AMC
USA(J)MO
USAMO
symmetry
geometry
geometric transformation
reflection

Hello. I was recently trying out some USAMO problems for the first time. Because this is one of the first USAMO problems for which I have formulated a solution, and because it is different from those which I have previously seen, I think that my solution is wrong. Would you guys mind finding a flaw? Thanks.

Alice and Bob play a game on a 6 by 6 grid. On his or her turn, a player chooses a rational number not yet appearing in the grid and writes it in an empty square of the grid. Alice goes first and then the players alternate. When all squares have numbers written in them, in each row, the square with the greatest number in that row is colored black. Alice wins if she can then draw a line from the top of the grid to the bottom of the grid that stays in black squares, and Bob wins if she can't. (If two squares share a vertex, Alice can draw a line from one to the other that stays in those two squares.) Find, with proof, a winning strategy for one of the players.

[b]Solution Sketch[/b]

Denote the rows as $ r_1, r_2, r_3 \dots r_6$, and the columns as $ c_1, c_2, c_3 \dots c_6$.

Denote Bob's move $ A$ as follows. After Alice fills in box $ (c_i, r_j)$, let Bob fill in $ (c_i, r_{7\minus{}j})$ with a rational number such that if in row $ r_j$, $ (c_a, r_j) > \dots > (c_i, r_j) > \dots (c_b, r_j)$ then in row $ r_{7\minus{}j}$ similarly $ (c_a, r_{7\minus{}j}) > \dots > (c_i, r_{7\minus{}j}) > \dots (c_b, r_{7\minus{}j})$ (Note that $ c_i$ is also implied to be able to be the largest or smallest of the rational numbers already written). In other words, Bob performs a symmetric placement relative to the rational numbers in the row opposite that of Alice's move.

Let Bob perform $ A$ for all except for his last move. Because the black squares are now symmetric around the horizontal axis dividing the grid into two $ 6$ by $ 3$ grids, only full column of black squares will permit Alice to win.

Because Bob goes second and there are an even number of squares, he fills in the last square. Since he can make it less than or greater than what is already the greatest number in that row, he can determine the black square in that last row and deny Alice the ability to draw a line.

However, if and only if Bob's last move is in either $ r_1$ or $ r_6$ and chooses between a square which would complete a column and one of the two squares adjacent to it can Alice still draw a line. Note that if his move is in either $ r_1$ or $ r_6$, because Bob is acting on symmetry, the other was completed by Alice. Also, because of symmetry, they each must have completed $ 3$ rows. This implies that Bob completed another row prior to the last. Because in this row he chose between two options where the black square went he can make it impossible for Alice to draw a line.

(P.S. Sorry if some of my wording is sketchy). Thanks.

[quote="SnowEverywhere"]Let Bob perform $ A$ for all except for his last move. Because the black squares are now symmetric around the horizontal axis dividing the grid into two $ 6$ by $ 3$ grids, only full column of black squares will permit Alice to win. [/quote]
This is not necessarily true. Consider, for instance, if in the 1,2,5, and 6th rows, the 3rd column is filled in, while in the 3rd and 4th rows, the 4th column is filled.

Also, your last paragraph doesn't make much sense to me.

Edit: Wait, nevermind, I misread the problem. (didn't think the line had to be straight  :oops: )

Edit2: Yeah, I think your proof works. However, the problem statement is still kinda weird, because it's possible for Bob to win even if the line isn't straight, but disconnected. I think by "line", the problem probably meant anything without taking the pen off the paper, but it should have been more clear about that. I still don't see why you need the last paragraph though, with the line has to be straight interpretation.

What if the last two squares on the board are $ (2,2)$ and $ (2,3)$, and the black squares are $ (1,2), (3,3), (4,3), (5,3), (6,3)$ (where $ (r,c)$ represents the $ r$th row, $ c$th column)?

But since the one in the first row is in the 2nd column, because of symmetry wouldn't the one in the sixth row have to also be in the 2nd column (unless I am forgetting some case which varies from symmetry and allows these choices).

Oh, OK, you're right.  But I'm still not convinced that this works.  Your solution makes sense for the most part, but the last part is really sketchy.  You're giving Bob a strategy that he should follow until the last move.  Then you correctly acknowledge that there are certain cases where he cannot win on the last move.  Your method for Bob in these cases is, "make a different move earlier in the game."  But how is Bob supposed to know when he's supposed to deviate from the strategy earlier in the game?

Yeah, I agree that the official solution is much better. My answer to your question is probably even sketchier than my last paragraph. Because when Bob fills out the second last row that he will complete, it is the third last row to be completed (by symmetry), if there are three in a single column already, he should take the hint  :lol: .

Oh by the way, thanks guys for giving such in-depth analyses of my solution.

Good idea.  I think that works, but to finish writing your proof, you should organize your proof a lot better.  Basically, here's what your solution should look like:

1)  Describe Bob's basic strategy: Each time Alice makes a move, Bob makes his move in the square that is the reflection across the midline of the board from the square Alice made her move in.  Bob makes a corresponding move unless Alice just finished a row, in which case he has two options. (You would need to describe this much more clearly than I'm doing.)

2)  Prove this: Each time Alice finishes a row and determines where the black square in that row will be located, Bob can finish a corresponding row and choose where the black square in that row will be located.  One of his two options will be the square in the same column as the black square that Alice just determined.

3)  Describe what Bob should do when he finishes the fourth (his second) row.  ("Taking the hint")

4)  Describe what Bob should do on his last move.

5)  Prove that Bob wins by following this strategy.

I'd describe this as "consolidating" your proof.  You have the ideas that lead to a correct solution, but you need to put them together in a rigorous way.  I'm not demanding that you write out a complete proof or even that you post anything more here.  I'm just saying that you should get the five steps above at least straightened up in your mind before you consider yourself finished with this problem.  And actually, if you're just starting to solve USAMO-level problem, you may need to practice writing out solutions.  I know I had to practice this.

All of that is assuming the problem actually requires the line to be straight.  I tend to interpret this problem statement as not requiring the line to be straight, since Bob still wins in this case and it's a slightly harder problem.

Thanks for your help. Because I have only ever written the AMC 10 and the AIME, I really had no idea as to how it could be structured.


Was it a specific word that you said that would disappoint her, or could any word have disappointed her?

[quote="perfect628"]A woman told me about herself, and i said nothing. Then she told me more descriptively about her life, and I said nothing. Then she told me for hours about her life, and I said a single word, and she was very disappointed. Explain.[/quote]

Lingomaniac88, it was a specific word. Kstan013, no :roll:.

I know the answer too. If no one guesses it, I can post a hint.

[hide="Maybe?"]At the end, the lady asked, "Do you care at all about what I'm saying?" and you said, "No."[/hide]

[hide="guess"]maybe he said "huh?" so the woman was very pissed[/hide]

[hide]You didn't speak the same language, and thus did not understand her? I'm not sure how that could be expressed in one word, though. Unless she specifically asked whether you understood the language.[/hide]

you said [hide]oui[/hide] and she has no ability to speak french or what lingomaniac said or you said  [hide]whatever[/hide]
or [hide]bye[/hide]

No, the narrator and the woman speak the same language. The narrator cares about what the woman has to say--in fact, more than the average person might.

Yes or no questions are allowed.

Yeah. mathnerd, pm me your answer so I can make sure it's right. It sounds as though you do know.

hmm, is this some sort of interview?  or perhaps an interrogation?

did the narrator and the woman know each other before this?

[hide="She said"]Will you be able to help me out?(Because you happen to be either a psychiatrist or a lawyer)[/hide]
[hide="So you said"] NO :rotfl: [/hide]
[hide="Or...."] You are a judge and she asked you whether you believe what she said and  whether u believe she's innocent so you said no.
Or u r a judge and she asked you if you were listening to her so you said no!! :rotfl: [/hide]

[quote="perfect628"]A woman told me about herself, and i said nothing. Then she told me more descriptively about her life, and I said nothing. Then she told me for hours about her life, and I said a single word, and she was very disappointed. Explain.[/quote]

you might have said "it,s all greek to me." :!:
but it was not a single word .Right :?:
you said that her language was not clear to you.

Two questions

Is the answer no for some reason?

Are the man and women related?

Are you a policeman? Was the woman a suspect? Was the single word "guilty"?

I'm not a policeman, the word was "guilty", there was no relationship, and the woman was a suspect. You've basically got it. Just say who I am.

[hide="You are a..."]
judge

At first I was thinking interrogation, but this actually makes more sense.
[/hide]

[quote="laughinghead505"]Or maybe its a one person jury. :)[/quote]

Like the one we had at mock trial today :rotfl:

Seriously, good job everyone :clap:


I've been trying the HMMT's problems. And I'm stuck on 1998 HMMT Algebra Problem 4.Here's the problem: r and s are relatively prime integers such that $ \frac{r}{s}\equal{}\frac{2(\sqrt{2}\plus{}\sqrt{10})}{5(\sqrt{3\plus{}\sqrt{5}})}$. Find r and s.
Any hints? (ONLY HINTS)

Compute $ \left(\frac {r}{s}\right)^2$ or $ \sqrt{3\plus{}\sqrt{5}}\equal{}\sqrt{\frac{6\plus{}2\sqrt{5}}{2}}$

hmm..that's what i've been trying so far...I'll report back after an hour

Ok well here's my solution that I somehow messed up on.  :oops:

Btw I know you said hints but this is an INCORRECT solution.  Maybe I just made an arithmetic error though.  So don't click poincare.

[hide="someone want to point out what I did wrong please"]
First I got rid of the fractions

$ 5r(\sqrt {3 \plus{} \sqrt {5}}) \equal{} 2s(\sqrt {2} \plus{} \sqrt {10})$
Next I squared both sides

$ 25r^2(3 \plus{} \sqrt {5}) \equal{} 4s^2(12 \plus{} 4\sqrt {5})$

Then I expanded and factored.
$ 75r^2 \plus{} 25\sqrt {5}r^2 \equal{} 48s^2 \plus{} 16\sqrt {5}s^2$
$ 75r^2 \minus{} 48s^2 \plus{} 25\sqrt {5}r^2 \minus{} 16s^2\sqrt {5} \equal{} 0$
$ 3(25r^2 \minus{} 16s^2) \plus{} \sqrt {5}(25r^2 \plus{} 16s^2) \equal{} 0$
$ (25r^2 \minus{} 16s^2)(3 \plus{} \sqrt {5}) \equal{} 0$
$ (25r^2 \minus{} 16s^2) \equal{} 0$
$ (5r \plus{} 4s)(5r \minus{} 4s) \equal{} 0$

But then.. I got 5/4 which isn't an integer??
[/hide]

EDIT: OH WAIT OH WAIT I JUST REALIZED SOMETHING REALLY STUPID. >_< I get what I did wrong now that solution is now correct.

[hide="part I messed up on and real answer"]
Uh don't mind that 5/4 part.  r = 4 and s = 5.
[/hide]

[quote="sunehra"]There is actually a really quick solution given kunny's hint. If you haven't found it yet:

[hide="Hint #1"]
What is $ \sqrt {6 \plus{} 2\sqrt {5}}$ ? (Simplify it.)
[/hide]

[hide="Hint #2"]
Factor $ \sqrt {2} \plus{} \sqrt {10}.$
[/hide][/quote]

wow, that's a neat quick way!

actually sqrt (3+ sqrt 5) = (sqrt 2 + sqrt 10)*2.5 so 2/2.5 = 4/5 so r=4 and s=5


If $ a,b,c\in \mathbb{C}$ such that $ a\plus{}b\plus{}c\neq 0,\ a^2\plus{}b^2\plus{}c^2\equal{}0$ and $ |a|\equal{}|b|\equal{}|c|\equal{}1$, then $ |a^3\plus{}b^3\plus{}c^3|\leq 7$.


Is it possible to draw the angle symbol in geometer's sketchpad?

Yes. Draw three points and arc them. Then hide the pts.

EDIT: Here's a picture

[url=http://mathprob.com/_private/get.php/c6c5992d7d4c0d51827a56e3dff9f310/angle_geometry_sketchpad.gif][img]http://mathprob.com/_private/get.php/c6c5992d7d4c0d51827a56e3dff9f310/angle_geometry_sketchpad.gif[/img][/url]

What's Geometer's Sketchpad?
Is it a Geometry Sketchpad?

http://www.dynamicgeometry.com/

It's Geo software.

Is there a free version I can download and use legally?

No but there are other free programs that can do almost the same things: kseg for example. Also you can use [[Asymptote (Vector Graphics Language)]].

I personally prefer Compass and Ruler - its easier to use and more interactive. Its more useful and lets you see what happens as you move points around. Even better, its free!  :D


Use the numbers 2,3,4, and 5 and operations of addition, subtraction, multiplication, division and exponentiation to generate the largest possible result! (Hint:)[hide]tetration[/hide]


If a matrix has $m$ columns and all of them are linearly independent; is there a simply proof to show there are also a maximum of $m$ linearly independent rows?

Consider the transpose of the matrix. Since it will have $m$ rows, there will be a maximum of $m$ pivots in an associated augmented matrix. But this implies there will be a maximum of $m$ pivot columns. The number of pivot columns is equal to the number of linearly independent columns, so the transpose contains a maximum of $m$ linearly independent columns, which is what you were looking for.

i'm pretty sure that:

dim (row space) = dim (column space)


The sequence of even integers is grouped in brackets as follows:

(2), (4, 6), (8, 10, 12) (14, 16, 18, 20), (22,  …), …

such that each bracket contains one more integer than the preceding bracket.

Show also that the sum of all the integers in the first n brackets is (n)(n+1)(n^2+n+2)/4 .

If we divide the entire sequence, $S$, by $2$ $S/2=(1),(2,3),(4,5,6)$...
In this sequence the first $n$ brackets contain the first $\frac{n(n+1)}{2}$ integers.
Since the sum of the first $n$ integers is $\frac{n(n+1)}{2}$, the sum of the first $\frac{n(n+1)}{2}$ integers is $\frac{(n(n+1)/2)(n(n+1)/2+1)}{2}$.

Multiplying this out we find that $\frac S2=\frac{(n^4+2\cdot n^3+3\cdot n^2+2n)}{8}$.
This means that $S=\frac{(n^4+2\cdot n^3+3\cdot n^2+2n)}{4}$.
Factoring the numerator we find that $S=\frac{n(n+1)(n^2+n+2)}{4}$ which is the result that we wanted.
If someone could put this in Latex that would be great.

Why don't [i]you[/i] put it in latex?  The only commands you need are \frac{a}{b} for $\frac{a}{b}$ and a^b for $a^b$.

[quote="Shogia"]
If someone could put this in Latex that would be great.[/quote]

Would be even greater if you put it in $\LaTeX$.  This time I went back and edited for you but $\LaTeX$ is not at all hard to learn and I guarantee that typing " ^ " is much easier than typing < sup > ... < /sup >.

Hmm I'll have to take note of that.  Thank you for the help.


inequalities
trigonometry
inequalities proposed

Let $ \Delta ABC$ be a right triangle of legs b,c and hypothenuse a. Suppose
that $ b > c$. Show that

$ a \minus{} b  < \frac{\sqrt{2}}{\sqrt{2} \plus{}  1} \cdot \frac {c^2}{a}$

Since $ b>c$, $ \tan B>1\equal{}\tan \frac {\pi}{4}$ so $ \sin B>\sin \frac {\pi}{4}\equal{}\frac {1}{\sqrt {2}}$

We have to prove $ \frac {a^2}{c^2}\minus{}\frac ac . \frac bc \equal{}\sec ^2 B\minus{}\sec B \tan B\equal{}\frac {1}{1\plus{}\sin B}< \frac {\sqrt {2}}{\sqrt {2}\plus{}1}$
which is obviously true


OH MAN I'M NOT ON THE FINAL LEVEL I GUESS I HAVE TO STOP POSTING ON G&FF NOW AND POST SOMEWHERE ELSE TO GET MY POST COUNT UP I MUST MAKE THE FINAL LEVEL.

I'm only on the second level!!! I choose NOT to spam OR post in the test forum anymore! If people are going to spam their way to the top, I say that they should be at level 1!

[quote="jhollenbeck"]The way this works is...to achieve level 1 $\pi$ness  you must have 3 or more posts.  To achieve level 2 $\pi$ness, you must have 31 or more posts.  This continues to 314, 3141, and 31415 posts.  The number of AoPsers that have achieved the different ranks are...
3 posts: 4589
31 posts: 1479
314 posts: 303
3141 posts: 15
31415 posts: 0

May AoPS live to see the final level of $\pi$ness be accomplished![/quote]

Why is there a "final" level. aren't 314159, 3141592, 31415926, etc. all levels of $\pi$ness?

It'd be cool if the number of people with 3 were 14159, 4159 with 31, 592 with 314, 59 with 3141, and 9 with 31415.

That could be the case in about 6 or 7 years... ;)

I have level 3 piness. This post is not spam.

[quote="joml88"]Whoa I'm a level 4 super saiyan...er I mean AoPSer. :ninja:[/quote]

.....

do you think this is a joke? :mad:


Prove: Given a $2^n$ by $2^n$ checkerboard with any one square deleted, it is possible to cover this board with L-shaped pieces.

I have one question: Any $m\times n$ L-shaped piece?
[code]   L
   L
   L
 LLL

4x3 L-shape piece[/code]

Sorry. That was a bit vaue.

And L shape like:

LL
L

all we have to do is prove that multiple L's can be split to make an L with dimensions twice as large.
Since the above is true, then we can take a 2 by 2 (which is missing a square) and split it into a L with dimensions twice as large. Thus there is a 2by2 in the corner missing a square, and that can be filled with a small L. The process can be repeated ad infinitum, by splitting a 2^n by 2^n into 4 2^(n-1) by 2^(n-1) squares

not a very good proof, but i dont know how to get diagrams on here to illustrate what i mean.

[hide]
We proceed by induction.  The base case, where $n=1$, is trivial.  One of the four squares of the $2\times 2$ square will be missing thus leaving an L shaped assortment of squares.  So just put an L on these three.

Now, assume that an $n\times n$ square with a unit square missing can be tiled. Now consider an $n+1\times n+1$ square with one square missing.  By drawing two lines, one down the middle horizontally and the other down the middle vertically, we divide the big square into 4 $n\times n$ squares.  One of these is missing a square, but the others are not.  Therefore, we place an L in the middle of the big square so that it takes up one square on each of these three big squares.  We are thus left with 4 $n\times n$ squares which are missing a square each.  By our inductive hypothesis, we can tile these.  This completes our induction.[/hide]

[quote="seamusoboyle"]Typo or major problem joml88? :D[/quote]

I accidentally put the base case is when $n=2$, is that what you're talking about?  I meant when it is a $2\times 2$ square...but, of course, $n$ equals 1 in that case.   :blush:


In a community of more than $ 6$ people, each member exchanges letters with precisely three other members of the community.  Prove that the community can be divided into two nonempty groups such that each member exchanges letters with atleast 2 members of his group.

Searching for "community exchanges letters" (or "community letters 3" or a variety of other things) yields
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103426

I have not carefully checked the solution there to make sure that it's correct.


[quote="mdk"]If tanx +cotx=a, prove that (tanx)^3+ (cotx)^3 a^3-3a[/quote]

[hide]Let $ \tan x = r$. Then $ \cot x = \frac{1}{r}$. Hence,
$ r+1/r = a$

Cubing both sides,
$ r^{3}+3(r+1/r)+1/r^{3}= a^{3}$

Plugging in $ r+1/r = a$,
$ r^{3}+3a+1/r^{3}= a^{3}$

$ r^{3}+1/r^{3}= a^{3}-3a$

Re-substitution proves the desired result.[/hide]


At Sandy's Skateboard Shop, eight of the employees are to be rewarded by a trip to the skateboard expo in St. Louis, Missouri. Sandy likes eahc employee equally, so she decides to determine who will go by having the 20 employees stand equally spaced on the circumference of a circle and successively count off 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, $ \ldots$ repeatedly. Each person who says '6' steps out of the circle and picks up a travel ticket. What is the number of degrees in the measure of the central obtuse angle between the first and last person who receives the trip?

1st will be the 6th person, last will be 19th person (count just). They form an angle of: 13*360/20=234 or other angle is 126 degrees.

That's incorrect...the first person is number $ 6$ and the last person is number $ 15$ (just list it out and cross out), so the angle is $ \frac {15 \minus{}6}{20}\cdot360^\circ \equal{} \boxed{162^\circ}$.

[b]Edit:[/b] Fixed a typo.

I think what math meant was that there are 11 spaces and 20 people so 360/20 = 18*11 = 198. We need to find the obtuse angle so 360-198 = 162.

No, I just typoed. Sorry! (The people [b]are[/b] number $ 6$ and $ 15$.)


I can't write math symbols when I'm tryin' to use \colorbox.

For example:

[code]\colorbox{red}{$$a+b$$}[/code]

Also, suppose I'm using

[code]\fcolorbox{color}{color}{tex goes here:
$\begin{aligned}
stuff
\end{aligned}$}[/code]

but the aligment stays in the same line where the text is written. I can't put down my alignment.

[quote="Null"][code]\colorbox{red}{$$a+b$$}[/code][/quote]Too many dollar signs. Use [code]\colorbox{red}{$a+b$}[/code]
looks like $ \colorbox{red}{\(a\plus{}b\)}$

[quote="Null"][code]\fcolorbox{color}{color}{text goes here
$\begin{aligned}
stuff
\end{aligned}$}[/code]
but the aligment stays in the same line where the text is written. I can't put down my alignment.[/quote]
Boxes like this only take one line so you must wrap up more than one line in a parbox
[code]\fcolorbox{red}{green}{\parbox{2.5cm}{text goes here: \\
$\begin{aligned}
1+2=3
\end{aligned}$}}[/code]
$ \fcolorbox{red}{green}{\parbox{2.5cm}{text goes here \\
\(\begin{aligned}
1\plus{}2&\equal{}3 \\
11\plus{}12&\equal{}23
\end{aligned}\)}}$

To reiterate what I said before:
[quote="stevem"]Try the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=empheq]empheq package[/url] which does this sort of thing well.[/quote]


Denote $ b_n\equal{}5^{2^n}\plus{}3^{2^n}\plus{}2^{2^n}$.  Find the largest integer $ B$ such that $ B | b_n$ holds for all positive integers $ n$.

[hide="Thought"]Let $ n \equal{} 1$. This gives us $ b_{1} \equal{} 5^{2} \plus{} 3^{2} \plus{} 2^{2} \equal{} 38$. Now, $ 38$ has four positive divisors: $ 1, 2, 19, 38$. Two qualifies as $ B$, since $ b_{n} \equal{} 5^{2^{n}} \plus{} 3^{2^{n}} \plus{} 2^{2^{n}} \equiv 1 \plus{} 1 \plus{} 0 \mod 2 \equiv 0 \mod 2$, so all that's left to figure out is if $ 19\mid b_{n}$ holds for all $ n$ (can you prove why or why not?). If it does, we can say that $ \max \{B\} \equal{} 38$.[/hide]

[hide] Here are the remainders $ \mod{19}$ for the sequences $ a_n\equal{}a_{n\minus{}1}^2$ for $ a_1\equal{}25,9,4$ with the brackets denoting the period:

$ [6,\minus{}2,4,\minus{}3,9,5],6,\minus{}2,...$

$ [9,5,6,\minus{}2,4,\minus{}3],9,5,...$

$ [4,\minus{}3,9,5,6,\minus{}2],4,\minus{}3,...$

Notice that the period lengths are equal.  Thus, it is obvious that for all $ n$, $ 5^{2^n}\plus{}3^{2^n}\plus{}2^{2^n}\equiv0\pmod{19}$ by checking all the values in the period.

Also, $ 5^{2^n}\plus{}3^{2^n}\plus{}2^{2^n}\equiv1\plus{}1\plus{}0\equiv0\pmod{2}$.  Since $ 5^{2^1}\plus{}3^{2^1}\plus{}2^{2^1}\equal{}38$, the largest possible divisor of all $ b_n$ is $ 38$.  Since we have shown that $ 2|B$ and that $ 19|B$, we have that $ B\equal{}38$.
[/hide]


I am trying to solve a problem and I was able to get so far and then did now how to proceed.   One of the problems we were suppose to solve was as follows:

Prove that the sequence is a Cauchy sequence.
sqrt(n^2 + 3n) - n       n = 1,2,3,.......

I found that the it converge to 3/2 (hopefully that is correct).  I know that the next step is to set up the following
|3/2 - (sqrt(n^2 + 3n) - n)| < 1/m

But then I got confused.  I am not sure where to proceed from here or if I am on the right path.

$ x_n \equal{} \sqrt {n^2 \plus{} 3n} \minus{} n \equal{} \frac{n^2\plus{}3n \minus{} n^2}{\sqrt{n^2\plus{}3n}\plus{}n}\equal{} \frac{3}{1\plus{}\sqrt{1\plus{}\frac{3}{n}}} \to L\equal{}\frac{3}{2}$

proving that $ x_n$ is a cauchy sequence means showing

$ \forall \epsilon > 0 , \exists N_\epsilon$ such that

$ n> m \ge N_\epsilon \to |x_n \minus{} x_m|< \epsilon$

this can be done by simply solving for the $ M_\epsilon$ s.t.

$ n\ge M_\epsilon \to |x_n \minus{} L | < \epsilon$ and then setting $ N_{2\epsilon} \equal{} M_\epsilon$

suffices to show this for each of the form $ \epsilon \equal{} \frac{1}{m}$

so we merely have to solve the equation

$ |x_n \minus{} L| \equal{} \frac{3}{2} \minus{} \frac{3}{1\plus{}\sqrt{1\plus{}\frac{3}{n}}} \equal{} 3 \left[ \frac{1}{2}\minus{}\frac{1}{1\plus{}\sqrt{1\plus{}\frac{3}{n}}} \right] \equal{} \frac{1}{m}$

$ \frac{1}{2}\minus{}\frac{1}{3m}\equal{} \frac{1}{1\plus{}\sqrt{1\plus{}\frac{3}{n}}}  \to \frac{6m}{3m\minus{}2} \minus{}1 \equal{} \sqrt{1\plus{} \frac{3}{n}}$

$ n\equal{}\frac{3}{\left( \frac{6m}{3m\minus{}2}\minus{}1 \right)^2 \minus{} 1} \equal{} \frac{3m\minus{}2}{\frac{12m^2}{3m\minus{}2}\minus{}4m}\equal{} \frac{9}{8}m \minus{} \frac{3}{2} \plus{} \frac{1}{2m}$

picking $ M_\frac{1}{m} \equal{}$ ceiling $ \left ( \frac{9}{8}m \minus{} \frac{3}{2} \plus{} \frac{1}{2m} \right)$

we are done

If you found that it converges then you are done.


These phone calls to Atari are from Mario and Luigi (Okay, so Luigi's voice doesn't sound right, but Mario rocks!!!!)

[youtube]O87eGFUx21o[/youtube]

 :rotfl:  :rotfl:  :rotfl:

ROFLMFAO, that was priceless...  :rotfl:

Lol, that was awesome.

Err, these weren't to atari, you know. I think you misinterpreted it. they were calling game crazy about buying the atari system.

Sorry, I made some slight errors but still that video is pretty awesome. :blush:


what is this place?

what do you do in here?

Is it open for everyone?

This is the forum for an online math contest that's currently dormant. I used to help run it, but only the league commissioner (mathjoker) has any idea what the future of NOML might be. My guess is that said future is pretty bleak until the beloved commissioner gets his PhD.

For the record, NOML was awesome, and it will be awesome again when it returns.


A point 'O' is situated on a circle of radius R. With centre O, another circle of radius 3R/2 is described. Inside the crescent shaped area intercepted between these circles, a circle of radius R/8 is placed. If the same circle moves in contact with the original circle of radius R, find the length of the arc described by its centre in moving from one extreme position to the other.

[hide]Answer: 7πR[/hide]

This one foxed me completely. Can anyone help?

i dont really get the problem, are the first two circles tangent and what are you trying to ask??

There is a lens-shaped area that's shared by the two circles. To each side of that region, there are areas that are inside one circle but not the other. I'd call those crescents.
I don't understand how the "solution" could fit either:[hide]If the smallest circle (radius=$ R/8$) went one full turn around the outside of the original circle (radius=$ R$), its center would be describing a circle of radius $ 9R/8$. That's $ 9\pi R/4$ in length, far less than $ 7\pi R$.[/hide]


geometry
circumcircle
analytic geometry
LaTeX
geometric transformation
reflection
geometry solved

let $\bigtriangleup   ABC$ and $D$ be arbitary point on side $BC$ and $O$ be the circumcenter of
$\bigtriangleup   ACD$ ,and $M$ is the midpoint of $BC$ , $P$ is a point such that $PD \perp AB$  and
$PM \perp OM$ prove that $PB \parallel AC$

easy with analytic geometry. just place the triangle in the cartesian plane. M  coordinates are calculated easily. now set coordinates for D. then O  also is calculated easily. so is P . with that then it is straightforward. too many numbers and no latex so any solution id try to write would be hard to understand. hopefully somone can write my solution for me.

The problem is actually wrong, since a condition was omitted (AB = AC). The correct statement:

[color=blue][b]Corrected problem.[/b] Let ABC be an isosceles triangle with AB = AC. Let D be an arbitrary point on the line BC, and let O be the circumcenter of triangle ACD. Also, let M be the midpoint of the segment BC, and let the perpendicular to the line AB through the point D meet the perpendicular to the line OM through the point M at P. Prove that PB || AC.[/color]

[i]Solution.[/i] We will use the extended butterfly theorem:

[color=blue][b]Lemma 1 (extended butterfly theorem).[/b] Let O be the center of a circle, and let A, B, C, D be four points on this circle. The lines AC and BD intersect at a point R, and the perpendicular to the line OR through this point R meets the lines AB and CD at the points X and Y. Then, the point R is the midpoint of the segment XY.[/color]

How could we apply this theorem to solve the problem? Well, first denote by L the point of intersection of the lines DP and AM, and by Q the point of intersection of the lines PM and AC.

We will use directed angles modulo 180°.

Since the triangle ABC is isosceles (with AB = AC), the midpoint M of its base BC must also be the foot of the altitude from its apex A. Thus, $AM\perp BC$, and consequently, < (AM; BC) = 90°. We also know that $PD\perp AB$, so that < (AB; PD) = 90°. Thus,

  < ALD = < (AM; PD) = < (AM; BC) + < (BC; AB) + < (AB; PD) = 90° + < (BC; AB) + 90°
              = 180° + < (BC; AB) = < (BC; AB) = < CBA.

Since the triangle ABC is isosceles (with AB = AC), we have < CBA = < ACB, and thus < ALD = < CBA = < ACB = < ACD. Hence, the point L lies on the circumcircle of triangle ACD.

Now, we can apply Lemma 1 to the circumcircle of triangle ACD with its center O, and to the four points A, C, L and D on this circle. The lines AL and CD intersect at the point M, and the perpendicular to the line OM through this point M meets the lines AC and LD at the points Q and P. Thus, after Lemma 1, the point M is the midpoint of the segment QP. In other words, the point P is the reflection of the point Q in the point M. On the other hand, since the point M is the midpoint of the segment BC, the point B is the reflection of the point C in the point M. Since reflection in a point maps any line to a parallel line, we thus have PB || QC. In other words, PB || AC. And the problem is solved.

  Darij


[quote="warut_suk"]What's the projection of point $(4,5)$ onto line
[color=red](a)[/color] $x+y=0$
[color=red](b)[/color] $x=y$[/quote]

Projection of a point on to a line is the intersection point of given line with the line perpendicular to it and passing through the given point.

Here, line perpendicular to $x+y=0$ and passing through $(4, 5)$ is $x-y+1=0$.  Solving them, we get

the projection as $( - \frac{1}{2},\,\frac{1}{2})$

Similarly, in the case of $x-y=0$, the perpendicular line is $x+y-9=0$.  Solving these two equations, we get the projection as $(  \frac{9}{2},\,\frac{9}{2})$


function
algebra
polynomial
search
partial fractions

HI,  :blush:
I need help on this one please.

Write down a generating function for the number of the specified quantity for all n by translating the following combinatorial problem depending on the unknown n into generating functions expressions in a variable q.

The number of non-negative integer solutions to the equation x_1+2x_2+3x_3+4x_4 = n = 20 with x_i >= 0

Thanks

You might get more responses if you post in one of the problem forums such as Highschool basics or the Intermediate forum.

The coefficient of $ x^n$ in the above generating function is precisely $ \sum_{a \plus{} 2b \plus{} 3c \plus{} 4d \equal{} n} x^{a \plus{} 2b \plus{} 3c \plus{} d}$.  This is because the first factor can contribute a power of $ x^a, a \equal{} 0, 1, 2, ...$ whereas the second factor contributes a power of $ x^{2b}, b \equal{} 0, 1, 2, ...$ and so forth.

Thanks a lot t0rajir0u,

How did you come up with this answwer... what is your starting line...
The other thing is that when  n=20, is this going to set some restrictions on the x_1, x_2, x_3, and x_4. I mean is 0<=x_1<=20 and 0<=x_2<=10, and 0<=x_3<=6, and 0<=x_4<=5

Am I right?

The generating function already takes that into account.  $ n \equal{} 20$ corresponds to the coefficient of the $ x^{20}$ term, and none of the higher-power terms in any of the factors contribute to it (by definition).

The starting point is the number of non-negative integer solutions to the equation $ x_1 \equal{} n$.  Clearly there's only one solution for every $ n$, so this generating function is

$ 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} ...$.

Then we can look at the number of non-negative integer solutions to $ x_1 \plus{} x_2 \equal{} n$.  Clearly we can have $ (0, n), (1, n\minus{}1), (2, n\minus{}2), ... (n\minus{}1, 1), (n, 0)$, which is a total of $ n \plus{} 1$ solutions, so the generating function is

$ 1 \plus{} 2x \plus{} 3x^2 \plus{} 4x^3 \plus{} ...$.

On the other hand, we can write the generating function as

$ (1 \plus{} x \plus{} x^2 \plus{} ... )(1 \plus{} x \plus{} x^2 \plus{} ...)$

because the first factor contributes a power of $ x^{x_1}$ and the second factor contributes a power of $ x^{x_2}$.  Indeed, you can verify that the two generating functions are the same.  :)

What do you mean by definition?

 :blush:

The definition of polynomial multiplication.

[b]Definition:[/b]  Let $ P(x) \equal{} \sum_{k \equal{} 0}^{n} p_k x^k$ and $ Q(x) \equal{} \sum_{k \equal{} 0}^{m} q_k x^k$.  Then

$ P(x) Q(x) \equal{} \sum_{k \equal{} 0}^{m \plus{} n} \left(\sum_{i, j \ge 0, i \plus{} j \equal{} k} p_i q_j \right) x^k$.

Hi t0rajir0u,

Is there any book to study for the generating functions, or website beside the one that you have already on your site?

In order to go from the generating function back to the sequence, we can set up a recurrence: $ \sum a_nx^n\equal{}\frac1{(1\minus{}x)(1\minus{}x^2)(1\minus{}x^3)(1\minus{}x^4)}$, so $ (1\minus{}x)(1\minus{}x^2)(1\minus{}x^3)(1\minus{}x^4)\sum a_nx^n\equal{}(1\minus{}x\minus{}x^2\plus{}2x^5\minus{}x^8\minus{}x^9\plus{}x^{10})\sum a_nx^n$
$ \equal{}\sum_n(a_{n\minus{}10}\minus{}a_{n\minus{}9}\minus{}a_{n\minus{}8}\plus{}2a_{n\minus{}5}\minus{}a_{n\minus{}2}\minus{}a_{n\minus{}1}\plus{}a_n)x^n \equal{}1$.

Once you generate ten terms, you can keep going with the recurrence. If you want an explicit form for the $ a_n$, decompose using partial fractions instead.

If you want a book- search for "generatingfunctionology". It is freely available for download.


I have a problem unsolved!!
who can help me??
given a sequence :
$ a_1 \equal{} 1 a_2 \equal{} 4;$;
$ a_{n \plus{} 1}$ = 6$ a_n$ - $ a_{n \minus{} 1}$+2
find n sastifying $ a_n$ is a square number!!
thanks for yours!!! :)

Let $ a_n \equal{} b_n \minus{} \frac 12$, then $ b_n \equal{} (\frac 94 \minus{} 3\sqrt 2 )(3 \plus{} \sqrt 2)^n \plus{} (\frac 94 \plus{} 3\sqrt 2)(3 \minus{} \sqrt 2)^n$.

but , you are not completed this problem!!!
I think we cann't solve it by your solution!!!
I need a solution :)


MO is such a loser state. Its in the middle of everything.

28th alphabetical

24th state to join union

27th pop. density

21st total area

and the list keeps going....

Don't forget us selling our state's biggest company to Belgium!

i'll take your word for it..... :roll: :ewpu:

[quote="#H34N1"]everything's oversized in kansas[/quote]

everything?  :huh:


algebra
polynomial
function
domain
complex numbers
real analysis
real analysis unsolved

There exists a sequence of complex numbers $a_n$ with the property: for any continuous complex function defined on $[0,1]$ which vanishes at 0, there exists a sequence $k_n$ such that $\sum_{i=1}^{k_n}{a_ix^i}$ converges uniformly to $f$ on the domain of definition.

$E$ the set of continuous functions defined on $[0;1]$ that vanish at $0$.
Take $f_1,f_2,..,f_n,..$ a sequence of function in $E$ that is dense in $E$.
(for example, an enumeration of polynomials with rational coefficients and null at $0$).

Now, it is an easy consequence of Stone-Weierstrass theorem that the set $T_n$ of polynomials
of the form $P(x)=c_nx^n+c_{n+1}x^{n+1}+..$, where $n$ is a fixed integer is dense in $E$.

Consider a sequence $(u_n)$ of positive integers that takes infinitely many times any integer.
Take a polynomial $P_1 \in T_1$ such that $|P_1-f_{u_1}|<\frac{1}{1}$, and $deg(P_1)=d_1$.
Take a polynomial $P_2 \in T_{d_1+1}$ such that $|(P_1+P_2)-f_{u_2}|<\frac{1}{2}$, and $deg(P_2)=d_2$.
.. and continue ..
Take a polynomial $P_{n+1} \in T_{d_n+1}$ such that $|(P_1+P_2+..+P_{n+1})-f_{u_n}|<\frac{1}{n+1}$, and $deg(P_{n+1})=d_{n+1}$.

If $Q_n=P_1+P_2+..+P_n$, the sequence of polynomials $(Q_n)$ is dense in $E$ and from here it
is easy to see that the $(a_i)$ that are the coefficients of the $P_k$ work.


If $ x \geq 1$ and $ y \geq 1$ then   $ 3\left(\frac{x^{2} \minus{} y^{2}}{8}\right)^{2} \plus{} \frac{xy}{x \plus{} y} \geq \sqrt{\frac{x^{2} \plus{} y^{2}}{8}}.$

We have $ \sqrt {\frac {x^2 \plus{} y^2}{8}} \minus{} \frac {xy}{x \plus{} y} \equal{} \frac {A}{B}$ where
\[ B \equal{} \sqrt {\frac {x^2 \plus{} y^2}{8}} \plus{} \frac {xy}{x \plus{} y}
\]

\[ A \equal{} \frac {x^2 \plus{} y^2}{8} \minus{} \frac {x^2y^2}{(x \plus{} y)^2} \equal{} \frac {(x \minus{} y)^2(x^2 \plus{} 4xy \plus{} y^2)}{8(x \plus{} y)^2}
\]
We must show $ \frac {A}{B}\leq 3\left( \frac {x^2 \minus{} y^2}{8}\right)^2$. Factoring out $ (x \minus{} y)^2$, it remains to show
\[ \frac {x^2 \plus{} 4xy \plus{} y^2}{(x \plus{} y)^2\sqrt {8x^2 \plus{} 8y^2} \plus{} 8xy(x \plus{} y)}\leq \frac {3}{64}(x \plus{} y)^2
\]
But this is now certainly true, since the denominator of the left hand side is at least 32 and $ 2(x^2 \plus{} 4xy \plus{} y^2)\leq 3(x \plus{} y)^2$.


The "Half Average" game is played as follows:
[b]1.[/b] I need at least 10-15 sign-ups.
[b]2.[/b] When I recieve a good number of sign-ups, I will announce the beginning of a round.
[b]3.[/b] Each round, everybody PM's me an integer between 0 and 100.
[b]4.[/b] Things are done as follows:
- When everybody has sent in a number, I will post all the numbers and the average.
- Anybody under half the average is out of the game.
- Anybody over or equal to half the average recieves the distance from them to half the average points.
- Anybody with 100 or more points is out.
[b]5.[/b] The next numbers are PM'd.

The winner is the last person remaining in the game.

[u][b]PLAYER LIST:[/b][/u]
Klebian

[b][u]ELIMINATED[/u][/b]
Footballrocks41237
Mnmath
Ignite168
CircleSquared
Bubka
BanishedTraitor
13375P34K43V312
Mustafa
Lingomaniac
Perfect628
AstroPhys
Junggi

I'd do the probability calculations, but I'm too lazy to get past the point that the expected half value is $24\frac{76}{101}$.
So I'll join in instead.

i'm in

 The message is too small. Please make the message longer before submitting.

I'm in
since there's an editing limit and I'm a mod of this forum, if you feel like updating the first post after the editing limit runs out, feel free to PM me what you would want the first post to look like (including all color/bolding/whatever tags)

To do this, I'd recommend saving the post in BBCode form so you don't have to retype tags when PMing me.

[quote="Ignite168"]/in

Guys I'm sending 100 evary time.[/quote]

Good thinking. You'll at least last two turns.
But I think my 50 is smarter.
What will make it interesting is, of course, alliances. (Remember thumbtacks and whatsitsname?)

[quote="Ignite168"]/in

Guys I'm sending 100 evary time.[/quote]
But then you'll get approximately 50 a round, getting you out in two rounds.

Oh man. You thought I was putting down 20, but I put down 19. My mind games seem to have paid off. I'm not going to tell you what I'm doing this time. I don't know if you will be able to survive.

[b][size=134]ROUND 5 RESULTS[/size][/b]
[hide="Submissions"]Junggi: 100
Klebian: No submission (100)
Mustafa: No submission (100)
Lingomaniac: 25
Perfect628: 24
AstroPhys: 19[/hide][hide="Half the average"]Half the average was about 30[/hide][hide="Under Half the Average"]Lingomaniac, Perfect628, AstroPhys[/hide][hide="Scores"]Junggi: 169
Mustafa: 134
Klebian: 97[/hide][hide="Over 100"]Junggi, Mustafa[/hide][hide="We have a winner!"]Congratulations, Klebian![/hide][hide="Game 2"]I might make a game 2.  If you would like there to be a game two, please say so.[/hide]

Good plan. Game two is a good idea. Btw, I didn't submit on purpose, to screw with the scores.

Wow. I forgot to submit, but I was considering submitting 100. Ha.

LOL that worked out nicely

no i didnt decode it

[quote="miyomiyo"][quote="junggi"]no i didnt decode it[/quote][url=http://simplythebest.net/scripts/DHTML_scripts/javascripts/javascript_122.html]Maybe this will help.[/url][/quote]
no it doesnt

If you start another game, I will join.

junggi, it does.

firefox gave me exactly 23 javascript errors when i tried to decode that crap so i'm going to say it doesn't help.

damn it me and klebian were in an alliance and planning to go both 100 in like round 5 or something but i only accounted for 2 100s in round 3 and got screwed over when a 3rd one popped up.

I'll be in 2. I should have won that...I almost went back to 41...


Let a,b,c be positive reals. Find the minimum value of \[\frac{a}{b+3c}+\frac{b}{8c+4a}+\frac{9c}{3a+2b}.\]

I was trying substitution:
$x=b+3c\\
y=8c+4a\\
z=3a+2b$
But after that it's hard to be always sure that $a,b,c$ are positive reals (although it seems to lead to nice solution)...

the substitutions are good. there's no flaw there

[quote="ondrob"]After this substitution it comes to:
$-\frac{1}{3}-\frac{1}{16}-\frac{9}{12}+\frac{1}{8}\cdot\frac{y}{x}+\frac{1}{2}\cdot\frac{x}{y}+\frac{1}{6}\cdot\frac{z}{x}+\frac{3}{2}\cdot\frac{x}{z}+\frac{1}{12}\cdot\frac{z}{y}+\frac{9}{16}\cdot\frac{y}{z}$
Can you please show me how to complete it? Maybe it's obvious, but I can't find the minimum.[/quote]
you can do it directly.and you'll find it just right :)

Sorry, but I still can't complete it  :blush: . Maybe using derivates, but it looks messy. Maybe if I knew for which a,b,c there's minimum I would know how to complete it.

[hide="Hint..."]Use AM-GM on the x/y and y/x terms.  Then do the same for the x/z and z/x, and y/z and z/y.[/hide]

[quote="dts"]Use AM-GM on the x/y and y/x terms.  Then do the same for the x/z and z/x, and y/z and z/y.[/quote]It will not help...

why not? [tex]\displaystyle \frac{x}{2y}+\frac{y}{8x} \ge 2\sqrt{\frac{1}{16}} = \frac{1}{2}[/tex] right?

Or we could do the whole thing together and get
[tex]
\displaystyle \frac{1}{3}-\frac{1}{16}-\frac{9}{12}+\frac{1}{8}\cdot\frac{y}{x}+\frac{1}{2}\cdot\frac{x}{y}+\frac{1}{6}\cdot\frac{z}{x}+\frac{3}{2}\cdot\frac{x}{z}+\frac{1}{12}\cdot\frac{z}{y}+\frac{9}{16}\cdot\frac{y}{z} \ge \frac{3\sqrt[6]{3}}{2}+\frac{1}{3}-\frac{1}{16}-\frac{9}{12} = \frac{3\sqrt[6]{3}}{2} - \frac{23}{48}[/tex].

[quote="paladin8"]why not? [tex]\displaystyle \frac{x}{2y}+\frac{y}{8x} \ge 2\sqrt{\frac{1}{16}} = \frac{1}{2}[/tex] right?
[/quote]

Yes. But thats when you consider that by itself. When you try to put all three of them together, you can't obtain equality.. So you have proven that your answer is a lower bound, but not the minimum.

yeah i realized that just now. but what if you did them all together like in the second part? that should work right?

You still can't have equality, for the same reason (now you have 6 things all equal to each other, rather than 3 sets of 2 things).

Do anyone know (or guess) a,b,c for which there's a minimum? After that we can use AM-GM in right way (maybe).
Also after this substitution to x,y,z not for all x,y,z positive reals are a,b,c positive reals, so minimum of
$-\frac{1}{3}-\frac{1}{16}-\frac{9}{12}+\frac{1}{8}\cdot\frac{y}{x}+\frac{1}{2}\cdot\frac{x}{y}+\frac{1}{6}\cdot\frac{z}{x}+\frac{3}{2}\cdot\frac{x}{z}+\frac{1}{12}\cdot\frac{z}{y}+\frac{9}{16}\cdot\frac{y}{z}$
could be smaller than minimum of
$\frac{a}{b+3c}+\frac{b}{8c+4a}+\frac{9c}{3a+2b}$

Time to kill this post.  I've attached a solution.


geometry
calculus
AMC
AIME
USA(J)MO
USAMO
AoPS Books

Hello, i was wondering about what books people on this site use besides AoPS books (such as for proof-based contests or more advanced contests).

I was also looking for some recommendations for some intro college topics + Highschool topics:

Algebra:
I'm going to get intermediate algebra in may, but after that - ???

Geometry - ??? I only have intro to geometry

Combinatorics: Im getting intermediate counting/probability, and i downloaded "concrete mathematics" and "generatingfunctionology" for future use - i dont know after that?

Number theory: I have "An introduction to the theory of numbers", and a few other ones i have not looked at.

Calculus: I downloaded Apostol vol. 1,2 ( i guess this covers linear algebra too).

General problem solving - after Aops 2, ???

How to write proofs: If anyone knows of a book that teaches you how to write a proof this would be great, because I have no clue how to write proofs.

If anyone has any recommendations, knowledge, or thoughts on the current books, it would be great. Thank you.

I think you should start with the books you already have! That's quite a bit to dig through already. Some of the books you've listed, like Apostol, might help introduce you to "college topics", but will have very little to do with "contest" mathematics. For example, it won't teach you how to solve many AIME or Olympiad style problems as these problems are usually based on elementary* techniques [elementary in this sense of the word means High School, not easy].

You should also know that learning how to write a proof doesn't require knowing "advanced" or college level mathematics. Proof is such a daunting word - but all it really means is to give a clear logical explaination of how you know something is true. The best method for learning how to do a proof is simply by practicing and learning to critique your own logic.

First make sure you've mastered the basic mathematical tools (use AoPS vols. 1 & 2 for this). Then try the Part II problems here - http://gauss.math.oakland.edu/mmpc/previousexams.html

Then after these, move on to the books [u]The Art and Craft of Problem Solving[/u] by Paul Zeitz and [u]Problem-Solving Strategies[/u] by Arthur Engel, and start trying to solve old USAMO problems.

If you want a book that will introduce you to some "advanced" mathematics and let you practice some simple techniques of proof at the same time, try a cheap little book called [u]Introduction to Graph Theory[/u] by Richard Trudeau.


geometry
3D geometry
sphere
analytic geometry
induction

I've been trying to understand example 2.4.5 on page 63
from The Art and Craft of Problem Solving.  (The planets problem)
I understand the main idea of the solution but at the end the
solution states that:

 "Given any location x, it must be private on some planet"

But that doesnt seem right, because there may be some
locations which are public on every planet.
(e.g. if there are two planets and the poles lie on the
 same tangent plane,  then both poles would be public)

I wonder if anyone has worked on this example from Zeitz's book
and maybe could help me out on this one...

For those who are interested an exact copy of Zeitz's
problem and solution follows below:

Problem statement

We are given n planets in space, where n is a positive integer. Each planet
is a perfect sphere and all planets have the same radius R. Call a point
the surface of a planet private if it cannot be seen from any other planet.
(Ignore things such as the height of people on the planet, clouds, perspective, etc.  Also, assume that the planets are not touching each other.)  It is easy to check that if n = 2, the total private area is 4*PI*R*R, which is just the total area of one planet. What can you say about n = 3? Other values of n?

Solution

A bit of experimentation convinces us that if n = 3, the total private area
is also equal to the total area of one planet. Playing around with larger
n suggests the same result. We conjecture that the total private area is
always equal exactly to the area of one planet, no matter how the planets
are situated. It appears to be a nasty problem in solid geometry, but must
it be? The notions of private and public seem to be linked with a sort
of duality; perhaps the problem is really not geometric, but logical . We need
some notation. Let us assume that there is a universal coordinate system,
such as longitude and latitude, so that we can refer to the same location
on any planet. For example, if the planets were little balls floating in a room,
the location north pole would mean the point on a planet which was closest to the ceiling.
Given such a universal coordinate system, what can we say about a planet
P which has a private point at location x? Without loss of generality, let x
be at the north pole. Clearly, the centers of all the other planets must lie
on the south side of the Ps equatorial plane. But that renders the north
poles of these planets public, for their north poles are visible from a point in
the southern hemisphere of P (or from the southern hemisphere of an planet that lies between). In other words, we have shown pretty easily that
If location x is private on one planet, it is public on all the other
planets.
After this nice discovery, the penultimate step is clear: to prove that
Given any location x, it must be private on some planet.
We leave this as an exercise (problem?) for you!

[quote="zabelman"]If you post the problem, I'll take a look at it.[/quote]

There cannot be any points which actually have area -- basically, there cannot be a region that is public on every planet.  It sounds sort of topological, actually.

[quote="zabelman"]Basically, the set of points that are not private on any planet is the border of the private region on each of the planets. So the private regions are open regions (meaning they do not contain their boundaries) but that does not affect the are of the region.[/quote]

I disagree... The set of points that are not private (public) is
not necessarily a border of any region at all. Take for instance
the case of two planets. Then the set of points that are not
private is not a border but two closed hemispheres.

In any case...I understand the general idea of the solution.
My point is that at the end, Zeitz (the author of the problem)
asks to prove something which I think is simply not truth.

That is,

"Given any location x, it must be private on some planet"

[quote="zabelman"]In my last post I wasn't saying that any public point is on the border of a private region. I was saying that if a point was public on EVERY planet then it was probably on the border of a private region somewhere. (Think of the edges of the hemispheres in your last post.)[/quote]

OK (sorry)...I get your point  ...

[quote="josejacobi"][quote="probability1.01"]I can non-rigorously prove by induction (it's a little weird to think about, but it seems to work).  Just say that a point is private.  Then try to make it public by adding another planet.  You will see that the same point on that planet is then private.[/quote]

Yes... that would certainly work, assuming the point is private...
But that is not what the argument is saying...According to the solution
you should prove that ANY point is private on some planet.

The point is that this last statement seems to me like a small flaw
at the end of the argument...[/quote]

What I said was that assuming that 2 planets works (it obviously does).  Then three planets will work for the following reason.  If the additional planet renders a previously private point public, then that same point (mapped to the new planet) will be private.  Therefore, all points will still be private on some planet.  I still need a good way to show that what I said is true.  In fact, I'm not even 100% sure, but what I visualize says it's correct.

[quote="probability1.01"]

What I said was that assuming that 2 planets works (it obviously does).  Then three planets will work for the following reason.  If the additional planet renders a previously private point public, then that same point (mapped to the new planet) will be private.  Therefore, all points will still be private on some planet.  I still need a good way to show that what I said is true.  In fact, I'm not even 100% sure, but what I visualize says it's correct.[/quote]

Again..your induction is proving that:

If a point is private on a planet then it will be private on some
different planet.

I can also see that this is correct. But that's not what the
solution is asking to prove (at the end).

The solution is asking to prove that ANY point must private
on some planet (which I think is not true)

[quote="josejacobi"][quote="STK"]You can look at it this way:
Take 3 equally large collinear spheres and put them on a horizontal flat surface and when the middle one is cut throught its axis, the left half and right half will be everything visible from the left and right spheres respectively. Since the point at the north pole doesn't belong to both hemispheres, this means the 2 spheres you described cannot have the entire point at the north pole visible from the other sphere.
This is probably a very unusualy interpretation, but I think it helps to understand the problem.
Besides that, a single point on the sphere is infinitely small and excluding it has no affect on the surface area.[/quote]

I don't know....

Are you trying to say that the north pole is a private
point if it can be seen from more than one planet....
What if there are only two planets?
in that case the north pole would be private or public?[/quote]

What I was trying to say was, if there were more than 2 planets and you marked the regions that were public from one planet to another - for example if there are 3 planets- then it's possible to cut one of the planets in half and say that each half marked the region that was public to one of the other two planets. Then which half contains the north pole?

Does this make sense?

[quote="josejacobi"][quote="probability1.01"]

What I said was that assuming that 2 planets works (it obviously does).  Then three planets will work for the following reason.  [b]If the additional planet renders a [i]previously[/i] private point public[/b], then that same point (mapped to the new planet) will be private.  Therefore, all points will still be private on some planet.  I still need a good way to show that what I said is true.  In fact, I'm not even 100% sure, but what I visualize says it's correct.[/quote]

Again..your induction is proving that:

If a point is private on a planet then it will be private on some
different planet.[/quote]
Incorrect.  I have said that if for n planets [b]all[/b] points are private on some planet, then it is impossible to add a planet to make it so that (non-trivial) points are public on all planets.  If all points satisfy that they are not public on all planets, then all points must be private on one planet (obviously).  We know that two planets satisfy:

[quote="The solution"]After this nice discovery, the penultimate step is clear: to prove that
Given any location x, it must be private on some planet.
We leave this as an exercise (problem?) for you![/quote]

I have shown that if we add a planet, all [b][i]previously[/i][/b] private points will either stay private or be considered private on the new planet.  We already have that all points are private somewhere (the base case, 1 planet).  I have shown that it will remain that way no matter how many more planets are added.

ANSWER TO probability1.01

OK...OK... I get your point. BUT, I could still argue that your inductive
hypothesis ONLY considered nontrivial points  (in fact, you mentioned them in your proof)... What about the trivial ones (like poles)? I guess they might be public on every planet.

I know that trivial points have area zero so we may just discard them.

The thing is...Shouldn't the solution have ended in something like this:

"Given any NON-TRIVIAL location x, it must be private on some planet. "

When you quote, delete the parts of the things that you are quoting that are irrelevant -- making the huge nested quotes is annoying and takes up space.

Sorry... I'm new at this...I'll try to fix it...

[quote="zabelman"]Basically, the set of points that are not private on any planet is the border of the private region on each of the planets. So the private regions are open regions (meaning they do not contain their boundaries) but that does not affect the are of the region.[/quote]

Is there a simple way to prove this?

I mean...Is there a way to prove that if a point is public
on every planet then it must lie on the border of the
public region of some planet?


How do you find the zeros of a polynomial of even degree above 4 whose graph lies above the x-axis?

There's no fool proof method for finding the (exact) complex zeroes, only some tricks for special cases.

hm could you give me an example of some of these tricks?

Examples:

P(x) = x^6 + x^4 + x^2 + 1 is always positive, for real x to find complex zeros note that

P(x) = (x^8 -1)/(x^2 - 1) and hence P(x) is zero when x^8  - 1 = 0 and x^2 - 1 is nonzero.

Q(x) = x^8 + 2x^4 + 1 factors as (x^4 + 1)^2 and so the zeros are the same as those of x^4 + 1 = 0.

R(x) = x^8 + 5x^5 + 5x^3 + 1 = 0 can be solved by dividing through by x^4:

x^4 + 5(x + 1/x) + 1/x^4 = 0

Let u = x + 1/x

Then u^2 = x^2 + 1/x^2 + 2,

(u^2 - 2)^2 = x^4 + 1/x^4 + 2

So x^4 + 1/x^4 = (u^2 -2)^2 -2, hence the equation is

(u^2 -2)^2 - 2 + 5u = 0

which can be solved for u by the quartic formula and then for u.

Don't go around looking for a solution (by algebraic methods) to the generalized quintic equation.  In the 1800's (I'm not sure exactly when) Abel proved that it's [i]impossible[/i] to find an algebraic solution to a general equation of degree >=5.  In other words, choose any equation of degree >=5, and (with almost 100 % likelihood) it is impossible through a finite number of algebraic operations (addition and subtraction of polynomials from both sides, plus root-taking operations and such) to solve the equation.  That can only be done in special cases.  If you're interested in reading more about it, there is a "general solution" to the quintic equation using something called Jacobian theta functions (you can also use Weirstrassian theta functions for that purpose) but it's quite high-level math and won't do anything for you on a high school math contest.

[quote="jli"]find all the factors of the coeficient of the one with the largest power

find the factors of the number at the end

divide them in all the possible ways and put a plus/minus in fron of the fractions and those are the possible roots[/quote]

That's partially correct, but the roots you get as a result are all the possible RATIONAL roots, and some or all of the numbers you get that way may not be roots at all.  The process helps not one bit in finding the irrational roots.  However, on math contests, the so-called "rational zero theorem" helps ALOT, because many times, the roots of polynomials in math contests problems are integers or fractions.

[quote="p_adic"]
R(x) = x^8 + 5x^5 + 5x^3 + 1 = 0 can be solved by dividing through by x^4:

x^4 + 5(x + 1/x) + 1/x^4 = 0

Let u = x + 1/x

Then u^2 = x^2 + 1/x^2 + 2,

(u^2 - 2)^2 = x^4 + 1/x^4 + 2

So x^4 + 1/x^4 = (u^2 -2)^2 -2, hence the equation is

(u^2 -2)^2 - 2 + 5u = 0

which can be solved for u by the quartic formula and then for u.[/quote]
that equation is symmetric, (having the same terms on either side), and a symmetric equations can be solved this way in generla

and i dont think you need that quartic?

goodyfresh741: can you please give me a link or something like that to a solution for quintic equations using Thetafunctions¿

[quote="goodyfresh741"]  In the 1800's (I'm not sure exactly when) Abel proved that it's [i]impossible[/i] to find an algebraic solution to a general equation of degree >=5. [/quote]

Isn't it Galois?  Of course, It is not very important...  ;)

I think Abel proofed it for rational polynomials, Galois in a more general case for fields. (But's possible that I'm wrong)

Yes, Abel proved it first, but Galois proved it for a more general class of polynomials over fields.

This guy (http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Ruffini.html) proved that the quintic is unsolvable by radicals around 1800 and sent it to Lagrange and others but no one seemed to care or believe him. I know about him because he was a northern Italian whose last name was Ruffini. My father is a northern Italian whose last name is Ruffini  :D We could be related..


function
AMC
AIME
algebra
functional equation

It is question 107 in AoPS 2 (chapter 7). I don't quite understand the solution in the solution book. I substitute y=0 and get f(x) = 1. However, now, when the solution given proceeds to plug in f(x) = 1 back into the problem, it seemed to plug in f(x) = 1 and y=0 for one of the y's (on the left hand side, in f(x+y)), but not for the other (on the right side, y+2). Then it proceeded to say that the equation had no solutions. I am a bit confused. Could anyone clarify please, thx?

Fierytycoon

I think you mean problem #108.  #107 is the one about f(ax)=a*f(x)

[color=cyan]In the future, would you mind actually posting the questions you're having trouble with?  I don't own AOPS, so I can't even try to help without the question.  [/color]

ok..hopefully Mr. Rusczyk won't mind if I post one... (if you do Mr. Rusczyk, just delete this post)

108. Find all solutions to the functional equation f(x) + f(x+y) = y + 2  (M&IQ 1991)

Fierytycoon

Ah I see. That functional equation has to hold for ALL values of x and y. So when you substitute y=0, you get f(x) = 1 for all x. This means f(x)=1, f(x+y)=1, f(anything)=1. So when we substitute this in we get 1+1=y+2 for all y, and this is obviously impossible.

Feel free to post problems/etc from the books.  That's a small part of why we built the forum - so those who have questions about what's in the books can ask them here...

This seems a bit strange, but here's what I did:

f(x)+f(x)=0+2 by putting in y=0
2f(x)=2
f(x)=1 for all x.

Thus f is the constant function f(x)=1, and there are no other solutions.

Oops, typoed f(x)=1 instead of 0. Fixed that now. But this function doesn't work either because of the y on the right hand side.

Ah yes, you are correct. I apologize for my error.

Oh, I see. For some reason, I simply did not realize that f(anything) = 1 (hey, I was tired :) ). thx for the help guys

Fierytycoon

The point TripleM was making (and that I didn't understand at first) is that the only function that could possibly satisfy the condition is f(x)=1 for all x. But it is still necessary to check to see if that function works. We get 1+1=y+2, which is false unless y=0, so in general f(x)=1 does not satisfy the given functional equation.

Yeah, its similar to that if and only if thingy again. In these questions you have to do it in two parts, 1) prove that if there are any solutions, they must be these ones and 2) prove that these ones actually work. Often worth the last mark in a question.

[quote]Often worth the last mark in a question.[/quote]

..or, rather the whole mark in an AMC/AIME situation.

Fierytycoon

Yes, but they don't ask questions like that on AMC/AIME. I think if it actually doesn't work on USAMO/IMO, it will cost you more than a point if you don't check it. If it does, it's probably only one point.

[quote]Yes, but they don't ask questions like that on AMC/AIME. [/quote]

Well, maybe not on the AIME, but on the AMC, there could be a series of possible choices, one with the false answer and another saying "no solutions".

Fierytycoon


function
calculus
integration
real analysis
real analysis unsolved

Let $I \subset \mathbb{R}$ be an open interval and $f_n:I\to \mathbb{C}$ a sequence of continuous functions such that $\sum_{n=1}^\infty\sup_{x \in I}|f_n(x)| < \infty$. Prove that the function $S(x):=\sum_{n=1}^\infty f_n(x)$ is continuos on $I$. (Hint: Use the Lebesgue's Dominated Convergence Theorem).

Why would anyone want to apply the Dominated Convergence Theorem? This is a classic theorem - perhaps the most central theorem of the whole course - for a pre-Lebesgue course on analysis. (The course I happen to be teaching this semester.) This is nothing more or less than the statement that the uniform limit of continuous functions is continuous, coupled with the Weierstrass M-test for uniform convergence of a series.

Thank you Kent Merryfiel; I agree unconditionaly with your opinion but the purpose of this problems is to give an application of Lebesgue's dominated convergence theorem (without any reference to Weierstrass M-test).


Do you know where I could find out how to translate mathematical expressions to java code?
For example, my computer science teachers says $ \sqrt{6}$ = Math.sqrt(6) in java.


Find the number of square units in the area of the region in the first quadrant which is bounded by $ x \equal{} 4, x \equal{} 0, y \equal{} 2,$ and $ y \equal{} 8$.

This rectangle has a width of $ 4\minus{}0\equal{}4$, and a height of $ 8\minus{}2\equal{}6$. So our desired area is $ \boxed{24}$.


Let $ m_a,$ $ m_b$ and $ m_c$ are medians of triangle $ ABC$ with area $ S.$ Prove that:
1)\[ m_am_b\plus{}m_am_c\plus{}m_bm_c\geq3\sqrt3S\]
2)\[ m_am_b\plus{}m_am_c\plus{}m_bm_c\geq3\sqrt3S\plus{}\frac{(a\minus{}b)^2\plus{}(a\minus{}c)^2\plus{}(b\minus{}c)^2}{8}\]

[hide="(1)"] Using $ 4m_a^2=2b^2+2c^2-a^2$ etc the inequality is equivalent to
\[ \sum_{cyc}\sqrt{(2a^2+2b^2-c^2)(2b^2+2c^2-a^2)}\ge 12\sqrt S\]
Let $ a=x+y,b=y+z,c=z+x$. Verify that $ 2a^2+2b^2-c^2=4y(x+y+z)$. Combining this with $ S=\sqrt{xyz(x+y+z)}$ the inequality becomes
\[ 4(x+y+z)(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})\ge 12\sqrt{3xyz(x+y+z)}\]
Or
\[ \sqrt{x+y+z}(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})\ge 3\sqrt{3xyz}\]
Which follows from AM-GM
\[ \begin{array}{c}\sqrt{x+y+z}\ge \sqrt3(xyz)^{\frac 16}\
\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\ge 3(xyz)^{\frac 13}\end{array}\][/hide]


number theory
prime numbers
prime factorization

Find all prime numbers $p, q, r$ (or their negative counterparts) which satisfy
$\frac{1}{p - q - r} = \frac{1}{q} + \frac{1}{r}$

Simplifying gives us

$(p-q-r)(q+r)=qr$

By the unique prime factorization theorem, this splits up to $8$ cases:

1. $p-q-r=1, q+r=qr$
2. $p-q-r=-1, q+r=-qr$
3. $p-q-r=q, q+r=r$
4. $p-q-r=-q, q+r=-r$
5. $p-q-r=r, q+r=q$
6. $p-q-r=-r, q+r=-q$
7. $p-q-r=qr, q+r=1$
8. $p-q-r=-qr, q+r=-1$

Possibilites 3 and 5 obviously give no solution ($q=0$ and $r=0$ aren't valid), and 4 and 6 give $q=-2r\iff r=\pm 1$ and $r=-2q\iff q=\pm 1$, but $\pm 1$ aren't prime. Therefore, we're left with 1,2,7 and 8.

Number 1 gives $r=\frac{q}{q-1}\iff q-1=\pm 1.$ Therefore $q=2\implies r=2\implies p=q+r+1=5.$ One solution is $(5,2,2)$

Number 2 gives $r=-\frac{q}{q+1}\implies q+1=\pm 1.$ Therefore $q=-2\implies r=-2\implies p=q+r-1=-5.$ Another solution is $(-5,-2,-2)$

Number 7 gives $q-(-r)=1$, which means that $q$ and $-r$ are consecutive primes, and the only two possibilities are (a) $q=3, -r=2\iff r=-2$, then $p=qr+q+r=-5$ and (b) $q=-2, -r=-3\iff r=3$, then $p=qr+q+r=-5.$ Another two solutions are $(-5,3,-2), (-5,-2,3)$

Similarly, from number 8 we get $(5,-3,2), (5,2,-3)$

Hence we have six solutions: $(5,2,2),(-5,-2,-2),(5,-3,2),(-5,3,-2), (5,2,-3),(-5,-2,3)$


Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 Locked!


Let $\Gamma$ be the incircle of an equilateral triangle $ABC$ of side length $2$ units.
(a) Show that for all points $P$ on $\Gamma$, $PA^2 +PB^2 +PC^2 = 5$.
(b) Show that for all points $P$ on $\Gamma$, it is possible to construct a triangle of sides equal to $PA,PB,PC$ and whose area is equal to $\frac{\sqrt{3}}{4}$ units.

a)[color=blue][b]Lemma[/b][/color] : let $P$ be an arbitrary point in plan,and $G$ be the centroid of triangle $ABC$ then We have:
$PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3PG^2$

[i]proof[/i].let $D$ and $A'$ be midpoints of $AG$ and $BC$ use stuart theorem in triangles $AGP,DPA',PBC$ and do same for $B$and $C$.do the rest of ir yourself.

so now your problem: in equilateral triangle(let side lenght be $a$) we have $O\equiv I\equiv G\equiv H\equiv \dots$ so from lemma we have:
$PA^2+PB^2+PC^2=3(\frac{\sqrt 3}{3})^2a^2+3(\frac{\sqrt{3}}{6})^2a^2=\frac 54a^2$for $a=2$ it will be $5$

Hey Mathx, can the 2nd part be solved by the similar idea as the one you have used in the 1st part????

Yes the 2nd part can also be solved using the same idea.
Moderators, please move this to the Geometry Solved Problems.


Find all the solution to x, y and z (x, y z belong to Real Numbers) so that

x + 1/y = y + 1/z = z + 1/x
and
xyz = 1

Thanks!

OBS+> How can I edit my text thus he stay better than x + 1/y = y + 1/z = z + 1/x, I want the 1 on y, the 1 on z and the 1 on x! How can I do it?
Thanks!

surround the 'math' parts with dollar symbols, to make fractions use the command \frac{1}{2} inside the dollar symbols, to get $ \frac{1}{2}$

so
\$x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x}\$

gives

$ x \plus{} \frac{1}{y} \equal{} y \plus{} \frac{1}{z} \equal{} z \plus{} \frac{1}{x}$

I've also come to realize some shortcuts. Anytime you're using functions (of latex) the first character after each function is immediately applied (and only the first) without braces.
That is if you did \frac ab, you'd get $ \frac ab$. This is the same as \frac{a}{b}. However there isn't a shorthand for $ \frac{ab}{c}$ since $ \frac abc$ is \frac abc which shows what I mean originally.
This works with exponents too --> e^123 is $ e^123$ which is not e^{123}, that is $ e^{123}$

=)

[quote="hvaz"]Find all the solution $ (x,y,z)\in\mathbb R^3$ such that

$ x \plus{} \frac 1y \equal{} y \plus{} \frac 1z \equal{} z \plus{} \frac 1x$
and
$ xyz \equal{} 1$
[/quote]

So $ z \equal{} \frac 1{xy}$ and so $ x \plus{} \frac 1y \equal{} y \plus{} xy \equal{} \frac 1{xy} \plus{} \frac 1x$, from where we immediately get $ y \equal{} \minus{} \frac 1{x \plus{} 1}$

And so $ (x,y,z) \equal{} (a, \minus{} \frac 1{a \plus{} 1}, \minus{} 1 \minus{} \frac 1a)$ for any real $ a\notin\{\minus{}1,0\}$

Hi all
see here
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311324


conics
parabola
symmetry
analytic geometry
graphing lines
slope
rotation

A container with parabolic cross-section and partly filled with water rests wth the vertex point on level ground.

The container is gently rolled over so that no water spills.

Prove that the centre of mass of the water stays at a constant distance from the point where the container touches the ground.

To be clear: the reference is to a parabolic cylinder, not to a paraboloid of rotation. (The claim would be false for the paraboloid of rotation.)

I'm trying to imagine this problem in more-or-less Archimedean terms. In fact, I wouldn't bet against Archimedes himself being able to solve it - quicker than any of us.

Since we're talking about a cylinder, this is essentially a two-dimensional problem. Start with a parabola (of the form $ y \equal{} ax^2$) cut off by a horizontal line at a certain height. The center of mass of this shape lies at a specific point on the axis of symmetry. (We could calculate its height, but I don't need to for the argument I'm going to make.)

Now, shear this figure. That is, send the point $ (x,y)$ to $ (x,y \plus{} mx).$

Note that this shear leaves the original line of symmetry fixed. Since we are leaving $ x$ fixed for all points, the $ x$ coordinate of the center of mass remains unchanged. Furthermore, each upward movement of a point on one side is exactly balanced by a downward motion on the other side - hence, the $ y$ coordinate of the center of mass remains exactly unchanged. That, is, we still have the exact same center of mass.

The top horizontal line, after the shear, is now a line of slope $ m.$ But the tangent line at the origin, which was the $ x$ axis, has now become the line $ y \equal{} mx$ and is still parallel to the top line.

The parabola $ y \equal{} ax^2$ has now become the parabola $ y \equal{} ax^2 \plus{} mx.$ It's not the same as the original parabola, but (and this is crucial), it is congruent to the original parabola. It's just that the vertex is no longer at the origin.

For the final step, take the sheared figure and rotate it by the angle necessary to turn lines of slope $ m$ into horizontal lines. We claim that this is precisely the form of the tipped container with re-leveled water. (After all, the top surface is parallel to the tangent line at the point of contact.) And the center of mass is just where the original center of mass landed after the shear and rotation. It is the same distance from the point of contact as the original center of mass was, in the direction of the axis of the parabola.

But do note this: the center of mass is the same distance from the point of contact as is was to begin with. It is no longer directly above that center of mass, and hence it is lower. We have lost potential energy in the tipping operation, and (if not further restrained) the container is unstable with respect to tipping further and spilling.


limit
function
algebra
polynomial
calculus
derivative
real analysis

Let $f(x)\in C^{\infty}$ on $(-a,a)$, and $f^{(n)}(x)$ converges uniformly on $(-a,a)$. It is known that $\lim_{n\to\infty}f^{(n)}(0)=1$. Find $\lim_{n\to\infty}f^{(n)}(x)$.

[i](5 points)[/i]

Answer:$e^x$.
Let's define $\displaystyle g(x)=\lim_{n\ti\infty}f^{(n)}(x)$. Since the sequence $f^{(n)}(x)$ converges uniformly we can differentiate $g(x)$ by parts and get $g'(x)=g(x)$ with initial condition $g(0)=1$. Which gives us the answer $g(x)=e^x$

Unfortunately, I see a very large family of other solutions. Start with the exponential and add functions like:
polynomials
$\sin(kx)$ for a constant $k<1$
$e^{kx}$ for a constant $k<1$
$\frac{e^x-1}{x}$
- anything with derivatives that converge uniformly to zero. Such functions are entire, and go to infinity slower than $e^x$ in all (complex) directions, but that's still a large family.

:huh:
The question is to find $\lim_{n\to\infty}f^{(n)}(x)$. Do you agree that it is $e^x$ for all your examples?

Oh, [i]that's[/i] what the question asked.

I suppose the fact that I don't use smilies confused you. I misread the question, that's all.


[hide="hint, although I am not sure how good of a hint this is..."]factor[/hide]

[quote="hien"]Find all positive integer roots of following equation:
$x^{6}+4x^{4}y+x^{3}(y^{4}-4y)-xy^{2}-y^{5}=2001$[/quote]
That looks freaky. :o
I don't know how to approach it. :ninja:

[quote="hien"]Find all positive integer roots of following equation:
$x^{6}+4x^{4}y+x^{3}(y^{4}-4y)-xy^{2}-y^{5}=2001$[/quote]

[hide="another hint"]

i think you can factor, and use the binomail theorum[/hide]

Why any body only show the hint of problem. Is there anybody have solved one susceffuly?


Let $ x,y,z >0 $ and $ x^2+y^2+z^2 =1 $ . Prove that :
 $ \frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y} \geq 2.\sqrt{\frac{x+y+z}{xyz}} $

[quote="Ghang Hwan"]Brute force

<(4,2,0)-(4,11)>+<(3,3,0)+(2,2,2)-2(3,2,1)>

first term is nonegative because of Muirhead.

second  term is nonegative because of Schur for n=1,(xy,yz,zx)

so Sum is nonegative[/quote]

  Your solution is OK but I like rather a solution without brute force  :)

Nttu can you post a solution without muirhead?

We can prove the general ineq :
 $ \frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y} \geq 2.\sqrt{\frac{(x+y+z)(x^2+y^2+z^2)}{xyz}} $
   Hint: Apply the lemma :
  $ (a+b+c)^2 \geq 4(ab.sin^2Z+ac.sin^2Y+bc.sin^2X) $ , where $ X,Y,Z $ are three angles of triangle $ XYZ $
  Let $ sinX = \sqrt{\frac{x(x+y+z)}{(x+y)(x+z)}} $

Easier? I think it is an unpredictable solution

Actually ,it is a beautiful solution.I thought of that inequality too,but because I'm lazy I haven't tried. Anyway,the inequality is very nice. ;)


With $ x,y,z > 0$ such that $ xyz=8$.  Prove that ${ \sum_{cyc}\frac {x^2 }{\sqrt{(1+x^3)(1+y^3)}}\geq \frac{4}{3}}$

[quote="shenyixin"]With $ x,y,z > 0$ such that $ xyz = 8$.  Prove that ${ \sum_{cyc}\frac {x^2 }{\sqrt {(1 + x^3)(1 + y^3)}}\geq \frac {4}{3}}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30952


Prove that if $ k>2$ then $ 2^{2^k\minus{}1}\minus{}2^k\minus{}1$ is composite.


Hint : Find a polynomial $q(x)$ has small degree such that $q(a)=0$ and write $f=qm+n$, where $\deg{n}<\deg{q}$.


integration
function
Euler
trigonometry
calculus
calculus computations

\[ \left\{\begin{array}{rcl} J(u): \equal{} \int_0^1 \! (u'^2(x) \plus{} u(x)) \, dx\to min \\
\int_0^1 \! u^2(x) \, dx \equal{} 1 \\
u'(0) \equal{} u'(1) \equal{} 0 \\
\end{array} \right.\]
Where function u has a continuous derivative.

Notice that if a function $ u$ satisfies $ \int_0^1 u^2 dx \equal{} 1$ and minimizes $ \int_0^1 u'^2 \plus{} udx$, then it also minimizes $ \int_0^1 u'^2 \plus{} u \minus{} \lambda u^2dx$ for any $ \lambda\in \mathbb{R}$ (why?).  Therefore, any candidate minimizer for the original problem must satisfy the [url=http://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation]Euler-Lagrange equation[/url] corresponding to the modified problem:
\[ 2u'' \minus{} 1 \plus{} 2\lambda u \equal{} 0 \\
u'(0) \equal{} u'(1) \equal{} 0 \\
\int_0^1 u^2dx \equal{} 1\]
It's a straightforward ODE to solve.  The boundary conditions imply either $ u \equiv \text{const}$ or $ \lambda \equal{} n^2\pi^2$, $ n \equal{} 1,2,...$ and $ u_n \equal{} \frac {1 \plus{} c_n\cos n\pi t}{2n^2\pi ^2}$ for some $ c_n$.  For the non-constant solutions we have $ J(u_n) \equal{} \frac {2 \plus{} c_n^2}{8n^4\pi^4}$, which are all greater than the solution $ J(u) \equal{} \minus{} 1$ with $ u\equiv \minus{} 1$, which is therefore the minimizer.

[quote="JoeBlow"]Notice that if a function $ u$ satisfies $ \int_0^1 u^2 dx \equal{} 1$ and minimizes $ \int_0^1 u'^2 \plus{} udx$, then it also minimizes $ \int_0^1 u'^2 \plus{} u \minus{} \lambda u^2dx$ for any $ \lambda\in \mathbb{R}$ (why?).
[/quote]

And which theorema did you use here?

[quote="JoeBlow"]Therefore, any candidate minimizer for the original problem must satisfy the [url=http://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation]Euler-Lagrange equation[/url] corresponding to the modified problem:
\[ 2u'' \minus{} 1 \plus{} 2\lambda u \equal{} 0 \\
u'(0) \equal{} u'(1) \equal{} 0 \\
\int_0^1 u^2dx \equal{} 1\]
[/quote]

 I see in wikipedia that Euler-Lagrange equation is without any additional integrals like this:
\[ \int_0^1 u^2dx \equal{} 1\]

[quote="eduardos88"]And which theorema did you use here?[/quote]No theorem needed here, it's entirely an straightforward argument.  Any function $ u$ satisfying $ \int u^2dx \equal{} 1$ satisfies $ \int_0^1 u'^2 \plus{} u \minus{} \lambda u^2dx \equal{} \minus{} \lambda \plus{} \int_0^1 u'^2 \plus{} u dx$.  Therefore any minimizer of $ \int_0^1 u'^2 \plus{} u dx$ is a minimizer of $ \int_0^1 u'^2 \plus{} u \minus{} \lambda u^2 dx$ for every $ \lambda$.

[quote="JoeBlow"]Therefore, any candidate minimizer for the original problem must satisfy the [url=http://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation]Euler-Lagrange equation[/url] corresponding to the modified problem:
\[ 2u'' \minus{} 1 \plus{} 2\lambda u \equal{} 0 \\
u'(0) \equal{} u'(1) \equal{} 0 \\
\int_0^1 u^2dx \equal{} 1\]
[quote="eduardos88"] I see in wikipedia that Euler-Lagrange equation is without any additional integrals like this:[/quote][/quote]The constraint $ \int_0^1u^2dx \equal{} 1$ is [i]not[/i] present in the modified optimization problem for any given $ \lambda$, it is only used to get a fix on $ \lambda$ once the problem is solved in terms of $ \lambda$.

The $ \lambda$ here plays essentially the same role as a Lagrange multiplier in the more familiar type of optimization you deal with in calculus.  Just as the multiplier has the effect of 'removing' the constraint in that case, so it does here.

Yeah, I had to understand that at once.
Thank you very much!


[quote="alex2008"]Let be $ a,b,c > 0$ such that $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 1$ . Prove that :

$ 3(a \plus{} b \plus{} c)^6 \minus{} 8(a \plus{} b \plus{} c)^4(ab \plus{} bc \plus{} ca)\ge 3$

With the same condition it can be proved that :

$ 3(a \plus{} b \plus{} c)^9 \minus{} 8(a \plus{} b \plus{} c)^7(ab \plus{} bc \plus{} ca)\ge 3$[/quote]
For the first:
$ (a\plus{}b\plus{}c)^3[3(a\plus{}b\plus{}c)^3\minus{}8(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)]\equal{}(a\plus{}b\plus{}c)^3[3(a^3\plus{}b^3\plus{}c^3)\plus{}9(\sum_{sym}a^2b\plus{}2abc)\minus{}8(\sum_{sym}a^2b\plus{}3abc)\equal{}(a\plus{}b\plus{}c)^3(3\plus{}\sum_{sym}a^2b\minus{}6abc)$
But $ \sum_{sym}a^2b \ge 6abc$
So we have to prove that $ (a\plus{}b\plus{}c)^3 \cdot 3\geq 3$,which is true.....

For the second:
Factor out $ (a\plus{}b\plus{}c)^6$,and it becomes like the first :wink:


w.l.o.g. $ \beta \ge \gamma$. Let E the second intersection between the circumcircle of XYD and $ \omega$. DE meets $ \omega$ on F. So 'cause DXYE is cyclic $ \angle FEY \equal{} \angle YAC \plus{} \angle BCA$, so $ \angle FEC \equal{} \angle BCA$ and then F is the symmetric  of A w.r.t. the axis of BC. Then 'cause E is the intersection between FD and $ \omega$ that are all fixed it's also fixed.

[quote="freemind"]Let $ \omega$ be the circumcircle of $ ABC$ and let $ D$ be a fixed point on $ BC$, $ D\neq B$, $ D\neq C$. Let $ X$ be a variable point on $ (BC)$, $ X\neq D$. Let $ Y$ be the second intersection point of $ AX$ and $ \omega$. Prove that the circumcircle of $ XYD$ passes through a fixed point.[/quote]
[hide="Solution"]
Consider two points on $ BC$ not equal to $ D$, $ B$, or $ C$.  Call them $ X$ and $ X'$.  Let $ AX\perp BC$ and $ X'$ be a random point.  Let $ AX$ and $ AX'$ meet the circle at $ Y$ and $ Y'$ respectively.  Allow the circumcircle of $ \triangle DXY$ to meet $ \omega$ at $ Z$.  Let $ \angle Y'ZY \equal{} \alpha$, so since $ \angle DXY \equal{} 90\implies \angle DZY \equal{} 180 \minus{} \angle DXY \equal{} 90$, which means that $ \angle Y'ZD \equal{} 90 \minus{} \alpha$.  Also, $ \angle Y'AY \equal{} \alpha$ and $ \angle AXB \equal{} 90$, so $ \angle DX'Y' \equal{} \angle AXX' \plus{} \angle XAX' \equal{} 90 \plus{} \alpha$.  This results in $ \angle DX'Y' \plus{} \angle DZY' \equal{} 90 \minus{} \alpha \plus{} 90 \plus{} \alpha\equal{}180$, so $ X'DZY'$ is cyclic, which means that the circumcircle of $ X'Y'D$ passes through $ Z$.  Yet, $ Z$ is fixed because it is the intersection between the circumcircle of $ XDY$ (which is fixed because $ X$, $ Y$, and $ D$ are fixed points) and $ \omega$, which is also fixed.  Hence, the circumcircle of $ \triangle X'DY'$ passes through a fixed point, as desired. [/hide]

Let $ X_1$ be a fixed point on $ BC$ and $ X_{2}$ be any point on $ BC$ ($ X_{1}, X_{2}$ are distinct from D). Let $ AX_{i}$ cut the circumcenter of the triangle $ ABC$ at $ Y_{i}$.
Let the circumcenter of the triangle $ DX_{1}Y_1$ touch the circumcenter of the triangle $ ABC$ at $ T$.

Then $ \angle TDX_{1}\equal{} 180^{\circ}\minus{}\angle{TY_1A}\equal{}\angle{TBA}\equal{}180^{\circ}\minus{}\angle{TY_{2}A}$, therefore the point $ T$ belongs to the circumcenter of the triangle $ DX_{2}Y_{2}$, and since the points $ X_{2}, Y_{2}$ are mobile, the problem is solved.

Suppose that the diagram is as given (the other situations are the same). Let $ AD\cap \omega \equal{} A,F$. Let $ \Gamma$ be the circle through $ D$, $ F$ that is tangent to $ BC$. Clearly $ \Gamma$ is fixed. Let $ \Gamma\cap \omega \equal{} E,F$ (so $ E$ is fixed). Now \[ \angle XYE \equal{} \angle AYE \equal{} \pi \minus{} \angle AFE \equal{} \pi \minus{} \angle AFD \equal{} \pi \minus{} \angle EDX\] so $ DXYE$ is cyclic. Hence the circumcircle of $ \triangle DXY$ passes through a fixed point, $ E$.

For the record, the circumcircle of DXY passes through D, a fixed point ;)

But we know what you mean.

If $E=(DXY)\cap \omega$ and $X'$ is another variable point in $BC$ and $Y'$ the intersection of $AX'$ with $\omega$ we have,
$\angle X'DE=AYE=180-AY'E=180-X'Y'E$
So $X'Y'DE$ is cyclic and hence we are done


geometry
3D geometry
algorithm
search
Support

hey what is a fast algorithm to find cuberoots of large integers upto 150 decimal digits...i wrote a code using my own large integer implementation and had a long division type method for computing cuberoots
...but it's getting TLE'D on SPOJ..time limit is 5 secs

can anyone suggest better methods
thanks

I got AC with time 0.67s in Java using a simple binary search. Something like newton-raphson would be even quicker.
(You can keep everything to integers by multiplying by 10^30 at the start. Thats still easily small even for binary search).

then it seems my big integer implementation is too slow (in c++ i.e)...coz binary search was the thing i implemented first :)
anyways i cant think of ways to speed  it up...so i think i shud  learn a bit of java/python...

The NTL for C++ is implemented very efficiently by Victor Shoup. It supports arbitrarily large integer arithmetic called ZZ.  I believe it has a cube root function.  At any rate, this is a great path to go if you are going to be doing a lot of large integer arithmetic in C++. See: http://shoup.net/ntl/doc/tour.html


i found no relationship between the line segments AP, BP, CP, DP, and EP is there any relationship???

can u help please?

I think $ PA\plus{}PC\plus{}PE\equal{}PB\plus{}PD$.


How many people on the state team were not there with a team and how many were there as individuals?

Last year, all 4 people were there with their school's team for CT. This year, someone else and I were at states as individuals.

My school and team sucks, we placed 6th out of 13 teams at the smallest chapter in CT, these are the CHAPTER scores for the 5 other people from my school. None of them went to states.

Teammates at Chapter
20/6
12/8
12/4

Individuals at Chapter
12/10
12/0

zero. For some time before chapter, I thought I might go individually to state, but my team did well.

Scores of the other 3:
21/16, 21/16, 22/14
And a 20 on team.

The other 3 nationals people were all on really good teams.

I'm in the hardest chapter in FL. We had 4 in countdown and 2 going to nats. My chapter only sends two teams and 4 individuals though...

My chapter had top 5 individuals and top 2 teams...one person going to nats is not on the school team even thought the team made it...so that means
5-8 in school yet 2 in state

I voted for the wrong one. I voted for 0 but it really is 2...

My team barely made states.

Chapter took 4, we were 4th.  We had a 58, and the 6th place team (our rivals w00t) got a 56.25.  So, it was close.

The closest someone got was 7th at our state. He wasn't on Honey Creek's main team, but he was #5.

[quote="Ignite168"][quote="XxHarryxX"]For my nationals team, I was the only one at state without a team. We were sixth out of 22, but our chapter only takes the top four.

Scores were 45 (me!), 29, 20, and 17, with a perfect team round. 47.75 isn't bad, but it makes you feel bad when you have an individual score 24...[/quote]

Wow your school owns our school at Chapter, we had not including me: 28, 18, 16 (Team) Then we had two individuals who scored: 22, 12[/quote]

Yeah, maybe you school, but we're toast at our chapter.  I don't know what the scores were, but one school had places 4,5,6 and 13 for their team, and probably a perfect team round. So... probably 55 or 60? That's not great compared to 47.75..

In chapter, our scores were
teammates
27/16 (me!!! :D )
23/16
23/14
22/14
18 on the team round
total score 56.75
1st place team
Individuals
22/14
23/12
19/16
20/12
experimentalists (6th graders)
19/12
15/14
... ( 2 more people) (6th graders)
lowest... 13/10 (6th grader)
sadly, none of our individuals made it because our chapter only brings 1 team and 4 individuals to State. cutoff for state was 36. so we got close.

Our team scored an 18/20 with an average of 26 making a total of 44 putting us 6th. From our chapter, it was top 3 teams and top 12 individuals not on a team. Top 4 countdown, top 4 written, and top 4 teams got trophies. Countdown was 8 people. 5th place only had 16/20 on the team round.


[quote="now a ranger"]since I need to improve and have fun at the same time, I found this website

check it out  :lol:

http://www.writing.com[/quote]

cool! heres another good site but it is mostly based on games/books/ etc.

[url]http://www.fanfiction.net[/url]


geometry
3D geometry
tetrahedron
circumcircle
geometry unsolved

Let $ ABCD$ be a tetrahedron with circumradius $ 1$. Let $ AB\equal{}a,CD\equal{}a',AC\equal{}b,BD\equal{}b',AD\equal{}c,BC\equal{}c'$. Prove that:
1) $ aa'\plus{}bb'\plus{}cc'\leq{8}$.
2) $ aa'\plus{}bb'\plus{}cc'\equal{}8$ iff $ ABCD$ is an isosceles (or equifacetal) tetrahedron(ie. $ a\equal{}a',b\equal{}b',c\equal{}c'$).

[b]1). [/b]If G centroid and O circumscenter of tetraedron ABCD, then: $ R^2 \minus{} \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} a'^2 \plus{} b'^2 \plus{} c'^2}{16} \equal{} |OG|^2\ge0$; (R=circumradius).
[b]2).[/b] If R=1, then $ 16 \ge a^2 \plus{} b^2 \plus{} c^2 \plus{} a'^2 \plus{} b'^2 \plus{} c'^2\ge 2(aa' \plus{} bb' \plus{} cc') \Rightarrow aa' \plus{} bb' \plus{} cc'\le8.$
[b]3).[/b] If ABCD equifacial (a=a', b=b', c=c'), then $ G \equal{} O$ and (1) $ \Rightarrow a^2 \plus{} b^2 \plus{} c^2 \equal{} 8.$

OK Mihai, but 2 remarks:
In your 2): $ 16\geq{a^2\plus{}b^2\plus{}\cdots}$.
Moreover it remains to prove that $ aa'\plus{}bb'\plus{}cc'\equal{}8$ implies that $ a\equal{}a',b\equal{}b',c\equal{}c'$ (it's easy with the help of your post).

[b]OK!, CORECTED![/b]
[b]PS.[/b] For (1) view: [b][i]Nathan Altshiller-Court:[/i] "MODERN PURE SOLID GEOMETRY"(sec.ed); p.64 (the Theorem 186.1)[/b]
or [b][i]Victor Théabult[/i] book: "Parmi les belles figures de la Géométrie dans l'espace(Géométrie du Tétraèdre)"[/b] at:
http://ilib.mirror1.mccme.ru/djvu/geometry/thebault.htm
or
http://www.math.ru/lib/author/Theabult


function
probability
integration
real analysis
real analysis unsolved

Take 3 points independently and randomly and with uniform distribution  in the interval $[0,1]$. Determine in function of the positive number $m$the probability that:
$min{|x_{1}-x_{2}|,|x_{2}-x_{3}|,|x_{1}-x_{3}|}>m$

For $m \geq \frac{1}{2}$ it is 0.  For $m \leq \frac{1}{2}$ it should be $6\int_{0}^{1-2m}\int_{x+m}^{1-m}\int_{y+m}^{1}dz\, dy\, dx = (1+m)(1-2m)^{2}$.  (The 6 corresponds to our imposition of an order on the three variables, or alternatively to the size of the sample space, $\int_{0}^{1}\int_{x}^{1}\int_{y}^{1}dz\,dy\,dx = \frac{1}{6}$.)

the official solution is $(1-2m)^{3}$ for m<=1/2

Erm, yes, I mis-integrated.  $(1-2m)^{3}$ it is indeed.


geometry
analytic geometry
graphing lines
slope
calculus
algebra
system of equations

Find the minimum value of $ \sqrt{x^2\plus{}y^2}$ if $ 5x\plus{}12y\equal{}60.$

 :)

[quote="pi^2/6"]Find the minimum value of $ \sqrt {x^2 + y^2}$ if $ 5x + 12y = 60.$

 :)[/quote]
Hey, where are the choices!?
[hide="eww I hate geometry"]Denote the minimum value of $ \sqrt{x^2+y^2}$ as $ r$. Then $ \sqrt{x^2+y^2} \leq r \Rightarrow x^2 + y^2 \leq r^2$.
Hmm... that reminds me of a certain shape...
Obviously, we want $ r$ to be minimal. To do this, the line $ 5x+12y=60$ must be tangent to the circle $ x^2+y^2=r^2$. The line's slope is $ -\frac{5}{12}$, so the radius touching it has slope $ \frac{12}{5}$. To find where they intersect, solve the system of equations
$ 5x+12y=60\&\\
y=\frac{12}{5}x$
Blah. Someone do the rest.
[/hide]

$ (5^2\plus{}12^2)(x^2\plus{}y^2)\ge (5x\plus{}12y)^2\equal{}60^2$ by cauchy. The min is $ \frac{60}{13}$ which occurs when $ (5,12)$ is parallel to $ (x,y)$.

well the answer choices kind of give the problem away because they are all fractions involving 5, 12, 13, and 60, which immediately reminds you of a 5-12-13 right triangle...

well , i think this kind of problem can easily be done by Fermat's theorem  :)
firstly we obtain $ x = \frac {60 - 12y}{5}$ , we need to find the minimum of $ f(y) = \sqrt {(\frac {(60 - 12y)}{5})^2 + y^2}$
$ f'(y) = {338y - 1440}{125} = 0 \leftrightarrow y = \frac{1440}{338}$ gives the minimum $ f(y) = \frac {60}{13}$

This problem can also be solved easily without calculus etc.  :)
[hide]
Condider the line $ 5x\plus{}12y\equal{}60.$
Then $ \sqrt{x^2\plus{}y^2}$ is the distance from the origin to a point on the line.[/hide]


I am a contractor and oftem deal with final contract numbers that have a predetermined % markup on them.
This may be a really silly question but I am puzzled on how to figure this out with a calculator in hand.

I am trying to determine the cost of the labor and material before they are marked up with 15%. If I know that the final contract dollar amount is $\$$ 4,800.00, and there is a 15% markup on this number, what was the cost of the labor and materials?
Is there an easy way to figure this out with a caculator?
Every time I take (4800-15%) I get 4080. Than when I take (4080+15%) I get 4692. I just don't get it.

Thanks for taking a look at this for me. I apologize if this is a really silly question.

John

If you take 15% off $x$,

$\implies 0.85*x$ and then you increase this by 15%

$\implies 0.85*x*1.15 \implies 0.9775x$

it is not what you originally started out with, i hope this clears up your misunderstanding

Well in relation to your question, it is quite long and i hope i havnt misread anything,

I am assuming 4800 is the price after the 15% markup

let $x =$original amount

$\implies 4800 = \frac{115}{100}*x$

$\implies x=4173.913043$

c=1.15x
x=c/1.15

Theres a big difference between percent decreases and percent increases.

by the  way (1-a)(1+a) is less than 1 for all nonzero real values of a.


calculus
derivative
function
real analysis
real analysis solved

Let $f:R->R$ a derivable function.We know that the set of points for which $f'(x)>0$ is dense in $R$.Is $f$ strictly increasing?

You like difficult questions, don't you? Somehow I remember that this problem or a very related one was discussed. The answer is no!

Dear [b]harazi[/b] I'm very interested in the example of such function.Can you demontrate it.

Don' ask me, my friend. Just look back and study the solution. It is not me who found it!

But has it been solved on the forum? I searched for it and found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2822&highlight=]this[/url] topic. As you can see, there's absolutely no answer.. Is there another topic?

Oh, I see, but the examples there do not concern this problem. Those are increasing functions with an infinite/dense set of $0$'s for the derivative. I think the only thing there which concerns this problem is Pierre's statement (which I will quote):

"There exist functions f : |R -> |R for which f' exists everywehre and such that, for each a< b, f is not monotonous on [a,b]. For such a function, it is clear that each of the sets P = {x / f'(x) > 0}, N = {x / f'(x) < 0} and Z = {x / f'(x) = 0} is dense on |R (the third one because, f' has the intermediate values property).
Such functions have been constructed by Denjoy (first half of 20th century), and there are others papers on them since then (but, this is not so easy...)."

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2822[/url] for an example.


conics
parabola
function
geometry
geometric transformation
reflection
analytic geometry

A parabola with equation $ y \equal{} ax^2 \plus{} bx \plus{} c$ is reﬂected about the $ x$-axis. The parabola and its reﬂection are translated horizontally ﬁve units in opposite directions to become the graphs of $ y \equal{} f(x)$ and $ y \equal{} g(x)$, respectively. Which of the following describes the graph of $ y \equal{} (f \plus{} g)(x)$?

$ \textbf{(A)}\ \text{a parabola tangent to the }x\text{ \minus{} axis}$
$ \textbf{(B)}\ \text{a parabola not tangent to the }x\text{ \minus{} axis} \qquad \textbf{(C)}\ \text{a horizontal line}$
$ \textbf{(D)}\ \text{a non \minus{} horizontal line} \qquad \textbf{(E)}\ \text{the graph of a cubic function}$

[hide]If $ f(x)\equal{}ax^2\plus{}bx\plus{}c$, then the reflection of the x-axis is $ \minus{}f(x)\equal{}\minus{}ax^2\minus{}bx\minus{}c\equal{}g(x)$.

Say that we translate $ f(x)$ by $ m$ units to the left, and $ g(x)$ by $ m$ units to the right.  Then the function is now $ f(x\plus{}m)\equal{}a(x\plus{}m)^2\plus{}b(x\plus{}m)\plus{}c$ and its reflection is $ g(x\minus{}5)\equal{}\minus{}a(x\minus{}5)^2\minus{}b(x\minus{}m)\minus{}c$.

Expanding,
$ f(x)\equal{}ax^2\plus{}2axm\plus{}am^2\plus{}bx\plus{}bm\plus{}c$
$ g(x)\equal{}\minus{}ax^2\plus{}2axm\minus{}am^2\minus{}bx\plus{}bm\minus{}c$

Therefore $ (f\plus{}g)(x)\equal{}4axm\plus{}bm$
Or, using $ xy$ notation,
$ y\equal{}(4am)x\plus{}bm$, which we see to be a line with slope $ 4am$ and y-intercept $ bm$.  Answer $ D$[/hide]

Why isnt the answer C?

Let $y=a(x-h)^2+k$. So after transformations, $f(x)=a(x-h)^2+k+5$ and $g(x)=-a(x-h)^2+k-5$. So $(f+g)(x)=2k$ which is a horizontal line...

[b]Horizontal[/b] translations would be $f(x)=a(x-h\pm5)^2+k$

[b]Solution:[/b] Writing this up for the sake of storage. The equations are $f(x)=a(x+5)^2 + b(x+5) + c$ and $g(x)=-a(x-5)^2 - b(x-5) - c$. Adding those two functions, we get our new function defined by the equation $(20a)x + 10b$. Clearly, this is answer choice $D$.


modular arithmetic
number theory
algebra proposed
algebra

Prove that for each natural number $ k$ number :

$ k^{2000} \plus{} (k \plus{} 1)^{2000} \minus{} k(k \plus{} 1)$ is complex (has at least 2 prime divisors)

From the expression, I think that the expression is divisible by $ k(k \plus{} 1)$. Yet, it is not true. However, it is always true that your expression is divisible by $ t \equal{} k^2 \plus{} k \plus{} 1$. The proof requires simple modular arithmetic.

($ k^{2000} \equal{} (k^2)^{1000} \equal{} ( \minus{} k \minus{} 1)^{1000} \equal{} (k \plus{} 1)^1000 \pmod{t}$, ...).

Moreover, the highest exponent for $ t$ is 1. That is:

$ \gcd(\dfrac{k^{2000} \plus{} (k \plus{} 1)^{2000} \minus{} k(k \plus{} 1)}{t}, t) \equal{} 1$

That's it!

P/S: This problem should be posted in the [b] Number Theory [/b] section.


[quote="ttd"]find necessary and sufficient condition of $p\in Z$ so that there exists $p\in Z$ such that both $f(x)$ and $f'(x)$  have distinct roots  when $f(x)=x^3+px+q$[/quote]

Perhaps it's very hard, but for the moment that's very unclear  :D
 :cool:

I think it must be:find condition of p so that there exists q!!

But maybe it's "Find a condition on $q$ such that there exists $p$"... ttd?