum.taf?function=CombinedDetail&ParentID=5503]Here[/url].

Oh, the mention regarding USACO was quite short. I can see why ppl in the US might want to cheat because for the 15 ppl each year who make the USACO camp, his/her chance of getting into Harvard has probably shot up to at least 70%. The strange thing is that I remember most cheaters in the USACO were non-US people!

[quote="Magnara"]Ok, well that was essentially my question.  I don't mean look up the individual problem, I just mean look up techniques.[/quote]

This is permissible.  I will change our Rules page to reflect that not only can calculators and computers be used, but books, old tests, and anything of that sort may be used as well.  If you are resourceful enough to find something helpful in a dusty old textbook that can lead you to answering a question correctly (while showing sufficient work), then more power to you.

This aspect of it will really help schools not as experienced in competitive mathematics compete with the best teams in the country.  After all, in "real life", you will have textbooks, calculators, etc. at your disposal to help you figure out problems...AHSIMC feels that its test should reflect this reality.

[quote="AHSIMC"]This aspect of it will really help schools not as experienced in competitive mathematics compete with the best teams in the country.  After all, in "real life", you will have textbooks, calculators, etc. at your disposal to help you figure out problems...AHSIMC feels that its test should reflect this reality.[/quote]

Aww, I love you guys.  That's what I think the biggest flaw in competetive math, it doesn't reflect that aspect of math.

[quote="webhappy"]So will there be an observers division?[/quote]

Not quite sure what you mean by this, but I'm guessing you're asking if you can get the test emailed to you as its emailed out to competing teams.

The answer to this is no.  However, the test document will be posted to our webpage as soon as the competition period is over.

Let me know if this isn't what you meant by your question.

[quote="Magnara"][Aww, I love you guys.  That's what I think the biggest flaw in competetive math, it doesn't reflect that aspect of math.[/quote]

I meant to respond to this a long time ago...thank you!  We think so too.

In high school, I always wished that there was a competition in which every single high school competed.  Such a competition would encompass local and regional rivalries - you'd be checking to see if you beat your rivals - but you'd be competing against the best elsewhere in the nation as well.

That is what we hope to achieve, and we hope you will participate!  Thank you again.

[quote="Magnara"]Well, one way that would probably facilitate that would be to have some sort of divisions sort of thing.  That would of course be after the contest (hopefully) takes off.  But to have subdivisions, and the winner from that division goes on to the next level would certainly encourage local competition.  There is more of a direct competetion with rivals than there would be otherwise.  I suppose there could also be a seperate division for those multi-geographical groups.

Just a suggestion.

Edit: Lol, I just read somewhere else you guys are doing that.  Beat me to the punch.[/quote]

Yes, as we progress with AHSIMC and our other initiatives, we certainly will have our eye on how to encourage rivalry between schools - particularly with schools that may not have a math team program already in place.

Currently, the best math team programs in the country already know who their rivals are.  While they may not be in the same division, they can certainly look forward to meeting each other in the latter rounds of our bracket-type system.

We can't proceed with any of this until after we conduct our first Exam.  We need to see how students respond to certain types of questions, get feedback from both students and sponsors, and improve our product.

Great suggestion!

Just wanted to mention in this thread as well, for the record, that if AoPS teams get their registration forms turned in to register@AHSIMC.com by June 1, you can simply list your AoPS user name under "faculty sponsor" and don't have to have a faculty sponsor in order to register.

Obviously, should you win, you will need to prove that you are high school students (or younger) enrolled in the United States before we can deliver your prize.

The sooner you register, the better (for us, anyway!)  Thanks!


function
real analysis
real analysis unsolved

Let $X$ be a topological space, and let $f: X \to \mathbb{R}$. Show that the set $C$ of all points $x \in X$ such that $f$ is continuous at $x$ is the intersection of a countable family of open subsets of $X$.

I can prove this for $X$ a measure space but I'm not sure how to for a general topological space.

Choose a function $f$. For each $x\in X$, let $V(x)=\inf_U (\sup\{|f(t)-f(s)|: s,t\in U\})$, where $U$ ranges over neighborhoods of $x$. We call $V(x)$ the variation of $f$ at $x$. $f$ is continuous at $x$ exactly when $V(x)=0$.

Now, can you show that $\{x: V(x)<\epsilon\}$ is open for each $\epsilon$?

This of course generalizes to functions from any topological space to any metric space.


I just made up a dumb question:

There is a digital clock being sold at a general store that constantly reduces its numbers by dividing by the greatest common factor of the hour number and the minute number. For example, if the clock was at 4:14, it would not show up as that: it would show up as 2:07. The clock still goes from 12:59 to 1:00, and a.m. and p.m. don't show up on the clock. If the clock starts 4:25, when will the clock read 1:01?

[hide]First let it reach 1:(something). The next reducable time will be 4:26, or 2:13. 1 minute later, it would 2:14, or 1:07. Thats two minutes past. When the clock reads n:n, it is definitely going to read 1:01, and since 1:(something) never reduces, we go to 2:02 (2:00 and 2:01 dont reduce). 1:07 to 2:02 is 53+2=55 minutes, including the 2 from the reducing times, we have 55+2=57 minutes past. [/hide]

If someone has a better solution than the one above (like a trick or something), plz post!

Seems like a pretty good solution.  That's the way I went about it.

im sorry, but wouldnt 1:01 be the same as 9:09, and 4:25 the same as 8:50, so the minimum time that must be waited in real time is 19 minutes?

No because the clock is constantly reducing. In other words, 8:50 doesn't exist. ( well it does, but not in the problem )

so, for example, if it were 12:08(real time) and 12:09 the minute after, would the clock go from 3:02 to 3:03 or from 3:02 to 4:03?

If it were 12:08, it would go down to 3:02, then 1:01(3:03) then 1:02....

NEW COUNTING PROBLEM

How many "existing" times on the digital clock are there? For example, 2:08 doesn't exist on the clock because it would be expressed as 1:04.

[hide]
$ 60 \cdot 12 \minus{} 262 \equal{} \boxed {458}$[/hide]

New problem:
If the clock starts at 12:00, it will eventually cycle itself. Find the period of this cycle.

Edit: Bah, AM/PM doesn't matter.

[quote="Yongyi781"][hide]
$ 60 \cdot 12 \minus{} 262 \equal{} \boxed {458}$[/hide]

New problem:
If the clock starts at 12:00 AM, it will eventually cycle itself. Find the period of this cycle.[/quote]

Actually, I didn't even do the existing time problem myself yet, but where did u get the 262? I was thinking of finding the factors/multiples of each number and then see when it can reduce....

Um, didn't I mention that there wasn't an am and a pm?  :)


Hi,

Do you guys know when Harvard will start accepting transfer students again? Just curious...

:wink: It most definitely would say it on their website.

Harvard may never admit undergraduate transfers again. It didn't last year, and announced last year that it wouldn't this year, and I have no reason to think it will go out of its way to admit undergraduate transfer students in any future year.


Two circles and three straight lines lie in the same plane. If neither the circles nor the lines are coincident, what is the maximum possible number of points of intersection?

My real question is what is a really good method to solve this.

[hide]theres a formula i think but i think of this intuitively.
draw a circle.
draw another circle. this one can intersect the first circle in max 2 points
draw a line. this line can intersect each circle max 2 times. that 4 intersections more
draw another line. this line can intersect each circle max 2 times again and the other line once. thats 5
the last line is the same and can create up  to 6 intersections

2+4+5+6=17[/hide]

[hide]
I came up with 17 too but differently.

The max intersections of three lines is three (triangle)
The max intersections of two distinct circles is two.
The max intersections of a circle with a triangles is six.

Thus if both circles intersect the triangle we have 12 intersections.
Total intersections is:
3+2+12=17[/hide]

[quote="abacadaea"][hide]that works too, but what if you have say, just 15 lines, or 5 lines and 3 squares.[/hide][/quote]
[hide]
for 5 lines and 3 squares I would essentially try to maximized the intersections with the square and lines.

Each line hits the square twice maximum.
Three squares and five lines equal 30 intersections.
three squares max intersect is 6.
I dont know the maximum intersection between five lines, but I would add that at the end. im guessing its 10
Max intersections is 30 +6 + 10=46.

[/hide]

What method would you use.

basically the one above.
2 squares can intersect at max 8 points.
the third square can create 16 more intersection points.
the first line can create 2 intersection points with each square. thats 6
the second line can create 6 with squares and 1 with the first line
third line can create 6 with squares and 2 with others
we have (8+16)+(6*6+15)=75

[quote="abacadaea"]
2 squares can intersect at max 8 points.
[/quote]

I cant seem to see how two squares can intersect in eight places.
could you describe this to me?

i cant really draw it, i cant use any drawing thing on the computer because this one hates me.

basically, you take two squares that coincide, and take on and rotate it some degrees. and there you go

ahh i see it now. lol!

I wonder if there is a way to generalize this to $ m$ polygons and $ n$ lines.


calculus
integration
function
trigonometry
geometry
rectangle
calculus computations

(1) Let $ f(x)$ be a continuous function defined on $ [a,\ b]$, it is known that there exists some $ c$ such that

\[ \int_a^b f(x)\ dx \equal{} (b \minus{} a)f(c)\ (a < c < b)\]

Explain the fact by using graph. Note that you don't need to prove the statement.

(2) Let $ f(x) \equal{} a_0 \plus{} a_1x \plus{} a_2x^2 \plus{} \cdots\cdots \plus{} a_nx^n$,

Prove that there exists $ \theta$ such that

\[ f(\sin \theta) \equal{} a_0 \plus{} \frac {a_1}{2} \plus{} \frac {a_3}{3} \plus{} \cdots\cdots \plus{} \frac {a_n}{n \plus{} 1},\ 0 < \theta < \frac {\pi}{2}.\]

The first statement is the Mean Value Theorem for integrals.  Graphically, the area under the graph of $ f$ on $ [a,b]$ is equal to the area of the rectangle of side lengths $ b\minus{}a$ and $ f(c)$ for some $ c\in [a,b]$

For each $ c\in]0,1[$ there exists $ \theta$, such that $ \theta \in ]0,\frac{\pi}{2}[$ and $ \sin(\theta)\equal{}c$ (follows from $ \sin(0)\equal{}0, \sin(\frac{\pi}{2})\equal{}1$ and the fact that $ \sin(x)$ is continuous.

And from $ (1)$ there exists $ c$ such that:

\[ f(\sin\theta)\equal{}f(c)(1\minus{}0) \equal{} \int_0^1f(x)dx\equal{} a_{0}\plus{}\frac{a_{1}}{2}\plus{}\frac{a_{3}}{3}\plus{}\cdots\cdots\plus{}\frac{a_{n}}{n\plus{}1}\]


Can anybody help me on this problem:
If 3 dices are thrown, then what is the probability of getting the total numbers as equal to or more than 10? How can I solve this problem?

I think i can help.

Let $S$ be our sample space (this is just the set of all of the different possible outcomes), then |$S$|=$6^3$.

Now, let $A$ be the set of all outcomes such that the sum of the 3 dice is greater than or equal to 10. Let $B$ be the set of all outcomes such that the sum of the 3 dice is less than or equal to 9. Then $A\cap B=\emptyset$, and $A\cup B=S$,i.e., $A$ and $B$ partition $S$.

Therefore $P(S)$=$P(A)$+$P(B)$, but $P(S)=1$ $\Rightarrow$ $P(A)=1-P(B)$. Now we  count the number of ways to make 9 or less.

