probability trigonometry geometry MATHCOUNTS FTW Alcumus AMC

First I would like to thank everyone again for their helpful advice on my question from the other topic I posted. Currently I am planning to save money for the counting & probability book. Well lets see..where to begin.. Okay, I'm a college student (1st year), and more than anything I want to change, and improve my skills in mathematics (I hope to teach it at college level one day). In my math class (Pre-Calc 2, mainly consists of trig) at college I find it pretty easy, I am doing the best in the class. Many of the other students find it difficult. All of the math classes I have been in since about 7th grade, I have gotten an A. However like all of my math classes since 7th grade(including my college one), they have all seemed to be about memorization. I feel like a parrot, and I ask myself what does it mean to be good at math. When I took the SAT, I found some of the questions very difficult...but all the math involved was very basic..this kind of hinted that something with my math skills were wrong. I tried the pre-tests for some of the intro books here, and I found them to be very difficult (and these books were recommended for middle schoolers!). I wrote to Mr. Rusczyk about how embarrassed I was about this, and he was very polite and helpful in his response (this is what largely influenced me to make an account). So what is being good at math about? Creativity? Thinking critically? I believe this is what I lack...I really hope these are skills that can be built. I want math to be more than memorization. For people who have this skill, do you find that it helps in other areas, besides math? Right now my problem is, that I also have Biology, and Psychology, along with the math. (These classes take up a lot of my time) How can I manage my time, so that I have time to build these skills? I would really appreciate any advice, and stories of people who had similar problems, and how building this skill helped them out, and changed their view on mathematics.

in my school, we do a lot of memorizing as well
personally, i think that memorizing equations is useless
if you happen to forget (which happens to the best of us) then you're screwed
the books i have proved the equations that i memorized before, and it showed me how to get the equation again if i happen to forget it

i have not taken the SAT, but i think that critical thinking would be important in those questions
a lot of problems out there require you to use what you know, and bend it so that you can get the answer to your question
seeing as how you're in college, i would suggest Art of Problem Solving 1 and 2, partially because those are the only ones i have..., but also because they are challenging and have great problems
of course, i don't know what level you're at, but i bought volume 1 in order to learn the material, and volume 2 in order to go beyond what i would regularly learn in school at my age and get better in mathematics overall, and i think that these two books would help you a lot if you're trying to get better at understanding the equations you memorized and getting better at mathematics overall
oh, BTW volume 1 has some trig for right triangles, while volume 2 goes into trig for nonright triangles, so you can have an idea of what the level of these two books are like =)

as how to manage your time, i don't have many suggestions
maybe cutting out some time you might use somewhere else that is not overly important?

my story is that i have always done good (well?) at math
i was always at the head of my class
however, through participating in mathcounts last year, i realized i still have a lot to learn that will not be taught at school in my grade
i started using FTW more, and later, alcumus
i must say, these two sources helped me a lot in speed, and ability to solve problems, as well as general knowledge
a few weeks back, i ordered the books i have now, and they are excellent too
i have learned a lot from the books that i would not have learned right now otherwise
even though the information i learned through these sources might not help me directly in mathcounts, i'm sure it will have an impact on my future

i hope what i have written will help you
good luck!

Thanks for your reply. Yeah I need to figure out how I can fit this math into my schedule.
By the way I am unfamiliar with all these tests, they just seem like a bunch of random acronyms.
Can someone tell me about these tests, for example I have seen USAMO, AMC, AIME, etc mentioned.
What is the level of difficulty of these tests? Like AIME<USAMO<SAT<AMC ? (I just made this up but could some one show the difficulty level like this?) Also what books will I have to complete (like intro probability) and get a good understanding of, to start being able to actually do decent on the exams. Finally, If you are at the level where you are doing good on the exams, will this carry over to being good at other areas in school? For example say you do good on the exams, and say the highest level of content they test is Pre-Calc, when you take calculus, even though you haven't seen the material will you be able to look at it with a new perspective where it comes almost easy, opposed to not having done any of the books/ tests?

For difficulty, it's SAT<AMC<AIME<USAMO, but they're only for people in high school or lower (you mentioned you were in college).

as for the acronyms

for the ones you put:
USAMO = United States of America Mathematical Olympiad
AMC = American Mathematics Competitions
AIME = American Invitational Mathematics Examination

some others you might see
mandelbrot = not an acronym
M&IQ = Mathematics and Informatics Quarterly (i have not seen this outside of my AoPS book, but since it's in there, i though i'd put it)
ARML = American Regions Mathematics League
MA(little sign i can't type) = Mu Alpha Theta

Thanks AIME15 and Vallon22!
AIME 15 do you think being able to do the USAMO problems, would help me out in college. I don't plan to take the test ( I guess I can't anyway)I can do most of the SAT problems pretty comfortably, I hope to get at a level where I could do most of the USAMO questions comfortably. Honestly I find some of the SAT questions difficult, so the USAMO most be extremely difficult :maybe:
Well AIME 15, Vallon22, or anyone else could you help me make a study plan (really basic).
For example, could you recommend me which books I should complete, and how far I should get on Alcumus before I'm ready to start working , and moving on to different exams?
So for instance, A demo recommendation: (Feel free to change the format, and to add or omit any ideas)

[b]Test:[/b] SAT
[b]Difficulty:[/b] Easy: Your foes cower and fail before your unstoppable onslaught, yet final victory will leave you wanting more.
[b]Preparation:[/b] Introductory Series/ Alcumus
[b]Target Score before moving on:[/b] 700-800

[b]Test[/b]: AMC
[b]Difficulty:[/b] Normal: Hordes of aliens vie to destroy you, but nerves of steel and a quick trigger finger give you a solid chance to prevail.
[b]Preperation:[/b] Intermediate Series
[b]Target Score before moving on:[/b] ???

[b]Test:[/b] AIME
[b]Difficulty:[/b]Heroic: Your enemies are as numerous as they are ferocious; their attacks are devastating. Survival is not guaranteed.
[b]Preparation:[/b] Classic Problem Solving series
[b]Target Score before moving on:[/b] ???

[b]Test:[/b] USAMO
[b]Preparation:[/b] ???
[b]Difficulty:[/b] Legendary: You face opponents who have never known defeat, who laugh in alien tongues at your efforts to survive. This is suicide.
[b]Target Score before moving on:[/b] ??? (What's the next Test?)

Halo is fun :P

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=561744066&t=241217]here[/url]; USAMO/AIME/AMC problems are MUCH MUCH MUCH harder than SAT math questions.

The Introduction series on AoPS should be enough to get you to the hard AMC level (although missing some topics from Intermediate Algebra).  Then, after the intermediate series you should be able to do most of the problems in AIME (someone correct me if this is wrong).  The AoPS Volume 1 and Volume 2 books are good for contest prep; many people say that after Volume 1 you should be able to pass the AMC's to the AIME, and after Volume 2 you should be able to score very highly (about 135-150) and do very well on AIME.

For USAMO, there are a number of choices.  As I've never been there (:( ) i can only share recommendations from others, which have mostly been ACoPS (the Art and Craft of Problem Solving by Paul Zeitz, great book (i have it) its good for learning how to approach problems, but might not necessarily help you to MASTER usamo problems and geometry section is quite weak) and Arthur Engel's book Problem Solving Strategies (really gets into harder problems).

If you can do most of the SAT problems (here I mean like 1-3 wrong at max), I suggest that you take a practice AMC exam under contest conditions (75 minutes, no calculator, 6 points for correct answer, 1.5 for blank).  You can find old AMC exams [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182]here[/url].  After taking a couple, you can find which areas you're weak in and study more in them.

The Are You Ready? and Do You Need This? tests can also help with this; also, they can help you decide if you really need that Introduction book, and maybe you should just skip to the Intermediate one.

So do you think by prepping with all of these books, it will help me out in college? For example if I take Calculus soon, is there anyway that these books will help me and possible give me an advantage to learning calc.

Um, I'm not sure of that.  To excel at calculus, I'm pretty sure that you just need to know algebra, geometry and precalc...as far as I know, calculus at "normal" school doesn't require much number theory/combinatorics.

AoPS has a new calculus class, you can see the class page [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php#coursedetails40]here[/url].  Try the Are You Ready test.

@MrMagician: It will help you very much in mathematics, but more importantly, it will teach you how to solve difficult problems (generally, not just confined to mathematical contexts). I don't know how rigorous your Calculus class is, so I can't really answer your second question. Regardless, it will probably still help.

@AIME15's post above the last post: I more or less agree, but I have to mention that books obviously can't give you everything, so problem solving skills and creativity play a large role. Exposure to many different types of problems is also crucial.

for me, alcumus has gotten pretty easy
i can do most of the problems it gives me with little difficulty

which is why i bought the books i have, AoPS 1 & 2
as for the other books, i don't have them, so i can't give you suggestions on them

however, my mathcounts proctor had some of the others and i took a quick look at them
the introductions were a little easy for me, so i would go for the intermediate, but i don't know what level you are at

if you wish to score high on the AMC and AIME, i would go straight to either AoPS 1 & 2, or the intermediate series

I am doing Calculus right now at my school and it is really easy even thought I am not all that good at math... (I hope to make the AIME this year, and I would be content with a 5 on AIME for now). If you did well thus far I don't think you should have any trouble with Calculus and doing math at AoPS helps in all situations I think and it's just really fun. I personally thought SAT math was easy (miss read the number line lol), but being able to easily do USAMO is a though goal, only about 500 people make it every year and many people don't even get a question right. I think MOP level is achievable from what I heard, but with a lot of practice and effort. So, don't be discourage, but if you want to be a MOP level problemsolver, it might take you years of practice, so I think you should do math for fun and maybe you will reach such level. I really think you should buy AoPS V1 it really does help.

My Question for you is if you love math, why are u taking trig in college. At most schools even the less talented kids atleast get to take AB calc in college. Secondly, if you can't do SAT math perfect USAMO is probably 4 years of hard work away. The USAMO equivalent in college is the Putnam- I guess you could shoot for that  :wink:

I didn't think you had to be good at math to like it. I'm taking trig to prepare for Calc. I'm trying to get better at math, and to do that you have to start somewhere. That somewhere for me happens to be trig. I was curious about these tests because (correct me if I'm wrong) they indicate a level of skill and achievement in math. Would you agree that a person who is good at math is good at problem solving? This is a weak area for me, and I need to improve it. These exams test problem solving abilities for a high shcool student. I don't think my problem solving abilities are at the level right now. I would like to get them above and beyond that level. I would like to get to the level, where I could participate and do decent on the putnam. How could I get to that level? Do you think it would be wise to start with the exams given in highschool and then work my way up? Or do you think this is a flawed approach?
BTW Jhangil what is MOP?

Many people are good in math but horrible at problem solving.  Problem solving is a lot harder to be taught than math, but is done pretty well in the Art and Craft of Problem Solving by Paul Zeitz.

Putnam-level is going to be tough to reach.  Out of 14 (or 12?), the median score is often zero, and is a test given to undergraduates.

MOP is the Math Olympiad Summer Program, in which potential/future IMO'ers qualify for by scoring well on the USAMO.

AIME15 do you think the books here are good resources to learn problem solving?
Should I wait to try the book by Paul Zeitz, and focus on reading the books here?

I don't really have a good grasp of your level yet, nor am I very experienced myself :P

And yes, the books here are [b]very[/b] good resources for problem solving.  They go more in-depth than your average textbook, much more in-depth.

Try one or two of the more recent tests [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44]here[/url] under full contest rules and see what you can get, and what you need help on.


inequalities
function
inequalities proposed

Prove for non-negative reals $ a$, $ b$, $ c$ and $ n \in \mathbb{Z^ \plus{} }$ the following inequality:
\[ \sqrt [n]{\frac {a^3 \plus{} abc}{3}} \plus{} \sqrt [n]{\frac {b^3 \plus{} abc}{3}} \plus{} \sqrt [n]{\frac {c^3 \plus{} abc}{3}} \ge \sqrt [n]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \plus{} \sqrt [n]{abc}.\]
(Actually, the inequality holds for real $ n\ge 1$.)

This is really obvious with Minkowski's Inequality: you can just multiply by the nth root of 3 to make it simpler first.

Sorry I haven't used Minkowskis in a while so I made a huge mistake

Rather let $ f(x) \equal{} \sqrt [n]{x}$. Then $ f$ is a concave, increasing function, and we have to prove that
$ f(a^3 \plus{} b^3 \plus{} c^3) \plus{} f(3abc) \plus{} f(0) \leq f(a^3 \plus{} abc) \plus{} f(b^3 \plus{} abc) \plus{} f(c^3 \plus{} abc)$.
WLOG let $ a \geq b \geq c$. Then $ a^3 \plus{} abc \geq b^3 \plus{} abc \geq c^3 \plus{} abc$, and $ a^3 \plus{} b^3 \plus{} c^3 \geq 3abc \geq 0$ by AMGM and the fact that $ a,b,c \geq 0$.
Now if $ a^3 \plus{} b^3 \plus{} c^3 \geq a^3 \plus{} abc$, then we also have $ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \geq a^3 \plus{}abc \plus{}b^3 \plus{} abc$ and $ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \equal{} a^3 \plus{} abc \plus{} b^3 \plus{} abc \plus{} c^3 \plus{} abc$ from whence we have that $ (a^3\plus{}b^3\plus{}c^3,3abc,0)$ majorizes $ (a^3\plus{}abc,b^3\plus{}abc,c^3\plus{}abc)$. The required inequality then follows from the HLP Majorization inequality.
If on the other hand $ a^3 \plus{} b^3 \plus{} c^3 \leq a^3 \plus{} abc$, then we have to do something different.


