linear algebra
matrix
linear algebra unsolved

Hi,

I wonder if anyone can at least give me some reference on where I could read more about following matrix recursions:

A_k = (P_k)*(A_(k-1))

I also know that the determinant for all P_k is the same.
I'm mostly interested in studying the convergence behaviour of these equations. For example, for which P_k is it true that the above recursion converges (from any given starting point) to the same value after a number of steps? And how fast does it converge?

Should I maybe study the error, and hence the matrix (P_k)*(P_(k-1))*...*P_0? Maybe it's eigenvalues?

Thanks in advance.

There was no need to post this a second time in a different forum. The first post was sufficient, and properly placed.

Oh sorry...I saw that the answers weren't that frequent in the other section, so I thought someone might answer sooner here. I also have no idea if it is solved/studied or not, so I thought it would be good to place it here also. Shall I remove from this section?


topology
linear algebra
matrix
function
calculus
derivative
complex numbers

Prove that $ SU(2)$ (the set of complex 2x2 unitary matrices with determinant 1) is a smooth subset of $ R^{8}$ of real dimension 3.

Here is my attempt to do it:

Let $ a, b, c, d \in C$ be the complex numbers in a matrix $ A \in SU(2)$ and let $ a \equal{} a_{1} \plus{} ia_{2}$ etc., where $ a_{i}, ..., d_{i}$ are real. The conditions that A is unitary and has determinant 1 are equivalent to a set of equations:

\[ (1) a\bar{a} \plus{} b\bar{b} \equal{} 1\]
\[ (2) c\bar{c} \plus{} d\bar{d} \equal{} 1\]
\[ (3) a\bar{c} \plus{} b\bar{d} \equal{} 0\]
\[ (4)ad \minus{} bc \equal{} 1\]

The equations (1) and (2) are in fact two real equations and equation (3) yields two real equations, one for the real part and one for the imaginary, so I've got 4 real equations.

Unfortunately, I don't know what to do next, because:

- I need 5 real equations to show that SU(2) has real dimension 3; equation (4) doesn't affect smoothness, because it follows from the condition that A is unitary that its determinant must be 1 or -1, so we needn't consider it, but then we have only 4 real equations

- if we nevertheless use the equation (4), we can write 5 functions $ f_{1}, ..., f_{5}$ corresponding to the real equations; the condition that $ SU(2)$ is smooth is then equivalent to the condition that the matrix with rows $ \nabla f_{1}, ..., \nabla f_{5}$ (differentiation is with respect to all 8 real variables) has maximum rank, i.e. 5, for $ a_{1}, ..., d_{2}$ satisfying the equations, but unfortunately this matrix has for example rank 4 when A is the identity matrix

How to solve this problem?

(I know that it can be shown that $ SU(2)$ is diffeomorphic to $ S^{3}$, but I want to show its smoothness and dimension directly)

A being unitary only implies $ |det(A)|\equal{}1$, so the last equation is really necessary.

But it doesn't affect smoothness and dimension (or does it?) - and including this equations leads to the $ \{ \nabla f_{i}\}$ matrix not having maximum rank, as I showed.

Of course it affects the dimension. $ U(2)\cong S^1\times S^3$ is has dimension four, $ SU(2)\cong S^3$ has only dimension three.


Opening, we have to prove that $(a^{3}+b^{3}+c^{3})+(b^{2}a+c^{2}b+a^{2}c) \geq 2(2^{b}+b^{2}c+c^{2}a)$

But $a^{3}+b^{2}a \geq a^{2}b$, by Am-Gm! (the same for the other parts)

Hi,

In a lot of topic, I see "SOS", what's SOS ?  :blush:

[quote="zweig"]Hi,

In a lot of topic, I see "SOS", what's SOS ?  [/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127


Kaya ate 60 cookies over a five-minute period. On average, how many cookies did she eat each minute?

1: Kaya is not watching her diet very well....
2: $ \frac{60}{5}\equal{}\boxed{12}$.


Let $x, y, z$ three non-negative real numbers. Prove that

\[ \frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\geq\frac{x}{x^{2}+yz}+\frac{y}{y^{2}+zx}+\frac{z}{z^{2}+xy}. \]

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=70283 .

  darij

Some problem of Mr.[[color=red]b]Cezar lupu[/b][/color] used Schur:
$x(a-b)(a-c)+y(b-c)(b-a)+z(c-a)(c-b)\ge 0$ with...


The AMC-10/12 dates are February 10 and February 25.  Please do not discuss the first set of tests until after 8 AM CST, Wednesday February, 11, and do not discuss the second set until after 8 AM CST, Thursday, February 26.

We intend to have Math Jams on those discussion dates to talk about the more challenging problems on the tests.

We will be posting similar restrictions for the AIME and the USAMO as we get closer to those testing dates.

May I ask that, if possible, the moderators be a bit more active on the day of the AMC12's and the day after to make sure nobody jumps the deadline and ruins the competitions for everyone.

Thanks.

sometimes their solutions are hard to understand, so they'll try to give us solns we can understand easier and different ways of approaching the problem.

Or, in my case, my teacher doesn't let us see the solutions, so it'd be nice to know the answers.

[quote="mathfanatic"]I'll do my best.  darkquantum and Yannik haven't been active in a while...[/quote]

Yes, darkquantum seems largely to have vanished. I PMed him once and he never read the message.

It's as if he was reading our minds, because he just posted something.

eh...maybe he's been reading the board the whole time and just didn't want to post.

[quote="churchilljrhigh"]why do we need a mathjam they give you the solutions anyway?[/quote]

Our goal with Math Jams and also our classes is not just to teach you how to understand somebody else's solution, but to help you to develop the insights that would allow you to solve each and every problem on your own.

There are very few sources for problem solving education that deal with helping students develop an understanding of how to unravel nonstandard problems that they may never have seen before.  This is the gap in education that we are trying to help fill.

[quote="Tare"]Um on the front of the test booklet it said like don't discuss problems anywhere including the internet, but it also said not to copy problems via fax machine or computer or copy machine or internet or that sort of a thing, so how is the math jam going to work without the problems?[/quote]

We just typed up the problems from one of our student's test papers.  We'll be able to discuss them tomorrow.  All is well.

*AHEM AHEM*  Not active for a while?  You should say that about darkquantum not me.  LoL.

its after 8 on wednesday. why are are the posts still locked?

They're unlocked now, and Mr. Crawford or Mr. Rusczyk may not have visited the site yet at the time.  You can't expect the mods to do it either because they are at school.


calculus
integration
derivative
function
trigonometry
calculus computations

Find the length of the curve (4x - x^2)^(1/2) with 0<x<3
I've tried using the equation to solve for the length but I've had problems solving the intergral. Thanks!

use $ s \equal{} \int_a^b \sqrt{1 \plus{} [f'(x)]^2} dx$

Take the derivative, combine the terms in the fraction, and separate out the root in the numerator and denominator. It appears to be an arcsin function.

See if you can work it out

$ y \equal{} \sqrt{4 \minus{} (x \minus{} 2)^2}$ is the graph of a circle with radius $ 2$ centered at $ x \equal{} 2$.  This should clue you in as to what kind of function to expect for the indefinite integral :)

There is no need. You have the following function:

$ y\equal{}\sqrt{4x\minus{}x^2}$
$ y^2\equal{}4x\minus{}x^2$
$ y^2\equal{}4x\minus{}x^2\minus{}4\plus{}4$
$ y^2\equal{}\minus{}(x\minus{}2)^2\plus{}4$
$ y^2\plus{}(x\minus{}2)^2\equal{}4$

Or, a circle with $ R\equal{}2$ and origin $ (2,0)$. Also, you need the arc length from $ [0,3]$, so it is wise to look at the point $ (3,y(3))$ and that is $ \sqrt{3}$. Now, if you connect the origin of the circle, the point $ (3,\sqrt{3})$ and the point $ (3,0)$. You will get a triangle, with the property that the radius is dislocated from the $ x$ axis by $ \phi$ radians. Also, from the triangle:

$ \sin(\phi)\equal{}\frac{\sqrt{3}}{R}\equal{}\frac{\sqrt{3}}{2}$

so

$ \phi\equal{}\frac{\pi}{3}$

or the angle of the arc we need to calculate:

$ \alpha\equal{}\pi\minus{}\phi\equal{}\frac{2 \pi}{3}$

And so the arc length is:

$ L\equal{}\alpha R\equal{}\frac{4 \pi}{3}$

I imagine the problem is an exercise in the general method of computing arc lengths; the answer (which is easy to arrive at geometrically) is less important than the correct evaluation of the relevant integral.

If for nothing, for answer checking after evaluating the integral.


Find all function $ f: \mathbb{N} \rightarrow \mathbb{N}$ such that $ f(2f(n))\equal{}n\plus{}1998$

Let $ n\equal{}2f(m)$, then $ 2f(m)\plus{}1998\equal{}f(2f(2f(m)))\equal{}f(2(m\plus{}1998))\equal{}f(2m\plus{}3996).$
It give $ f(m)$ is even, if m>3996 even. But for odd n $ f(2f(n))\equal{}n\plus{}1998$ odd.
Therefore for odd n  $ f(n)\le 1998$.


Assuming we're supposed to simplify this.

[hide="Solution"]
$ (1\plus{}\sqrt x)^{100}\equal{}\sum_{k\equal{}0}^{100}\binom{100}{k}x^\frac{100\minus{}k}{2}$
$ (1\minus{}\sqrt x)^{100}\equal{}\sum_{k\equal{}0}^{100}(\minus{}1)^k\binom{100}{k}x^\frac{100\minus{}k}{2}$

Adding these will remove all terms where $ k$ is odd and double all those where $ k$ is even. That is:

$ (1\plus{}\sqrt x)^{100}\plus{}(1\minus{}\sqrt x)^{100}\equal{}2\sum_{k\equal{}0}^{50}\binom{100}{2k}x^{50\minus{}k}$

Then $ \frac12(1\plus{}\sqrt x)^{100}\plus{}\frac12(1\minus{}\sqrt x)^{100}\equal{}\boxed{\sum_{k\equal{}0}^{50}\binom{100}{2k}x^{50\minus{}k}}$.

Can't really be simplified any more.
[/hide]


anyone know when we can expect to find out the USAMO qualifiers?

i'm guessing about two weeks after the second AIME.
if the AMCdirector or anyone else remembers from last year how long it took i'd appreciate it.

Somehow, I think we found out a week after the second AIME. They'll also email you.

I found out on April 13 last year... I don't know how long it had been posted by then though.

Sounds bout right. My email says it was sent out April 14th. And I recall the qualifiers were posted two days before. So I'm thinking the 12th?

[quote="DPopov"]I was wondering what they changed the criteria to for participation in the USAMO....i remember that there was discussion about this but I can not find it.[/quote]

No. Qualification and participation have not changed, at least the teacher's manual and AMC site still look the same.

[quote="DPopov"]There was somethign that apparently changed from "permanent resident" to "legal resident" I think....i remember seeing a discussion about it. Also the teacher's manual says that the results will be back on the 28th of March.[/quote]

I'd thought it was always "legal resident" b/c I remember a few MOPpers who were not permanent residents. (I could be wrong though.)

As for March 28, that makes sense since just about everything is 2 weeks early this year.

Yeah, March 28...
It says in the USAMO Teacher Manual or whatever that "The selection of the USAMO participants will be made on or before Monday, March 28. All selected students and their School Manager will receive a letter of invitation, to be mailed out during the week of March 28th to April 1st, or before.

March 28=AIME I scores

"legal resident".....I hope it's like that, should I email the AMC and ask about it? I am not a permanent resident possessing a green card but our "petition" was approved so we are waiting for our adjustment of status and we should receive our green card within two years (the process takes a while, also I have lived here for 10 years).

[quote="billzhao"]The AIME scores ... any idea when our schools are expected to receive them?

For the AMC's, the AMC A and AMC B scores came out at the same time. Will it be the same situation for the AIME I and AIME II's ?[/quote]

Yes.

Does anyone know of anyone else who had a similar problem and how it worked out? I am going to email them anyway but I was just wondering. :?

Not confirming my qualifying but IF I did qualify (which I hope I did since I'm a sophomore) then I wanted to know if I would be allowed to OFFICIALLY PARTICIPATE due to my lacking of a green card and citizenship. That is what I wanted to email them about. Not my qualification of USAMO.

[quote="darktreb"]Since when did they send out e-mails confirming USAMO qualification? I don't ever remember getting one ... though I think I've just blanked it out of mind. I've been using the UNL site to find my qualification confirmation for as long as I can remember....[/quote]

They've done that for many years, according to some people. I know for sure they did that last year. :-)

And correct me if I'm wrong, but you also receive a letter of invitation.

Or you could (quickly beyond belief) get a green card ;)


Let $AA_{1},BB_{1},CC_{1}$ be the altitudes of the triangle $ABC$.
$AA_{2}\perp BC(A_{2}\in BC),BB_{2}\perp CA(B_{2}\in CA),CC_{2}\perp AB(C_{2}\in AB)$.
Prove that the perpendiculars to line $BC$ passing through $A_{2}$, to line $CA$ passing through $B_{2}$ and to line $AB$ passing through $C_{2}$ are concurrent.

Do you mean the perpendicular line of $BC$ through $A_2$ is an altitude of triangle ABC?   :?

The only way I see to make this problem make sense is this:

[color=blue][b]Problem.[/b] Let ABC be a triangle, and X, Y, Z the feet of its altitudes from the vertices A, B, C, respectively. Let X', Y', Z' be the feet of the perpendiculars from the points A, B, C to the lines YZ, ZX, XY. Prove that the perpendiculars to the lines BC, CA, AB through the points X', Y', Z' concur at one point.[/color]

[i]Solution.[/i] For the sake of simplicity, we will only consider the case when triangle ABC is acute-angled.

Since < BYC = 90° and < BZC = 90°, the points Y and Z lie on the circle with diameter BC. Thus, the quadrilateral BZYC is cyclic, so that < ZYC = 180° - < ZBC. Similarly, < XYA = 180° - < XBA. Since < ZBC = < XBA, we thus have < ZYC = < XYA. Thus, the line CA is the external angle bisector of the angle XYZ. Similarly, the line AB is the external angle bisector of the angle YZX. Thus, the point A, being the point of intersection of these two lines CA and AB, must be the center of the excircle of triangle XYZ opposite to its vertex X. Hence, the point X', being the foot of the perpendicular from the point A to the line YZ, is the point where this excircle touches the side YZ of triangle XYZ. Similarly, the points Y' and Z' are the points where the excircles of triangle XYZ opposite to its vertices Y and Z touch the sides ZX and XY of triangle XYZ. Thus, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=17214 , the parallels to the internal angle bisectors of the angles ZXY, XYZ, YZX through the points X', Y', Z' are concurrent.

Now, the internal angle bisector of the angle XYZ is perpendicular to the external angle bisector of this angle XYZ, i. e. to the line CA. Hence, a parallel to the internal angle bisector of the angle XYZ is a perpendicular to the line CA. Similarly, a parallel to the internal angle bisector of the angle YZX is a perpendicular to the line AB, and a parallel to the internal angle bisector of the angle ZXY is a perpendicular to the line BC. Thus, the perpendiculars to the lines BC, CA, AB through the points X', Y', Z' are the parallels to the internal angle bisectors of the angles ZXY, XYZ, YZX through the points X', Y', Z', and as we noted above, these parallels are concurrent. Hence, we conclude that the perpendiculars to the lines BC, CA, AB through the points X', Y', Z' are concurrent.

  Darij


[quote="José"][quote="Tony2006"]este fue un problema del selectivo de yucatan 2005:

Encuentra todas las soluciones reales del sistema de ecuaciones

$x^3=2y-1$, $y^3 = 2z-1$ y $z^3 = 2x-1$

pongan sus ideas o soluciones  :P !![/quote]

[hide="ayudita"]factorizar $x^3+y^3$ y sustitución [/hide][/quote]

como puse en el otro post, si quieres ayudar ayuda bien no?....

el objetivo es probar que $x=y=z$.

[hide="una ayuda"]Procedamos por contradicción:
Suponga que $x>y$  entonces $x^3>y^3$ entonces $2y-1>2z-1$ entonces.... completen hasta llegar a una contradicción  :D [/hide]

$Tipe$

Ok ahora si eso si es de mucha ayuda!!
gracias!!
 :D


I just rated a post $ 1$, and somehow that caused the overall rating of that post to go from $ 1$ cube to $ 1.5$ cube. What...in the world?

Someone else rated the post as well. The average after your rating went to 1.5 cubes. It just didn't show the previous voted before you rated it.


Prove that: $ 1.1\plus{}1.2\plus{}2.5\plus{}...\plus{}F_nP_n\equal{}\frac13(F_{n\plus{}1}P_n\plus{}F_nP_{n\plus{}1})$


Welcome to the Pytheagle Mathematical Competition (PMC)!

[b]If you want to join, post your name on the forum. DO NOT PM ME.[/b]

Here are some things you need to know before taking the test:

1) Calculators are not permitted
2) You may take as much time as necessary
3) You may not use any resources other than yourself
4) Each question is out of 5 points and is judged by the 3 C's (completeness, correctness, and clarity)
5) Solutions are required
6) It is due by 12:00:00 EST August 3rd.
7) ANY SUSPICION AS TO CHEATING WILL RESULT IN DISQUALIFICATION

There will be 4 rounds of PMC. The dates are hereby unknown and the difficulty is random.

You have 2 choices as to how to send me your solutions:

1) Make a PDF and PM/E-mail it to me
2) PM the solutions USING THE HIDE TAG

Have fun!

Did I mention people could also sign up to be part of the grading commitee :?:

Post any questions you have on the forum.

[quote="Problem number 2"]Annie can do a job in 3 hours, Bob can do it in 4, Carter can do it in 5, and Daniel
can do it in 6. [b]Each person works on the job individually for 23 minutes.[/b] Then, they
work together. How long will they have to work together to finish the job?[/quote]

Do you mean that they split the time up and work alone or each person works on there on job?

Oh and I join

Each person takes turns working for 23 minutes.

[quote="5849206328x"]I'll volunteer for grading.  I need a way to waste time and it'll be interesting as well I suppose.[/quote]

Sign up for grading...

Notes to my grading:

If you bash, you better do it correctly, since bashing makes me ANGRY so i will grade your paper more strictly.

Bump. Is the test too hard? Is that why people aren't joining?

Can people PLEASE post some feedback? I don't care if it's EXTREMELY negative or ANYTHING, I just want to know how to improve this. This is my first tournament and I don't know what to improve where. It will become popular once I get some experience.

I think I'd classify it as High School basics.

On question 4, can k be imaginary, because if not, then
[hide] There are no solutions or possible values of k that work, i think :maybe: [/hide]

i sign up

[color=white][size=9]tho i will most likely fail[/size][/color]

I will join  :lol:
[quote="pythag011"][quote="5849206328x"]My best feedback would probably be to be patient.  It's only been about a day or two since you opened sign-ups :wink:[/quote]

This + "You may take as much time as needed."

When you are given such a constraint in competitions (for example, usamts), typically people don't submit until they need to.[/quote]

I agree with that, but I also think you picked a pretty bad weekend to start the contest. Lots of people are going to AMSP (I think its this week?) and some are going to NSF nats.

I need his email!! I was out all week all I need is to send it!

I won't have time to do it, so I drop out. Could you post the solutions sometime?

Can you post solutions to the problems at least?

Credit goes to Rd5493 for providing excellent solutions.

[hide="1)"]In order for us obtain a perfect square as a sum, we must add together powers of $ 2$ and $ 3$ so that they themselves are perfect squares.  This means that both $ m$ and $ n$ must be even.  We can then think of the problem as $ a^2\plus{}b^2\equal{}p^2$ where $ a$ and $ b$ are the values of $ 2^m$ and $ 3^n$ respectively.  The smallest such values for $ m$ and $ n$ in this case are $ 4$ and $ 2$ respectively, making the statement $ 2^4\plus{}3^2\equal{}p^2$ which makes $ p\equal{}5$.  To make sure we have found the smallest possible case, we can check all other cases that exist such that $ m\plus{}n< 6$.  Since $ m,n \ge 1$, we only have a few cases to check.  Since $ m$ and $ n$ also have to be even, we have only $ 2$ cases to check, both of which result in numbers that are not perfect squares, so our answer is that the smallest possible value of $ p$ occurs when $ m\equal{}4,n\equal{}2$.  So our answer is $ \boxed{p\equal{}5}$[/hide]

[hide="2)"]Our first step involves adding together all $ 3$ equations to produce
$ 27(a\plus{}b\plus{}c)\equal{}162$
$ a\plus{}b\plus{}c\equal{}6$

We then can multiply this equation by $ 9$ to set it equal to one of the given equations.
$ 9a\plus{}9b\plus{}9c\equal{}54\equal{}8a\plus{}10b\plus{}9c$

Simplifying, we get:
$ 9a\plus{}9b\equal{}8a\plus{}10b$
$ a\plus{}9b\equal{}10b$
$ a\equal{}b$

Knowing this, we can then change our simplified equation from $ a\plus{}b\plus{}c\equal{}6$ to $ 2a\plus{}c\equal{}6$.  Then by multiplying the second equation given by $ 2$, we get
$ 18a\plus{}16b\plus{}20c\equal{}90$, which can be set equal and simplified to 15(2a+c):
$ 34a\plus{}20c\equal{}90\equal{}30a\plus{}15c$
$ 5c\equal{}\minus{}4a$
$ \frac{\minus{}4}{5}a\equal{}c$

This makes our addition statement now read
$ 2a\minus{}\frac{4}{5}a\equal{}6$
$ \frac{6}{5}a\equal{}6$
$ a\equal{}5$

This means that $ b\equal{}5$ and that $ c\equal{}\minus{}4$
Thus $ \boxed{abc\equal{}(5)(5)(\minus{}4)\equal{}\boxed{\minus{}100}}$[/hide]

[hide="3)"]Annie can do the job in $ 3$ hours, which means that she will do $ \frac{1}{3}$ of the job in an hour.  In the same manner, a job that can be done in $ h$ hours, can do $ \frac{1}{h}$ of the job in an hour.  The amount of the job that it takes Annie, Bob, Carter, and Daniel to do in one hour is $ \frac{1}{3},\frac{1}{4},\frac{1}{5},$ and $ \frac{1}{6}$.  Since each person will work for $ 23$ minutes alone, we can multiply each fraction by $ \frac{23}{60}$ and then add them together, which gives us a total of $ \frac{437}{1200}$ of the job done after each works alone.  This means that we still have $ \frac{763}{1200}$ of the job left to do together.  Combined, they can do $ \frac{19}{20}$ of the job in an hour.  We multiply $ \frac{19}{20}$ by the amount of time they will need to complete the job, $ \frac{x}{60}$.  We can then set up a proportion:
$ \frac{763}{1200}\equal{}\frac{19x}{1200}$
$ 763\equal{}19x$
$ \boxed{x\equal{}\frac{763}{19}}$ or $ \boxed{x\equal{}40\frac{3}{19}}$ minutes[/hide]

[hide="4)"]As $ (x\minus{}2k)$ is a factor of $ x^4\plus{}3x^2\plus{}kx\plus{}6$, then $ 2k$ must be one of the roots. Then, \[ 16k^4\plus{}12k^2\plus{}2k^2\plus{}6\equal{}0\] Simplifying, we get \[ 8k^4\plus{}7k^2\plus{}3\equal{}0\] Now plug in $ \alpha$ for $ k^2$. We now have the quadratic $ 8\alpha^2\plus{}7\alpha\plus{}3$.  We find the roots to be negative, but the trivial inequality states it is not possible. Thus our answer is $ 0$.[/hide]

[hide="5)"]By a well-known trigonometric formula stating $ A\equal{}\frac{1}{2}ab\text{sinC}$, we get an answer of $ \frac{375\sqrt{2}}{2}$.[/hide]

[hide="6)"]By Vieta’s, the answer is $ \minus{}\frac{7}{2}$.[/hide]

hmmm, for #1, wouldnt it be -5? (-5)^2=5^2, -5<5

anyways, sry for not PMing the answers. I thought I failed miserably, but it turns out I got 5 of the questions right! :oops:

@ Thunder: Actually, I think that $ p$ had to be positive.  If not, then your solution would be the correct answer.

[quote="pytheagle"]Um... I'm sorry but PMC is canceled. Only 3 people (including modularmarc) participated, and the graders didn't grade any of them completely. Next time, I have to remember to leave the grading to myself....

Sorry, I will restart PMC with new problems later. I hope it's not horrible then...[/quote]

Woops, missed the solutions. Also, when does "later" mean? Again, thanks for the problems.

A question: Can I compile a PDF version of the solutions?


inequalities quadratics function calculus derivative AMC

darkquantum wrote:let a=x^2+2

FInd the minimum value of a/sqrt[a-1]

[hide]

a/sqrt(a-1) = a*sqrt(a-1)/(a-1)

So then because b^2 - 4ac>=0

then a^2 - 8 >=0

a^2>=8

a>=2 :rt2:

So we plug in 2 :rt2:  for a to get

(2*sqrt(2 -  :rt2:  ))/( :rt2:  - 1)

2 * sqrt(2 :rt2:  - 2) + 2 * sqrt( 2 -  :rt2:  ) ???[/hide]

Sorry, Chinaboy, but to my knowledge, that is not correct. What I don't understand is how you got b^2-4ac>=0. Can you explain where that came from, since we don't ahve a quadratic here?

ooooo.... oops... yah... I see the mistake. thanks

Ok then... we find the vertex (substituting a for y to get y = x^2 + 2)

So the formula for finding the vertex is c - b^2/4a
y value for the vertex is 2 so then
wouldn't the answer be 2??? Or am I going at the wrong direction???

Does finding the minimum value of some function necessarily mean finding the minimum value of the variable and substituting it in?

What do you mean by finding the minimum value of the variable? In this case, there is no minimum for x....

And, I believe 2 is the correct solution. Here is my solution, in spoiler:

[hide]

We are trying to find the minimum of (x^2+2)/(sqrt[X^2+1]). So, we can rewrite the numerator as X^2+1+1/sqrt[x^2+1], which after we divide, is sqrt[x^2+1]+1/sqrt[x^2+1]. Then, we apply the AM-GM inequality, to get that the minimum value of (x^2+2)/(sqrt[X^2+1]) is 2. [/hide]

Now it makes sense why the title is an easy inequality. thanks.

Well, maybe its not THAT easy, at least to some people.

But for others, I guess it's pretty easy.

Sorry, I didn't make much sense.. but I think what Chinaboy was doing was finding the smallest value of a and then substituting that into a/sqrt(a-1). That won't work for many problems.

Um, he's actually not finding the minimum value. See, by finding the turning point, wat he did was find the value of x for which the expression in minimized, and then plugging it in to see wat that minimal value is.

ANd you are rite, finding minimum values for varaibles does not work, since in most cases, THERE IS NO MINIMUM VALUE FOR THE VARIABLES!!!

Well, when you solve quadratics, most people remember the formula [u][i]that will find spit out the correct value for x that corresponds to minimum/maximum value for f(x)[/i][/u].  This formula is easier to remember, since it's -b/2a, I believe...

Good.  I'm losing my algebra...it's been too many years of other stuff.

haha.
learn the differentials and multivariate, forget the multiplication table.

NIce way to get ahead in life!  :D

Hint: [hide]the expression is obviously always positive.  Therefore, square it, and minimize that new expression, then take the square root of your result.[/hide][/hide]

This looks like a job for chain rule and quotient rule...

For all of you calculus nutters, consider multivariable calculus in the context of the following problem (shamelessly stolen from Thomas Mildorf and the Mandelbrot Problem Corner)

Let $a, b, c$ be positive reals. Determine the minimum possible value of

$\displaystyle \frac{a^3}{4b} + \frac{b}{8c^2} + \frac{1+c}{2a}$

EDIT: Changed the $a^2$ to an $a^3$. My mistake, sorry guys.

You'd just take the partial derivatives with respect to each variable, set them equal to 0, and solve the simultaneous equations to find critical points, which you'd then test to see which gives the global minimum, right?  I'd work out the math, but it's late where I live and my brain is not working well right now  :blush:

[quote]You'd just take the partial derivatives with respect to each variable, set them equal to 0, and solve the simultaneous equations to find critical points, which you'd then test to see which gives the global minimum, right?[/quote]

You're probably right.

Have fun with that.

[quote="darktreb"]For all of you calculus nutters, consider multivariable calculus in the context of the following problem (shamelessly stolen from Thomas Mildorf and the Mandelbrot Problem Corner)

Let $a, b, c$ be positive reals. Determine the minimum possible value of

$\displaystyle \frac{a^2}{4b} + \frac{b}{8c^2} + \frac{1+c}{2a}$[/quote]

There is no minimum there; the extremum $\sqrt[4]{\frac{1}{2}}$ is never obtained.  In the original, we have $a^3$ instead of $a^2$, as in

$\frac{a^3}{4b} + \frac{b}{8c^2} + \frac{1+c}{2a}$

[quote="treegoner"]Why derivative ? It is just AM -GM inequality  :P
  $\sqrt {a-1} + \frac{1}{\sqrt{a-1}} \geq 2$
  The equality holds when $a=2$ or $x = 0$[/quote]

This is best solution!

Yeah, thanks Thomas for pointing that out. I was looking at it a moment ago and realized something would be missing. I've made the correction to the problem posted earlier.


geometry
3D geometry
search
number theory open
number theory

You could try searching as this has been posted too many times before. For example, here: http://www.mathlinks.ro/viewtopic.php?search_id=177336861&t=33776

EDIT: Here is another one: http://www.mathlinks.ro/viewtopic.php?search_id=1026196883&t=134862

Thank you!!! what a nice proof, simpler than I thought!!


superior algebra
superior algebra unsolved

a) Find
           \[ \left [\mathbb{Q} (\sqrt{5},\sqrt{7}): \mathbb{Q}\right], \left [\mathbb{Q} (\sqrt{a},\sqrt{b}): \mathbb{Q}\right]  \]

           \[ \left [\mathbb{Q} (\sqrt{5},\sqrt [3] {7}): \mathbb{Q}\right], \left [\mathbb{Q} (\sqrt{a},\sqrt [3]{b}): \mathbb{Q}\right],  \]

   where $a, b\in\mathbb{Q_{+}}^{*}$.

b) Prove that \[ \mathbb{Q} (\sqrt{5},\sqrt{7})=\mathbb{Q}(\sqrt{5}+\sqrt{7})  \] and

                    \[ \mathbb{Q} (\sqrt{5},\sqrt [3] {7})=\mathbb{Q}(\sqrt{5}+\sqrt [3] {7}).  \]

a)
Asuming only the nontrivial $a,b$ each:
The degree is $4$ for the first two except iff $ab$ is a rational square.
For the second two, it's $6$ with no further condition.

b) $\frac{1}{\sqrt{5}+\sqrt{7}} = \frac{\sqrt{7}-\sqrt{5}}{2}$, thus $\sqrt{5}-\sqrt{7}$ is contained. From that point on it's trivial.
For the second one: similar reasoning.

Note that there is http://www.mathlinks.ro/Forum/viewtopic.php?t=27089 which pwns most of the things you posted recently ;)


search
function
number theory unsolved
number theory

Find all positive integers n such that n^3 + 3^3 is a perfect square.

I ve just determined what vales the last digit of n can take but i dont think this is useful.So i dont have any idea.What about you??

Iris Aliaj please STOP posting the same problems over and over again. I deleted many duplicates of yours, so please keep track of your posts, or at least use the search function to search for problems you (or others for that matter) posted, OK?

[quote="grobber"]Haven't you posted this already?[/quote]
It was solved already?

We know it. But we are so lazy to write complete solution, that why I asked whether it was already solved.


Let $ a,b,c>0$.Prove that $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \geq \frac{a\plus{}b\plus{}c}{^3\sqrt{abc}}$

[quote="SUPERMAN2"]Let $ a,b,c > 0$.Prove that $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {a \plus{} b \plus{} c}{^3\sqrt {abc}}$[/quote]

By AM-GM we have :

$ 3(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} )\equal{}\sum ( \frac {a}{b} \plus{} \frac {a}{b} \plus{} \frac {b}{c})$

$ \geq 3\sum \frac{a}{\sqrt[3]{abc}}$

Done ! :)

I THINK IT IS ALSO VERY EASY IF WE USE MUIRHEAD INEQUALITY :wink: ..NICE EXERCISE!!! :)


How do you solve this with balls and urns?:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158301[/url]

Strictly speaking, balls and urns only applies to situations where the coefficients are $ 1$, by which I mean equations like

$ a\plus{}b\plus{}c\plus{}... \equal{} n$

Whereas the coin problem depends on solving the equation

$ a\plus{}5b\plus{}10c\plus{}25d \equal{} n$

Which is quite different and not solvable with the same technique.

Er... would the extended version by ordinary boy work?:
[quote="ordinary boy"]I get an answer of $ 93$ using an extended form of balls and urns.
$ 2\binom{8}{2}\plus{}\binom{6}{2}\plus{}\binom{5}{2}\plus{}4\binom{3}{2}\equal{} 93$[/quote]

He didn't generalize balls and urns.  If you read his reasoning carefully, he just used some mods and did some simple casework, which is usually a decent approach for problems of this type.  It doesn't generalize readily, though.


How do I prove by induction that every positive integer less than n! can be written as the sum of distinct divisors of n!, where n is a positive integer?

Check for small $ n$ ($ 1$, $ 2$, ...), Assume true for $ n$. Take $ 0 < x < (n\plus{}1)!$. Then $ x \equal{} (n\plus{}1)y \plus{} z$, with $ y < n!$ and $ z \leq n$. So $ y \equal{} \sum_{d \in Y} d$, $ z \equal{} \sum_{e \in Z} e$, for some subsets $ Y$, $ Z$ of the divisors of $ n!$. Therefore $ x \equal{} \sum_{d \in Y} (n\plus{}1)d \plus{} \sum_{e \in Z} e$, using distinct divisors of $ (n\plus{}1)!$, since $ e \leq n < (n\plus{}1)d$ for all $ d \in Y$ and $ e \in Z$.


[quote="The Captain"]I found the following problem interesting and want to share it with you :)
Prove that the number of ways we can partition a natural number $n$ with $2,3,4$($3$ is necessarily used)is equal to the number of triangles of natural sides and circumference $n$ .[/quote]

Your problem is wrong and I think your meant was:  ;)

Prove that the number of triangles with natural sides with circumference $n$ is equal to the number of ways that we can partition a natural number $n-3$ to the numbers $2,3,4.$  :D

[quote="The Captain"]We are saying the same thing.[b]ehsan2004[/b]. :) So if you feel more comfortable with your notion,there's no problem.[/quote]

 ;)

[quote="fesghelbahal"]Hey ehsan2004 you are wasting your time and also the others time.
I have a simple solution but i'm busy and I can't describe it full.But here is a hint :
you can find a $1-1$correspondence between the number of triangles and the partitions satisfying the problems condition.notice that you have to partition each number $4$ to numbers $3$ and $1$ and notice that the number of [u]$3$[/u]s is greater then the number of [u]$1$[/u]s...[/quote]

 :clap2:


We have a number greater than $1000$.
Say we perform these operations on it:
1. Add $9$ to it.
2. Interchange the last two digits of it the new number.
3. Interchange the first and third digits of the new number.
4. Double the new number.

If the resultant number is equal to the original number, what is the original number?

[hide]First, we would need to translate the number into math. Assuming that the number is four digits, it can be written as $1000a+100b+10c+d$. Going step by step, we get $1000a+100b+10c+(d+9)$. Here, I'm just going to hope that d<0 and change it to $1000a+100b+10(c+1)+(d-1)$. Next, interchanging the last two digits, we get $1000a+100b+10(d-1)+(c+1)$. Interchanging 1 and 3, we get $1000(d-1)+100b+10a+(c+1)$. Doubling that, we get $2000(d-1)+200b+20a+2(c+1)$, which gives the equations $2000(d-1)+200b+20a+2(c+1)=1000a+100b+10c+d$. This can be reduced to $1999d+100b=980a+8c+1998$. We are now looking for answers that are digits. First of all, d must be even, because all of the other numbers are even. Doing some casework starting with d=2 quickly gives the answer [b]2052[/b][/hide]


[quote="GameBot"]Each wheel on Sally's new bicycle has a diameter of 60 cm. How many revolutions does a wheel make with Sally pedaling a distance of one kilometer? Use $ \pi \equal{} 3.14$ and round your answer to the nearest revolution.[/quote]

Every revolution is $ 60\pi\equal{}188.4$ cm. $ 1 km\equal{}1000 cm \approx 5$ rotations.

[quote="BOGTRO"]Every revolution is $ 60\pi\equal{}188.4$ cm. $ 1 km\equal{}1000 cm \approx 5$ rotations.[/quote]

One kilometer does not equal 1000 cm.

ok first $\pi = 3.14$ and diameter is $60$

$(3.14)(60) = 188.4$

A kilometer is $1000$ meters and a $100$ centimeters is a meter so $100*1000 = 100,000$

then we do $100000/188.4$ is about $530.8$

so the answer is $531$ revolutions

Ha, no wonder I always got this problem wrong. So 1km=100000cm, and same as above.


probability
percent
probability and stats

If an employee shrinks his work 30 percent of the time, what is the probability that he will be caught if his boss checks on him 4 times at random?


Hi,
Is there a formula to find the answer to this problem?
In the sequence 13, 10, 1, 3, ... each term, after the first, is formed by adding the tens digit and three times the units digit of the previous term.  How many distinct perfect squares will appear in this sequence?

Thanks,
Joshim5

Through recursions, but as far as mathcounts is concerned, you can find it out constructively; for example, the sequence can never reach 9+3*9=36, etc.

It could probably be proven (quickly or long) that every square less than 36 is included (or so), which gives the 5


1) Find $\inf \left\{ \frac{\phi(n)}{n}| n \ge 2, n \in \mathbb{N}\right\}$.

2) Evaluate $\prod_{i=1}^{\infty}\frac{2^{i}-1}{2^{i}}$.

(Disclaimer :  I have not solved these, so I have no idea of how difficult they are.)


real analysis
integration
limit
calculus
function
Support
topology

f is $ \mathcal{L}([a,b])$, Lebesgue integrable.

$ \frac{1}{h}\int_0^h | f(x\plus{}t) \minus{} f(x) | dt$ is decreasing when h approach 0 ?

Is it right?

This is from the equation.
$ \int_{\minus{}\infty}^{\plus{}\infty} [ \lim_{h\to 0} 1/h \int_0^h | f(x\plus{}t) \minus{} f(x) | dt ] dx\equal{}
\lim_{h\to 0} \int_{\minus{}\infty}^{\plus{}\infty} [ 1/h\int_0^h | f(x\plus{}t) \minus{}f(x) | dt ] dx$

This is lim and integral interchange. That book says this is because Levi Lemma.

If this ,we shall prove $ \frac{1}{h}\int_0^h | f(x\plus{}t) \minus{} f(x) | dt$ is decreasing when h approach 0.

How to prove the decreasing of $ \frac{1}{h}\int_0^h | f(x\plus{}t) \minus{} f(x) | dt$?

Thanks

No way. It hasn't the slightest desire or obligation to be decreasing in general.

[quote="water2011"]
This is from the equation.
$ \int_{ \minus{} \infty}^{ \plus{} \infty} [ \lim_{h\to 0} 1/h \int_0^h | f(x \plus{} t) \minus{} f(x) | dt ] dx \equal{} \lim_{h\to 0} \int_{ \minus{} \infty}^{ \plus{} \infty} [ 1/h\int_0^h | f(x \plus{} t) \minus{} f(x) | dt ] dx$

This is lim and integral interchange. That book says this is because Levi Lemma.

[/quote]

can you explain why the equation holds? Condition is $ f\in \mathcal{L}([a,b])$.

Uniform integrability (or whatever the right term in English is) is sufficient to justify swapping of the integral and the limit on a finite interval. The criterion is that for every $ \varepsilon > 0$ there is $ \delta > 0$ such that for every set $ E$ of measure $ < \delta$ and every $ n$, $ \int_E |f_n|\le \varepsilon$. This statement is almost as useful as the dominated convergence theorem (as a matter of fact, it implies dominated convergence theorem). I'm not sure to whom we should attribute it; if your book says Levi, perhaps, it is indeed Levi.

On the other hand, what puzzles me is how do you know that the limit exists a.e. :? I mean, I know it is $ 0$ a.e., but from a completely different argument.

You say $ \lim_{h\to 0} \int_{ \minus{} \infty}^{ \plus{} \infty} [ 1/h\int_0^h | f(x \plus{} t) \minus{} f(x) | dt ] dx\equal{}0$?

In particular, but I rather mean $ \lim_{h\to 0}
\frac{1}{h}\int_0^h | f(x\plus{}t) \minus{} f(x) | dt$. How do you know it exists?

[quote="fedja"]In particular, but I rather mean $ \lim_{h\to 0} \frac {1}{h}\int_0^h | f(x \plus{} t) \minus{} f(x) | dt$. How do you know it exists?[/quote]

Good question.

if $ \lim_{h\to 0} \frac {1}{h}\int_0^h | f(x \plus{} t) \minus{} f(x) | dt\equal{}0$, x is Lebesgue point.

You mean the limit may not exist.

$ r_n\in\mathcal{Q}$,$ f(x)\minus{}r_n$ is in $ \mathcal{L}([a,b])$.
$ \lim_{h\to 0}\int_x^{x\plus{}h}|f(u)\minus{}r_n|du\equal{}|f(x)\minus{}r_n|\quad a.s.$
$ E(r_n)$ is the exceptional set, $ m(E(r_n))\equal{}0$.
$ E\equal{}(\cup_{n\equal{}1}^\infty E(r_n))\cup E(|f|\equal{}\plus{}\infty),m(E)\equal{}0$,

We can prove $ [a,b]\minus{}E$ is Lebesgue point set.

$ \forall x\in [a,b]\minus{}E, \forall \varepsilon >0,\exists \delta>0$,when $ |h|<\delta$ ,we have
$ \frac {1}{h}\int_0^h | f(x \plus{} t) \minus{} f(x) |dt<\varepsilon$.

Now according the definition of lim ,we got the answer of your question.

[quote="fedja"]Yeap, that is the standard argument. I just had an (I hope, false) impression that the formula you mentioned was aiming at proving that almost every point is a Lebesgue point (indeed, to prove that the limit of the right hand side is $ 0$ is much easier: just approximate $ f$ by a continuous function with compact support in $ L^1$ and you'll be done in no time; also, the convention is to write formulae in such a way that the information about the left hand side is derived from that about the right hand one). So I was worried a bit about the possibility that you might be reading some fallacious argument :)[/quote]

Yes , what I mentioned (swapping of lim and integral equation) is from another method to prove almost every point is a Lebesgue point .
Can you say more details of "approximate $ f$ by a continuous function with compact support...."?

Let's denote the integral on the right by $ I_h(f)$. Then $ I_h(F\plus{}G)\le I_h(F)\plus{}I_h(G)$, $ I_h(F)\to 0$ as $ h\to 0$ for every continuous function $ F$ with compact support and $ I_h(G)\le 2\|G\|_{L^1}$ for any $ G$. So, in order to show that $ I_h(f)$ is eventually $ <\varepsilon$, you can just represent $ f\equal{}F\plus{}G$ where $ F$ is continuous with compact support and $ \|G\|_{L^1}<\frac \varepsilon 3$.

[quote="fedja"]Let's denote the integral on the right by $ I_h(f)$. Then $ I_h(F \plus{} G)\le I_h(F) \plus{} I_h(G)$, $ I_h(F)\to 0$ as $ h\to 0$ for every continuous function $ F$ with compact support and $ I_h(G)\le 2\|G\|_{L^1}$ for any $ G$. So, in order to show that $ I_h(f)$ is eventually $ < \varepsilon$, you can just represent $ f \equal{} F \plus{} G$ where $ F$ is continuous with compact support and $ \|G\|_{L^1} < \frac \varepsilon 3$.[/quote]

You mean $ I_h(f) \equal{} \frac{1}{h}\int_0^h |f(x\plus{}t)\minus{}f(x)|dt$.
Then why $ f$ can be expressed as $ f\equal{}F\plus{}G$?
That use what theorem?

No, my $ I_h$ is the integral of your $ I_h$, i.e., a number, not a function.

I use regularity of Lebesgue measure plus approximation of arbitrary functions by elementary step-functions. By regularity of measure I mean the statement that for every Borel set $ E$, there exist a closed set $ F\subset E$ and and open set $ G\supset E$ such that $ m(G\setminus F)$ is arbitrarily small. Then we can construct a continuous function $ f$ (continuous descent from $ F$ within $ G$) such that ${ f-\chi_{{}_{\scriptstyle E}}}$ is small in $ L^1$.

Someone says the swapping is due to Beppo Levi,namely MCT.
If that, we shall prove $ \frac{1}{h}\int_0^h |f(x\plus{}t)\minus{}f(x)|dt$ increasing when $ h \to 0$.
Is it right?

[quote="fedja"]In the Monotone Convergence Theorem, we need to prove that it is either increasing and bounded from below by an integrable function, or decreasing and bounded from above by an integrable function.[/quote]

I don't know that Monotone convergence theorem you mentioned.

Can you give more details of your MCT?

The Monotone Convergence Theorem is the theorem you referred to here:
[quote="water2011"]Someone says the swapping is due to Beppo Levi,namely MCT.[/quote]
You did imply that you know the theorem already.


Let a,b,c are possitive real nubers such that $ a\plus{}b\plus{}c\equal{}3$. Prove the inequality:
$ \sqrt{a^3\plus{}b^2\plus{}c}\plus{}\sqrt{b^3\plus{}c^2\plus{}a}\plus{}\sqrt{c^3\plus{}a^2\plus{}b} \geq 5$

[quote="hedeng123"][quote="AngelHolder"]Let a,b,c are possitive real nubers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove the inequality:
$ \sqrt {a^3 \plus{} b^2 \plus{} c} \plus{} \sqrt {b^3 \plus{} c^2 \plus{} a} \plus{} \sqrt {c^3 \plus{} a^2 \plus{} b} \geq 5$[/quote]

Using Holder Inequality

$ (\sqrt {a^3 \plus{} b^2 \plus{} c} \plus{} \sqrt {b^3 \plus{} c^2 \plus{} a} \plus{} \sqrt {c^3 \plus{} a^2 \plus{} b})^2$ $ \sum (a^3 \plus{} b^2 \plus{} c)^2\geq (\sum(a^3 \plus{} b^2 \plus{} c))^3$

we only prove

$ \frac {(\sum(a^3 \plus{} b^2 \plus{} c))^3}{\sum (a^3 \plus{} b^2 \plus{} c)^2}\geq 25$

The inequality  holds, but I can not prove it.[/quote]

>f:=numer(subs({a=3*x/(x+y+z),b=3*y/(x+y+z),c=3*z/(x+y+z)},csum(a^3+b^2+c,[a,b,c])^3-25*csum((a^3+b^2+c)^2,[a,b,c]))):

>csymprove(f);

                   F1=f(x,y,z)+f(x,z,y) is symmetric for {z,x,y}

                               DEGU = 3

                               DEG = 3

                              ITS = [ee]

                            IS = 1, N = 1

                        The inequality holds.

                              ITS = [ee]

                            IS = 1, N = 1

                        The inequality holds.

                              ITS = [ee]

                            IS = 1, N = 1

                        The inequality holds.

                              ITS = [ee]

                            IS = 1, N = 1

                        The inequality holds.

                     F1 is psd!(SYM3(1))

                            它是平凡非负的!

                             TIME = 0.360

                  F2=f(x,y,z)f(x,z,y) is symmetric for  {z, x, y}

                               DEGU = 6

                               DEG = 6

                              ITS = [ee]

                            IS = 1, N = 1

                        The inequality holds.

                              ITS = [ee]

                            IS = 1, N = 1

                        The inequality holds.

                              ITS = [ee]

                            IS = 1, N = 1

                        The inequality holds.

                              ITS = [ee]

                            IS = 1, N = 1

                        The inequality holds.

                              ITS = [ee]

                            IS = 1, N = 1

                        The inequality holds.

                              ITS = [ee]

                            IS = 1, N = 1

                        The inequality holds.

                              ITS = [ee]

                            IS = 1, N = 1

                        The inequality holds.

                    F2  is psd!(SYM3(1))

                            它是平凡非负的!

                             TIME = 1.843

                   The original form is psd!(SYM32)

[quote="AngelHolder"]Let a,b,c are possitive real nubers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove the inequality:
$ \sqrt {a^3 \plus{} b^2 \plus{} c} \plus{} \sqrt {b^3 \plus{} c^2 \plus{} a} \plus{} \sqrt {c^3 \plus{} a^2 \plus{} b} \geq 5$[/quote]

we prove the stronger:

$ \sqrt {a^3 \plus{} b^2 \plus{} c} \plus{} \sqrt {b^3 \plus{} c^2 \plus{} a} \plus{} \sqrt {c^3 \plus{} a^2 \plus{} b} \geq 3\sqrt{3}.$

both sides of the square

$ \iff \sum {\left( {a^3  \plus{} b^2  \plus{} c} \right)}  \plus{} 2\sum {\left(\sqrt {a^3  \plus{} b^2  \plus{} c} \sqrt {b^3  \plus{} c^2  \plus{} a} \right)}  \ge 27,$

by CS, we have

$ \sum {\left( {\sqrt {a^3  \plus{} b^2  \plus{} c} \sqrt {b^3  \plus{} c^2  \plus{} a} } \right)}  \ge \sum {\left( {a^2  \plus{} b^{\frac{5}{2}}  \plus{} c^{\frac{3}{2}} } \right)},$

so we just need to prove that

$ \sum {\left( {a^3  \plus{} b^2  \plus{} c} \right)}  \plus{} 2\sum {\left( {a^2  \plus{} b^{\frac{5}{2}}  \plus{} c^{\frac{3}{2}} } \right)}  \ge 27$

$ \iff \sum {\left( {a^3  \plus{} 2a^{\frac{5}{2}}  \plus{} 3a^2  \plus{} 2a^{\frac{3}{2}}  \plus{} a} \right)}  \ge 27,$

by AM-GM, we have

$ \sum {\left( {a^3  \plus{} 2a^{\frac{5}{2}}  \plus{} 3a^2  \plus{} 2a^{\frac{3}{2}}  \plus{} a} \right)}  \ge \sum {9a^2 }  \ge 3\left( {a \plus{} b \plus{} c} \right)^2  \equal{} 27,$

done.

It's weaker than my old result. See problem and solution here (#12)  :) :
http://www.maths.vn/forums/showthread.php?t=26809&page=2

[quote="AngelHolder"]Let a,b,c are possitive real nubers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove the inequality:
$ \sqrt {a^3 \plus{} b^2 \plus{} c} \plus{} \sqrt {b^3 \plus{} c^2 \plus{} a} \plus{} \sqrt {c^3 \plus{} a^2 \plus{} b} \geq 5$[/quote]
Actually, the more general inequality holds:

Let $ a,$ $ b,$ $ c$ be positive real numbers such that $ a\plus{}b\plus{}c\equal{}3.$ If $ k \ge 0,$ then
$ \sqrt{a^3\plus{}kb^2\plus{}c}\plus{}\sqrt{b^3\plus{}kc^2\plus{}a}\plus{}\sqrt{c^3\plus{}ka^2\plus{}b} \ge 3\sqrt{k\plus{}2}.$

The proof of this inequality, see here: http://can-hang2007.blogspot.com/2010/01/inequality-34-n-t-hieu-v-q-b-can.html


aha i love them they're so cute!! lolz

  :ddr:  :starwars:  :10:

[quote="1=2"]My favorite ones are the ones that are not listed in the list. Only one found so far. :([/quote]

Actually, there are two.
[img]http://www.artofproblemsolving.com/Forum/images/smiles/mrgreen.gif[/img]
[img]http://www.artofproblemsolving.com/Forum/images/smiles/happy.gif[/img]

If you are wonder these are the two sites:

[url]http://www.artofproblemsolving.com/Forum/images/smiles/mrgreen.gif[/url]
[url]http://www.artofproblemsolving.com/Forum/images/smiles/happy.gif[/url]

is there anyway to get the santa green and happy face as avatras in the classroom? like whats their avatar  id

[quote="herefishyfishy1"]Guess what? I voted 1! I hate when people randomly spam all the shoutboxes with random emoticons! IT'S AN OUTRAGE![/quote]
I think its funny
I like all the emotes They're cute :D  :D  :D


function
real analysis
real analysis theorems

Hello everyone, this issue reads:

Find the orthogonal trajectories of the family of curves $ y \equal{} C_1e^{\minus{}x}$

Well, by definition, the steps are these:

First solve for $ C$. and $ C \equal{} \displaystyle\frac{y}{e^ {\minus{}x}}$

differentiate the equation:

$ y * (y\prime) \plus{} Ce^{\minus{}x} \equal{} 0$, you get rid of the constant C and finally going to be that, by definition,

$ y_2' \equal{} \minus{} \displaystyle\frac{1}{y'}$

But I have problems to find the answer of the book is

$ \displaystyle\frac{y^2}{2} \equal{} x \plus{} C$

Thank you very much,

greetings

You have found a differential equation for your functions $ y$. Solve it.

Hello, what happens is that if I replace

$ C \equal{}\frac{y}{e^{\minus{}x}}$ in $ y * (y\prime)\plus{}Ce^{\minus{}x}\equal{} 0$ I have:

$ (y\prime) \equal{} \minus{}1$

however, by the definition I quote:

$ y_{2}' \equal{}\minus{}\frac{1}{y'}$ be the solution I'm looking for is integrated on both sides:

$ y_2' \equal{} 1 \Longrightarrow{ \displaystyle\frac{dy}{dx} \equal{} 1}  \Longrightarrow{dy \equal{} dx}$

$ y \equal{} x \plus{} C$

But I have problems to find the answer of the book is

$ \frac{y^{2}}{2}\equal{} x\plus{}C$

I do not understand why I do not care.

What is wrong?

Greetings

Ah- I missed something. Rewrite that equation entirely in terms of the new functions $ y_2$. Mixing in the old $ y$ is bad.

Now, what do we know about the old curves $ y\equal{}Ce^{\minus{}x}$? There, $ y'\equal{}\minus{}y$. At the particular point we're looking at, $ y_2$ and $ y$ have the same value, so the differential equation becomes $ y_2'\equal{}\frac1{y_2}$. That's an equation you can solve.

No. I do not understand

Why do you say that $ y'=-y$

to me is that $ (y\prime) =-1$

$ y = C_{1}e^{-x}$

I'll apologize, I take to see why this other equation does not agree with the book:

Find the family of curves of

$ \cos(y) = Ce^{ - x}$

I have to derive:

${ cos(y)  = Ce^- x}$
$ -sen(y)y' =  - Ce^{- x}$

eliminating the C is:

$ cos (y) = sen(y)y'$

Now, by definition

$ y_2 = - \displaystyle\frac{1}{y'}$ where is that:

$ y_2 = -tan(x) \Longrightarrow{  \displaystyle\frac{dy}{dx}  = -tan(x)} \Longrightarrow{dy = -tan(x).dx}$

integrand becomes:

$ y = Ln(Cos(x)) + C$

But the answer of the book is:

$ sen(y) = K.e^{-x}$

where does?

Thank you very much

Greetings :)


1. In acute triangle $ABC$ point $O$ is circumcenter, segment $CD$ is a height, point $E$ lies on side $AB$ and point $M$ is a midpoint of $CE$. Line through $M$ perpendicular to $OM$ cuts lines $AC$ and $BC$ respectively in $K$, $L$. Prove that $\frac{LM}{MK}=\frac{AD}{DB}$

Let $M_{a}$ and $M_{b}$ the midpoints of CB and CA, respectvely.
Is easy to see that the quadrilaterals $MOM_{a}L$ and $MOKM_{b}$ are cyclics.
So, we easy angle chasing, we get $\angle{LOM}= \angle{B}$, and $\angle{KOM}= \angle{A}$.

now, we have that $\frac{LM}{KM}= \frac{tgB}{tgA}= \frac{AD}{DB}$


How many squares are there in total in a normal 8 by 8 checkerboard. (try using sequences and series to help)?

Hint:[hide]How many centers are possible?  Also helpful: [tex]\displaystyle \sum_{i=1}^{n}i^2 = \frac{n(n+1)(2n+1)}{6}[/tex][/hide]

An other hint :

[hide]

a(n+1) = a(n) + (n+1)(n+2) - (n+1) by looking how many squares are added.[/hide]

Answer in spoiler:

[hide]First, we find how many squares in total of 1 (1x1) there are = 64. Next, we look for how many 2's there are. THere are 7 that use the first row, and then discounting those, another 7 that use the second, etc, etc till we get 7^2 = 49 for this total. We can continue counting in this manner and find that the number of i squares there are = (9-i)^2. So the total is 204[/hide]

In these types of problems, I always just add up the perfect squares, going up until n-squared, where n is the side length, in squares.  For this, I would add up 1+4+9+16+25+36+49+64 which equals 204.
Matter of fact, this works in 3-d, just with perfect cubes.  If you had a 4 by 4 by 4 cube, there are 1+8+27+64 equals 100 cubes that can be made from it.

Yep and for perfect cubes the summation formula's [n(n+1)/2] :^2:.

[quote="trickdacy"]Matter of fact, this works in 3-d, just with perfect cubes.  If you had a 4 by 4 by 4 cube, there are 1+8+27+64 equals 100 cubes that can be made from it.[/quote]

This problem was on a state MathCounts sprint round problem, and "What is the sum of all the perfect cubes less than 100" was a MathCounts countdown round problem.  I wonder if this can be applied to 4-dimensional objects... [url=http://mathworld.wolfram.com/Four-DimensionalGeometry.html]hmm...[/url]


can anyone please post a link to
1)SCHAUM'S 3000 solved problems in organic chemistry(e-book)
2)problems in organic chemistry by I.L.FINAR(e-book)

I don't know those links, but I have experience in Organic Chemistry/Synthesis teaching, and in my opinion those type of books are not the best for real learning. Even for practice, I would recommend those problems that appear in some common textbooks (Carey, Solomons, Volhart, etc) that have no solution in the book. Those are the best from a pedagogical point of view.
I also have quite a collection of problems (many by my own), we could discuss them here. And of course, how about the problems I have already posted?

well thanks .as u could have got from my posts i am still a high school student.i  am studying organic from morrison & boyd and needed some conceptual questions to test my concepts .ur questions are great and most imp. creative.i woul be very happy if u could pm me set of conceptual questions if it's possible or we could continue discussion here .it would be very nice if u could start organic from  very base that is alkane but with challenging questions.

You could refer IL Finar's Textbook of Organic Chemistry,not the Problems Book.The former is really good and the one by CK Ingold is also very good for mechanisms.However,I doubt if online versions are available.
Sathej


There is an account called "Beat_AIME15".

And it seems to be a user we all know...

EDIT: Completely ignore this thread for now.

Apparently, it's my friend Danny...i'll ask him at school.

I told him to make an account on AoPS, he's on my MC team.


symmetry
combinatorics unsolved
combinatorics

In each line and column of a table $ (2n\plus{}1)\times (2n\plus{}1)$ are written arbitrarly the numbers $ 1,2,...,2n\plus{}1$.It was constated that the repartition of the numbers is symmetric wrt the main diagonal of this table.Prove that all the numbers on the main diagonal are distinct.

There are 2n+1 of each number. After removing, the main diagonal, there must be an even quantity of each number, by the symmetry constraint. Thus, at least one of each number must be on the main diagonal.


Show that simple 3-connected graph has a circuit with a chord.

Use Dirac's theorem, which states that if $ G$ is a $ k$-connected graph (with $ k \geq 2$), and $ S$ is a set of $ k$ vertices in $ G$, then $ G$ has a cycle including $ S$ in its vertex set.

Just choose a vertex $ A$ with three neighbours $ B,C,D$ and use the above theorem for $ S \equal{} \{B,C,D \}$, then eventually add $ A$ to the cycle.

Pierre.


geometry
circumcircle
ratio
geometric transformation
dilation
analytic geometry

I'm not sure if this is an advanced level problem and I only half understand the proof so I just wanted to post this to see if anyone here could prove something called Euler's line.

Prove that the orthocenter, centroid, and circumcenter of any triangle are colinear and that the centroid divides the distance between the orthocenter and circumcenter in the ratio 2:1.

The medial triangle is a dilation away from being the original triangle, with the dilation having the centroid as center.  The circumcenter of the big triangle is the orthocenter of the medial triangle.  Thus you get directly that the orthocenter, centroid and circumcenter are colinear in that order, and that 2OG = HG.

(This was quite sketchy, but I'm assuming you've worked with these things enough to follow.  Otherwise, I could go through it in more detail if you needed.  This is Advanced level, I would say.)

On another note, you can prove it to yourself by construction.  I did this activity in geometry class a few years ago and I can scan it when I get back to school (on vacation now, no scanners at home) if you are interested.  Also, go to:

http://people.ucsc.edu/~erowland/eulers.html

This has a proof that is pretty easy to understand (topic wise) but involves a lot of use of variables.

Also, if you have trouble understanding this then try doing it for one triangle w/integral side lenghts instead of working w/variables.  You would be doing it the same way, just w/o all of the manipulations w/variables (aka easier).  This will help you understand the proof.

Also, if you happen to know barycentric coordinates ....


I'm looking for "Speeding up the Pollard rho method" by P.L Montgomery on the net. Can i find it in pdf file or some other files that i can open and view on my computer?

most probably you are looking for this
http://www.jstor.org/view/00255718/di970601/97p0023n/0
you can view it if you have registration to JSTOR

Well, i don't have a registration.  So, that article is not free.


Just try different ways. I can't draw it, unfortunately, but the highest I came up with was $ \boxed{14}$.

Since I posted this, I know the answer is $ \boxed{14}$, but I'm not really sure how that is drawn.

I believe ernie is correct. There are $ 7$ points in the diagonal, each having $ 2$ segments, so the answer is $ 7\cdot2\equal{}\boxed{14}$.

Or we note that there are 7 vertical and 7 horizontal segments, and usually 7+7=14.


[quote="Altheman"]If
\[ \frac {x^4 \plus{} x^2}{2} \equal{} x^3
\]
, then
\[ \frac {x^4 \plus{} x^8 \plus{} x^{12} \plus{} x^{16}}{4} \equal{} x^{10}
\]
. (This isn't for you experienced problem solvers who are lurking in the HSB forum haha)[/quote]

Can I also say
\[ \frac {x^1 \plus{} x^{23} \plus{} x^{456} \plus{} x^{78910} \plus{} x^{11121314}}{5} \equal{} x^{2008}
\]
:maybe:

Happy 2008 :)

P.S: This is not trolling, but an indirect way to confirm my answer

How are these polynomials derived from the orginal?
I noticed the average of the exponents on the LHS equals the exponent on the RHS for those in the former post, but what about the latter post?

[quote="mountain.dew"][hide="hint"]Assuming x doesn't equal 0, divide both sides of each equation by the thing on the right.[/hide][/quote]

Well there is a way of doing it assuming that x is 0. Manipulate the first equation until you get a polynomial equal to 0 which is factorable.

Altheman, what are you trying to show us exactly?
The above can be rearranged to be a quartic where you find 1 and 0 as solutions for x... but obviously you're trying to tell us something else, right?

No it is just an exercise to show you to not overthink problems. It has a real nice form and it looks like oh you can apply one of those random algebraic manipulations. But then it is really pretty simple.

Also I never said that it was supposed to be hard. It is a 'HSB' level proof.

I liked it because it has a nice form.

There are other similar looking problems that are a bit less trivial.

The idea that I was thinking about was how to show generally that P(x)=0 implies that Q(x)=0

at first, I was thinking just do P(x) divides Q(x), but that doesn't take multiplicity of roots into account.

Also,

Follow-up: Show that the converse of the original problem is not true, that is the second polynomial doesn't imply the first one.


I think last years were easy now that I take a second glance at it...I want to know if this is getting easier each year or just easy for last years since I plan on taking it this year too, and I only took it once before.

2 years ago was more difficult than last year's test, but overall i dont think the difficulty is changing that much.  If you look back at tests from a couple years back, they're about the same as last year's.


Does anyone here play Shoddy Battle? It's a pokemon simulator (diamond/pearl), written in Java, that allows you to battle with people online. Rather than play through the actual game you just build your team through a series of drop-down menus and checkboxes. It's fun to play for about fifteen minutes at a time (there's no way I would ever have the time to actually play the real Pokemon game)

http://www.shoddybattle.com

My username there is the same as the one here

sounds like fun to me, have to sign up tho. does it require an email address? if not, okay! user will be (probably) regigas1337.

[quote="randomdragoon"]Actually, registration is (a little too) simple. They only require you to enter a username and password, which makes me wonder how they haven't been spammed yet.

Oh, by the way, make sure you use the official server; sometimes it says there are 0/100 users online but that's usually a glitch... though other times the server is legitimately down.[/quote]
i signed up, my username is TempestRyu
i saw you in official server the other day hehe.


Hi,

How do you create an animation of a KaleidoTile picture and upload it as your avatar? I know how to use Kaleidotile but I don't know how to create animations.

Thanks,
ThinkFlow

Well I think you use photoshop and a few other programs.

Sorry, maybe I wasn't clear about what I was asking. :blush:

What are the steps I need to take in order to produce an animated GIF file based on a KaleidoTile design? What programs do I need (I don't have PhotoShop) and how do I use them to produce an avatar like, for instance, nikeballa96's avatar?

Thanks,
ThinkFlow

Hm.. I think you'll have to pm either nike or isabella. She's making one for me.

Well, how do you make an animated GIF file?

Generally, it requires some sort of photo editing program. I use the animation tool from paint shop pro.


calculus
trigonometry
modular arithmetic
number theory

[b]Given the equation: sin (kx) = sin (x)
Find the value of k for which this equation and the equation cos (3x) = cos (2x) have, within the range (0, 360] (degrees), one and only one common solution[/b]
Thanks all for help :)

Hello;I think you should consider the points x=2π/5,4π/5,6π/5,8π/5,putting them in euation 1.And find out k.couldn't find an elegant solution.EXTREMELY SORRY!!! :blush:  :blush:

This isn't a calculus problem and doesn't belong in the calculus forum, so after I finish replying, I'll find somewhere to move it. Consider it to be a number theory problem.

Angles will be in degrees.

When is $ \sin a\equal{}\sin b?$ When either $ a\equiv b\pmod{360}$ or $ a\equiv 180\minus{}b\pmod{360}.$

When is $ \cos a\equal{}\cos b?$ When either $ a\equiv b\pmod{360}$ or $ a\equiv \minus{}b\pmod{360}.$

So the equation $ \sin kx\equal{}\sin x$ can be written as $ kx\equiv x\pmod{360}$ or $ kx\equiv 180\minus{}x\pmod{360}.$ That gets us two families of solutions:

$ x\equal{}\frac{360j}{k\minus{}1}$ or $ x\equal{}\frac{180\plus{}360j}{k\plus{}1}$ for $ j\in\mathbb{Z}.$

The equation $ \cos 3x\equal{}\cos 2x$ can be solved as follows:

$ 3x\equiv 2x\pmod{360}$ which implies $ x\equiv 0\pmod{360}$ or $ x\equal{}360n.$

or

$ 3x\equiv \minus{}2x\pmod{360},$ which implies $ 5x\equiv 0\pmod{360}$ or $ x\equal{}\frac{360n}5.$

That second equation includes the first.

So, when do solutions coincide?

Either $ \frac{360n}5\equal{}\frac{360j}{k\minus{}1}$ or $ \frac{360n}{5}\equal{}\frac{180\plus{}360j}{k\plus{}1}.$

Take the first equation, divide by $ 360$ and multiply by $ 5(k\minus{}1)$ to get $ (k\minus{}1)n\equal{}5j.$

This always has $ n\equal{}0,j\equal{}0$ as a solution. We also have solutions whenever $ 5\,|\,n$ (but that's the same place on the circle). If $ 5$ doesn't divide $ n,$ then we would need $ 5\,|\,(k\minus{}1)$ or $ k\equiv 1\pmod{5}.$ Then $ j\equal{}\frac{(k\minus{}1)n}5,$ and as $ n$ ranges over all integers not equivalent to $ 5,$ then $ j$ will always be an integer.

Now let's look at the other equation. This time, divide by $ 180$ and multiply by $ 5(k\plus{}1).$ That leaves $ 2(k\plus{}1)n\equal{}5\plus{}10j.$

If $ 5$ divides $ k\plus{}1,$ we get no solution, as one side is divisible by $ 10$ and the other side is $ \equiv 5\pmod{10}.$ But if $ 5$ doesn't divide $ k\plus{}1,$ then we would have $ 5\,|\,n,$ which gets us back to $ x\equiv 0\pmod{360}.$

So:

If $ k\not\equiv{1}\pmod{5},$ then the only solution in the circle is $ x\equiv0\pmod{360}.$ However, if $ k\equiv 1\pmod{5},$ then $ \{0,72,144,216,288\}$ and their equivalents mod 360 are all solutions.

The question asked for the $ k$ that produce a unique solution in the circle; that would be $ \{k: k\not\equiv{1}\pmod{5}\}.$


geometry
parallelogram
geometry unsolved

$ABCD$ is a quadrilateral such that $AB$ is not parallel to $CD$, and $BC$ is not parallel to $AD$. Variable points $P, Q, R, S$ are taken on $AB, BC, CD, DA$ respectively so that $PQRS$ is a parallelogram. Find the locus of its center.

It is easy to construct a parallelogram $ PQRS$ as the problem states, beginning from an arbitrary point $ P$ on the side-segment $ AB.$

Through the point $ P,$ we draw two lines parallel to $ AC,\ BD,$ which intersect the side-segments $ BC,\ AD$ of the given quadrilateral $ ABCD,$ at points $ Q,\ S,$ respectively.

If now, $ R$ is the point on $ CD,$ such that $ QR\parallel BD\parallel PS,$ it is easy to show that $ RS\parallel AC\parallel PQ,$ from $ \frac {AS}{SD} \equal{} \frac {AP}{PB} \equal{} \frac {CQ}{QB} \equal{} \frac {CR}{RD}.$

$ \bullet$ In our problem now, if we consider the non-convex quadrilateral $ BCAD,$ it is easy to show that the midpoint $ M$ of $ QS,$ as the center of the parallelogram $ PQRS,$ lies on the segment $ XY,$ where $ X,\ Y$ are the midpoints of  $ AC,\ BD$ respectively,

as a direct application of the [b][size=100]ERIQ theorem[/size][/b] ( = [b]E[/b]qual [b]R[/b]atios [b]I[/b]n [b]Q[/b]uadrilateral ), because of $ \frac {BQ}{QC} \equal{} \frac {DS}{SA}$ and $ \frac {BY}{YD} \equal{} \frac {QM}{MS} \equal{} \frac {CX}{XA}.$

Hence, the wanted locus is the segment $ XY$ and the solution is completed.

Kostas Vittas.

PS. When the point $ P$ is taken outwardly to the side-segment $ AB$ then, all the vertices of the parallelogram $ PQRS,$ are also outwardly to the side-segments of $ ABCD$ and in this general case, we can say that the locus of $ M$ is the line segment $ XY.$

[quote="vittasko"]PS. When the point $ P$ is taken outwardly to the side-segment $ AB$ then, all the vertices of the parallelogram $ PQRS,$ are also outwardly to the side-segments of $ ABCD$ and in this general case, we can say that the locus of $ M$ is the line segment $ XY.$[/quote]
See the schema [b][size=100]t=134116(a)[/size][/b].

But, we can easy to construct a parallelogram $ PQRS$ as the problem states, without $ PQ\parallel AC\parallel RS$ and $ QR\parallel BD\parallel PS,$ as it seems in the schema [b][size=100]t=134116(b)[/size][/b].

In this general case, the problem has a meaning only when the points $ P,\ Q,\ R,\ S,$ lie on the side-segments of $ ABCD$ and then, the wanted locus is the area of the parallelogram $ XX'YY'$ with $ XX'\parallel YY'\parallel CD$ and $ XY'\parallel X'Y\parallel AD,$ including its side-segments.

I am surprised because such that problem was a contest problem and I wonder about its official solution.

Kostas Vittas.


AMC AIME USA(J)MO USAMO geometry trigonometry HMMT

From now until the AMC, I think I can finish 1 AoPS book in addition to taking a class online (precalc). Which book do you think would help me the most to do well on the AIME, and help me advance to the USAMO? I am open to all suggestions except I am done with the intro series. I will also be doing AIME problems to prepare for the AMC.

well if your done with the intro series, then the only AoPS book left is Volume 2 "and beyond" I guess.
Since AMC has all types of problems and Volume 2 goes over most types of those problems for AIME level+ (i think), you would be better off doing volume 2 than any intermediate series book which focuses in one area (i think)

Have you started/done with any of the Intermediate Books? If not, I think doing those books before Vol. 2 is the better thing to do. Vol 2 doesn't go nearly as in depth as the Intermediate books do. (rrusczyk suggested me the same thing.) For trigonometry, I'd probably get the Precalculus book and just look over the trigonometry parts. (Also, I think you need to do Interm Algebra before doing Precalc.)

Just keep doing AIME problems, a lot of problems. Also do the problems of like HMMT, PUMaC, ARML, etc, they also help a lot. Finally, do lots of lots Geometry as well; AIME usually has lots of Geometry, I guess Intro to Geometry from AoPS and Challenging Problems in Geometry (also the geometry sections from Vol. 2) serve best for that purpose (Geometry Revisited is more for Olympiad, correct me if I am wrong.)

Good Luck!


On a past  Mathematical Olympiad  the maximum  possible  score  on a  problem was 5. The  average score of boys  was 4,  the average  score  of girls  was 3.25, and the overall  average  score  was  3.60. Find the total number  of participants, knowing  that  it was  in the range  from  31  to 50.

Let $b$ be the number of boys, and let $g$ be the number of girls, then we have $4b+3.25g=(b+g)3.6 \Rightarrow b=\frac{7}{8}g$, so $g$ must be a multiple of 8, the only value that results in the sum being between 31 and 50 is $g=24, \ b=21$, so there were $45$ participants.


function
real analysis
real analysis unsolved

suppose $ f : \Bbb R \to \Bbb R$ is meaurable and everywhere finite function such that $ \alpha\circ f$ is integrable for all $ \alpha : \Bbb R \to \Bbb R$ continuous  functions. Prove that $ f \in L^\infty$.

[color=green]That's /circ, not /circle. -Moderator[/color]

As stated, there are no such functions; $ 1\circ f$ is never integrable. We should require that $ \alpha(0)\equal{}0$.
It's also necessary that $ f$ is supported on a set of finite measure, or alternately we restrict $ \alpha$'s behavior near zero even further. The proof of that is essentially the same idea as for the question asked here.


percent
geometry
3D geometry
prism
probability
rectangle
algebra

PLEASE SEND ME YOUR ANSWERS NOW. I WILL NOT REMOVE THE PROBLEMS, BUT PLEASE SEND ME YOUR ANSWERS IN THE NEXT FEW MINUTES.

Mock AMC D

1) A + B = 7, B + C = 8, A + C = 11. What is A + B + C?
a) 63 b) 64 c) 65 d) 66 e) 67

2) A circle centered at the origin is tangent to the graph 25(x-1) - 36y = 900. Find the minimum radius of this circle.
a) 5 b) 5.5 c) 6 d) 6.5 e) 7

3) Calculate the number of divisors of 2^2004.
a) 2003 b) 2004 c) 2005 d) 2006 e) 4010

4) Given that a is 30% of r and b is 50% of r, what percent of r is a*b?
a) 15 b) 1.5 c) 15r d) 1.5r e) NOTA

5) If a rectangular prism has edges of length 5, 12, and 84, what is the greatest distance between any two of its vertices?
a) sqrt(7081) b) 60sqrt(2) c) 85 d) 84 e) NOTA

6) What is the highest power of 14 that divides into (500!)?
a) 14^82 b) 14^71 c) 14^94 d) 14^343 e) 14^2

7) If a regular polyhedron has 12 more edges than faces, find the number of vertices.
a) 10 b) 11 c) 12 d) 13 e) 14

8) A fish eats magic powder at 12:00 AM. On every hour after 12:00 AM (1:00 AM, 2:00 AM, 3:00 AM), the fish triples in size instantaneously. If it started at 12:00 AM with a length of 2 inches, find its length at 6:30 AM that same day.
a) 1458 ft b) 4374 ft c) 486 ft d) 121.5 ft e) 192 ft

9) What is the greatest prime factor of 1 + 2 + 3 + 4 +  + 2003 + 2004 ?
a) 4009 b) 2003 c) 401 d) 211 e) NOTA

10) A triangle has sides of length 5, 6, and 7. Find the greatest integer less than the area of its circumcircle.
a) 36 b) 37 c) 38 d) 39 e) 40

11) Twelve children are to be separated into 4 groups of 3. Find the probability that Marcus, Maxim, and Meep are in the same group.
a) 1/12 b) 2/11 c) 1/120 d) 1/10 e) NOTA

12) Rectangle ABCD and point P are drawn. The distances from P to A, B, and C are 12, 13, and 14 respectively. Find the distance from P to D.
a) 168/13 b) 15 c) sqrt(365) d) sqrt(171) e) NOTA

13) The polynomial x^8-16 can be factored into n smaller polynomials with integer coefficients. Find the maximum value of n.
a) 1 b) 2 c) 3 d) 4 e) 5

14) Find the period of the graph of y = 12 sin(4pix/7) + 13 cos(2pix/5+pi/2).
a) 35 b) 7 c) 7/2 d) 5 e) NOTA

15) In one turn of a game called Take My Bling, a player rolls a six-sided dice twice. If he rolls a number n twice, the host of the game must give him n dollars. If he rolls two different numbers m and n, then he must pay the host of the game m*n dollars. Calculate the expected winnings in dollars for the host to the nearest cent in one turn of Take My Bling.
a) -7.19 b) 7.19 c) -9.72 d) 9.72 e) 0

16) Billy the Cyclops wants to paint a sphere hanging from his ceiling from a cord of length 6 meters. Standing directly under the center of the sphere with his eye a distance of 13 meters from the center of the sphere, he stares up and 100*pi square meters of the area of the ceiling is blocked from his vision. If it takes 1 gallon of paint to paint 1 square meter of sphere, find the least whole number of gallons of paint that Billy should buy.
a) 314 b) 315 c) 79 d) 80 e) 100

17) Given a horizontal line and a point 6 units above the line in the coordinate plane, a locus is drawn such that every point on the locus is the same distance from the line as it is from the point. If an isosceles right triangle ABC is constructed with right angle C located at the lowermost point of the graph, and A and B are located on the graph, find the area of ABC.
a) 144 b) 72 c) 36 d) 18 e) 9

18) What is the coefficient of x^3*y^4*z^6 in (x+y+z)^13?
a) 9009 b) 12012 c) 36036 d) 60060 e) NOTA

19) ABC has AB = 75, BC = 78, CA = 51. DEF fully contains ABC and has DE parallel to AB, EF parallel to BC, and FD parallel to CA with a distance of 3 between corresponding sides. What is the area of DEF?
a) 2490 b) 2493 c) 2496 d) 2499 e) 2502

20) In a two-person game, one person says an integer, and the other person must say an integer greater than or equal to 1/3 of the previous persons integer, but less than the previous persons integer. They go back and forth following this rule until one person says 1, and that person loses. If Johnny says 2004, what number should Bernie counter him with to guarantee a win for himself?
a) 1820 b) 1821 c) 1822 d) 1823 e) 1824

21) Let A and B be diametrically opposite on a circle. Choose points C and D on the tangents to the circle at A and B respectively, such that AC = 11, BD = 13 and CD is tangent to the circle as well. Find the area of the circle.
a) 143*pi b) 144*pi c) 145*pi d) 72*pi e) NOTA

22) Evaluate sum(k = 1 to 9999) [log(sin(pi*k/10000))] to the nearest integer given that log(2) ≈ 0.30103.
a) -3005 b) -3006 c) -3007 d) -3008 e) -3009

23) An ellipse has major axis of length 122 and minor axis of length 120. A circle centered at a focus of the ellipse is tangent to it. What is the sum of all possible diameters of the circle?
a) 200 b) 221 c) 222 d) 244 e) 256

24) 17 points are spaced at equal intervals around a unit circle. From one point, segments are drawn to every other point. Find the sum of the squares of the lengths of these segments.
a) 4.25 b) 8.5 c) 17 d) 34 e) 68

25) Given: a, b, c, and d are real numbers and a + b + c + d = 36. Find the maximum value of 2a+3b+6c+24d.
a) 125 b) 150 c) 175 d) 200 e) 225

I just found out I can't participate in this.  Oh well.

Overally, it was easy.  I think we will have a couple of 130+ people (certainly not me):)

[quote="beta"]That was definitely the hardest one. My score is like 20 points lower than Mock AMC B score[/quote]

Huh?  Maybe I was exaggerating when I said it was really easy, but I don't think it was the hardest one.
BTW I thought 16 was easy but I couldn't get it.

I took it early and I thought it was pretty hard... then I got my score back and yeah, it was hard.

I got 68 on AMC A.  I cannot do worse on this one.  It is very likely that you all did better than I did.


There is a game between John and Jack.
In the game there are $2$ cards $\"0\"$ and $\"1\"$.
Each player chooses a card between the two cards and each time there are two cards
and there is no connection between one choosing to the other.
When the player chooses the card $\"1\"$ he wins.
John begins the game, chooses a card and if it isn't $\"1\"$ Jack chooses a card and if he don't win, again John chooses
and so on.
Find the probability such that John will win.

The probability that John wins is $1/2$ that he picked a 1-card on his first turn, plus $1/8$ that he picked a 1-card on his second turn (after neither he nor Jack picked a 1-card on their first turns), plus $1/32$, etc.  So the probability that John wins is $1/2+1/8+1/32+1/128+...$, which equals $2/3$.

There are about 10000000000000000000000000 questions of this type.
[hide="Trivial"]For each pick, the probability of picking a card is 1/2.
John always goes on the (2k+1)th turn and his probability of picking "1" is 1/2*(1/4)^k.
The total probability is:
1/2+1/2*(1/4)^1+1/2*(1/4)^2...
this is infinite geometric series with common ratio less than 1.
Hence the required probability is:
1/2 * (-1) / (1/4 - 1)
= 2/3. :)[/hide]

[hide="Cool Solution"]For each 2 turns, starting with the turn with an odd index, Jack picks after John.Because the probability of picking each card is 1/2, Jack's probability is 2 times less than John's. Hence the total probability of John winning is 2 times bigger than Jack's.
But the probabilities add up to 1. :P
Hence the answer is 2/3. :gleam: [/hide]

Let's say John and Jack are picking out of two different piles of cards, so that there is probability $p$ each turn that John picks a winning 1-card, and probability $q$ each turn that Jack picks a winning 1-card.  Let $0<p<1$ and $0<q<1$.  If John picks first, and John and Jack alternate, define $q$ in terms of $p$ so that John and Jack are equally likely to win the game.

[quote="rem"]Looking back at my Cool Solution ;) , the ratio of Jack's winning probability to John's is q. The probabilities are equal, so q = 1.
OR p = 0.
However, both cases are impossible,because p,q are larger than 0 and less than 1, so the result of the problem will never exist. :([/quote]

I don't really understand your logic.  How did you get that "the ratio of Jack's winning probability to John's is q?"  Try $p=1/3$ and $q=1/2$; John and Jack are equally likely to win.

:wallbash: I forgot to substract q and p from 1:
Looking back at my Cool Solution, the ratio os the 2 probabilities is:
p / ((1 - p)q). the top is p, the probability that John picks the winning card, the bottom is the probability of the fact that John doesn't pick the winning card, but Jack does.The probabilities are equal, so:
p = q - pq.
q = p + pq.  :gleam:

Now a different problem: what is you have the same conditions for n people, and for each preson i, the probability he picks the winning card is m(i), and again, the probability of winning for each person is the same?
The cool thing is you get (n - 1) equations in the system, so there are still infinitely many solutions.

What if there are 2^z people and we have play-offs, nad the probabillity of winning the championship is the same for each person. Now that is a beautiful problem!!!!!!!!


Reviving this thread since the national result came out few days ago
anyone got national award?
I got $\$250$ :D

[quote="mws21"]Has anyone took this year's KSEA National Math Competition?[/quote]i have taken the KSEA competition in 2002~2005, but I wasn't able to go to 2006 because of ICTM (Illinois Council of Teachers of Mathematics) competition and 2007 because of Science Olympiad.. :(

Oops, sorry, didn't see this here. I failed the competition. Totally. Like usual.


I was just curious, the question just came to my mind. Who here has been in Philly, and if so when and what did you do. I am asking because I live here.

I want to go there someday, as it is only a four hour drive from my town, but we haven't got time yet.

I have, I saw a Sixers game, and I also got lost in downtown Philly in 1:00 AM and the cops had to help us out :?

I've been through the airport there (I think, I was very sleep-deprived) but have never actually seen any of the city.  I would love to though.

Umm I visited all the stuff there like the liberty bell and those museums or whatever a few years ago. And.. last summer I went to Dorney - which is in allentown, thats near philly, right?

I've been to Philadelphia. I think it was in 2000 or something.

i just went 2 or 3 wks ago
was vacation up North, toured UPenn

[quote="hello"]Juicybooty, u were close yto were I study, I go to Drexel University which is like a block away from UPENN. Any of u ever been in the national museum of american jewish history? maybe u forgot if u even have been but nevertheless.[/quote]

do you happen to have any involvement with Drexel University's math forum: http://mathforum.org/

  I use to love going here as a kid, and solving math problems.  Give my regards to whoever's in charge of the website, or who thought of it for that matter.  :lol:

I am definetly involved in it, I am a math doctor there since February, unfortuneatly I do not get paid. Many some aopsers can tell the administrators there how good I am and they will pay me. LOL LOL LOL.

I was born there and go back at least once every two years...

I been there 4 times in the last 4 years though

I love it

where do u live now? where in philly did u live?

I went to philly on a weekend during school.  i visited drexel college because i went there with my frieind's family and my friend's parents went to Drexel.   :)

does eating a philly cheesesteak at penn state count for anything?  :D

[quote="hello"]Was it in Philadelphia? Penn State is not in Philly though?[/quote]
it's as close as i've been  :P

I live in Blue Bell, which is like 30 minutes away.  Thus, I've been there a lot.  Actually, I have an internship at Fox Chase Cancer Center that starts in August.  I'll be driving down to Philly a couple days a week as soon as that begins.

Nice, my internship will be in King of Prussia.