I plan on taking AP Physics next year, and the physics teacher told me, "AP Physics will apply all of the math you know and then some."  Will I actually learn any useful math in their besides vectors?

For an AP Mechanics class it helps to have a good understanding of Calculus. It's not absolutely necessary, but it helps elucidate some things that you otherwise have to do a lot of hand-waving at. Other than that (and possibly vectors), there's not much [i]new[/i] math that needs to be introduced.

There is some (probably) new and interesting math at work behind E&M, but most physics teachers won't go into great detail about it. You can learn the physical laws and how to apply them without understanding why they are true. In reality it's not possible to really [i]get[/i] E&M until you take multivariable calculus.

I suppose I should be grateful that I'm taking multivariable first....

Our physics teacher taught us the entire multi curriculum in 2 weeks. He then covered a good deal of differential equations in another week or so. Admittedly, I have not learned much physics (and am greatly unprepared for the AP) -- but like my teacher said, "If you want to learn math, be a physics major."

[quote="ComplexZeta"]That's silly. If you want to learn math, be a math major and take eight math classes in the same quarter. (Not that [i]I[/i] would do a thing like that... no wait, I do.)[/quote]

Man, I can't wait to do that.

What year are you and where do you go to school?

Whoa-- how do you take so many? Don't you need to take other classes like English, science, and social studies? I don't think I could handle the work.

Also, doesn't each class build on top of a previous class?

I think I need to take a history class and an ethnic studies class at some point in my undergraduate career. That might be negotiable though. (I would like to learn something about those subjects anyway, so I won't try to negotiate.) The [url=http://www.ccs.ucsb.edu]College of Creative Studies[/url] gives students very few general education requirements and more freedom to study whatever they are interested.

Most of the classes I am taking do indeed build off of previous classes (most of which I haven't taken and will not take). It's pretty amazing what you can pick up from taking more advanced classes though. In the fall, I thought I knew just about none of the undergraduate material (although I was being thrown into graduate classes), and now I'm convinced that I know almost all of it despite not having taken most of the classes. One part of it is confidence: I read a lot of textbooks in a rather careless manner in high school (I didn't usually do many problems since they seemed so simple) and assumed that it would never be enough to learn the course material carefully, but apparently I absorbed much more than I had previously thought. Going into a class that moves very quickly and assumed a prerequisite that technically shouldn't be needed is very difficult though. I quickly became overwhelmed by my graduate algebra class in the fall and decided that it would be better to audit it. Now it's the third quarter, and I'm still auditing it. I'm not learning the course material as well as the people who are taking it, but I'm definitely getting a lot out of it.

[quote="churchilljrhigh"]WHat are the different AP PHysics classes? I'm taking Honors Physics next year and might take AP PHysics.[/quote]

Don't waste your time in honors physics or Physics B AP... if you have a decent grasp on calc, then go straight to Physics C AP, otherwise, you will be extremely bored to death. I go to a really ghetto school, one that doesn't offer Physics B or Physics C, so I took the class Independent Study, aka, Learn it on my own...

The Physics C AP exam for me is on monday, *crosses fingers*, I need a 5 in both mech and E&M or I get nothing, YAY!

I never took any other physics, just C AP, it's really easy stuff. Mechanics is common sense, E&M is a bit less intuitive, but it makes sense nevertheless.

I do have a firm grasp on Multivar Calc, though I don't think it helps anyone unless you are truely concerned with why exactly Guass' Law holds etc. The class can be done without any multivar knowledge.

[quote]Don't waste your time in honors physics or Physics B AP... if you have a decent grasp on calc, then go straight to Physics C AP, otherwise, you will be extremely bored to death.[/quote]

I completely agree, but my parents dont seem to understand this. Although I was offered the chance to take C next year (junior year), my parents are forcing me to take honors instead. I believe this is a big mistake. How can I convince them otherwise? OR, if I really [i]should[/i] take honors, how can I be convinced?

zableman, I have seen your math abilities around this forum and I'd say that you could EASILY get a 5 on Physics C, honors would be a gastly waste of your time...

I don't know your parents, and I don't know what would work for them. Do they understand physics? If they do understand physics (or math) then they should be able to understand that you are far passed the physics honors level. My parents aren't really involved in my decisions, so I never ran into this problem.

Maybe somebody else has some ideas.

lol, thanks alot. Maybe I'll just make them read this conversation! Anyway, good luck on the exam on monday!

i was actually going to suggest that, lol.

[quote]Don't waste your time in honors physics or Physics B AP... if you have a decent grasp on calc, then go straight to Physics C AP, otherwise, you will be extremely bored to death.[/quote]

I respectfully disagree, as a future physics major (not far off). I have not taken either Physics AP class, but I've taken the B test and I'm taking both Cs Monday - and I know the curriculum. Mathematically, B is boring, or at least not very exciting, but physically, it is interesting and glorious, especially if you haven't had any sort of a physics class before. There are some results in B that will completely amaze when seen for the first time (assuming they are derived properly so the student fully grasps what they mean). Furthermore, B covers some topics that are not covered on either C exam, but yet are important. Niether C test covers fluid mechanics, thermal physics, thermodynamics, waves, optics, or atomic/nuclear physics, but I'd argue that some of that is extremely interesting. I studied B on my own last year in a month before the AP test and found it interesting enough that not only did I decide to go back and study C on my own, but I decided to major in physics. Don't skip B just because you want a more mathematically intense approach. The physics alone is enough to make it interesting.

see, this is where opinions clash, and nobody is wrong... confuted has some very good points, and it all depends on what type of a person you are and what your interests are...

Personally, without the math, I find physics quite boring, the elegant nature of how mathematics ties into the real world is what fascinates me... the "here is a fluid dynamics equation.... now here is a problem, plug in the numbers" is incredibly boring and if I ever wanted that again, I'd take Chem AP again...

Like I said, it's all a matter of tastes, and what works for you.

It depends on what grade you are in as to which course you should take. I never had a basic physics; I went straight to AP Physics B. As long as you're a math enthusiast (which you probably are since you are on this forum), the math in Physics B should be absolutely no problem. I agree with Matt's points. Also, B is kind of a base for C, and you will probably find the course interesting.

This year, I took AP Physics B as a sophomore and I will probably take Physics C as a senior. So if you have this option, I would HIGHLY recommend it (don't even consider wasting your time in an Honors' course).

[quote="Spoon"]Personally, without the math, I find physics quite boring, the elegant nature of how mathematics ties into the real world is what fascinates me...[/quote]
And why is it any less elegant to describe nature with algebra than it is to describe it with Calculus? Sure, perhaps Calculus is more challenging, but I'd argue that the simplest approach is the most elegant. If it can be done with algebra, do it with algebra. Don't use a tool bigger than the job.

I personally find calculus yields more elegant proofs of physical properties, just because algebra can get so messy.  What would taken half a page or even a page of algebra can be done often in 3 or 4 lines with calculus.

I personally would advocate for taking some level of physics before taking physics C -- my biggest problem with physics C was a difficulty seeing the relationship between the calculus and the actual physical properties of the world.  That is, I knew how to integrate to find the moment of inertia, but I didn't know what the moment of inertia actually [i]was[/i], in real-life terms.  Taking Physics B (or Honors physics, or even just plain old regular physics) before taking C will help avoid this problem.

[quote]and if I ever wanted that again, I'd take Chem AP again... [/quote]

I deeply disagree. I took the chem AP exam this morning. Several people I know did nothing but memorize a massive list of formulas. They failed.

what does that have to do with anything? I don't know about your class, but no basis for anything was ever taught in my AP curriculum... We never knew why anything worked, just that it did... And it comes down to knowing how to do a problem based on the fact that something just works. Most of the time (not all, but most), it involves using some equation.

"Just memorized equations"... If you understand how they work, and how to do each problem, then why couldn't one get a 5? I got a 5 last year in chem... and are you saying it's possible to get a 5 without memorizing the equations? I sure would like to see you derive the dH/dT = cm (this is a more mathematical form of the equation that is typically presented) during the test, just to use it for a problem. If you don't know the equations, you aren't getting a 5.

[quote="jelyman"]OFF TOPIC HERE

Just curious, Spoon, did you ever do MAO?[/quote]

I've been President of my MAO chapter for two years running... this year I personally didn't do so well at Florida state convention... though I did manage to snag 4th in L&D.

I make too many errors... why do you ask?

[quote="tetrahedr0n"]Im not claiming that anyone can score a 5 without knowing the formulas. However, I disagree with your statement that all there is to chem AP is knowing a large number of formulas. If this was the case, a majority of testakers would score perfect on the written part, where a very large number of formulas are given on the test. Thats not to say there aren't questions where all is required is a formula. There are also many conceptual questions where there aren't any numbers involved, just ideas and concepts.[/quote]

I never said that one could pass just knowing the equations in the first place... I just said that the class was boring because nothing could be derived, there was no math involved beyond plugging numbers into given equations.

[quote="Spoon"]
I never said that one could pass just knowing the equations in the first place... I just said that the class was boring because nothing could be derived, there was no math involved beyond plugging numbers into given equations.[/quote]

That's because most students don't know enough calculus to get into the derivations of things in high school. (At some of the specialized math and science schools, like IMSA and Montgomery Blair, they offer more advanced classes that do do that) This is why, the difference between science in high school vs. science in college is like night and day. When I took physics in college it was one of the most enlightening experiences of my life! Don't let your high school experience with Chemistry, Physics, or Biology taint you against it. Just consider it a foundation on which to learn more.


AMC
AIME
LaTeX
function
AMC 10
analytic geometry
Vieta

Let's do a problem marathon with AIME-Level problems. I thought it would be fun, interesting, and at the same time great preparation for next year's AIME! Rules (according to me) : You must post 2 problems after you solve the previous problems. The problems should not be multiple-choice and may be proof-problems as well. Have fun!

Let's start with something relatively easy.

Problem #1: Prove that 1234321 in base x is the product of two perfect squares, where x > 4.

I suppose that what you are asking for is some x (such that x>4) for which this is true?

If not do you desire to show that for some set of x the statement holds true?

could you change this to AMC10-12 marathon, please? i think AMC will work much better and faster :lol:

Okay. I'll go ahead and create a few pretty easy problems:

Problem 2: Prove that any number can be written uniquely in the following form:

a1*1! + a2*2! + a3*3!.... + an*n! given that a1 is less than or equal to 1 and, in general, the nth coefficient (an) is less than or equal to n and nonnegative.

Problem 3 (Should be interesting): How many distinct symmetric 6 sided figures with lattice point coordinates can be constructed so that the coordinates of each point are given by (x, y) where x and y are less than or equal to 5?

EDIT: Clarification: The above should read: where x and y are between -5 and 5

Just to clarify:

This is an AIME marathon, so it should be AIME-level problems. But, hard AMC-12 probs. are also welcome.

-rts2007

Might it be best to move this thread over either to High School Basics or Intermediate Topics?
It might get more exposure somewhere else, as well as being slightly better classified.

Ok. That's fine. But how do you move threads? Do you ask the moderators to move them or simply just create a new topic in High School Basics/Intermediate?

-rts2007

I don't really know. I'm new... If it doesn't get moved in a sufficient amount of time, we can just restart it over in the intermediate section, with my problems or whoever has posted most recently.

PM a moderator of this forum to move it.  I suggest Intermediate Topics or Pre-Olympiad, seeing as this is AIME, which is a Pre-Olympiad exam.

Thanks worthawholebean!

Now, do you mind adding a problem to this?

-rts2007

Here's a problem:

Euler's formula states that for a convex polyhedron with $ V$ vertices, $ E$ edges, and $ F$ faces, $ V\minus{}E\plus{}F\equal{}2$.  A particular convex polyhedron has $ 32$ faces, each of which is either a triangle or a pentagon.  At each of its $ V$ vertices, $ T$ triangular faces and $ P$ pentagonal faces meet.  What is $ 100P\plus{}10T\plus{}V$?

1993 AIME Problem 10

By pure application of logic and mental math I get 255...
I don't have time to go through and construct the linear equations in terms of T and P, since Java remains resolute in smiting me at all possible times.

You'd think that determining the intersections of several hundred line segments would be easy...

EDIT:
Oops... 250.

I'll go ahead and post now.

[quote="cstan"]Never mind, I suppose you mean that 1234321 is written in base x itself? (This would make sense from our constraint on x).

Well... obviously 1234321 is equal to 1+2x+3x^2+4x^3+3x^4+2x^5+x^6 for base x

Then, factoring:

1234321 (base x) = (1+x)^2*(1+x^2)^2
Producing the desired result.

Should I go ahead and post AMC level problems or AIME problems?[/quote]
hmmm, i'm not very good at factoring... but is there a fast way to factor this?

Factoring is (usually) pretty easy if the equation is simple and if you know a few tricks.

By Vieta's
the coefficient of the costant (1) over the coefficient of the highest power (1) times (-1)^n where n is the highest power gives the product of the factors.

This prompts us to try (x +/- 1)

Vieta's can basically be shown as follows.
Suppose we have a quadratic with roots a and b.
Then
f(x) = (x-a)(x-b)

foiling, we have x^2 - (a+b)x + ab
illustrating the property that the last term is the product of the roots.

Now lets look at a cubic (since 3 is odd).
We have (x^2 - (a+b)x + ab)(x-c).
Since we only care about the last term (and first in nontrivial cases), we find that it is equal to
-abc.

We can show that in general this will equal a*b*c*...*n*(-1)^n
but it isn't really necessary to do so on this forum.

Vieta's has many other interesting properties, all of which I suggest you explore on your own,
and all of which are highly intuitive.