[quote="xbatox"]i know this question can be kinda ... ridiculous? hehehe but... i was wondering.. if there's goingg to be an election of "miss imo 2008"[/quote]

Just search for "girl" or similarly in the sub-forums of the [url=http://www.mathlinks.ro/index.php?f=87]IMO[/url] and there were some topics about in in the past. But we came to the conclusion that you may not want to discuss this issue in the math section of this forum. Compare with  [url=http://www.mathlinks.ro/viewtopic.php?p=556804#556804]old threads[/url] as this one.


What question numbers correspond to which questions? I only remember a couple...

Great!!
When I originally posted this, I thought that people would put together their stuff on the test here... this was the day after the AIME.

try to look around the forum(this AMC forum) and the contest resources section

and yes, our speedy posters (props to joml88 and chess64) have gotten the AMC series online the day after :D


calculus
integration
algebra
polynomial
quadratics
partial fractions
calculus computations

The textbook explains for an integral such as this:
\[ F(x) \equal{} \int\dfrac{x \minus{} 2}{(x \plus{} 1)(x^2 \minus{} 5x \plus{} 2)^2}
\]
I can divide up the integrand into partial fractions like this:
\[ \dfrac{A}{x \plus{} 1} \plus{} \dfrac{B_1x \plus{} C_1}{x^2 \minus{} 5x \plus{} 2} \plus{} \dfrac{B_2x \plus{} C_2}{(x^2 \minus{} 5x \plus{} 2)^2}
\]
I understand how this would work for linear factors in the denominator of the integrand, but I can't grasp why the single constant (arbitrarily $ A$ in the above example) in the numerator of a partial fraction has to be replaced with a polynomial of the form $ Bx \plus{} C$.. (so if a factor is a cubic instead of a quadratic, would the numerator of the partial fraction be of the form $ Dx^2 \plus{} Ex \plus{} F$? Why?)

Also, why should there be additional partial fractions when a factor is repeated? I guess what I'm looking for is just how to derive these fractions from the original integrand. Thanks for any tips :)

partial fractions can only be done if the degree of the numerator is[b] [u]strictly less[/u][/b] than the degree of the denominator.
i particularly don't feel like explaining it now,  but you can read

http://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx

essentially, if you have something like

$ \frac {\textbf f(x) }{ (x \plus{} 2)^2(x^2 \minus{} x \plus{} 1)^3(2x^3 \minus{} x^2 \plus{} 4x \minus{} 7)^2(6x^4 \minus{} 3x^2 \plus{} 21x \minus{} 2)^3}$

where $ \textbf f(x)$ is a polynomial (in $ x$) of degree $ 25 ( \equal{} 2\plus{}6\plus{}6\plus{}12\minus{}1)$ or less,  then you can  break it down into

$ \frac {A}{x \plus{} 2} \plus{} \frac {B}{(x \plus{} 2)^2} \plus{} \frac {Cx \plus{} D}{x^2 \minus{} x \plus{} 1} \plus{} \frac {Ex \plus{} F}{(x^2 \minus{} x \plus{} 1)^2} \plus{} \frac {Gx \plus{} H}{(x^2 \minus{} x \plus{} 1)^3}$

$ \plus{} \frac {Ix^2 \plus{} Jx \plus{} K}{2x^3 \minus{} x^2 \plus{} 4x \minus{} 7} \plus{} \frac {Lx^2 \plus{} Mx \plus{} N}{(2x^3 \minus{} x^2 \plus{} 4x \minus{} 7)^2} \plus{} \frac {Ox^3 \plus{} Px^2 \plus{} Qx \plus{} R}{6x^4 \minus{} 3x^2 \plus{} 21x \minus{} 2}$

$ \plus{} \frac {Sx^3 \plus{} Tx^2 \plus{} Ux \plus{} V}{(6x^4 \minus{} 3x^2 \plus{} 21x \minus{} 2)^2} \plus{} \frac {Wx^3 \plus{} Xx^2 \plus{} Yx \plus{} Z}{(6x^4 \minus{} 3x^2 \plus{} 21x \minus{} 2)^3}$

[quote="undefined117"]I understand how this would work for linear factors in the denominator of the integrand, but I can't grasp why the single constant (arbitrarily $ A$ in the above example) in the numerator of a partial fraction has to be replaced with a polynomial of the form $ Bx \plus{} C$.. (so if a factor is a cubic instead of a quadratic, would the numerator of the partial fraction be of the form $ Dx^2 \plus{} Ex \plus{} F$? Why?)[/quote]

This is basically polynomial division, disguised somewhat.  Suppose your original fraction is $ \frac {P(x)}{Q(x) \cdot (\textrm{other stuff})} \equal{} \frac {R(x)}{Q(x)} \plus{} \textrm{more other stuff}$, where we want to know what we can about $ R(x)$.  Multiply both sides through by $ Q(x)$ and write $ P(x)$ in the form that you get from dividing by $ Q(x)$ (some quotient times $ Q(x)$ plus some remainder); grouping all the terms divisible by $ Q(x)$, the leftover part on the left is just the remainder when $ P(x)$ is divided by $ Q(x)$, and on the right it's $ R(x)$.  So these two things must be equal.  (This is just like taking the entire equation modulo $ Q(x)$, if you will.)  In particular, $ R(x)$ can be a polynomial of any degree less than that of $ Q(x)$.

ehh.. that [i]looks[/i] kinda complicated.. well, just confusing :blush:

I've got the hang of this partial fractions thing now, but I think I'll need some more time to think about JBL's post before I actually understand it :lol:

Thank you both for the help!

One point to emphasize: the total number of unknown coefficients must equal the the total degree of the denominator. (This was implicit in both JBL's and misan's posts; I'm just stressing it.)

Something I say in class: Partial fractions is not a technique of integration. It is a technique of algebra that happens to have uses in integration.


function
logarithms
trigonometry
limit
complex numbers
real analysis
real analysis unsolved

By Riesz/Fischer theorem we know that $ \sum_{n>0} \frac{e^{inx}}{n}$ is the fourier series of a function in $ L^2[\minus{}\pi,\pi]$.  Show that this series is not the series of a piece-wise continous function.

(even a stronger result seems to hold: the series is not series of any Reimann integrable function).

In this particular case, we can find out whatever we want by explicit computation. We have that $ \sum_{n\equal{}1}^{\infty}\frac{z^n}{n}\equal{}\minus{}\log(1\minus{}z)$ for $ |z|<1.$ Using Abel's Theorem to evaluate the boundary values and employing a few trigonometric identities, we get that:

For $ 0<x<2\pi,$ $ \sum_{n\equal{}1}^{\infty}\frac{e^{inx}}{n}\equal{} \minus{}\ln\left(2\sin\frac{x}2\right)\plus{}i\left(\frac{\pi\minus{}x}2\right).$

The imaginary part is piecewise continuous, but the real part has a logarithmic singularity at zero.

---

There should be methods that don't depend on such explicit computation, and could then be used in other cases that resemble this in some important way.

[quote="Kent Merryfield"]We have that $ \sum_{n \equal{} 1}^{\infty}\frac {z^n}{n} \equal{} \minus{} \log(1 \minus{} z)$ for $ |z| < 1.$ Using Abel's Theorem to evaluate the boundary values and employing a few trigonometric identities, we get that:

For $ 0 < x < 2\pi,$ $ \sum_{n \equal{} 1}^{\infty}\frac {e^{inx}}{n} \equal{} \minus{} \ln\left(2\sin\frac {x}2\right) \plus{} i\left(\frac {\pi \minus{} x}2\right).$
[/quote]

can you show this?

I dont see how Abel's theorem allows here...as far as i can see we ahve
$ \sum\frac {r^ne^{in\theta}}{n} \equal{} \minus{} \log(1 \minus{} re^{i\theta})$
by Abels theorem we know
\[ \lim_{r\rightarrow 1^ \minus{} } A_rf(\theta) \equal{} \frac {1}{2}(f(\theta^ \plus{} ) \plus{} f(\theta^ \minus{} ))
\]
.

I dont see how Abel's theorem implies that  $ \sum\frac {e^{in\theta}}{n} \equal{} \minus{} \log(1 \minus{} e^{i\theta})$,
because if we have this, then the $ \minus{} \log(1 \minus{} e^{i\theta})$ shows the discontinuity at zero.  (we might not need to even find the exact function).

(although it would be helpful to see what trignometric identities does Kentmerryfield uses to find the exact function~ maybe henry23 has a solution to this?)

There are several "Abel theorems". The one Kent is referring to is that if the series converges, then the sum equals to the limit of its Abel-Poisson sums, which can be found from other considerations (in this case, from the Taylor expansion of the analytic in the open unit disk function $ \log(1\minus{}z)$).


Let $D$ be the set of positive reals different from $1$ and let $n$ be a positive integer.
If for $f: D\rightarrow \mathbb{R}$ we have $x^n f(x)=f(x^2)$,
and if $f(x)=x^n$ for $0<x<\frac{1}{1989}$ and for $x>1989$,
then prove that $f(x)=x^n$ for all $x \in D$.

Cute ...

By induction we have $f(x^{2^n}) = f(x) * x^{(2^n-1)n}$

if $1/1989 \leq x < 1$ there exists n  ${x^{2^n} < 1/1989}$ and then $f(x^{2^n}) = x^{2^n n} * f(x) * x^{2^n-1}$ and then $f(x) = x^n$

The same way, if $1 < x < 1989$ ther existe n such as $1989 < x^n$


geometry
rectangle
perimeter
parallelogram
cyclic quadrilateral
geometry unsolved

ABCD is a cyclic quadrilateral and M is a point of the plan.
Prove that the sum of distances from the point M to the 4 sides of ABCD is constant if and only if M is inside or on the sides of ABCD and that ABCD is a rectangle.

- If M is inside or on the sides of ABCD and ABCD is a rectangle, the sum of distances from the point M to the 4 sides of ABCD is the half perimeter of ABCD so it is constant.
- If M is any point of the plan such as the sum of distances from M to the 4 sides of the quadrilateral ABCD is constant then M must be inside or on the sides of ABCD.
This situation is similar to the pb http://www.artofproblemsolving.com/Forum/viewtopic.php?t=247350
that I had given " a solution " , so ABCD is a parallelogram then a rectangle because ABCD is cyclic.


I once lived in 1 room with 5 other girls, and i locked them out of their room on purpose for a few hours while they desperately needed to get back in

which forced them to climb up through the window which was not easy to open.

That is not half as weird as what I did.

I once, at the age of six, hatched an ingenious plan and locked myself inside a room, so that I couldn't get out even if I wanted to, and somebody else had to climb up a tree, land in the backyard, forced open the hatchlock of the back door and rescue me. :rotfl:


function
real analysis
real analysis solved

Does there exist a partition of the set X=[0,1) with two classes A,B
and a continuous function from X to X such that f(x) is from B for every
x from A and f(x) is from A for every x from B.

The original problem had also the case X=[0,1[.It cant exist a function from X to X with that property because if f is continuous and defined on a compact it has a fixed point wich cant be with the conditions f(x)is in A fopr x in B etc

It's simple! We take f(x)=(x+1)/2

A=all the intervals of the form  [1-1/2^n;1-1/2^(n+1)) with n odd and B
all the intervals of the same form but with n even.


khob nataieje oLampiada ham ke Omad ..:D ma ke ghAbool nashodim :(   ishalla sale dg ...vali olampiad computer mashallla aaali boood allame tabatabayi 9 ta ghabooli dad..be hamashon tabrik migam be bachehaye riaziam tabrik miGam hamchenin baghie;)

Double posting! Please delete your post by pushing delete button.


Thank you very much

Do you can generalize this problem?


Saying that 4 < x < 9 is equivalent to saying what about x ?

A. 0 < x < 5

B. 0 < x < 65

C. 2 < x < 3

D. 4 < x < 9

E. 16 < x < 81

Surely it is D, since it cannot be any of the others!
Are you sure you are not missing a square or a root sign for example from the question?

uuhh..... Maybe this problem is incomplete????? Because saying x is obviously equivalent to saying x...

It looks like D if this is the correct question...

perhaps the question wants to know what saying:
\[4<\sqrt{x}<9\]
is equivalent to?

This is not clear but more than 1 tutor claimed D as the answer and this is good enough for me.


[quote="manjil"]If a polynomial in $ x$ always assumes integer values for all real $ x$ then prove that the co efficients of the polynomial are all rational.[/quote]

A polynomial is a continuous function. If it takes only integer values, it must be constant and equal to an integer.

And ... you are not in the good forum.


In trapezoid ABCD, line segment AB is parallel to line segment CD, AB=7 units, and CD=10 units. Segment EF is drawn parallel to line segment AB with E lying on line segment AD and F lying on line segment BC. If BF:FC=3:4, what is EF? Express your answer as a common fraction.

Thanks.

Edit: Yup, there was a typo. Thanks to Rofl etc.

Is there a typo in the statement of the problem?

I wondered because no matter how I draw it, $ BC$ is longer than $ FC$.

What does MC stand for?

MC is mathcounts ;) not a line segment

Also, I think you mean $ BF: FC = 3: 4,$ not $ BC: FC = 3: 4$ gogators :)

[hide="Solution"]
Let $ EF = x.$  When we draw $ \overline{EF},$ we divide $ ABCD$ into two smaller trapezoids. We will call the height of the top trapezoid $ h_1$ and the height of the bottom one $ h_2.$  Since we know point $ F$ divides $ BC$ into lengths in the ratio $ 3: 4.$ Using this, we have $ 3h_2 = 4h_1\Rightarrow h_2 = \frac {4h}{3}.$
The area of the big trapezoid can be expressed in two ways. The first is
\begin{align*}[ABCD] & = \frac {\left(h + \frac {4h}{3}\right)(7 + 10)}{2} \\
& = \frac {7\cdot17h}{6} \\
& = \frac {119h}{6}.\end{align*}
The second is expressed as the sum of the areas of the two smaller trapezoids.
\begin{align*}[ABCD] & = [ABEF] + [EFCD] \\
& = \frac {h(x + 7)}{2} + \frac {\frac {4h}{3}(x + 10)}{2} \\
& = \frac {h(x + 7)}{2} + \frac {4h(x + 10)}{6} \\
& = \frac {3h(x + 7) + 4h(x + 10)}{6} \\
& = \frac {h(7x + 61)}{6}.\end{align*}
Setting these two expressions equal, we can find $ x$
\begin{align*}\frac {119h}{6} & = \frac {h(7x + 61)}{6} \\
119 & = 7x + 61 \\
x & = \boxed{\frac {58}{7}.}\end{align*}
Quite a long solution, looking for a better one.[/hide]

MathCounts!

EDIT:1000th post! W00T!
and beaten!

Um, cgyao15, could you lend an explanation also?

I shall post the solution, but I understood RoFlLoLcOpT's solution better than the official one.
[hide="Official Solution"]In trapezoid ABCD, AB || CD, AB = 7 and CD = 10. EF || AB and BF:FC = 3:4. So what is EF? EF gets proportionately closer in size as it moves down the segments AD and BC. 10 – 7 = 3 which is the amount EF increases in size as it moves down AD and BC to become DC. Since BF:FC = 3:4 BF = 3/7 BC. 3/7*3=9/7.
9/7+7=9/7+49/7=58/7.[/hide]

@gogators i was answering a question, about what is MC.

I was just asking if you had an explanation.


The sum and the product of two purely periodic decimal fractions $a$ and $b$ are purely periodic decimal fractions of period length $T$. Show that the lengths of the periods of the fractions $a$ and $b$ are not greater than $T$.
[i]Note.[/i] A [i]purely periodic decimal fraction[/i] is a periodic decimal fraction without a non-periodic starting part.

It posted before.
$T=lcm(T_{1},T_{2})$, were $T_{1},T_{2}$ are periods a and b. It is obviosly $T_{1},T_{2}\le T$.

Right. I found it.
http://www.mathlinks.ro/Forum/viewtopic.php?t=103851

Anyway, how to prove that if $x=\frac{a}{b}$ purely periodic with $(b,10)=1$, then $b|10^{T}-1$, where T is period?

If the problem just give the condition about the product and not the sum, it is still enough, right?

Here is a nice proof [same as in NT concepts,...]
Consider the polynomial
$P(x)=(10^T -1)(x-a)(x-b)$
According to the problem , the coefficients are integers so by rational root theorem denominator of $a$ and $ b $ divides $10^T-1$ so their period divides $T$. $\blacksquare$

It directly follows from the fact $(R_n,R_m)=R_{[m,n]}$ where $R_n$ is nth repunit


geometry
trigonometry
AMC
AIME
parallelogram
geometric transformation
homothety

Equilateral $\triangle ABC$ is inscribed in a circle of radius 2. Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13$, and extend $\overline{AC}$ through $C$ to point $E$ so that $AE=11$. Through $D$, draw a line $l_1$ parallel to $\overline{AE}$, and through $E$, draw a line ${l}_2$ parallel to $\overline{AD}$. Let $F$ be the intersection of ${l}_1$ and ${l}_2$. Let $G$ be the point on the circle that is collinear with $A$ and $F$ and distinct from $A$. Given that the area of $\triangle CBG$ can be expressed in the form $\frac{p\sqrt{q}}{r}$, where $p$, $q$, and $r$ are positive integers, $p$ and $r$ are relatively prime, and $q$ is not divisible by the square of any prime, find $p+q+r$.

[hide]Notice that $\angle{E}=\angle{BGC}=120^\circ$ because $\angle{A}=60^\circ$. Also, $\angle{GBC}=\angle{GAC}=\angle{FAE}$ because they both correspond to arc ${GC}$. So $\Delta{GBC}$~$\Delta{EAF}$. $[EAF]=\frac12\cdot11\cdot13\cdot\sin{120^\circ}=\frac{143\sqrt3}4$. $[GBC]=\frac{BC^2}{AF^2}\cdot[EAF]=\frac{12}{11^2+13^2-2\cdot11\cdot13\cdot\cos120^\circ}\cdot\frac{143\sqrt3}4=\frac{429\sqrt3}{433}$

The answer is $\boxed{865}$[/hide]

[quote="me@home"]I was sooo stuck on this problem, that is an amazing solution in how short it is...[/quote]
I sort of was too. I didn't think of this way until afterwards. I had to use vectors/coordinates to do it

[quote="me@home"]Yes, don't you hate that when during the test you can't do the problem and then you thi;nk about it later and it has a simple solution...[/quote]
Yes I hate that, it happens a lot to me

[quote="me@home"]you qualified though, right?[/quote]
No, made 6 stupid mistakes on the AIME II and only got a 7

wow, so you worked 13 problems solidly!
I did something similar, realy did 11-ish but made 4 stupid mistakes, only 4 problems right...

[quote="Chocoboom"]O this question was horrible. Quick question, how do you know $[EAF]=\frac12\cdot11\cdot13\cdot\sin{120^o}=\frac{143\sqrt3}4$ ?[/quote]
We know $AE=11$, and since $AEFD$ is a parallelogram, $EF=AD=13$. $\angle{E}=120^\circ$ because $\angle{A}=60^\circ$ and $EF||AD$. Then I used the formula: $[ABC]=\frac12\cdot a\cdot b\cdot\sin{C}$

[quote="SamE"]I did this AIME for practice and got all of them, making stupid CALCULATION mistakes on two of the problems. This was one of them. If you want to see a really long solution that still gets to the same answer (using similar triangles, Power of a Point and homothecy on the way),
[hide]We easily calculate $AB=BC=AC=2\sqrt3$. Let $H$ lie on $AD$ such that $AH=12$ and let $I$ lie on $AE$ such that $AE=12$. By SAS, $\triangle AHI$ is equilateral. Because of all of the nice angles, it also passes through the midpoint of $ED$ at $J$, where $HJ=5.5$ and $IJ=6.5$. (Just draw a few pictures.)

Using nat mc's calculations, we get $AF=\sqrt{443}$, so $AJ=\sqrt{443}/2$. Let the intersection of $AG$ and $BC$ be $K$. Now, consider the homothecy that takes $\triangle AHI$ to $\triangle ABC$. This takes line $AG$ to itself, so it takes $J$ to $K$. Since the proportionality constant is $12/2\sqrt3=2\sqrt3$, $AK=\frac{\sqrt{433}}{4\sqrt3}$. We also use this to get $BK=11\sqrt3/12$ and $CK=13\sqrt3/12$. By Power of a Point on $K$, $KG=\frac{BK\cdot CK}AK=\frac{429/144}{\sqrt{433}/4\sqrt3}=\frac{429\sqrt{3}}{36\sqrt{433}}$. Now, dropping altitudes from $A$ and $G$ to $BC$, we find that the former of these is 3, while the latter is the height of our triangle. Similar triangles gives us $\displaystyle h=\frac{3GK}{AK}=\frac{\frac{429\sqrt{3}}{12\sqrt{433}}}{\frac{\sqrt{433}}{4\sqrt3}}=\frac{429}{433}$. Now, since the area is equal to $\frac12 bh$, this is $\frac122\sqrt3\frac{429}{433}=\frac{429\sqrt3}{433}$, and $429+3+433=865$, as desired.[/hide][/quote]
Not to be bias or anything, but umm I like my solution better  :P

Yeah, yours is 2 steps long. Mine is around 20. But they both get the same answer, don't they?

But time is of such importance on the AIME!!!  :!:  :roll:

I did something like SamE, I just took this AIME for practice at home. After about 2 hrs. I just decided to stop because the problems kept getting scarier, but I was satisfied with a 6 (3 stupid mistakes).

let $\angle BAG=x=\angle BCG$, then $\angle CAG=60-x=\angle CBG$, $\angle BGC=120=\angle ADF$
clearly, $BC=2\sqrt{3}$, law of cosines gives $AF=\sqrt{433}$, law of sines gives $\frac{\sqrt{433} }{\sin120}=2R=\frac{ 2*\sqrt{433*3} }{3}=\frac{11}{\sin x}=\frac{13}{\sin60-x}$
$[CBG]=\frac{BC^2*\sin x*\sin60-x}{2\sin 120}=\frac{4\sqrt{3}*11*13*9}{4*3*433}=\frac{429\sqrt{3}}{433}$

now that i look at it, i just used that they were similar triangles without noticing and using trig  :P

What I did, and what was surprisingly easy, was using coordinates.  Make A the origin, and drop the altitude from A, and make that the x-axis.  Then all you have to do is find the x-coordinate of G, subtract 3 (the x-coordinate of BC), and that's your height of the triangle.  The base is easy to find.  The x-coordinate of G is easy.  Just find the coordinates of the opposite point of the parallelogram which is surprisingly not ugly at all, then find solutions with $(x - 2)^2 + y^2 = 4$.

[hide="Motivation"]
Wow it took me about 10 minutes of chasing to figure out that CGB ~ FEA and GA bisects CGB. From there its just law of cosines and then the sine formula for area.
[/hide]

[hide]
 Let $AG \cap BC = F$. Now, observe that $AF$ passes through the midpoint $M$ of $DE$. Through the ratio lemma, we obtain that \[ \frac{\sin{\angle{GAD}}}{\sin{\angle{EAG}}} = \frac{\sin{\angle{MAD}}}{\sin{\angle{EAM}}} = \frac{13}{11} \]. This gives that $\frac{BF}{FC} = \frac{11}{13}$. Using this and the fact that $BC = 2\sqrt{3}$, $BF = \frac{11\sqrt{3}}{12}$ and $CF = \frac{13\sqrt{3}}{12}$. Pythagoras gives that \[ AF = \frac{\sqrt{1299}}{12} \], and PoP gives \[  FG= \frac{143\sqrt{3}}{12\sqrt{433}} \] Now its just an area ratio: \[ \frac{[ABC]}{[BGC]} = \frac{AF}{FG} = \frac{433}{143} \]. The answer is now $3\sqrt{3}$ divided by this, or $\frac{429\sqrt{3}}{433}$ [/hide]

This one took way too long :noo:
[hide=Solution]Note that $AB=2\sqrt3$. The condition boils down to $\frac{[ABG]}{[ACG]}=\frac{11}{13}$. Since $\angle AGB=\angle AGC=60^{\circ}$, from the sine area formula we have $\frac{BG}{CG}=\frac{11}{13}$. Let $BG=11k$ and $CG=13k$.
By LoC on $\triangle BGC$, we have
$$12=k^2(11^2+11\cdot13+13^2)=433k^2\implies k^2=\frac{12}{433}$$
Thus, $[CBG]=\frac12 (11k)(13k)\sin 120^{\circ} = \frac{\sqrt3}{4}\cdot 143\cdot \frac{12}{433}=\frac{429\sqrt3}{433}\implies\boxed{865}$.

[hide=sketch(to be fully written soon)]Establish the side length of the triangle as $2\sqrt3$. Ascertain $\frac{AD}{AE} = \frac{CG}{BG}$, then since $\angle CGB = 180-60 = 120^\circ$, we can use LOC on the triangle $CGB$ to finish.[/hide]

Bruh sob this took too long for a 2006 12

It's easy to see $BC=2\sqrt{3}$ by equilateral triangle circumradius properties and LoC in $\triangle ADF$ gives $AF=\sqrt{433}$ since by construction, $ADFE$ is a parallelogram so $\angle D=180^\circ - \angle A = 120^\circ .$ Angle chasing in cyclic quad $ABGC$ and LoS, $$\sin BCG = \sin DAF = 11 \cdot \frac{\sqrt{3}}{2\sqrt{433}}$$ $$\sin GBC = \sin FAE = 13 \cdot \frac{\sqrt{3}}{2\sqrt{433}}.$$ By LoS in $\triangle BGC,$ $$BG = \sin BCG \cdot \frac{2BC}{\sqrt{3}}=4\sin BCG=44 \cdot \frac{\sqrt{3}}{2\sqrt{433}}.$$ Now, we compute the area of $\triangle BCG$ as $$\frac{1}{2} \cdot BG \cdot BC \cdot \sin GBC=\frac{1}{2} \cdot 44 \cdot \frac{\sqrt{3}}{2\sqrt{433}} \cdot 2\sqrt{3} \cdot 13 \cdot \frac{\sqrt{3}}{2\sqrt{433}}$$ which, believe it or not, simplifies to the answer.

[color=#00f]Part 1: Restatement and ratio[/color]

Rephrase the problem to get rid of $F$:\\

Let $ADE$ be a triangle such that $AD=11,AE=13,\angle DAE=60$. Let $M$ be the midpoint of $DE.$ Let $B$ and $C$ be points on $AD$ and $AE$ so that $AB=AC=2\sqrt{3}$, and let $AM$ meet $(ABC)$ at $G$.\\

Suppose that $AG$ and $BC$ intersect at $P$.