3: {(1,1,1)}, i.e., one way.
4: {(2,1,1),(1,2,1),(1,1,2)}, 3 ways
5: {(2,1,1),(1,2,2),(2,1,2),(3,1,1),(1,3,1),(1,1,3)},6ways.
6: 3+6+1=10
7: 6+3+3+3=15
8: 3+6+3+6+3=21
9: 1+6+6+6+3+3+=25, giving:

25+21+15+10+6+3+1=81. Therefore $P(B)=\frac{81}{6^3}=\frac{81}{216}$ $\Rightarrow$ $P(A)=1-\frac{81}{216}=\frac{135}{216}$.
The case counting above is a well known problem from enumerative combinatorics; they are known as a composition of n with k parts.However we are working with the set {1,2,...6} ,thats why the n=9 case is 25 and not $\binom{8}{2}=28$; we lose the three compositions (7,1,1),(1,7,1),and (1,1,7). When we are working in {1,2,3.....} , the composition of n with k parts is given by $\binom{n-1}{k-1}$. Of course $\binom{n-1}{k-1}$ holds for the cases of 3 through 8, this is becuase you cant make a 3-part composition of those numbers with a number greater than 6.
Hope this was helpful.


trigonometry
limit
algebra
function
domain
analytic geometry
graphing lines

46. Find the exact value of the series:

$\sum_{k=0}^\infty{\cot^{-1}(k^{2}+k+1)}$

Observations:
[hide]
The problem can be converted to:

$\sum_{k=0}^\infty{\tan^{-1}\frac{1}{k^{2}+k+1}}$

but I'm not sure if that helps or where to go with it.

[/hide]

Ack, no luck so far, I'm trying to make it work with the telescoping hint but I just don't see how the terms ever end up "touching" each other, if you will... I'm not sure how to get rid of the $\cot^{-1}$ functions.

[hide="deriviation of inverse cotangent addition formula"]
The tangent addition formula tells us that

$\tan(x+y)=\frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}$.

Reciprocate both sides to get

$\cot(x+y)=\frac{1-\tan{x}\tan{y}}{\tan{x}+\tan{y}}$.

Now substitute $x=\cot^{-1}a$, $y=\cot^{-1}b$. This implies that $\tan{x}=\frac{1}{a}$, $\tan{y}=\frac{1}{b}$.

$\cot(\cot^{-1}a+\cot^{-1}b)=\frac{1-\frac{1}{ab}}{\frac{1}{a}+\frac{1}{b}}$.

To finish, multiply top and bottom of the RHS by $ab$ and take the inverse cotangent of both sides.

$\cot^{-1}a+\cot^{-1}b=\cot^{-1}\frac{ab-1}{a+b}$[/hide]

Cool! But couldn't you also have $\lim_{x\to{\infty}}\tan^{-1}x=\frac{3\pi}{2}$?

No.  Look at the graph $y=\arctan x$. :wink:

[quote="Karth"]Hint:

[hide]It is a misconception that Galileo first invented the ...[/hide][/quote]

I dont get it?

He wants you...
[hide]
telescope!
[/hide]
and he's right based on the recently posted solution.

*Edit* Oh yeah, because for tangent you tend to consider the domain from positive to negative pi... rats.

[quote="paladin8"][quote="bthings_2000"]*Edit* Oh yeah, because for tangent you tend to consider the domain from positive to negative pi... rats.[/quote]

It's actually not even that much; I think the range of $\arctan{x}$ is traditionally $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.[/quote]
Not just traditionally. It's the only way to have arctan be a function.

[i]Only[/i]?  There's nothing wrong with defining the range of $\arctan$ to be anything of the form $(a, a+\pi)$ for an arbitrary real $a$.

[quote="worthawholebean"]Not just traditionally. It's the only way to have arctan be a function.[/quote]
If you think of tan as slope, you can see that any half of the unit circle works.

Ehhh, never mind then. I'm... muddled today. I need sleep.


Stie cineva unde as putea gasi rezolvarile problemelor de pe listele scurte ONM din 2005-2008?(m-ar interesa cls a 12a mai mult) :D
(un document decarcabil ar fi perfect...)

Si eu am nevoie de ele, mai ales de cele din 2008 (cls a 10 a) :D


Is it possible to have the scenario wherein the displacement of a particle induces further displacement? Say displacement of a particle (by some unit) disrupts it's immediate environment in such a way that it causes it to displace again, causing another disruption and another displacement and so on? Are "loops" like this feasible? How fast would such a particle move?

I believe a moving particle with a charge will create a magnetic field and be itself affected by that magnetic field. Or I may have just made that up.

If you drop a magnet down a metal tube, you will observe the magnet not to fall at anywhere near the gravitational field strength predicts. This is due to Lenz's law.

As for a positive feedback, I doubt this exist because it would seem to violate the conservation of energy law if a particle's acceleration as a function of displacement is always positive within a closed system.

The velocity is also limited by the speed of light which might suggest that no system can exist...

If you have a pendulum with a sheet of metal in the plane of the swings which travels through a magnetic field, the pendulum will slow down very fast - the time periods observed will not agree with the predictred value based on a normal pendulum. This is another example of negative feedback for the displacement of the pendulum. With no friction at the connection and no air resistance it will still slow down fast. This again is because there is an energy transfer from the motion of the pendulum to eddy currents in the electrons of the metal sheet as they are a subject to a force due to their motion perpendicuar to a magnetic field. This example isnt so similar though...

[quote="degreestudent"]I believe a moving particle with a charge will create a magnetic field and be itself affected by that magnetic field. Or I may have just made that up.[/quote]:). No, that won't work.

Interesting thought.

A fellow received a Nobel Prize for a similar thought.  He was explaining how an electron pair tunneled and moved without resistance in a superconductor.

Other than at those very small dimensions and special conditions, the movement of the environment takes so much energy (remember all is at inverse square forces) that the effect on the object itself is too small to sustain the movement.


MATHCOUNTS
geometry
AMC
AIME
ARML
AMC 12
trigonometry

does anyone have this book who frequents this forum? I'm new to problem solving.. getting into the game a bit late I suppose but I find it fun and enjoyable for the most part.

I'm looking to step up though and I realize now I have to take my learning into my own hands, so Ive got to learn a bunch of thigns myself I was never taught.

The text seems to be something that would be aimed directly at me, but if anyone could shed some light on how the book is and how helpful it really is thatd be cool..

Sorry if this doesnt belong here.

I have it!

The book is really awesome, they teach you stuff that would normally come in AMC, MATHCOUNTS, or even a little bit could help you in AIME.

The only thing i regret is my parents didn't order the solution manuel =(

some chapters go through basic Algebra II stuff (for AMC) and some go through modular arithmetic, and tricks and shortcuts (for MATHCOUNTS) and there is a good amount of geometry thrown in

[quote="Glen Mann"]I'm looking to step up though and I realize now I have to take my learning into my own hands, so Ive got to learn a bunch of thigns myself I was never taught.[/quote]

umm if you want to take learning into your own hands (possibly due to the lack of consistancy in your math class) then i would advise you to get one of the books from

[url]http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php[/url]

(except for the last 2)...they focus on a single topic and provide more detail.

I [i]have[/i] to take it into my own hands. In school I am getting a mark of 104% in math, then I come on here and middle schoolers can do harder stuff than I can :|

well I am in 11th grade now, taking the highest level math course I can for my grade here in Canada, and like I said I am getting 104. The course is a snore, I barely ever go.

But then I go and practice mathcounts drills and I get maybe half of them right and some questions literally leave me stumped, which to me is really embarrassing coming on here. I just started getting into problem solving this year so I know I'm not going to be the best and most of you have years upon years of practice so of course youre better than me, but I really want to step up my game for next year, surprise the kids in the mathclub and such  :lol:

Most of the users with a higher number of posts (and many others) have a copy of the AoPS books.

The other texts are also great (the best textbooks I can think of).

If you really want to do well, then take the some of the classes.  That is where I have learned most of what I know about advanced math.

In addition, for later reading, you might try The Art and Craft of Problem Solving by Paul Zeitz, which I am using right now.

Well Vol 1 seems like it'd be perfect for you, but if you're looking for help with ARML, AMC12, and AIME mainly, then it would be way too easy. I felt I should've bought Volume II instead after going through it but for basics, it seems great but I didn't really learn anything new other than sorta obscure facts that I could derive given time as I had already gone through a full round of MathCounts training.

[quote="now a ranger"][quote="davidyko"]I did the whole of AoPS Vol. I last year (8th grade), and I can tell you, it was great. That and some of Vol. II raised my AIME score from a 1 to a 5. I learned loads of stuff I would have learned later or never learned at all.[/quote]

just out of curiousity, what math course did you take in school in 8th grade?[/quote]

Geometry. Was a snore. Skipped Algebra 2, now taking Precalc Honors, which is still a snore.  :P

EDIT: On second thought, maybe it was Vol. II that taught me most of the AIME stuff. But I still learned some from Vol. I.

[quote="Glen Mann"]But then I go and practice mathcounts drills and I get maybe half of them right and some questions literally leave me stumped, which to me is really embarrassing coming on here. I just started getting into problem solving this year so I know I'm not going to be the best and most of you have years upon years of practice so of course youre better than me, but I really want to step up my game for next year, surprise the kids in the mathclub and such  :lol:[/quote]

Yeah I'm probably one of the best if not the best at math in my school but at competitions sometimes I don't even get the question.

I'm at least in the top 5 at math in my town(in schoolat least, I'm in 8th grade), and I screw up most math competitions. My town thinks I'm unbelievable, as my town is stupid.

Hmm I'm starting AMC next year in preparation for AIME does anyone recommend any specific topic? I went over i think the 2006 10A and 10B they were okay but not as hard as MATHCOUNTS (the problems were shorter thank god  :D ) and mostly focused on Algebra 2. I'm taking Algebra 2 this year but is there anything else I should look over?

For the AIME? Trig, the Geometry stuff in Vol. II...um, the Polynomial stuff in Vol. II, and some other stuff...in Vol. II.
You need some Precalc stuff, in short.


This link: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13369

has a post by JBL, linking to the following:  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13654

This second link reports "Sorry, but only administrators can read topics in this forum."

Unfortunately, links to other topics within the forum will have unreliable results in the near future.  In order to merge the sites, we had to move the topic numbers so they wouldn't overlap.  We'll see if we can find a clever way to fix this, but I unfortunately think it will be pretty hard.


so yeah...
my teacher is trying to have me learn logarithms for easier ways of multiplying 1234*5678 with using mental math and stuff,
but i dont get how they work...

can somebody please help me?

I seriously doubt that this is in the right forum...  :maybe:


In the sequence 13, 10, 1, 3, ... each term, after the first, is found by adding the tens digit and three times the units digit of the previous term. How many distinct perfect squares will appear in this sequence?


We've all seen Polya's classic "[i]How To Solve It[/i]"; but recently I found about the book:

 [i]How To Solve It: Modern Heuristics[/i] by Zbigniew Michalewicz

I was wondering how this book is. Has anyone seen or heard anything about it?

Is it of similar style to Polya's book, as informative? Does it offer [i]different[/i] strategies and perspectives then Polya's book (so that I'm not buying just another version of Polya's book)?

Thanks for all your help  :) ,

Sunchips

Actually the book you told is completely different, it isn't even mathematical. It's a Computer Science book related to some Algorithms for Problems not completely solvable (At least not solved in Polynomial time yet) (In CS they are called NP-Complete problems)


How many squares are pictured?

[img]7215[/img]

[hide]
Forgetting about the smaller squares, we are looking for the number of squares in a $4 \times 4$ square, which is $16+9+4+1 = 30$.

Using the smaller squares, we can either have $1 \times 1$ squares or $2 \times 2$ squares. We just have to count the number of these now. There are $20$ of the former and $5$ of the latter.

In all, we have $30+20+5 = \boxed{55}$.[/hide]

Pretty sure you missed a few, ubemaya. I got 59...

I got $60$ but its wrong. I believe no right answers so far. No, 67 is not right. :(

i just did it again and i got 63 this time... :huh:

I have a solution for 64. There are 16+9+4+1=30 squares in the biggest square if you remove the cross. Inside the cross, you have 20 1x1 squares and 9 2x2 squares. There are 4 1x1 squares that are outside of the cross. Finally, there is 1 4x4 square in the very center of the big square. Total is 30+20+9+4+1=64.

You did not count the 3 by 3 squares of the little squares in the middle.

You double counted the 4 by 4 square in the middle.

So that would be +4-1 to your answer and that would be 67.  I am not trying to oppose you, but I want to make sure.

Whoa.  :stink: A stumper indeed. Then again, i didn't try...

[quote="ProtestanT"]i got 67

(16+9+4+1)*2+8-1=67[/quote]

unless someone posts a solution for 64,  i will stay with 67 because it makes the most sense and is the most logical.

the big 16 has 30 squares, the little 16 has 30 squares, 4 squares are counted twice, but 4 new squares of teh same size are uncounted. there are also 8 little squares taht extended out from the middle, and you subtract 1 for counting the big middle block twice

Now that I read the solutions, I get 67 as well.

What was the source of this? It must be pretty hard to take so many AoPSers like 3 days to solve it.

Okay, I agree with 67. I left out some and counted one twice.

This was a 23 on a National Sprint Round!!!

What year? and yes i have 67 as well, I had 63, but I forgot the four 3by3 squares so 67=63+4, if it was 64, then they overcounted the middle square, and then did not count the four 3by3 squres

are u sure about the answer?that u didnt misread or get this on the internet, because not everything on the internet is always right..

I am absolutely positive...

I just checked again

well, if 64 was the mathcounts answer, then it just screwed up a mathcounts national competition because 64 is obiviously not the correct answer, and ya what year was it?

BUMP:

This was #22 on the 1999 National Test. Does anyone have the answer key?

And now I get 63... ARGH.

i got 63 and my key says the answer is 67...
EDIT: now i got 67 after lookin at others solutions

My answer book says that it is $67$.
I have the test.


How do I get this equation: 2x+3y=6 into the y=mx+b form?

can you show me the steps as to how you got that?

What you have is standard form. The trick to converting it is to remember that what you do to one side, you have to do to the other. You know that you want $y$ all alone, so the first thing that you have to do is subtract $2x$ from each side. This gives you $3y=-2x+6$. Then, you divide everything by three so that the $y$ variable doesn't have a coeffecient. That gives you $y=-2x/3+2$

This should be in high school basics or classroom math.

anyways, basically for any problem like this just put all terms other than y on the RHS. Then divide by the coeeficient of y on both sides.


[quote="huang"]Suppose you have $ n$ nonnegative real numbers whose sum is $ 3$. Prove that if the sum of their squares is strictly greater than $ 1$, you can choose $ 3$ of those numbers such that their sum is strictly greater than $ 1$.[/quote]

Consider $ x_i$ ordered in decreasing values.
Suppose $ x_1\plus{}x_2\plus{}x_3\leq 1$

We obviously have $ x_3\leq\frac{1}{3}$
We also have $ x_1^2\plus{}x_2^2\plus{}x_3^2\leq(1\minus{}2x_3)^2 \plus{} x_3^2 \plus{} x_3^3$ (3 numbers greater than $ x_3\geq 0$ whose sum is max 1 reach the max of the sum of squares when the biggest is the biggest possible).

So  $ x_1^2\plus{}x_2^2\plus{}x_3^2\leq 1 \minus{}4x_3\plus{}6x_3^2$

We also have $ \sum_{k>3}x_k^2 \leq \sum_{k>3} x_3x_k$ $ \leq x_3\sum_{k>3}x_k$ $ \leq x_3((\sum_k x_k)\minus{}x_1\minus{}x_2\minus{}x_3)$ $ \leq x_3(3\minus{}3x_3)$

So $ \sum x_k^2\leq (x_1^2\plus{}x_2^2\plus{}x_3^2)\plus{}\sum_{k>3}x_k^2$ $ \leq (1 \minus{}4x_3\plus{}6x_3^2)\plus{}x_3(3\minus{}3x_3)$ $ \equal{}1\plus{}3x_3^2\minus{}x_3$ $ \equal{}1\plus{}3x_3(x_3\minus{}\frac{1}{3})$ $ \leq 1$

So $ x_1\plus{}x_2\plus{}x_3\leq 1$ $ \implies$ $ \sum x_k^2\leq 1$

And $ \sum x_k^2> 1$ $ \implies$ $ x_1\plus{}x_2\plus{}x_3> 1$

Q.E.D.


Factor:
\[ P(x,a,b,c) \equal{} \sum_{cyc} (x \minus{} a)(x \minus{} b)(a^2 \minus{} b^2)\]


What calculator is given at the Mathcounts National competition?

Do they give it before or after the competition?

looks like TI 84SE+

very likely that they'll give it to you when you sign in

how do you know that it will be a TI 84 Silver Edition Plus?

thats what its been the last few years and its the rumour on the street

you mean the "virtual street", right? :rotfl:

Yes, they've always been giving TI-84 Plus Silver Editions.

I wish they'd change the prize or hand them out a few days early.

Most people who make it to nationals already have a good calculator.

However:  Who'd say no to a nice free calculator (worth 100$ \$$+)?

it will probably be a silver edition, you get them when your coach registers, along with all the other stuff:
hats
nametags
sponsor stuff
etc...


function algebra domain conics parabola trigonometry

If you have a function of two variables, you can graph the surface and its contour plot that contains level curves. How do you find the amount of extreme values based on the level curves?

If you have a specified domain and range,  how do you find the amount of minimums and maximums?

For the first question:

imagine you have some surface.  Consider yourself standing on top of one of the maxima while you are very, very small.  What will you see?  The surface will fall away from you in all directions.  Now imagine the level curves -- they will all be encircling you and getting closer and closer to the point on which you are standing.

Now, if you have a single point as a maximum (or minimum), you will always get this pattern.  However, this technique is not perfect:  imagine that you have the surface caused when you drop a rock in a lake, with waves heading outward from a central point.  Each wave has maximal points, but those points are all in a continuous circle -- they are totally non-identifiable on a contour plot.

With functions of two variables, does there need to be a minimum between any two maximums?

An interesting question -- what do you think?  (I'm not just going to give you the answers to everything without making you try a little bit ;) )

Remember, you have a lot of flexibility in 3-D.

[quote="Mathhelp"]Is the answer yes to my question?  There is an extreme value theorem also in 3-d so I think the answer is yes.[/quote]

Well, I don't happen to know about such a theorem, but I can give you something to imagine that will answer your question:

Think about a camel with two humps on its back.

See if you can answer your question with that.  (Post here with you answer and an explanation, or if you need more help.)

I can't think of any examples.  What is that conjecture under your name?

[quote="zscool"][quote]I have no idea.[/quote]

keep thinking, you can do it!

how about this, try to visualize or draw yourself a picture of something that does not have a min between two maxes, that might help convince you of the correct answer[/quote]

Why is the Poincare conjecture not a theorem?

[quote="zscool"][quote]I have no idea.[/quote]

keep thinking, you can do it!

how about this, try to visualize or draw yourself a picture of something that does not have a min between two maxes, that might help convince you of the correct answer[/quote]

I see a minimum between the two maximum.   How could something have an official name if it's not a theorem?

[quote="JBL"][quote="Mathhelp"]I visualize two maxima.

[quote="JBL"]Visualize the surface that is a two-humped camel's back -- how many maxima does it have?[/quote][/quote]

Good.  Is there a minimum between them?

With respect to your other questions: the conjecture under his (or anyone else's) names here is a measure of the number of posts that they have made.  There are numerous other threads about this lying around -- the 7 questions are the [url=http://www.claymath.org/millennium/]Clay Mathematics Institute's 7 Millenium Prize Problems[/url].  The Poincare Conjecture is still a conjecture because mathematicians aren't satisfied that anyone has proved it yet.[/quote]

There's definitely no minimum?

Why does this thread have an arrow next to it?   What rests on the Riemann hypothesis being true?

[quote="JBL"]Is it really a minimum?  If you go across the camel's back (perpendicular to its spine), you decrease from that point.  A minimum is a point such that no matter which direction you go, iyou go up.

Why shouldn't it have a name?  Fermat's Last Theorem had a name for years before anyone proved it.  I have a name, and no one has ever proven me.  Things have names, proven or not, generally relating to their importance.  For example, these 7 questions are very important for a variety of reasons.  The Navier-Stokes Equations (which are unsolved) would be immensely important in fluid dynamics and related fields.  The Riemann Hypothesis, unproven but suspected to be true, is an immensely important piece of mathematics -- many, many other things rest on its being true and huge chunks of mathematics would be shattered if it's false.  So why not give them names?[/quote]

Well, stretch your imagination a little bit and imagine that the camel's sides, instead of forming up to make its stomach, just continued downwards forever.  There wouldn't be any "bowl" shape for in which we could have a local minimum.

For another example, take a downwards facing parabola and slide it along a sine curve so that the plane the parabola is in is perpendicular to the plane the sine curve is in, and the vertex of the parabola is always on the sine curve.  This will give you a shape with lots of maxima (places in which the curve falls away in every direction) but no minima (places in which the curve rises up in every direction).  If you happen to have 3-D graphing software, the curve I just described has equation z = sin(x) - y2

I don't know what you mean about the arrow question.  I don't see any arrow.

ComplexZeta (or a variety of other people) can answer that question better than I can.  The claymath website I linked to might have some information, as might [url=http://mathworld.wolfram.com/RiemannHypothesis.html]Mathworld[/url].  Note that it has a large number of things which are "equivalent to" the Riemann Hypothesis.  Lots of mathematicians have gone on writing theorems "on the condition that the Riemann Hypothesis is true," which normally would be bad form except that (1) it's an immensely powerful statement and (2) just about everybody is pretty sure that it is true, they just can't prove it.

Isn't there an extreme value theorem though for functions of two variables?


algebra
polynomial
linear algebra
matrix
inequalities
linear algebra solved

Let $A\in M_n(K)$ where $K$ is commutative field

$rank(A)=r$

Prove that the characteristic polynomial of $A$ is of the form

$x^n+a_{n-1}x^{n-1}+...+a_{n-r}x^{n-r}$

This is just the fact that the algebraic multiplicity of a root of the characteristic polynomial is greater than or equal to the geometric multiplicity. The matrix being of rank $r$ is the same as $0$ being a root of geometric rank $n-r.$ This in turn means that $0$ is a root of the characteristic polynomial of multiplicity at least $n-r,$ which is the same as saying that $x^{n-r}$ is a factor of the polynomial.

Or were you looking for us to prove that (algebraic mulitplicity) $\ge$ (geometric multiplicity)?

It also readily from the Schur theorem, stating that every matrix with rank $r$ is similar to the uppertriangular matrix with rank $r$, thus there are $r$ non-zero elements and $(n-r)$ zero elements on the main diagonal of this uppertriangular marix

To invoke the Schur theorem, you must have the existence of eigenvalues. To be sure of eigenvalues, we would need an algebraically complete field. All Moubinool said was that $K$ was a field, not that it was algebraically complete.

That's too bad - the Schur theorem is one of my favorite tools.

Kent ,are you meaning by geometric multiplicity the multiplicity in the minimal polynom ? i have never heard that from my teacher  :?

The characteristic polynomial can be written as $P_A =(-1)^n \left( X^n - \sigma_1 X^{n-1} + \ldots + (-1)^{n-1} \sigma_{n-1} X + (-1)^n \sigma_n \right)$, where $\sigma_k$ is the sum of all diagonal minors (i.e., minors whose elements on the diagonal are also on the diagonal of $A$) of order $k$.

$\textrm{rank} A = r$ tells us  that $\sigma_{k+1}=\ldots=\sigma_{n}=0$ as the minors involved are zero.

Hence, the conclusion.


modular arithmetic
induction
number theory proposed
number theory

Let $ k$ and $ n$ be positive integers. Prove that
\[ (n^4\minus{}1)(n^3\minus{}n^2\plus{}n\minus{}1)^k\plus{}(n\plus{}1)n^{4k\minus{}1}\]
is divisible by $ n^5\plus{}1$.

[quote="yuyang"]Let $ k$ and $ n$ be positive integers. Prove that
\[ (n^4 \minus{} 1)(n^3 \minus{} n^2 \plus{} n \minus{} 1)^k \plus{} (n \plus{} 1)n^{4k \minus{} 1}\]
is divisible by $ n^5 \plus{} 1$.[/quote]

[u][b]Lemma [/b][/u]: $ (n\plus{}1)(n^4\minus{}n^3\plus{}n^2\minus{}n)^p\equal{}(\minus{}1)^p(n\plus{}1)\pmod{n^5\plus{}1}$
Easy to prove thru induction on $ p$ :
True for $ p\equal{}0$ : $ (n\plus{}1)(n^4\minus{}n^3\plus{}n^2\minus{}n)^0\equal{}(\minus{}1)^0(n\plus{}1)\pmod{n^5\plus{}1}$
Induction step : $ (n\plus{}1)(n^4\minus{}n^3\plus{}n^2\minus{}n)^p\equal{}$ $ (n^5\minus{}n)(n^4\minus{}n^3\plus{}n^2\minus{}n)^{p\minus{}1}$ $ \implies$ $ (n\plus{}1)(n^4\minus{}n^3\plus{}n^2\minus{}n)^p\equal{}$ $ \minus{}(n\plus{}1)(n^4\minus{}n^3\plus{}n^2\minus{}n)^{p\minus{}1}\pmod{n^5\plus{}1}$
$ \implies$ $ (n\plus{}1)(n^4\minus{}n^3\plus{}n^2\minus{}n)^p\equal{}$ $ \minus{}(n\plus{}1)(\minus{}1)^{p\minus{}1}\equal{}(\minus{}1)^p(n\plus{}1)\pmod{n^5\plus{}1}$

[u][b]Demo [/b][/u]:
Let $ A\equal{}(n^4\minus{}1)(n^3\minus{}n^2\plus{}n\minus{}1)^k\plus{}(n\plus{}1)n^{4k\minus{}1}$

$ n^{k\plus{}1}A\equal{}(n^5\minus{}n)(n^4\minus{}n^3\plus{}n^2\minus{}n)^k\plus{}(n\plus{}1)n^{5k}$
$ n^{k\plus{}1}A\equal{}\minus{}(n\plus{}1)(n^4\minus{}n^3\plus{}n^2\minus{}n)^k\plus{}(n\plus{}1)(\minus{}1)^{k}\pmod{n^5\plus{}1}$
$ n^{k\plus{}1}A\equal{}\minus{}(n\plus{}1)(\minus{}1)^k\plus{}(n\plus{}1)(\minus{}1)^k\pmod{n^5\plus{}1}$
$ n^{k\plus{}1}A\equal{}0\pmod{n^5\plus{}1}$

and since $ \gcd(n^{k\plus{}1},n^5\plus{}1)\equal{}1$ : $ A\equal{}0\pmod{n^5\plus{}1}$
Q.E.D.

Multiply the expression by $ n$ and notice that $ n^5 \minus{} n\equiv \minus{} (n \plus{} 1)\pmod{n^5 \plus{} 1}$ so we now must prove that

$ (n \plus{} 1)(n^{4k} \minus{} (n^3 \minus{} n^2 \plus{} n \minus{} 1)^k)\equiv0\pmod{n^5 \plus{} 1}$

which is obvious as $ n^4 \minus{} n^3 \plus{} n^2 \minus{} n \plus{} 1|n^{4k} \minus{} (n^3 \minus{} n^2 \plus{} n \minus{} 1)^k$.


+ give a,b,c is a real number  find a,b,c so that
    a 2 +b 2 =15c
    b 2 +c 2 =15a
    c 2 +a 2 =15b :)

It's more than not hard: it's easy :). a, b, c>=0. Assume, for example, that a>c; then from the first 2 equations we get c2>a2=>c>a (because a, c>=0), so we have a contradiction, so we must have a=c, so a=b=c=0 or a=b=c=15/2.

I think  this problem is very easy and this new problem
    find a,b,c is real number so that
              a 2 +b 2  =23c 2
              c 2 +b 2  = 27a
              a 2 +c 2  = 29b ^3  :)  :(

find a,b,c is real number so that a^b+b^c+c^a=(abc)^abc :?

are my problem hard? why  do you don't solve it?


calculus
integration
logarithms
calculus computations

integrate
$ \int \frac{1}{2x(\sqrt{x}\plus{}1)}dx$

hello, set $ \sqrt{x}\equal{}t$, from here you will get
$ dx\equal{}2tdt$.
Sonnhard.

oh yes.....so silly of me to ask a silly q... :D

thanks

hello, to compare the result, i have got
$ \int\frac{1}{2x(\sqrt{x}\plus{}1)}\,dx\equal{}\frac{1}{2}\ln(x)\minus{}\ln(\sqrt{x}\plus{}1)\plus{}C$.
Sonnhard.

Hello,

I was wondering how to get

$ \int\frac{1}{t^{2}\plus{}t} \equal{} \ln(\frac{t}{t\plus{}1})$

Thanks.

hello, observe that
$ \frac{1}{t^2\plus{}t}\equal{}\frac{1}{t}\minus{}\frac{1}{t\plus{}1}$
Sonnhard.

oh, ok. I understand it now.

Thanks!! :)

hello, ok i hope you can solve the problem now.
Sonnhard.


trigonometry
quadratics
LaTeX
algebra
quadratic formula
polynomial
Rational Root Theorem

[quote="JBL"]Not easily, in general.  Cubics are solvable via a number of methods -- I particularly like one using sines -- although there is a reasonably complicated formula (the cubic formula, analogous to the quadratic formula) that also suffices.  If you're looking for "nice" roots, you can apply the rational root theorem.  In this case, it says that all rational roots must be of the form $\pm \frac{a}{b}$ where a divides 30 and b divides 6.  (In general, it says that the numerator must divide the constant term while the denominator must divide the leading coefficient.)  In this case (although not in all), this method will work nicely.[/quote]

Solving a cubic using sines?  I'm interested.  Please explain.

[edit: You know what I meant!  :P ]

Cubic.  Use the fact that $\sin(3\theta) = 3sin\theta - 4\sin^3\theta$.  First, you take your cubic,
$ax^3+bx^2+cx+d=0$ and set $y = x+e$, with an appropriate choice of e, in order to transform it to $a'y^3+c'y+d'=0$.  Then, set $z = fy$ for an appropriate choice of f, in order to multiply through by something in order to transform it to $3z - 4z^3 = d''$.  In this case, the solutions are so nice that this method would take much too long.  But it would work!  (Perhaps I'll write it up in Latex here in a minute, if I find insufficient other activities for my time.)

[quote="joml88"][quote="JBL"]Not easily, in general.  Cubics are solvable via a number of methods -- I particularly like one using sines -- although there is a reasonably complicated formula (the cubic formula, analogous to the quadratic formula) that also suffices.  If you're looking for "nice" roots, you can apply the rational root theorem.  In this case, it says that all rational roots must be of the form $\pm \frac{a}{b}$ where a divides 30 and b divides 6.  (In general, it says that the numerator must divide the constant term while the denominator must divide the leading coefficient.)  In this case (although not in all), this method will work nicely.[/quote]

Solving a quadratic using sines?  I'm interested.  Please explain.[/quote]
No, he said solving a [b]cubic[/b] using sines  :D
More info about rational root theorem is here
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=rational+root+theorem&t=17958[/url]

So, just for the record, applying the method to this particular problem is excruciatingly messy and doesn't in fact result in the nice solutions that it should, unless you just plug it into a calculator.  Basically:

In the first step, we set $x=y-\frac{25}{18}$ to make our equation look like:
$-\frac{3190}{243} - \frac{427y}{18} + 6y^3 =0$.
In the second step, we set $y = \frac{\sqrt{427}}{9}z$ and multiply the equation through by $\frac{972}{854\sqrt{427}}$ and switch the sides of some terms to make it
$3z-4z^3 = -\frac{6380}{427\sqrt{427}}$
Now, setting $z = \sin\theta$ in this, we can find the exact values for theta by taking an arcsin.  Dividing by 3, we can then take the sine to find x, multiply to find y, and add to find x.  By the time you finish, the clever student who started plugging in small integers has already had lunch :)

One of the three solutions I get from this method is $-\frac{25}{18}-\frac{\sqrt{427}}{9}\sin{\left(\frac{1}{3}\arcsin\frac{6380}{427^{3/2}}\right)}$, which most of us are familiar with in the slightly more compact form [hide]-2[/hide]

[quote="**********"]How does one go about solving this?
What values of x satisfy: 6x^3 + 25x^2 + 11x = 30?
thanks for your help[/quote]

you could use an interative method to find the roots to a certain number of decimal places;;

6x^3 + 25x^2 + 11x = 30
11x = 30 - 6x^3 - 25x^2
x = (30 - 6x^3 - 25x^2)/11

call x_(n+1) = (30 - 6(x_n)^3 - 25(x_n)^2)/11
set n = 0 and x_n = -2, (a nearby root I think)
continue the interation...

i mean, the thing turns out to factor nicely, into (x+2)(x+3)(6x-5) = 0

I'm pretty bad at factoring, and I was wondering if there was any way that that particular expansion could be seen quickly.

I tried synthetic division, and that factored the thing for me, but i didnt try that at first...so i wonder if synthetic division is something that you would always try first, to see if the thing factors nicely?

[quote="**********"]so i wonder if synthetic division is something that you would always try first, to see if the thing factors nicely?[/quote]

Absolutely -- you should always check for nice roots by synthetic division, with your choices being guided by the rational root theorem.  There isn't any other way to "see" the nice roots than to test numbers and see if they work.


I run 64-bit Kubuntu 8.10 with Asymptote v. 1.70.

From both Kile and from the command line, asy does not run and returns the following:

[code]asy: error while loading shared libraries: libreadline.so.5: cannot open shared object file: No such file or directory[/code]

I've checked /lib/ , and libreadline.so.5 is present. Reinstalling libreadline5 did not solve the problem. What "shared object file" does libreadline.so.5 try to open? Or is the system failing to find libreadline.so.5 in its proper place?

meh I think I solved it: it's the 64-bit.

Installing lib32readline5 takes care of this particular error message, but then asy complains that libgc1c2 (C/C++ garbage collector?) cannot be found.

I compiled Asymptote from source, and now it works fine.


calculus
derivative
limit
Functional Analysis
advanced fields
advanced fields unsolved

Let $X$ and $Y$ be Banach spaces and let $W$ be the set of invertible bounded linear operators from $X$ to $Y$.  Then $W$ is open in the uniform operator topology. Define $G$ on $W$ by $G(A) = A^{-1}$. $G$ is differentiable on $W$ and the derivative of $G$ at $A$ is given by $G'(A)T =-A^{-1}TA^{-1}$, where $T$ is a bounded linear operator.

For some reason I find this problem hard. I can't do it  :blush:

hmm. I don't know how to define the derivative on a banach space

but that does not stop me from doing this

$G'(A)T = \lim_{h \to 0}\frac{G(A+hT)-G(A)}{h}= \lim_{h \to 0}\frac{(A+hT)^{-1}-A^{-1}}{h}$. Now, we know that for $B$ an operator with sufficiently small norm (less than one), $I-B$ is invertible with inverse $I+B+B^{2}+\cdots$, so $A-AB$ is invertible with inverse $A^{-1}+BA^{-1}+B^{2}A^{-1}+\cdots$, so if $L = AB$ has sufficiently small norm, $A-L$ is invertible with inverse $A^{-1}+A^{-1}LA^{-1}+(A^{-1}L)^{2}A^{-1}+\cdots$.
Thus $(A+hT)^{-1}-A^{-1}= A^{-1}(-hT)A^{-1}+(A^{-1}(-hT))^{2}A^{-1}+\cdots$; dividing through by $h$ and setting $h = 0$ yields $-A^{-1}TA^{-1}$.

Let $F\colon U\to V$ be a map between Banach spaces. The Frechet derivative of $F$ at a point $x\in U$ is a linear map $S\colon U\to V$ such that $\parallel F(x+z)-F(x)-S(z)\parallel /\parallel z\parallel \to 0$ as $z\to 0$.

If one requires only that $\parallel F(x+tz)-F(x)-tS(z)\parallel /t\to 0$ as $t\to 0$, for any fixed $z\in U$, then one has the definition of Gateaux derivative. The proof given by MysticTerminator uses the Gateaux definition, but it would work with Frechet all the same.


Soit $a,b,c$ des nombre positifs. Montrer que inegalites suivant.

(1) $a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 6.$

(2) $\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\geq \frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}.$

Excuse moi si je ne parle pas bien en Francais.Pour premiere inegalite nous utilisons l'inegalite evidente a+1/a>=2,equivalente avec (a-1)^2>=0.Pour inegalite(2) nous utilisons l'inegalite evidente 1/a+1/b>=4/(a+b),
equivalente avec (a-b)^2>=0. :)

[quote="stancioiu sorin"][quote="mathmanman"]Quelques corrections (sur la langue ;)) :

@[b]Kunny[/b] : "Soi[b]en[/b]t ... des nombre[b]s[/b] ... Montrer [b]les[/b] inégalités [b]suivantes[/b] ...".

@[b]Stancioiu sorin[/b] : La première inégalité que tu utilises découle (en anglais) de : AM-GM. En français : AM-GM -> IAG (c'est à dire : Inégalité Arithmético Géométrique).
Et la deuxième : HM-GM -> IHG (Inégalité Harmonico Géométrique).

:)[/quote]
   Excuse moi, mais je ne sais pas tres bien la langue Francaise.Ne parlons d'accents! :blush: J'apprendrai cette langue et puis j'ecrirai en Francais.Mais le solutions sont bonnes, n'est-ce pas? ;).<Auf wiedesen!>(en Allemagne)[/quote]
    Es-tu connais une <site> ou je peux apprendre le Francais?Je dois apprendre cette langue parce que je veux que je connaisse trois langues: l'Anglais, le Francais et ll'Allemande.

[quote="stancioiu sorin"]
   Es-tu connais une <site> ou je peux apprendre le Francais?Je dois apprendre cette langue parce que je veux que je connaisse trois langues: l'Anglais, le Francais et ll'Allemande.[/quote]
Non, je ne connais pas de sites pour apprendre le français..
Mais, je pense que tu apprendras vite si tu continues à parler régulièrement dans la partie française du forum! ;)
Et, si tu veux, je peux corriger toutes tes fautes pour que tu progresses :
"Es[b]t-ce que[/b] tu ...? ... je veux [b]connaître[/b] trois .... [b]Allemand[/b]."

:)

Merci.Je veux que tu corriges me.Je dois apprendre ces langues parce que dans notre jours il faut connaitre 2-3 langues et, pour exemple(pour moi) la mathematique.Aussi, je ne sais pas d'ou je peux utiliser les accents.Peux-tu aider moi? :)

[quote="mathmanman"] Montrer [b]les[/b] inégalités [b]suivantes[/b] ...".

[/quote]

Je n'ose pas vraiment corriger un Français... mais est-ce que ne c'est pas : "Montrez les inégalités suivantes?"

Kunny, on apprend le français au Japon :!:  :| ?.
Si je n'avais pas mes bandes dessinées du Japon, que je ne peux pas acheter en anglais ou néerlandais, j'aurais oublié tout! :lol:

[quote="stancioiu sorin"]Merci. Je veux que tu [color=red]me[/color] corriges. Je dois apprendre ces langues parce que d[color=red]e[/color] no[color=red]s[/color] jours il faut connaitre 2-3 langues et, p[color=red]a[/color]r exemple (pour moi) la mathematique. Aussi, je ne sais pas ou je peux [color=red]prendre[/color] les accents. Peux-tu [color=red]m'[/color]aider? :)[/quote]
Ok! :) (j'ai corrigé tes fautes en rouge :) )
Et, pour les accents, regarde ici : http://www.asciitable.com , dans les "extended ASCII codes". Par exemple, pour "é", tu fais : ALT+130.

[quote="fredbel6"]Je n'ose pas vraiment corriger un Français... mais est-ce que [color=red]c[/color]e [color=red]n[/color]'est pas : "Montrez les inégalités suivantes?"[/quote]
Non, pas forcément, en fait, les deux sont bons. ;)

[quote="fredbel6"]
... j'aurais [color=red]tout[/color] oublié![/quote]

:)

J'ai oublié tout n'est pas correct aussi?

Quand on exige quelque chose :"Montrer" est correct? :o

[quote="fredbel6"]J'ai oublié tout n'est pas correct aussi?[/quote]
Non...

[quote="fredbel6"]Quand on exige quelque chose :"Montrer" est correct? :o[/quote]
Ah, si tu le vois comme ça.. alors, oui, l'impératif est bien : Montre[b]z[/b].
Mais bon, quand on demande de résoudre un problème, on n'exige pas, si? :)
Enfin bref, toujours est-il que les deux sont utilisés! ;)


integration
calculus
calculus computations

Evaluate \[ I\equal{}\int_{0}^{1}\frac{4x\plus{}3}{\left(x^2\minus{}x\plus{}1\right)^2} dx\]


Does exits many $ x;y;z \in Z$ such that $ x^3\plus{}3y^3\equal{}5z^3$?

Under a rational transformation this equation corresponds to an elliptic curve:
\[ v^2 \equal{} u^3 \minus{} 6075/4\]
of rank 1 over the rationals. Therefore, there exists infinitely many rational points on this curve and thus there are infinitely many solutions to the given equation.


function
integration
probability
calculus
derivative
real analysis
probability and stats

If the random variable X is normally distributed, how is the random variable Y = 1/X distributed? How would we go about figuring this out?

Thanks!

Define a function $ \Phi(x) \equal{} \frac1{\sqrt {2\pi}}\int_0^xe^{ \minus{} t^2/2}\,dt.$

For $ x > 0,$ $ \Phi(x) \equal{} P(0 < X < x);$ for $ x < 0,$ $ \Phi(x) \equal{} \minus{} P( x < X < 0).$

Note also that $ \Phi'(x) \equal{} \frac1{\sqrt {2\pi}}\,e^{ \minus{} x^2/2}.$

Suppose $ y > 0.$ We have

$ P(Y > y) \equal{} P\left(\frac1X > y\right) \equal{} P\left(0 < X < \frac1y\right) \equal{} \Phi\left(\frac1y\right).$

Let $ g(y)$ be the probability density function for $ Y.$ This is the derivative of the cumulative distribution function or the negative of the derivative of the complementary distribution function. That is, for $ y > 0,$

$ g(y) \equal{} \minus{} \frac {d}{dy}\,P(Y > y) \equal{} \minus{} \Phi'\left(\frac1y\right)\left( \minus{} \frac1{y^2}\right) \equal{} \frac1{y^2\sqrt {2\pi}}\,e^{ \minus{} 1/(2y^2)}.$

If you chase through the details, we also have that for $ y < 0,$ the density is

$ g(y) \equal{} \frac1{y^2\sqrt {2\pi}}\,e^{ \minus{} 1/(2y^2)}.$

For $ y$ near zero, this density is extremely small (and we may safely say that $ g(0) \equal{} 0.$) For large $ y,$ this density is approximately $ \frac1{y^2\sqrt {2\pi}},$ which is too large for $ Y$ to have an expected value.

Note: when you said "normally distributed" I assumed you mean "standard normal," meaning mean zero and standard deviation one. The answer would vary in a predicable fashion for other cases, with the same techniques in use.

[quote="Kent Merryfield"]For $ x > 0,$ $ \Phi(x) \equal{} P(0 < X < x);$ for $ x < 0,$ $ \Phi(x) \equal{} \minus{} P( \minus{} x < X < 0).$[/quote]
Isn't something wrong here?

You have $ f_X(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{\frac{-(x-\mu)^2}{2\sigma^2}}$

$ h(x)=1/x=y$
$ h^{-1}(y) = 1/y = x$

$ f_Y(y) = f_X(x)|\frac{dh^{-1}}{dy}| = f_X(1/y)|\frac{dh^{-1}}{dy}| = \frac{1}{\sqrt{2\pi}\sigma}e^{\frac{-(1/y-\mu)^2}{2\sigma^2}}\frac{1}{y^2}$

I do not recognize that as a familiar distribution.

I fixed the mistake that limsup noted.

[quote="brunojoyal"]I do not recognize that as a familiar distribution.[/quote]
It's not.

brunojoyal's method is quicker and more direct than the one I used. The textbook I use when I teach this course shows both methods, but I find that the quicker direct method is also more confusing and harder for students to remember. It's a tradeoff.


A while back, someone posted videos created by AoPS on SFFT, and I can't seem to find them now...does anyone have a link?

[url]http://www.artofproblemsolving.com/Wiki/index.php/Simon%27s_Favorite_Factoring_Trick[/url]

From the handy-dandy AoPSWiki.  The video links are at the bottom of the page.


geometry
geometric transformation
reflection

Hi everyone,

Given three circles with the same radius that intersect at a common point, how can we prove that the radius of the circle that is defined by the other three points of intersection is the same as the radius of the original three circles?

I tried analytical geometry (way too messy) and some Euclidean geometry but I am completely stuck.

[hide]
Let the centers of the circles be $A,B,C$, the common intersection point be $P$, and the other intersection points be $X,Y,Z$ where $X$ in the intersection of circles $B$ and $C$, etc.

We first find $\angle YXP=\angle YZP$. To prove this, it's equivalent to $\angle YCP=\angle YAP$ (each angle is doubled from the first equation to get the second). But this is clearly true since quadrilateral $YAPC$ is a rhombus.

Similarly, $\angle ZXP=\angle ZYP$. It follows that
\[\angle YXZ+\angle YPZ=\angle YXP+\angle ZXP+\angle YPZ \]

\[=\angle YZP+\angle ZYP+\angle YPZ=180^\circ \]
So if we reflect $P$ over the line $\overline{YZ}$ to $Q$, then $\angle YXZ$ and $\angle YQZ$ are supplementary, so quadrilateral $XYQZ$ is cyclic. Therefore, the circumradii of $\triangle YPZ$, $\triangle YQZ$, and $\triangle YXZ$ are all equal, as desired.
[/hide]

Thanks for the reply.

I don't see how you get $\angle YXP+\angle ZXP+\angle YPZ =\angle YZP+\angle ZYP+\angle PYZ$.  Because $\angle YXP=\angle YZP$ and $\angle ZXP=\angle ZYP$, the above implies that $\angle PYZ = \angle YPZ$ which I don't know how to prove.

actually, that was a typo. the third angle should still be $\angle YPZ$.

Ah okay everything makes sense now.  Thanks a lot for your solution!


Because I'm getting tired of all the mudslinging between liberals and conservatives on this site, [url=http://www.jibjab.com]here's a website with a humourous cartoon lampooning both Bush and Kerry equally.[/url]  Hope everyone enjoys it (but if you don't have fast internet, it will probably take about 5 minutes to download-it's worth it, though)!  Feel free to post what you thought about it!

I posted this a few times, but no one listened  :(

This is a great cartoon, everyone, go watch it!

The best internet cartoon I've ever seen.

This is really good.  I was watching it in the computer lab here at mathcamp and everyone came over to see it.

Im going to watch it in the mathcamp computer lab right now. who are you dandelion?

Amusing, if a bit abrasive at times...that's the LAST time I click on a link in here with a parent in the room...

In my opinion, the real stars of the film were Bill and Hillary Clinton.  Their appearance was both unprecented and alarmingly funny.

[quote="JSRosen3"]I posted this a few times, but no one listened  :([/quote]

It's because nobody likes you.  :P

Anyways, I got countless suggestions and promptings to watch it, and now that I've seen it, I'm the one annoying people on AIM to see it.


In $ \triangle$  $ ABC$,the perpendiculars drawn from B and C to the opposite sides intersect each other at D.Prove that $ AB^{2} \plus{} CD^{2}$ = $ AC^{2} \plus{} BD^{2}$.

Thank you.

The perpendiculars drawn from $ A,B,C$ to opposite sides meets $ BC,CA,AB$ at $ I,H,K$ respectively.
By Pythagorean theorem, we have:
$ AB^2\plus{}CD^2\equal{}AI^2\plus{}BI^2\plus{}CI^2\plus{}DI^2\equal{}(AI^2\plus{}CI^2)\plus{}(BI^2\plus{}DI^2)\equal{}AC^2\plus{}BD^2$


[quote="kunny"]Telescope![/quote]

hubble or arecibo ?  :rotfl:

$ \cot^{\minus{}1}\left[\dfrac{(k\plus{}1)k\plus{}1}{(k\plus{}1)\minus{}k}\right]\;\equal{}\;\cot^{\minus{}1}k\minus{}\cot^{\minus{}1}(k\plus{}1)$


integration
calculus
derivative
function
quadratics
calculus computations

Find the interval $ [a,b]$ so that the value of $ \int_a^b(2\plus{}x\minus{}x^2)\,dx$ is maximum.

If it was a multivariable calculus question, it's easy because we can put $ f(a,b)\equal{}\int_a^b(2\plus{}x\minus{}x^2)\,dx$ and after integration and calculating partial derivatives for $ a,b$ and letting them equal to zero, the rest follows.

How does one solve this without multivariable calculus?

For this particular function, $ f(x)\equal{}2\plus{}x\minus{}x^2,$ there is an interval on which the function is positive, and outside that interval the function is negative. What do we have if we integrate over precisely that interval, and why does that work?

Incidentally, the multivariable calculus approach would lead you to the same place. Suppose $ f$ is a continuous function with $ f(x)\equal{}0$ only at $ x\equal{}x_0$ and $ x\equal{}x_1,$ and $ f(x)>0$ only for $ x\in(x_0,x_1).$

$ G(a,b)\equal{}\int_a^bf(x)\,dx.$

$ \frac{\partial G}{\partial a}\equal{}\minus{}f(a).$

$ \frac{\partial G}{\partial b}\equal{}f(b).$

So the only critical points are $ a\equal{}x_0$ or $ x_1$ and $ b\equal{}x_0$ or $ x_1,$ and we only get a positive value of $ G(a,b)$ when $ a\equal{}x_0$ and $ b\equal{}x_1.$

Transforming given integral gives us
\[ \int_{a}^{b}{\left(2 \plus{} x \minus{} x^{2}\right)\mathrm{d}x} \equal{} \int_{b}^{a}{\left(x \minus{} 2\right)\left(x \plus{} 1\right)\mathrm{d}x} \textrm{.}
\]
Examining the quadratic function graph we conclude that integral value is maximal when $ a \equal{} \minus{} 1$ and $ b \equal{} 2$, and equals $ 9/2$.
(Because of $ a \leqslant b$, we examine the upside-down graph of integral function.)


Let $ a,b,c$ are three sides of a triangle such that $ a\plus{}b\plus{}c\equal{}1$.Prove that
$ \frac{1}{1\minus{}2(a^2\plus{}b^2\plus{}c^2)}\plus{}\frac{9(a^2b\plus{}b^2c\plus{}c^2a)}{a^2\plus{}b^2\plus{}c^2}\ge 6$

[quote="vo thanh van"]Let $ a,b,c$ are three sides of a triangle such that $ a \plus{} b \plus{} c \equal{} 1$.Prove that
$ \frac {1}{1 \minus{} 2(a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {9(a^2b \plus{} b^2c \plus{} c^2a)}{a^2 \plus{} b^2 \plus{} c^2}\ge 6$[/quote]

Nice  :lol:
$ a,b,c$ are three sides of a triangle so:$ 4(ab \plus{} bc \plus{} ca) \ge (a \plus{} b \plus{} c)^2 \equal{} 1$
We have:$ ab \plus{} bc \plus{} ca \ge \frac {1}{4}$
And:$ a^2b \plus{} b^2c \plus{} c^2a \ge \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2}.abc \equal{} 9abc(a^2 \plus{} b^2 \plus{} c^2)$
Use "p;q;r" Ineq
$ < \equal{} > \frac {1}{4q \minus{} 1} \plus{} 81r \ge 6$
But:$ r \ge \frac {4q \minus{} 1}{9}$ - by shur :)
So:$ 81r \ge 9(4q\minus{}1)$
By Am-Gm:
$ \frac {1}{4q \minus{} 1} \plus{} 9(4q \minus{} 1) \ge 6$
We have done :)

[quote="caubetoanhoc94"][quote="vo thanh van"]Let $ a,b,c$ are three sides of a triangle such that $ a \plus{} b \plus{} c \equal{} 1$.Prove that
$ \frac {1}{1 \minus{} 2(a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {9(a^2b \plus{} b^2c \plus{} c^2a)}{a^2 \plus{} b^2 \plus{} c^2}\ge 6$[/quote]

Nice  :lol:
$ a,b,c$ are three sides of a triangle so:$ 4(ab \plus{} bc \plus{} ca) \ge (a \plus{} b \plus{} c)^2 \equal{} 1$
We have:$ ab \plus{} bc \plus{} ca \ge \frac {1}{4}$
And:$ a^2b \plus{} b^2c \plus{} c^2a \ge \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2}.abc \equal{} 9abc(a^2 \plus{} b^2 \plus{} c^2)$
Use "p;q;r" Ineq
$ < \equal{} > \frac {1}{4q \minus{} 1} \plus{} 81r \ge 6$
But:$ r \ge \frac {4q \minus{} 1}{9}$ - by shur :)
So:$ 81r \ge 9(4q \minus{} 1)$
By Am-Gm:
$ \frac {1}{4q \minus{} 1} \plus{} 9(4q \minus{} 1) \ge 6$
We have done :)[/quote]
Ok,it's the same my solution.:)


I saw the some awards given to the student who got the highest score on his/her school on AMC-10 and AMC-12. Is this applying to every school who's taking the competition?

I know the certificate for qualifying AIME is given to everyone who passed to AIME but I wondered about this case.

Thanks. :P

For an AMC-10 certificate, i believe you must atain a score of at least 100.

I've never gotten anything, neither pins nor certificates.
I got 100+ on 2 AMC 10s and on last year's AMC 12.
I only got a letter for nominating a teacher for that Edith May Sliffe award (sp?). He recently got a check for $400 in the mail.


Assume we have a very large concentration of a strong acid like HCl. Say concentration is 10^4 M. Obviously amount of H+ is almost equal to 10^4 since a strong acid undergoes almost complete dissociation. And canculating PH by our usual formula we get PH=-4.

1) Can PH be negative by definition.
2) Am I making any conceptual error above? If so what.

Please when you are giving the answer explain it with some clarity and not just yes or no.

~ CS

Of course the pH can be negative. You just illustrated that.

no that's wrong....the formulae for pH holds only for samll concentrations that is for say $ 10^{x}$ where $ x leq 0$ theoritically....so no such thing called pH=-4

It's true that the common used formula for pH holds exactly only for not very high concentrations (for which we must substitute concentrations by activities), but its also true that there is nothing that prevent $ pH \equal{} \minus{} \log(a_{H^\plus{}})$ to be negative. A 10M HCl solution have pH = 0. For a 12 M solution the pH is negative. However, for the extremely high concentration of Cyber-Shreyas the concentration of H+ would not be the same as the acid concentration: this because for an acid to ionize completely there must be some molecules in the medium that can accept the H+: water molecules. For such high concentrations there is no enough water to accept all H+, and therefore a considerable proportion of H+ stays together with chloride ions.

As usual carcul is right. :D The formula for PH is not log[H+], it is log of activity of H+. http://en.wikipedia.org/wiki/PH ... I guess PH can be negative but we usually don't compare the strengths in that case, we like comparing on the other borders.


summer program
Mathcamp
AMC
USA(J)MO
USAMO

That deserves a major *whack*, tetrahedr0n.

Eh, don't worry danielkim (ooo...very kool first name btw) I don't think anyone - barring alumni - has been accepted to [b]2005[/b] MathCamp yet. But you'll hear about it on this forum as soon as it happens. ^_^

That's almost as evil as the people who went around saying "USAMO results are out!"  Wait make that all people excluding me :P.

[quote="danielkim"]I was just wondering if anyone got accepted yet.[/quote]

It's gonna be Friday at the earliest, folks.

[quote="bubala"]Will you send an e-mail to those who don't get accepted, so that if you're not accepted you'll know you need to make other summer plans?[/quote]

Yes -- there will be three categories, accepted, rejected, and waitlist.

[quote="pilot"]It looks like two eyes and a nose.[/quote]

Very much so. It's just a "curious" smiley.

[quote="sonar"][quote="MithsApprentice"]Very much so. It's just a "curious" smiley.[/quote]

I've always felt that it was a "bemused" smiley.[/quote]

That too. Yet I was discussing the relevance in context.

We are pretty much done with the grading.  We still need to make our decisions, though.

[quote="bubala"]so would u say that we'd know by tomorrow.  I'm pretty sure I'm going to be borderline, so I'm very anxious.[/quote]

It's possible, but I can't promise that.

ah whatever, I try not to stress about things that I am (as of now) powerless to control.  :lol:

yea me neither.. .although if i had the control, i would definitely give myself an acceptance.

Why do I have a feeling that most people who applied would give themselves acceptance? :D


Or you can proceed this way:

Multiply left side by 2:
$2\left(\binom{n}{1}^2+2\binom{n}{2}^2+\ldots+n\binom{n}{n}^2\right)=\\\binom{n}{1}^2+2\binom{n}{2}^2+\ldots+n\binom{n}{n}^2+\binom{n}{1}^2+2\binom{n}{2}^2+\ldots+n\binom{n}{n}^2=\\\binom{n}{1}^2+2\binom{n}{2}^2+\ldots+n\binom{n}{n}^2+\binom{n}{n-1}^2+2\binom{n}{n-2}^2+\ldots+n\binom{n}{0}^2=\\n\left(\binom{n}{1}^2+\binom{n}{2}^2+\ldots+\binom{n}{n}^2\right)=\\n\binom{2n}{n}$

It flows immediately from this.

Last $=$ is a well-knoen identity. You can prove it using some countings or by finding coefficient to $x^n$ in polynomials $(x+1)^{2n}$ and $(x+1)^n(x+1)^n$. (Yes, they're the same => these two coefficients must have same values).


logarithms
real analysis
real analysis unsolved

Let $\left\{ a_{n}\right\}_{n \geq 1}$ be a sequence with $a_{1}\in \mathbb N^\ast$ given, defined by $a_{n+1}= \mu \left( a_{n}\right)+6$, where $\mu (x)$ is the smallest natural number $n$ such that $x|n!$. Prove that the sequence $\left\{ s_{n}\right\}_{n \geq 1}$ defined by \[s_{n}= \left( \sum_{k=1}^{6n}\frac{a_{k}}{k}\right)-t \ln n\] converges for exactly one value of $t$.

[i]Note.[/i] Yeah, it's [i]probably[/i] $t \ln n$ instead of $t \ln t$.

[i]Costel Chiteş, Adrian Stoica[/i]

Sorry, something must be wrong -- $t \ln t$ is constant with respect to $n$.

I'm quite sure it is $t\ln n$ (and that the key to the solution is to note that the sequence $a_{n}$ is bounded and, therefore, eventually periodic).


counting
derangement
search
combinatorics unsolved
combinatorics

Find the no. of derangements of an n-tuple.

This is equivalent to the following well known problem:
Find the search function.


[i] Taken fron Sharygin's geometry book[/i]

 In the exterior of a triangle $ABC$ we take the [color=red]squares[/color]  $BCDE , ACKL , ABMN$ and the parallelograms $EBMP , DCKS$. Prove that the triangle $APS$ is right and isosceles.

 [u]Babis[/u]

SORRY :
[b]
                     quadrat = square [/b]  Babis

I shall use the complex numbers. Define the point $X$ with affix $x$. Thus,
$d=c+i(b-c)\ ,\ e=b+i(b-c)\ ,\ k=c+i(c-a)\ ,\ m=b+i(a-b)\ .$
$p+b=m+e=2b+i(a-c)\Longrightarrow p=b+i(a-c)\ .$
$s+c=d+k=2c+i(b-a)\Longrightarrow s=c+i(b-a)\ .$
$s-a=(c-a)+i(b-a)\ ,\ p-a=(b-a)+i(a-c)\Longrightarrow$
$s-a=i(p-a)\Longrightarrow AP=AS\ \ \wedge\ \ AP\perp AS\ .$

Of course Virgil's solution with complex numbers is very nice and easy.

 I'm waiting a pure synthetic solution. Only equal triangles , and the proof is ready. But first try to find two triangles each of which is equal to $\triangle ABC$. Then,  the other two triangles with sides ... are equal

 Babis

Stergiu, please see http://www.mathlinks.ro/Forum/viewtopic.php?t=78705
and read the my last reply "Only for Robpal".[hide="Here is the syntetical proof much wished by the my friend Stergiu (briefly)."]
$\triangle ACN\sim\triangle ALB$ $\Longrightarrow$ $NC=LB$ and $NC\perp LB$.
$\triangle BMP\equiv \triangle ABC$ $\Longrightarrow$ $PB=AL$ and  $PB\parallel AL$.
$\triangle CSK\equiv \triangle ABC$ $\Longrightarrow$ $CS=AN$ and $CS\parallel AN$.
The quadrilaterals $ALBP$ and $ASCN$ are parallelograms $\Longrightarrow$
$AP=LB$, $AP\parallel LB$ and $AS=NC$ and $AS\parallel NC$ $\Longrightarrow$
$AP=AS$ and $AP\perp AS$.[/hide]


AMC
USA(J)MO
USAMO
AIME
function
articles
AMC 12

OK, so I qualified for USAMO this year through a series of miracles. By some stroke of fate, I managed a 101.5 on AMC 12. I got insanely lucky and got a 7 on AIME. I got a 10 on the USAMO, probably because I spend too much time playing games and therefore got Problem 4.

So naturally, I'm kinda interested in making it again (and optimally making MOP) next year. So I've been practising. Yeah, I feel reasonably certain that I can get at least one problem on any USAMO, and that'll go up to 2 or 3 by USAMO time next year. Plus, I've gotten 7s or 8s on all the AIMEs I've done (which should go up by next year...) So I started working on AMCs. I did the 2001 AMC 12, taking 45 minutes, and I got... a 63.8. Ow. So yeah, that'd be OK if the score increased proportionally with the time. But we all know it doesn't. I guess if I had my TI-83+ instead of a four-function, and 90 minutes instead of 45, I'd maybe get 100. But I want to be fairly sure of a 120 instead of possibly getting 100.

So yeah. Basically, if you know you can do the AIME and USAMO, how does one prepare for AMC?

I have almost the exact question. I can do many medium to hard AIMEs and a few USAMO questions(although I am not confident I can do one from every year). But AMC is annoying, I got 97 on AMC12A. Yesterday I did AHSME 1980 for practice and i got 105. I answered 20 questions and got 3 wrong. It's not that bad but it took me the whole hour and half to do 20 questions and I alway make 2-3 careless mistakes.
 How should I improve?

[quote="JSRosen3"]Have you read the article on AoPS about how to avoid making stupid mistakes? You can find it at [url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php[/url]. It's very useful.[/quote]

You took the words right out of my fingers.  :)  That's a great article, and everyone preparing for next year's AMCs ought to read it, reread it, and reread it again.

[quote="beta"]How should I improve?[/quote]

Practice.  A lot.  And don't just practice to the extent of "do I know how to do this?"  Practice to the extent of, "If I got the wrong answer, I'm counting it as wrong, even if I just wrote down the wrong thing at the end."  And then do every AMC problem you've ever seen according to that standard.  If you know more than one way to do a problem, do it both ways.  And do every problem until you get it right.  Even if that means doing the same logic and the same calculations twice.

Also, never look at the answers in advance.  Ever.  Except for the (very small) class of problems where it is absolutely necessary to solve.

[quote="beta"]I read that page. The problem is: Some AMC problems are designed to trick you(like this year's AMC12A #7). Also I need to work on speed,  it seems that I can't do computations(solving equations) very fast because I can usually see the tricks to do problems(at least first 20 problems of AHSME) right away.[/quote]

I typically hate to recommend this to a student who can already work some USAMO problems, but it might be worth your time to bootcamp yourself with AMC/AHSME problems.  Also, if you feel you would like outside help organizing your approaches to such problems (organization breeds speed), then you might consider our AMC-12 problem series.

Good sources of practice problems include Mu Alpha Theta (good for speed training), ARML (good for problem solving training), Harvard/MIT tournament questions, and Mandelbrot problems.

Just a note on the AMC-12 Problem Series--it is much harder than I expected. It focuses on all the hardest AMC 12/AHSME problems, so I'd recommend it even if you're already able to do AIME questions or have gotten a high AMC 10/12 score in the past.

Thanks for the suggestions.  I'll be helping my students prepare for the AMCs this year, and hopefully for the AIME, too.  These suggestions will be useful.

[quote="takuanitromars36"]How did you manage to score so well on the AIME and on the USAMO? One of my "friends" got a 118 on the AMC 12, and she did not qualify for the USAMO. Is it just that you might have been careless on the AMC 12? To be honest, I got 0.5 higher than you on the AMC 12 this year, but I scored significantly lower than you on the AIME.[/quote]

Actually, I think it's because the more difficult problem solving questions are the ones I'm better at. I'm better at proofs than at AIME-type questions, and better at AIME-type questions than easy AMC questions. Also, I prepared NONE for the AMC, a little for the AIME, and a LOT for USAMO. So...

Edit: GAAAAAAAH! I misused an apostophe: I typed "one's" instead of "ones!" If anyone saw...and told people... I can never show my face in public again...

I think I'm slipping. I did the same thing just the other day, on a different forum.

Ok, I got 3 on both Mock AIMEs. Either I overestimated my AIME skill(since I have never taken an AIME before under real AIME pressure) or I am slipping. What really bothers me is that I solved some of problems I didn't get in the 3 hours easily(7, 8 on AIME I, 4, 7, 11 on AIME II) after the test.
Help on AIME preparation

I think the content of the AIME and AMC are similar enough such that if you prepare well for AIME, you will have the knowledge for AMC. Anything not covered on AIME and on AMC is probably trivially easy. USAMO is completely different, and just because you're prepared for USAMO doesn't mean you're prepared for the preliminary ones.

Yan Zhang

im practicing aime problems right now and i've been getting like only the first six or seven, and maybe number 9 or 10.  I think i need more practice, but i think that is good enough for usamo.

I need some help with preparation also.


consider the matrix

1 0 2
2 0 4
0 0 0

the rank of it is obviously 1, but the nullity seems to be 1 too since the general solution is t(-2,0,1) for all real t, but then we can't get rank(A) + nullity(A) = n where n is 3 in this case...
what's the problem?

EDIT:
Never mind, I forgot that we should include s(0,1,0) - guess i need more sleep... please delete this topic

[quote="mathwizarddude"]Never mind, ... please delete this topic[/quote]
When I'm teaching and I ask for responses from the class, there are times when I get more in the lecture from a wrong answer than I ever would from a right answer. This is a perfect illustration of that principle, and I'll let it stand, as is.


geometry
geometric transformation
homothety
LaTeX
geometry proposed

The circle $\zeta_{1}$ is inside the circle $\zeta_{2}$, and the circles touch each other at $A$. A line through $A$ intersects $\zeta_{1}$ also at $B$, and $\zeta_{2}$ also at $C$. The tangent to $\zeta_{1}$ at $B$ intersects $\zeta_{2}$ at $D$ and $E$. The tangents of $\zeta_{1}$ passing thorugh $C$ touch $\zeta_{2}$ at $F$ and $G$. Prove that $D$, $E$, $F$ and $G$ are concyclic.

Consider the inversion of pole $C$ and power $CG^2=CF^2$. It fixes $\zeta_1$, and it turns $\zeta_2$ into a line which must be tangent to $\zeta_1$ in the point which is the intersection between $\zeta_1$ and $AC$. This point is $B$, so $\zeta_2$ is turned into the line $DE$. This means that the points $D,E$ are fixed by the inversion (since they lie both on $\zeta_2$ and on the image of $\zeta_2$), so $CD=CE=CG=CF$, meaning that $E,D,F,G$ all lie on a circle centered at $C$.

Consider the homotety of center $A$ which tranforms ${\zeta_1}$ to $\zeta_2$ then $DE$ will transform to the tangent of ${\zeta_2}$ at $C$ parallel to $DE$. Hence $CD=CE$ and $\angle{CEB}=\angle{CAE}$ and from here $CD^2=CE^2=CB\cdot{CA}=CG^2=CF^2$ and we are done.

I was in the contest. I'm not that good at geometric transformation, so I straightforwardly calculated the distances from C to those points, and I found them equal. :)
Sorry for my ignorance concerning Latex!


Solve the system of equations:

$ x \plus{} y \plus{} \sqrt {xy} \equal{} 28$
$ x^2 \plus{} y^2 \plus{} xy \equal{} 336$

hello, from the first equation we have $ x\plus{}y\equal{}28\minus{}\sqrt{xy}$, squaring both sides and using the second equation we get $ xy\equal{}64$. Plugging this in both equation and symplifiying we get
$ x^2\plus{}y^2\equal{}272$
$ x\plus{}y\equal{}20$
Inserting $ y\equal{}20\minus{}x$ in the equation above and solving  $ x^2\minus{}20x\plus{}64\equal{}0$ we get $ x_1\equal{}4$, $ x_2\equal{}16$, from here we can calculate $ y_1\equal{}16$ and $ y_2\equal{}4$.
Both solutions fulfill our system.
Sonnhard.


Two circles($O_{1}$,$R_{1}$) and ($O_{2}$,$R_{2}$) intersect at $A$ and $B$($O_{1}$ and $O_{2}$ are not on the same side of $AB$.The line $m$ through $B$ meets ($O_{1}$) and ($O_{2}$) at $C$ and $D$ respectively ($B$ $\in$ [$CD$] ).$M$ is the midpoint of $CD$ the line $AM$ intersects the circle ($O_{2}$) at the second point $P$.The perpendicular to $OM$ at $M$ intersects the line $AC$ at $Q$ .
Prove that the line $PQ$ passes through a point independent of $m$.

Oh,I know this problem,It's in the team selection test of The Hanoi university of education,isn't it?
So $O$ is $O_1$.
This is a nice problem by Mr.TranVanTan.The solution is (I myself solve it  :lol: ) :
We denote that:
$R$ is the inersection of the line $AP$ and the circle$(O_1)$($R\neq A$)
$S$ is the intersection of two lines $BR$ and $MQ$
From the [color=green]butterfly theorem [/color]result that:$MS=MQ$.[color=red](1)[/color]
In other hand,It's easy to show that $MP=MR$(since $MC=MD$)[color=red](2)[/color]
From [color=red](1)[/color] and [color=red](2)[/color] we have $PQ,RS$ are parallel. $\longrightarrow \widehat{APQ}=\widehat{ARB}=\frac{1}{2}\widehat{AO_1B}=constant$
So if we denote $I$ is the intersection of the line $PQ$ and the circle $(O_2)$($I\neq P$) then $I$ is independent of $m$.
[color=red]Q.E.D [/color]


[quote="NikolayKaz"]Let $ a_0 \equal{} 0,91$ and $ a_k \equal{} 0,99...900...01$ ( there are $ 2^{k}$ nines and $ 2^{k} \minus{} 1$ zeros)

Find

$ \lim_{n \rightarrow \plus{} \infty} (a_0 \cdot a_1 \cdot ... a_n)$[/quote]

Let $ x\equal{}\frac{1}{10}$ , then $ a_k\equal{}1\minus{}x^{2^k}\plus{}x^{2^{k\plus{}1}}$ .

So $ (1\plus{}x\plus{}x^2)a_0a_1...a_n\equal{}(1\plus{}x\plus{}x^2)(1\minus{}x\plus{}x^2)(1\minus{}x^2\plus{}x^4)...(1\minus{}x^{2^n}\plus{}x^{2^{n\plus{}1}})$
$ \equal{}1\plus{}x^{2^{n\plus{}1}}\plus{}x^{2^{n\plus{}2}}$

$ \Rightarrow \lim_{n \rightarrow \plus{} \infty} (a_0 \cdot a_1 \cdot ... a_n)\equal{}\lim_{n \rightarrow \plus{} \infty}\frac{1\plus{}x^{2^{n\plus{}1}}\plus{}x^{2^{n\plus{}2}}}{1\plus{}x\plus{}x^2}\equal{}\frac{100}{111}$


This are two lemmas i encoutered while prooving this theorem
EVERY NATURAL NUMBER CAN BE WRITTEN AS A SUM OF FOUR SQUARES.
However you may prove yours i want help with two lemmas
1)For odd numbers of the form 3 in modulo 4,it comprise of 3 even numbers and 1 odd
2)For odd numbers of the form -1 in modulo 4,it comprise of 1 even and 3 odd numbers.
3)for even numbers well i have not devised a lemma.

[quote="21 7 15"]T
1)For odd numbers of the form 3 in modulo 4,it comprise of 3 even numbers and 1 odd
2)For odd numbers of the form -1 in modulo 4,it comprise of 1 even and 3 odd numbers.
.[/quote]
Isnt $ \minus{}1$ modulo $ 4$ and $ 3$ modulo $ 4$ the same ,
Maybe you meant $ 1$ modulo $ 4$

Even squares $ (2m)^2 \equal{} 4m^2$ are $ 0 \pmod{4}$; odd squares $ (2m\pm 1)^2 \equal{} 4(m^2\pm m) \plus{} 1$ are $ 1 \pmod{4}$. Hence if $ n$ is $ 3 \pmod{4}$, the only way this may occur is $ 3 \equal{} 1\plus{}1\plus{}1\plus{}0$, while if $ n$ is $ 1 \pmod{4}$ the only way is $ 1 \equal{} 1\plus{}0\plus{}0\plus{}0$.
Similarly, if $ n$ is $ 2 \pmod{4}$ the only way is $ 2 \equal{} 1\plus{}1\plus{}0\plus{}0$. However if $ n$ is $ 0 \pmod{4}$ then we may have $ 0 \equal{} 0\plus{}0\plus{}0\plus{}0$, but also $ 0 \equiv 1\plus{}1\plus{}1\plus{}1 \pmod{4}$, for example $ 84 \equal{} 1^2 \plus{} 3^2 \plus{} 5^2 \plus{} 7^2 \equal{} 4\cdot 21 \equal{} 0^2 \plus{} 2^2 \plus{} 4^2 \plus{} 8^2$.


[quote="Jakey"]Hi, can anyone help me with these? thanks and please show workings...

1) For which integers k does x^k have the largest coefficient in the expansion of (x+3)^50.

2) Determine which is larger sqrt 10 + sqrt 17 or sqrt 53.

3) In a race of 1760 meters, A beats B by 330 meters, A beats C by 460 meters. If the contestants continue to run at the same uniform rates, by how many meters will B beat C?[/quote]

[hide="Hint for 1"]
Start with the coefficient of $ x^0$ and consider the quotient of the coefficients of two consecutive powers of k. When do you have a maximal quotient?
[/hide]

[hide="2"]
\begin{align*} 170 & > 169 \\
2\sqrt {170} & > 26 \\
10 + 2\sqrt {170} + 17 & > 26 + 10 + 17 \\
\left(\sqrt {10} + \sqrt {17}\right)^2 & > \left(\sqrt {53}\right)^2 \\
\sqrt {10} + \sqrt {17} & > \sqrt {53} \end{align*}
[/hide]

[hide="Hint for 3"]
Compute the quotient of the velocities of A and B and A and C, respectively. How can you compute the quotient of the velocities of B and C from these quotients?
[/hide]

[quote="Jakey"][quote]Start with the coefficient of  and consider the quotient of the coefficients of two consecutive powers of k. When do you have a maximal quotient? [/quote]

i seem to get your logic here bertram, but i only reached this point: the quotient of say consecutive coefficients which are for example: (50Ck)3^k and 50C(k-1) 3k-1 is:

k/3(51-k). now what do i do? anyone help me please?[/quote]

Hm. I got something different from yours:
\begin{align*}
(x+3)^{50}&=\sum_{i=0}^{50} a_ix^i \\
a_i &= \binom {50} i 3^{50-i} \\
\frac{a_{i+1}}{a_i} &= \frac{\binom {50} {i+1} 3^{50-i-1}}{\binom {50} i 3^{50-i}} \\
\frac{a_{i+1}}{a_i} &= \frac{50-i}{3i+3}
\end{align*}

A very important aspect is that the RHS is strictly decreasing in i; this means that $ a_i>a_{i+1}\Rightarrow a_k > a_{k+1} (k\ge i)$ (why?). What can we then say about the relation between that $ a_i$ and all the following $ a_k$'s? How does this help us to find the maximum?

DISCLAIMER: It's 1:45 a.m. and I'm le tired. So if I'm saying complete nonsense please correct me :(

hmm...i know why we got different results, i started from x^50 rather than x^0. But anyhow, i seem not yet to understand, would you rather show me the solution, i really tried but here's all that i could think of:

a) the denominator, ai is always greater than a(i+1).
b) we need to get the maximum(?) quotient to...(?)...

i'm confused. its because the answer ought to be 12. if this is the case, then at some point, the value of the coefficients are increasing->decreasing and vice versa...sorry if i disturb you too much bertram, i really appreciate your efforts to help:D

He's posted these problems twice? Anyone get the idea its homework?

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=174947[/url]

I got 12 with my method too, so I hope it really works :D

[hide="Solution"]
Set $ f(i) \equal{} \frac{a_{i\plus{}1}}{a_i} \equal{} \frac{50\minus{}i}{3i\plus{}3}$ as I showed above. Suppope $ a_i \equal{} \max (a)$ and get $ f(i\minus{}1) \equal{} \frac{a_i}{a_{i\minus{}1}}\ge 1$ and $ f(i)\equal{}\frac{a_{i\plus{}1}}{a_i}\le 1.$ Using our formula for f, we see that these inequalities reduce to $ \frac{47}4 \le i \le \frac{51}4\Rightarrow i \equal{} 12\;\;\;\Box$
[/hide]

[quote="Jakey"]thanks bertram...by the way simonM, this is not homework, im practicing for a contest[/quote]

Doesn't explain why you posted it twice

[quote="Jakey"]thanks bertram...by the way simonM, this is not homework, im practicing for a contest[/quote]

Well I'll keep my fingers crossed for you. This problem is actually similar to a problem I did recently (n, a, b given, find the largest term in the expansion of $ (a\plus{}b)^n$), so I kind of knew the solution :P

[quote="Jakey"]Hi, can anyone help me with these? thanks and please show workings...

1) For which integers k does x^k have the largest coefficient in the expansion of (x+3)^50.

2) Determine which is larger sqrt 10 + sqrt 17 or sqrt 53.

3) In a race of 1760 meters, A beats B by 330 meters, A beats C by 460 meters. If the contestants continue to run at the same uniform rates, by how many meters will B beat C?[/quote]

I don't really understand bertram's solution
but I think it's just because I don't really know what the huge huge huge bracket means :oops:
[hide="try"]
for 1     (coe of x^49)x3x49/2= (coe of x^48)

so the coe of the previous term times 3 and then a (r/s) would equal the coe of the next term, where r is the degree of the the previous term, and r+s =51
3r/s > 1 for r>12  as 12/39x3 =12/13
so the coe keeps increasing up till x^12 and diminishes from x^11 onwards

for 3   B beats C by 130 in a 1430 race
so if it is a 1760 race, it would be (130/1430)x1760=160
[/hide]

$ \binom nk \equal{} \frac{n!}{k!(n\minus{}k)!}$ is the amount of different ways you can choose a subset of size k from a set of size n. I also saw $ C^n_k$ on the forums, but the first is the notation I'm used to ;)

The way this "huge bracket" was used is the binomial theorem: $ (a\plus{}b)^n\equal{}\sum_{i\equal{}0}^n \binom ni a^i b^{n\minus{}i}$

[quote="bertram"]$ \binom nk \equal{} \frac {n!}{k!(n \minus{} k)!}$ is the amount of different ways you can choose a subset of size k from a set of size n. I also saw $ C^n_k$ on the forums, but the first is the notation I'm used to ;)

The way this "huge bracket" was used is the binomial theorem: $ (a \plus{} b)^n \equal{} \sum_{i \equal{} 0}^n \binom ni a^i b^{n \minus{} i}$[/quote]

 :D thanks :blush: