AMC
AMC 10
AMC 10 A
AMC 12
geometry
trigonometry
AIME

10A: 126.5 (I'm pretty sure I got one wrong)

And be honest  :roll:

AMC 10 A: 150

its that much better when your teammate does it too  :lol:

Man, I really screwed up.

AMC 12A: 113.5 (2 careless mistakes :()

I plan to take revenge by getting a 150 on AMC 10 B, I hope. :D

AMC 12A: 136

I beat my score from last year, plus I think the test was a bit more difficult this year (especially questions 17-20)

JB

[quote="dakyru"][quote="amirhtlusa"]AMC12 A: 150 :D[/quote]

Seriously?   :)[/quote]
yup!

AMC 12A :  123

Yeah, 12:59 maximizes the sum of the digits!

AMC 12A: 133.5

I missed #10  :oops: and geometry once again pwned me...

What reallly stinks is if you bubble in the circle of the wrong question.  Like bubbling the ans to #14 in line 15 or something.  That happened to me once. (but not this time)

Ew. I got 138 on 10A as a 10th grader...

Things to learn from this test:

1. 0 is an integer.
2. 18/3 is 6, and 2*6 is 12.

Yeah...problems 7 and 10.

[quote="krustyteklown"]I'd have to say alot of those people are wrong[/quote]

Then which ones are right?   ;)

heh, how bout those saying the official word from AMC

the floor disappearing would be a huge announcement, and you can always check the teacher's manual for AMC 2006, cuz those include policies on further examinations

Well I'd guess that the full scorers are correct at least since they're easily verifiable on the AMC website.

For some reason, our post office screwed up and couldn't send us the AMC 10A and AMC 12A, so now I have to (and the rest of our school) take the B test. Sob  :(

111.5
amc12A

14right, didn't bother to look at the rest :P

Stupid, stupid me.  I got the pizza problem wrong (my pizza had 9 slices instead of 8  :oops: ).  And I was compelled for some reason unknown to me to change my answer on the rectange question from the right one to the wrong one when 30 seconds was called.  I forgot to grid in the two circle-external tangent question.  And I stupidly answered $\$$90 without thinking a little more for the goats, cows, sheep, llamas, or whatever farm animal question.

It stunk.  [size=75]101.5[/size]

I took the AMC 10 A today (it doesn't count, because the date already passed because the post office messed up [grr...])... Perfect score! :D  :D

Dude, Mr. Rusczyk and Mr Lehoczky, you guys are awesome! Thanks so very much for AoPS 1 and 2, as they really helped me (they are the only books I studied...) to get a perfect score on a (fake, but real...) AMC 10!!!!!

There are people here upset about getting 120"s. I got 100.5  :blush:
Need to study harder. :oops:

123.

Answered 19, 1 wrong.

For the $\sqrt{120-\sqrt{x}}$ problem, I forgot to include $120^2$.  :wallbash_red:

138. And it looks like i'm not alone in struggling with #17 :P

[quote="BTBasham"]I am the worst ever ;\
118 on the AMC 12 A
Not only did I miss the function one which was incredibly easy...
I missed problem #2.[/quote]
Basham!!!
I missed #3 on the 10A and scored highest in Oregon... lol
[quote="Zellex"]123.

Answered 19, 1 wrong.

For the $\sqrt{120-\sqrt{x}}$ problem, I forgot to include $120^{2}$.  :wallbash_red:[/quote]
I missed that one too... same mistake as you

Please don't bump extremely old threads.


linear algebra
matrix
number theory unsolved
number theory

Let be given a sequence (an) , n  \in N* satisfying the conditon :
 0<a_(n+1) - a_n <=2001
Prove that there are infinitely many pairs positive integers (p,q)
such that  if p<q then a_p is divisor of a_q

Thus, from any 2003 consecutive numbers, at least one is a term of the sequence. We first prove that we can find such a pair. Let A the matrix wich has the first line 1,2,...,2003, the second line 2003!, 2*2003!,...,2003*2003! , the third P, 2P,...,2003P where P=2003!*2*2003!*...*2003*2003! and so on. Each line has a term of the sequence.If we consider the first 2004 lines, we will find that there is a column that contains 2 terms of the sequence. Those terms, let them be a_p_1 and a_q_1 have the property that a_p_1|a_q_1. Then take the matrix B which has the same property as A, but the first line is M,M+1,...,M+2002 where M>max{a_p_1,a_q_1}. Similarly, we find a_p_2 and a_q_2 and so on.


When I run the attached program there are no compile errors. However, when I run it, it says "There was an error opening this document. The file cannot be found."

Immediately afterward, it says [DocOpen("%bm.pdf");[FileOpen("%bm.pdf")]  Cannot execute the command.

My latex file and picture are both in my documents. What should I do?

You repeat the word 'it' but don't say what 'it' is.
[quote]when I run it, it says[/quote]
[quote]Immediately afterward, it says [/quote]
So start again and say exactly what you are doing.
Also
1. Is myimage.png in the same directory as LaTeX3.tex?
2. Is a file created? Check this by noting how many pages are created.
3. If the pdf file LaTeX3.pdf is created can it be opened directly eg from Windows Explorer (assuming you are using Windows)?

1. I don't know what a directory is. I checked it up on dictionary.com and I don't understand its definition: "a list of files contained in external storage." myimage.png and LaTeX3.tex are both in my documents.
2. The pdf isn't created.
3. The pdf isn't created.

What I Did:
1. Saved an image titled "myimage.png" as well as the LaTeX3.tex file in my documents.
2. Compiled the LaTeX document
3. Pushed F5 in TeXnicCenter
4. Saw adobe reader open with a message box saying "There was an error opening this document"
5. Clicked okay on the box
6. Saw another message box in TeXnicCenter saying "[DocOpen("%bm.pdf");[FileOpen("%bm.pdf")] Cannot execute the command"

The output box on the bottom is attached

A directory is a folder. So My Documents is a directory.
The problem is that the pdf is not being created and the screenshot that you posted says that:
pdfTeX error: (file myimage.png): libpng: internal error
It looks like the image that you used is faulty so try another one. The one attached works correctly; right-click on it then click on Save Picture As. If that image works then you know your image is faulty and will need recreating.

Actually.... How would I have known that the picture is faulty? How could I have fixed the problem?

Also, is there an easier way to show you what is appearing the output box? I'm getting another error with another picture and pdf and I'm stuck again...

That picture seems fine BUT it has been processed by the forum so it may have been automatically corrected. Zip the png and upload that so it remains the original faulty image and then I might be able to see what the problem is. How did you create the image, maybe that process is faulty?

[quote="aufha"]Also, is there an easier way to show you what is appearing the output box?[/quote]
Copy the output and post it here directly; an image is not necessary. :)

i_like_pie, I can not highlight the output box with the cursor or ctrl-a for some reason. Can you?

stevem, how do I zip it the image? Originally I had renamed an existing bmp with a ".png" at the end. I copied and pasted the image on microsoft paint on to a new file and saved it as a png. It works now...

What happened exactly?

I'm not familiar with TeXnicCenter; I use Kile in Linux (which is absolutely awesome). Maybe TeXnicCenter is just not as awesome. :D

[quote="aufha"]i_like_pie, I can not highlight the output box with the cursor or ctrl-a for some reason. Can you?[/quote]Just open the log and copy from that. In TexnicCenter, go to Tools, Windows Explorer and you'll find a log file with the same name as the tex file. Most of the information is in there.
[quote="aufha"]stevem, how do I zip it the image? Originally I had renamed an existing bmp with a ".png" at the end. I copied and pasted the image on microsoft paint on to a new file and saved it as a png. It works now...

What happened exactly?[/quote]What happened is you can't just change the name from bmp to png. They are very different formats and you must convert one to the other which you did by using Paint. Your original image was not a png file (whatever its name was) which is why it didn't work.

If you have Windows XP then you can zip a folder and put files in it using the instructions at [url=http://www.worldstart.com/tips/file-management/zip-fldrs-xp.htm]Zip Your Folders in XP[/url].


Given a regular n-gon $ ABCDEF$.... and a point $ P$ in its plane ,we know intuitively that the sum $ PA \plus{} PB \plus{} PC \plus{} PD \plus{} PE...,$ is miminum when $ P$ coincides with the center of the polygon , but how to prove it geometrically?

[hide="Sketch"] For $ n$ even, it suffices to prove this for pairs of opposite vertices, which is the triangle inequality.  For $ n$ odd, it suffices to prove this for a triplet of a given vertex and the two opposite vertices, which reduces to the case $ n \equal{} 3$ after scaling.  Do you know how to do this case? [/hide]


calculus
integration
trigonometry
calculus computations

Let $P(x) = \sqrt{(x-a)(b-x)}$, with $a \neq b$.  Integrate

$\int P(x) \, dx$

$\int \frac{1}{P(x)}\, dx$

edit: i meant to put this in computations; can someone move it?

Using the same substitution, I get:

$ \int\sqrt{(x-a)(b-x)}\,dx=\left(\frac{b-a}{2}\right)^{2}\int\sin^{2}{2t}\,d(2t)$. The rest should be easy.


algebra
polynomial
linear algebra
matrix

What is the Jordan Canonical form of [0,-1,1;1,2,-1;0,0,1].
Would it be possible to explain the process? Thanks

$\left(\begin{array}{ccc}0 &-1 & 1 \\ 1 & 2 &-1 \\ 0 & 0 & 1 \end{array}\right)$
So find your eigenvalues the normal way:
det$\left(\begin{array}{ccc}0-t&-1 & 1 \\ 1 & 2-t &-1 \\ 0 & 0 & 1-t \end{array}\right)$ = 0 (Find the roots of the characteristic polynomial)
$= (t-1)^{3}$, thus you have all the same eigenvalues.  Plug your $t$s back into the original matrix:
$\left(\begin{array}{ccc}-1 &-1 & 1 \\ 1 & 1 &-1 \\ 0 & 0 & 0 \end{array}\right)$
Using Gaussian Elimination, I found that you get your eigenspace to be: $E_{1}= span\{(-1,1,0),(1,0,1)\}$ The number of Jordan Blocks for each eigenvalue corresponds to the dimension of $E_{\lambda}$. Thus the Jordan Form would be:
$\left(\begin{array}{ccc}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)$
Remember the multiplicity of the eigenvalue corresponds to the dimension of the generalized eigenspace, not the eigenspace.

Another way to think about the dimension of the eigenspace is if you have learned the difference between the characteristic polynomial and the minimal polynomial. Though the characteristic polynomial gives you $(x-1)^{3}$, the minimal polynomial is $(x-1)^{2}$ since $(A-I)^{2}=0$. So you know that the dimension is only 2 instead of what initially looked like 3.

Let me follow this up by mentioning a few things.  If you suppose that your matrix represents a linear transformation $T = L_{A}$ (the transformation of multiply by $A$), then you can say some more stuff other than just doing the computation. Let $T_{1}$ be the restriction of $L_{A}$ to the generalized eigenspace.  Then suppose you had an ordered basis $\beta_{1}$ (a Jordan Canonical Basis) for $T_{1}$.  I won't go into finding this basis.  But your eigenvalue $\lambda_{1}$ had multiplicity 3, so it follows that $\dim(K_{\lambda_{1}})=3$ where $K_{\lambda_{1}}$ is your generalized eigenspace.  So you end up with a Jordan Block $A_{1}= [T_{1}]_{\beta_{1}}= \left(\begin{array}{ccc}\lambda_{1}& 1 & 0 \\ 0& \lambda_{1}& 0 \\ 0 & 0 & \lambda_{1}\end{array}\right)$.  Then if you were to have more eigenvalues, $\lambda_{2},\ldots$ you keep putting more of these Jordan Blocks along the diagonal: $A_{2},\ldots$  This is how you build it up.  Since your eigenvalues were 1's, things looked a little silly.


A horse is urged to pull a wagon. The horse refuses to try, citing Newton's third law as his defence:"'The pull of the horse on the wagon is equal but opposite to the pull of the wagon on the horse.' If i can never exert a greater force on the wagon than it exerts on me, how can I ever start the wagon moving?" asks the horse. How do you reply?

Well it is a good question since it forces one to think, but if one has grasped Newtonian physics enough it should not be difficult.

The wagon will certainly act on the horse with a "normal" force by Newton's third.

But consider the wagon alone. Which forces act on it? Gravity and the horse-force and by the second law it will accelerate, since there is a nonzero net force.

Considering the system horse-wagon, we see that the wagon don't move relative the horse so the net force between the is zero

You have to distinguish between forces and consider object alone.
[url=http://en.wikipedia.org/wiki/Newtons_law]Please read this article[/url]

Please post more comments on this thread, its "old" but still useful

So by proving that the wagon moves and the wagon does not move relative to the horse, we can deduce that the horse and the wagon both accelerate?

It's a fairly simple question at first glance, but fiendishly difficult to explain  :P

My friend says something about the friction between the horse's hoofs and te ground being greater than the friction between the wagon's wheels and the ground. Does that come into play?

My ninth grade physics curriculum included this exact question. (Just for the record, the curriculum is "T.I.M.E. for Physics First" and isn't all that good.)

To answer your question: Relative to itself, the horse and the wagon are an action-reaction pair, and they stay a fixed distance from one another. However, the pair is still moving relative to the ground due to friction, but this is an entirely different set of forces.

I hope my (somewhat rudimentary) explanation makes sense.


algebra
polynomial
function
algebra proposed

Given positive integer $ n$. Find all functions $ f: \mathbb{R}\to\mathbb{R}$ such that, for all distinct real numbers $ x_1,x_2,\dots,x_{n\plus{}1}$, $ \sum_{i\equal{}1}^{n\plus{}1}\frac{f(x_i)}{\prod_{i\neq j}x_i\minus{}x_j}\equal{}0$.

For distinct $ x_1, ... x_{n\plus{}1}$ the problem condition is equivalent to the condition that

$ \left| \begin{array}{cccc} f(x_1) & f(x_2) & \hdots & f(x_{n\plus{}1}) \\
 x_1 & x_2 & \hdots & x_{n\plus{}1} \\
 \vdots & \vdots & \ddots & \vdots \\
 x_1^n & x_2^n & \hdots & x_{n\plus{}1}^n \\ \end{array} \right| \equal{} 0$

so the answer is all polynomials of degree at most $ n$ satisfying $ f(0) \equal{} 0$.


algebra polynomial calculus derivative algorithm

[quote="car202"]There is no P(x) in Z which takes same value infinitely many times.

Is this true? What is the proof?[/quote]

Let $ P(x)$ be of degree $ n$ and assume that the same value $ \equal{}k$ does exist.

Then $ P(x)\minus{}k\equal{}0$, which is also a polynomial, has infinitely many distinct roots , contradiction (because a polynomial of degree $ n$ has at most $ n$ distinct roots).

Formally speaking, the only such polynomial is $ P(x)\equiv C\in\mathbb{Z}$

Right, so the original assertion is false :P

By P(x) in Z do you mean all the coefficients are integers?
Also, does the fundamental theorem of algebra not apply to constant (polynomial)?

Of course it applies: the statement of the theorem begins, "If $ f(x)$ is a non-constant polynomial ...."   :wink:

Actually, I think there is the constant polynomial, and then there is a zero polynomial.  The reasons for this are that the derivative of a linear polynomial is a constant, but never 0, and also any non-zero constant polynomial obeys the fundamental theorem of algebra: it has 0 roots.

Just my 2 cents.  :)

Wait, so 0 can't be considered as a constant (polynomial)? Also how does it have to do with derivatives?

The definitions are a matter of convention.  The question is not what counts as a polynomial or what counts as a constant but what degree you assign the zero polynomial.  If you say that the degree of the zero polynomial is $ \plus{} \infty$, then the fundamental theorem does in fact remain true.  In other contexts, such as stating the division algorithm, it may be more convenient to assign the zero polynomial degree $ \minus{} 1$ or even $ \minus{} \infty$.

In my experience, the most common statement of the fundamental theorem of algebra is that every non-constant polynomial over an algebraically closed field (read: $ \bf C$) has a root.  That it has exactly $ n$ is (again in my experience) usually taken as a corollary of this result, with good reason (you have to talk about this "with multiplicities" thing, and you have to make this more detailed exception in the constant case).


The digits 1,2,3,4,5,6,7, and 9 are used to form four two-digit prime numbers, with each digit used exactly once. What is the sum of these four primes?
(A) 150
(B) 160
(C) 170
(D) 180
(E) 190

[hide] The only digits that can be the unit digit of a 2 digit prime are $1,3,7,9$. The other ones $2,4,5,6$ must be the tens digit. $10(2+4+5+6)+(1+3+7+9)=10\cdot17+20=\boxed{E)190}$ [/hide]

Wow, lotgreengrapes, your solution is very elegant. You would think you would need to find the actual primes first, but your solution made it clear that you don't.

[quote="Klebian"][quote="sapphyre571"]wow lotrgreengrapes, that was easy. I actually found the primes, [hide="which are"] 53, 67, 29, and 41, whose sum is 190.[/hide][/quote]

In fact, the primes could also be

[hide]59, 67, 23, and 41. [/hide][/quote]

There are 4 possibilities for the primes in fact
[hide]
41 23 67 59
41 53 67 29
61 23 47 59
61 53 47 29[/hide]


conics
ellipse
Putnam
geometry
advanced fields
advanced fields unsolved

Let $E$ be an ellipse with foci $F_1,\ F_2$. Let $P\in E$. The distance from $O$ (the center of $E$) to the tangent in $P$ is denoted by $d$. Show that the value of $PF_1\cdot PF_2\cdot d^2$ is constant when $P$ moves on $E$.

[i]Putnam 1976/B4[/i]

is'nt it pointless? it's solvable just  with high school knowledge.

well usually analytic geometry goes in here, because the advanced section / geometry forums are for IMO-like problems :P

Well, it's not very important for me, you know :). The moderators can just delete it if it seems pointless..


In the game $ \emph{Cover It Up}$ two standard six-faced dice are rolled and their sum determined. The player can then cover that number on the game board or any two numbers that have that sum. For example, if you roll a 3 and 5, you can cover the 8, the 1 and 7, the 2 and 6, or the 3 and 5. What is the probability of being able to cover the 9 on the first roll of the two dice? Express your answer as a common fraction.

[asy]size(101);
picture p;
draw(p,(0,0)--(1,0)--(1.5,sqrt(3)/2)--(1,sqrt(3))--(0,sqrt(3))--(-.5,sqrt(3)/2)--cycle,linewidth(1));
for(int i = 0; i<3;++i)
for(int j = 0; j<3;++j)
add(shift(i*2*right)*shift(j*sqrt(3)*up)*p);
label("4",(.5,sqrt(3)/2)); label("6",(2.5,sqrt(3)/2)); label("3",(4.5,sqrt(3)/2)); label("7",(.5,3*sqrt(3)/2)); label("2",(2.5,3*sqrt(3)/2)); label("9",(4.5,3*sqrt(3)/2)); label("1",(.5,5*sqrt(3)/2)); label("8",(2.5,5*sqrt(3)/2)); label("5",(4.5,5*sqrt(3)/2));[/asy]

To cover up a "9" the sum of your die needs to be 9 or more.

Prob: 10/36=[b]5/18[/b]


Let $\triangle ABC\ (AB<AC)$ with the incenter $I$. The bisectors $[AI,\ [BI,\ [CI$ meet the opposite sides in the points $D,\ E,\ F$ respectively. I note the middlepoint $N$ of the side $[BC]$, the intersection $P\in BC\cap EF$, the point $Q\in (AC),AQ=AB$ and the intersection $R\in AB\cap DQ$.
Prove that the points $P,Q,R,N$ belong to the same circle.

$\displaystyle{PD=\frac{2abc}{b^2-c^2}}$, $\displaystyle{DN=\frac{a(b-c)}{2(b+c)}}$.
Hence $\displaystyle{PD\cdot{DN}=\frac{{a^2}bc}{(b+c)^2}}$.
From $\displaystyle{\triangle{RBD}}$ we have
$\displaystyle{RD=\frac{ab}{(b+c)}}$ and
$\displaystyle{BD=DQ=\frac{ac}{b+c}}$.
Consequently$\displaystyle{RD\cdot{DQ}=PD\cdot{DN}}$ and we are done!

It is a nice metrical solution Sailor, see the general case http://www.mathlinks.ro/Forum/viewtopic.php?t=57444
Maybe solve it.


when i did it a couple years ago they gave me five problems and i had 25-30 minutes to do them.  They never gave it to you one at a time.  Ask colts18 she did MOEMS this year.

For the past 5 years, I remember it as you get all five questions and get a huge block of time to work on it. It's rather easier this way!  :)

Yeah, that's what I did when I did MOEMS, but my brother said that it was meant to be done the other way, so I was unsure. Thanks :D

It is up to the school how they administer it.  They can either give the questions one at a time, each for the prescribed number of minutes, or add together all of the times and give the whole thing at once.  Most schools do the latter.

I've never even heard of this competition until it appeared on the AoPS forum.  :|

[quote="texas137"]I have a different question - I am considering MOEMS for a small group of 6th graders next year, but we'll probably have fewer than 10. It says the team score is the sum of the top 10 scores after the fifth contest. If we had, say, 5 kids who did five contests, would the team score be the top 10 out of 25 scores, or the top 10 scores per contest, in which case we would only have half as many scores as a large group?[/quote]

It is the top 10 students' totals over the 5 contests.  But, I would encourage you to go ahead anyway, even with too few kids to have a competitive team.  There are individual awards for top 50%, and I think 5% and 2% or something like that, plus certificates for everyone who participates.  I did it last year with only 4 kids, and it was definitely worth it.

Have Fun!

[quote="sirorange"]Sorry, but I am just wondering how the actual olympaids work. I know you are given five questions, each with a number of minutes listed near them, however do you do all five problems in a certain time limit?

Or, are you given problem one, have the listed number of minutes to work on them, have that taken away, and then given problem two, have the listed number of minutes to work on them, and have that taken away, and so on?[/quote]
I've done these for a few years, and every time we had a 24-30minute time period to work on the 5 questions. However, depending on the school/school district the teacher may make you do one problem in the time it says on the sheet. It IS much easier when you get them in one time block because you can skip a question that's too hard (usually not the case in MOEMS but still) and come back to it, and you can check all your answers at the end.


I have a choice of taking the SAT I or the SAT II this month but I don't know which one to pick. I want to take the SAT I one last time because I really think that I can get a 2400 on it. I want to take the SAT II's because  I want to have more than an SAT I score, and I think after lots of practice that I can get 3 790's.

Which would you take, and why?

I would need to know a lot more than that before deciding, but here are some possibilities:

Reasons to take SAT I again
- You think you can push yourself into the 2300+ range or get 100+ points higher than you did last time.
- You want to sit through another grueling 5 hours.

Reasons to take SAT II
- You haven't done any yet.
- You picked the wrong subjects last time, so you got bad scores.
- You actually studied this time.

I'm perfectly capable and willing to do all these things, so it's a little hard.

it depends on your previous scores...

[quote="paladin8"]
I would [b]need to know a lot more[/b] than that before deciding, but here are some possibilities:
[/quote]


Find for every real number $ a$ all real pairs $ (x;y)$, such that

$ x^{2}+y^{2}=25$

$ x+y=a$.

Some two graphs give the answer, which tells that we should classify the value of $ a=\pm 5\sqrt{2}$.

or u can solve by Vie-te 's theorem , turn it into
$ x+y=a$
$ xy=\frac{a^{2}-25}{2}$


Say, I wanna ask you something about the games of the wsop,if you dont mind..
I saw that they play holdem!!
But im a Omaha high low player…is there a competition of this game?
Both at the [url=http://www.win-wsop.com]wsop[/url]?  Can I play them

There are plenty of other places to play poker online.  Look around, you'll find SOMETHING.


function
calculus
derivative
real analysis
real analysis unsolved

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

Let f(x) be a convex function on [a,b].
Prove that:
1) f is continuous on [a,b].
2) f is both left-derivatiable and right-derivativable on [a,b].
3) f'- (x) <= f'+ (x).

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

No matter what definition you use, you can't show it's continuous on [a,b]. The best we can do is show continuity on (a,b). We can easily build examples continuous on (a,b) and discontinuous in a and b: f:[0,1]->R, f(x)=0 if x<>0,1 and f(0)=f(1)=1 clearly works.

I think it can be done by geometric reasoning (each "piece" of the graph being "under" the chord with the same endpoints).

Anyway, I only know one definition: for all t $\in$ (0,1) and all x,y $\in$ [a,b] we have f(tx+(1-t)y)<=tf(x)+(1-t)f(y).

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

another one (i know only those two):
f((x+y)/2) <= (f(x)+f(y))/2
are they equivalent on R? (seems difficult to prove or disprove)

how do you prove that the function is continuous on $(a,b)$? i remember seeing this on ML some time ago, but I can't find it now

[quote="darij grinberg"][u][b]The author of this posting is : liyi[/b][/u]
another one (i know only those two):
f((x+y)/2) <= (f(x)+f(y))/2
are they equivalent on R? (seems difficult to prove or disprove)[/quote]
mid-convexity, as it is called I think, implies convexity only if you add some condition (like continuity, local boundedness, mesurability, ...) and this is a good exercice which I had at an oral exam in the past  :P

this last exercise should be related to non-continuus solutions of $f(x+y) = f(x) + f(y)$, and i think working on hamel basis could give some bad mid-convex function.

in the general case (full-convexity),in order to prove the continuity you just have to observe that it has left and right derivatives (which, by the way, are the same except for a countable set).

But about the problem in the first post?

solution is quite long and boring, and not difficult.. just some extimation..
it's been posted a couple of days ago, in a thread opened by grinny.

I found the topic you are talking about. But jmerry proved there a different thing.

I was asking about this one:

[quote]Let $f(x)$ be a convex function on $[a,b]$. (in the sense that $t f(p) + (1-t) q \geq f \left( tp+(1-t)q \right)$)

Prove that:

1) $f$ is continuous on $(a,b)$.

2) $f$ is both left-differentiable and right-differentiable on $(a,b)$. (is $f$ right-differentiable in $a$ or left-differentiable in $b$?)

3) $f^{\prime -} (x) \leq f^{\prime +} (x)$.[/quote]

which is quite different.

I remember that I saw a proof by grobber (?), but, as I said, I can't find it anymore.

uhm..
as far as i can remember, it was quite easy and algebraically tricky to show that the function $F(x,y) = \frac{f(y)-f(x)}{y-x}$ is increasing in both variables, hence you've got points 2 and 3.
the point 1 follows from this (you have a condition like $f(x) + f'^-(x)h \le f(x) \ge f(x) + f'^+(x)h$, from which continuity follows..
hope i won't need to make computations :)


Here's a little one that shouldn't be too hard - but anyway i can't figure out how to do it :?

A(0,0), B(1,3), C(4,3)
These 3 are the three corners in the triangle. How do i find the lenght of a (which i define as |BC|)?

[hide]
Distance Formula?

Or, if you graphed it you could see that B and C are exactly 3 units apart
 [/hide]

I know it is 3, but i want to calculate it, and graphs aren't welcome in this task :)

What is the distance formula?

Distance formula: the distance between the points $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. It can be easily proved using the Pythagorean Theorem.

Ah great, don't think i've heard of that one before. Would it be too much to ask for the prove of this formula? If it has something to do with drawings just forget about it, but if you could do it with letters and symbols only, then i would like to see it (i often learn a lot from proves).

It would probably be better with a diagram, but I'll try.

Let there be a line segment between points $(x_1,y_1)$ and $(x_2, y_2)$. Let the length be $d$. Draw the lines $y=x_1$ and $y=y_2$. We now have a right triangle with the original line segment as the hypotenuse. The lengths of the 2 sides are $x_2-x_1$ and $y_2-y_1$. The Pythagorean Theorem tells us that $d^2=(x_2-x_1)^2+(y_2-y_1)^2$. So we have $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

Ok, the prove wasn't too easy to understand, but i think i've got the point, so guess that was everything i needed to do this task. Thx for the help.

Sorry my explanation wasn't very good. :( Like I said, it would've been better with a diagram. Maybe drawing one would help you understand it. :)

I guess the problem was that english isn't my first language, and you are surely better than me in english though my guess is that english isn't your first language neither. But at least it is better than mine (i don't know much english math words).

No, English wasn't my first language, but I live in the US. AoPS/MathLinks will probably help your English though, as well as your math. :)

Um, aren't we on AoPS/MathLinks right now? :huh: Depending on which one you use, of course...

Ah yes, AoPS is a shorting of Art of Problem Solving :)  Any part of the site where you would refer to if i want to know some more math words?

Anywhere that you feel most comfortable. :)


How many nitrate ions are there in 3.54 moles of Rhodium(|||) Nitrate

3.54*6.023*3=6.394*10^24 or 6.39*10^24?

I confused, please tell me which one is the most appropriate answer, thank you!

6.39, because the given data also have 3 significant figures.


This was a brainteaser that was in my agenda book...but it got my teachers stumped!

A bottle of juice sells for $\$$1.00. The juice costs 90 cents more than the empty bottle. How much money would you expend for the empty bottle?

P.S. But why would you need an empty bottle?

[hide][b]5 cents[/b]. If x+y=100 and x-y=90, then by elimination, 2x=190, and x=95. Therefore, y=5.[/hide]

[hide][u][b]Another way[/b][/u]:

b->cost of  empty bottle; j->cost of juice; and so:
Cost of a full bottle of juice = b+j

Now, 100 = b+j; and j = b+90; giving:
100 = 2b+90; so that:
b= 5 and j = 100 - 5 = 95.[/hide]

is this a brainteaser, coz it's like more of math...
[hide] let bottle=x,juice=x+90. x+x+90=100(cents)
2x=10,x=5[/hide]
oh by the way if someone's already posted this answer,sorry coz i didn't read any other posts... :D

Here's a similar problem by Ramond Smullyan:

A bottle of wine cost ten dollars. The wine was worth 9 dollars more than the bottle. How much was the bottle worth?

P.S. You need an empty bottle to put the juice in!

Easy question :D:P (Refering to D3m0n Shad0w's question)

[hide]The solution is obviously 5, because 5+5+90 = 100 = 1.00 Dollar :)[/hide]

I'm not sure...
[hide]50 cents[/hide]   :D

nutz_for2.718281828 wrote:

[quote]A bottle of wine cost ten dollars. The wine was worth 9 dollars more than the bottle. How much was the bottle worth?[/quote]

[hide]Puzzle Solution :

50 cents

Explanation:
Br the problem, twice the cost of the bottle is 900 cents (9 dollars) less 1000 cents (10 dollars); giving:

The cost of the bottle =  (1000 - 900)/2 = 50 cents[/hide].

The bottle was 10 dollars...that's exactly what you told us...


I have this new 3d modeling program called Wings3d (http://www.wings3d.com) and I think i'm getting the hang of it.

It has this 'tesselate' function which can split a polygon into triangles or quadrangles.

But it's not doing something i want, and i'm wondering if it's possible to tesselate a triangle into quadrangles.

Do you mean just break a triangle into quadrilaterals?  Because that's easy -- pick a point on each side of the triangle and join each of them to some point in the interior (the same point for all of them).  If you mean something more complicated, then I'm not sure.

Is that the only way? (assuming it's an equilateral)

Of course not -- break it into a bunch of triangles, and then each of them can be broken into 3 quadrilaterals.  There may also be ways to break it into 5 or 7 quadrilaterals, but you should just try drawing some pictures to see.


If the two diagonals are drawn across a square floor with square tiles, they pass through a total of 29 square tiles. How many square tiles are in the entire floor?

They both pass through exactly one square, so diagonal passes through 15 squares. There are 225 tiles.


geometry circumcircle conics hyperbola Asymptote function parameterization

P,Q are two points in  triangle ABC, which PAB=QAC  , PCB=QCA and P,O,Q are in a line ( O is  circumcenter of  triangle ABC)
Prove that a circle which passing through image of P,Q on AB,AC,BC is tangent to circumcircle of medial triangle.

  :?:

P, Q are isogonal conjugate points WRT the $ \triangle ABC.$ Let the line $ k \equiv PQ$ meet the circumcircle (O) of the $ \triangle ABC$ at points S, T. Let s, t be the Simson lines of the $ \triangle ABC$ with the poles S, T. Since $ O \in ST \equiv PQ,$ S, T are diametrically opposite points of (O), $ s \perp t,$ and s, t meet at the orthopole L of k WRT the $ \triangle ABC,$ which lies on its 9-point circle (N). The isogonal conjugate of the line $ k \equiv ST \equiv PQ$ passing through the circumcenter O WRT the $ \triangle ABC$ is a hyperbola $ \mathcal K$ with asymptotes s, t, i.e., a normal hyperbola. $ P \in k$ means $ Q \in \mathcal K$ and $ Q \in k$ means $ P \in \mathcal K,$ the isogonal conjugate points P, Q are the intersections of $ k \cap \mathcal K.$ The Simson lines s, t cut in half the segments HS, HT from the orthocenter H of the $ \triangle ABC$ to their poles S, T. Since $ (N) \sim (O)$ with similarity center H and coefficient 1/2, the midpoints U, V of the segments HS, HT lie of the 9-point circle (N) and the lines $ UV \parallel ST \equiv PQ \equiv k$ are parallel. Because of the right $ \angle ULV \equal{} 90^\circ,$ N is the midpoint of the segment UV. Let the Simson lines s, t meet the line $ k \equiv ST \equiv PQ$ at points X, Y. Since the collinear points $ P,\ Q \in \mathcal K$ lie on the hyperbola and $ X \in s,\ Y \in t$ on its asymptotes, the segments XY and PQ have a common midpoint K (this is trivial to prove). Since $ XY \equiv ST \parallel UV,$ the right $ \triangle LXY \sim \triangle LUV$ are centrally similar with similarity center L and so are their circumcircles. Thus $ K \in LN$ and the circumcircle (K) of the $ \triangle LXY$ is tangent to the 9-point circle (N) at L. Since P, Q are isogonal conjugate points WRT the $ \triangle ABC,$ their pedal triangles have a common circumcircle centered at the midpoint K of the segment PQ. By the  [url=http://mathworld.wolfram.com/FonteneTheorems.html]2nd Fontene theorem[/url] (or the problem [url]http://www.mathlinks.ro/viewtopic.php?t=173756[/url]) , the circumcircles of the pedal triangles of all points on the line k pass through a fixed point. Two of these pedal triangles, the pedal triangles of S, T, are degenerated and so are their circumcircles, identical with the Simson lines s, t, intersecting at the orthopole L of the line $ k \equiv UV \equiv PQ,$ L is the fixed point. Thus the common circumcircle of the 2 pedal triangles of P, Q centered at K passes through L, hence it is identical with (K), which is tangent to (N) at L. (The posted problem is actually the 3rd Fontene theorem.)

[hide="Show details"]
Let $ \mathcal L(x, y, ...) \equal{} a \plus{} bx \plus{} cy \plus{} ...$ be an arbitrary linear function of its parameters (no mixed, quadratic, or higher power terms). WLOG, let coordinates of the vertices B, C of a $ \triangle ABC$ be (-1, 0) and (+1, 0) and denote $ b \equal{} \tan B,\ c \equal{} \tan C.$ Let y = mx + q be equation of an arbitrary line k and P a point on this line, with coordinates (u, v) = (u, mu + q). Equation of the line BP is

$ y \equal{} \frac {mu \plus{} q}{u \plus{} 1}(x \plus{} 1) \equal{} \tan(\phi) (x \plus{} 1)$

Let P* be the isogonal conjugate of P WRT the $ \triangle ABC.$ BP* is then isogonal of the line BP WRT the $ \angle B,$ having slope $ \tan(B \minus{} \phi) \equal{} \frac {\tan B \minus{} \tan \phi}{1 \plus{} \tan B \tan \phi},$ so that equation of the line BP*

$ y \equal{} \frac {b(u \plus{} 1) \minus{} (mu \plus{} q)}{(u \plus{} 1) \plus{} b(mu \plus{} q)} (x \plus{} 1) \equal{} \frac {\mathcal L_1(u)}{\mathcal L_2(u)}(x \plus{} 1)$

and similarly, equation of the line CP* is $ y \equal{} \frac {\mathcal L_3(u)}{\mathcal L_4(u)}(x \minus{} 1).$ To get the locus of P*, we eliminate the arbitrary parameter u. Calculating u from both equations,

$ \frac {\mathcal L_5(x, y)}{\mathcal L_6(x, y)} \equal{} u \equal{} \frac {\mathcal L_7(x, y)}{\mathcal L_8(x, y)}$

which is clearly equation of a conic. Thus the isogonal conjugate of an arbitrary line WRT a $ \triangle ABC$ is a conic $ \mathcal K.$

The line k generally intersects the triangle sidelines BC, CA, AB at X, Y, Z. Reflections of BX, CX in bisectors of the $ \angle B,\ \angle C$ are the lines BA, CA, intersecting at A, so that A is isogonal conjugate of X and similarly, B, C are isogonal conjugates of Y, Z.  On the other hand, isogonal conjugates of A, B, C are undefined, because any points on the opposite sidelines will do. Thus the conic $ \mathcal K$ passes through A, B, C. Generally, it intersects the triangle circumcircle at one other point (it may be tangent to the circumcircle at one triangle vertex).

Let D be pole of the Simson line $ d \perp k.$ To find D, draw a normal to k from one triangle vertex, say A, intersecting its circumcircle (O) again at A', and then draw a normal to BC from A', intersecting the circumcircle at D. Isogonal conjugate of any point on the triangle circumcircle, including D, is at infinity. WLOG, let D be on the circumcircle arc AB opposite to C. Reflect AD, BD in bisectors of of the $ \angle A, \angle B$ and let D* be their intersection.

$ \angle D^*AB \plus{} \angle ABD^* \equal{} \angle BAD \plus{} \angle A \plus{} \angle B \plus{} \angle DAB \equal{}$
$ \equal{} \angle A \plus{} \angle B \plus{} \pi \minus{} \angle ADB \equal{} \angle A \plus{} \angle B \plus{} \angle C \equal{} \pi,$

$ AD^* \parallel BD^*,$ D* is at infinity. Since $ AA' \perp k,\ A'D \perp BC,$ we have $ \angle AA'D \equal{} \angle YXC.$

$ \angle D^*AC \equal{} \angle BAD \equal{} \pi \minus{} (\angle ADB \plus{} ABD) \equal{}$
$ \equal{} \angle C \minus{} \angle AA'D \equal{} \pi \minus{} (\angle XCY \plus{} \angle YXC) \equal{} \angle XYC,$

Thus $ AD^* \parallel YX \equiv k$ and the isogonal conjugate D* of D is the point at infinity of the line k, $ D^* \in k,$ which means $ D \equal{} (D^*)^* \in \mathcal K.$ The pole D of the Simson line $ d \perp k$ is the remaining intersection of the conic $ \mathcal K$ with the triangle circumcircle. If $ k \parallel BC,$ then $ D \equiv A$ and $ \mathcal K$ is tangent to the triangle circumcircle at A.

The line k can intersect the triangle circumcircle at 2 points, touch it at 1 point, or it does not have to intersect it at all. Therefore, the conic $ \mathcal K$ has 2, 1, or 0 points at infinity and it is a hyperbola, parabola, or an ellipse, respectively. Let the line k intersect the triangle circumcircle at P, Q. The isogonal conjugate of k is a hyperbola $ \mathcal K.$ The isogonal conjugates of P*, Q* of P, Q are at infinity, but at the same time $ P^*,\ Q^* \in \mathcal K,$ hence P*, Q* are directions of the hyperbola asymptotes.

Assume now that the line k passes through the triangle circumcenter O, so that PQ is a circumcircle diameter. Let the reflections AP*, AQ* of AP, AQ (P*, Q* are at infinity) in the bisector of the $ \angle A$ meet the triangle circumcircle again at P', Q'. Since AP, AP' are isogonals of AB, AC, PBCP' is an isosceles trapezoid and $ PP' \parallel BC.$ Constructing the pole of the Simson line with the direction AP', we just draw a normal to BC at P', identical with a normal to PP' at P', which intersects the triangle circumcircle at Q. Thus Q is the pole of a Simson line q with the direction $ AP' \equiv AP^*$ parallel to one hyperbola asymptote and similarly, P is the pole of a Simson line p with the direction $ AQ' \equiv AQ^*$ parallel to the other hyperbola asymptote. Since $ AP \perp AQ,$ their reflections $ AP' \perp AQ'$ in bisector of the $ \angle A$ are also perpendicular, and so are the Simson lines $ q \perp p$ and the hyperbola asymptotes.

The triangle orthocenter H is the isogonal conjugate of its circumcenter O. Since $ O \in k,$ it follows that $ H \in \mathcal K.$ If we rotate the line $ k \equiv PQ$ around the triangle circumcenter O (if we take a family of all lines through O), we obtain a family of all rectangular hyperbolas $ \mathcal K \equal{} k^*$ (with all possible directions of major axes and perpendicular asymptotes) passing throuh the triangle vertices A, B, C. But the triangle orthocenter H, being the isogonal conjugate of the circumcenter O, lies on all of them. Thus the orthocenter H lies on any rectangular hyperbola passing through A, B, C.

Let H' be orthocenter of the $ \triangle DBC.$ Since $ D,\ B,\ C \in \mathcal K,$ it follows that $ H' \in \mathcal K.$ $ \angle BH'C \equal{} \pi \minus{} \angle D \equal{} \pi \minus{} \angle A \equal{} \angle BHC,$ both orthocenters H and H' are on the reflection (O') of the circumcircle (O) in BC and DH' = AH. In addition, $ DH' \parallel AH,$ so that AHH'D is a parallelogram inscribed in the hyperbola $ \mathcal K.$ This means that the parallelogram diagonal intersection, identical with the midpoint L of HD, is the hyperbola center and the intersection of its asymptotes. The circumcircle diameter $ PQ \perp AA'$ cuts the chord AA' at its midpoint M. Perpendicular to BC from M is the midline of the 2 parallel lines $ AH \parallel A'D$ (both perpendicular to BC) and since L is the midpoint of HD, L lies on this perpendicular to BC from M. Similar argument can be repeated for the remaining triangle vertices B, C, hence by definition, L is the orthopole of the line $ k \equiv PQ$ WRT the $ \triangle ABC.$ The 9-point circle (N) of the $ \triangle ABC$ is centrally similar to its circumcircle (O) with the center H and coefficient 1/2, which means $ L \in (N).$

The Simson line $ d \perp k$ with the pole D also cuts the segment HD at its midpoint $ L \in (N)$ and the Simson lines $ p \perp q$ with the poles P, Q cut in half the segments HP, HQ at their midpoints $ U,\ V \in (N).$ The $ \triangle LUV \sim \triangle DPQ$ are centrally similar with center H and coefficient 1/2 and their corresponding sides parallel, $ LU \parallel DP,\ LV \parallel DQ,\ UV \parallel PQ.$ Since $ PP' \parallel BC,$ we have $ \angle AA'D \equal{} \angle YXC \equal{} \angle QPP',$ AD = QP', ADQP' is an isosceles trapezoid, $ AP' \parallel DQ.$ But $ q \parallel AP',$ hence $ q \parallel DQ \parallel LV$ and similarly, $ p \parallel DP \parallel LU.$ In conclusion, the Simson lines p, q with the poles P, Q pass through the midpoints U, V of HP, HQ and they are parallel to LU, LV, hence they are identical with the lines LU, LV. Since they are parallel to the hyperbola $ \mathcal K$ asymptotes and meet at the hyperbola center L, they are identical with its asymptotes.

[b]Lemma[/b]: Let a line meet a hyperbola at points P, Q and its asymptotes at U, V. Then the segments PQ and UV have a common midpoint K.

Let L be the hyperbola center (intersection of its asymptotes). Using a parallel projection (which preserves ratios of line segments on the same line), project the $ \triangle LUV$ to an isosceles (or even isosceles right) $ \triangle LUV'$ with LU = LV'. The hyperbola $ \mathcal H$ with asymptotes LU, LV goes into a hyperbola $ \mathcal H'$ with asymptotes LU, LV', the line UPQV into a line UP'Q'V' intersecting the hyperbola $ \mathcal H'$ at P', Q'. UP' = V'Q' by symmetry, hence UP = VQ and UV, PQ have a common midpoint K.
[/hide]


AMC
AIME
analytic geometry
factorial
geometry
parallelogram
USA(J)MO

How did you do? How did you think you would do? What do you think of how you did? etc

I luckily manged not to screw up this round and got a 25 :) for a 97 total for the year. I also got 2 commended this round, so 3 total. Last year, I got 96 total with 5 commended.

The last option being just to see how many people choose it  :P  :wink:

I got a 25, but no commends  :( for a total of 99.

Yay, I got a 5 for my nonsimplified answer, x(1-x)/(x^2+(1-x)^2). So 100 for the year, 1 commended.

I got 25 and 2 commends, for a total of 100 with 6 commends (...an evidence of my unmitigated perfectionism). I think I improved much this year. I would not be able to do the problems of Round 4 had I not done the previous rounds. I hope I can one day do these problems as fast as AIME problems...

'Tis time to face the upcoming (failure at) USAMO...

hate all of ya who brute forced #5 with mathematica
my #5 got 2 pts docked for incompleteness so 23 this round and 98 total

Yay, the unique 10 commended  :lol:, with a 98.

ooh i like this comment for my #5

"I do not see how we can determine P3E=P3F simply by 'observing and measuring the figure'"

hehe

I got a 23, a grader gave me a 3 on one of them because ... well I don't actually understand why.  So maybe it'll go up .. at the moment I have a 93 for the year, which is a big improvement from my 36 last year  :D

I like my grader's comment for number 5:
"Not bad for a coordinate solution!!"  :lol:

EDIT: hurray, 3 to 5 commended!!
so that's a 5 [b]5 5[/b] 5 5

I got [b]5[/b]5544 for a total of 23 in round 4 (I thought I would have done a lot worse).  This gives me a grand total of 92.

Yay, I got a 25! (not factorial)

So my grand total is 0+0+23+25=48 with no commends.

oh SKILLZ, i got a 0. for a total of

19 + 0 + 0 + 0 = guess what. and i have 1 commended solution. oh well i have 4 more years.

My number 5 solution was really really dull.  And I didn't use Mathematica or anything else.  So I'm glad that I got a 5.

My solution to number 4 was really simple, so I got a commended.  And I think it was different from what anyone else posted.

Anyway:
5 5 5 [b]5[/b] 5 = 25
97 overall with 1 commended

55[b]5[/b][b]5[/b]5 = 25

Year total: 99. I'm satisfied with that.

[b]@ calc rulz[/b]: You should make a poll. It would be very annoying if another user made a poll. Then I'd have to look at two topics, and I'm too lazy to do that.

So 55[b]5[/b]55 for 25. Total of 98, two commends. At least I reached my goal.

Here's my favorite comment:

Round 4 Problem 2: "Although correct, your proof is extremely convoluted"

5 [b]5[/b] 5 5 5

98 for the year
3 commendeds

For some reason my upper-bound/computer brute forced solution got a commended.  :rotfl:

- 5 3 5 -

Still no commendeds. I had hardly any time to do this round because of school, yuck.

5[b]5[/b]5[b]5[/b]5

I was rather disappointed that my parallelogram solution didn't get a commend, but oh well.

[quote="worthawholebean"]
I had hardly any time to do this round because of school, yuck.
[/quote]

My thoughts exactly.
5 5 2 5 - =17
total 86, so not too bad.

Argh, 54[b]55[/b]5 this round for 99 total.  I apparently lost a point on #2 for not explaining my last paragraph enough and for not simplifying my answer from $ 2\cdot 858 \minus{} 1$, which I did on purpose because I decided that writing the answer as $ 2\cdot 858 \minus{} 1$ was more relevant to the problem - kind of like writing $ 4\cdot 2007 \minus{} 1$ for last year's USAMO #4.  Not one of my better ideas.  :lol:

Still, I'm very happy with 11 commended solutions for the year!  :)

That's kind of a lame docking...as I said, I got a 5 for leaving my answer as $ \dfrac{\mu(1\minus{}\mu)}{\mu^{2}\plus{}(1\minus{}\mu)^{2}}$ on #3.

5 [b] 5 [/b] 5 [b] 5 5 [/b]

for 8 commendeds, 100/100

uh yeah my 95 last year was pretty disappointing... 1 point off from gold

anyway my large number of commended solutions surprises me

[quote="CatalystOfNostalgia"]That's kind of a lame docking...as I said, I got a 5 for leaving my answer as $ \dfrac{\mu(1 \minus{} \mu)}{\mu^{2} \plus{} (1 \minus{} \mu)^{2}}$ on #3.[/quote]I got a point off on number 3, because in the problem it said to express the value in terms of $ \mu$, so I put $ 0\cdot \mu \plus{} \frac {1}{2}$ (because I thought we HAD to express it in terms of $ \mu$...
The grader took a point off; he said it was not fully simplified. I did it on purpose... :( AWEHGAHWEGHAWEG

^ wow that point-docking is kinda ridiculous. Meh...

I got a 25 this round-- my first 25 ever, and this is my third year. stupid mistakes get me every time...

so I squeaked in with a 96 : D

How do you "brute-force" number 5 with Mathematica? I'm not really familiar with the program.

I brute forced with java... calculated the lengths of the diagonals, then used nested for loops to find a close set of coefficients. got the correct ones immediately, all that was left was some trig manipulation. it took me only less than half a page to go through the trig, so it wasn't bad at all. i feel bad about my 25 though... like i didn't deserve it. :9 oh well. heh

How do you find  "close set of coefficients"...like 6 loops running from 0 to 10 each? How did you use the trig?

well i expressed the diagonals d_1 through d_6 in pretty simple terms using cosine, assuming d_1 = 1. I also calculated s using a pretty nifty similarity.

Then I had the expression c = c_1 * d_1 + c_2 * d_2 + ... c_6 *d_6

then a for loop with for(int c_1 = -10; c_1 < 11; c_1++) {
//nested loop
}

and it would break and print the coefficients if Math.abs(c - s) < 0.00001

ahhh. i drew it in autocad, and had all the approximate lengths...i should've wrote a program...

5 5 5 5 5

Totally surprised with this, this was my best round yet, ended up with a 93. So pretty much if I didn't make stupid mistakes on Round 1, then I could have gotten gold...meh

I'm happy. Got a total of 100!!! Yippie!

Does anyone know when we can select the prizes?

I received scores of 5-5-5-5-1 for the problems, for a total of 21 for the fourth round and a grand total of 91, approximately the score I expected to get. (My actual fourth-round score mattered very little to me: a score of 5 would have given me a 75 total and the maximum score of 25 would have given me a 95 total.)

Can anybody explain how to brute force number 5 with Mathematica?

Oh, and I got my point back for that unsimplified answer ($ 0 \cdot \mu \plus{}\frac{1}{2}$) for number 3.

Do the same thing mathclass did, except in Mathematica.

lol everyone who posted scored 90+ :) I scored a 67 which I am really happy about... :P This is my first year XD

5  5  5  5  -

Dont have Mathematica to brute force number five. Had no idea how to actually solve it. Puts me at an 82 for the year. Beats my 62 from last year and my 14 from freshman year. Going for gold next year.

5 4 5 5 -
I got docked a point on number 2 because I didn't include a transcript of the program I used to find the answer.  That's okay, though, because I was hoping for a 3.

[quote="mathclass"]I got 70, 87, 96 in freshman, sophomore, and junior year. -_- unlike ubemaya and kyyuanmathcount and catalystofnostalgia... freshmen who all beat me this year. you guys suck :P[/quote]

 :o    :(

Yeah let's say I lost a point on Round 1

[b]5[/b] [b]5[/b] 5 [b]5[/b] 5
100 on the year, 6 commended.
That's weird: I got 3 commended on round 1 and none on rounds 2 or 3. It's kind of all or nothing.

Now that I read them again, it makes sense that these were commended. On 1, I actually took the time to draw (on a computer) the pictures, and it looks nice. Only 6 people were commended on that problem, the lowest for the year. On 2, my solution was quicker than most (I added 1 at the beginning). On 4, mine was like three lines, but I bet they were just trying to get a lot of different solutions commended, and mine sounded different.

By the way, what you guys are saying to do with Mathematica, I did with a TI-83+ (had it work by absolute values).

100 is an improvement over last year's 99 (stupid calculator told me $ \pi/2$ and $ 3\pi/2$ worked on that trig problem) and sophomore year's 84 but my commendations continue to decline (10, 7, now 6).

Also, I like how the website calls me a "Gold Winner winner".

[hide="round 4"]0 :rotfl: [/hide]
[hide=" year total"]0 :rotfl: [/hide]

[hide]that's right, i didnt take it :rotfl: [/hide]

This was my first year, as an 8th grader  :(

I got 5-5-5-5-5 (I was expecting around 22!)

Check the round 4 discussion thing that emton made to see my proof of #5, its pretty nasty  :P

I got a grand total of 92!!


invariant
Hi
game
infinite chessboard
combinatorics
IMO
IMO 1993

On an infinite chessboard, a solitaire game is played as follows: at the start, we have $n^2$ pieces occupying a square of side $n.$ The only allowed move is to jump over an occupied square to an unoccupied one, and the piece which has been jumped over is removed. For which $n$ can the game end with only one piece remaining on the board?

[hide="hint I (solvable cases)"]By induction with some small cases you can prove for $ n \not \equiv 0 (3)$[/hide]
[hide="hint II (unsolvable cases)"]$ n \equiv 0 (3)$ use the associative and commutative $ *$ relation of $ \{1,a,b,ab\}$ with $ a*b \equal{} ab, a*ab \equal{} b, b*ab \equal{} a$ [/hide]

PS: I have been there, and learn that nice solution after the IMO  :blush:

[u]Solution.[/u] Clearly, cases with $ n \equal{} 1,2$ are solvable, and a $ 1\times 3$ "tile" can be deleted if there exists an empty and full square to the left and right of an end of the tile.  Hence, a $ n\times n$ square can be reduced to a $ n\bmod 3\times n\bmod 3$.  If $ 3\nmid n$, then we are done.

Assume to the contrary that a $ 3n\times 3n$ square can be reduced to one point.  Consider the function $ f: \mathbb{N}^2\rightarrow \mathbb{Z}_3$ such that $ f(a,b) \equal{} a \plus{} b\bmod 3$.  Let $ f_0,f_1,f_2$ be the cardinality of the set of preimages of $ 0,1,2$, respectively.  Since initially $ f_0 \equal{} f_1 \equal{} f_2 \equal{} 3n^2$ and each step reduces two and increases one of $ f_0,f_1,f_2$ by 1, $ f_0\equiv f_1\equiv f_2\pmod 2$ at all times.  Ergo, $ (f_0,f_1,f_2)\neq (0,0,1), (0,1,0),(1,0,0)$.

And the conclusion follows. $ \Box$

[quote="anonymous1173"]Isn't it just a special case of Russia 1992 Grade 9 problem 4,
Given an infinite sheet of square ruled paper. Some of the squares contain a piece. A move consists of a piece jumping over a piece on a neighbouring square (which shares a side) onto an empty square and removing the piece jumped over. Initially, there are no pieces except in an m x n rectangle (m, n > 1) which has a piece on each square. What is the smallest number of pieces that can be left after a series of moves?
I think it is very interesting.[/quote]

Yes, it is.

Dieter Gronau writes in his report on IMO'1993:

Erst während der Klausur wurde bekannt, daß die 3. Aufgabe, die von Finnland eingereicht worden war, im vergangenen Jahr bei der All-Republiks-Olympiade der GUS in Alma-Ata, an der wohl alle Nachfolgestaaten der UdSSR, außer dem Baltikum, teilnahmen, gestellt worden war. Keiner der Leiter dieser Delegationen informierte die Jury. Selbst der russische Delegationsleiter (Prof. Vavilov) informierte die Jury nicht, obwohl er erst kürzlich einen Artikel in Quant über diese Olympiade veröffentlichte. Einer der Stellvertreter, die die Aufgaben erst während der Klausur erhalten, deckte diesen Sachverhalt auf. Es gab eine lange und sehr kontroverse Diskussion in der Jury. Einerseits wird dieser Sachverhalt nicht vom Reglement überdeckt und jeder hatte Zugang zu diesem Artikel. Andererseits empfanden es viele Delegationsleiter als unakzeptabel im Hinblick auf die Bestrafung der Teilnehmer hier nichts zu unternehmen. Allerdings lassen die Ergebnisse der Teilmehmer aus diesen Ländern nicht darauf schließen, daß die Schüler die Aufgabe gut kannten. Die Anträge auf Abstimmung in der Jury reichten von der namentlichen Verurteilung des russischen Delegationsleiters bis zur Ignorierung des Sachverhaltes. Schießlich wurde folgende Erklärung von der Jury mehrheitlich angenommen (13 Ja, 10 Nein, 46 Enthaltung): Die Jury der 34. IMO mißbilligt die Inaktivität der Leiter einiger Delegationen hinsichtlich der 3. Aufgabe.

The case for a $3n \times 3n$ square can be solved nicely by using a simple colouring/invariance argument: colour the squares of the infinite chessboard with 3 colours $(a,b,c)$ such that you have a bottom-left-to-top-right diagonal of colour $a$ next to a bottom-left-to-top-right diagonal of colour $b$ next a bottom-left-to-top-right diagonal of colour $c$ and so on. Now let the total number of pieces on squares coloured with colour $a$ be $A$; define $B,C$ similarly.

Now any move keeps the parities of $A-B, B-C, C-A$ the same: e.g. if you move from a square of colour $a$ to one of colour $c$ then $A$ goes to $A-1$, $B$ goes to $B-1$ and $C$ goes to $C+1$ and so $A-B$ stays the same while $B-C$ decrease by $2$ and $C-A$ increases by $2$. At the start the values of $A-B, B-C, C-A$ are all $0$ (there are the same number of pieces $(=3n^2)$ on each type of square). While in the configuration we are trying to get to, two of $A-B, B-C, C-A$ are odd. So the configuration can't be reached.

[hide=solution]
Note that there will be a total of $n^2-1$ moves if we are going to reach one piece remaining.

[b]Lemma[/b]: If $n$ is divisible by 3, it is impossible to reach a state with only one piece on the board.

Assume otherwise.

Color the chessboard like so:

ABCABC...
BCABCA...
CABCAB...
..................

Then, note that if $n$ is divisible by three, there are an equal number of squares on A, B, and C, due to each three by three square containing 3 A's, 3 B's, and 3 C's. Let the number of pieces on A, B, and C each be $3k$

Also, each move removes two pieces of different colors and adds one piece of the third color.

Let an A-move remove pieces on B and C and add one on A. Define B-moves and C-moves similarly, and assume there are X, Y, and Z A-moves, B-moves, and C-moves, respectively. Assume wlog that we will end with a piece on block C. This means that

$$-X+Y+Z=3k$$
$$X-Y+Z=3k$$
$$X+Y-Z=3k-1$$

There is no integral solution to this system of equations (Add the latter two and divide by 2 to get $X$ which is clearly not an integer), so there is no way of reaching one piece if $3|n$ $\Box$

Now, we must find algorithms to prove it is possible to end with one piece if 3 does not divide $n$

1. $n=3n+1\implies n=3n-1$

Note that we can remove three pieces in a row in the following manner, where X is a piece and O means no piece:

XX___OOX___OXX____ XO
XX--->XX__---> XO__--->XO
_X____ X______  O______O
(call this a 3-removal)

If we have a $3n+1$ by $3n+1$ block, we can simply go around the board in this manner and remove every block on the perimeter, leaving a $3n-1$ by $3n-1$ block.

2. $n=3n+2\implies n=3n-1$

Consider a $3n+2$ by $3n+2$ block. First, we use $3n$ 3-removals to remove a $3$ by $3n$ block from the corner. This leaves us with a $3n-1$ by $3n+1$ block attached to a 2 by 3 block. Note that we can completely remove this 2 by 3 block, like so:

XXXX____XXOO____XX___X
XXXX --->_XXOO --->XX--->X
X________XOXX____XX___X
...

with the last step involving a 3-removal. This means we now have a $3n-1$ by $3n+2$ rectangle. Now, we can do pretty much the same thing. Remove a $3(n-1)$ by $3$ block using $n-1$ 3-removals, which leaves a $3n-1$ by $3n-1$ block connected to a 2 by 3 block, which we can eliminate.

It's easy to prove that 1 by 1 blocks and 2 by 2 blocks work. Thus, we can always reduce a square whose side lengths are not divisible by 3 to a 2 by 2 square, so $n=3k+1, 3k+2$ satisfies for nonnegative $k$. $\blacksquare$

[/hide]

I'll prove that $3|n$ then the statement is false
first color the chessboard as below
let $S$ the sum of occupied square
firstly $S=0$
note that in each move $S$ by $2\omega^i$
so at the end $S=2a+2b\omega +2c\omega^2=\omega^i=2a'+2b'\omega$
[b][u][color=#600]case(1)[/color][/u][/b]:$b'=0 \implies 2a'=1$ a contradiction
[b][u][color=#600]case(2):[/color][/u][/b] $b' \neq 0 \implies \omega \in Q$ a contradiction
and we win :D

The answer is all $n$ not divisible by $3$.
First, we show the task is impossible when $3 \mid n$. Color the board periodically red, green, blue as shown: \[ \begin{bmatrix} 		R & G & B & R & G & \dots \\ 		G & B & R & G & B & \dots \\ 		B & R & G & B & R & \dots \\ 		R & G & B & R & G & \dots \\ 		G & B & R & G & B & \dots \\ 		\vdots & \vdots & \vdots & \vdots & \vdots & \ddots 	\end{bmatrix} \] Note that every move changes the parity of the number of tokens of each color. When $3 \mid n$, there are an equal number of tokens of each color, so it is impossible to arrive at one token of one color and none of the others.
We now devise an algorithm for $3 \nmid n$. This requires two parts:
[b][color=red]Claim:[/color][/b]  	Suppose we have the following configuration, 	where a $\ast$ represents a cell with a token 	and a $\cdot$ represents an empty cell: 	\[ 		\begin{bmatrix} 			& \ast & \\ 			& \ast & \\ 			\cdot & \ast & \ast \\ 		\end{bmatrix} 	\] 	Then we can remove the three cells in the center column.
[i]Proof.[/i]  	Jump the lower-right corner, then the upper counter, 	then the lower-left counter. $\blacksquare$
Using this, we can show $n=4$ is possible in the following way: \begin{align*} 	\begin{bmatrix} 		\ast & \ast & \ast & \ast \\ 		\ast & \ast & \ast & \ast \\ 		\ast & \ast & \ast & \ast \\ 		\ast & \ast & \ast & \ast \\ 	\end{bmatrix} 	&\mapsto 	\begin{bmatrix} 		\ast \\ 		\ast & \ast & \ast & \ast \\ 		\ast & \ast & \ast & \ast \\ 		\ast & \ast & \ast & \ast \\ 	\end{bmatrix} 	\mapsto 	\begin{bmatrix} 		\ast \\ 		\ast \\ 		\ast & \ast & \ast & \ast \\ 		\ast & \ast & \ast & \ast \\ 	\end{bmatrix} \\ 	&\mapsto 	\begin{bmatrix} 		\\ 		\\ 		& \ast & \ast & \ast \\ 		\ast & \ast & \ast & \ast \\ 	\end{bmatrix} 	\mapsto 	\begin{bmatrix} 		\\ 		\\ 		\\ 		\ast & \ast & \ast & \ast \\ 	\end{bmatrix} 	\mapsto 	\begin{bmatrix} 		\\ 		\\ 		\\ 		\ast & \phantom{\ast} & \phantom{\ast} & \phantom{\ast} 	\end{bmatrix} \\ \end{align*}
This lets us to reduce an $(n+3) \times (n+3)$ square to an $n \times n$ square in the following way:
[asy] 	size(5cm); 	filldraw(scale(7)*unitsquare, invisible, red); 	label("$n \times n$", (3.5, 3.5), red); 	for (int i=0; i<3; ++i) { 		dot( (0.5, 7.5+i), green); 		dot( (1.5, 7.5+i), blue); 		dot( (2.5, 7.5+i), green); 		dot( (3.5, 7.5+i), blue); 		dot( (4.5, 7.5+i), green); 		dot( (5.5, 7.5+i), blue); 		dot( (7.5+i, 0.5), green); 		dot( (7.5+i, 1.5), blue); 		dot( (7.5+i, 2.5), green); 		dot( (7.5+i, 3.5), blue); 		dot( (7.5+i, 4.5), green); 		dot( (7.5+i, 5.5), blue); 	} 	draw(shift(6,6)*scale(4)*unitsquare, black); 	dot((6.5, 6.5), red); 	for (int i=0; i<=3; ++i) { 		for (int j=0; j<=3; ++j) { 			if (i+j>0) { 				dot((6.5+i, 6.5+j), black); 			} 		} 	} [/asy]
Remove the green and blue dots shown in alternating waves, then repeat the strategy with $n=4$ to remove all black dots.


The integers from 1 to $ 2008^2$ are written on each square of a $ 2008 \times 2008$ board. For every row and column the difference between the maximum and minimum numbers is computed. Let $ S$ be the sum of these 4016 numbers. Find the greatest possible value of $ S$.

[hide]Consider the 2008 rows. We compute the difference between the maximum and minimum numbers in each row. Let $ S_1$ be the sum of the 2008 numbers. Clearly, the maximum value is obtained when the numbers $ 2008^2,2008^2\minus{}1,2008^2\minus{}2,\ldots,2008^2\minus{}2007,1,2,3,\ldots,2008$ contribute to the sum. Hence the greatest value of $ S_1$ is $ 2008^2\plus{}2008^2\minus{}1\plus{}2008^2\minus{}2\plus{}\ldots\plus{}2008^2\minus{}2007\minus{}1\minus{}2\minus{}3\minus{}\ldots\minus{}2008\equal{}8092352448$. Analogously, the greatest possible sum of the differences between the maximum and minimum numbers in each column is the same. Now we prove that this is attainable. Look at the table below (which is the upper leftmost $ 6\times6$ subboard):
\[ \begin{array}[b]{cccccc}
2008^2&1&\minus{}&\minus{}&\minus{}&\minus{}\\
2&2008^2\minus{}1&\minus{}&\minus{}&\minus{}&\minus{}\\
\minus{}& \minus{} & 2008^2\minus{}2 & 3& \minus{} & \minus{}\\
\minus{}&\minus{}&4&2008^2\minus{}3& \minus{} & \minus{}\\
\minus{}&\minus{}&\minus{}&\minus{}&2008^2\minus{}4&5\\
\minus{}&\minus{}&\minus{}&\minus{}&6&2008^2\minus{}5\\
\end{array}
\]
In this table, we get the greatest possible value of $ S$, which is $ 2\times8092352448\equal{}16184704896$.[/hide]

There can be another solution, with the known of the existence of $ \max{f(2n)}$, where $ 2n\times 2n$ is the size of our square board.

Let $ f(n),n\in\mathbb{N}^*$ denotes the greatest value of the sum of those differences mentioned for the $ n\times n$ board. We just consider even $ n$. Since for every $ n\in\mathbb{N}^*$, there are exactly $ (4n^2)!$ ways to arrange $ 4n^2$ positive integer $ 1,2,...,4n^2$ into an $ 2n\times 2n$, which is a limited number, there exists $ f(2n)$.

Now, consider $ i,j\in{1,2,3,...,n}$ suppose that $ i$ and $ j$ are in the same column or row, since $ \min{(i)}$ must be the smallest number in that row(or column) and $ \max{(j)}$ must not be the greatest one, if we rearrange these numbers in another way such that $ i,j$ are in two different row(column), we can obtain a greater sum. Therefore, $ n$ numbers $ i\in{1,2,3,...,n}$ must be arranged on a diagonal of the board.

Similarly, $ n$ numbers $ i\in{n^2,n^2-1,...,n^2-n+1}$ must be on the other diagonal.

Finally, the left numbers is arranged freely.

With that way of arranging, the greatest sum is expressed by:
$ f(n)=\sum_{i=1}^{n}(n^2-(2k-1))+n(n^2-1)=2n^2(n-1)$.

At $ n=2008$, we got the desired result, the same as joh's result :lol: .


trigonometry
analytic geometry
geometry
circumcircle
HMMT
exterior angle
trig identities

Triangle $ ABC$ has $ \angle C \equal{} 60^{\circ}$ and $ BC \equal{} 4$.  Point $ D$ is the midpoint of $ BC$.  What is the largest possible value of $ \tan{\angle BAD}$?

$ \textbf{(A)} \ \frac {\sqrt {3}}{6} \qquad \textbf{(B)} \ \frac {\sqrt {3}}{3} \qquad \textbf{(C)} \ \frac {\sqrt {3}}{2\sqrt {2}} \qquad \textbf{(D)} \ \frac {\sqrt {3}}{4\sqrt {2} \minus{} 3} \qquad \textbf{(E)}\ 1$

[hide="Solution"]
Let the length of AC be x, the measure of BAD be j, and the measure of CAD be k.

Project points D and B onto AC; their distances from C are 1 and 2, respectively.

We then have that: $ \tan k \equal{} \frac{\sqrt{3}}{x\minus{}1}$ and $ \tan (j\plus{}k) \minus{} \frac{2\sqrt{3}}{x\minus{}2}$.

Now, let $ \tan k \equal{} y$. By the tangent addition formula:

$ \frac{2\sqrt{3}}{x\minus{}2} \equal{} \frac{\frac{\sqrt{3}}{x\minus{}1} \plus{} y}{1\minus{}\left( \frac{\sqrt{3}}{x\minus{}1} \right) y}$

Multiplying through by the denominator, and then by $ (x\minus{}1)(x\minus{}2)$ and simplifying gives:

$ y \equal{} \frac{\sqrt{3} x}{x^2 \minus{} 3x \plus{} 8} \equal{} \frac{\sqrt{3}}{\left( x \plus{} \frac{8}{x} \right) \minus{} 3}$

By AM-GM:

$ y \leq \frac{\sqrt{3}}{2 \sqrt{x \cdot \frac{8}{x}} \minus{} 3} \equal{} \frac{\sqrt{3}}{4 \sqrt{2} \minus{} 3}$
[/hide]

Admittedly, the coordinate solution is simple. However, the geometric argument has some nice reasonings.

[hide]Let $ A'$ be on ray $ AB$ so that the circumcircle of $ \triangle A'BM$ is tangent to $ AC$ (assume for sake of contradiction that $ A$ and $ A'$ are distinct.)If $ A'$ is between $ A$ and $ C$, then let $ AM$ intersect the circumcircle again at $ X$. Then $ \angle MA'B \equal{} \angle MXB$. By the exterior angle theorem, $ \angle MXB\ge \angle XAB \equal{} \angle MAB$. Hence $ \tan \angle MA'B\le\angle \tan \angle MAB$. We can similarly apply this argument when $ A'$ is beyond $ A$ on ray $ AC$. Hence $ A'$ is our desired point.

Hence we may use the triangle similarity produced by this tangent circle to simplify these relationships. In particular, we have
\[ \triangle AMC\sim\triangle BAC\implies \frac {AM}{BA} \equal{} \frac {MC}{AC} \equal{} \frac {CA}{CB}
\]
Suppose that $ AM \equal{} x$. Then
\[ \frac {x}{BA} \equal{} \frac {2}{AC} \equal{} \frac {CA}{4}
\]
Hence $ AC \equal{} 2\sqrt {2}$ and $ BA \equal{} x\sqrt {2}$. Let $ \theta \equal{} \angle MAB$. By the law of cosines,
\[ \cos \theta \equal{} \frac {3x^2 \minus{} 4}{2\sqrt {2}\cdot x^2}\qquad x^2 \equal{} 4(3 \minus{} \sqrt {2})
\]
\[ \cos\theta \equal{} \frac {8 \minus{} 3\sqrt {2}}{(2\sqrt {2})(3 \minus{} \sqrt {2})}
\]
Then we apply the identity $ \tan^2\theta \equal{} \minus{} 1 \plus{} \cos^{ \minus{} 2}\theta$. Now it is clear that the numerator is going to retain that $ 4\sqrt {2} \minus{} 3$ factor so we can skip the computation and just choose the answer $ \boxed{D}$. . It is easy to verify by working backwards that D is the answer.[/hide]

There was a similar problem on the HMMT Guts round last year, involving the same first two steps as Altheman (but less computation). Sadly, I saw that you had to draw the tangent circle, but messed up some other part of my diagram and got that <BAD < 30.

[quote="The Zuton Force"][hide="Solution"]
Let the length of AC be x, the measure of BAD be j, and the measure of CAD be k.

Project points D and B onto AC; their distances from C are 1 and 2, respectively.

We then have that: $ \tan k \equal{} \frac {\sqrt {3}}{x \minus{} 1}$ and $ \tan (j \plus{} k) \minus{} \frac {2\sqrt {3}}{x \minus{} 2}$.
[/hide][/quote]
How'd you get that? I don't really know anything about projections.
Edit: Oh, never mind. Once you draw in $ DD'$ and $ BB'$ they're 30-60-90 right triangles.

Hmmm I don't know why I didn't see this during the test (I had like 35 minutes!) but here is some geometry:

[hide]
We know from geometry that the maximum is attained when the circumcircle $ \omega$ of BAD is tangent to CA. Let X be the point on $ \omega$ such that BX || CA. Let Y be the projection of D onto XB, and let Z be the projection of B onto AC. It is now possible to calculate CA (from CA*CA = CD*CB) and CZ, so we know AZ, so we know XB (just twice AZ). We also know BY and YD. Thus, we know YD/YX, which is the desired tangent.
[/hide]

Illustration of why the circle is tangent at A.  [hide]For any choice of $ A$ s.t. $ ABC$ is a triangle as described, a circle can be constructed passing through points $ A$, $ B$, and $ D$.  Then $ \angle{BAD} \equal{} \frac {1}{2}\angle{BOD}$.  Since $ \angle{BOD} < 180^\circ$, $ \angle{BAD} < 90^\circ$.  Therefore $ \tan{\angle{BAD}}$ is maximized when $ \angle{BAD}$ is maximized, which occurs when $ \angle{BOD}$ is maximized.  $ \angle{BOD}$ is maximized when point $ O$ is as close to line segment $ BD$ as possible, which occurs when the circle is of minimum possible radius.  This, in turn, occurs when line $ AC$ is tangent to the circle.[/hide]

Sorry, I don't understand how we applied AM-GM here...

[hide=solution]
[asy]unitsize(12mm); pair C=(0,0), B=(4 * dir(60)), A = (8,0), D=(2 * dir(60)); pair E=(1,0), F=(2,0); draw(C--B--A--C); draw(A--D);draw(D--E);draw(B--F); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F); label("\(C\)",C,SW); label("\(B\)",B,N); label("\(A\)",A,SE); label("\(D\)",D,NW); label("\(E\)",E,S); label("\(F\)",F,S); label("\(60^\circ\)",C+(.1,.1),ENE); label("\(2\)",1*dir(60),NW); label("\(2\)",3*dir(60),NW); label("\(\theta\)",(7,.4)); label("\(1\)",(.5,0),S); label("\(1\)",(1.5,0),S); label("\(x-2\)",(5,0),S);[/asy]
Let $x = CA$. Then $\tan\theta = \tan(\angle BAF - \angle DAE)$, and since $\tan\angle BAF = \frac{2\sqrt{3}}{x-2}$ and $\tan\angle DAE = \frac{\sqrt{3}}{x-1}$, we have

\[\tan\theta = \frac{\frac{2\sqrt{3}}{x-2} - \frac{\sqrt{3}}{x-1}}{1 + \frac{2\sqrt{3}}{x-2}\cdot\frac{\sqrt{3}}{x-1}}= \frac{x\sqrt{3}}{x^2-3x+8}\]
With calculus, taking the derivative and setting equal to zero will give the maximum value of $\tan \theta$. Otherwise, we can apply AM-GM:

\begin{align*} \frac{x^2 - 3x + 8}{x} = \left(x + \frac{8}{x}\right) -3 &\geq 2\sqrt{x \cdot \frac 8x} - 3 = 4\sqrt{2} - 3\\ \frac{x}{x^2 - 3x + 8} &\leq \frac{1}{4\sqrt{2}-3}\\ \frac{x\sqrt{3}}{x^2 - 3x + 8} = \tan \theta &\leq \frac{\sqrt{3}}{4\sqrt{2}-3}\end{align*}
Thus, the maximum is at  $\frac{\sqrt{3}}{4\sqrt{2}-3} \Rightarrow \mathbf{(D)}$.

[hide=alternate sketch]
Equivalently, we seek the value of $\tan \angle BAD$ when $\angle BAD$ is maximized. (Clearly it can't exceed $60^{\circ}$, so we're fine.) Let the circumcenter of $\triangle ABC$ be $O$. Note that the locus of $D$ as $C$ moves is the circle with diameter $OB$. Call this circle $\omega$. Refer to the attached diagram. Clearly $AD$ must be tangent to $\omega$ in order to maximize $\angle BAD$. The rest of the problem is now purely computational with the use of Power of a Point and the tangent sum formula.
[/hide]

I still don't understand how to do this problem.

[hide=solution]I set $x$ to be the distance from $A$ to the base of the altitude from $B$. Then, I got $\tan(k)=\frac{\sqrt{3}}{x+1}$ and $\tan(j+k)=\frac{2\sqrt{3}}{x}$ I ended up getting $\tan(\theta)=\frac{\sqrt{3}}{x+3}$ with an answer of $\frac{\sqrt{3}}{3}$ by setting $x=0$. Just wondering where I made a mistake in this solution. Thanks.

[b]Solution:[/b] Dropping altitudes from $B$ and $C$ at $E$ and $F$ respectively, we can see that, if we let $AC = x$, then $AF = x-2$. Using some easily computed values, we have $\tan\angle BAF = \frac{2\sqrt{3}}{x-2}$ and $\tan\angle DAE = \frac{\sqrt{3}}{x-1}$. Using the $\tan$ subtraction formula and optimizing, we get our answer of $\boxed{\bf{\text{(D)}}}$.

[i]Comments:[/i] Dropping altitudes make finding the value of the angle much easier.


What does it mean if I have not received any e mail?

We emailed everyone whose application was received on time.  So it could mean either of two things:

1) You applied on time and for some reason the email didn't go through to you.

2) You applied late and we haven't looked at your application yet.

If you think you might be in the first situation, please let us know ASAP!


If $p$ and $q$ are odd integers, prove that the equation
\[ x^2 + 2px + 2q = 0 \]
has no rational roots.

(From Eotvos competition 1907)

[hide]

If a quadratic has rational roots, its discriminant $b^2-4ac$ must be a perfect square.

The discriminant for the equation is $4(p^2-2q)$.
Since $4$ is a perfect square, in order for the above expression to be a perfect square $p^2-2q$ must be a perfect square.

Look at $p^2-2q$ in $\mod 4$.
Since $p$ is odd, $p^2\equiv 1 \mod4$.
Since $q$ is odd, $2q\equiv 2\mod4$.
But then $p^2-2q\equiv 3 \mod4$, but the only perfect square $\mod 4$ are $0, 1$. Thus, $p^2-2q$ can't be a perfect square, so the quadratic equation has no rational roots.
[/hide]


Hi, this is from my BC calc exam review for first semester.
A gym will admit 5000 members at maximum. The cost for 100 members to join is \$200 per year. For every 20 additional members who join, the cost decreases \$5 per year. How many members should join to maximize the gym's income?
Thank you

Unless I'm reading this wrong, there is a problem: after 800 people join, the cost is $ \$0$, which doesn't make any sense.  So after 900 members, the gym isn't earning money anymore.  Tell me if I'm misinterpreting something.

[quote="JRav"]Unless I'm reading this wrong, there is a problem: after 800 people join, the cost is $\$0$, which doesn't make any sense.  So after 900 members, the gym isn't earning money anymore.  Tell me if I'm misinterpreting something.[/quote]

but won't the gym make an extra $1600 after 800 people join ??

What is the source of this problem?  I know you said from a BC calc exam review, but from where in particular?  A question your teacher made up, an old exam?

i was told by my teacher that this exam was made up by a very experienced teacher in the county...


[hide]Adding up all the equation, we get $6(x_1+x_2+x_3+x_4+x_5)=186$, or $x_1+x_2+x_3+x_4+x_5=31$.

Then the rest is easy- just compare the original equations with $x_1+x_2+x_3+x_4+x_5=31$.

$x_4=17$ and $x_5=65$, so $3x_4+2x_5=3(17)+2(65)=\boxed{181}$
[/hide]

[quote=kante314][quote=pi271828][quote=OlympusHero][hide=Solution]Adding gives $6(x_1+x_2+x_3+x_4+x_5)=6 \cdot 31$, so $x_1+x_2+x_3+x_4+x_5=31$. This means $x_4=17, x_5=65$, so the answer is $51+130 = \boxed{181}$.[/hide][/quote]

olympus how long did it take you to scroll through the "algebra" topics to get to this one

anyways, the sketch is add everything, solve for the sum then subtract[/quote]

this is an aime problem :P[/quote]

w h a t

this seems so trivial...

[quote=pi271828][quote=kante314]

this is an aime problem :P[/quote]

w h a t

this seems so trivial...[/quote]

i think it is a verrrrrrry old aime lol

[hide=Solution]
Summing and dividing by $6$,
$$x_1+x_2+x_3+x_4+x_5=31$$
Therefore, $$x_4=17 \quad \text{and} \quad x_5=65$$
Thus, $$3x_4+2x_5=\boxed{181}$$

don't think i've seen an aime problem this easy

[hide=Solution]Add the equations. We have $$6(x_1+x_2+x_3+x_4+x_5)=6(1+2+4+8+16) \implies x_1+x_2+x_3+x_4+x_5=31$$. Subtracting this equation from equations 4 and 5 respectively, we have $x_4=17$ and $x_5=65$, meaning that the desired answer extraction is $3 \cdot 17+2 \cdot 65=\boxed{181}$.[/hide]

If we add up all the equations, $6(x_1+x_2+.....+x_6)=186$, so $x_1+x_2+.....+x_6=31$, now comparing equations, $x_4=17$, and $x_5=65$, so we have $\boxed{181}.$


Express $ \dfrac1{0.0001}$ as a power of 10.

Well, $ \dfrac1{0.0001}$ is $ 10,000$ which can be expressed as $ 10^4$ since a number that starts with a one and has ONLY zeros following it (let's just say it's $ n$ zeros) means that it can be expressed as $ 10^n$. In this case "$ n$" is 4, which means the answer is $ 10^4$.


[quote="Interval"]In a survey of 400 teenage shoppers at a large mall, 240 said they
shopped at Abernathy’s, 210 said they shopped at Bongo Republic,
and 90 said they shopped at both stores. How many of the teenage
shoppers surveyed did not shop at either store?[/quote]
[hide="PIE"]240+210-90+x=400. x=40.[/hide]


You did the arithmetic right, but check your final answer again, A+MATH.

yup...A+MATH, there was a little decision making problem....

it is the best to chop off 4 legs..

little bloody...

is starfish good?( I'm talking about it as food )

never tried it in my life

[quote="Treething"]Albatross thinks starfish legs are a delicacy. So, he decides to farm starfishes, chop off their legs, cook them, and sell them.

If you cut off one starfish leg, it takes two hours to grow back.
If you cut off two starfish legs, it takes 225 minutes to grow back.
If you cut off three starfish legs, it takes one-third of a day to grow back.
If you cut off four starfish legs, it takes 410 minutes to grow back.
If you cut off all five starfish legs, it takes ten hours to grow back.

To maximize productivity, how many legs should Albatross chop off of each starfish?
How about to minimize productivity?

No starfish were harmed in the making of this problem, as starfish legs grow back ;) .[/quote]

[hide="solution"]it takes the least amount of time to grow back if you cut 4 legs off[/hide]

[hide="If a competitor named Quorl decided to use immoral business tactics"]He would offer to work for free for Albatross, then cut off three legs.[/hide]


calculus
integration
number theory unsolved
number theory

Find all pairs of positive integers $ (m,n)$ such that $ \sqrt{m}\plus{}\frac{2005}{\sqrt{n}}\equal{}2006.$

You're missing $ (m,n) \equal{} (1605^2,25); (2001^2,401^2); (2005^2,2005^2)$.

[hide]$ \sqrt {m} \plus{} \frac {2005}{\sqrt {n}} \equal{} 2006$

$ \frac {2005}{\sqrt {n}} \equal{} 2006 \minus{} \sqrt {m}$

$ \dfrac{2005^2}{n} \equal{} 2006^2 \plus{} m \minus{} 4012\sqrt {m}$

Since the left side is rational, the right side must also be rational. Thus, $ \sqrt {m}$ must be rational.
Since $ m$ is an integer, this means that $ m$ must be a integral perfect square. So let $ m \equal{} p^2$ for some positive integer $ p$.

Plugging this into the original equation yields: $ p \plus{} \frac {2005}{\sqrt {n}} \equal{} 2006$.

Since $ p$ and $ 2006$ are integers, $ \frac {2005}{\sqrt {n}}$ must also be an integer. Thus, $ n$ must also be an integer perfect square. So let $ n \equal{} q^2$ for some positive integer $ p$.

This yields the equation: $ p \plus{} \dfrac{2005}{q} \equal{} 2006$.

Since $ \dfrac{2005}{q}$ must be an integer, $ q \mid 2005 \equal{} 5 \cdot 401$.

Thus, $ q \equal{} 1, 5, 401, 2005$. These yield $ p \equal{} 1, 1605, 2001, 2005$ respectively.

Therefore, the solutions are: $ (m,n) \equal{} (1,1); (1605^2,25); (2001^2,401^2); (2005^2,2005^2)$. [/hide]


algorithm
linear algebra
matrix
linear algebra theorems

Greeting to all,

I am looking for an algorithm (possibly recursive) to compute determinant of nth order matrix.
Can someone please suggest.

Thanks in advance,

Regards,

-Zapp

The simplest reasonably fast method: Use Gaussian elimination to row reduce to triangular form. Most row operations leave the determinant unchanged; row interchanges negate the determinant.
The determinant of an upper tringular matrix is the product of the diagonal entries.


calculus
integration
trigonometry
real analysis
real analysis unsolved

find K so that

$\int^{x}_{a}f(t)dt+K = \int^{x}_{b}f(t)dt$

$f(x)= sin^{2}x; a=0, b=2$


geometric transformation
geometry
homothety
projective geometry
geometry proposed

Let $ (O_a)$ be the circle tangent to $ AB,AC$ and denote similarly $(O_b),(O_c).$ Then, $ A_1 \equiv (O_b) \cap (O_c),$ $ B_1 \equiv (O_a) \cap (O_c)$ and $C_1 \equiv (O_a) \cap (O_b).$ $ B_1$ and $ C_1$ are the insimilicenters of $ (O_a) \sim (O_c)$ and $ (O_a) \sim(O_b).$ Then by Monge-d'Alembert theorem, it follows that $ A'' \equiv BC \cap B_1C_1$ is the exsimilicenter of $ (O_b) \sim (O_c).$ Analogously, intersections $ B'' \equiv A_1C_1 \cap AC$ and $ C'' \equiv A_1B_1 \cap AB$ are the exsimilicenters of $ (O_a) \sim (O_c)$ and $ (O_a) \sim (O_b)$ $ \Longrightarrow$ $ A'',B'',C''$ are collinear on the homothety axis of $ (O_a),(O_b),(O_c).$ Now, based on Desargues theorem, the triangles $ \triangle ABC$ and $ \triangle A_1B_1C_1$ are perspective through $A''B''C''$ $\Longrightarrow$  $ AA_1,BB_1,CC_1$ concur.

thanks.my solution is exactly like you and the theory that i writed in title is Deasarge.but i want another solution without this theory.is there anybody that can help me?

[quote="MJ GEO"]3 circles are tangent to each other and the sides of $ ABC$.if  the tangency of circles with each other named $ A_1,B_1,C_1$ prove that $ AA_1,BB_1,CC_1$ concur at one point.[/quote]
Another interesting result:

[color=Darkblue]Given $ \triangle ABC$ and three circles $ (O_1)$, $ (O_2)$, $ (O_3)$ externally tangents to each other pairwise and $ (O_1)$ tangents to $ AB$ and $ AC$, similarly for $ (O_2)$, $ (O_3)$. Let $ A_1$ is the tangency point of $ (O_2)$ and $ (O_3)$, $ I_a$ is the $ A\minus{}$ excircle center. Similarly for $ I_b,$ $ I_c$ and $ B_1$, $ C_1$. Then $ I_aA_1,$ $ I_bB_1,$ $ I_cC_1$ are concurrent.[/color]

anybody? i am sure you can find another solution  :wink:


inequalities
Cauchy Inequality
inequalities proposed

I always impressed with your solution, but I can't solve.What on earth how would you make inequalities problems,
Sir Vascile Cirtoaje.

kunny

I belief that my solution in the following thread will  also work for your problem, Vasc!
http://www.mathlinks.ro/Forum/viewtopic.php?t=62458
That's easy,right???

I think the inequality is also valid for $xyz=1$. :lol:

[quote="Vasc"]I also am impressed with Japan, kunny. Last week I visited Tokyo (4 days) and Osaka (2 days).
Very beautiful towns!
The Japoneses are very hard working. :o[/quote]

Oh! Really? I also would like to Europe. Certainly Japanese work very hard.

Did you enjoying trip in Japan?After about a month you can see cherry blossoms in bloom.

kunny

Indeed, your solution is correct. A stronger inequality follows under the hypothesis $x^2y^2+y^2z^2+z^2x^2=3$.
In this case the problem reduces to
$\sum\frac1{1+b+c} \leq 1$  for $ab+bc+ca=3$.
The proof is not difficult.

[quote="Vasc"]Let $x,y,z$ be non-negative numbers such that $xy+yz+zx=3$. Then,

$\frac1{1+x^2+y^2}+\frac1{1+y^2+z^2}+\frac1{1+z^2+x^2}\leq 1$.[/quote]

This is basically hungkhtn's approach from a similar problem:

We show the equivalent inequality
$\frac{a^2+b^2}{1+a^2+b^2}+\frac{b^2+c^2}{1+b^2+c^2}+\frac{c^2+a^2}{1+c^2+a^2}\ge 2$
Observe that
$\frac{a^2+b^2}{1+a^2+b^2}-\frac{(a+b)^2}{2+(a+b)^2}=\frac{(a-b)^2}{(1+a^2+b^2)(2+(a+b)^2)}$
and by Cauchy inequality
$\sum_{cycl}\frac{(a+b)^2}{2+(a+b)^2} \ge \frac{4(a+b+c)^2}{6+\sum_{cycl}(a+b)^2}= \frac{4(a+b+c)^2}{2\sum_{cycl} ab+\sum_{cycl}(a+b)^2}=2 .$

[quote="Vasc"]A possible generalization is the following:
Let $x,y,z$ be non-negative numbers such that $xy+yz+zx=3$. If $r>0$, then
$\frac1{r+x^2+y^2}+\frac1{r+y^2+z^2}+\frac1{r+z^2+x^2}\leq max(\frac1{r},\frac{3}{r+2})$.[/quote]

Vasc, you are so great. Let me typset it again
Let $x,y,z$ be non-negative numbers such that $xy+yz+zx=3$. If $r>0$, then
\[ \frac1{r+x^2+y^2}+\frac1{r+y^2+z^2}+\frac1{r+z^2+x^2}\leq \max\Bigl\{\frac1{r},\frac{3}{r+2}\Bigl\}. \]

[quote=luofangxiang]$\sum\frac1{1+b+c} \leq 1$  for $ab+bc+ca=3$.
[img]http://s14.sinaimg.cn/middle/0018zOAxgy71HHP6Zlz0d&690[/img][/quote]

My proof
$\left( {\prod {\left( {1 + a + b} \right)} } \right)\left( {1 - \sum {\frac{1}{{1 + b + c}}} } \right)$$ = \left( {1 + b + c} \right)\left( {1 + c + a} \right) - \sum {\left( {1 + a + b} \right)\left( {1 + b + c} \right)} $
$ = 1 + \sum {\left( {a + b} \right)}  + \sum {\left( {a + b} \right)\left( {b + c} \right)}  + \prod {\left( {a + b} \right)}  - 3 - 2\sum {\left( {a + b} \right)}  - \sum {\left( {a + b} \right)\left( {b + c} \right)} $
$ =  - 2 - 2\sum a  + \left( {\sum a } \right)\left( {\sum {ab} } \right) - abc = \sum a  - abc - 2 \ge 3 - 1 - 2 = 0$

[url]http://blog.sina.com.cn/s/blog_547541ce0102wg47.html[/url]

[quote=scpajmb]
The proof is wrong!
[/quote]
I don't agree with you! I think, the luofangxiang's proof is true and very nice.


induction
superior algebra
superior algebra unsolved

Given a field $F$ and a finite extension $E$ of $F$ such that the number of fields between $E$ and $F$ is finite, show that $E$ is simple.

When $E$ is a finite field, the assertion reduces to showing that a finite extension of a finite field is simple. This, however, follows immediately from the fact that the multiplicative group $F^*$ of $F$ is cyclic, so $F=E(g)$ for every generator $g$ of this group.

Assume now that $E$ is an infinite field. We can prove the proposition by induction on $n=[F: E]$:

For $n=1$ there's nothing to prove. Suppose now that we've shown the sentence to be true for values smaller than $n$ of the dimension $[F: E]$. Let $G$ be a maximal intermediate field lying between $E$ and $F$ but different from both (if no such field exists, then $F=E(\theta)$ for any $\theta\in F\setminus E$, and we're done). By the induction hypothesis, $G=E(\alpha)$ for some $\alpha\in G\setminus E$. Also, since $G$ is maximal, for any $\beta\in F\setminus G$ we have $F=G(\beta)$, which means that $F=E(\alpha,\beta)$. The aim will be to prove that $F=E(\alpha+t\beta)$ for some $t\in E$.

Look at all intermediate fields of the form $E(\alpha+t\beta)$ for $t\in E$. Since there are infinitely many $t$ but only finitely many fields, some two of them coincide: $E(\alpha+t\beta)=E(\alpha+s\beta)$ for $s\ne t\in E$. This means that $\alpha+s\beta\in E(\alpha+t\beta)$, and since $\alpha+t\beta\in E(\alpha+t\beta)$, we find that $\beta=\frac{(\alpha+t\beta)-(\alpha+s\beta)}{t-s}\in E(\alpha+t\beta)$. We now have what we wanted: both $\alpha$ and $\beta$ are contained in $E(\alpha+t\beta)$, so $F=E(\alpha,\beta)=E(\alpha+t\beta)$.


Let $ a,b,c,d,k$ be positive real numbers. Prove that the following inequality holds
$ (\frac{a}{b})^k\cdot\frac{a-b}{b+c}+(\frac{b}{c})^k\cdot\frac{b-c}{c+d}+(\frac{c}{d})^k\cdot\frac{c-d}{d+a}+(\frac{d}{a})^k\cdot\frac{d-a}{a+b}\geq0$

$ {\frac {a \left( a\minus{}b \right) }{b \left( b\plus{}c \right) ^{t}}}\plus{}{\frac {b
 \left( b\minus{}c \right) }{c \left( c\plus{}d \right) ^{t}}}\plus{}{\frac {c \left( c\minus{}d
 \right) }{d \left( d\plus{}a \right) ^{t}}}\plus{}{\frac {d \left( d\minus{}a \right) }{
a \left( a\plus{}b \right) ^{t}}}$
let t=-1,is hold,Learning your.

$ {\frac {a \left( a\minus{}b \right)  \left( b\plus{}c \right) }{b}}\plus{}{\frac {b
 \left( b\minus{}c \right)  \left( c\plus{}d \right) }{c}}\plus{}{\frac {c \left( c\minus{}d
 \right)  \left( d\plus{}a \right) }{d}}\plus{}{\frac {d \left( d\minus{}a \right)
 \left( a\plus{}b \right) }{a}}
\geq0$
BQ

Dear [b]xzlbq[/b].
it is given that  $ k$  is positive.
why you get that  $ t\equal{}\minus{}1$???
I don`t undestand... :?:

Let $ a,b,c,d$ be positive real numbers. Prove that the following inequality holds
$ {\frac {a \left( a \minus{} b \right) \left( b \plus{} c \right) }{b}} \plus{} {\frac {b \left( b \minus{} c \right) \left( c \plus{} d \right) }{c}} \plus{} {\frac {c \left( c \minus{} d \right) \left( d \plus{} a \right) }{d}} \plus{} {\frac {d \left( d \minus{} a \right) \left( a \plus{} b \right) }{a}} \geq0$
BQ

[quote="marin.bancos"]Dear [b]xzlbq[/b],
What do you mean? There is no relation between my inequality and your remarks. Do you have a solution to my problem? Please, post it.
For new inequalities you have the possibility to post new topics![/quote]
[color=cyan][size=200]Thank you for the reminder。[/size][/color]
BQ

Just keep it simple,  I may use some notations. Hopefully, it does not really irritate some people.

The original inequality can be expressed as

\[ \sum^{a,b,c,d}_{cyc}\frac{(a\minus{}b)(a^k\minus{}b^k)}{b\plus{}c}\plus{}\sum^{a,b,c,d}_{cyc}\frac{a\minus{}b}{b\plus{}c} \ge 0\]

Hence, it is enough to check that

\[ \sum^{a,b,c,d}_{cyc}\frac{a\minus{}b}{b\plus{}c} \ge 0\]

The above inequality is equivalent to

\[ \frac{a\plus{}c}{b\plus{}c}\plus{}\frac{b\plus{}d}{c\plus{}d}\plus{}\frac{a\plus{}c}{d\plus{}a}\plus{}\frac{d\plus{}b}{a\plus{}b} \ge 4\]

Now, using the inequality  $ \frac{1}{x} \plus{} \frac{1}{y} \ge \frac{4}{x\plus{}y}$ appropriately. We get

\[ \left\{ \begin{array}{cc}
\frac{a\plus{}c}{b\plus{}c}\plus{}\frac{a\plus{}c}{d\plus{}a} \ge 4\frac{a\plus{}c}{a\plus{}b\plus{}c\plus{}d} \\ \\
\frac{b\plus{}d}{c\plus{}d}\plus{}\frac{b\plus{}d}{a\plus{}b} \ge 4\frac{b\plus{}d}{a\plus{}b\plus{}c\plus{}d} \end{array}
\right.\]

We finish our proof here.

[b]Dear great math[/b],
Can you explain why the original inequality is equivalent to
$ \sum^{a,b,c,d}_{cyc}\frac{(a\minus{}b)(a^{k}\minus{}b^{k})}{b\plus{}c}\plus{}\sum^{a,b,c,d}_{cyc}\frac{a\minus{}b}{b\plus{}c}\geq 0\ \  ?$
Thanks!

[quote="marin.bancos"]Let $ a,b,c,d,k$ be positive real numbers. Prove that the following inequality holds
$ (\frac {a}{b})^k\cdot\frac {a - b}{b + c} + (\frac {b}{c})^k\cdot\frac {b - c}{c + d} + (\frac {c}{d})^k\cdot\frac {c - d}{d + a} + (\frac {d}{a})^k\cdot\frac {d - a}{a + b}\geq0$[/quote]
As a little remark, we see that
$ \frac{(a^k-b^k)(a-b)}{b^k(b+c)} \ge 0.$
Therefore,
$ \frac{a^k}{b^k}\cdot \frac{a-b}{b+c} \ge \frac{b^k}{b^k}\cdot \frac{a-b}{b+c} =\frac{a-b}{b+c}.$
According to this result, it suffices to prove that
$ \frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b} \ge 0.$
Since $ \frac{a-b}{b+c}=\frac{a+c}{b+c}-1,$ the inequality is equivalent to
$ \frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{d+a}+\frac{d+b}{a+b} \ge 4.$
By the Cauchy-Schwarz Inequality, we have
$ \frac{a+c}{b+c}+\frac{c+a}{d+a} \ge \frac{4(a+c)}{b+c+d+a}.$
Similarly,
$ \frac{b+d}{c+d}+\frac{d+b}{a+b} \ge \frac{4(b+d)}{c+d+a+b}.$
By adding these two inequalities, the conclusion follows.

Thank you [b]can_hang2009[/b]. Thank you [b]great math[/b]. Your solutions to this problem are great.

[u]My (detailed) solution to the last inequality[/u]

$ (\forall)a,b,k > 0,\ \ (\frac {a}{b})^k\cdot(a - b)\geq a - b,\ \Leftrightarrow\ (\frac {a}{b})^k\cdot \frac {a - b}{b + c}\geq \frac {a - b}{b + c}$

Taking into account this kind of inequality, we get:
$ (\frac {a}{b})^{k}\cdot\frac {a - b}{b + c} + (\frac {b}{c})^{k}\cdot\frac {b - c}{c + d} + (\frac {c}{d})^{k}\cdot\frac {c - d}{d + a} + (\frac {d}{a})^{k}\cdot\frac {d - a}{a + b}\ \geq\ \frac {a - b}{b + c} + \frac {b - c}{c + d} + \frac {c - d}{d + a} + \frac {d - a}{a + b}\geq 0.$

My different approach for this inequality (Vasc's problem) is the following:
$ \frac {a - b}{b + c} + \frac {b - c}{c + d} + \frac {c - d}{d + a} + \frac {d - a}{a + b}\geq0 \ \ \Longleftrightarrow\ \ (\frac {a - b}{b + c} + 1) + (\frac {b - c}{c + d} + 1) + (\frac {c - d}{d + a} + 1) + (\frac {d - a}{a + b} + 1)\geq4\ \ \Longleftrightarrow$
$ \Longleftrightarrow\ \frac {a + c}{b + c} + \frac {b + d}{c + d} + \frac {c + a}{d + a} + \frac {d + b}{a + b}\geq4$

For this inequality:
$  LHS = \frac {(a + c)^2}{(a + c)(b + c)} + \frac {(b + d)^2}{(b + d)(c + d)} + \frac {(c + a)^2}{(c + a)(d + a)} + \frac {(d + b)^2}{(d + b)(a + b)}\stackrel{Cauchy - Schwarz/Titu's\ lemma}{\geq}$
$ \geq\frac {[(a + c) + (b + d) + (c + a) + (d + b)]^2}{(a + c)(b + c) + (b + d)(c + d) + (a + c)(d + a) + (b + d)(a + b)} = $
$  = \frac {[2(a + b + c + d)]^2}{(a + c)(a + b + c + d) + (b + d)(a + b + c + d)} = \frac {4(a + b + c + d)^2}{(a + b + c + d)(a + b + c + d)} = \frac {4(a + b + c + d)^2}{(a + b + c + d)^2} = 4 = RHS$

In fact, all our solutions have the same core!


modular arithmetic
calculus
integration
advanced fields
advanced fields unsolved

[hide="#1"]Prove that $ x^2 \minus{} d y^2\equal{}\minus{}1$ has no solution if $ d \equiv 3 \pmod{4}.$

Here I'm citing a theorem on Pell's equation that says that:

[i]All positive solutions of $ x^2 \minus{} dy^2\equal{} \pm 1$ are to be found among $ x\equal{}h_n, y\equal{}k_n$, where $ h_n/k_n$ are the convergents of the expansion of $ \sqrt{d}.$ If $ r$ is the period of the expansion of $ \sqrt{d},$ and if $ r$ is even, then $ x^2 \minus{}dy^2\equal{}\minus{}1$ has no solution, and all positive solutions of $ x^2\minus{}dy^2\equal{}1$ are given by $ x\equal{}h_{nr\minus{}1}, y\equal{}k_{nr\minus{}1}$ for $ n\equal{}1,2,3,\cdots.$[/i]

So I think that I just need to prove that the period of the expansion of $ \sqrt{d}$ is even, where $ d \equiv 3 \pmod{4}.$ But how am I supposed to do this?
[/hide]

[hide="#2"]Prove that the sum of the first $ n$ natural numbers is a perfect square for infinitely many values of $ n.$

Here I think I need to relate this some how to Pell's equation. But how?
[/hide]

[hide="#3"]
Prove that $ n^2 \plus{} (n\plus{}1)^2$ is a perfect square for infinitely many values of n.

Here there is a hint given back in the book that says use the following two theorems:

Theorem 1. Let $ a,b,c$ be integers with not both $ a$ and $ b$ equal to 0, and let $ g\equal{}g.c.d.(a,b).$ If $ g	\nmid c$ then the equation\[ ax\plus{}by \equal{} c \]
has no solution in integers. If $ g\mid c$ then this equation has infinitely many solutions. If the pair $ (x_1, y_1)$ is one integral solution, then all others are of the form \[ x\equal{}x_1\plus{}\frac{kb}{g}, \quad y\equal{}y_1 \minus{} \frac{ka}{g}\]
where $ k$ is an integer.

Theorem 2. If r is the length of the period of the expansion of $ \sqrt{d},$ we have for $ n\geq 0,$\[ h_{nr\minus{}1}^{2} \minus{} d k_{nr\minus{}1}^{2} \equal{} (\minus{}1)^{nr} q_{nr} \equal{} (\minus{}1)^{nr}.\]

I don't know how these two theorems are related to this problem.
[/hide]

[hide="#4"]
Show that if $ d$ is divisible by a prime number $ p, p \equiv 3 \pmod{4},$ then the equation $ x^2 \minus{} dy^2 \equal{} \minus{}1$ has no solution in integers.
[/hide]

For $ \#1$ just use modulus. $ d = - 1 mod4$ is equivalent to: $ x^2 + y^2 = - 1 mod4$ which is impossible.
For #$ 2$,    $ \sum_1^n i = \frac {(n)(n + 1)}{2}$ and $ \frac {n(n + 1)}{2} = k^2$ $ \iff$ $ 2k^2 = n^2 + n$ $ \iff$ $ 4n^2 + 4n + 1 - 8k^2 = 1$ $ \iff$ $ (2n + 1)^2 - 8k^2 = 1$ has an infinite number of solutions. But it is a Pell equation. (And the term $ 2n + 1$ if substituted by $ x$ would make no difference since $ x$ is necessarily odd.
#$ 3$ seems doable with the same trick I used for #$ 2$.
#$ 4$ the equation would be solvable iff there exist $ x,y$ such that: $ x^2+1=dy^2$ this means $ x^2+1=0 mod p$ $ p=-1 mod 4$ then it is easy to show $ p$ must divide $ x$ and $ 1$ contradiction.

The last comes from the lemma: if $ p|a^2+b^2$ and $ p=-1mod4$ then $ p|a$ and $ p|b$


Let $ p, q$ be two consecutive odd prime numbers. Prove that $ p\plus{}q$ is a product of at least $ 3$ natural numbers $ > 1$ (not necessarily dierent)

[hide]
Let $ p\plus{}q\equal{}2r$.

We know $ r\equal{}\frac{p\plus{}q}{2}$, so $ r$ must be between $ p$ and $ q$. Since $ r$ lies between two consecutive primes, $ r$ cannot be prime, so $ r$ has at least two prime factors. Thus we can express $ p\plus{}q$ as $ 2ab$ where $ a,b>1$, which is a product of three natural numbers greater than 1, as desired.[/hide]


function
real analysis
real analysis unsolved

Let $ f : Q \rightarrow R$ be a function. Prove that there exists a sequence of continuous functions $ f_n : Q \rightarrow R$ which converges to $ f$.

    This exercise is in the Baire Theory.

Did you mean poitwise convergence? If yes, then just take $ f_n (x): \mathbb{Q}\to \mathbb{R}$ continous function such that $ f_n (r_k)\equal{}f(r_k)$ for $ k\le n$ where $ r_n$ is $ n$-th rational number.