Since $DM/EM=1$, by Ratio Lemma, $$\frac{BP}{CP}=\frac{13}{11}.$$ Now, we turn our focus to $\triangle ABC.$\\

[color=#00f]Part 2: Bashing and finish[/color]

Let $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Note that $P=(0:11:13),$ so by cevians, $G=(t:11:13)$. However, since $G$ lies on $(ABC)$, we have $$11t+13t+143=0\rightarrow t=-\frac{143}{24},$$ so $G=(-143/24:11:13).$ This means that $$\frac{[GBC]}{[ABC]}=\frac{143/24}{\frac{143}{24}+11+13}=\frac{143}{433},$$ so our answer is $$\frac{143}{433}\cdot 3\sqrt{3}=\frac{429\sqrt{3}}{433}\rightarrow 865.$$

[quote=asdf334]um what the similar triangle literally jumps out at you because of how painfully similar the skinny triangles look

Angle chasing gives that $\triangle CBG\sim \triangle ADF$. Now we are essentially done; we have
\[AF^2=13^2+11^2-2\cdot 13\cdot 11\cdot \cos{120^{\circ}}=433\]
and since $BC^2=12$ the answer is just
\[\frac{12}{433}\cdot \frac{1}{2}\cdot \sin{120^{\circ}}\cdot 13\cdot 11=\frac{429\sqrt{3}}{433}\implies \boxed{865}.\][/quote]

wait this is smart

but what is the motivation for keeping point $F$, the first thing I did after seeing the problem is getting rid of $F$ using the midpoint since it seems like a very "unnatural" point idk

although it seems like point $F$ does actually help


Un problema muy interesante:

Hallar todos los primos "p" tales que:

$p+1=2x^{2}$
$p^{2}+1=2y^{2}$

con $x, y$ enteros.

[color=blue]Haciendo unos rápidos cálculos he llegado a que [b][i]p[/i]|[i]y[/i]+[i]x[/i][/b] o [b][i]p[/i]|[i]y[/i]-[i]x[/i][/b]. Voy a esforzarme para obtener observaciones más útiles.[/color]

Restando obtenemos:
p(p-1) = 2(y+x)(y-x)
y asi, p|(y-x), 2(y+x)|(p-1)  o  p|(y+x), 2(y-x)|(p-1).
En el primer caso, p <= y-x < 2(y+x) <= p-1.
En el segundo caso, 2(y-x)+1 <= p <= y+x, o sea  y+1 <= 3x.
Ademas, (2x²)² = (p+1)² < 2(p²+1) = 2(2y²), y luego  x² < y.
Entonces,  (y+1)² <= (3x)² <= 9(y-1) y, en particular,  y <= 5.
p=7.

[quote]Entonces, $(y+1)^{2}\leq  (3x)^{2}\leq  9(y-1)$[/quote]
Como aseguras eso?? No sera: $(y+1)^{2}\leq  (3x)^{2}\leq  9(y)$?

Como son enteros:    x² < y     ~     x² <=  y-1


geometry
geometric transformation
reflection

nice game http://laser.narr.as/
solutions http://www.raymer.org/games/laser/solutions/index.php
I'm lvl 18 :)

my math teacher used to make us play the first game when we're done with our work.  :D  i was the first one to finish all of the levels

I like the game. The first ten-fifteen levels are really easy, but then it gets quite challenging and interesting. However the game sometimes starts to glitch and sometimes the objects get left somewhere off the board and they cannot be reached so you have to restart the level.
I am on level 21 now.


algebra
function
domain
LaTeX
real analysis
real analysis unsolved

Consider
Code:
T =-d^{2}/dx^{2}
an operator on
Code:
L^{2}(R)
with domain in
Code:
C_{0}^{\infinity}
. What is the adjoint of T? and is it essentially selfadjoint? what happens if we change the domain in
Code:
L^{2}(R)
?

Please , if you know the answer, can you explain to me in detail. At least first part.

thanks a lot!
mikiduta

Haven't you made a thread about the exact same thing [url=http://www.mathlinks.ro/viewtopic.php?t=264492]here[/url]?

Also, $ \text{\LaTeX}$ please!
Consider the operator $ T = - \frac {d^2}{dx^2}$ on $ L^2(\mathbb R)$ with domain in $ C_0^\infty$.
What is the adjoint of $ T$? Is $ T$ essentially self-adjoint? What happens if we change the domain to $ L^2(\mathbb R)$?


(PVT) Let $ a,b,c,d$ be nonnegative real numbers, no two of which are zero. Prove that
$ \sum \frac{a^2(b\plus{}c\plus{}d)}{(a\plus{}b)(a\plus{}c)(a\plus{}d)} \le \frac{3}{2}.$
:)

Dear Can_hang...I expanded it that finally got

$ ab(c\plus{}d)^2(a\minus{}b)^2\plus{}ac(b\plus{}d)^2(a\minus{}c)^2\plus{}ad(b\plus{}c)^2(a\minus{}d)^2\plus{}$
$ bc(a\plus{}d)^2(b\minus{}c)^2\plus{}bd(a\plus{}c)^2(b\minus{}d)^2\plus{}cd(a\plus{}b)^2(c\minus{}d)^2\geq0$

but I don't think it a good solution...I'm looking forward to your solution. :)


hm i can't find out what i'm doing wrong. Let's say we had a triangle $ ABC$ where $ \angle ABC \equal{} \theta$. So $ \vec{AB}\plus{}\vec{BC}\equal{}\vec{AC}$. Then
\[ |\vec{AB}\plus{}\vec{BC}|^2\equal{}|\vec{AC}|^2\]
\[ |\vec{AB}|^2\plus{}|\vec{BC}|^2\plus{}2\vec{AB}\cdot\vec{BC}\equal{}|\vec{AC}|^2\]
\[ AB^2\plus{}BC^2\plus{}2(AB)(BC)\cdot\cos(\theta)\equal{}AC^2\]
where did i mess up?  :huh:

If $ \angle ABC \equal{} \theta$ then the angle between $ \vec{AB}$ and $ \vec{BC}$ is actually $ 180 \minus{} \theta$ (vectors have direction).

I attempt to explain something similar [url=http://www.mathlinks.ro/viewtopic.php?search_id=1223026205&t=158064]here[/url].
After rereading it, though, it is a bit convoluted :oops:. But the reason is exactly with t0rajir0u stated, the direction of the vectors matter.
If someone could provide a more concise explaination, then please do.  :)


$ 3\plus{}\sqrt{3\plus{}\sqrt{3\plus{}\sqrt{3\plus{}...}}}\equal{}$

I absolutely don't know how to solve this. I know there is most likely a genius way to solve this.

[hide="hint"]
crudely speaking $ \infty$ is quite rich u can't take anthing away from it  :D
so set $ x \equal{} 3\plus{}\sqrt{3\plus{}\sqrt{3\plus{}\sqrt{3\plus{}...}}}$ assuming it converges.
so $ (x\minus{}3) \equal{}\sqrt{3\plus{}\sqrt{3\plus{}\sqrt{3\plus{}...}}}$
this gives $ (x\minus{}3)^{2}\equal{} 3\plus{}\sqrt{3\plus{}\sqrt{3\plus{}\sqrt{3\plus{}...}}}\equal{} x$
now solve the quadratic $ (x\minus{}3)^{2}\equal{} x$[/hide]

Okay, so when I solve it I get $ \frac{7\pm\sqrt{13}}{2}$.

I don't think $ 7\minus{}\sqrt{13}$ over $ 2$ works, since there is already a 3 in the equation to stat with, and that would give an answer less than 3, which isn't possible.

So I guess the solution $ \equal{}\frac{7\plus{}\sqrt{13}}{2}$?

Haha interesting way to solve it, in that hint you showed. I really didn't think of that.

And we should prove 1st it's converges. So for example:
$ x_{1}\equal{}3$ and $ x_{n\plus{}1}\equal{}3\plus{}\sqrt{x_{n}}$ and only we should show that sequence $ \{x_{n}\}$ is increasing and bounded, hence it's converge. And now we can find the limit.


geometry
trigonometry
area of a triangle
Heron\u0027s formula
trig identities
Law of Cosines

There is a classic problem in which one is given the distances from a point inside an equilateral triangle to its vertices, and asked to find the side of the triangle. If anyone can recall the full canonical solution, I would be much obliged, thanks.

Look at the geometry-college thread http://mathforum.org/epigone/geometry-puzzles/fongpholjan , especially at the posting of "Anonymous" (= vinoth_90_2004).

  Darij

One classic way to do it :

Let ABC be the equilateral triangle  and point P (say inside of the trianble).
(Let PA=p , PB=q, PC= r  are given)
Take a point D outside ( say otherside of line BC) such that  angle DCP = 60 and DC=PC=r .( Now the  triangle DBC ==  Triangle PAC) so  DB= p

Then the triangle  PBD has sides p,q,r and thus you can find the angle BPD, now angle BPC= (60+ angle BPD) . now just solve the traingle  BPD (You know the two sides  q and r and  the angle between them).

P.S. (Sorry Did not check the above post)    this is the same method as mentioned  in the above post..

I was in class yesterday and Long Cheng posted a hardcore solution to this problem using areas. (I did it on our contest with the law of cosines as per above.) You just draw a-b-s, a-c-s, b-c-s triangles on the sides in a certain way, resulting in a figure composed of 3 a-b-c triangles, whose areas you calculate with Heron's formula; and 3 equilateral triangles of sides a, b, and c. This figure has twice the area of the big triangle, which is s2 :sqrt: 3/4.

Can you post wiht a little more details. Thanks.

(Added later: Ok I see it. This gives the same result as the above
 or   (side)^2 = (p^2+q^2+r^2)/2 + (2* :sqrt: 3)(area of triangel formed by p,q,r)


Anyone can help me out by explaining the probability of an equiangular triangle if each angle is a multiple of 12. I remember the answer is like 1/19, but I couldn't figure out how to get that answer.

The angles are 12x, 12y and 12z, where x+y+z=15. So we need to find the number of ordered pairs (x,y,z).
First, there is only one case where x,y,z are the same (5,5,5). Also, if two are the same and one is different, we can have from (1,1,13) to (7,7,1), which would be 7, but we already counted (5,5,5) so it's 6. Now, for the rest, think of having 15 objects placed in a row and you have two dividers you can place in the 14 spaces between them, to split it up into 3 numbers. But note that if all three numbers are different, then they can be permuted 6 ways, if two are the same, they can be permuted three ways, and if all are the same (5,5,5) there is only one permutation. So the answer is $ 1 \plus{} 6 \plus{} \frac {{14 \choose 2} \minus{} 1 \minus{} 3 \cdot 6}{6} \equal{} 19$ permutations, of which (5,5,5) is only one.


abstract algebra
group theory
induction
number theory
relatively prime
superior algebra
superior algebra unsolved

Show that a finite abelian group that is not cyclic contains a subgroup which is isomorphic to $ \mathbb{Z}_{p}\oplus\mathbb{Z}_{p}$ for some prime p.

Do you know how to show that abelian groups are a direct product of their Sylow subgroups?  From there it suffices to solve the problem for $ p$ groups, which is not hard.

It's good to provide some background on what you do or don't know for questions like this.  Otherwise you will tend to get a lot of responses like "It is obvious from the structure theorem". :)

I am just starting to read about finite abelian group. I do not know how to prove that abelian groups are a direct product of their Sylow subgroups. It's not easy to see that as clearly as you said.

[hide]
A group is called an elementary abelian $ p$-group (given a prime $ p$) if it's abelian and $ x^p\equal{}1$ for every element $ x$ in the group. Now every noncyclic elementary abelian $ p$-group contains a subgroup isomorphic to $ \mathbb{Z}_p^2$: just take $ a\neq 1$ and $ b\neq <a>$ and use the subgroup generated by $ a,b$.

Now suppose $ G$ is a finite abelian group which does not contain a subgroup isomorphic to $ \mathbb{Z}_p^2$. We show $ G$ cyclic by induction on the order; for the trivial group, it's obvious.

For the inductive step, any nontrivial $ G$ has an element of order $ m>1$. If $ p$ is a prime divisor of $ m$, then $ G$ has an element of order $ p$. Now let $ \varphi$ be the $ p$th-power map (i.e. $ \varphi (x)\equal{}x^p$). As $ G$ abelian, $ \varphi$ is a homomorphism. Denote its kernel by $ K$ and its image by $ H$.

$ K$ is the elementary abelian $ p$-group consisting of the identity and all elements of order $ p$, and $ H\simeq G/K$ so $ |H|\equal{}\frac{|G|}{|K|}<|G|$ as $ G$ contains an element of order $ p$. By induction, $ H$ is cyclic (certainly all subgroups of $ H$ are subgroups of $ G$ and $ H$ is abelian), so we can write $ H\equal{}<h^p>$ for some $ h\in G$. Also $ K$ is cyclic as it's elementary abelian and also cant contain a subgroup isomorphic to $ \mathbb{Z}_p^2$, so we can write $ K\equal{}<k>$ for some element $ k$ of order $ p$.

For any $ g\in G$, we can find $ a$ such that $ g^p\equal{}h^{pa}$ so $ gh^{\minus{}a}\in K\Rightarrow g\equal{}h^ak^b$ for some $ b$. As a result, $ G$ is generated by $ h$ and $ k$. If $ k$ is generated by $ h$, then $ G\equal{}<h>$ and we're done.

Otherwise, $ <h>$ and $ <k>$ intersect in a proper subgroup of $ <k>$, therefore trivially. So $ G\equal{}<h>\times <k>$. $ h$ can't have order divisible by $ p$, for if its order were $ pr$, then $ <h^r>\times <k>$ would be isomorphic to $ \mathbb{Z}_p^2$. So the orders of $ h$ and $ k$ are relatively prime, implying $ G$ is generated by $ hk$.
[/hide]


calculus
derivative
function
trigonometry
LaTeX
calculus computations

The function f is called the hyperbolic sine function,and g is called the hyperbolic cosine. Until the last step, it was possible that we were dealing with functions which don't really exist. The formula for g^-1 defines a function, though, and we could work backwards from it to get properties of f and g. Find formulas for f and g as follows:

 1. Show that the function h= f + g is its own derivitive, and that h(0)= 1.

 2. Show that h(x)/e^x is constant and evaluate the constant to see that its 1.
  Hence h(x)=e^x. Similarly show that f - g is the function -e^-x (or use the
  relation above).
 3. Solve this system of two equations in two unknowns.

 I'm sorry if you can't make out the different variables in this problem since I
 don't know how to use the Latex program. Any help provided would be appre-
 ciated. If somone could explain what they did in their problem it would be even
 a bigger assistance.

You forgot to include the definition of $f$ and $g$.

I'm sorry I forgot to mention a big component of this problem:

Suppose f is a function which has the property that [i]f[/i][b]"[/b]= [i]f[/i]. Suppose also that [i]f[/i](0) and [b]f[/b]'(0)=1. Let g=[b]f[/b]'(the derivitive of [b]f[/b])[/b]

This is really not that hard.
1. $h'=f'+g'=f'+f''=f'+f=h$
2. $(e^{-x}h(x))'=\dots=0$ by the product rule. $(e^{x}(f(x)-g(x)))'=\dots=0$ also by the product rule. And so on.

Thank you for your reply mlok but I think you didn't completely answer the question. The question askes not just to show these function but also explain how you came about with the answers that you did. Just writing h' equals h isn't
enough. Also could you explain the last part 'Solve this system of two equations and two unknowns.' Thank you in advance.

Can mlok or anyone else that replies on these forums explain:

From Part 1 how h(0)=1

From Part 2 how h(x)/e^x is a constant and evaluate the constant to see that it is one.
From Part 3 show the system of two equations and two unknowns.

When I write 'show' I mean that I'd Like to see all the step that lead you to the answer of the problem. Just giving me an answer doen't help me.

We know:
$f=f''$,
$g=f' \Rightarrow g'=f''$, and
$f(0)=0$ and $f'(0)=1$, I think you were wrong when you said they both were one.  That doesn't fit the description of the hyperbolic $\sin$ function.
$f(x)=\text{h}\sin{x}=\frac{e^{x}-e^{-x}}{2}$
$g(x)=\text{h}\cos{x}=\frac{e^{x}+e^{-x}}{2}$

$1.$ Show that the function $h= f+g$ is its own derivative, and that $h(0)= 1$.
[hide]$h=f+g$
$\&h'=f'+g'$; Differentiate
$h'=f'+f''$; Substitute for $g'$
$h'=f'+f$; Substitute for $f''$
$h'=g+f=f+g$; Substitute for $f'$
$\boxed{h'=h}$
$h(x)=f(x)+g(x)$

$h(x)=f(x)+f'(x)$; Substitute for $g=f'$
$h(0)=f(0)+f'(0)=0+1$; By what was given [$f(0)=0$ and $f'(0)=1$]
$\boxed{h(0)=1}$[/hide]

$2.$Show that $\frac{h(x)}{e^{x}}$ is constant and evaluate the constant to see that its $1$. Hence $h(x)=e^{x}$. Similarly show that $f-g$ is the function $-e^{-x}$ (or use the relation above).
[hide]$\text{Let }y=\frac{h}{e^{x}}$
$y'=\frac{h' \cdot e^{x}-h \cdot e^{x}}{e^{2x}}$
$y'=\frac{h' \cdot e^{x}-h \cdot e^{x}}{e^{2x}}$; Substitute $h'=h$
$y'=\frac{h \cdot e^{x}-h \cdot e^{x}}{e^{2x}}$
$y'=0$

Since $\frac{d}{dx}(\frac{h(x)}{e^{x}})=0$, we can assume that before we differentiated, $y$ was a constant or $0$.  Because $y(0)=\frac{h(0)}{e^{0}}=\frac{1}{1}=1$, we can assume that $y$ was not $0$  $\forall$  $x$.  Not to imply the function can never be $0$, but this obviously does not hold true $\forall$ cases (but if you evaluate it using $f$ and $g$ as given you see that it is obviously never $0$).

$y=\frac{h(x)}{e^{x}}$
$y=\frac{f(x)+g(x)}{e^{x}}$
$y=\frac{\frac{e^{x}-e^{-x}}{2}+\frac{e^{x}+e^{-x}}{2}}{e^{x}}$
$y=\frac{\frac{2e^{x}}{2}}{e^{x}}$
$y=1$
$\frac{h(x)}{e^{x}}=1 \iff \boxed{h(x)=e^{x}}$

$f(x)-g(x)=\frac{e^{x}-e^{-x}}{2}-\frac{e^{x}+e^{-x}}{2}=\frac{-2e^{-x}}{2}=\boxed{-e^{-x}}$[/hide]

I tired to prove as much as I could and lay the steps out as much as possible.. I'm not sure what part $3$ was asking for.


1) Let $x,y,z\geq0$ such that $x^2+y^2+z^2=xyz$, prove that
\[ (x-1)(y-1)(z-1)\geq8.  \]
2) Let $x,y,z\geq0$ such that $x^2+y^2+z^2=xyz$, prove or disprove that
\[ x+y+z+xyz\geq\frac43(xy+yz+zx).  \]

part 1): from am-gm we have that
$xyz=x^2+y^2+z^2\geq3\sqrt[3]{(xyz)^2} \\\Rightarrow \dfrac{x+y+z}{3}\geq\sqrt[3]{xyz}\geq3 \\\Rightarrow x+y+z\geq9$
it is easy to see that the problem is equivalent to $(x^2+y^2+z^2-xy-xz-yz)+(x+y+z-9)\geq0$,
but this is trivial since every term in the $LHS$ is nonnegative.

and $x+y+z+xyz \ge xyz=x^2+y^2+z^2\ge xy+yz+zx \ge \dfrac{2}{3}(xy+yz+zx).$


In a group of nine people, show that there is a group of four people who know every other people in the group or there is a group of three people who don't know any other in the group.

[hide]
This is the same problem as coloring K9 red and blue, and either getting a red K4 or a blue K3, so I will use that instead.

Suppose there is such a coloring without any red K4s or blue K3s

Look at any vertex, there are 8 edges coming out of it. If there are four blues coming out from it, then look at those vertices that get the blues. If there are any blue edges we have K3 with the two they connect and the one original vertex. But if they are all red that makes a red K4.  So there must be less than 4 blues coming from any one vertex.

Ok now suppose there are at least 6 reds coming out from the vertex. In those 6 we know that there must be a red or blue K3 (this is pretty well known) so if its blue were done, and if its red then those combined with the original vertex make a red K4. So there must be less than 6 reds. Meaning that at any vertex there is less than 4 but greater than 2 blue edges, so thats three blue edges at any given vertex.

each edge connects two vertices, so thats (9x3)/2 blue edges, but you can't have half an edge so thats a contradiction.[/hide]


For what values of $ a$ is

$ 4 \cdot a^{2} \plus{} 1$ not a prime.

[hide="Solution"]We can use the identity of Sophie Germain.

$ x^4 \plus{} 4y^4 \equiv (x^2 \plus{} 2y^2 \plus{} 2xy)(x^2 \plus{} 2y^2 \minus{} 2xy)$

The given expression is of the form $ x^4 \plus{} 4y^4$, and so can be factored, when $ a$ is a perfect square.

However, when $ a \equal{} 1$, one of the factors turns out to be 1, so the number is in fact prime.

Thus $ 4a^2 \plus{} 1$ can be factored and therefore is not a prime if $ a$ is a perfect square not equal to 1.

[b]EDIT: Proof that this is the only time that this happens[/b]
A sum of 2 squares $ x^2 \plus{} y^2$ can be factored only if $ 2xy$ is also a square.

The given expression is equivalent to $ (2a)^2 \plus{} 1^2$ and so can be factored only if $ 4a \equal{} 2^2 \cdot a$ is a square.

This only occurs when $ a$ itself is a square.[/hide]

That's wrong. For example, take $ a\equal{}6$.

Oh. I don't know then...

I know that if $ a$ is a square then the expression is not prime, but this is obviously only a one way implication. It is easy to see that if $ a^2$ ends in a $ 1$ or a $ 6$ then it would not be prime since $ 4a^2 \plus{} 1$ would end in a $ 5$. However I am not sure of other cases now.

Well if a>1 then all $ a \equiv 1 \pmod 5$ results in a composite number.

It's not known for [b]any[/b] irreducible polynomial of degree greater than $ 1$ whether it takes infinitely many prime values or not, so this question is open.

Well so far we have that the expression is non-prime if either one or both of the following is true:

$ a \equiv 1$ or $ 4 \bmod {5}$
$ a$ is a perfect square

However, these are not the only cases when the expression is not prime (Example: $ a\equal{}30$.)

So are there any further conditions we can specify that would make the expression non-prime if $ a$ satisfies these conditions?

Plenty.  If $ P(x)$ is any irreducible polynomial and $ a$ is an integer such that $ P(a) \neq 0, 1, \minus{}1$, then $ P(kP(a) \plus{} a)$ is composite for sufficiently large $ k$.

sorry if I am stupid its just this right?
If $ f(x)$ is a polynomial with integer coefficients then for all integers $ a$ $ f( a \plus{} f(a))$ is a multiple of $ f(a)$ ,is that large $ k$ essential ..

We can plug in some values for $ a$ and try to form a pattern  and proving it.
Maybe this is useful.
Like I plugged values for $ a$ till 10
The values for $ f(a) \equal{} 4a^{2} \plus{} 1$ are $ 5,17,37,65,101,145,197,257,325,401$ for $ a \equal{} 1,2,3,4,5,6,7,8,9,10$ respectively
Except for the condition $ a$ is congruent to $ 4$or $ 1$ $ mod5$ and for $ a$ to be a perfect square ,
the remaining values are primes.
So it is likely that $ f(a)$ is prime for all values for $ a$ except for those where it is a perfect square,1 mod 5, 4 mod 5.
So we have to prove this conjecture .(assuming it is true ) :starwars:

Unrelatedly Nice question Harekrishna :clap2:

[quote="srinath.r"]If $ f(x)$ is a polynomial with integer coefficients then for all integers $ a$ $ f( a \plus{} f(a))$ is a multiple of $ f(a)$ ,is that large $ k$ essential ..[/quote]
Yes it is.  It can happen that $ f(a \plus{} f(a)) \equal{} f(a)$ is prime.

[quote="t0rajir0u"]Plenty.  If $ P(x)$ is any irreducible polynomial and $ a$ is an integer such that $ P(a) \neq 0, 1, \minus{} 1$, then $ P(kP(a) \plus{} a)$ is composite for sufficiently large $ k$.[/quote]
Can you please explain on that large $ k$ by giving an example .?


AMC
AIME
USA(J)MO
USAMO
probability
geometry
3D geometry

Of the national competitors that took the AIME, what were your scores? Post AIME score and rank at nationals.

AIME I:7
AIME II(unofficial which stinks cause I would have made it to USAMO):10

Rank? dunno

I didn't make it nats though

[quote="sirorange"]AIME:8

Rank at nats:14[/quote]That must suck being 14th

Hey, my laser was defunctional, besides, sterling whooped you!

yeah, yeah, whatever, anyways. . . .

Who absolutely adored that mongrel probability problem on the AIME I with the cube? I know I did.

I loved that one.  It was cool, not too hard if you think and I got it right!

[quote="sirorange"]Yeah, it does, considering I was sick that day, and spilled water on myself during the competition, and am considerably good at countdown.[/quote]

What does spilling water have to do with the competition?

I got a 7 too one the  AIME I, which ones did you get?
did you take the AIME II at all?

I got 5th in Indiana

[quote="nat mc"]YEah, AMC and AIME are nothing like MC. Usually, no one practices, and its not a "sport" like Mathcounts is.[/quote]

Usually no one practices?

[quote="nat mc"]YEah, AMC and AIME are nothing like MC. Usually, no one practices, and its not a "sport" like Mathcounts is.[/quote]

Are you ****ing kidding me?

NOONE PRACTICES? WHAT!?  :o I'm practicing, and I haven't even taken it yet. I don't know if my high school even has it, but I can talk to the math team person around here.

Well, I'm practicing NOW! but not then. . . .

Well, I'm sure I'll practice some before next year, but I didn't practice before my first time taking it... I probably should have, maybe I would have stood a chance of making it to USAMO.

[quote]AIME I: 7 :([/quote]

Please don't act like that. A 7 on the AIME, especially in 8th grade, is extremely good. I'm in 9th, and I got a 3 (although I expected more like a 5 or 6).

[quote="JesusFreak197"]
[quote]AIME I: 7 :([/quote]

Please don't act like that. A 7 on the AIME, especially in 8th grade, is extremely good. I'm in 9th, and I got a 3 (although I expected more like a 5 or 6).[/quote]

See complaining about your high scores thread in AMC forum :P.  But he has a right to be disappointed if he expected better.

[quote="JesusFreak197"]I'm disappointed as well, but I would say that he has less right to be disappointed about a 7 than I have to be disappointed about a 3, especially when he's a year younger. He should instead be happy that he did better than most of the other 8th graders who took the AIME.[/quote]

Well, I think he can be disappointed if he wants, but that disappointment is best converted into more practice instead of telling people he's disappointed.  But he hasn't really complained much so give him a break ;).  Anyhow, it's of no use to sulk.  Go to kalva and.. PRACTICE! :D

Yup, I mean, a three still IS better than average. . . .But you still definetly have the right to be disappointed, but for not practicing a three is impressive.