Problem 5: Show that n^4 + 4^n is composite for all n>1.

Problem 6: What is the sum of the first n fibonacci numbers?

These problems are in keeping with easier AIME problems (though the first is far more difficult than the second).

Woah, woah! Hold on! I hope I'm not being too bossy but you're supposed to solve the previous problems before posting 2 more. Just to clarify! :D And how do you do Problem #2 and 3? (Not to mention 4 and 5)!

-rts2007

I hate to solve my own problems....

Okay, the first, note that if we have a1*1! + a2*2! +... an*n!
With a1, a2, ... an maximized...
then the above is equal to an+1*(n+1)! -1

Creating an unambiguous mixed radix number system. While a full inductive proof is necessary to show this, it is elementary.

Number 2 suggestion:
Casework...
This might be too computationally difficult to ask though...

AIME15's:
You create linear equations in T and P by realizing you over count each edge thrice with each triangle, five times with each pentagon, and that we under count by half by counting this way.

The radix of a number system is the number of values the digits of a number are allowed to have.
Thus in base 10 the radix is 10, and in base n the radix is n.

In a mixed radix number system, the allowable digits varies between digits. For our problem, the radix of the nth digit was n.

Moderators,

Sorry to bother again, but please move to Pre-Olympiad.

Thank you

rts2007

It's supposed to be an AIME marathon, not a proof marathon which it has turned into. I'll post a question at AIME level.

Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface. The larger tube has radius 72 and rolls along the surface toward the smaller tube, which has radius 24. It rolls over the smaller tube and continues rolling along the flat surface until it comes to rest on the same point of its circumference as it started, having made one complete revolution. If the smaller tube never moves, and the rolling occurs with no slipping, the larger tube ends up a distance x  from where it starts. Find x when it is expressed in the form $ a\pi \plus{} b \sqrt{c}$.

Label the center of the big circle $ A$ and smaller circle $ B$, with the point of tangency of the big circle with the ground $ T_A$ and $ T_B$ for the smaller circle (this is the starting position before the big circle rolls). Let $ P$ be the projection of $ B$ on radius $ AT_A$. $ \angle ABP\equal{}\theta\equal{}\arcsin[(72\minus{}24)/(72\plus{}24)]\equal{}\arcsin(48/96)\equal{}30^{\circ}$.

The length of arc on the smaller circle that the big circle rolls is $ 2\pi*24*(120/360)\equal{}16\pi$.

Let the point of tangency of the big circle with the ground after it has just finished rolling on the small circle be $ T_A'$. Let the point of tangency of the big circle with the ground after it has completed one revolution be $ T_R$.
The length of segment $ T_A'T_R\equal{}2\pi*72\minus{}16\pi\equal{}128\pi$

$ x\equal{}128\pi\plus{}\sqrt{96^2\minus{}48^2}\equal{}128\pi\plus{}96\sqrt3$.

[quote="cstan"]Okay, the first, note that if we have a1*1! + a2*2! +... an*n!
With a1, a2, ... an maximized...
then the above is equal to an+1*(n+1)! -1[/quote]
How are those maximized? I'm not following. :huh:

Problem 5: Show that n^4 + 4^n is composite for all n>1.
[hide="I have seen this before."]$ n^4 \plus{} 4^n \equal{} n^4 \plus{} 2^{2n} \plus{} 2*n^2*2^n \minus{} n^2*2^{n \plus{} 1} \equal{} (n^2 \plus{} 2^n)^2 \minus{} (n*2^{(n \plus{} 1)/2})^2$
Now factor this as a difference of squares and we see the case when n is even is trivial.[/hide]
Problem 6: What is the sum of the first n fibonacci numbers?
[hide]$ F(n \plus{} 2) \equal{} F(n) \plus{} F(n \plus{} 1)$; $ F(1) \equal{} F(2) \equal{} 1$
Use characteristics equation $ t^2 \minus{} t \minus{} 1 \equal{} 0$ and summing we get a geometric series... kinda messy to type those out but you get the idea...[/hide]

I believe you are missing a $ (5!)^3 \cdot n$ and an $ n$ at the end of your problem.

Yes, that brings back so many bad memories :)

I won't answer this since I've already seen the answer and don't want to ruin the fun  :D

This is no more than your average intermediate counting problem (even though i got it wrong because i am TERRIBLE at counting :wacko: )

[quote="rts2007"]Thanks crazypianist for pointing that out. Here's another problem (from the 2009 AIME I):

The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings. At meetings, committee members sit at a round table with chairs numbered from  to  in clockwise order. Committee rules state that a Martian must occupy chair  and an Earthling must occupy chair , Furthermore, no Earthling can sit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sit immediately to the left of an Earthling. The number of possible seating arrangements for the committee is . Find .

-rts2007[/quote]
I believe you've left out some info that were in LaTeX when you copied and pasted...
could you include them? (too lazy to look it up plus for the convenience of others too...) :P

The complete problem:

The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings. At meetings, committee members sit at a round table with chairs numbered from $ 1$ to $ 15$ in clockwise order. Committee rules state that a Martian must occupy chair $ 1$ and an Earthling must occupy chair $ 15$. Furthermore, no Earthling can sit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sit immediately to the left of an Earthling. The number of possible seating arrangements for the committee is $ N\cdot (5!)^3$. Find $ N$.

Is there a way to do this with generating functions?
It seems like there would be.

I've already solved this one, so I suppose it's not fair to answer it...


[quote="illusiat"]1. The sum of 2 numbers is 15, and their product is 16. What is the sum of the reciprocals of the 2 numbers? Express ur answer as a common fraction.[/quote]
Weird. I got a different answer than 1/240.

$ x\plus{}y \equal{} 15$
$ xy \equal{} 16$

$ \displaystyle{(x\plus{}y)/(xy)\equal{}15/16}$
$ \displaystyle{x/xy\plus{}y/xy\equal{}15/16}$
$ \displaystyle{1/y\plus{}1/x\equal{}15/16}$

I'll post more answers if they aren't up by the next time that I'm on.
For your problem---
4. Whats the value of sqrt(20+sqrt(20+sqrt(20+....)))
You have all that equal x. Then, since you have x as sqrt(20+sqrt(20+sqrt(20+....))), you can say that:
 x = sqrt((20) + x)
Why? Because x equals all that infinite stuff and by doing so, you get rid of it.
Now you square both sides--
x^2 = 20 + x
Subtract and factor, which looks like this:
x^2-x-20=0
(x+4)(x-5)
(x+4)=0   (x-5)=0
[u][b]x = -4       x = 5 [/b][/u]
I'm not sure if you only want the positive answer to that question. It was posted on the forums before, relatively recently, perhaps you can find that thread and see what others had to say about it?
EDIT: Your answer key only asks for 5. So, just 5. I guess the problem wants you to reject it since it's negative. :)

And this is how you hide things-
[ hide = Click here for the answer ] ANSWER [ / hide ]
(Take out the spaces. That ends up looking like this.)
[hide="Click here for the answer"] ANSWER [/hide]

[hide="Click here for the answer to #6"] 20ft+20ft+30ft=70ft
70ft*12=840in
840in/20=42[/hide]

[hide="Answers to 3 and 7"]3. Since it is the average of two numbers, if the sum is even, the average is an integer so the integer after the decimal point is zero. If the sum is odd, the average is a number ending in .5, so the integer after the decimal point would be 50. We know that one integer is 50, and to make one number appear in the average, the difference must be 1 (then there will be .5 after the integer in the average). 51 wouldn't work because the average would be 50.50, making 50 appear twice in one number. 49 is the othr choice and works because the average is 49.50.

7. The faster pump empties the tank in 5 hours. Together in one hour, they pump (1/5)+(1/8). Let the hours it would take them to finish the job be x.
((1/5)+(1/8))*x=1
(13/40)x=1
x=40/13
5-(40/13)=25/13[/hide]

[quote="The_Scintillator"]
I guess the problem wants you to reject it since it's negative. :)
[/quote]
You have to reject $ x\equal{}\minus{}4$, since the square root sign implies the positive root. If something said $ \sqrt{9}$, $ \minus{}3$ is not an answer.

um. next time, could you READ to see if others actually answered the question before answering the exact same one the exact same approach?

another approach is fine. but yeah...
this may be considered as spam, but this is to reduce spam..

[hide="3"]Let one of the integers be $ a$ and the other be $ b$.

We are given that $ \frac{a\plus{}b}{2}\equal{}a\plus{}\frac{b}{100}$.
$ \frac{a}{2}\equal{}\frac{49b}{100}$
$ a\equal{}\frac{49b}{50}$
Since $ 49$ and $ 50$ have no common factors, $ b$ must be $ 50$ in order for $ a$ to be an integer.  We thus have that $ a\equal{}49$, the desired value.[/hide]

[hide] For number 4) Whats the value of sqrt(20+sqrt(20+sqrt(20+....)))
Say X=sqrt(20+sqrt(20+sqrt(20+....))). Square both sides and X^2=20+sqrt(20+sqrt(20+sqrt(20+....))).
Subtract both equations so X^2-X=20, cancelling out the endless square roots. You can subtract 20 from both sides and do the quadratic equation if you want or you can guess and check and get X=5. Basically in general,for sqrt(N+sqrt(N+sqrt(N+....)))=X, remember that X^2-X=N.[/hide]


AMC
AIME
inequalities
number theory
relatively prime

The decimal representation of $m/n$, where $m$ and $n$ are relatively prime positive integers and $m < n$, contains the digits 2, 5, and 1 consecutively, and in that order.  Find the smallest value of $n$ for which this is possible.

Im studying for the AIME, and I want to know the solution so could someone please do this problem?

[hide="Solution"] WLOG $\frac{m}{n}$ starts with the digits $0.251...$ (or else you can just multiply and subtract until it does).  Then you want to solve the inequality

$251n < 1000m < 252n$
$n < 250(4m-n) < 2n$

The minimum value of $4m-n$ is $1$, hence we need $250 < 2n \implies 125 < n$.  Since $4m-n = 1$ we need $n+1$ to be divisible by $4$, and this first occurs when $n = \boxed{ 127 }$.

(This gives $m = 32$ and the fraction $\frac{32}{127}\approx 0.2519685...$) [/hide]

Why will having $251$ immediately after the decimal point minimize n? I don't understand the wiki explanation for this (http://www.artofproblemsolving.com/wiki/index.php/2003_AIME_I_Problems/Problem_14). I understand that $\frac{10^k m - nX}{n} = \frac{p}{q}$ is true, but I don't understand why $q\leq{n}$ is true.

It is because either $10^km-nX$ and $n$ are relatively prime, or they are not. If they are relatively prime, then $q=n$, and if they are not, then $q$ is some divisor of $n$, so $q \le n$.

[quote=macandcheese]Why will having $251$ immediately after the decimal point minimize n?.[/quote]

If we have $\frac{m}{n} = .abc...xyz251...$ with minimum n, we can multiply by 10 repeatedly (which keeps the denominator the same) and have $abc...xyc.251...$ and then subtract $abc...xyz$ to get $.251...$ with the same denominator as the original.

amctrivial grind.

WLOG $\frac{m}{n}=0.251\ldots$ because if not, then we can multiply by a power of $10$ and then subtract by a bunch of ones to get $m<n$.

We can find that $n\le 450$ as $\frac{113}{450}$ starts with $0.251$.

If $m=\frac{n}{4}$, then $\frac{m}{n}$ starts with $0.25$.

If $m=\frac{n}{4}+1$, then $\frac{m}{n}$ can't start with $0.251$ as $\frac{1}{n}>0.002$.

If $m=\frac{n+3}{4}$, assuming that's an integer, then $\frac{3}{4n}<0.002\implies n>375$.

If $m=\frac{n+2}{4}$, assuming that's an integer, then $\frac{1}{2n}<0.002\implies n>250$.

If $m=\frac{n+1}{4}$, then $\frac{1}{4n}<0.002\implies n>125$.

So $n>125$ and $n\equiv 3\pmod4$. $n=\boxed{127}, m=\frac{127+1}{4}=32$ works, so we are done.


it is the time that i am allowed to release this hard problem
assume that $f:\mathbb{Z}^2 \rightarrow \mathbb{Z}^2$ and we know that there exists kthat for each x,y in $\mathbb{Z}^2$ (x,y is two points )  $\frac{1}{k} \leq \frac{|f(x)-f(y)|}{|x-y|}\leq k$ (| x-y|means the euclidean distance between x and y) prove that there exists r such that any circle with radius r in the plane has at least one point from the image.

I may be wrong, but I think that it is not that hard.
I mean, suppose for some circe C(C,R) , it is not true, then we can consider x is the origin, since f(x) = x - C will also satisfy the conditions. Now take the inequality for Origin, And you can probably get some R that will always satisfy the condition. I didn't make all calculations, this is just a guess

I am very stupid, sorry for the above post, I didn't realize that x will not change. Sorry again.


[quote="Inspired By Nature"]I'm having some trouble with the cross product, its not as easy as the dot product:
if $\overrightarrow{v}=\left( \begin{array}{ccc}3 & 8 & 2 \end{array}\right)$ and $\overrightarrow{w}=\left( \begin{array}{ccc}2 & 7 & 3 \end{array}\right)$, find $\overrightarrow{v}\times\overrightarrow{w}$.
Please explain all your steps, thanks in advance. :)[/quote]

this is just a formula:

$\begin{tabular}[c]{|ccc|}i&j&k\\ 3&8&2\\ 2&7&3\\ \end{tabular}$

where $i$, $j$, and $k$ are the unit vectors in the directions or the positive $x$, $y$ and $z$ axes.

This can easily be derived using the definition of the dot product, namely that it makes the vector that is perpendicular to $u$ and $v$.

A nice way:

$\begin{tabular}[c]{|ccc|cc}i&j&k&i&j\\ 3&8&2&3&8\\ 2&7&3&2&7\\ \end{tabular}$

Now multiply down the first diagonal, top left down, then add that to the second, next over and the third.  Multiply down the diagonal from the top right toward the left and subtract from previous stuff.  Repeat 2 more times moving to the left each time.  The resulting sum is your cross product.


Yes the third one is undefined because x is supposed to equal -9 and 10/9.

i made a typo for the third one. :blush:  Sorry.  Check the post again.  I just edited it

[quote="math92"]If x=2 then evaluate:
2x-7

6x+5

Solve this equation:
9x-1=-x+9[/quote]

[hide]
1. -3
2. 17
3. x=1[/hide]

[hide]
$1.\ -3$
$2.\ 17$
$3.\ x=1$
too lazy to post solutions :sleeping:[/hide]
WOW MY 1100th POST :wow: :wow:


function
real analysis
real analysis unsolved

Show that 	the function satisfying following conditions:
        $f(x)=0$  if  $x=0$ and  $f(x)=e^{\frac{-1}{x^{2}}}$ if $x$ is not equal to $0$
is not solution of the differential equation
             $y^{(n)}+P_{1}{(x)}y^{(n-1)}+...+P_{n}{(x)}y=0$
where $P_{i}$  is from  $C[-1;1]$ for all $i=1,2,3,...,n$

The unique solution $y(x)$ of the differential equation s.t.
$y(0)=...=y^{(n-1)}=0$ is $0$ (Cauchy Lipschitz).
$f$ est $C^{\infty}$, $f(0)=...=f^{(n-1)}=0$ ...........

In math mode, use \dots or \cdots rather than ...


geometry
incenter
angle bisector
Locus
Locus problems
IMO Shortlist
IMO Longlist

Circle $k$ and its diameter $AB$ are given. Find the locus of the centers of circles inscribed in the triangles having one vertex on $AB$ and two other vertices on $k.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

We have to find the locus when all three points are variable?

In this case, the locus is the entire disc bounded by $k$ (except maybe for the actual circle; that can be reached only when the two points on it coincide). To see this: fix a point $X$ on $AB$ and a line $\ell_X$ through $X$, and make the two points $Y,Z$ on $k$ move so that $\ell_X$ is the angle bisector of $\angle YXZ$. Let $\{P,Q\}=\ell_X\cap k$. When $Y,Z\to P,Q$, we get the incenters $\to P,Q$ respectively, so, by continuity, the entire segment cut by $k$ on $\ell_X$ belongs to the locus.

The only problem could arise when the incenter is $X$. This happens iff $X,Y,Z$ are collinear, so maybe we should disregard $X$, but it can be obtained as the incetner of another triangle constructed in a similar fashion by fixing a point $X'\ne X$ on $AB$, and taking $\ell_{X'}=AB$.


let $(O)$ be a circle and $TA$ ,$TB$ two tangant on $(O)$ and let $N$ the midpoint of $AB$ ,$M$ be the midpoint of  $\frown{BA}(arc)$ and $(l)$ be an arbitary line such that  passes from $M$ and meet $TA$,$TB$,$(O)$ at $D$,$E$,$F$ respectively and $S= AE\cap BD$ prove that $S,N,F$ are coliner

This was solved at http://www.mathlinks.ro/Forum/viewtopic.php?t=28195 .

  darij


your inequality can be changed into
S \leq 3 \sqrt 3 /4 R 2
so it's easy now  :D


prove that (5^125)-1/(5^25)-1 is not prime.


[quote="April"]Let be given a triangle $ABC.$ Prove that: $\frac{r}{R}+\frac{(b-c)^{2}+(c-a)^{2}+(a-b)^{2}}{16R^{2}}\le\frac{1}{2}$[/quote]

after we use: $\sum{ab}=p^{2}+r^{2}+4Rr$ and $\sum{a^{2}}=2p^{2}-2r^{2}-8Rr$ we obtain:

$\frac{r}{R}+\frac{(b-c)^{2}+(c-a)^{2}+(a-b)^{2}}{16R^{2}}\le\frac{1}{2}\Longleftrightarrow \frac{p^{2}-3r^{2}-4Rr}{8R^{2}}\leq \frac{1}{2}$
$\Longleftrightarrow p^{2}\leq 4R(R+r)+3r^{2}$ , which is true because of the well known distance : $IH^{2}=4R(R+r)+3r^{2}-p^{2}$, which of course is $\geq 0$

[quote]$\boxed{IH^{2}=4R(R+r)+3r^{2}-p^{2}}$[/quote]
Sorry for my scupidity. But I don't know the relation. Can you prove this for me? Thank you very much :D

You have to ues equality as following:
$a\vec{IA}+b.\vec{IB}+c.\vec{IC}=\vec{0}$.
The problem can prove by another way that easy to understand. The work will be writed clearly in next my post of this forum.
So i be ready to post in tomorow

[color=blue][b]hien[/b][/color], I'm waiting for your proof. :D

From above inequality, we infer instantly [b]Gerresen's[/b] inequality:
$4R^{2}+3r^{2}+4Rr\geq p^{2}$

Thanks, [b]hien[/b]
And I use: $\begin{cases}a\overrightarrow{IA}+b\overrightarrow{IB}+c\overrightarrow{IC}=0\\ aIA^{2}+bIB^{2}+cIC^{2}=abc\\ aHA^{2}+bHB^{2}+cHC^{2}=4R^{2}(a+b+c)-\left(a^{3}+b^{3}+c^{3}\right)\end{cases}$ to prove the relation:
\[\boxed{IH^{2}=4R^{2}-\frac{a^{3}+b^{3}+c^{3}+abc}{a+b+c}}\]
And you?  :wink:

Yes! this is my approach but we can get it buy using the fact:
the side lengths $a,b,c$ of triangle are roots of equation:
$t^{3}-2pt^{2}+(p^{2}+r^{2}+4Rr)t-4pRr=0$


geometry
AMC
AIME
trigonometry
angle bisector
similar triangles

In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$, and $\overline{AD}$ bisects angle $CAB$. Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$, respectively, so that $AE = 3$ and $AF = 10$. Given that $EB = 9$ and $FC = 27$, find the integer closest to the area of the quadrilateral $DCFG$.

This was the first actual AIME taken in real condition so I wasn't sure when the problem meant..

"find the integer cloest to the area of quadrilateral DCFG."

When I estimated, I got around 148.6 something so I put 149. But the answer is 148. I didn't think I did any calculations wrong other than maybe setting $\sqrt{91}$ (you'll know what I mean if you do this. I used angle bisector theorem on $\triangle AEF$ and $\triangle ABC$ and just subtracting area) as 9.5. But other than that, I don't think I made a huge mistake and clearly, 149 is cloest to 148.6.

Any idea on how to "accurately" solve this problem as for calculation? The problem itself was pretty easy and straightforward even though it was #10 (I thought #10's were way harder..)

P.S. I didn't put any comment on the first post in case moderators want to add this problem to Resources page. I'm not in charge or taking care of AIME parts but just for you to know.  :lol:

I get your method. So, in this type of problem, is it recommended that I try to find the area using $\frac{1}{2} \sin C ab$ rather than by simple $\frac{1}{2} bh$ since this gives you square roots?

I'll have to go back to my paper to see where I screwed up but this made me mad.. (I thought I got #10 then off by 1!)

its a matter of preference.  some people will be able to plow through messy expressions, while others will just be intimidated/disgusted.  i would much rather keep things as simple as possible, especially for these estimation problems.

Obviously using Pythagorus we get $BC=35$, from here, we can do angle bisector theorem, so we have $12x+37x=35$, and so we have $BD=\frac{60}{7}$ and $DC=\frac{185}{7}$, so since we know that $\triangle {ABD}=\frac{360}{7}$ and $\triangle {ABC}=210$, we see that $\triangle {ADC)=\frac{1110}{7}}$, so now all we have to do is find the area of $\triangle {AFE}$, so we can use angle bisector theorem again: $3x+10x=\sqrt{91}$, so $EF=\frac{10\sqrt{91}}{13}$, so now we can use similar triangles, since $\triangle {AEF}$ is similar to $\triangle {ADC}$ we have $\frac{10}{37}=\frac{AE}{\sqrt{217.5}}$.....but by looking at this, I don't think this is a good way at all :(

That's why it was problem for not using what alan did. I did same way as you did but then the wording "integer closet to it.." took me off. But again, if it was floored value, I could've got it right since I had 148.68 which will be 148 by floor function.


I have a test tomorrow and  I badly need help with these problems
http://ucsub.colorado.edu/~yyan/mid3practice.pdf
All the problems on the first page of the pdf file and just pick any problem from the next pages (I only need one becausae I think once I see how they are done Ill be able to figure out the other ones)

The problems are pretty straight foward, so if you could help me out that would be great, my test is tomrrow so plz help quick

Thanks a ton

1a too easy, but if you cannot do 1a, sorry but you will fail test

1b $M'$ means elements not in $M$. Which ones? Now what are elements of $N$? $M'\cap N$ means elements in both $M'$ and $N$

1c $n(U)$ means total elements. $n(A')$ means number of elements not in $A$ Now how many are in $A$? This is $n(A)$. You can do $n(B)$. $n(A'\cup B')$ means number of elements either not in $A$ or not in $B$. But what does this tell you? It means $A'\cup B'$ is same as $(A\cap B)'$ because if an element is either not in $A$ or not in $B$ then it is not in $A\cap B$ (A and B). You can do $n(A\cap B)$ Last, use $n(A\cup B)=n(A)+n(B)-n(A\cap B)$

2a Easy. $0!=1$, $3!=3\cdot 2\cdot 1$, $n!=n\cdot(n-1)!=n(n-1)\cdot(n-2)!=n(n-1)(n-2)\cdot(n-3)!$, and you do simplify

2b What is combinations? You can do this super easy one

3a OK think 5 spaces like _ _ _ _ _. How many choices can you choose first digit? You have {1,2,3,5,6,7,8,9,0} because 4 cannot be first digit. You have 9 choices.
OK now put in 9 in first space like 9 _ _ _ _. How many choices can you choose second digit? You have 9 choices because it cannot be same as first digit.
OK now put in 9 in second space like 9 9 _ _ _. Repeat and you can do problem.

3b If you can do 3a, you can do 3b, same rule, multiplication rule

4a One word solve this problem: combinations!

4b Product of combinations

5 Same method as 4b

6a OK if numbers are less than 600, what should be the hundreds digit? How many choices in tens digit and units digit? What does it mean divisible by 5? What do numbers divisible by 5 look like? Find probability

6b How many face cards in deck? How many non-face cards in deck? 4b can help you here. Then what is 8 cards chosen from a deck of 52?(Hint: combinations). Then find probability.

6c What is expected value formula? $\sum X\times P(X)$

7 & 8 Apply correctly expected value formula and you will get answer

[hide="All number answers to check if you got 100 I did"]1a B

1b {2,4}

1c 71, 64, 98, 37

2a dakyru's answer

2b $n$

3a $9^5$

3b $6\times 5\times 2\times 5$

4a $\binom{10}{4}$ or 10 choose 4

4b $\binom{11}{5}\times\binom{10}{2}$

5 $\binom{9}{1}\times\binom{8}{4}$

6a $\frac{2\times 5\times 1}{2\times 5\times 5}$

6b $\frac{\binom{12}{5}\times\binom{40}{3}}{\binom{52}{8}}$

6c 0.62 $\sum X_ip_i$

7 $\$14 $C\times\frac{1}{8}+(-2)\times\frac{7}{8}=0$
8 $7\times\left(\frac{5^3}{6^3}\right)+(-10)\times\left(1-\frac{5^3}{6^3}\right)$[/hide]

I was going to post more, but it seems 10000th User has pretty much covered everything.  Just remember:

$\displaystyle{n\choose k} = \frac{n!}{k! \cdot (n-k)!}$

That should help on some of the ones involving combinations.

[quote="chess64"]$\{n\} \subset \{k\}$ means that $\{n\}$ is a subset of $\{k\}$?[/quote]

Yes.


AMC AIME probability geometry rectangle rotation perimeter

1.  v, w are roots of z^1997 = 1 chosen at random. Find the probability that |v + w| >= √(2 + √3).
2.  Find the area of the largest equilateral triangle that can be inscribed in a rectangle with sides 10 and 11.
3. An a x b x c box has half the volume of an (a+2) x (b+2) x (c+2) box, where a ≤ b ≤ c. What is the largest possible c?
4.  R is a 6 x 8 rectangle. R' is another rectangle with one vertex on each side of R. R' can be rotated slightly and still remain within R. Find the smallest perimeter that R' can have.
5.  The triangle ABC has AB = 1995, BC = 1993, CA = 1994. CX is an altitude. Find the distance between the points at which the incircles of ACX and BCX touch CX.
6.  A 150 x 324 x 375 block is made up of unit cubes. Find the number of cubes whose interior is cut by a long diagonal of the block.
7.  ABCD is a parallelogram. ∠BAC = ∠CBD = 2 ∠DBA. Find ∠ACB/∠AOB, where O is the intersection of the diagonals.

Problem 5
Not entirely sure, the numbers seem kind of messy so I probably did something wrong...

[hide]
You can use Heron's formula to find out the area of the $\bigtriangleup ABC$, which in this case is $\sqrt{2991\times998\times997\times996}$. Divide that by length $AB$ to find the altitude $CX$.We can use Pythagorean's theorem to find out the lenghts of the line segments $AX$ and $BX$. We can then use the fact that the area of a triangle equals $\frac{1}{2}$ the perimeter $\times$ the radius of the inscribed circle to find out the length of the radius of each of the two circles. The distance between the points where each circle touches $CX$ should be the difference in the lengths of their radii.

[/hide]

[quote="Max300"]I like how this topic is titled "Aime easy problems" when all of the problems you posted are like number 14's and 15's on actual AIME's.   ;)[/quote]Haha, I had that exact thought.  :roll:

I agree... and honestly I was disappointed since I need to practice the easier ones to make sure I have my basics down (read: practice preventing stupid mistakes)

Number 5 though, although it's a number 15, is very easy actually. Use Heron's formula and $rs=A$ but to make it very easy, replace the numbers with $x$, $x+1$, $x+2$ not to get bogged down with the numbers.

Remember we are just looking at the difference of their radii  :)

[quote="scorpius119"]These aren't the most difficult of 14's and 15's, but EASY PROBLEMS?!?!?!??!?!?!?!?!?!?!?!!!?!??!?!??!?!?!??!??!?

[hide="1"]
Let $\theta=\frac{2\pi}{1997}$ and $\omega=e^{\theta i}$. Since $|\omega^j|=|\omega|^j=1$, $|v+w|=|1+\omega^k|$ for some positive integer $k<1997$.

$|1+\omega^k|^2\geq 2+\sqrt{3}\Rightarrow 2+2\cos k\theta \geq 2+\sqrt{3}\Rightarrow \cos k\theta \geq \frac{\sqrt{3}}{2}$.

If $k\theta$ is a first-quadrant angle, then $k\theta \leq \frac{\pi}{6}$. Isolating the variable, $k\leq \frac{1997}{12}$. There are 166 positive values of $k$ that satisfy this.

If $k\theta$ is a fourth-quadrant angle, then $(1997-k)\theta$ is a first-quadrant angle, and both or neither of these will satisfy the inequality (because the cosines are equal). So there are 166 more solutions for a total of 332.

Since there are 1996 integers $k$ to choose from, the probability is $\boxed{\frac{83}{499}}$.
[/hide][/quote]
14s and 15s shoudn't be scary. They are like 1's, but with more math.


geometry
circumcircle
trigonometry
incenter
cyclic quadrilateral
angle bisector
geometry unsolved

$a$, $b$, $c$ are the sides of a triangle, such that $a+b=2c$. Prove that the centres of the inscribed and circumscribed circles are concyclic with the midpoints of the sides $a$ and $b$.

Let $M,N$ midpoints of sides $CA$ and $CB$ respectively and $CD$ be internal bisector of $C$ and $E$ be the intersection of $CO$ and the circumcircle of $ABC$ . assume that $CD$ intersect circumcircle of $ABC$ at $T$ . we know that $CB=2CN$ , $CA=2CM$ , $CE=2CO$ , and if we prove $CT=2CI$ then
$I,C,M,N,O$ are on the same circle . we know that $CT=\frac{a+b}{2\cos{C/2}}$ and $CI=\frac{ab\cos{C/2}}{p}(p=\frac{a+b+c}{2})$
now $a+b=2c$ and $p=\frac{3c}{2}$ . we must prove that : $4ab\cos^{2}{C/2}=3c^{2}$ but we have : $2\cos^{2}{C/2}=1+\cos{C}$ , $2ab\cos{C}=a^{2}+b^{2}-c^{2}\longrightarrow 2ab+a^{2}+b^{2}=4c^{2}\Longrightarrow (a+b)^{2}=4c^{2}$ which is true :)

[quote="khashi70"]Let $M,N$ midpoints of sides $CA$ and $CB$ respectively and $CD$ be internal bisector of $C$ and $E$ be the intersection of $CO$ and the circumcircle of $ABC$ . assume that $AD$ intersect circumcircle of $ABC$ at $T$ . we know that $CB=2CN$ , $CA=2CM$ , $CE=2CO$ , and if we prove $CT=2CI$ then
$I,C,M,N,O$ are on the same circle . we know that $CT=\frac{a+b}{2\cos{C/2}}$ and $CI=\frac{ab\cos{C/2}}{p}(p=\frac{a+b+c}{2})$
now $a+b=2c$ and $p=\frac{3c}{2}$ . we must prove that : $4ab\cos^{2}{C/2}=3c^{2}$ but we have : $2\cos^{2}{C/2}=1+\cos{C}$ , $2ab\cos{C}=a^{2}+b^{2}-c^{2}\longrightarrow 2ab+a^{2}+b^{2}=4c^{2}\Longrightarrow (a+b)^{2}=4c^{2}$ which is true :)[/quote]

Sorry, but could you explain the used formulas ($CT=\frac{a+b}{2\cos{C/2}}$ and $CI=\frac{ab\cos{C/2}}{p}$). (I mean the proof or the name of the theorem(s) you used.)

And some notes:
Let $I$ be the centre of the inscribed circle (and $O$ be the centre of the circumscribed circle).
Assume that $CD$ intersect circumcircle of $ABC$ at $T$. (Not $AD$.)

Yes i can explain :) . assume that $CD$ be internal bisector then by the sin law :
$\frac{AD}{\sin{C/2}}=\frac{CD}{\sin{A}}=\frac{AC}{\sin{D}}$ and $\frac{BD}{\sin{C/2}}=\frac{CD}{\sin{B}}=\frac{BC}{\sin{180-D}}$
now we get $\frac{AD}{BD}=\frac{\sin{B}}{\sin{A}}=\frac{AC}{BC}=\frac{b}{a}\longrightarrow AD=\frac{bc}{a+b}, BD=\frac{ac}{a+b}$
put in the first sin law we get that :
$\frac{AD}{\sin{C/2}}=\frac{CD}{\sin{A}}\longrightarrow CD=\frac{AD\sin{A}}{\sin{C/2}}=\frac{bc\sin{A}}{(a+b)\sin{C/2}}=\frac{ab\sin{C}}{(a+b)\sin{C/2}}\Longrightarrow CD=\frac{2ab\cos{C/2}}{a+b}$
 . now $BI$ is the internal bisector of $\angle ABC$ so we have :
$\frac{CI}{ID}=\frac{CB}{BD}=\frac{a+b}{c}\longrightarrow \frac{CI}{CD}=\frac{a+b}{a+b+c}$. now we use $CD=\frac{2ab\cos{C/2}}{a+b}$ and we get :
$CI=\frac{ab\cos{C/2}}{p}$ . now draw two perpendicular $TL$ and $TK$ from $T$ to $AC$ and $BC$ respectively . we can easily prove that :
$CL=CK$ and $AL=BK$ and we get that $CL=CK=\frac{a+b}{2}\Longrightarrow CT=\frac{CL}{\sin{(90-C/2)}}=\frac{a+b}{2\cos{C/2}}$

Let $M,N$ midpoints of sides $CA$ and $CB$ respectively
Let $w$ be te circumcircle of $\triangle CMN$. Clearly $O\in w$

Let $D\in AB$ such that $AD=AM$. Then $BD=AB-AD=BN$.

So the triangles $ADM,BDN$ are isosceles. $I$ should belong to the perpendicular bisectors of their bases $MD,ND$.
Hence $I$ is the circumcenter of $\triangle MND$

From $\angle MDA = 90-\frac{A}{2}$ and $\angle NDB = 90-\frac{B}{2}$ we get

$\angle MDN = 180-(\angle MDA+\angle NDB) = \frac{A+B}{2}$

$\angle MIN = 2\angle MDN = A+B=180-C$, hence $I\in w$

[color=green][u]Note[/u]
$\frac{AC}{AD}= \frac{BC}{BD}= 2\Rightarrow AD$ is the internal bisector of $C$[/color]

Denote by $O$ circumcenter and by $I$ incenter of triangle $\triangle ABC$.

Extend $CI$ to meet circumcircle of triangle $ABC$ at point $L$. Then it is well known that $LI=LB=LA$.

Applying Ptolomeys equation to cyclic quadrilateral $ALBC$ we get:

$AL.a+BL.b=CL.c$ or equivalently $2AL(a+b)=2CL.c$.

After dividing both sides by $2c=a+b$ we get:

$2AL=CL=CI+IL$ and since $IL=LA$ it follows that $AL=C$ or $LI=CI$ so $I$ is midpoint of $CL$.

Since $OC=OL$ it follows that $OI$ is perpendicular to $Cl$ or $\angle OIC=90$.

If we denote midpoints of $AC$ and $BC$ with $M$ and $N$,resp.,then it is obvious that angles $\angle OMC$ and $\angle ONC$ are right.
Now we can easily conclude that circle with diameter $CO$ contains points $M,O,I$ and $N$. (also $C$).

NOTE:Almost same question was given in St.Petersburg city M.O. There, problem was to prove that $\angle OIC$ is right if and only if $a+b=2c$.
This version of problem was given in one regional competition in Bosnia and Herzegovina.

First some notations: $M_{a}$ midpoint of $a$ and $M_{b}$ midpoint of $b$. $CI$ intersects $M_{a}M_{b}$ at $J$ then using bisector theorem we get: $M_{a}J = \frac{a}{4}$ and $M_{b}J= \frac{b}{4}$.

Let $K$ be the intersection of the angle bisector of $C$ and $AB$: $\frac{CI}{IK}=2$ and $CJ=JK$ then $CJ=\frac{m}{2}$, $JI=\frac{m}{6}$, $IK= \frac{m}{3}$.

Also: $(CK)^{2}=(CB)(CA)-(AK)(BK)$ $\Rightarrow$ $m^{2}=ab-\frac{ab}{4}=\frac{3m}{4}$ $\Rightarrow$ $(M_{a}J)(JM_{b})=(CJ)(JI)$ $\Rightarrow$ $M_{a}, C, M_{b}, I, O$ $are$ $concyclic$

see the 8th post from top [url=http://www.mathlinks.ro/viewtopic.php?p=424169#424169]here(INMO 2006,1)[/url]


conics hyperbola Asymptote geometry circumcircle parabola ellipse

There is a line $l$ in a plane and triangle $ABC$ in plane.
If $K$ moves on line $l$ what is the locus of isogonal conjugate of $K$?

Oh I forgot to say if $AB<>AC$ and $l$ is perpendicular bisector of $BC$ I have solved the problem wants points $P$ that $<ABP=<ACP$
I proved the locus is a hyperbola passing through $A,B,C,H,A',A\"$(H orthocenter)(A' is apoint that A'BAC is parrallelogram)(A" is a point that A"BC=A"CB=90)
and it's asymptotes are line parrallell to internal bisectors and external bisectors of $BAC$ that are drawn from midpoint of BC.(On the figure M is midpoint of BC)

Oh, but it's always a circumconic conic, man. I don't think it always turns out to be a hyperbola, but it's definitely a circumconic.

Since $K$ moves on a line, it means that the map $BK\to CK$ is projective, so $BK'\to CK'$ is also projective ($K'$ is the isogonal conjugate). This means that the locus of the intersection of $BK',CK'$ (which is $K'$, of course) is a conic passing through $B,C$ (well, if the line is parallel to $CA$, for example, the point never actually reaches $B$, but it comes arbitrarily close).

I have drawn dynamic sketches in which the locus is a hyperbola and some in which the locus is an ellipse. Moreover, the locus is the circumcircle when the line is the line at infinity. So in general, the answer is: a circumconic. Besides, you could always employ the ugly stuff: determinants and barycentrics and all that, in case you're not convinced :).

[quote="grobber"]I have drawn dynamic sketches in which the locus is a hyperbola and some in which the locus is an ellipse.[/quote]

Just a note about this: The locus is

- a hyperbola if and only if the line l intersects the circumcircle of triangle ABC;
- a parabola if and only if the line l touches the circumcircle of triangle ABC;
- an ellipse if and only if the line l doesn't meet the circumcircle of triangle ABC.

The explanation of this all is that a conic is a hyperbola, a parabola, or an ellipse if and only if it meets the line at infinity at two, one, or no points, respectively. Now, since our circumconic is the isogonal conjugate of the line l, while the line at infinity is the isogonal conjugate of the circumcircle, the number of common points of the circumconic with the line at infinity equals to the number of common points of the line l with the circumcircle.

  Darij


Situation:
Fishermen/farmers produce food, workers produce non-food items such as shirts, and poets and artists produce works of art.

However, of all the food that the farmers produce, they also consume 7/13 of it, 2/10 of the non-food items, and 1/10 of the work of arts.

Similarly, workers, poets and artists’ consumption are shown on the table.

Question, how are you going to price the items so that there is equilibrium? That is, the income of each group is equal to their expenditures.

The associated Markov chain has as matrix: $ A\equal{}\begin{pmatrix}7/13&5/13&1/13\\2/10&3/10&5/10\\1/10&4/10&5/10\end{pmatrix}$ where $ m_{i,j}$ is the probability of the transit $ i\rightarrow{j}$ and $ 1\equal{}farmer,2\equal{}worker,3\equal{}artist$.
The equilibrium is given by $ \lim_{n\rightarrow\infty}A^n\equal{}\begin{pmatrix}l&m&n\\l&m&n\\l&m&n\end{pmatrix}$ where $ l\approx0.2422,m\approx0.3602,n\approx0.3975$ and $ l\plus{}m\plus{}n\equal{}1$.
Let $ a,b,c$ be the prices of unit of food,non food,picture; $ a,b,c$ are defined up a factor.
It remains to solve the system in $ a,b,c$: $ (1\minus{}l)a\equal{}lb\plus{}lc,(1\minus{}m)b\equal{}ma\plus{}mc,(1\minus{}n)c\equal{}na\plus{}nb$; fortunately the matrix of this system is non invertible.   QED


algebra
polynomial
calculus
integration
quadratics
number theory
algebra unsolved

Show that there do not exist polynomials $p(x)$ and $q(x)$ each having integer coefficients and of degree greater than or equal to 1 such that \[ p(x)q(x) = x^5 +2x +1 . \]

Well, another idea is to consider the polynomial modulo some prime...

Well, I am thinking that 3 and 5 are good for this...
$x^5+2x+1 \equiv 3x+1 (mod\ 5)$

Well, I am thinking that if we can prove that the factorization is unique mod 5, the problem finishes...

[quote="tµtµ"]A far from smart, but working, solution (a clever one may use Eisenstein ??) may be :

Since $P(x) = x^5 +2x + 1$ has no integer root (which should be $\pm 1$) we would have, wlog, $degree(p) = 2$ and $degree(q) = 3$

$p$ and $q$ may be written (by taking$-p$ and $-q$ if needed) :
 $\left ({x}^{2}+bx+c\right )$ and $\left ({x}^{3}+m{x}^{2}+nx+o\right )$

Expanding $\left ({x}^{2}+bx+c\right )\left ({x}^{3}+m{x}^{2}+nx+o\right )$ gives
$c = o = \pm 1$, $m = -b$, $m^2-n = -c$, $m-n = 2$ which has no solution.[/quote]

That's right but the text says $deg(p),deg(q)\geq1$ so you have to check the other case I think.

[quote="rushil"]Show that there do not exist polynomials $ p(x)$ and $ q(x)$ each having integer coefficients and of degree greater than or equal to $ 1$ such that $ p(x)q(x) \equal{} x^5 \plus{}2x \plus{}1$ .[/quote]

Ummm......From my source, the INMO-1999 Question no. 3 is
Show that there do not exist polynomials $ p(x)$ and $ q(x)$ each having integer coefficients and of degree greater than or equal to $ 1$ such that $ p(x)q(x) \equal{} x^3 \plus{}2x \plus{}1$ .

Note that it's a cubic instead of a fifth power equation........
Can someone confirm my doubt?

No, it is a fifth degree polynomial, not a cubic. As for the solution, P(4) = 1033, a prime. Hence by Cohn's criterion P is irreducible

Obviously, $x^5+2x+1$ has no integral roots.  Therefore, a linear factor is ruled out.   If $x^5+2x+1$ is reducible over $\mathbb{Q}$, then it has a quadratic factor $x^2+px+q \in \mathbb{Z}[x]$.  Now, $x^5+2x+1$ divided by $x^2+px+q$ gives the remainder \[0=\left(p^4-3p^2q+q^2+2\right)x+\left(p^3q-2pq^2+1\right)\,.\]

We now work in $\mathbb{F}_5$.  As \[p^3q-2pq^2+1=0\,,\,\,\,\,\,(*)\] we have $p \neq 0$, $q \neq 0$, and $p ^2 \neq 2q$.  However, according to Fermat's Little Theorem, we have $p^4=1$, so $0=p^4-3p^2q+q^2+2=1-3p^2q+q^2+2$.  That is, \[q\left(q-3p^2\right)=-3\,.\,\,\,\,\,(**)\]  Combining (*) and (**), we get \[-1=pq\left(p^2-2q\right) = -2pq\left(q-3p^2\right) = -2p(-3)=p\,.\]  Plugging $p=-1$ into (**) yields $(2q-3)^2+3=4\left(q^2-3q+3\right)=0$.  This is a contradiction since $-3$ is a quadratic nonresidue modulo $5$.  So, $x^5+2x+1$ is irreducible over $\mathbb{Q}$, and hence, over $\mathbb{Z}$.

[quote=pco][quote=Bubu-Droid]What's wrong in my solution pls tell:

On the contrary lets say there exists two polynomials $p(x)$ and $q(x)$ with integer coefficients such that $p(x)q(x) = x^5 +2x +1$.

Then let $p(x)q(x) = z(x) = x^5 + 2x + 1$, then $z(x)$ also has integer coefficients.

The polynomial $z(x)$ has no root in integers(which should be $\pm 1$).

So $z(x)$ has root in rational, so the coefficients are also not of integers, which leads us to a contradiction that $p(x)$ and $q(x)$ do not exist.[/quote]
Exactly the same proof could be applied to $x^4+2x^2+1$

But $x^4+2x^2+1=(x^2+1)(x^2+1)$
So just backtrace your proof in order to get your error :)[/quote]

ohk bro tysmm got it broo! should i delete my wrong solution?


\textnormal{Problem 3:}

\textnormal{Problem:}
\left(\frac{-2a^{4}b}{a^{-3}b^{2}}\right)^{-3}

\textnormal{Which I turned into:}
\left(\frac{a^{-3}b^{2}}{-2a^{4}b}\right)^{3}

\textnormal{The answer I have is:}
\left(\frac{b^{3}}{-8a^{21}}\right)

\textnormal{The book says:}
\left(\frac{-b^{3}}{8a^{21}}\right)

\textnormal{I don't understand why the negative sign is infront of the character 'b' instead of the eight.}

\frac{\left(3x^{2}\right)^{-2}}{x^{3}x^{-4}x^{0}}

\textnormal{What I got:}
\frac{-9}}{x^{3}}

it starts breaking up in improper places.. even though the code is right..

[img]http://www.mathlinks.ro/Forum/latexrender/pictures/377ce70324709e4a8cd72aefb0c23434.gif[/img]

i don't even understand how to make the latex code show.. that's a /img tag i used

maybe someone could help you in the $\LaTeX$ forum

$\textnormal{Problem 3:}
\textnormal{Problem:}
\left(\frac{-2a^{4}b}{a^{-3}b^{2}}\right)^{-3}
\textnormal{Which I turned into:}
\left(\frac{a^{-3}b^{2}}{-2a^{4}b}\right)^{3}
\textnormal{The answer I have is:}
\left(\frac{b^{3}}{-8a^{21}}\right)
\textnormal{The book says:}
\left(\frac{-b^{3}}{8a^{21}}\right)
\textnormal{I don't understand why the negative sign is infront of the character 'b' instead of the eight.}
\frac{\left(3x^{2}\right)^{-2}}{x^{3}x^{-4}x^{0}}
\textnormal{What I got:}
\frac{-9}}{x^{3}} $
To get the $\text{\LaTeX}$ to show, just put $\$ around it, like
$\$x^2+5x+6=0$\$ would be the code for
$x^2+5x+6=0$.
Does that make sense?
BTW, it doesn't matter where you put the negative sign in a fraction. Like
$-\frac{a}{b}=\frac{-a}{b}=\frac{a}{-b}$

[quote]\textnormal{I don't understand why the negative sign is infront of the character 'b' instead of the eight.}[/quote]
Because you put it there!
\left(\frac{$\boxed{-}$b^{3}}{8a^{21}}\right)

[quote]it starts breaking up in improper places.. even though the code is right..[/quote]

The code isn't correct, you put an extra "}"

\frac{-9}$\boxed{\}}${x^{3}}

And here are some tips:

You don't need to write so many \right and \left just write the stuff out

And, instead of posting it as an image you just put dollar signs (shift+4) around your code. i.e. take this line of your code:
\left(\frac{a^{-3}b^{2}}{-2a^{4}b}\right)^{3}
And put dollar signs around each side and get:
$\left(\frac{a^{-3}b^{2}}{-2a^{4}b}\right)^{3}$


1. What is the greatest possible two-digit number for the sum of the first $ n$ positive integers?

2. What is the remainder of $ 8^{6}\plus{}7^{7}\plus{}6^{8}$ divided by $ 5$?

3. What positive integer value of $ x$ will satisfy the equation $ (x^{2}\minus{}6x\plus{}9)^{4}\equal{}5^{16}$?

Thanks.

[hide="#1"]
as you see on the side pane(maybe) the sum of the first n positive integeres is $ \frac {n(n \plus{} 1)}{2}$
now, you find the maximum value of this less than 100 and you get 91...
$ \frac{13(13 \plus{} 1)}{2} \equal{} 91$[/hide]
[hide="#2"]
$ 8^6$ would have the same remainder with respect to $ 5$ as $ 3^6$
$ 7^7$ would have the same remainder with respect to $ 5$ as $ 2^7$
and $ 6^8$ would have the same remainder with respect to $ 5$ as $ 1^8$ or $ 1$
if you know the pettern for remainders with respect to $ 5$, you know it repeats every 4 terms so...
$ 3^6$ has the same remainder with respect to $ 5$ as $ 3^2$ and this is easily computed to be 4
$ 2^7$ has the same remainder with respect to $ 5$ as $ 2^3$ and this is easily computed to be 3

then $ 4 \plus{} 3 \plus{} 1 \equal{} 8$
[/hide]

1. Use the equation n(n+1)/2<100
n(n+1)<200
here you can easily do it out from guess and check.
It is 13 which inserted into the first equation gets 91.

2. Just find the units digits of each and then add them up and divide that by 5.

8^1=8
8^2=4
8^3=2
8^4=6
8^5=8 It repeats every 4 so 8^6=4

7^1=7
7^2=9
7^3=3
7^4=1
7^5=7 It repeats every 4 so 7^7=3

6^1=6
6^2=6.....

It is always 6, so 6^8=6

6+3+4=13

13/5 has a remainder of 3.

3. Notice that to the fourth is a fourth of 16. So x^2-6x+9 has to equal the quartic root of 5, or 625.
x^2-6x+9=625
x^2-6x-616=0
(x-28)(x+22)=0
Two solutions are 28 and -22, the positive solution is 28.

Edit: you beat me to it steven :mad:  :)

ahh but i only did that by reserving *lol*  :lol:

[quote="Ihatepie"]steven, isn't the first one 91? 1-13.

I'm probably completely wrong XD[/quote]

The answer key says 91.  However, I don't get why it's 91 instead of 13.  (I got 13.)

Thanks.

omg, I beat steven and he got it wrong  :o
 :D

Well the reason that it is 91 and not 13 is because it says what is the greatest possible sum of the numbers. So you add up 1-13 to get 91.

OK OK OK ITS 91 AT LEAST GIVE ME A CHANCE TO EDIT

Re: stevenmeow
 :o you actually made [b]two[/b] mistakes? And didn't answer #3?  :o

Edit 2: lol you said 8 when we were dealing with mod 5
Edit 3: renumbered my edit :oops:

Re: Ihatepie
[quote="Ihatepie"]
So x^2-6x+9 has to equal the quartic root of 5, or 625.
[/quote]
Not quartic root...you mixed it up. Actually, you could factor the perfect square trinomial.
(x-3)(x-3) = $ \pm$ 625
(x-3)= $ \pm$ 25       or      (x-3) = $ \pm$ 35i (discard this one)
x=28 or x= -22


Between every two digits of the number 14641 k zeros are inserted. Prove that the obtained numbers for different k>=1(e.e. 104060401, 1004006004001, 10004000600040001,etc.) are all perfect squares.

[hide="Proof"]\begin{eqnarray*}a^4+4a^3+6a^2+4a+1 &=& (a^4+2a^3+a^2)+(2a^3+4a^2+2a)+(a^2+2a+1)\\ &=& (a^2+2a+1)(a^2+2a+1)\\ &=& (a+1)^4\end{eqnarray*}

The numbers in question are obtained by putting $ a=10, 100, 1000,...$ into the above expression. Moreover, thus is proven they all are not only squares but perfect fourth powers.[/hide]


[b]Let[/b] $n\in N,\ n\ge 2,$ [b]the ecuation[/b] $\sqrt[n]{x+1}-\sqrt[n]{x}=\frac{1}{n},\ x\in [0,1].$ [b]Prove that the ecuation has a only zero[/b]
$x_n\in (0,1)$ [b]and[/b] $\lim_{n\to\infty} x_n=\frac {1}{e-1}.$

$f(x) = (x+1)^\frac{1}{n} - x^\frac{1}{n}$ is decreasing strictly on $[0,1]$. $f(0)= 1, f(1) = 2^\frac{1}{n}-1<\frac{1}{n}$. The uniqueness follows easily.

You solved the simple part. But limit ??? This is the problem !

We can show that $\frac{1}{\left(1+\frac{1}{n-1}\right)^n-1}\leq x_n\leq \frac{1}{\left(1+\frac{1}{n}\right)^n-1}$. Try it yourself.

I am sorry, I don't se the last your post.

[quote="levi"][b]Let[/b] $n\in N,\ n\ge 2,$ [b]the ecuation[/b] $\sqrt[n]{x+1}-\sqrt[n]{x}=\frac{1}{n},\ x\in [0,1].$ [b]Prove that the ecuation has a only zero[/b]
$x_n\in (0,1)$ [b]and[/b] $\lim_{x\to\infty} x_n=\frac {1}{e-1}.$[/quote]

Is the limit x tending to infinity or n tending to ?

[quote="levi"]Sorry, it is evidently, $n\rightarrow \infty$ This affair hinder you to solve ?[/quote]

Of course not, but your problems are some times too specific about certain conditions and methods.  So I thought I should get it clear from you.


Let ABC be a triangle. Prove that:

3/2*S \geq r* \sum a*sinB*sinC

We write a*sinB*sinC as absinC*sinB/b=2S/2R. We do the same for the others and we have (3/2)*S>=3r/R, which is true because R>=2r.


floor function
induction
algebra unsolved
algebra

Find great positive integer $k$ such that $2^{k}|\lfloor{(1+\sqrt{3})^{2n}}\rfloor+1$  :)


ratio
geometry
function
Mafia
email
geometric transformation
USAMTS

[quote="perfectnumber628"]this could be like, the reverse of "answer a question, ask a question".  the first person posts an answer, then the next person posts a question that has that answer.  (we could get really creative with this...) for example:

Answer: a large furry animal, native to china, that eats bamboo

and so the next person would post something like:

Question: What is your history teacher like?  :lol:

and after you post a question, post another answer for the next person.

i'll start:

Answer: chaotic noise and disorder[/quote]
I found this thread a while ago. It's pretty funny. We should start it again. I'll start this time.

Saturn's rings

What did you throw at the prime minister?

101 dalmations

What did you spend your entire English class drawing in your notebook?

Flouride

What did you dump an entire box on your brother's head?

Atop an asteroid in outer space

What will nutz say if we ask him to recite the digits of $e$?

a

This is the first letter of the alphabet?

Penguin

What is written in Xantos C. Guin's signature? lol

Apple

What did you slice, mix with sugar, top with dough, bake, bury, and say "In a few months I will have an apple pie tree!"

"Time to jump off the Brooklyn bridge!"

What did you say when all of your fellow citizens say when they were going to lynch you?

whoops, sorry.

Who did you give antibiotics to?

My pet fish

In times of extreme desperation and famine, who does one turn to for food?

Microsoft

What's the sloppiest programmer's job related to?

werecats

What's the answer to question #163 on my test?

Chicken.

What's my brain made of?

stupid school project

what do you like to do for fun?

tomatoes

What ball did I use for the tennis match?

w00t

What did you say 10 times in your speech?

bananas

What does Gwen Stefani know how to spell?

Moo!

how do you scare away annoying people?

insanity

What did you at 2:00 AM yesterday?

Entropy

What is my middle name?

Alfred Hitchcock

what name did i misread as adolf hitler?

acops

What book caused me to jump out a third story window in frustration?

Floop.


geometry
vector
function
integration
3D geometry
calculus
calculus computations

[hide="Volume"]$ \text{Find the volume generated when the area between the curve }$ $ y^2 \equal{} x \plus{} 4$ $ \text{ and the line sharing its x and y intercepts is revolved about the x axis.}$[/hide]
[hide="Velocity"]Given that the path of a particle is described by \[ y(t)\equal{}\frac{2}{3}t^3\] \[ x(t)\equal{}t^2\minus{}3\] Find the magnitude of the velocity vector when $ t\equal{}5$[/hide]

 :D [/b]

Show us your setup. What function do you integrate, and from where to where? If you do that, someone here can probably spot the mistake.

$ V \equal{} \displaystyle \pi\int^{0}_{\minus{}4}\left[(\sqrt {x \plus{} 4}^2 \minus{} \left(\frac {x}{2} \plus{} 2\right)^2\right]\,dx$
The second one I just thought was a good question

$ V\equal{}\int_{\minus{}4}^0 \pi y^2\ dx\equal{}\pi \int_{\minus{}4}^0 (x\plus{}4)\ dx\equal{}8\pi$.

Yeah, I just used washers.

$ \displaystyle \minus{}\pi \int^0_{\minus{}4}y^2\,dx\equal{}\minus{}\pi\int^0_{\minus{}4}\left[\frac{x^2}{4}\plus{}2x\plus{}4\right]\,dx\equal{}V_2$

$ V_2\equal{}\minus{}\pi\left(\frac{x^3}{12}\plus{}x^2\plus{}4x\right|^0_{\minus{}4}\equal{}\minus{}\frac{16\pi}{3}$

$ V_1\plus{}V_2\equal{}V_T\equal{}\frac{8\pi}{3}$

Maybe I just can't add?

Or perhaps a little bit simpler, we could have subtracted the volume of a cone with a radius of two and height of four from the first integral.

If I did that on the BC test, would I bee docked? And in general, if you make a note and use a geometric equation, what would the scorer's reaction be on the extended response?

[quote="igiul"]If I did that on the BC test, would I be docked? And in general, if you make a note and use a geometric equation, what would the scorer's reaction be on the extended response?[/quote]

I'm sure it would be fine as long as it was clear what you were doing.


Let ABC be a triangle inscribed in (O; R) circle and let (I; r) be a circle inscribed in ABC triangle.
Prove that: $ OI^{2}\equal{} R^{2}\minus{}2Rr$.

Thank for help me soon. Wish you good luck and happy.

http://www.mathlinks.ro/Forum/viewtopic.php?t=45138

  Darij


search
inequalities
inequalities unsolved

[u][b]The author of this posting is : harazi[/b][/u]
____________________________________________________________________

I do not find it so difficult. Vasc has already found much stronger results and something tells me that this problem has already been discussed.

Oh,yes
And now there is a diffirent ineq:

Show that:
$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}} \le \frac{5}{4}\sqrt{a+b+c}$

I solved that one very long time ago. Posted by Fedor Becharev, firstly I gave a wrong solution and after that Namdung said it is correct. Search for it.

Thank you very much if you tell me where the topic is?

Solved problems section. Search for author Fedor Becharev and something like Lovely Crux inequality;

http://www.mathlinks.ro/Forum/viewtopic.php?t=1459

  darij

[quote="Fedor Petrov"]I think, Fedor is Bakharev, not Becharev.[/quote]

Anyway, what happened to Fedor Bakharev ? I have not seen him on the forum for some time now.


Let $p$ and $p+2$ be prime. Prove that there is $n \in \mathbb{N}$, which satisfies $n = k*(k+1)*...*(k+a)$ (for some natural numbers, $a,k \geqslant 1$) and a remainder $r$ of $\frac{n}{p(p+2)}$ is $r = p^2+p-1$

you say that the remainder obtained by dividing $n$ to $p(p+2)$ is $p^2+p-1$????

If we find $n=1(\mod p+2), n=-1(\mod p)$ then we are done. Let $p(x)=\prod_{i=1}^{p-1}(px+i)$. By Wilson theorem $(p-1)!=-1(\mod p)$, so $p(x)=0(\mod p)$, for every $x$. $(p,p+2)=1$ so we can find $x_0,y_0\in \mathbb{Z}$ such that $px_0-(p+2)y_0=1$. First assume they are both positive. So $f(x_0)=\prod_{i=2}^{p}((p+2)y_0+i)=i!(\mod p+2)=\frac {-1}{p+1}=1(\mod p+2)$. So we found $x_0p+1,p-2$ with that property. In fact we just determine those  $x_0,y_0$. $y_0=\frac {kp-1}2, x_0=y_0+k$, with $k$ odd satisfy, and of course they are positive.


How do I simplify the following?

$\sqrt[3]{5 + 2\sqrt{13}} + \sqrt[3]{5 - 2\sqrt{13}}$

     -- Pawel

[hide="click text"] Set the expression equal to x. Then cube the expression. Simplfiy, and look for where you can replace with another x. Then solve the resulting equation. [/hide]

I would just like to add that I've seen this problem in many books and on many contests.  I would suggest learning how to do it.  The solution method is brilliant.

This problem is certainly beyond the realm of introductory problems.  I'll move it to the Intermediate Forum.

Please remember that the Introductory forum is not for heavier competition problems.  The first 20 or so AMC 12 problems are fine for the Introductory forum, but anything harder should be posted in Intermediate.

[quote="fuzzylogic"]
\and $ab=-3$.
[/quote]
maybe im being really stupid/lazy but how do you multiple them easily?
Did you just split the cube roots and multiply (x+y)(x-y)?

[quote="henryyangrui"]
The problem is addition, not multiplication. ($\sqrt[3]{5+2\sqrt{13}}*\sqrt[3]{5-2\sqrt{13}}$)[/quote]
no he was just answering my question as to how to multiply them.. im pretty stupid

and thats a nice way paladin


induction
number theory unsolved
number theory

Prove that there doesn't exist two positive integers $ x\ne y$ satisfying $ x^{2007} \plus{} y! \equal{} y^{2007} \plus{} x!$ where $ n! \equal{} 1\cdot 2\cdots n$

Suppose $ x>y$  so $ x^{2007}\minus{}x!>y^{2007}\minus{}y!$

To prove that is the same to prove that $ x^{2007}\minus{}x!>(x\minus{}1)^{2007}\minus{}(x\minus{}1)!$  and is sufficient to just prove that $ x^{2007}>x!$

So we use induction.

Suppose $ (x\minus{}1)^{2007}>(x\minus{}1)!$. Multiply both sides by $ x$.
$ x (x\minus{}1)^{2007}>x (x\minus{}1)!\equal{}x!$
$ x^{2007} >(x\minus{}1)^{2007} x>x!$

And we start at $ x\equal{}2$ and is true for all $ x>1$
Follows that (multiplying both sides of the second equation by $ x$)

$ x^{2008}\minus{}(x\plus{}1)!>(x\minus{}1)^{2007} x\minus{}x!>0\Longrightarrow x^{2008}>(x\plus{}1)!\Longrightarrow x^{2007}>x!$

The equation will never be equation unless $ x\equal{}y$

[quote="EulerIntegral"]
To prove that is the same to prove that $ x^{2007} \minus{} x! > (x \minus{} 1)^{2007} \minus{} (x \minus{} 1)!$  and is sufficient to just prove that $ x^{2007} > x!$
Can you give more detail? (explain, please)

$ x > y$  so we can rewrite $ y$ as $ x \minus{} 1,x \minus{} 2,x \minus{} 3$ etc. So

$ x^{2007} \minus{} x! > (x \minus{} 1)^{2007} \minus{} (x \minus{} 1)! > \cdots > (x \minus{} k)^{2007} \minus{} (x \minus{} k)! \equal{} y^{2007} \minus{} y!$

This is induction. if $ P(n) > P(n \minus{} 1)$ so $ P(n) > P(n \minus{} k)$ so $ P(n) > P(m)$ if $ n > m$.

But there is a BIG flaw on this proof.

$ x^{a\plus{}1}>x (x\minus{}1)^a$  not $ x^{a}>x (x\minus{}1)^a$
$ x^{2008}>x!\Longrightarrow x^{2007}>(x\minus{}1)!$

And the induction is messed up.
I will rethink this and correct my "proof".
Sorry for the wrong solution.

look here

http://www.mathlinks.ro/Forum/viewtopic.php?p=848062#848062

Daniel


graph theory
combinatorics proposed
combinatorics

Let $G$ be a bipartite graph. Let $A$ and $B$ be the two shores of $G$ (this means that $V\left(G\right) = A\cup B$, that $A\cap B = \emptyset$, and that every edge of $G$ has one endpoint in $A$ and one endpoint in $B$). Assume that the bipartite graph $G$ has a perfect matching from $A$ to $B$. Furthermore, assume that $A \neq \varnothing$.

Show that there exists a vertex $v\in A$ such that for every edge $e$ of $G$ satisfying $v\in e$, there exists a perfect matching of $G$ which contains the edge $e$.

[quote="fedja"]Take the minimal (with respect to inclusion) non-empty subset $ A'$ of $ A$ such that the cardinality of the set of its neighbors in $ B$ equals the cardinality of $ A'$ (it exists because the full set $ A$ satisfies this property), choose any vertex in it, and recall Hall's lemma ;).[/quote]

Nice, this was the proposed solution (or at least the person who gave me the problem told me so; the problem comes from some graph theory course). Just a remark: When I wrote "perfect matching from $ A$ to $ B$", I meant a set of vertex-disjoint edges which cover all of $ A$ - but not necessarily all of $ B$. The problem does not require $ \left|A\right|$ to be $ \equal{} \left|B\right|$. But your solution can easily be adapted to the general case: If there is no subset $ A^{\prime}$ satisfying your condition, you can take $ A$ itself as $ A^{\prime}$, and while it does not satisfy the condition, the proof will still work with it.

My solution does not make use of Hall's theorem:
Let $ A \equal{} \left\{a_1,a_2,...,a_n\right\}$ and $ B \equal{} \left\{b_1,b_2,...,b_m\right\}$ with $ n\leq m$. We can WLOG assume that $ M \equal{} \left\{\left\{a_1,b_1\right\},\left\{a_2,b_2\right\},...,\left\{a_n,b_n\right\}\right\}$ is a perfect matching of $ G$.
Now, assume that the problem is wrong. Then, for each $ i\in\left\{1,2,...,n\right\}$, there exists an edge $ e$ of $ G$ satisfying $ a_i\in e$ such that no perfect matching of $ G$ contains the edge $ e$. In other words, for each $ i\in\left\{1,2,...,n\right\}$, there exists some $ u\left(i\right)\in\left\{1,2,...,m\right\}$ such that the edge $ \left\{a_i,b_{u\left(i\right)}\right\}$ exists in the graph $ G$ but is not contained in any perfect matching of $ G$. Obviously, $ u\left(i\right)$ must be in $ \left\{1,2,...,n\right\}$ (because if $ u\left(i\right) > n$, then we can replace the edge $ \left\{a_i,b_i\right\}$ in our perfect matching $ M$ by a new edge $ \left\{a_i,b_{u\left(i\right)}\right\}$, and obtain a new perfect matching which, unlike we claimed, does contain the edge $ \left\{a_i,b_{u\left(i\right)}\right\}$). Hence, we get a map $ u: \left\{1,2,...,n\right\}\to\left\{1,2,...,n\right\}$. Thus, a sufficiently large power of the map $ u$ (where, of course, "power" means composition with itself) must have a fixed point; let $ f$ be a fixed point of some power of $ u$. Let $ p$ be the minimal positive integer satisfying $ u^p\left(f\right) \equal{} f$ (where $ u^p$ means $ \underbrace{u\circ u\circ ...\circ u}_{p\text{ times }u}$). Then, the numbers $ u^0\left(f\right)$, $ u^1\left(f\right)$, ..., $ u^{p \minus{} 1}\left(f\right)$ are pairwise distinct (anything else would contradict the minimality of $ p$). Modify the perfect matching $ M$ by replacing the edges $ \left\{a_{u^k\left(f\right)},b_{u^k\left(f\right)}\right\}$ for $ k\in\left\{0,1,...,p \minus{} 1\right\}$ by the edges $ \left\{a_{u^k\left(f\right)},b_{u^{k \plus{} 1}\left(f\right)}\right\}$ for $ k\in\left\{0,1,...,p \minus{} 1\right\}$. As a result you get a perfect matching of $ G$ that contains quite a lot of edges which, according to our assumption, should not belong to any perfect matching. Contradiction, qed.

Is this correct?

  darij

Yes, it is fine, though it certainly seems inspired by one of the proofs of Hall's lemma :). I initially thought of something like "suppose there is such an edge for each vertex; let's show that those edges form a perfect matching themselves". It obviously wasn't true, so I had to consider chains similar to yours, but that immediately reminded me of the inductive proof of Hall's lemma, so I got lazy and decided to use the lemma itself :P.


I am working through 'contest problem book v',  ( ashme's 1983-1988 ).  how hard are these compared to modern amc-12's?

Could some one suggest a score correction factor, to gauge the difficulty of these things? I relize these older exams ARE harder, but I'm having difficulty figuring out if I'm making progress, when my average scores are lower than that of the modern amc 12's

Well, recently (means this year) the first 15 problems are considerably easier, while the last five are harder than previous AMCs.  I don't know about old AHSMEs tho.  But I believe the AMC director said that since points given for blank answers were lower now, they'll make the first 15 easier so that more people can complete those I guess...

yeah, but the ones in the 80's have 30 questions, each worth less, yet I find the early questions harder. I have no idea how to make the scores translate to modern ones =(


This is my first time to ARML.. and the arml website says absolutely nothing about time limits..

Does anybody know?

Individual: 4 rounds of 2 questions each (total of 8 problems); problems are in increasing difficulty; 10 minutes for each round
Team: 10 problems; 20 minutes; whole team of 15 works on them
Power: 10 problems (all related to the same overall theme); mostly proof based; increasing difficulty; 60 minutes; all 15 work on them
Relays: groups of 3 people; each person get's their own problem; person #2 needs person #1's answer and person #3 needs person #2's answer; if you get the answer within 3 minutes you get 4 pts; 2 pts for getting in within 6 minutes (but later than 3 minutes, of course)

Calculators are permitted on team and power rounds; not permitted on individual and relay rounds.

The total posible number of points for a team is 240.

120 points on the individual round at 1 point per question. 8 questions times 15 people is 120 points.

40 points on the team round: 4 points per questions, 10 questions.

40 points on the power round. Usually, the instructions specify how these points are divided. Usually there are some early parts of a power round that are pretty much straightforward working out of examples. If you're not the sort of person who's going to writing a really difficult proof, then you can still make yourself useful by working out the examples.

40 points on the relays. There are two relays, and your 15-member team will become five 3-member relay teams.

Organization matters for the team round and the power round. You can't just put 15 people in a room without a plan and expect it to all work out.

The ARML site, http://www.arml.com , has scoresheets for each of the last several years. Look at those to get an idea of how many points top teams, A/B borderline teams, and middle-B teams get on each of the parts.

I have a few questions about the competition.

Regarding the super relay -- Does it take place between the tiebreaker round and the awards ceremony?  Is there a time limit?  Can a team submit more than one answer?

Also, I assume anyone can attend the awards ceremony.  Are any of the competition rounds also open to the public?

Well, our coaches say that for team and power rounds, each team gets a classroom to itself, so spectators obviously wouldn't be allowed there.

I don't know; it wouldn't be much of a spectacle, would it?  Nothing like MATHCOUNTS countdown (unless of course, audience members tried working the questions based on their being read aloud).

I'm pretty sure super relay is between tiebreakers and awards.  There's no time limit--if there is one, then in the ones I've been at, they extended it until there was a winner.  It does stop after someone turns in the right answer, though.  And you only get one shot at answering.

As far as I know, spectators don't attend.  Whether they'd be allowed to, I have no idea.


My question is in regard to this site: http://www.cogito.org/Interviews/InterviewsDetail.aspx?ContentID=16865
[quote]My first semester I took Honors General Chemistry, a class geared towards the chemistry majors. The whole premise of this class is making the transition from “plug-and-chug chemistry” to the more analytical concept-based thinking I mentioned earlier. [/quote]Should we take intro undergrad chemistry after ap chemistry even if the latter lets you skip the former?

Would it be better to take intro undergrad chemistry over the summer rather than take a post-ap chem class the summer after ap chem like a class in chemical structure and reactivity?[ provided you are interested in chemistry]

1. Depends how much of a baller you are
2. Depends if you're premed or not

Why does it depend on how much of a baller you are or if you are premed?
I checked up baller on urbandictionary.com, and I'm not sure that I understand.
"A thug that has "made it" to the big time?"


Evaluate:
$ C^a_k\plus{}C^{a\plus{}1}_k\plus{}\ldots\plus{}C^{b}_k$
where $ b\geq a\geq k$ and $ C^n_k\equal{}\frac{n!}{k!(n\minus{}k)!}$

Hint: compare it to the "hockey-stick identity."


geometry
circumcircle
analytic geometry
graphing lines
complex numbers
geometry proposed

The problem can be solved by using scalar product of vectors.
I hope to see a sinthetical solution.. I will also look for one.

i will post later the sinthetical solution from mr. Enescu's book. my solution is a bit computational: just take
$OE \perp CD \Longleftrightarrow OC^{2}-OD^{2}= EC^{2}-ED^{2}$, and express these only in $a,b,c$, that is simple because they are only medians and the result follows.

Let $ G$ be the centroid of triangle $ ABC$, $ N$ the midpoint of $ BC$ and $ M$ the midpoint of $ CA$.

It is true that $ GE\parallel{}AB$, so $ OD\perp GE$. This implies that $ OE\perp CD$ if and only if $ O$ is orthocenter of trianble $ DEG$. This is equivalent to $ GO\perp DE$. On the other hand, $ O$ is orthocenter and $ G$ is centroid of triangle $ DMN$. So, $ GO\perp DE$ if and only if $ NG$ is median and altitude from $ N$ in triangle $ DMN$. This is equivalent to $ DN\equal{}NM$, and to $ AB\equal{}AC$.

:)

Can someone give a clearer proof since the posters below say that this implies even though I did not get why.

Complex numbers :D

Set $O$ as the origin of the complex plane, and let $a, b, c, d, e$ be the coordinates of points $A, B, C, D, E$ respectively. Then we have \[d=\frac{a+b}{2} \text{ and } e=\frac{a+c+d}{3}=\frac{3a+b+2c}{6}.\] Using the real product of complex numbers, if $R$ is the circumradius of triangle $ABC$, then we have \[a\cdot a=b\cdot b=c\cdot c = R^2.\] Also note that $CD\perp DE \iff (d-c)\cdot e = 0$, i.e. \[(a+b-2c)\cdot (3a+b+2c)=0.\] The last relation is equivalent to \[3a\cdot a+a\cdot b+2a\cdot c+3a\cdot b+b\cdot b+2b\cdot c-6a\cdot c-2b\cdot c - 4c\cdot c=0,\] which implies \[a\cdot b=a\cdot c (*)\] Now, observe that \[AB=AC \iff |b-a|^2=|c-a|^2.\] This yields \[(b-a)\cdot(b-a)=(c-a)\cdot(c-a)\] or \[b\cdot b-2a\cdot b+a\cdot a=c\cdot c-2a\cdot c+a\cdot c.\] Hence, we have \[a\cdot b=a\cdot c (**)\] From $(*)$ and $(**)$ we have $CD\perp OE\iff AB=AC$. $\Box$

[hide="Coordinate bash"]
[hide="If"]
Let's have $O(0,0),A(0,2a),B(-2b,2c),C(2b,2c)$ for reals $a,b,c$. Note that $a^2=b^2+c^2$.

Since $D$ is the midpoint of $\overline{AB}$, $D$ is $(-b,a+c)$. $E$ is centroid of $\triangle{ACD}$ so $E$ is $\left(\frac{b}{3},a+c\right)$.

The slope of $\overline{OE}$ is just $3\frac{c+a}{b}$.

The slope of $\overline{CD}$ is $\frac{2c-(a+c)}{2b-(-b)}=\frac{c-a}{3b}$. Thus, the product of the slopes of $\overline{OE}$ and $\overline{CD}$ is $\left(3\frac{c+a}{b}\right)\left(\frac{c-a}{3b}\right)=\frac{c^2-a^2}{b^2}=-1$.

Thus, $\overline{OE}\perp\overline{CD}$.[/hide]

[hide="Only if"]Let $O(0,0), A(0, 2a), B(2b, 2c), C(2d, 2e)$ where $a^2=b^2+c^2=d^2+e^2$.

Midpoint of $\overline{AB}$ is $D(b, a+c)$. Therefore, centroid of $\triangle{ACD}$ is $E\left(\frac{b+2d}{3},\frac{3a+c+2e}{3}\right)$.

Then the slope of $\overline{OE}$ is $\frac{3a+c+2e}{b+2d}$ and the slope of $\overline{CD}$ is $\frac{a+c-2e}{b-2d}$. $\overline{OE}\perp\overline{CD}$ iff slopes multiply to $-1$, so $\overline{OE}\perp\overline{CD}$ iff

$\frac{3a+c+2e}{b+2d}\cdot\frac{a+c-2e}{b-2d}=-1$

which is equivalent to $3a^2+4ac-4ae+c^2-4e^2=4d^2-b^2$ or

$3a^2+b^2+c^2-4(d^2+e^2)=4ae-4ac$.
The left side is equal to 0, so we have $4ae=4ac$ or $2ae=2ac$. This of course is equivalent to $b^2+(a-c)^2=d^2+(a-e)^2$ which is equivalent to $AB=AC$.
[/hide][/hide]

Can somebody write a solution with complex numbers? I could not understand tc1729's solution. I get another things I'm my complex bashing.

[hide = Detailed Complex Bash Solution]
WLOG, let the circumcircle of $ABC$ be the unit circle in the complex plane. We can set
$$O = 0$$
$$A = 1$$
$$B = b, |b| = 1$$
$$C = c, |c| = 1$$
$$D = \frac{1+b}{2}$$
$$E = \frac{1+c+\frac{1+b}{2}}{3} = \frac{3+b+2c}{6}$$
Notice that $CD \perp OE \implies \frac{c-\frac{1+b}{2}}{\frac{3+b+2c}{6}}$ is pure imaginary. Or
$$\frac{6c-3-3b}{3+b+2c} = -\overline{\left(\frac{6c-3-3b}{3+b+2c}\right)}$$
Notice that $|b| = |c| = 1$, so $\overline{b} = \frac{1}{b}$ and $\overline{c} = \frac{1}{c}$.
$3 + b + 2c = 0$ only if $b = c = -1$, since $|b| = |c| = 1$. However, then $ABC$ is not a nondegenerate triangle.
$$-\overline{\left(\frac{6c-3-3b}{3+b+2c}\right)} = -\frac{\overline{6c-3-3b}}{\overline{3+b+2c}} = -\left(\frac{\frac{6}{c}-3-\frac{3}{b}}{3+\frac{1}{b}+\frac{2}{c}}\right) = -\left(\frac{6b-3bc-3c}{3bc+c+2b}\right)$$
Now we prove the iff statement. Note that if
$$\frac{6c-3-3b}{3+b+2c} = -\frac{6b-3bc-3c}{3bc+c+2b}$$
Then
$$-(6c-3-3b)(3bc+c+2b) = (6b-3bc-3c)(3+b+2c)$$
$$-(18bc^2+6c^2+12bc-9bc-3c-6b-9b^2c-3bc-6b^2)=(18b+6b^2+12bc-9bc-3b^2c-6bc^2-9c-3bc-6c^2)$$
$$-18bc^2+9b^2c-6c^2+6b^2+3c+6b = -3b^2c-6bc^2+6b^2-6c^2+18b-9c$$
$$12bc^2-12b^2c-12c+12b = 0$$
$$bc^2-b^2c+b-c=0$$
$$(bc-1)(c-b)=0$$
If $b = c$, then $ABC$ is degenerate, so $bc = 1$ , or $c = \frac{1}{b}$. This implies that $AB = AC$. Since our steps were reversible, the statement $c=\frac{1}{b}$ will imply perpendicularity. $\blacksquare$
[/hide]

I will show the $AB=AC$ case, the other one can be done similarly (I swear I didn't read #2 :D)
[quote=pohoatza]In a given triangle $ABC$, $O$ is its circumcenter, $D$ is the midpoint of $AB$ and $E$ is the centroid of the triangle $ACD$. Show that the lines $CD$ and $OE$ are perpendicular if and only if $AB=AC$.[/quote]
[b]Solution:[/b]
[b][color=#f00]Lemma:[/color][/b] In isosceles $\Delta ABC$, where $AB=AC$, let $M_A,M_C$ be the midpoints of $BC,AB$. Let $X$ be a point on $\overline{BC}$, such, $BX:XC$ $=$ $1:2$. Let $H$ be the orthocenter of $\Delta ABC$. Then, $XH \perp CM_C$
[b][color=#00f]Proof:[/color][/b] We'll use vectors,
$$\begin{cases} \stackrel{\longrightarrow}{HX}=\frac{-a}{6} ~ \i - \frac{a}{2} \tan \frac{A}{2} ~ \j \\\ \stackrel{\longrightarrow}{M_CC}=\frac{3a}{4} ~ \i - \frac{a}{4} \tan \left( 90^{\circ}-\frac{A}{2} \right) ~ \j \end{cases} \implies \stackrel{\longrightarrow}{HX} \cdot \stackrel{\longrightarrow}{M_CC} =\frac{-a^2}{8}+\frac{a^2}{8} \tan \frac{A}{2} \tan \left( 90^{\circ}-\frac{A}{2} \right) =0 \qquad \blacksquare$$
Back to the Problem: Let $M_A,M_B$ be midpoints of $BC,AC$, then $BDM_BM_A$ and $DM_ACM_B$ are parallelogram. If $M_AM_B$ $\cap$ $CD$ $=$ $X$ and $M_AD$ $\cap$ $BM_B$ $=$ $Y$, then, $X,Y$ are midpoints of $M_AM_B$, $M_AD$. $BM_B$ $,CD$ and $AO$ are concurrent, say at $Z$, which is centroid of $\Delta ABC$ and $\Delta DM_AM_B$. $ADOM_B$ is cyclic $\implies$ $O$ is orthocenter of $\Delta DM_AM_B$. Now apply [b][color=#f00](Lemma)[/color][/b]

Quite straightforward complex bash as usual set center as 0.Its easy to see that $AB=AC$ is equivalent to $a^2=bc$.$CD\perp OE$ is nothing but $\dfrac{2c-a-b}{2ab-ac-ab}=-\dfrac{3a+2c+b}{3bc+2ab+ac}$.Therefore \begin{align*}(3bc+2ab+ac)(a+b-2c) &=(3a+2c+b)(2ab-ac-bc) \\ \Longleftrightarrow 3abc+2a^2b+a^2c+3b^2c+2ab^2+abc-6bc^2-4abc-2ac^2 &= 6a^2b-3a^2c-3abc+4abc-2ac^2-2bc^2+2ab^2-abc-b^2c \\
\Longleftrightarrow 4a^2c-4c^2b &=4a^2b-4b^2c	\\ \Longleftrightarrow	a^2&=bc 	 \end{align*}
So we are done.$\blacksquare$


Lisa has 10 friends and 34 marbles. What is the minimum number of additional marbles she needs so that she can give each friend at least one marble and no two friends receive the same number of marbles?

If every friend gets a different amount of marbles, then Lisa needs at least $ 1\plus{}2\plus{}3\plus{}\cdots\plus{}10\equal{}55$ marbles. $ 55\minus{}34\equal{}\boxed{21}$.


A parabola has vertex at (4,-5) and has two x-intercepts, one positive and one negative.  If this parabola is the graph of $ y\equal{}ax^2\plus{}bx\plus{}c$, which of $ a$, $ b$, and $ c$ must be positive?

(A) only a
(B) only b
(C) only c
(D) a and b only
(E) none

[hide="Each coefficient has a geometric connection click for hint"]$ a$ tells you about the concavity.
$ b$ tells you about the vertex.
$ c$ tells you about the y-intercept.[/hide]

[hide]Well considering that the vertex is below the x-axis, the fact that it has x-intercepts shows that it opens upward, which implies that $ a$ is positive. $ c$ is negative because the vertex is shifted downward.
$ \frac{\minus{}b}{2a}\equal{}4$, and because $ a$ is positive, $ \minus{}b$ must be positive as well, implying that $ b$ is negative.

So the answer is $ \boxed{A}$[/hide]


For $ a,b,c\ge0$  prove
$ \frac{1}{3}(x^3\plus{}y^3\plus{}z^3)\ge xyz\plus{}\frac{3}{4}|(x\minus{}y)(y\minus{}z)(z\minus{}x)|$

Presume x+y+z=1.

  Inequality becomes
\[ \frac {(x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2}{3}\ge \frac {3}{2}|x \minus{} y\parallel{}y \minus{} z\parallel{}z \minus{} x|
\]
But
\[ \frac {(x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2}{3}\ge \sqrt [3]{|x \minus{} y|^2|y \minus{} z|^2|z \minus{} x|^2}
\]
so it suffices to prove that
\[ |x \minus{} y\parallel{}y \minus{} z\parallel{}z \minus{} x|\le \frac {8}{27}
\]
wich is cleary true.

[quote="dima ukraine"]For $ a,b,c\ge0$  prove
$ \frac {1}{3}(x^3 \plus{} y^3 \plus{} z^3)\ge xyz \plus{} \frac {3}{4}|(x \minus{} y)(y \minus{} z)(z \minus{} x)|$[/quote]

  $ \frac {1}{3}(x^3 \plus{} y^3 \plus{} z^3)\ge xyz \plus{} \frac {1}{\sqrt[3]{2}}|(x \minus{} y)(y \minus{} z)(z \minus{} x)|$ it is also true.

[quote="Marius Mainea"][quote="dima ukraine"]For $ a,b,c\ge0$  prove
$ \frac {1}{3}(x^3 \plus{} y^3 \plus{} z^3)\ge xyz \plus{} \frac {3}{4}|(x \minus{} y)(y \minus{} z)(z \minus{} x)|$[/quote]

  $ \frac {1}{3}(x^3 \plus{} y^3 \plus{} z^3)\ge xyz \plus{} \frac {1}{\sqrt [3]{2}}|(x \minus{} y)(y \minus{} z)(z \minus{} x)|$ it is also true.[/quote]
$ EMV$ of hungkhtn works here

Using Am-Gm, we can prove the stronger:
$ x^3\plus{}y^3\plus{}z^3 \geq\ 3xyz \plus{} \frac{3\sqrt[]{3}}{\sqrt[3]{4}}|(x\minus{}y)(y\minus{}z)(z\minus{}x)|$


Define the number of permutations  f for which
f :{1,2,..,n} -> {1995,1996} and which satisfy the condition :
 f_(1) + f_(2) + ... + f_(n) is odd

I think there are 2n-1 such permutations. Because for each of f(1), f(2), ..., f(n - 1) you can choose whether it's even or odd (this can be done in exactly 2n - 1 ways) and then there is only one possibility for the value of f(n) to assure that the given sum is of the desired parity.


Αν είναι $ 0<p\leq a,b,c,d,e\leq q$ νδο.

$ (a\plus{}b\plus{}c\plus{}d\plus{}e )( \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d}\plus{}\frac{1}{e})\leq 25 \plus{} 6( \sqrt{\frac{p}{q}}\minus{}\sqrt{\frac{q}{p}})^2$

Δες τη λύση μου στο ποστ του Αλέξανδρου . Η ιδέα είναι ακριβώς ίδια (αν έχεις αμφιβολίες πες μου να γράψω πλήρως τη λύση) :wink:

http://www.mathlinks.ro/viewtopic.php?t=165314&start=100

Αν δε σου γίνομαι πολύ πιεστικός, μήπως μπορεις ρε Σιλουανέ να βάλεις ακριβώς τη λύση σου??? :blush:

P.S.: το φαντάστηκα ότι ήταν ειδική περίπτωση αυτής αλλά με χάλασαν λίγο τα ριζικά :oops:

Είναι θέμα από τη USAMO του 1977.

http://www.kalva.demon.co.uk/usa/usoln/usol775.html

Στέλιος

[quote="meinstein"]αλλά με χάλασαν λίγο τα ριζικά :oops:[/quote]

Όταν η ρίζα μας χαλάει την ξεριζώνουμε  :P  Θέτουμε $ p \equal{} m^2,\ q \equal{} n^2$

Επειδή τώρα η συνάρτηση είναι κυρτή σε κάθε μεταβλητή της θα έχουμε ότι το μέγιστο επιτυγχάνεται όταν $ k$ μεταβλητές είναι ίσες $ m^2$ και $ 5 \minus{} k$ ίσες με $ n^2$

Επομένως αρκεί νδο $ (km^2 \plus{} (5 \minus{} k)n^2)(\frac {5}{p^2} \plus{} \frac {5 \minus{} k}{q^2})\leq 25 \plus{} 6(\frac {m}{n} \minus{} \frac {n}{m})^2$

που ισοδυναμεί με $ \frac {(m \minus{} n)^2(m \plus{} n)^2(k \minus{} 2)(k \minus{} 3))}{m^2n^2}\geq 0$ που ισχύει για $ k \equal{} 0,1,2,3,4,5$ αρα QED


analytic geometry
vector
linear algebra
matrix

I'm having a little trouble trying to figure out the next step in this problem:

Find the coordinate vector of P relative to the basis S={p1,p2,p3},
where, P = 2 - x + x^2, p1 = 1 + x, p2 = 1 + x^2, and p3 = x + x^2.

I got as far as setting the problem up as
2 - x + x^2 = c1(p1) + c2(p2) + c3(p3)

and substituting the equations into their respective slot, and then combining like terms on the right side of the equation (so something like (c1+c2) + x(c1+c3), so on so forth). The problem i now have is that, when using vector bases, I could easily set up a matrix and use gaussian elmination to find my coefficients...I'm unsure what way to continue to do the same with polynomials.

Thanks in advance for any help you can provide.

I think you already solved the problem. Comparing coefficients gives

$\begin{eqnarray*}c_{1}+c_{2}&=& 2 \\ c_{1}+c_{3}&=&-1 \\ c_{2}+c_{3}&=& 1 \\ \end{eqnarry*}$

which is easy to solve.


[quote="krustyteklown"]if you have a nearby school or center offering it, it is ok, you might have to pay a small fee and run it through wiht the guys at UNL first[/quote]
yeah i would recomend you go to another school...

Is there a quick way to check if the AMC 12 B is being offered in a given area, and if so how to contact the test center?

Uh....oh yeah, what's UNL? And if I do find a school nearby, what do I do? Contact the school and ask if I can take it?

university of nebraska at lincoln, they're the ones that adminster the AMC's
you should contact the school first, as they might have talked to UNL about offering the B test to other students

Uh, who do I exactly contact if I want to see if I can take the AMC test B from another school? Is there a specific phone number or contact e-mail?

*EDIT*

Sorry for not making it clear, I meant that, who do I contact from UNL?

If I were to buy a set of 10 booklets for the AMC Test B, who would be an acceptable proctor for that test?

[quote="pkerichang"]Is it still possible to register for 12B???(my school only offers A)[/quote]

Check with Chapman, that's where I'm taking it. It might not be too late to register through them.


I am unable to solve this problem.
I do not know the way of which I should go about solving it.

{2/5x - 1/6y = 7/10
{3/4x - 2/3y = 19/8

we eliminate y from the equation and soon get x=0.5. then, solving for y, we gey y=-3.

Also, you can multiply the first equation by 30 and the second one by 24 to get rid of the fractions.


One side of a given triangle is $ 18$ inches.  Inside the triangle a line segment is drawn parallel to this side forming a trapezoid whose area is one-third of that of the triangle.  The length of this segment, in inches, is:

$ \textbf{(A)}\ 6\sqrt {6} \qquad \textbf{(B)}\ 9\sqrt {2} \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 6\sqrt {3} \qquad \textbf{(E)}\ 9$

The largest triangle that is in the larger triangle but not any part of the trapezoid has area of ratio $ 1\minus{}\frac13\equal{}\frac23$ to the larger triangle's. Let the length of the desired segment be $ x$ such that $ \left(\frac x{18}\right)^2\equal{}\frac23\implies x^2\equal{}216\implies x\equal{}\boxed{\textbf{(A)}\ 6\sqrt6}$.


let's x, y, z be the real number satisfying xyz=1.
prove that x^2 :( /(1-x)^2+y^2/(1-y)^2+z^2/(1-z)^2>=1 :( [/code]

The correct forum where to post this would be
http://www.mathlinks.ro/index.php?f=52

[Moderator: Indeed, moved]

[quote="voong"]
let's x, y, z be the real number satisfying $ xyz\equal{}1.$
prove that $ x^2  /(1\minus{}x)^2\plus{}y^2/(1\minus{}y)^2\plus{}z^2/(1\minus{}z)^2>\equal{}1$ :( [/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113999


calculus
algebra
function
domain
induction
calculus computations

You dont require calculus to prove that e^x<or=e^x^2

Take a  :| look at what I have found out.

case1: for +ve values of x

e^x<e^x^2{since e>1}
now add x to both sides

x+e^x<e^x^2+x

therefore we can safely conclude

e^x<x+e^x^2 :|

case 2: forx=0

e^x=1=e^x^2 which proves the equality

What happens for -ve values of x?   :|

This doesn't hold when the value of $ x$ is between $ 0$ and $ 1$.

Eg/ $ x = \frac {1}{2}$

$ e^{\frac {1}{2}} < e^{\frac {1}{4}}$, a contradiction.

its cool to get my name out there and all, but i think hes right.  $ x^2 > x>0 \to e^{x^2} < e^x$

[quote="kenn4000"]its cool to get my name out there and all[/quote]

 :rotfl:

[quote]  $ x^2 > x > 0 \to e^{x^2} < e^x$[/quote]

x=2?  :maybe:

didnt you see the invisible 1 > ?

i didnt mean always, i just meant it was possible up there =P

Ya look i think i have got it wrong

Yes, for fractional values of x [between 0 and1]e^x^2<e^x.And i appologise straight away.
But then none of you have actually mentioned how to solve it

Well it's a pretty trivial problem.

Just show $ x^{2} > x \;\;\; \forall x \in ( \minus{} \infty, 0) \cup (1, \infty)$

For the case where x is negative, we have $ x^{2} > 0$ whereas $ x < 0$. For the other case, divide both sides to get $ x > 1$, which is self-evident (the answer matches the domain for that specific case).

You could also use induction or compare growth rates if you're willing to burn time.