Is a 2 above average  :?  :blush:? Lol, I got that on a practice competition. (I made a few careless mistakes.)

Well depends which year you practiced on. . .

*Scratches beard* I remember 1983, when dinosaurs ruled the earth and we carved our answers into stone tablets. . .Seriously though, anyone know

The old AIMEs (early 1980s) are usually easier than the AIMEs today.  Since the average score is usually around 2 now, a 2 on 1983 would probably be a little below average if it was today.


geometry
IMO Shortlist
projective geometry
geometry proposed

Let convex quadrilateral $ABCD$ and $X$ is point insider such that : $ADX$ and $BCX$ are similar. The circle $(ADX)$ meet $(BCX)$ at $X,W$. Circle $(ABW)$ meet $(DCW)$ at $W,Y$. Prove that: $YA=YB$.

I need all geometry from $MO,TST$ of $RUSSIAN (2003)$ and $IMO$ Shortlist (2004).can you help me ?

Apply an inversion of pole $W$. In general, let the image of point $P$ be called $P'$.

It's easy to see (simple angle chases) that $\angle X'WD'=\angle X'WC',\ \angle D'WA'=\angle C'WB'$. We also have $Y'=A'B'\cap C'D'$. Now I'll use a bit of projective geometry:

Let's assume we keep everything fixed except for the points $C'D'$, which move s.t. $\angle X'WD'=\angle X'WC'$ at all times. The mapping $C'\mapsto D'$ is projective (from the line $X'B'$ to the line $X'A'$), and the intersection of the lines $X'B',X'A'$, which is $X'$, maps to itself. This means that the mapping is, in fact, a perspectivity, i.e. the line $C'D'$ passes through a fixed point as $C',D'$ move along in the described manner. Since $C'D'$ can be $\perp WX'$, it means that this point lies on the perpendicular at $WX'$ in $W$ (call this line $\ell$). We can put $C'=WA'\cap XB',D'=WB'\cap XA'$, and, in this case, $C'D'\cap A'B'$ lies on $\ell$, so $C'D',A'B',\ell$ are always concurrent. However, $Y'=A'B'\cap C'D'$, so we have found $WY'\perp WX'\iff WY\perp WX$, and from here the onclusion follows.

The problem derive from the application of the following lemma:

lemmaRL160205.

Let e(E) anf f(F) be two circles with centers E and F intersecting at X and W. Let P on e(E) and Q on f(F) be two points s.t. <PWX = <QWX. If L and M are the antipodes of X wrt e(E) and f(F) then c(PWQ) cuts LM at its midpoint Y and YP = YQ

Proof.

It is a well known and easy to prove fact that L,W and M are collinear points. Let Y=c(PWQ)^LM.
If G is the center of c(PWQ), as GE_|_PW, GF_|_QW and EF_|_XW, then <GFE = <XWQ = <XWP =<GEF, from which GE = GF. This means that EY and WF are simmetric wrt the axis of EF (or YW, since the two segment are parallel), so EY=FM and EY//FM. Then LM=2EF=2YM. As YG and WX intersect each other on c(PWQ) then YP=YQ.

Problem
Apply two times the above lemma one with {P,Q} ={A,B} and the other with {P,Q} ={C,D}  and you have the thesis.


Let $n, d, k$ be positive integers such that $d < n < 4d$. For what values of $k$ does the equation $4nd-n^{2}= k$ have a unique solution?

$-n^{2}+4nd-k=0 \\ n_{1,2}=\frac{-4d+-\sqrt{16d^{2}-4k}}{-2}=d-+\sqrt{4d^{2}-k}$
There shall be just one solution, so $4d^{2}-k=0 \Leftrightarrow d=\frac{\sqrt{k}}{2}$.
Therefore, $k$ has to be an even square number. The solutions for $n, d$ are:
$d=\frac{\sqrt{k}}{2}\\ n=\sqrt{k}$

[quote="kathi;)"]There shall be just one solution, so $4d^{2}-k=0$ [/quote]

Could you tell me why $4d^{2}-k=0$ if there is only one solution?
Isn't it possible that $4d^{2}-k=a^{2}$ for some fixed positive integer $a$? How can we say that a unique solution for $n$ guarantees a unique solution for $k$?

PS: That should be $2d$ in the second line...

[hide="i hope this is right"]
true, in the second line there should be

$n_{1,2}=2d+-\sqrt{4d^{2}-k}$

and $k\leq 4d^{2}$
now we have two cases

first:

$d<2d+\sqrt{4d^{2}-k}<4d\\-d<0<\sqrt{4d^{2}-k}<2d\\ 4d^{2}-k<4d^{2}\\ k\in N$
this is obviously a truth

second:

$d<2d-\sqrt{4d^{2}-k}<4d\\-d<-\sqrt{4d^{2}-k}<2d$

a)$\\ \sqrt{4d^{2}-k}>-2d$
and this is obviously a truth

b)$\\ \sqrt{4d^{2}-k}<d\\ k>3d^{2}$

from all this we have: $k \in <3d^{2},4d^{2}]$. Since $n\in N$, there is just one solution $k=4d^{2}$
[/hide]


logarithms
number theory
number theory unsolved

Given [tex]n\in N^*[/tex]
Put [tex]\sum\limit_{k=1}^n\frac{2^k}{k}=\frac{p_n}{q_n} ;(p_n;q_n)=1[/tex]
Let f(n) be the hightest degree k of 2 for wich [tex]2^k|p_n[/tex]
Prove that [tex]f(n)\ge \frac{n-3}{2}[/tex]

If i don't make mistake with my calculation then we have $f(14)=5$,and then ,because $n-\frac {\ln n}{\ln 2}\geq 5\forall n\geq 14$we have $f(n)=5\forall n\geq 14$.

nalpaction: I think that $f(14)=12$  :?
Hong Quy: what is the source of this problem???

I think it is related to a difficult problem from Kvant. I remember Peter Scholze presented a very nice approach and probably that one could shed some light into this issue. But I do not find anymore his posts.  :(

Found it!!! It is a little weaker result, I think, but it's already 99%! The post is called More Kvant number theory and the author is Darij Grinberg. You can find it in the proposed problems number theory section. Sorry, I don't know to create links.  :blush:

Thanks, Harazi :)
[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=kvant&t=16471]here[/url]

Hope someone will show me how to make these miracles links. Probably I should have asked Valentin himself today, in "Jardin de Luxembourg"... :D  :D

Hard problem  :D
In fact with Faddeev proof, it's easy to prove that $f(n)>n-\log_2n-2$

This problem with same solution as Hong Quy wrote can be found in "Impresje liczbowe" L.Kurlyandchik-even with same notations:)


geometry
3D geometry
tetrahedron
pyramid
prism
ratio

Given a regular tetrahedron (a triangular pyramid) with an edge of 8. Find the area of a cross section that contains the point of concurrency of the four altitudes of the pyramid.  :ninja:

[quote="rnwang2"]Given a regular tetrahedron with an edge of 8, find the area of the cross section that contains the point of concurrency of the four altitudes.  :ninja:[/quote]
A cross section is not defined by one point, I think you need more information. :P

[quote="rnwang2"]I was under the impression that a cross section of a prism is the intersection of the prism with a plane [b]parallel[/b] to the plane of the base.[/quote]
Then, the cross section would be a triangle parallel to the base but it would be formed by the midpoints of the "non-base" side. [hide]Using the midline theorem, the sides of the triangle would be 4 and then the area would be $\frac{2\sqrt{3}\cdot4}{2}=4\sqrt{3}$ I think[/hide] :P

it is (3/4)^2*base=9\sqrt{3}...obviously this makes two similar tetrahedrons, and the ratio of the heights is 3:4...[why?], the rest follows from similarity


here is something you might enjoy!

Let there be an increasing series (a_n) with a_0=a_1=47 which satisfies:

(a_(n+1))^2+(a_n)^2+(a_(n-1))^2- a_n*(a_(n+1))*(a_(n-1))=4 for any n>=1. Prove that for any natural number n we have:

i) 2+a_n is a perfect square
ii) 2+sqrt(2+a_n) is a perfect square.

bye bye!

It can be shown that if a 2 +b 2 +c 2 -abc=4, then there exist x,y,z so that x+y+z=0 and a=2ch x, b=2ch y, c=2ch z, where ch x=(e x+e -x)/2. If we put a 0=a 1=2ch x, we obtain inductively that an=2ch nx. Our problem becoms to show that e nx/2+e -nx/2 and e nx/4+e -nx/4 are integers. Because it is known that if z+z -1 is an integer then so is  z n +z -n, for all n, we only have to prove that  e x/2+e -x/2  and e x/4+e -x/4  are integers. But we know that e x+e -x=47 from where we obtain that  e x/2+e -x/2=7 and e x/4+e -x/4=3.

Nice soln! A lot of stuff can be solved with that (if a^2+b^2+c^2-abc=4 then a, b, c have form (x+1/x)/2). Here's a related thing: if positive a, b, c satisfy a^2+b^2+c^2+abc=4 then a, b, c are 2cosA, 2cosB, 2cosC where A, B, C are the angles of a triangle).


[quote="anirudh"]what does $\lceil$ and $\rceil$ do? :blush:[/quote]
It's called the ceiling function (or least integer function). What it means is the least integer greater than or equal to the number. Examples:
$\lceil \pi\rceil=4$
$\lceil \sqrt2\rceil=2$
$\lceil 11\rceil=11$
$\lceil-4.6\rceil=-4$

[hide]using the binomial theorum, i got:

$198\sqrt{6}+485$, which after lot of work calculating, is $969.999\dots$, whose ceiling is $970?$[/hide]

EDIT: edited :blush:
EDIT2: hidden :blush:

[quote="anirudh"]using the binomial theorum, i got:

$198\sqrt{6}+485$, which after lot of work calculating, is $969.999\dots$, whose ceiling is $969?$[/quote]
Actually the ceiling of [hide]969.999 is 970[/hide]. Also hide your solution. Also, can we use calculators?

[quote="Phelpedo"]Find $\left\lceil (\sqrt{3}+\sqrt{2})^{6}\right\rceil$.

Obviously, no calculators allowed.[/quote]

[hide="Hint"]Add it (w/o the ceiling) to $(\sqrt3-\sqrt2)^{6}$ noting that that expression is a lot smaller than 1.[/hide]

NO CALCULATORS!

Let's say that this problem can be done in under 45 seconds. In other words, you don't have the time to figure out $\lceil 198\sqrt{6}\rceil$ by hand (anirudh, it's pretty obvious that you used a calculator...).

[quote="robinhe"][hide="Hint"]Add it (w/o the ceiling) to $(\sqrt3-\sqrt2)^{6}$ noting that that expression is a lot smaller than 1.[/hide][/quote] Very nice, robinhe. :D


counting distinguishability MATHCOUNTS search

Hey Everyone,

Can you all explain to me the concept of distinguishability? Also, give me all the formulas for distinguishable balls and boxes, indistinguishable balls and boxes, distinguishable boxes but indistinguishable balls, and indistinguishable boxes but distinguisable balls? Please show [b]how to derive the formula's[/b].

Thanks in advance :lol: !

the idea about distinguishability problems is to [b]think[/b], after that it is just applying the fundamentals of counting

the two cases with distinguishable boxes are much simpler than the other two cases, both distinguishable is by far the simplest, indistinguishable balls and distinguishable boxes is an application of the balls and urns method.

the other two cases, if asked in mathcounts, the numbers are small enough for you to list them out

Search before posting: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244341


I've been looking for a blog FAQ section. Is there one? Can we make this board into one? Things like:
How to make a blog.
What is considered reportable in a blog,
Where to report a blog,
How to upload pictures into a blog,

All that good stuff. Many, many thanks. :)

A general FAQ can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=280499]here[/url]. Some of it may be helpful.

Thanks very much!
But I would also like to know what people are getting their blogs reported for.
Is it just spam or rudeness?
I want to abide by the rules, but I dont' really know what they are...........

Sometimes people advertise on blogs. This is considered spamming.


How many integers are there in the solution set of $ |x\minus{}2| \leq 5.6$?

$ \minus{} 5.6\le x \minus{} 2 \le5.6$
$ \minus{} 3.6\le x \le7.6$

$ 7 \plus{} 3 \plus{} 1 \equal{} \boxed{11}$ integers.


quadratics
real analysis
complex numbers
algebra
polynomial
sum of roots
number theory proposed

Find all non-negative integers $m,n,p,q$ such that \[ p^mq^n = (p+q)^2 +1 . \]

if $m,n\geq 1$
if $\frac{1+p^2+q^2}{pq} \in \mathbb{N}$ it is $3$ so $p^{m-1}q^{n-1}=5$ and we find the solutions (m,n,p,q)=(2,2,5,1) and (2,1,5,13) (2,1,5,2)
if m=0
then $q^n=(p+q)^2+1$ which is  solvable.

[quote="Albanian Eagle"]$q^n=(p+q)^2+1$ which is  solvable.[/quote]
Can you show me a solution of this part?,I have found a sloution,but I used the expretion$a^2+1=(a-i)(a+i)$ :blush: are ther other solutions,without complex numbers ?:)

[quote="xirti"]The main ideea is that the ecuation $q^n=(p+q)^2+1$ is quadratic in p, an if it has solution p it has also $2q-p$ because forom Viete the sum of roots is $2q$. Hence we can work with that root $\le q$. For big enough values of $p,q,n$ (greater than 3 or 4 is enough) we can prove that $q^n > (p+q)^2+1$. The details can be filled easily.[/quote]
sum of roots is -2q<0.

[quote="Albanian Eagle"]
if $\frac{1+p^2+q^2}{pq} \in \mathbb{N}$ it is $3$ [/quote]

can you show that please.

[quote="Albanian Eagle"]of course
show by infinite descent that if $p^2+ q^2 = apq-1$ then a must be $3$[/quote]
You havn't answered my question about second part!

[quote="Tiks"][quote="Albanian Eagle"]of course
show by infinite descent that if $p^{2}+q^{2}= apq-1$ then a must be $3$[/quote]
You havn't answered my question about second part![/quote]

Yeah...

But I thought that's pretty classic :wink:  :(

You Can Also See The Answer In The Following Link :

[code]http://www.irysc.com/forum/t6886-post90551/#post90551[/code]

Post #8 Is An Answer And Post #11 Is Another Answer To This Question. :)

I find a solution without using the fact: $\frac{p^2+q^2+1}{pq}=3$, which of course solves easily the problem.
We see that $(p+q)^2+1\ge 1$, so $p,q\ne 0$.
First, if $m=0$ or $n=0$ we get the following equations $q^n=(p+q)^2+1$ and $p^m=(p+q)^2+1$, respectively. But for a result of V. Lebesgue [url]http://www.math.leidenuniv.nl/~jdaems/scriptie/Catalan.pdf[/url] we know these equations have no non-zero solutions. Therefore we may assume that $m,n\ge 1$. So $m+n\ge 2$. If $m+n=2$ we get the equation $pq=(p+q)^2+1$ and it's easy to see that there are no solutions for $p,q>0$. Then $m+n>2$.

We have $p^{m}q^{n}=p^2+2pq+q^2+1$, so we infer that $p\mid q^2+1$ and $q\mid p^2+1$, and also if $(m,n,p,q)$ is a solution then $(n,m,q,p)$ is a solution too. Then we assume without loss of generality that $p<q$ (it's clear that $p=q$ is impossible). If $q=p^2+1$ we would have $p\mid 2$, so we found the pair $(2,5)$. Replacing into the equation we get $2^{m}5^{n}=50$, hence $m=1$ and $n=2$. Then we have the quadruple $(1,2,2,5)$. Now we assume that $q\ne p^2+1$. So $q\le \frac{p^2+1}{2}$. We check for $p\le 4$ if there are solutions and we don't find a new pair apart from $(2,5)$. From now on we suppose that $q>p\ge 5$.

We have $p^{m+n}<p^{m}q^{n}=(p+q)^2+1\le (p+\frac{p^2+1}{2})^2+1=\frac{(p+1)^4}{4}+1$, so $4p^{m+n}\le (p+1)^4$. If $m+n\ge 4$ we have $4p^4\le (p+1)^4$ but $3p^4>4p^3+6p^2+4p+1$ for all $p\ge 5$, hence $m+n<4$. So we have found $2<m+n<4$. Therefore $m+n=3$.

On the other hand $5q^n\le p^{m}q^{n}=(p+q)^2+1<(2q)^2+1$, then $5q^n\le 4q^2$, if $n\ge 2$ we get $5q^2\le 4q^2$, impossible because $q>0$, hence $n=1$ and so $m=2$.

Finally we have the equation $p^{2}q=(p+q)^2+1=p^2+2pq+q^2+1$. If we see this equation as a quadratic equation in $q$ we get $q^2+(2p-p^2)q+(p^2+1)=0$. In order to have $q\in \mathbb{Z}$ the discriminant of last equation must be a perfect square, so $(2p-p^2)^2-4(p^2+1)=n^2$. Therefore $p^4-4p^3-4=n^2$. But we have $p^4-4p^3-4>(p^2-2p-5)^2$ and $p^4-4p^3-4<(p^2-2p-2)^2$ for $p\ge 5$ (the reader can check these inequalities). Then $n=p^2-2p-3$ or $n=p^2-2p-4$. In the first case we get the equation $2p^2-12p-13=0$ and this equation doesn't have integer solutions. In the second case we get $p^2-4p-5$, so we find $p=-1$ and $p=5$. Since $p=5$ is only valid we get $q^2-15q+26=0$, then we find $q=2$ and $q=13$. Since $q>p$ we have $q=13$. Therefore we have the quadruple $(2,1,5,13)$. Hence all the solutions are $(1,2,2,5), (2,1,5,2), (2,1,5,13)$ and $(1,2,13,5)$.

I also comment that there is another solution for the equation $p^{2}q=(p+q)^2+1$ if we see it as a quadratic equation in $p$.


AMC USA(J)MO USAMO calculus probability function MATHCOUNTS

hmmm... This may not belong in the "games" forum...

anyways: [hide="solution"] This assuming that the cookies weren't ordered(e.g. the first cookie has 12 possible choices), there is a 1/12 chance that you pick a 1 on your first draw and a 1/11 chance that you pick a 2 on your second draw. Multiplying these gets us:[b]1/132[/b][/hide]

Since batteredbutnotdefeated did not post a new problem, I will:
[b]A water tank is originally 4/15 the way full.  When 12 gallons of water are added, the water tank is now 3/5 full.  If a car running on this hydrogen(starting at a full tank) has a constant mileage of 20 miles per gallon, how many miles can the car go before running out of gas?[/b]

3/5-4/15= 5/15
12 gallons of water = 1/3 of the whole water tank
36 gallons of water = the whole water tanks
36*20=720
720 miles
maybach i dont really know a harder question can u just post em?

I kinda decided this thread should die......................................................................

But you want some difficult problems?  Here's are two to  ponder:

Since it's just two questions, I don't think there's a copywriter's violation.

That's difficult?! What's that competition?

7. Take everything as a function of w

$ z\equal{}0.625w,y\equal{}\frac{5w}{12},x\equal{}\frac{5}{16}w$

The value is $ \frac{83}{48}$ if I'm calculating well.

8. The angle between 8 and 9 is 30 degrees, and in 15 minutes the hand goes 7,5 degrees from 8. The angle is 157.5 degrees.

That was the National Mathcounts Target Round for 2003 Problems 7 and 8.

Hard for an average middle schooler like me that is.........

They weren't that bad though I guess.

Yes, "ULTRA HARD" doesn't really describe MATHCOUNTS...

Here's one, taken from #10 in ARML. Should be harder.

If 6arctan(x)+4arctan(3x)=pi , compute $ x^2$.

6arctan(x)+4arctan(3x)=pi
6arctan(x)=pi-4arctan(3x)
divide by 4,
(3/2)arctan(x)=arctan(1)-arctan(3x)
tan((3/2)arctan(x))=(1-3x)/(1+3x)
hmm, w/e. by the Chuck-Jesus Theorem, the answer is 2.
If Tellah and Yang are playing a game of tictactoe, what is the probability that Obama is reelected? Express your answer in simplest radical form.

I hope i have correctly interpreted your placing this thread in the games forum.

Where would it be moved to?  The fun factory?

I kinda thought it would get locked by now.

Here's one that I solved:

A triangle $ ABC$ has incenter $ I$. The incircle is tangent to $ AB$ at $ T_C$, $ BC$ at $ T_A$, and $ AC$ at $ T_B$. Each of the line segments $ IT_A$, $ IT_B$ and $ IT_C$ are extended in the direction of $ I$ until they intersect the triangle again at points $ D$, $ E$, and $ F$, respectively.

Show that there will always be one side of ABC that contains two of the three points $ D$, $ E$, and $ F$.

This one actually [i]is[/i] hard, it took me 1 hour to get a proof, and even then only because I was lucky with a theory of mine.

[quote="maybach"]Here's how the game works.  First, I post a problem.  Then, the person after that should post a problem they think is harder along with the solution to the previous problem.  This will continue until, for 24 hours, a problem cannot be solved.  At that point, the person who made the "stumper"  will get 100 points for every full day since the round has started.  Once that has happened, a new round will be started and we start with an easy problem.  Obviously, we need rules regulating something like this, so here they are:
-Each problem must have a MODERATE increase in difficulty.  We ca n't have 2+2 be one problem and a USAMO problem the next.
-Please, don't post problems involving calculus or advanced trigonometry.  The problems should get more involved and take more thinking, but the subjects can't be too advanced.
-You judge the difficulty of the problem.  But please use common sense.
-If you want to post a new problem, you MUST post the solution to the previous one.  I'll start with an easy one to get us off:

[b]A cookie jar has 12 cookies each having one of the numbers 1,2,3,4,5,6,7,8,9,10,11, and 12.  No cookie has more than 1 number used and all numbers are used once.  What is the probability that you will select a one, then a two?  Express  your answer as a common fraction.[/b][/quote]

You probably won't find a number 6 (USAMO) that doesn't use more than basic trig..

Strange, mine doesn't use trig at all...

Here's one from China's TST:

Triangle $ ABC$ has sidelengths $ \alpha$, $ \beta$, and $ \gamma$ that add up to $ 2\pi$. Show that $ \sin \alpha$, $ \sin \beta$, $ \sin \gamma$ also form a triangle.

Or something like that...

If that's too hard, try this one:

There is a jar of 12 perfectly circular cookies, each numbered 1 to 12 (one number per cookie), and with a radius equal to the number printed on it.

You take two cookies out at random. What is the probability that they will both fit, without overlapping or breaking up any cookies, on a perfectly circular dish of radius 15?

[quote="Poincare"]

You probably won't find a number 6 (USAMO) that doesn't use more than basic trig..[/quote]

Trig != hard.

Hard  != trig.

Come on, can we put this out of my misery and move this topic?

[quote="james4l"]Trig != hard.

Hard  != trig.[/quote]

Agreed. I mean, just [i]look[/i] at the Riemann zeta hypothesis!

But yeah, let's move this topic.


Find the smallest value of $ c$ such that for all $ x_i > 0$, $ i \equal{} \overline {1,n}$, $ x_{k \plus{} 1}\ge x_1 \plus{} x_2 \plus{} ... \plus{} x_k$ and:
$ \sqrt {x_1} \plus{} \sqrt {x_2} \plus{} ... \plus{} \sqrt {x_n}\le c.\sqrt{x_1 \plus{} x_2 \plus{} ... \plus{} x_n}$

@[b]mathVNpro:[/b] Did you mean find $ c$ in term of $ n$ or a general $ c$ for all $ n \in \mathbb {N}^*$?


find all the bejective functions $f: \mathbf{R} \longrightarrow \mathbf{R}$such that :

 $\forall x$ we have $f(x)+f^{-1}(x)=2x$

i didn't know but i have a new solution so i post this problem to know if  there other solutions :D


This is an interesting question I came up with when folding a sheet of paper.  I tried to make the lengths nice in the hope that the answer will be somewhat nice, but I haven't gotten an answer yet.

A rectangular piece of paper $ ABCD$ has $ AB \equal{} CD \equal{} 12$ and $ BC \equal{} DA \equal{} 16$.  Folding point $ B$ onto point $ D$ results in pentagon $ DAEFC$.  Find the area of the heptagon formed by folding point $ C$ onto point $ E$.

Draw perpendicular bisector to the diagonal and let it intersect side BC at X. Let O be the center of the rectangle. Then OX satisfies $ 12^2\plus{}(16\minus{}y)\equal{}y^2$. The answer is just $ 12*16\minus{}0.5y*\sqrt{12^2\plus{}16^2}$.


Note that 1681 = 41^2. How many zerors would you need to add in between 6 and 8 to make the new number (e.g., 1600...081) also a squared number?

Have A Great Day!

Great that you've factored the right side! Now, can we factor the left side?

[hide="Hint"]Yes, we can!

We know $ 16 0 \ldots 000 = 2^{n + 6} 5^{n + 2}$, where $ n$ is the number of $ 0$'s in the [i]original[/i] number (ie, 16, n 0's, 81).

We therefore factor it by adding two more parameters, $ a$ and $ b$:

$ (2^{a} 5^{b}) \cdot \left( 2^{n + 6 - a} 5^{n + 2 - b} \right)$

We can pre-suppose that $ n + 2 - b \leq b$ to avoid a redundant list of factors.[/hide]
[hide="Solution Part 1"]Observe that one of the following must be true:

$ 2^{a} 5^{b} - 2^{n + 6 - a} 5^{n + 2 - b} = (x + 9) - (x - 9) = 18$
$ 2^{n + 6 - a} 5^{n + 2 - b} - 2^{a} 5^{b} = 18$

Since the $ 18$ is not divisible by $ 5$, the two expressions in $ a$ and $ b$ cannot both be divisible by $ 5$. This means that, in either case, we must have either $ n + 2 - b = 0$ or $ b = 0$. However, since we pre-supposed that $ n + 2 - b \leq b$, we can just consider the case when $ n + 2 - b = 0$ (including the other half of the cases would be redundant).

Therefore, we can simplify to one of these two cases:

$ 2^a 5^{n + 2} - 2^{n + 6 - a} = 18$
$ 2^{n + 6 - a} - 2^a 5^{n + 2} = 18$[/hide]
[hide="Solution Part 2"]Let's find another convenient divisibility condition. Since $ 5 \equiv 1 \pmod{4}$, let's take the system in mod $ 4$

$ 2^a - 2^{n + 6 - a} \equiv 2 \qquad \text{or} \qquad 2^{n + 6 - a} - 2^a = 2 \equiv - 2$

Note that the two possibilities simplify to the same modulo-4 condition.

We know that $ 2^k \equiv 1, 2, 4 \pmod{4}$ if $ k = 0$, $ k = 1$, and $ k \geq 2$, respectively. The only way we can make the difference of two powers of $ 2$ be equivalent to $ 2 \pmod{4}$ is if one is divisible by $ 4$ and the other is divisible by $ 2$ but not $ 4$; ie, the exponent on $ 2$ is $ 1$.

We must therefore have that either $ a = 1$ or $ n + 6 - a = 1 \implies a = n + 6 - 1$. We thus have one of four cases:

$ (1) \qquad 2 - 2^{n + 5} 5^{n + 2} = 18$
$ (2) \qquad 2^{n + 5} 5^{n + 2} - 2 = 18$
$ (3) \qquad 2^{n + 5} - 2 5^{n + 2} = 18$
$ (4) \qquad 2 5^{n + 2} - 2^{n + 5} = 18$[/hide]
[hide="Solution Part 3"]In (1), we have $ 2^{n + 5} 5^{n + 2} = - 16 = - 1 \cdot 2^4$, which does not lead to a solution.

In (2), we have $ 2^{n + 5} 5^{n + 2} = 20 = 2^2 \cdot 5^1$, which does not lead to a solution.

In (3), we divide by $ 2$ to get the equivalent form $ 2^{n + 4} - 5^{n + 2} = 9$. However, observe that $ 2^{n + 4} - 5^{n + 2} < 2^{n + 4} - 4^{n + 2} = 2^{n + 4} - 2^{2n + 4} \leq 0 < 9$, so there are no solutions.

In (4), we divide by $ 2$ to get the equivalent form, $ 5^{n + 2} - 2^{n + 4} = 9$. Observe that $ n = 0$ yields a solution, which is the given one. Let's see if there are any others:

Divide by $ 2^{n + 4} = 4 \cdot 2^{n + 2}$ to get the equation:
$ \frac {1}{4} \cdot \left( \frac {5}{2} \right)^{n + 2} - 1 = \frac {9}{4} \left( \frac {1}{2} \right)^{n + 2}$

In this form, LHS is strictly increasing and the RHS is strictly decreasing, which means that they can intersect at most once, and there can be no other solutions.

Therefore, if there is a solution, it must correspond to $ n = 0$. Double-check that $ 1681 = 41^2$ is indeed a square, and therefore must be the only square-number of this form[/hide]

[i]Don't worry about the length of this -- it's actually pretty fast to do out.[/i]

Off topic:
TZF are you able to receive pm's?


analytic geometry
vector
geometry
parallelogram
rhombus
angle bisector

Find the equation of the bisector of angle A of the triangle, the coordinates of whose vertices are[b] A(4,3) ,[/b] [b]B(0,0) and C(2,3)[/b]. Please give its solution. Thanks a lot for your help.

hello, we have $ \vec{AC}\equal{}\left(\begin{array}{c}\minus{}2\\0\end{array}\right)$ so we get the unit vector as
$ \frac{\vec{AC}}{|\vec{AC}|}\equal{}\left(\begin{array}{c}\minus{}1\\0\end{array}\right)$ analog we get
$ \frac{\vec{AB}}{|\vec{AB}|}\equal{}\left(\begin{array}{c}4/5\\3/5\end{array}\right)$
so our equation is given by
$ \vec{x}\equal{}\left(\begin{array}{c}4\\3\end{array}\right)\plus{}t\left(\begin{array}{c}\minus{}1/5\\3/5\end{array}\right)$ with $ t \in \mathbb{R}$.
Sonnhard.

[asy]pair A=(4,3), B=(0,0), C=(2,3);
draw (A--B--C--cycle);
dot ("$A$",A,NE);
dot ("$B$",B,SE);
dot ("$C$",C,NW);

pair D=(-2,0);
draw (C--D--B, dashed);
dot ("$D$",D,SW);[/asy]
$ ABDC$ is parallelogram.
$ D$ is $ ( \minus{} 2, 4)$.
So $ y \minus{} 3 \equal{} \frac {3 \minus{} 0}{4 \minus{} ( \minus{} 2)}(x \minus{} 4) \iff \boxed{y \equal{} \frac {1}{2}x \plus{} 1}$.

[quote="Dr Sonnhard Graubner"]hello, we have $ \vec{AC} \equal{} \left(\begin{array}{c} \minus{} 2 \\
0\end{array}\right)$ so we get the unit vector as
$ \frac {\vec{AC}}{|\vec{AC}|} \equal{} \left(\begin{array}{c} \minus{} 1 \\
0\end{array}\right)$ analog we get
$ \frac {\vec{AB}}{|\vec{AB}|} \equal{} \left(\begin{array}{c}4/5 \\
3/5\end{array}\right)$
so our equation is given by
$ \vec{x} \equal{} \left(\begin{array}{c}4 \\
3\end{array}\right) \plus{} t\left(\begin{array}{c} \minus{} 1/5 \\
3/5\end{array}\right)$ with $ t \in \mathbb{R}$.
Sonnhard.[/quote]
Your $ \vec{AB}$ is wrong. This is the equation of the external angle bisector, though.

@Kouichi: $ \overline{AD}$ is a diagonal of the parallelogram, which does not generally bisect $ \angle A$: for example, look at $ \overline{BC}$ in your diagram.

hello, sorry, you have right, $ \vec{AB}$ must be $ \left(\begin{array}{c} \minus{} 4 \\
\minus{} 3\end{array}\right)$, so our equation is given by
$ \vec{x}\equal{}\left(\begin{array}{c} 4 \\ 3\end{array}\right)\plus{}t\cdot\left(\begin{array}{c}\minus{}9/5\\\minus{}3/5\end{array}\right)$.
thank you for your hint.
Sonnhard.

I don't think vectors are basics level (are they?) Thanks!

No they aren't. If you want a simple solution, just use the angle bisector theorem to find where it meets BC and use those two points to find the equation.

[quote="Bictor717"]
@Kouichi: $ \overline{AD}$ is a diagonal of the parallelogram, which does not generally bisect $ \angle A$: for example, look at $ \overline{BC}$ in your diagram.[/quote]
Sorry, I mistook a rhombus for a parallelogram.

[asy]pair A=(4,3), B=(0,0), C=(2,3);
pair D=(-1,3);
draw (A--B--C--cycle);
dot ("$A$",A,NE);
dot ("$B$",B,SE);
dot ("$C$",C,N);
dot ("$D$",D,NW);

draw (C--D--B, dashed);

pair E=(-5,0);
draw (D--E--B, dashed);
dot ("$E$",E,SW);[/asy]

$ ABED$ is rhombus.
$ D$ is $ ( \minus{}1, 3)$ and $ E$ is $ (\minus{}5, 0)$.
So $ y \minus{} 3 \equal{} \frac {3 \minus{} 0}{4 \minus{} ( \minus{} 5)}(x \minus{} 4) \iff \boxed{y \equal{} \frac {1}{3}x \plus{} \frac{5}{3}}$.


calculus
limit
integration
function
derivative
calculus computations

Here is a problem given at a regional math contest in Romania. No one solved it and unfortunately i have to come up with a proof and present it in front of the math teachers at my school. And this is all happening tomorrow!! Any help would be more than appreciated.
Let $ \ f: (0,\infty)\rightarrow R, f(x)\equal{}e^\frac{1}{x^2}$ and F its antiderivative.
Compute $ \displaystyle\lim_{x\rightarrow 0, x>0} F(x).$

I think the result is $ \minus{}\infty$ but cannot prove it.

Assume that $ 0 < x < 1.$ We can write

$ F(x) \equal{} F(1) \minus{} \int_x^1e^{1/t^2}\,dt.$

Now we know that $ e^{1/t^2}$ tends to $ \plus{} \infty$ very rapidly as $ t\to 0.$ We could get the limit we need with a much slower rate of divergence. Here's one possible estimate:

$ e^{1/t^2}\ge \frac1{2t^2}.$

Hence $ F(x)\le F(1) \minus{} \int_x^1\frac1{2t^2}\,dt \equal{} F(1) \plus{} \frac12 \minus{} \frac1{2x}.$

Since that tends to $ \minus{} \infty$ as $ x\to0^ \plus{} ,$ we have the result you wanted.

There would be many similar methods differing in detail.

I have come up with another solution. I don't know whether it's correct and it's rather incomplete, but that is why i am asking your help. Here it goes:

We have $ F'(x)=f(x)>0\Rightarrow F$ is a monotonically increasing function.
This implies that $ \displaystyle \lim_{x\rightarrow0^{+}}F(x)\neq\infty$.

$ \displaystyle \lim_{x\rightarrow0^{+}}F(x)=\displaystyle\lim_{x\rightarrow\infty}F(\frac{1}{x})$.

Let's assume that there exists $ k\in(0,\infty)$ so that $ k=\displaystyle\lim_{x\rightarrow\infty}F(\frac{1}{x})$.
We get:

$ \displaystyle\lim_{x\rightarrow\infty}F(\frac{1}{x})=\displaystyle\lim_{x\rightarrow\infty} \frac{F(1/x)\cdot e^x}{e^x}\overbrace{=}^{L'Hopital}\displaystyle\lim_{x\rightarrow\infty}F(1/x)+\displaystyle\lim_{x\rightarrow\infty}f(1/x)\Rightarrow\displaystyle\lim_{x\rightarrow\infty}f(1/x)=0$

and this is clearly a contradiction.

Only two more options remain for the limit: $ -\infty$ or it does not exist.
But since F is monotonic and continuous, the function must have a limit in 0 (if anyone knows how to prove this, he or she is more than welcomed).

The derivative of $ F\left(\frac1x\right)$ is $ \minus{}\frac1{x^2}f\left(\frac1x\right)$. Be careful with the chain rule when making substitutions like this.
The function poses a problem because it blows up extremely rapidly at zero; something tame like a nonzero limit doesn't imply anything interesting.

Ouch, that is a big mistake. I worked with x instead of 1/x first, but then I could not apply L'Hopital and then just replaced it without doing the math. My bad.
But I am still wondering about the last part. Can anyone prove the following problem?
Let $ f: (a,\infty)\rightarrow R,a\in R$ be a monotonic and continuous function. Prove that $ \displaystyle\lim_{x\rightarrow a,x>a }$ exists.

That depends on whether you want to include $ \pm\infty$ in "exists".

The limit is the least upper bound of the values if it's decreasing, and the greatest lower bound of the values if it's increasing.


Has anyone ever done ACSL (see http://www.acsl.org for more info)? Made it to the All-Star Contest?

If so, can anyone give me tips on programming/short answer preparation? I'm a high school student, and most likely competing in the Intermediate division since I'm a beginner.

Right now, for preseason, my school is teaching us the short answer topics. So far, we've covered Number Systems (easy, since I've done lots of NT problems with those) and LISP Programming. Next time, we are doing Recursive Functions (which I am pretty familiar with due to the amount of math I'm doing

Also, I'm kinda new to LISP Programming. Any good sites/resources to help me practice? I've also heard it's probably one of the most basic computer programming languages, is that so?

Yeah, we did it last year not knowing what to expect.  Some problems were fairly straightforward and others were insanely hard because we didn't study.

I'd say you don't need to know much about Lisp beyond the basics though.

My teacher suggested Visual BASIC for people new to programming. I just recently got started on it.

Just wondering, but would any other more advanced languages be effective for the Intermediate Division?

VB will be fine if you like using it.  I personally don't, so I'd recommend a different language, but that's personal preference.

Our school does ACSL every year and always makes it to the All-Star contest. go on the acsl webpage and do the practice problems... that's the best way to prepare, i found.
We code in java. :)

Wait, sorry to revive but...how do you qualify for the ACSL all-star?

It was my first contest today and I took the senior division (because I am fairly decent at programming).

The short-answer questions were a bit annoying...especially recursive functions. But I had to make do with only about an hour or so of just reading some stuff to study and 25 minutes to take it...even in the senior division, the short-answer questions are fairly simple, just arithmetic-intensive.

If your school does well over the season, your school will qualify for the all-star meet.
Current rankings are posted on the ACSL website over the year.

Yeah...it looks from online that your school has to be in the top 3...and that definitely isn't happening.

My school has actually made All-Star for both intermediate and senior several times, and in some years, got n the top places.

But this year, our intermediate team might get pwned due to lack of programming participation. :(

And programming is killing my brain now. T_T

Is it really top three? I thought it was more... there are definitely more than 3 teams per division that come to the All-Star meet.

Do we have to have a team in order to partcipate in ACSL?
Or can we participate as an individual? Because no one else at my school seems to be interested in this + they are terrible at programming.

also, how does the difficulty of ACSL problems compare with USACO programming tasks?

i dunno about your first question, but ACSL is really easy compared to USACO for sure...
at least in my experience, the only difference between the junior and senior division is the amount of test cases to consider
the programming assignments in acsl are kind of the like the ones you'd see in a high school class.

[quote=walnutwaldo20][quote=Dafidz]PYTHON!![/quote]
If you look right in front of you, you will see the average mindless 8-year-old that happens to only know python in its natural habitat.[/quote]

I died a little bit inside

even c++ is better than that and c++ is a limited language that is non universal


A sequence converges in the discrete topology if and only if it's eventually constant.


[quote="Mathematicus"]I just can't quite figure this out  :maybe: ... Could someone show me how this is simplified?

$\sqrt{1-\cos^{2}x}$[/quote]

[hide]
a basic trig formula is:
$\sin^{2}\theta+\cos^{2}\theta=1$
Therefore, $\sin^{2}\theta=1-\cos^{2}\theta$
The question is then $\sqrt{\sin^{2}x}$, which in other words is $|\sin x|$[/hide]

Thanks guys. It seems pretty obvious now... :wallbash: