The MOSP Invitation Criteria are:

Invitation to June 4 - July 1 MOSP including the Team Selection Test:
9th, 10th, 11th, 12th graders with a USAMO score greater than or equal to 25.
(Must also be eligible for the US IMO team and able to attend IMO.)
There are 15 in this group.  (Incidentally, all happen to be 10th, 11th, 12th graders.)

Invitation to June 11 - July 1 MOSP
9th, 10th, 11th grader with a USAMO score greater than or equal to 19.
(must also be eligible for the US IMO team.)
There are 15 in this group.  (Incidentally, all happen to be 10th and 11th graders.)

Invitation to June 11 - July 1 MOSP  ("9th grade MOSP", or "Red Group MOSP")
9th grader with a USAMO score greater than or equal to 9.
There are 24 in this group.

Total 54 students.

The selection criteria are based on the following, appearing in Section VIII, page, 6 of the USAMO Teachers' Manual:
"The top 12 scoring students ont he USAMO will be invited to attend the MOSP program regardless of grade.  In addition to the top 12 USAMO winners, 12-18 high-scoring USAMO participants, who are current juniors and below, will also be invited to participate in the MOSP.

...we will inivte an additional (approximately 30) ninth-grade USAMO participants for a MOSP program."

Emails have been sent about an hour ago.  Paper mail envelopes will go out in tomorrow's mail.  Phone calls will be made this week, if possible.  Not receiving a phone means only that we were not able to contact you by phone, do not wait by the phone.  Your first notification will be ny email.  The paper mail is the official invitation, please allow up to a week for delivery, depending on where you live.

Steve Dunbar
AMC Director

Thank you Mr. Dunbar!  Will there be a list of invited or participating on the site?

YES!!!!!!!!!!!!!  This made all of my brace pains suddenly go away... :D

NOOOO!

GRRRR!

No offense, AMCDirector, but there are a lot of people right know who love you and a lot more who hate you.  Personally, I understand that you're just doing your job, but I'm still angry, as I scored 2 points below the red MOP cutoff.

Three questions, though.  (1) Will there be more detailed scoring (as in names, not just numbers) on the AMC site? (2) Will our papers be returned to us? (3) If Question 2 is true, will our papers have comments on them, so that we can see our errors and improve next year?

BTW, congrats, Carlos.  One Davidson freshman in red MOP (you) and one Davidson freshman out (me).  What are brace pains?

The answer to your questions are No, No, and....

you guessed it: NO.

[quote="Teki-Teki"]What are brace pains?[/quote]

Pains of having braces :P

b/c i've taken the USAMO before (and done equally badly... WTF???  :( )

[quote="Teki-Teki"]bubala - no offense, but how do you know?[/quote]

Well unless you go to MOP maybe you can ask for score distribution/comments from grader.

I mean you can trust the USAMO graders on not making mistakes.

I thought you did get detailed score reports, but not until later.

Congrats Noah Arbesfeld for making Red group with a 5 point cushion ... and you were worrying that you wouldn't make it!

beta - I'm not saying I don't trust the graders, I'm just saying that I would like to see my detailed score reports and comments if possible.  I mean, it might be kinda helpful to know where I got those 7 points and where my proofs went wrong.

[quote="MysticTerminator"]No. You never see *anything* about the USAMO. Regardless of whether you go to MOP, etc. - end of story. As probably noted in another thread, they didn't even announce the scores until a mistake a few years ago (and continue to not do so for the TST).

This has caused me some consternation as well, but eh.[/quote]

Is there any way for colleges (or other such institutions) to verify a score?  It seems like a decent USAMO score (but not MOP-good) could be a useful tool for determining acceptance into some programs.

Teki, I think you're overreacting, please calm down.

I'm sure we've all seen, many times on this very forum, the main reason they don't release the scores - they don't want people questioning them. I'm sure that if they did release the scores, they would be inundated with emails about various problems and if they were missing the papers or if they were drunk when they were grading, etc.

[quote="MysticTerminator"] if they were missing the papers or if they were drunk when they were grading, etc.[/quote]

Heh, that's what I almost thought at first.  :oops:  :D

Bah. Looks like I'll just have to get a perfect score next year.

[quote="MysticTerminator"]new development - RSI people do not attend MOP[/quote]

Oh that explains it. No wonder my score of 193847812390 didn't get me into MOP.

[quote="Teki-Teki"]MysticTerminator - I agree, that does cause major consternation.  I can think of four good reasons off the top of my head.

1. The reports are in our files.

2. They probably have electronic copies so it would take them like 20 minutes total

3. It would benefit us.

And the number one reason for doing anything:

IT WOULD MOST LIKELY BENEFIT OUR COUNTRY'S PERFORMANCE SOMEWHAT AT IMO.

Who wouldn't want the U.S. to win?  (other than the canadians, indians, chinese, russians, koreans, and so on ad infinitum et absurdum)

Okay, who IN THE U.S. wouldn't want the U.S. to win?

Please don't say "I would," Mr. Dunbar.[/quote]

Teki-Teki, no offense, but relax. With all due respect, if you make MOP you have a decent chance to ask about your paper (Mr. Feng among other people has probably reviewed your performance to get an idea for your strengths and weaknesses). If you don't make MOP ... well, to put it bluntly, you're not exactly their foremost concern.

It's not THAT easy to provide that information that you are seeking. I am a grader for WOOT, and if it were not for the fact that we are required to write comments on the solutions that we grade, we would not do it usually. In a vast majority of solutions, we don't have to take any notes to determine the score of a solution, it is very obvious whether or not someone has satisfied certain aspects of the rubric. It's also not in the best interests of time constraints to take notes on a solution and write up suggestions for improvement (and that does take significant time if done properly)

I do agree that releasing score distribution would not be too difficult. Again, it's not the biggest deal. As you become more experienced with Olympiad problem solving, you will become a lot better at predicting your score. The last three years I have been 6/6 in terms of predicting my score on USAMO problems.

Finally, the "marginal cost" of providing this extra information would be insignificant, in my opinion. I highly doubt that it would actually make a difference in future IMO scores. There's nothing magical about the comments of a USAMO grader ... compared to simply reviewing the solutions posted to the problems, which has far more potential to help you improve. If you think you got something and didn't get the score you expected, post what you did here and ask around.

EDIT: Let me give a little background information on why specific scores are released AT ALL. Until 2002, you were only informed of your score range (0-6, 7-13, 14-20, 21-27, 28-34, 35-42). That year, USAMO was centralized in Boston and essentially the opinions of the masses prompted the USAMO directors to start releasing people's actual scores. Basically, we were all like "it's ok if we got a zero, we want to know".

Finally (the second finally of this post!), enough people e-mail/call/harass the AMC people about various AMC things as it is (just look at this year's USAMO selection). Releasing specific scores about each problem will probably cause some people to call in and be like "what do you mean I got a 2??? I deserved 7!!!" It's really the last thing AMC wants to bother to put up with.


I am new to physics and i don't know much about it i just know things like basic mechanics like little anout impulse,force , momentum etc. Can someone plz help me find a way to study?

thanking you,
Poincare

Read the textbook, read the problems, read the solutions, and repeat.  Then when you feel comfortable, try the practice problems.

But i don't have a textbook, i am a fifth grader, my school doesn't even teach basic mechanics!

Then head to your library and take out a book.  That is the best way to learn unless you are in a class.


If $a,b,c$ are sides of a triangle prove that :
$\sum \frac{1}{\sqrt{a}+\sqrt{b}-\sqrt{c}} \geq 3 \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{a+b+c}$

\[
\sum {\frac{1}
{{\sqrt a  + \sqrt b  - \sqrt c }}}  \geqslant \frac{9}
{{\sum {\sqrt a } }} \geqslant 3\frac{{\sum {\sqrt a } }}
{{\sum a }}
\]
both cauchy.


trigonometry
integration
quadratics
logarithms
calculus
calculus computations

Find the general solutions of

(i) $ y' \equal{} (e^y \minus{} x)^{\minus{}1}$,
(ii) $ y' \equal{} x^2 (1 \plus{} y^2)$,
(iii) $ y' \equal{} \cos^2 x \cos^2 2y$,
(iv) $ y' \equal{} (x \minus{} y)^2$,
(v) $ (e^y \plus{} x)y' \plus{} (e^x \plus{} y) \equal{} 0$.

(i) We can rewrite the given form as ;$ y'e^y \minus{} xy' \equal{} 1$.
     And this is $ (e^y \minus{} y(x \minus{} 1))' \equal{} 1$.
     So, $ e^y \minus{} y(x \minus{} 1) \equal{} x \plus{} C$, the general solution is; $ e^y \minus{} y(x \minus{} 1) \minus{} x \plus{} C \equal{} 0$ or $ x \equal{} \frac {e^y \plus{} y \plus{} C}{y \plus{} 1}$. ($ C$ is constant)
(iv)Differentiate once more, and we get $ y'' \equal{} 2(x \minus{} y)(1 \minus{} y'),\quad (x \minus{} y) \equal{} \frac {y''}{2(1 \minus{} y')}$.
    Substitute this to the given form : $ y'' \equal{} 2(1 \minus{} y)\sqrt {y'}$. Let $ y' \equal{} u$, and seperate the variables.
    $ \int\frac {du}{2(1 \minus{} u)\sqrt {u}} \equal{} \int{dx}$

    $ \frac {1}{2}\int\left ( \frac {\sqrt {u}}{1 \minus{} u} \plus{} \frac {1}{\sqrt {u}} \right )du \equal{} x$.

    $ \frac {1}{2}\int\frac {\sqrt {u}}{1 \minus{} u}du \equal{} \frac {1}{2}\int\frac {t}{1 \minus{} t^2}\cdot{2tdt} \equal{} \int\left ( \frac {1}{1 \minus{} t^2} \minus{} 1 \right )dt \equal{} \tanh^{ \minus{} 1}(t) \minus{} t \plus{} C$, so

    $ \frac {1}{2}\int\left ( \frac {\sqrt {u}}{1 \minus{} u} \plus{} \frac {1}{\sqrt {u}} \right )du \equal{} \tanh^{ \minus{} 1}(\sqrt {u}) \minus{} \sqrt {u} \plus{} \sqrt {u} \plus{} C \equal{} x$.

    Rewrite for $ u$; $ u \equal{} \left ( \tanh(x \plus{} C) \right )^2 \equal{} 1 \minus{} \frac {1}{\cosh^2(x \plus{} C)} \equal{} \frac {dy}{dx}$.

    Thus, we get finally $ y \equal{} x \minus{} \tanh(x \plus{} C_1) \plus{} C_2$. ($ C_1,C_2$ are constant)

(v) $ (e^y \plus{} x)y' \plus{} (e^x \plus{} y) \equal{} 0$  can be rewritten by $ (e^y \plus{} xy \plus{} e^x)' \equal{} 0$.
    So, the general solution is $ e^y \plus{} xy \plus{} e^x \plus{} C \equal{} 0$ ($ C$ is constant).

[quote="akech"]For ii) and iii) you can separate the variables and just integrate. Since these are basic homework problems, I encourage you to show what you have done so far on each problem because then someone will not bother to redo what you have already done.[/quote]
Point taken - for (i), (ii) and (iii) I just want to check my answers:

(i) I get a different answer from J.Y. Choi for this one:
 $ \frac {dy}{dx} \equal{} (e^y \minus{} x)^{ \minus{} 1}$, so $ \frac {dx}{dy} \equal{} e^y \minus{} x$. Now we treat $ x$ as the dependent variable and $ x \equal{} x(y)$. So $ x' \plus{} x \equal{} e^y$, $ e^y x' \plus{} e^y x \equal{} e^{2y}$, and $ (e^y x)' \equal{} e^{2y} \Rightarrow e^y x \equal{} \frac {1}{2} e^{2y} \plus{} C$.

Treating this as a quadratic in $ e^y$, $ \frac {1}{2} (e^y)^2 \minus{} xe^y \plus{} C \equal{} 0$, and the solutions of this quadratic are $ e^y \equal{} x \pm \sqrt {x^2 \minus{} 2C}$.

Thus I get $ \boxed{y \equal{} \ln \left( x \pm \sqrt {x^2 \minus{} 2C} \right)}$.

(ii) $ y' \equal{} x^2 (1 \plus{} y^2) \Rightarrow \int \frac {1}{1 \plus{} y^2} \, dy \equal{} \int x^2 \, dx$ and after rearrangement I get $ \boxed{y \equal{} \tan \left( \frac {1}{3} x^3 \plus{} C \right)}$.

(iii) Rearranging I get $ \int \sec^2 y \, dy \equal{} \int \frac {1}{2}(\cos 2x \plus{} 1) \, dx$. The final answer I get is $ \boxed{y \equal{} \frac {1}{2} \tan^{ \minus{} 1} \left( \frac {1}{2} \sin 2x \plus{} x \plus{} D \right)}$

Got not much of a clue for (iv) and (v), checking J.Y. Choi's solution now.

I think I get number (iv), thanks.

[quote="J.Y.Choi"]So, the general solution is $ e^y \plus{} xy \plus{} e^x \plus{} C \equal{} 0$ ($ C$ is constant).[/quote]

Is it possible to express the solution for (v) in the form $ y \equal{} f(x)$?

hello, for (i), we have
$ (e^y\minus{}y(x\minus{}1))'\equal{}e^yy'\minus{}y\minus{}xy' \ne y'(e^y\minus{}x)$
the solutions are
$ y(x)\equal{}\ln(x\minus{}\sqrt{x^2\plus{}C})$
ore
$ y(x)\equal{}\ln(x\plus{}\sqrt{x^2\plus{}C})$
Sonnhard.

hello, for (ii) i get
$ y(x)\equal{}\tan(\frac{x^3}{3}\plus{}C)$ the same solution as above.
Sonnhard.

hello, for (iv) setting $ x \minus{} y(x) \equal{} u(x)$ we get the new equation
$ 1 \minus{} u(x)^2 \equal{} u'(x)$ with the solution
$ u(x) \equal{} \tanh(x \plus{} C)$
from here we get easily $ y(x)$.
Sonnhard.


what are the main types of proofs you have to know for olympiad problems?

so far i have:

1. contradiction
2. contrapositive
3. induction

Techniques are more important than which general method you use.  Many contradiction proofs can be phrased as a straight proof without much extra effort.


Hi!

Let $p \geq 5$ be a prime number. Show that 24 divides $p^2 - 1$.

:)

[hide="ansewr"]ferstely prove that 3 divides $p^2 - 1$.  8 divides $p^2 - 1$.
and deduce that 24 divides $p^2 - 1$.[/hide]

[hide]
FLT tell us that $3|p^2-1$ . Since all prime has the form $2k+1$ hence $p^2-1=4k(k+1)$ which is obvious that it is divisible by $8$ . Hence we are done .
[/hide]

[hide="My less-cryptic answer"]We want to prove 24 divides $p^2-1$ for all primes $p\ge 5$. First, $p$ is odd.
Now, $p^2-1=(p-1)(p+1)$
Because $p$ is prime larger than 3, it cannot be divisible by 3
$p-1,p,p+1$ are three consecutive integers, one of them must be divisible by 3. It cannot be $p$. So either $3|p-1$ or $3|p+1$, and so $3|(p-1)(p+1)$
Because $p$ is odd, $p-1$ and $p+1$ are both even. One of them must be divisible by 4. So $8|(p-1)(p+1)$.
Therefore, $24|p^2-1$[/hide]

I love this problem!  :lol:
[hide="Elegant, in my opinion"]Every prime $p \ge 5$ can be written in the form:
$p = 6k \pm 1$
So $p^2 - 1 = 36k^2 \pm 12k + 1 - 1 = 36k^2 \pm 12k$
[u]Case 1[/u]  (basically considering $k \equiv 0 \mod 2)$
If $k$ is of form $2n$, then we have:
$p^2 - 1 = 144n^2 \pm 24n$, which is divisible by $24$.
[u]Case 2[/u](basically considering $k \equiv 0 \mod 2)$
If $k$ is of form $2n + 1$, then we have:
$p^2 - 1 = 144n^2 + 144n + 36 + 24n \pm 12$
$= 144n^2 + 168n + 48$ or
$= 144n^2 + 168n + 24$, which are both divisible by 24.
Edit: "voilà!" :) [/hide]

Well, thanks to all of you!

I'm glad you liked it 1234567890 :)
By the way, it is "voilà" :P

Last comment .. $p$  need not be prime,  any odd number greater than 3 and not divisible by 3 (for p)  will do!
Edited Later  to add notdivisible by 3.

To what are you referring to, Gyan? The original problem, or someone's solution?

[quote="10000th User"][quote="Gyan"]Last comment .. $p$  need not be prime,  any odd number greater than 3 will do![/quote]
Counterexample: $9^2-1$ is not divisible by 24[/quote]

Change it to "any odd number not divisible by 3" and it works :P

[quote]Change it to "any odd number not divisible by 3" and it works[/quote]
Yes :blush: .. Edited the original. :)


Mr. O'Neil always leaves an $ 18\%$ tip, computed on the price of a meal before taxes and rounded to the nearest penny. The state in which he lives adds a sales tax to all meals of $ 6\%$ and rounded up to the next whole number of cents. The menu price for Mr. O'Neil's meal is $ \$12.95$. What is the total number of dollars he pays for the meal including the tip and taxes? Express your answer as a decimal rounded to the nearest cent.


algebra
nullstellensatz
3D geometry
IMO
IMO 2007
IMO Shortlist
Gerhard Woeginger

Let $ n$ be a positive integer. Consider
\[ S \equal{} \left\{ (x,y,z) \mid x,y,z \in \{ 0, 1, \ldots, n\}, x \plus{} y \plus{} z > 0 \right \}
\]
as a set of $ (n \plus{} 1)^{3} \minus{} 1$ points in the three-dimensional space. Determine the smallest possible number of planes, the union of which contains $ S$ but does not include $ (0,0,0)$.

[i]Author: Gerhard Wöginger, Netherlands [/i]

[quote="Valentin Vornicu"]Let $ n$ be a positive integer. Consider
\[ S = \left\{ (x,y,z) \mid x,y,z \in \{ 0, 1, \ldots, n\}, x+y+z > 0 \right \}\]
as a set of $ (n+1)^{3}-1$ points in the three-dimensional space. Determine the smallest possible number of planes, the union of which contains $ S$ but does not include $ (0,0,0)$.[/quote]
answer is 3n,isnt it

Yup, many people got the value and the example (including me), but failed to show it's the extreme value.  :(
I think this problem is as hard as yesterday's problem based on what I heard from the China IMO Team.

Let there be k planes:

        $ a_{1}x \plus{} b_{1}y \plus{} c_{1}z \equal{} d_{1}$,
         ......
        $ a_{k}x \plus{} b_{k}y \plus{} c_{k}z \equal{} d_{k}$,

and consider polynomial $ P(x,y,z) \equal{} (x \plus{} y \plus{} z)\prod_{i \equal{} 1}^{k}{(a_{i}x \plus{} b_{i}y \plus{} c_{i}z \minus{} d_{i})}$.
We look at subsets of set {0,1,2,...,n} and there exists $ t_{1},t_{2},t_{3}$ such that $ t_{1} \plus{} t_{2} \plus{} t_{3} \equal{} k \plus{} 1$ such that coeficient of $ x^{t_{1}}y^{t_{2}}z^{t_{3}}\neq 0$ (if it's doesn't exists then $ t_{1} \plus{} t_{2} \plus{} t_{3}$ is smaller and we get sharper inequality). Using [url=http://www.mathlinks.ro/viewtopic.php?t=19496]the Combinatorial Nullstellensatz[/url] we get condradiction unless $ k\geq 3n$. We can easily construct equaility case...

This is solution of student from Serbia Marko

I haven't used the Nullstellensatz much, so I may just be missing something here, but where exactly does the contradiction arise?

For example, why is it not possible that $ k=3n-1$ and  $ t_{1}=2n-1, t_{2}=t_{3}=\frac{n}{2}$ (in which case the corresponding set $ S_{1}$ is too large to fit inside $ \{0, \dots n\}$)?

[quote="vedran6"]Let there be k planes:

        $ a_{1}x \plus{} b_{1}y \plus{} c_{1}z \equal{} d_{1}$,
         ......
        $ a_{k}x \plus{} b_{k}y \plus{} c_{k}z \equal{} d_{k}$,[/quote]

This solution was given by Radchenko Danylo, from the Ukraine team: Let
$ P(x,y,z) \equal{} \prod_{i \equal{} 1}^{k}{(a_{i}x \plus{} b_{i}y \plus{} c_{i}z \minus{} d_{i})} \minus{} \alpha\prod_{j \equal{} 1}^{n}{(x \minus{} j)(y \minus{} j)(z \minus{} j)}$
where $ \alpha$ is choosen so that P(0,0,0)=0. Then, if $ k < 3n$ then coefficient at $ x^{n}y^{n}z^{n}$ is not zero, so from [url=http://www.mathlinks.ro/viewtopic.php?t=19496]the Combinatorial
Nullstellensatz[/url] it follows that for $ S_{1},S_{2},S_{3} \equal{} \{0,1,...,n\}$ There must be an $ x_{0}\in S_{1}, y_{0}\in S_{2}, z_{0}\in S_{3}$, so that $ P(x_{0},y_{0},z_{0}) \neq 0$. Contradiction.

This is a special case of a result of Alon and F\"uredi from 1993. It says that if $ n_{1},\ldots,n_{k}\geq 0$, then the minimal number of hyperplanes in $ \mathbb{R}^{k}$ needed to cover all but one points of the grid $ 0 \leq x_{i}\leq n_{i}$, $ 1 \leq i \leq k$, is $ n_{1}+\cdots+n_{k}$. This is proven in exactly the same way by the polynomial method. Another special case was discussed in the mathlinks forum long ago, where I showed the same solution: http://www.mathlinks.ro/viewtopic.php?search_id=1233358814&t=20525 . Strange then that this appeared on IMO, all the more that quite a lot of ML members know the combinatorial nullstellensatz from the numerous discussions here.

Well, you dont need CNS: Just consider the finite differences (P(x+1,y,z)-P(x,y,z) and similarly for y and z) and take it n times in each direction. by induction, you see that they are zero on rectangular parallelepipeds except for one vertex. thus, at the end, we get something nonzero on (0,0,0). on the other hand, if the degree was too small(less than 3n), it must be the zero polynomial(observe that finite differences decrease the degree by at least one). contradiction.

Thus, i think the point with this problem is not CNS but the idea of writing down that polynomial. Knowing CNS doesnt really help.

[quote="Peter Scholze"] Knowing CNS doesnt really help.[/quote]

Except that it solves the problem as it becomes instantly clear which polynomial to write down :roll:. Really, what you wrote is essentially CN in slight disguise, without the redundant full generality (but I agree that a more adequate description would be "the polynomial method of Alon" instead of "the CN").

Has anyone found a combinatorial solution?

I have a simple combinatorial solution for the two-dimensional case, but I don't get it generalized to the three-dimensional case needed here. Maybe someone can help?
My idea for the 2-D case is as follows. Let S_{m,n} be the (n,m)-"cube", ie, the collection of all pairs (x,y) with x in 0,1,...,m and y in 0, ..., n, but excluding (0,0), and let T_{m,n} be the boundary (all such points with x in {0,m} and y in {0,n}. So T_{m,n} has 2m+2n-1 points. Now consider a collection of "hyperplanes" (lines) not containing (0,0) and covering all points  in S_{m,n}. We want to show that there are at least m+n lines needed. Now, if there is a line that covers more than two points from the boundary, then it is a boundary line, hence one of the two boundary lines through (m,n). In that case, the other lines lines cover S_{m,n-1} or S_{m-1,n}, and the desired result follows by induction. On the other hand, if all lines cover at most two boundary points, then there are at least (2m+2n-1)/2 lines, hence at least n+m lines.
Can perhaps a similar idea work in 3-D?

I couldn't spent much time today on the problems but my idea was a recursion after proving it for two dimensions
I think I got a solution in that direction but I'm gonna check it and post again tomorrow. :)

[quote="mszew"]How about the cube with $ \{0,1,\dots,n+m\}$ without the cube with $ \{0,1,\dots,m-1\}$ and $ m\in \mathbb N$?

Is it still $ 3n$ ?[/quote]

Probably yes... However, you need to adjust your notation slightly: to cover the cube with $ \{0,1,\ldots,n+m\}$ without the cube with $ \{0,1,\ldots,m\}$, you need(?) at least $ 3n$ planes. The CN argument can be modified to show that you need at least $ 3n-3m$ planes, which is probably not sharp unless $ m=0$ (the original problem). More generally, to cover an $ n_{1}\times \cdots \times n_{k}$ grid without an $ \ell_{1}\times \cdots \times \ell_{k}$ subgrid, one needs at least $ \sum_{i=1}^{k}\max (n_{i}-2\ell_{i}+2, 0)$ hyperplanes. Perhaps this bound may be improved to $ \sum_{i=1}^{k}(n_{i}-\ell_{i}+1)$ in the case that each $ n_{i}> \ell_{i}> 0$, but I can't show this at the moment.

Let me demonstrate the version with $ 3n$ and $ 3$ dimensions (the general case being only notationally different). Let $ a_{i}x+b_{i}y+c_{i}z = d_{i}$ be the planes, $ 1 \leq i \leq s$; and suppose $ s < 3n-3m$. Let $ P(x,y,z) : = \prod_{i=1}^{s}(a_{i}x+b_{i}y+c_{i}z-d_{i})$, and let $ Q(x,y,z)$ be the polynomial, of degree $ \leq m$ in each variable $ x,y,z$, obtained from $ P(x,y,z)$ modulo the relations $ \prod_{j=0}^{m}(x-j) = \prod_{j=0}^{m}(y-j) = \prod_{j=0}^{m}(z-j) = 0$ (that is, we replace successively each $ x^{m+1},y^{m+1}, z^{m+1}$ with linear combinations of smaller powers using these relations, until there is no monomial left with partial degree $ > m$). By construction, $ Q(x,y,z) = P(x,y,z)$ for $ x,y,z \in{0,\ldots,m}$. Let as well $ R(x,y,z)$ be the polynomial, again of degree $ \leq m$ in each variable, obtained from the polynomial $ S(x,y,z) : = \prod_{j=m+1}^{n+m}(x-j)(y-j)(z-j)$ modulo these same relations. Once again, $ S(x,y,z)$ and $ R(x,y,z)$ take the same values on $ \{0,\ldots,m \}^{3}$.

Consider the polynomial $ T(x,y,z) : = P(x,y,z) R(x,y,z)-Q(x,y,z)S(x,y,z)$. The first summand, $ PR$, has total degree $ s+\deg{R}= s+3m < 3n = \deg{S}\leq \deg{QS}$, and in $ QS$ the partial degrees in $ x,y,z$ are all $ \leq n+m$. The contradiction follows from the combinatorial nullstellensatz, since $ T(x,y,z)$ vanishes identically on $ \{0,1,\ldots,n+m\}^{3}$.

The question you pose remains open to me, but [needless to say] I might well have missed the key that could make your question an easy prey to a modified CN argument.

In another thread, Darij implied that a solution using [i]Combinatorial Nullstellensatz[/i] would obtain only 1 point.  Is this true (readers' opinion, of course)?  If his claim is true, would providing a proof of [i]Combinatorial Nullstellensatz[/i] during the exam obtain 7 points (if there are no mistakes)?  I am curious if/how such theorems are accepted during such competitions.  Also, I would like to express my opinion about this problem; In my opinion, this is a very beautiful problem, but knowing this theorem trivializes it.

It seems that approaching the plane case will not give the student any points (unless he can somehow successfully relate it with the 3-dimensional case).

[quote="boxedexe"]In another thread, Darij implied that a solution using [i]Combinatorial Nullstellensatz[/i] would obtain only 1 point.  Is this true (readers' opinion, of course)?  If his claim is true, would providing a proof of [i]Combinatorial Nullstellensatz[/i] during the exam obtain 7 points (if there are no mistakes)? [/quote]

Surely, if you prove everything you use during the exam (e.g. a clean proof of the combinatorial nullstellensatz, but not the full version is needed), you'll obtain 7 points :). You might initially solve the problem with CN (rather: note that it's a direct consequence), and then describe the polynomial proof in such self-contained manner as to even leave no sign that you [i]knew[/i] the theorem. (E.g. as in Peter's proof above). About demonstrating that this problem is a consequence of a theorem called [i]combinatorial nullstellensatz[/i] (and stating that theorem), I wouldn't be surprised if you receive just 1 point (probably this is what is going to happen; those people there are very jealous about theorems trivializing their problems... ). But that would be a stupid thing to do: if you knew that CN does it, why not at least sketch a proof of CN? (this is problem 6 on IMO, after all...) Or, if you don't remember a proof or don't want to mention CN, you at least know that the problem admits an algebraic solution, and you know which polynomial to consider to obtain a contradiction -- and may be able to carry the idea to a self-contained solution avoiding explicit mention of CN -- again, such is Peter's solution above (and I suspect that the authors' solution is along similar lines, if not exactly the same! For this is simply an algebraic combinatorics problem, it is the way it was created! )

The comment in the brackets restrains me from admiring such problem at IMO. Not only it is merely a special case of the algebro-combinatorial result of Alon and Furedi stated above in this thread, but a similar problem was stated and solved on this forum in the link I provided above! It is my opinion that there may well not be a natural combinatorial solution.

I have two separate comments.

[b] 1. [/b]

[quote="shunka"]I think anyone that solves the problem (using CN or not) should get 7 points.

Knowledge should be rewarded.

It is not the fault of the competitors that the Committee made such a poor choice in the problem selection phase.

Whether they knew about CN or they did not know about it, it seems that choosing the problem was their error in either case.[/quote]

I agree completely. However, I think that, in reality, if you use CN as frivolously as you use, e.g., the cosine theorem in a geometry problem, they're going to reward you with 1 or 2 points (problems 3 & 6 are usually graded harsher on IMOs then the other problems). Absolutely this problem is a poor choice by the Committe. The more I think about a combinatorial approach, the more I realize that this is totally not a combinatorial problem: it was [i]invented[/i] (i.e. discovered) in 1993 (Alon - Furedi) as an application of the (algebraic) polynomial method of N. Alon, and I don't believe that it has a combinatorial meaning, at least in the general case (for the 3-D case the proposers may have found a combinatorial solution, I don't know yet; but I doubt it would generalize. I actually find it quite likely that all known proofs are algebraic!). Those planes and their incidence with the lattice points have absolutely no "natural" combinatorial meaning, which is why this problem is so difficult to organize combinatorially [i]ad hoc[/i]. Some people call such problems [i]algebro-combinatorial[/i], some put them in the realm of [i]combinatorial geometry[/i] (both are reasonable: the known solutions are algebraic, while the objects of study, planes and integral points on them, are geometric)... And some would call such problems [i]artificial[/i].

But [or [i] moreover[/i]] such questions abound in the literature on combinatorial geometry/combinatorial group theory/algebraic combinatorics. This is not a generalization, but a (more interesting, actually) finite analog: for a finite abelian group $ A$ with identity $ 0$, what is the minimal number of cosets [of subgroups $ H < A$] whose union is $ A \setminus \{0\}$? For the answer and proof, and similar questions, here is an interesting article for the interested reader: http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3Amath%2F0411244 .

[b]2. [/b] Above was proposed the following generalization for which I proved a weaker version:

[i] To cover the cube with $ \{0,1,\ldots,n+m\}$ without the cube with $ \{0,1,\ldots,m\}$, we need at least 3n planes. [/i]

I showed that we need at least $ 3(n-m)$ planes. But actually, the proof shows more:

[i] Suppose that: (1) we have covered all points of the cube with $ \{0,1,\ldots,n+m\}$ except possibly some points of the cube with $ \{0,1,\ldots,m\}$; and (2) we have [b]not[/b] covered at least one point of the cube with $ \{0,1,\ldots,m\}$. Then we have used at least $ 3(n-m)$ hyperplanes. Similarly for the obvious generalization. [/i]

Indeed, all that we used about the polynomial $ Q$ constructed in the proof that it was not the zero polynomial. This follows from the assumption that $ Q$ does not vanish on the cube with $ \{0,1,\ldots,m\}$. But it suffices to assume that $ Q$ does not vanish at just one point to reach the conclusion. Again, this is probably not sharp in general.

That said, I believe one can prove the sharp result conjectured by mszew as follows. I use the notation for my proof of the special case above. Instead of constructing $ Q$ and $ R$ as we did, construct ("kinds of inverses") polynomials $ Q, R$ of partial degrees all $ \leq m$ as follows: (i) $ PQ$ takes the value $ 1$ on the cube with $ \{0,1,\ldots,m\}$ (here we use with full force that $ P$ does not vanish on this cube, i.e. that the union of our planes is exactly the cube with $ \{0,1,\ldots,n+m\}$ minus the cube with $ \{0,1,\ldots,m\}$); and $ RS$ as well takes the value $ 1$ identically on the cube with $ \{0,1,\ldots,m\}$. Then $ R$ is a concrete polynomial (with concrete coefficients independent of the given covering), because $ S : = \prod_{j = m+1}^{n+m}(x-j)(y-j)(z-j)$ is. And I think we may show that the coefficient of $ (xyz)^{m}$ in $ R(x,y,z)$ is nonzero. If this is the case, then we may apply CN to the polynomial $ PQ-RS$ and conclude the proof! Thus, modulo my computational claim, we have two different generalizations of the initial problem, still in dimension $ 3$ and with cubes rather than general parallelepipeds!

But the generalizations to such questions are countless.

[quote="darij grinberg"][quote="boxedexe"]In another thread, Darij implied that a solution using [i]Combinatorial Nullstellensatz[/i] would obtain only 1 point.[/quote]

I didn't say this! I said that the things I had done (generalizing to rectangular parallelpipeda instead of cubes, proving it in the 2D case, and writing "looks like Cauchy-Davenport" on the scratch paper) are likely to get you 1 point. But this seems to be not the case.

  Darij[/quote]
Sorry, it seemed to me that you were implying that a solution using such a theorem would obtain 1 point.  Apparently, I misunderstood what you wrote.

Im PER1, today, i knew that i have 0 in question 6; back to the problem, i find my mistake and now i present the combinatorial solution:
Induction on $ m+n+p$:"If S={(x,y,z), such that $ 0\leq x\leq m$;$ 0\leq y\leq n$;$ 0\leq z\leq p$,$ x+y+z>0$}, then is neccesary $ m+n+p$ planes that cover S but not O(0,0,0)".
Prove: If $ m+n+p=1$, is trivial.
Suposing that for $ m+n+p=N$, is true, then for a configuration with $ m+n+p=N+1$, we have:
Denote $ A, B, C$ the corners of the parallelepiped, and let $ A_{1}, B_{1}, C_{1}, O_{1}$ their respectives opposite corners of $ A,B,C,O$. There are at least n planes "cutters" that cut OB(if two points of OB and S are in a plane, then O belong to the plane). We take a plane "cutter" by each one of this points(of OB and S)
Consider the figure G formed for the points of S, such that $ x=0, x=m, z=0, z=p$, as 2m+2p rows of (n+1) poimts, for each plane "cutter" and each row(distinct to OB), at most they share a point(Either, O belongs to the plane), then we can chose a point of every row(distinct to OB) which not belong to this n planes "cutters", and denote this points as "cheveres points".
If a plane is O_1B_1AC_1, we are done by the inductive hippotesis, similarly, if a plane is O_1B_1CA_1.
Either, each plane parallel to OB have at most 2 "cheveres points", then there are at least $ m+p(=[(2m+2p-1)/2])$ planes parallel to OB, hence is neccesary $ m+n+p=N+1$ planes for cover S but not O. Finally is sufficient m+n+p planes, consider the plnes:P_r={x+y+z=r}, 0<r<=m+n+p. For m=n=p, the answer is 3n.

[quote="billzhao"]http://www.imo2007.edu.vn/images/Category/Category_1610107.JPG

Well, it looks like that the intended solution is the CN-motivated one. No surprise there ...[/quote]

Thanks, Yufei, for pointing this out. That makes it impossible for those who don't know CN to solve the problem.
A problem is bad if it places the contestants in unequal situation.
I think this is not olympic-type problem, it is rather guess-what-I-am-thinking problem.

Let's look at the planes introducted from anywhere but not of origin (in fact, any of these planes must pass through based on a condition of the problem) and let's take projections of these planes on to a plane one from x-y, y-z or x-z planes. Denote by [i][b]n[/b][/i](d) the number of the points of the set S the plane d passes through. Then [i][b]n[/b][/i]([i]f[/i](n +1) - [i]f[/i](n)) <=(2n +4)/3 over this projection obviously (if it is not obvious then induct on n easily gives the result we want) then it is okay that f is a second degree polynomial (because informs the number of the introductory points this plane has depending on the variable k). So [i]f[/i](n) <=(n^2)/3 +n +1.
The union of the planes must contain the set of points S so [b]U[/b]d >=n^3 +3n^2 +3n [hide]d informs the planes[/hide] so easily Sum(d) >=n^3 +3n^2 +3n.[hide]d informs the planes[/hide] We know and conclude easily that number of sets is determined Sum(S)/[i]f[/i](n) and it is >= 3n so the minimum number of the planes is 3n openly.

vess: I think you are making mszew's ques harder.  To cover {0,1,……,n+m} without {0,1……m-1} means(necessarily) to cover {m-1,m,m+1,…m+n} without {m-1}. As already shown, the least num is 3(n+1). But these 3n+3 planes can be easily found, they are x,y,z=m,m+1……n+m

I think it is a very hard problem  I like this problem and its solution it is nice
  I saw a problem a few days ago:
      "Let n be a positive integer. Consider S={(x,y):-1<x,y<n+1  x,y are integer} as a set of (n+1)^2-1 points in the two-    dimensional space. Determine the smallest possible number of lines, the union of which contains S but does not include (0,0)."
  It can be proven by induction but its generalized from m,n and we induct on (m+n).I think this problem is nice too. :lol:  :lol:

In case D.Radchenko didn't write the proof of cns and just quoted it, the fact that he recieved 7pts for the problem would imply it's totally ok to use this theorem without proving it - so I wonder if anyone knows or if he himself is around

Here is a proof that looks simpler that the official solution, even though it uses the same principle. It is similar to a 'A short proof of Combinatorial Nullstellensatz' given by Mateusz Michalek. I'd be really grateful if someone could check this for mistakes.

Let $n_1, \ldots, n_d \in \mathbb{N}$. Let
            \[S(n_1, \ldots, n_d) = \{(x_1, \ldots, x_d)\, | \, 0 \leq x_i \leq n_i,\, x_1 + \ldots + x_d > 0\}.\]
            Consider a polynomial $P$ that [i]solves[/i] $S(n_1, \ldots, n_d)$, that is, $P(x_1, \ldots, x_d) = 0$ for all $(x_1, \ldots, x_d) \in S$ and $P(0, \ldots, 0) \neq 0$. We show by induction that $\deg(P) \geq n_1 + \ldots + n_d$. Assume w.l.o.g. that $n_1 > 0$. Apply polynomial division to write
            \[P = P'(x_1 - n_1) + R,\]
            such that $\deg(R) \leq \deg(P)$ and $R$ does not contain $x_1$. It is easy to see that
            \[R(x_1, \ldots, x_n) = R(n_1, x_2, \ldots, x_n) = P(n_1, x_2, \ldots, x_n).\]
            This means that $R(x_1, \ldots, x_d) = 0$ for all $(x_1, \ldots, x_d) \in S(n_1, \ldots, n_d) \cup \{0\}$. Clearly then $P'$ solves $S(n_1 - 1, \ldots, n_d)$. By induction we are done since
            \[\deg(P) \geq \deg(P - R) = \deg(P'(x_1 - n_1)) = \deg(P') + 1.\]

I realized that my solution is the same as https://artofproblemsolving.com/wiki/index.php?title=2007_IMO_Problems/Problem_6

How about a slightly twisted version of this problem: what if you remove all internal points of the cube and disallow any plane to cover a whole face? What is the minimal number of planes then?

Looks like the polynomial solution might not work any more: it is trivial to show that the number is still 2n in 2 dimensions, and yet you can easily come up with a polynomial of a smaller degree that covers all of the external points (think circles). So what do we do now?

Not sure if that's in the spirit of the forum, but I'd offer a generous bounty for a solution of the twisted version above. Just get in touch with me directly.

[quote=Peter Scholze]There appears to be a flaw in your argument: You first consider n plane "cutters" and the "cheveres points" that the imply. However, you forget that there might be additional plane "cutters" that could contain quite a few of those "cheveres points".

Note that my solution directly solves the more general problem given here, in the generalized form:

 Suppose that: (1) we have covered all points of the cube with $ \{0,1,\ldots,n+m\}$ except possibly some points of the cube with $ \{0,1,\ldots,m\}$; and (2) we have not covered at least one point of the cube with $ \{0,1,\ldots,m\}$. Then we have used at least $ 3n$ hyperplanes. Similarly for the obvious generalization.[/quote]

 This is the Field Medalist  :o

We claim the answer $3n$. As an example take a union of planes $x+y+z=c$ for $c=1,2,\dots ,3n.$

Assume FTSOC that planes $a_ix+b_iy+c_iz+d_i=0$ for $i=1,2,\dots ,k<3n$ cover $S;$ consider polynomial $$P(x,y,z)=(x+y+z)\prod_{i=1}^{k} (a_ix+b_iy+c_iz+d_i)$$ and set $\mathcal A=\left \{ 0,1,2,\dots ,n\right \} .$ Since $d_i\neq 0,$ at least one monom $x^py^qz^r$ of $P$ (with $p+q+r\leq k+1$) has a nonzero coefficient, hence by the Combinatorial Nullstellensatz there exist $s_{1,2,3}\in \mathcal A$ such that $P(s_1,s_2,s_3)\neq 0.$ Since $P(0,0,0)=0,$ we get the obvious contradiction, as desired.

The answer is $3n$. For the construction, use the planes $x=1, 2, \ldots, n$, $y=1, 2, \ldots, n$, and $z=1, 2, \ldots, n$. Now we show that no number of planes $k<3n$ works. Assume for contradiction that a configuration with $k<3n$ planes exists. Let these planes be $a_ix + b_iy + c_iz + d_i=0$ for $1 \le i \le k$. Define \[ f(x, y, z) = \prod_{i=1}^k (a_ix + b_iy + c_iz + d_i) \ \text{and} \ g(x, y, z) =  \prod_{i=1}^n (x-i) \prod_{i=1}^n (y-i) \prod_{i=1}^n (z-i).  \]
Furthermore, let \[ P(x, y, z) = f(x, y, z)-\frac{f(0, 0, 0)}{g(0, 0, 0)} g(x, y, z). \] Note that $P$ vanishes on $\{0, 1, \ldots, n\}^3$. However, the coefficent of $x^ny^nz^n$ in $P$ is $-\frac{f(0, 0, 0)}{g(0, 0, 0)}$ which is nonzero, so the combinatorial Nullstellensatz yields a contradiction, which completes the proof.

50 mohs simulator.

[b][color=pink]Claim:[/color][/b] A construction with $3n$ planes works.
[i]Proof.[/i] Taking $x + y + z = i$ for $i = 1, 2, \dots, 3n$ works. Taking $x = i, y = i, z = i$ for $i = 1, 2, \dots, n$ does as well. $\blacksquare$

[b][color=pink]Claim:[/color][/b] $3n$ is optimal.
[i]Proof.[/i] FTSOC there exists a construction with $N < 3n$ planes. Let the planes be of the form $a_ix + b_iy + c_iz + 1 = 0$.
Define \begin{align*} A(x,y,z) &= \prod_{i=1}^n (x-i)(y-i)(z-i) \\ B(x,y,z) &= \prod_{i=1}^N \left( a_i x + b_i y + c_i z + 1 \right). \end{align*}
Both of these polynomials vanish entirely on $S$ with $A(0, 0, 0) = (-1)^{n} (n!)^3$ and $B(0, 0, 0) = 1$.
However, then $A - (-1)^{n} (n!)^3 \cdot B(0, 0, 0)$ is nonzero, and vanishes on all of $\{0, 1, \dots n\}^3$, contradiction of combinatorial nullstellensatz since it has nonzero coefficient $x^ny^nz^n$. $\blacksquare$


complex analysis
complex analysis unsolved

Let D be a subset of C consisting of one element.  Prove that D is closed.

Ta

Hint: start by considering a sequence $(z_{n})_{n\ge 1}$ such that $z_{n}\in D$ for all $n$.


induction
number theory proposed
number theory

Prove that for every $n\in N$ there is number witch digits are all $1$ or $2$ and witch is divisable with $2^n$.

This is my solution but it is ugly one

[hide]
We find that nubmers for $n=1,...,10$ they are$2, 12, 112, 2112, 22112, 122112, 2122112, 12122112, 212122112, 1212122112$
Now see that $1212122112=2^{10}\cdot (4k+1)$ so we construct number with induction

from that we have base
assume that for $n=k$ there is $k$-digit number witch is divisable with $2^k$ but not with
$2^{k+1}$ and witch is equal $2^k(4l+1)$ , add to that number $1\underbrace{00..00}_k$ and that number is now divisable with $2^{k+1}$ and it is $2^{k+1}(4m+1)$ so it has needed atributes.... [/hide]

Hyper well-known AND posted in the wrong forum. Well done.

Since you posted it in the Unsolved problems section WITH your solution!

It should be in the proposed problems section. I moved it.

Ok...
I posted it there because I didn't like my solution, but you're matbe right....

I don't understand one thing, and I didn't understand it while I was solving this problem (I couldn't prove it) so it took me about an hour to solve it... here it is
Why $2^nk\equiv_{2^{n+1}} 2^n$ when $k$ is odd

:blush:  :blush:  :blush: OOOps... I get it now... thanks for link......


[hide="Easy"]\begin{eqnarray*}&& \sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\\ &\iff& |\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1\\ &\iff& 2\leqslant\sqrt{x-1}\leqslant 3\\ &\iff& 5\leqslant x\leqslant 10\end{eqnarray*}[/hide]


P1:
Do there exist polynomials p(x) and q(x) with integer coefficients such that
p(x), p(x) - q(x) and p(x) + q(x) are all squares, but p(x) is not a multiple of
q(x) or vice versa?

P2: There is an infinite sequence of distinct positive integers a1, a2, a3, ...
such that the real polynomial p(x) satisfies p(a1) = 0, p(a2) = a1, p(a3) = a2, ... . What are the possible values for the degree of p(x)?

P3: p(x) is a polynomial with integer coefficients. The sequence of integers a1, a2, ... , an (where n > 2) satisfies a2 = p(a1), a3 = p(a2), ... , an = p(an-1), a1 = p(an). Show that a1 = a3.

P4: Show that there is no real polynomial of degree 998 such that :
p(x)2 - 1 = p(x2 + 1) for all x.

P5: Find all polynomials p(x) with real coefficients such that for some n > 0
p(x+2) - p(x+3) + 2p(x+4) - 2p(x+5) + ... - n p(x+2n) + n p(x+2n+1) = 0 holds for infinitely many real x.

P6:
Let A be the set {2,7,11,13}. A polynomial f with integer coefficients
possesses the following property: for each integer n there exits p in A such that p/f(n). Prove that there exits p  in A such that p/f(n) for all integers n.

P7: Prove that  polynomial:P(x)=[(x-a1)(x-a2)...(x-an)]2 +1 is irreduciblefor all distinct integer numbers:a1,a2, ..an

Poland 2003:
The polynomial W(x)=x4 -3x3 +5x2 -9x is given. Determine all pairs of diffrent
integer a,b satisfying the equation: W(a)=W(b).
Ive a soln, and you? :D

Try to solv.I've some soln for these problems.

SOLUTION for P7!

lets consider that there exist P1,P2 from Z[x] so that P=P1*P2, P1 and P2 have equal dominatnt coeficients with 1. Because p(x)>0 for any x fom R we have that P1 and P2 don't have real roots and p1(x)>0 and p2(x)>0 for anyR.
1) lets consider that degree p1<degree p2 and so degree p1<n and degree p2>n. Now we take x=a1, then x=a2 and so on .. x=an we get that
p1(a1)p2(a2)=1;..;p1(an)p2(an)=1. because p1(x)>0 and p2(x)>0 we get that p1(a1)=p2(a1)=1 ... p1(an)=p2(an)=1 (1)

the polinomial with degree <n p1(x)-1 has n roots a1,a2,a3..,an so it is identic nul. So p1(x)=1 and p2(x)=p(x) so it is irreductioble.
2) if degree p1=degree p2=n we get from relations (1) that p1(x)-1=(x-a1)(x-a2)..(x-an)=p2(x)-1 and we have that p1(x)=p2(x)=(x-a1)(x-a2)..(x-an)+1 which give p(x)=[(x-a1)(x-a2)..(x-an)+1]^2 which is fals!!!

so from 1) and 2) we get that p is irreductible!!!

cheers mate!

About P5:

We know that if a polynomial has degree >0 then it becomes strictly monotonic on the interval (c,infinity) for a large enough c. If our expression is 0 for an infinity of values of x, then it is a polynomial with an infinity of roots, so it's=0 for any x. We choose an x such that x>c. Let's assume that P isn't constant and becomes strictly increasing on (c,infinity), because it's the same for strictly decreasing. Then all the diffrences in the left part are negative, so we get that the expression is smaller than 0 for that specific x, but we know that it is 0 for any x, so contradiction, so P must be a constant polynomial. I guess the solution is: P(x)=a=consta.nt

And about P6

Let's assume that there are integer numbers a,b,c such that: 2 does not divide f(a), 7 does not divide f(b), 11 does not divide f(c). Let r be a residue modulo 13. According to the Chinese remainder Thm, we can find xr=a(mod 2)=b(mod 7)=c(mod 11)=r(mod 13). Then 2, 7, 11 don't divide f(xr) so 13 must divide f(xr)=f(r)(mod 13), so f(r)=0(mod 13), but this can be done for any residue r modulo 13, so 13/f(n) for any integer n. So either 2, 7, 11 or 13 divides f(n) for any integer n.

Hope it's correct.Bye :D

P2

It's obvious that p can't be a constant polynomial, because p(a1)=0 so it would have to be p(x)=0, but then ai=0 for any i, which is impossible. So degree p>=1.

It's a well known fact that an infinite sequence of distinct natural numbers must tend to infinity, so the sequence an tends to infinity. Because p has positive values (the terms af our sequence) for arbitrarily large values of x(also the terms of our sequence, which we found tends to infinity), then limit of p(x) as x tends to infinity is infinity (because it's either infinity or -infinity). Another fact which can be easily found (if we assume that deg p>1) is that p(x) is a) strictly increasing on (c, infnity) for a large enough c; and b) p(x)>x on the interval (d, infinity) for a large enough d, so from a) and b) we found that p(x) is strictly increasing and >x on the interval (c+d, infinity) (we may assume c and d positive).

Take a(n0), a term of our sequence, such that a(n0)>c+d. Such a term exists, because an -> infinity. Then p(a(n0))=a(n0-1)>a(n0)>c+d; p(a(n0-1))=a(n0-2)>a(n0-1)>a(n0)>c+d... and so on, until we find a1>a(n0)>c+d so p(a1)=0>a1, but a1 is positive, so contradiction, so the degree of p must be 1. Such a p with degree 1 exists, for example p(x)=x-1: the sequence is 1, 2, 3, ....

This is for P3 (I know you posted this a long time ago, but the problems are still there, aren't they? :D).

I used the fact that m-n/p(m)-p(n) for any integer numbers m and n. I'm not going to post the rest of it (I'm too lazy), but this idea works out just fine.


Ceva informatii despre onm? Am auzit ca luni este sosirea elevilor ....tot luni pe la ora 6 festivitatea de deschidere si apoi marti proba...:(( trebuie sa avem si noi o zi libera intre ziua in care sosim si ziua probei...


When we simplify radicals, we're really trying to pull out perfect squares, so let's do so here:

$ \sqrt {2} \plus{} \sqrt {98}$

$ \sqrt {2} \plus{} \sqrt {49*2}$

$ \sqrt {2} \plus{} 7\sqrt {2}$

$ 8\sqrt {2}$


[i]Let $ a ; b ;c$ be fixed positive numbers  . Solve the following system of equation :

  $ ax \minus{} aby \plus{} \frac{1}{xy} \equal{} bc^2 \ ; \ abz \minus{} bc^2 x \plus{} \frac{1}{zx} \equal{} a  \  ; \ bc^2 y \minus{} az \plus{} \frac{1}{yz} \equal{} ab$[/i]

Let $ x\equal{}ab(xyz\plus{}u),y\equal{}a(xyz\plus{}v),z\equal{}bc^2(xyz\plus{}w)$, then
\[ a^2(u\minus{}v)\plus{}c^2dw\equal{}0\]
\[ b^2c^2(w\minus{}u)\plus{}dv\equal{}0\]
\[ c^2(v\minus{}w)\plus{}du\equal{}0\], were $ d\equal{}\frac{1}{xyz}$
Diskriminant is $ \minus{}d[a^2c^2\plus{}a^2b^2c^2\plus{}b^2c^6\plus{}c^2d^2]\not \equal{}0$.
Therefore $ u\equal{}v\equal{}w\equal{}0$ or \[ abyz\equal{}1;axz\equal{}1;bc^2xy\equal{}1\]
or $ xyz\equal{}\frac{1}{abc}$ or $ x\equal{}\frac 1c, y\equal{}\frac{1}{bc},z\equal{}\frac{c^2}{a}.$


taghatoe 2 dayereye C1 va C2 ra AB begirid. R rooye kamane AB az C1  va T   rooye kamane AB az C2 gharar darand.  AR va BR dar C va D ba O2 barkhord mikonand . AT va BT dar Q va P  ba O1 barkhord mikonand . (PR,TD)=E . (QR,TC)= F . sabet konid :      AE.BT.BR=AT.BF.AR

avval az hame biaid yekhorde hokme masalaro taghir bedim,khob bebinid tebghe ghovvate noghte $ T$ nesbat be $ C_1$ midoonim:

$ TP.TB\equal{}TA.TQ\Rightarrow \frac{AT}{BT}\equal{}\frac{TP}{TQ}$

hamchenin tebghe ghovvate noghte $ R$ nesbat be $ C_2$ midoonim:

$ RB.RD\equal{}RA.RC\Rightarrow \frac{BR}{AR}\equal{}\frac{RC}{RD}$

khob hala hokme masale az ma khaste sabet konim:

$ AE.BT.BR\equal{}AT.BF.AR$

$ \Rightarrow AE.\frac{BR}{AR}\equal{}BF.\frac{AT}{BT}$

$ \Rightarrow AE.\frac{RC}{RD}\equal{}BF.\frac{TP}{TQ}$

khob pas kafie sabet konim $ \boxed{AE.RC.TQ\equal{}BF.TP.RD}$

----------------------------------------------
khob az inja be bad yekhorde zavie bazi dare,vase hamin behtare ghabl az goftane rahe hal yekhorde sheklo roshan konim ta eshtebahi pish nayad,khob farz konid $ R$ tooye dayere $ C_2$ va $ T$ ham tooye dayere $ C_1$ bashe...

khob too dayere $ C_1$ zaviehaye $ \angle ARP,\angle ABP$ zaviehaye mohatie rooberoo be kamane $ AP$ az in dayeran pas ba ham barabaran yeni $ \angle ARP\equal{}\angle ABP$...

khob hala az tarafi too dayere $ C_2$ zaviehaye $ \angle ABT,\angle ACT$ zavayaye mohati rooberoo be kamane $ AT$ hastan pas ba ham barabaran yeni $ \angle ABT\equal{}\angle ACT$...

khob hala az do rabete bala natije mishe:

$ \angle ARP\equal{}\angle ACT\Rightarrow RP\parallel TC$

be tarighe moshabeh natije mishe $ RQ\parallel TD$,khob pas [b]4zelie $ ERFT$ motavaziolazlas[/b]...

hala ba tavvajoh be dayere $ C_1$ darim $ \angle APR\equal{}\angle AQR$,hamchenin az movazi boodane $ TE,QR$ natije mishe $ \angle TQR\equal{}\angle ATE$ pas darim:

$ \angle APE\equal{}\angle ATE$ yeni [b]4zelie $ APTE$ mohatie[/b]...

khob az mohati boodane $ AETP$ natije mishe $ \angle TAE\equal{}\angle TPE$ amma ba tavajoh be $ C_1$ darim $ \angle BPR\equal{}\angle BAR$ hamchenin ba tavvajoh be $ C_2$ darim $ \angle BAC\equal{}\angle BDC$ pas darim:

$ \angle TAE\equal{}\angle RDC$ (1)

az tarafi tebghe dayere $ C_2$ darim $ \angle ACD\equal{}\angle ABD$ va tebghe $ C_1$ darim $ \angle ABR\equal{}\angle AQR$ va dar nahayat tebghe tavazie $ TD,QR$ natije mishe $ \angle AQR\equal{}\angle ATE$ pas darim:

$ \angle ATE\equal{}\angle RCD$ (2)

$ (1),(2)\Rightarrow \triangle ATE\sim\triangle DCR$

be tarighe moshabeh natije mishe:

$ \triangle RBF\sim\triangle TQP$

khob hala ba tavvajoh be tashabohe mossalasai ke goftim darim:

$ \triangle ATE\sim\triangle DCR\Rightarrow \frac{DR}{AE}\equal{}\frac{RC}{ET}\Rightarrow DR.ET\equal{}AE.RC$

$ \triangle RBF\sim\triangle TPQ\Rightarrow \frac{BF}{QT}\equal{}\frac{FR}{TP}\Rightarrow BF.TP\equal{}QT.FR$

khob hala tarafeine ravabete fogho dar ham zarb konid mishe:

$ DR.ET.BF.TP\equal{}AE.RC.QT.FR$

khob amma 4zelie $ ETFR$ motavaziolazlas pas $ RF\equal{}TE$ pas tasavie bala be tasavie zir tabdil mishe:

$ DR.BF.TP\equal{}AE.RC.QT$

yeni hamoon chizi ke mikhastim...


Successive discounts of $10\%$ and $20\%$ are equivalent to a single discount of:

$\textbf{(A)}\ 30\% \qquad
\textbf{(B)}\ 15\% \qquad
\textbf{(C)}\ 72\% \qquad
\textbf{(D)}\ 28\% \qquad
\textbf{(E)}\ \text{None of these}$

Looking at an example, say, 100.

10% of 100 is 10 and 20% of 90 is 18.  90-18 = 72

72 is 72% of 100.

Thus the answer is $ \boxed{C}$

You're math is correct, but they are asking for the discount percent, so it should be D 28%

Alternatively, let x be the original price. 10% discount is the same as 9/10 and for 20% it will be 8/10. The ending price is equivalent to x(9/10)(8/10) = 72x/100 which is a 28% discount, D.

Assume something costed 100 dollars. Then with each successive discount, it would cost 90 dollars, then 72 dollars. This amounts to a total of 28 dollars off, so the single discount would be 28%

Thus, the answer is D. 28%


Three players each have a red card, blue card and green card. The players will play a game that consists of three rounds. In each of the three rounds each player randomly turns over one of his/her cards without replacement. What is the probability that, at the end of the game, one card of each color was turned over in each of the three rounds? Express your answer as a common fraction.

$ \frac{1}{18}$ See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=397269299&t=228576]here[/url]


How did you go from the second to last step to the last step, surge?

Really what the question is asking for is $\ x^5 + ( \frac{1}{x} )^5$ so we can use the factorization $\ a^5 + b^5 = (a+b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4)$.  Applying that to this problem gives you $\ x^5 + ( \frac{1}{x})^5 = (x+ \frac{1}{x} )(x^4 + \frac{1}{x^4} - ( x^2 + \frac{1}{x^2} ) + 1)$.  The squared and fourth terms in the second half of the factorization can be found easily as shown above.  Also, $\ x + \frac{1}{x}$ can be found by squaring it and substitution.

[quote="WindSlicer"][quote="surge"]I went to the second step to the last step by having three steps in between.  ;)[/quote]

You misread my question :)[/quote]

Indeed, I did. I just plugged the answer back into the original equation and noticed that it won't change for any value of $x$. Since the raising to the 5th power and adding the reciprocal will eliminate the difference in signs.

[quote="José"][quote="Hokkage"]A much nicer question would be $x^2+\frac{1}{x^2}=3$, find $x^4+\frac{1}{x^4}$[/quote]

And what about $x^2+\frac{1}{x^2}=3$, find $x^n+\frac{1}{x^n}$[/quote]See http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=x%2B1%2Fx&t=47110
[quote="Ravi B"]Okay, here is a derivation of the formula.  By the quadratic formula, one solution of $x + 1/x = a$ is
\[ x = \frac{a + \sqrt{a^2 - 4}}{2} \, .  \]
The other solution is the reciprocal
\[ \frac{1}{x} = \frac{a - \sqrt{a^2 - 4}}{2} \, .  \]
Raising both to the $b$th power, we get the formula
\[ x^b + \frac{1}{x^b} = \frac{1}{2^b} \bigl[ {(a + \sqrt{a^2-4})}^b + {(a - \sqrt{a^2-4})}^b \bigr] \, .  \]
This formula can already be considered a solution to your original problem, although you might object to the square roots.  To get rid of them, we can apply the binomial theorem to both parts.  Fortunately, half the terms, the ones involving an odd power of $\sqrt{a^2 - 4}$, will cancel out.  We are left with terms involving an even power of $\sqrt{a^2 - 4}$, which gets rid of the square roots.  Each such term will appear twice, once in each part.[/quote]

[quote="mysmartmouth"][quote="frt"]That's exactly the same as what Zenith Xan did... :roll:[/quote]

If you go back and look at keeyounglee's recent posts, most of the time he copies other people's answers.  I wonder... :D[/quote]

Well, maybe he's spamming a bit, or maybe he just wants to show his work.......


complex numbers
real analysis
real analysis unsolved

For a measurable space (X,B) we can define "complex-valued charges" as maps

v : B --> C  (C = complex numbers)

such that v(null set) = 0 and

v( U E_i ) = sum v(E_n)

for each countable family of mutually disjoint measurable sets (whose union
has finite charge), and the series converges absolutely.

I want to prove some properties of such a thing. For one, I want to show
that any such charge can be represented as v = (m_1 - m_2) + i (m_3 - m_4),
where m_i are positive, sigma-additive (real-valued) measures. My idea is
to consider v = Re(v) + i Im(v), and prove that Re(v) and Im(v) are signed
measures. Then from some results in my book I would know that Re(v) and
Im(v) can be decomposed into a difference of positive measures.

Is this true though? I cannot prove it.


geometry
circumcircle
geometric transformation
homothety
ratio
Euler
function

Let $ A_0,B_0,C_0$ respectively be the midpoints of $ BC,CA,AB$. Consider the homothety with center $ G$, ratio $ k \equal{} \minus{} \frac {1}{2}$. We have $ \mathcal {H}(G,k): \triangle ABC\mapsto \triangle A_0B_0C_0$, which implies that $ \triangle ABC\sim \triangle A_0B_0C_0$. Now consider the homothety with center $ O$, ratio $ r \equal{} 2$. We have $ \mathcal {H}(O,r): \triangle A_0B_0C_0\mapsto \triangle A_1B_1C_1$, which implies that $ \triangle A_1B_1C_1 \cong \triangle A_0B_0C_0\Longrightarrow \triangle A_1B_1C_1\sim \triangle ABC$. Let $ I$ be center of homothety that turns $ \triangle ABC \mapsto \triangle A_1B_1C_1$. It is easy to see that $ \mathcal {H}_{I} \equal{} \mathcal {H}_O\circ \mathcal {H}_G \Longrightarrow$ the ratio of this homothety is  $ \minus{} 1$  and $ I\in OG$- which is the $ Euler$ line wrt $ \triangle ABC$, further, $ I$ is also the center of the $ 9$- point circle $ (\mathcal {E})$ wrt $ \triangle ABC$. Or in other word, $ \mathcal {H}(I, \minus{} 1)$ is a symmedian through $ I$, let me denote it by $ \mathcal {S}(I)$. It is so obvious that $ \mathcal {S}(I): (\mathcal {E})\mapsto (\mathcal {E})$. Therefore $ (\mathcal {E})$ is also the $ 9$- point circle wrt $ \triangle A_1B_1C_1$.
Our proof is completed then.

[b]REMARK-[/b] In my solution, I have considered $ G\equiv S$.

[quote="Moonmathpi496"]Let $ ABC$ be a triangle, $ O$ its circumcenter, $ S$ its centroid, and $ H$ its orthocenter. Denote by $ A_1, B_1$ and $ C_1$ the centers of the circles circumscribed about the triangles $ CHB, CHA$ and $ AHB$ respectively.

Prove that the triangle $ ABC$ is congruent to the triangle $ A_1B_1C_1$ and that the nine-point circle of $ ABC$ is also the nine-point circle of $ A_1B_1C_1$.[/quote]
For any point, let the lowercase letter denote the complex number corresponoding to it. Let $ O$ be the center of the complex plane. Then $ |a|\equal{}|b|\equal{}|c|\equal{}R$, where $ R$ are the circumradius of the triangle.
Since $ h \minus{} 2o \equal{} a\plus{}b\plus{}c$ and $ o \equal{} 0$ we see: $ h \equal{} a\plus{}b\plus{}c$.
Since the circumcenter of the triangle with points in $ (\minus{}a,\minus{}b,c)$ must be $ O$ (since $ |\minus{}a| \equal{} |\minus{}b| \equal{} |c| \equal{} R$) the circumcenter of a triangle with points $ (\minus{}a\plus{}a\plus{}b,\minus{}b\plus{}a\plus{}b,c\plus{}a\plus{}b) \equal{} (b,a,a\plus{}b\plus{}c) \equal{} (b,a,h)$ must be $ a\plus{}b$. But then $ c_1 \equal{} a\plus{}b$ and likewise $ a_1 \equal{} b\plus{}c, b_1 \equal{} a\plus{}c$.
Let $ O_9$ be the nine point circle. It is wellknown that $ 2o_9 \equal{} o \plus{} h \equal{} a\plus{}b\plus{}c$. Consider the function $ f(z) \equal{} 2o_9 \minus{} z$, which is a reflection around $ O_9$. Since $ 2o_9 \equal{} a\plus{}b\plus{}c$ we see that $ f(a) \equal{} a_1, f(b) \equal{} b_1, f(c) \equal{} c_1$. Hence $ \triangle A_1B_1C_1$ is obtained by reflection $ \triangle ABC$ around $ O_9$. From here it is obvious that $ \triangle ABC$ is congruent to $ \triangle A_1B_1C_1$ and that it has the same nine point circle :)

[quote="Moonmathpi496"]Let $ ABC$ be a triangle, $ O$ its circumcenter, $ S$ its centroid, and $ H$ its orthocenter. Denote by $ A_1, B_1$ and $ C_1$ the centers of the circles circumscribed about the triangles $ CHB, CHA$ and $ AHB$ respectively.

Prove that the triangle $ ABC$ is congruent to the triangle $ A_1B_1C_1$ and that the nine-point circle of $ ABC$ is also the nine-point circle of $ A_1B_1C_1$.[/quote]

My solution:
Let $ \omega$ be the circumcircle of $ \triangle ABC$. It is well known that the reflection of $ H$ on $ BC$ lies on $ \omega$. This reflection maps $ O \to A_1$. Similarly for $ B_1$ and $ C_1$. So, A homothety with center $ O$ and ratio $ \frac 12$ sends $ \triangle A_1B_1C_1 \to \triangle A_0B_0C_0$. Where $ A_0B_0C_0$ is the medial triangle of $ \triangle ABC$. So, $ \triangle A_1B_1C_1 \cong \triangle ABC$
Now it is clear the radius of the nine point circles of $ \triangle ABC$ and $ \triangle A_1B_1C_1$ are equal. So it is enough to show that the nine point center of the triangles are the same.

The nine point center $ N$, of $ \triangle ABC$ is the midpoint of segment $ HO$. $ O$ is the orthocenter of $ \triangle A_0B_0C_0$, so it is also the orthocenter $ \triangle A_1B_1C_1$. We shall show that $ H$ is the circumcenter of $ \triangle A_1B_1C_1$. We have $ CH \parallel C_1O$ also, $ CH\equal{}C_1O$. So $ C_1HCO$ is a parallelogram and $ C_1H\equal{}CO\equal{}R$. Analogously we can prove this for the other vertices. Which implies that $ H$ is the circumcenter of $ \triangle A_1B_1C_1$. From which the result follows.

$ \mathbb{QED}$

Homothety $ \mathcal{H}(G, \minus{} \frac {_1}{^2})$ takes $ \triangle ABC$ into its medial triangle $ \triangle A'B'C'$ and homothety $ \mathcal{T}(O,2)$ takes $ \triangle A'B'C'$ into the triangle formed by the circumcenters of $ \triangle HBC,\triangle HCA$ and $ \triangle HAB,$ i.e. reflections of $O$ about $BC,CA,AB.$ The coefficient of composition $ \mathcal{H} \circ \mathcal{T}$ is $ \minus{} 1,$ which implies that $ \triangle ABC$ and $ \triangle A_1B_1C_1$ are centrally symmetric, and the circumcenter of one them is  the orthocenter of the other $\Longrightarrow$ $\triangle ABC$ and $\triangle A_1B_1C_1$ have the same nine-point circle.

Luis, does it follow directly from the homothety relations or should we prove like I did in my solution?
I mean when writing formal solution (like in a contests) how should we describe that  the circumcenter of one , the orthocenter of the another ?

[quote="Moonmathphi496"]how should we describe that the circumcenter of one, the orthocenter of the another ?[/quote]
Homothety between triangles takes orthocenters into orthocenters, centroids into centroids, circumcenters into circumcenters, etc. Note that the circumcenter of ABC is the orthocenter of its medial triangle.

In Moonmathpi's solution, I do not understnad  some things  1.why  the proerty of reflection of orthocentres map  $ O$ to $ A_1$.
2. Why should it be because of the above a homothety of centre $ O$ and ratio $ \minus{}1/2$ map $ A_1B_1C_1$ to $ A_0B_0C_0$?
3. WHy does 2. imply that the trinagle $ ABC$ will be congruent to $ A_1B_1C_1$? If the medial trinagle of a particular trinagle is  homothetic to some trinagle by ratio $ 1/2$, does it mean the other tinagle is congruent to the original one?
I am sorry for these doubts but it will be really helpful if someone helps me out.
Thanks in advance. Please dont ignore this.

I think that the main idea in Moonmathpi's solution is the same as mine proof- the first one. You can see the answer for all you question in my solution.
Best regard

[quote="Agr_94_Math"]In Moonmathpi's solution, I do not understnad  some things  1.why  the proerty of reflection of orthocentres map  $ O$ to $ A_1$.
2. Why should it be because of the above a homothety of centre $ O$ and ratio $ \minus{} 1/2$ map $ A_1B_1C_1$ to $ A_0B_0C_0$?
3. WHy does 2. imply that the trinagle $ ABC$ will be congruent to $ A_1B_1C_1$? If the medial trinagle of a particular trinagle is  homothetic to some trinagle by ratio $ 1/2$, does it mean the other tinagle is congruent to the original one?
I am sorry for these doubts but it will be really helpful if someone helps me out.
Thanks in advance. Please dont ignore this.[/quote]
Here are the answers to your questions:
1. Let the reflection of $ H$ on $ BC$ be $ D$. Then It is very well known (or just prove that these two triangles are congruent as they have two equal angles and an equal side $ BC$) that $ D$ lies on $ BC$. It implies that the circumcircle of $ \triangle BHC$ and $ \triangle BDC$ (and so of $ \triangle ABC$ are equal and their center lies on midperpendicular of $ BC$. So, $ O \to A_1$.

2. It is just clear from the reflection. Just have a look at the diagram carefully.

3. Yes. Simply homothety transforms figures to similar figures and the only change (apart from displacement is the length of the sides). And here each side of medial triangle is half of the original and triangle. And then you know that of you double the side lengths you get a another one. So they are directly congruent.

I hope that I have successfully answered your questions. :)

[quote="Luis González"]Homothecy between triangles takes orthocenters into orthocenters, centroids into centroids, circumcenters into circumcenters, etc. Note that the circumcenter of ABC is the orthocenter of its medial triangle.[/quote]
Thanks I just forgot that. :blush:  The last part of my solution is completely useless. It could certainly be made far shorter.

[quote="Moonmathpi496"]Let $ ABC$ be a triangle, $ O$ its circumcenter, $ S$ its centroid, and $ H$ its orthocenter. Denote by $ A_1, B_1$ and $ C_1$ the centers of the circles circumscribed about the triangles $ CHB, CHA$ and $ AHB$ respectively.

Prove that the triangle $ ABC$ is congruent to the triangle $ A_1B_1C_1$ and that the nine-point circle of $ ABC$ is also the nine-point circle of $ A_1B_1C_1$.[/quote]
We have $ 2R_{AHB}\equal{}\frac{AH}{sin \angle ABH}\equal{}\frac{AH}{sin \angle ACH}\equal{}2R_{AHC}$
Similarly we get $ R_{AHB}\equal{}R_{AHC}\equal{}R_{BHC}$
On the other hand, $ \angle C_1BC\plus{}\angle B_1CB\equal{}\angle C_1BH\plus{}\angle HBC\plus{}\angle HCB\plus{}\angle B_1CH$
$ \equal{}\angle HBA_1\plus{}\angle HCA_1\plus{}180^o\minus{}\angle BHC\equal{}360^o\minus{}\angle BHC\minus{}\angle BA_1C\plus{}180^o\minus{}\angle BHC\equal{}180^o$
Therefore $ BC_1//CB_1$ and $ BC_1\equal{}CB_1$ so $ BCB_1C_1$ is a parallelogram $ \Rightarrow BC\equal{}B_1C_1$
Similarly we get $ AC\equal{}A_1C_1, AB\equal{}A_1B_1$ then $ \Delta ABC\equal{}\Delta A_1B_1C_1$
But $ M, N, P$ are the midpoints of $ AH, BH, CH$, also are midpoint of $ B_1C_1, C_1A_1, A_1B_1$ then $ (MNP)$ is the 9 points circle of triangle $ ABC$ and $ A_1B_1C_1$


calculus
derivative
integration
trigonometry
calculus computations

Questions from my Calc review packet.
School's about to start and I can't figure out how to do these problems!

1. Calculate the derivative with respect to x of:
$ xf(x) \minus{} G(f(x))$, if $ G'(x) \equal{} f^{ \minus{} 1}(x)$ :
2. Sketch the graph and label all critical and inflection points for:
$ f(x) \equal{} x \plus{} (x \minus{} 8)^{2/3}$
 3. Show that the point $ (3/5, 4/5)$ is the closest point on the circle $ x^{2} \plus{} y^{2} \equal{} 1$ to the point $ (3, 4)$.
4. Evaluate the integral of $ \int cos^{4}(x^{7}) sin(x^{7}) x^{n}\,dx$ where $ n$ is an integer you may select.
5. Show that $ \int_0^1 g(x)\,dx \equal{} \int_0^1 g(x^{2}) 2x\,dx$
6. Show that $ \int_0^1 g(x)\,dx$ does not necessarily equal $ \int_0^1 g(x^{2})\,dx$ by finding two examples, one where the former integral is larger, one where the latter integral is larger. (Use both graphs/explanations and explicit formulae)
7. Which of the previous two integrals is larger if $ g(x)$ is increasing on $ (0, 1)$

6.
Construct $ g(x)\equal{}x$, then the value of the finite integral is $ \frac {1}{2}(1\minus{}0)\equal{}\frac {1}{2}$. Then,  $ g(x^2)\equal{}x^2$, the value of the integral is $ \frac {1}{3}(1\minus{}0)\equal{}\frac {1}{3}$.

Then,
construct $ g(x)\equal{}\sqrt {x}$, then the value of the finite integral is $ \frac {2}{3}(1\minus{}0)\equal{}\frac {2}{3}$, and $ f(x^2)\equal{}x$, then the value of the finite integral is $ \frac {1}{2}$.
I am sorry if there exist some of my words that are wrong.

[hide="Number 1"]$ [xf(x) \minus{} G(f(x))]' \equal{} f(x) \plus{} xf'(x) \minus{} G'(f(x)) \cdot f'(x)$

$ \equal{} f(x) \plus{} xf'(x) \minus{} f^{ \minus{} 1}(f(x)) \cdot f'(x) \equal{} f(x) \plus{} xf'(x) \minus{} xf'(x) \equal{} f(x)$[/hide]

[hide="Number 5"]Making the substitution $ x \equal{} u^2$, $ dx \equal{} 2udu$ we get

$ \int_0^1g(x)\,dx \equal{} \int_0^1 2u g(u^2)\,du$[/hide]

7. From the example above, $ g(x)$ can be larger than $ g(x^2)$, oir $ g(x^2)$ can be larger than $ g(x)$.
I am sorry if there exist some of my words that are wrong.
5. Look $ g(x)$, substitute $ u^2\equal{}x$, then $ dx\equal{}2udu$, when $ x\equal{}0$, $ u\equal{}0$ and when $ x\equal{}1$, $ u\equal{}1$, and we are done.

I am sorry if there exist some of my words that are wrong.

4. Take $ n\equal{}6$, then
\[ I \equal{} \int \cos^4(x^7)\sin^4(x^7)\ x^6\,\mathrm{d}x
\]
Put $ t\equal{}x^7$, then $ \mathrm{d}t\equal{}7x^6\mathrm{d}x$
Hence,
$ I\equal{}\dfrac{1}{7}\int \cos^4t\sin^4t \,\mathrm{d}t \equal{} \dfrac{1}{7\times 2^4}\int (2\cos t\sin t)^4\mathrm{d}t$
$ \ \equal{} \dfrac{1}{7\times 16}\int \sin^42t\mathrm{d}t$
$ \equal{}\dfrac{1}{112}\left(\dfrac{3t}{8}\minus{}\dfrac{\sin4t}{8}\plus{}\dfrac{\sin8t}{64}\right)$
$ \equal{}\dfrac{1}{112}\left(\dfrac{3x^4}{8}\minus{}\dfrac{\sin4x^4}{8}\plus{}\dfrac{\sin8x^4}{64}\right)\plus{}C$

3. Show that the point $ (\frac{3}{5},\frac{4}{5})$ is the closest point on the circle $ x^2\plus{}y^2\equal{}1$ to the point (3,4).

$ x^2\plus{}y^2\equal{}1 \to y\equal{}\pm \sqrt{1\minus{}x^2}$

$ F(x,y)\equal{}d^2((x,y),(3,4))\equal{}(x\minus{}3)^2\plus{}(y\minus{}4)^2\equal{}x^2\minus{}6x\plus{}9\plus{}y^2\minus{}8y\plus{}16$
$ \equal{}x^2\minus{}6x\plus{}25\plus{}1\minus{}x^2 \pm 8\sqrt{1\minus{}x^2}\equal{}26\minus{}6x \pm 8\sqrt{1\minus{}x^2}\equal{}F_\pm (x)$

min of F will be our desired point.

$ F_\pm '(x)\equal{}\minus{}6 \pm \frac{\minus{}8x}{\sqrt{1\minus{}x^2}}\equal{}\minus{}\frac{6\sqrt{1\minus{}x^2} \pm 8x}{\sqrt{1\minus{}x^2}}\equal{}0$

$ 6\sqrt{1\minus{}x^2}\equal{} \mp 8x \to 36\minus{}36x^2\equal{}64x^2 \to x\equal{}\pm \frac{3}{5}$

replugging into $ F_\pm '(x)$ we see that

$ x\equal{}\frac{3}{5}$ is the critical point to $ F_\plus{}$

$ x\equal{}\minus{}\frac{3}{5}$ is the critical point to $ F_\minus{}$

comparing the two results, we find the plus to be a min and the minus to be a max


The probability of having a computer assignment is $ 7/8$. The probability of having a textbook assignment is $ 4/5$. What is the probability of having either a computer assignment or a textbook assignment, but not both? Express your answer as a common fraction.

Right, you posted so many problems it could be considered spam; try grouping them in the same thread next time.
[hide]Probability of computer but not textbook: 7/8 * 1/5 = 7/40
Probability of textbook but not computer: 1/8 * 4/5 = 4/40
Sum = 11/40[/hide]

where did they all come from. did you make them up?

These originate from previous mathcounts problems.

H34N1, try to not spam as much:

Slightly different solution:
[hide]
ways you can get both: 7/8*4/5= 28/40
ways you can get neither: 1/8*1/5=1/40

add them 29/40.

1-29/40 to find ways you get one but not the other:

11/40.[/hide]


Sigur nu $a_{1}=\frac{1}{2}$ ?Ca nu e bine definit cu 0 cum spui tu....

pentru $<1$ se considera sirul $b_{n+1}=b_{n}+1/2 b_{n}$
apoi se demonstreaza ca  $a_{n}<b_{n}\leq 1$

sirul $b_{n}$ inseamna defapt la limita echivalent cu
o astfel de partitionare a segmenetului (0,1):

$(0,1)=(0,1/2)+(1/2,1/2+1/2^{2})+(1/2^{2},1/2^{2}+1/2^{3})+$
$(1/2^{3},1/2^{3}+1/2^{4})+...$
literalmente :
adauga tot timpul jumatate din distanta ramasa pan la a ajunge la 1
pornind de la pasul precedent


function
real analysis
topology
real analysis unsolved

Let $ X$ be a metric space. If $ x\in X$ and $ E\subset X$, define the distance from $ x$ to $ E$ as $ \rho(x,E)\equal{}inf\{\rho(x,y): y\in E\}$.

(i) Show that the function $ x\to\rho(x,E)$ is uniformly continuous on $ X$ for each $ E\subset X$, and that $ \bar{E}\equal{}\{x: \rho(x,E) \equal{} 0\}$.

(ii) Deduce from (i) that if $ F$ is a closed subset of $ X$, there exist open sets $ G_n$ ($ n\in \mathbb{N}$) such that $ F\equal{}\cap_{n\equal{}1}^\infty G_n$

(iii) If $ E$ and $ F$ are disjoint closed subsets of $ X$, show that $ f$, defined by

$ f(x)\equal{}\frac{\rho(x,E)}{\rho(x,E)\plus{}\rho(x,F)}$

is a continuous real-valued function on $ X$, with $ 0\leq f\leq 1$, and $ E\equal{}f^{\minus{}1}(\{0\}), F\equal{}f^{\minus{}1}(\{1\}).$ Deduce that there exist disjoint open sets $ U$ and $ V$ with $ E\subset U$ and $ F\subset V$.

I) To Show $ x \to p(x,E)$ is uniformly continuous note that $ p(x,y) \leq p(x,z) \plus{} p(z,y)$ so that
$ \mid p(x,y) \minus{} p(z,y) \mid \leq p(x,z)$ so taking infimum over $ y \in E$ we get uniform continuity.
Now if $ p(x,E) \equal{} 0$ then given $ \epsilon_n \equal{} 1/n$ we get that there exists $ y_n \in E$ such that
$ p(x,y_n) < 1/n$ this implies $ y_n \to x$ so $ x \in$ closure(E)

II) Let $ F$ be a closed set so $ F$ = closure($ F$). Let $ G_n \equal{} \{x : p(x,F) < 1/n \}$ then $ G_n$ is open
and $ F \equal{} \cap G_n$


Let $a,$ $b$ and $c$ are positive numbers. Prove that

$\left(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\right)^{2}\ge 3\left(\frac{a^{3}}{b}+\frac{b^{3}}{c}+\frac{c^{3}}{a}\right).$

Sorry argady, here is the solution. This inequality is created by toanhocmuonmau

1.Let a,b,c be positive numbers. Prove that:
$(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a})^{2}\ge 3(\frac{a^{3}}{b}+\frac{b^{3}}{c}+\frac{c^{3}}{a})$

Solution.
WLOG we may assume that c=max(a,b,c).
The inequalty can be written to the following form:
$f(a,b,c)=[(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a})^{2}-(a+b+c)^{2}]-[3(a^{2}+b^{2}+c^{2})-(a+b+c)^{2}]-3[(\frac{a^{3}}{b}+\frac{b^{3}}{c}+\frac{c^{3}}{a})-(a^{2}+b^{2}+c^{2})] \ge 0$
$\leftrightarrow f(a,b,c)=M(a-b)^{2}+N(a-c)(b-c) \ge 0$
With:
$M=(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}+a+b+c)\frac{a+b}{ab}-2-\frac{3(a^{2}+b^{2}+ab)}{ab}$
$N=(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}+a+b+c)\frac{b+c}{ac}-2-\frac{3(b^{2}+c^{2}+bc)}{ac}$
We have:
$M.ab \ge 2(a+b+c)(a+b)-2ab-3(a^{2}+b^{2}+ab)=2ac+2bc-a^{3}-b^{3}-ab \ge 0$
$N.ac=a^{2}+\frac{a^{2}c}{b}+\frac{b^{3}}{c}+\frac{bc^{2}}{a}+\frac{c^{3}}{a}+ab-ac-b^{2}-2c^{2}-bc$
Using AM-Gm inequalty:
$\frac{b^{3}}{2c}+\frac{b^{3}}{2c}+\frac{4c^{2}}{27}\ge b^{2}$
$a^{2}+\frac{c^{2}}{4}\ge ac$
$\frac{a^{2}c}{b}+\frac{bc^{2}}{a}+\frac{64c^{3}}{125a}\ge \frac{12c^{2}}{5}> (2+\frac{1}{4}+\frac{4}{26})c^{2}$
$\frac{c^{3}}{4a}+ab \ge 2\sqrt{bc^{3}}> bc$
$(1-\frac{64}{500}-\frac{1}{4})\frac{c^{3}}{a}\ge 0$
So we have $N \ge 0$, the inequality is solved.
The equality hold for a=b=c only.

To Argady and everyone:
Can you please help me to find the greastest constant k such that the following inequalty holds for any positive numbers a,b,c.
$(\frac{a^{k+1}}{b^{k}}+\frac{b^{k+1}}{c^{k}}+\frac{c^{k+1}}{a^{k}})^{2}\ge 3(\frac{a^{k+2}}{b^{k}}+\frac{b^{k+2}}{c^{k}}+\frac{c^{k+2}}{a^{k}})$
I can prove it with k=1, 2 and 3, so I think it's true for all $k \ge 1$, can you help me to find the greatest k!

Very strong method, 10maths_tp0609. Thank you!
I already saw it somewhere, but I did not assume, that it's strong up to such degree.


Prove that the sum of $9999$ consecutive squares [color=red]cannot be [/color]a perfect power

Squares mod 3 are 0, 1, 1.  So the sum of 3333 sets of 3  (modulo 3) are 0 (mod 3).

Squares mod 9 are 0, 1, 4, 0, 1, 4, 0, 1, 4.  The sum of 1111 sets of 9 modulo 9 is 6 mod 9.

If the perfect power is m^k with $k \ge 2$ then it contains a factor of 3 but doesnt contain a factor $3^2$.  Contradiction


Find all functions $ f: Q^\plus{} \to Q^\plus{}$ satisfying $ f(x\plus{}1)\equal{}f(x)\plus{}1$ and $ f(x^3)\equal{}f(x)^3$ $ \forall x \in Q^\plus{}$

I know a solution to this one from the book [i]functional equations a problem solving approach[/i] by Venkatachala,but its rather complication.I will wait if someone else post a solution.....

Dimitris

If it is too complicated,could you write down the hints to solve it?(some basic hints and the step to prove).Thanks a lot.

[hide="step 1"]
Show that $ f(r\plus{}n)\equal{}f(r)\plus{}n$ where $ n\in \mathbb N,r \in \mathbb Q^\plus{}$.
[/hide]
[hide="step 2"]
$ \left(\frac{p}{q}\plus{}q^2\right)^3\equal{}\left(\frac{p}{q}\right)^3\plus{}3p^2\plus{}3pq^3\plus{}q^6$. [/hide]

Is it enough???

It is a very diffucult problem i think....

[hide="solution"]
$ f(1) = f(0) + 1$ and $ f(1) = f(1)^3, f(0) = f(0)^3$, from which we conclude that $ f(0) = 0, f(1) = 1$
from Dimitris X's hints

[b]step 1[/b] show $ f(r + n) = f(r) + n$

From the first condition $ f(x + n) = f(x + n - 1) + 1 = f(x + n - 2) + 2 = ... = f(x) + n$
also $ f(n) = n$ for $ n\in \mathbb{N}$

[b]step 2[/b]

$ f(( \frac {p}{q} + q^2)^3) = f((\frac {p}{q})^{3} + 3p^{2} + 3pq^{3} + q^{6} ) = f(\frac {p}{q})^{3}) + 3p^{2} + 3pq^{3} + q^{6} \qquad(1)$

but we also have
$ f(( \frac {p}{q} + q^2)^3) = (f (\frac {p}{q}) + q^2)^3\qquad(2)$

Equation (1) and (2) gives

$ 3q^2f(\frac {p}{q})^2 + 3q^4f((\frac {p}{q})^) = f(3p^2) + f(3pq^3)$

$ \therefore f(\frac {p}{q})^2 - (\frac {p}{q})^2 = q^2(f(\frac {p}{q}) - \frac {p}{q})$

$ \therefore (f(\frac {p}{q}) - (\frac {p}{q}))(f(\frac {p}{q}) + (\frac {p}{q}))=q^2(f(\frac {p}{q}) - \frac {p}{q})$

So either $ f(\frac {p}{q}) - \frac {p}{q} = 0$ or $ f(\frac {p}{q}) = q^2 - \frac {p}{q}$

The second case, however, leads to obvious contradictions. For example $ q^2 - \frac {p}{q} = f(\frac {p}{q}) = f(\frac {2p}{2q}) = 4q^2 - \frac {p}{q}$

So the only solution is $ f(x) = x$[/hide]


6 red balls, 3 blue balls and 3 yellow balls are arranged in a line.How many possibilities such that any 2 neighbouring balls are different color.

[hide="do not click this"]
First we put 6 red balls in line. We will put 5 of remaining balls in order to separate all the red ones. We can do this in $ \frac {5!}{2!3!}$ ways. Now we have 11 balls in the line which makes 12 "gaps" to place a ball, but 4 of them are not valid since they are around 2 balls of the same color as remaining one so we have 8 possibilities. And not to forget, there are equal chances that the remaining ball is either blue or yellow so we need to multiply everything by 2 which makes total of $ \frac {5!}{2!3!}*8*2 \equal{} \boxed{160}$ ways to arrange the balls.[/hide]

[hide]Set up the reds first. There are two things we can do:
[u]
Way 1:[/u] Make a line of blue and yellow marbles and interweave it one-to-one with the line of red marbles. We can either start the line of red marbles first or the line of blue/yellow marbles first (2 possibilities), and there are 6-choose-3 ways of arranging the yellow-blue marble line (20 possibilities), for a total of 40 ways to carry out this case.

[u]Way 2:[/u] There are 2-ways to create a yellow-blue pair: YB and BY (2 possibilities). Represent such a pair as C. We then need to fill the five spaces in R_R_R_R_R_R with 2 Y's, 2 B's, and one C. There are 5!/(2! 2! 1!) ways of doing this (30 possibilities), for a total of 60 ways to carry out this case.

Therefore, there are 40+60=100 arrangements in total.[/hide]

I thought it was 100 too at first, but I think it's not correct.


Hi, everyone
I have a question which is about the derivation of the normal equations. wonder if anyone who could help, Thx in advance.

Function $f(x_1, ...,x_m)= f(x)= \Vert b-Ax \Vert_2^2$ is differentiable function of $m$ variables. It has minimun only when $\bigtriangledown f =(\partial f / \partial x_1, ..., \partial f / \partial x_m)^T=0$

Calculate  $\bigtriangledown f$?

$f(x)=<b-Ax|b-Ax>$. Then  if you compute $f(x+h)-f(x)$ you will find:
 $f(x+h)-f(x)=-2<h|A^T(b-Ax)> + o(|h|)$.
Hence the gradient of $f$ is $2(A^TAx-A^Tb)$, I think.

Hi, Thx for your help, but I dont understand how you come up with the answer. Could you explain?

1. Are you saying that $f(x)= \Vert b-Ax \Vert_2^2$ can be written as $f(x)= <b-Ax \vert b-Ax>$, and what is $\vert$ in the middle part
2. I dont get this part ...$f(x+h)-f(x)=-2<h|A^T(b-Ax)> + o(|h|)$, what is $o(|h|)$?
3. Total lost how you got the answer with $2(A^TAx-A^Tb)$

Sorry for asking more. Thx again.

1/ $<x|y>$ is the usual euclidian scalar product, $<x|x\geq |x|_2^2$
2/ just use that $<Ax|y>$ is equal to $<x|A^Ty>$ and that $<x+y|z>$ is equal to $<x|z>+<y|z>$. As regards $o(|h|)$, it is the Landau notation. Try to google it, it will surely explain this better than me
3/ it is the definition of the gradient: if $f(x_0+h)-f(x_0)=<h|y_0>$ then $y_0$ is the gradient of $f$ at the point $x_0$.

Hope this will help ..

Hi, I was following what you said, but I couldnt figure it out. Could you help a bid more?

Knowing that $f(x)=<b-Ax \vert b-Ax>$ then compute $f(x+h)-f(x)$ I am lost...

$f(x+h)-f(x)= <b-A(x+h) \vert b-A(x+h)>- <b-Ax \vert b-Ax>$... how to use the fact you gave to show it is equal to $-2<h \vert A^T(b-Ax)>+o(\vert h \vert)$

Could you show few steps? Thx alot.


[quote="anncar"]Prove that each prime divisor of 2p−1, where p is a prime, is greater
than p. (Notice that it follows from this problem that the number of primes
is infinite!)[/quote]

Erm, $ p\equal{}5$

There's a typo in the problem, because I can prove that all proper divisors of 2p-1 are [b]less [/b]than p.

It's fairly intuitive, since the minimum prime number is 2, and $ \frac{2p\minus{}1}{2}<p$.

[quote="anncar"]sorry guys.. Its 2^p - 1. Thats 2 raised to the power 'p'.[/quote]

Take a prime $ q\le p.$ If $ q\equal{}2$ then obviously $ q\not|2^p\minus{}1,$ so WLOG $ q>2.$ Now let $ k>0$ be chosen with $ q|2^k\minus{}1$ and $ k$ minimal. From Fermat's little theorem, we know $ k\le q\minus{}1<p$ and $ q|2^p\minus{}1\iff k|p.$ But since $ q>2,$ obviously $ k>1;$ so $ 1<k<p$ and since $ p$ is prime $ k\not|p\implies q\not|2^p\minus{}1$ qed.

the vertical line without the line going through it and the vertical line with the line going through it- what do they mean?

[quote="anncar"]thanks. I'm guessing the vertical line with the other line going theough it means 'does not divide'? and the vertical line just means "divides'?[/quote]

exactly.


How can we prepare 2-methyl-1-phenylpropene from phenylacetylene, [i]without breaking the C-C bond[/i]?

[hide="Maybe..."]
1. Hydration of phenylacetylene yields acetophenone.
2. Reduce it with Na/Hg, H2O to 1-phenylethanol.
3. Heat with acid and finally Heck Reaction with CH3Cl.
 :maybe: [/hide]

[quote="Nayel"]1. Hydrate the phenylacetylene in presence of peroxide to get 2-phenylethanal.[/quote]

I am not familiar with this procedure. Could you describe it in some detail, not forgetting the mechanism?

However, assuming this first step then the rest looks ok. Can you propose still another path?

Doesn't that follow from anti-Markovnikov rule? :maybe:

We can also perform that first step using some ruthenium catalyst instead of peroxide as wikipedia says [url=http://en.wikipedia.org/wiki/Markovnikov%27s_rule]here[/url].

Not sure if this'll work since I'm fairly inexperienced with orgo, but...

1. Treat phenylacetylene with CH3MgBr in an acidic aqueous solution to yield 1-phenylprop-1-en-1-ol.
2. Add sulfuric acid and heat to make 1-phenylpropyne
3. Add CH3Cl to produce 2-methyl-1-phenylpropene

*fingers crossed*

err cant edit the post, but change the first step to adding the CH3MgBr in THF and then adding water+acid.

CH3MgBr is a strong nucleophile and also a strong base. The proton of terminal alkynes is slightly acidic (pKa = 25). When you add a Grignard reagent to a terminal alkyne (in an ethereal solvent) an acid-base reaction simply occurs:

$ R\minus{}C \equiv CH \plus{} R'MgX \longrightarrow R\minus{}C \equiv C^\minus{}\,\, ^\plus{}MgX \plus{} R'H$.

If you add aqueous acid to this solution you will simply regenerate the alkyne:

$ R\minus{}C \equiv C^\minus{}\,\, ^\plus{}MgX \plus{} H_3O^\plus{} \longrightarrow R\minus{}C \equiv CH \plus{} (other stuff)$.

o so they only react to form alcohols with alkenes?

then never mind then  :oops:

[quote="nayel"]Well, heating 1-phenylethanol in presence of an acid would produce 1-phenylethylene. But I am doubtful about the Heck reaction part... I'm not sure if it would replace both of the hydrogen atoms with CH3 :maybe:

However, I think I found another way:

1. Hydrate the phenylacetylene in presence of peroxide to get 2-phenylethanal.
2. Add $ \ce{CH3MgBr}$ in presence of water to get 1-methyl-2-phenylethanol.
3. Oxidize the sec-alcohol with PCC to get 1-phenylpropanone.
4. Again add $ \ce{CH3MgBr}$ in presence of water to get 2-methyl-1-phenylpropanol-2.
5. Heating with acid should produce 2-methyl-1-phenylpropene.

Is it OK?[/quote]

In step 2, you're not using a zinc bromide...methyl magnesium bromide would just add twice...no need for steps 3 and 4.

Why would CH3MgBr add twice to an aldehyde?

Lol, looooong time....

1. NaNH2, MeI
2. H2, lindlar
3. CF3CO3H (get the epoxide)
4. MeI in basic medium
5. Quench and dehydrate

What exactly is "4. MeI in basic medium" ?

Me = Methyl = CH3 <--- orgo abbreviation

MeI = CH3I = Iodomethane

EDITED HERE

1. NaNH2, MeI
2. H2, lindlar
3. CF3CO3H (get the epoxide)
4. MeMgBr
5. Quench and dehydrate

Another proposal:

1. disiamyl borane in THF
2. H2O2, OH-
3. CH3MgBR
4. H+ workup
5. CrO3 oxidation
6. P(Ph)_3CH2 reacting in a Wittig reaction
7. H2SO4 to shift the double bond.

When writing a proposed synthesis, don't forget to indicate the reagent and expected product in each step. It is not the job of whoever is correcting an exercise to think in the place of the student. Write what you think, don't leave just clues.

heh sorry about that. not quite sure if i can name them correctly.

phenylacetylene + 1.disiamyl borane in THF, 2.H2O2,OH- -> 1-phenylethanal
phenylethanal + 3. CH3MgBR, 4. H+ workup -> 1-phenyl-2-propanol
2-phenylpropanol + 5. CrO3 -> 1-phenyl-2-propanone
2-phenylpropanone + 6. P(Ph)_3CH2 -> 2-methyl-1-phenyl-2-propene
2-methyl-1-phenyl-2-propene + 7. H2SO4 -> 2-methyl-1-phenylpropene

Well, I was thinking of going like this...

phenylacetylene + B2H6, H2O2, OH- => phenylethanal
phenylethanal + [O], KMnO4 => phenylethanoic acid
phenylethanoic acid + EtOH, H+ => ethyl phenylethanoate
ethyl phenylethanoate + 2 eq. CH3MgBr => 2-methyl-1-phenylpropan-2-ol
2-methyl-1-phenylpropan-2-ol + H2SO4 => 2-methyl-1-phenylpropene

Yes, both those syntheses look ok to me. You just have to be carefull on that oxidation,

phenylethanal + [O], KMnO4 => phenylethanoic acid.

A milder oxidizing agent would be better (why?).

would KMNO4 form an extra carbonyl at the benzylic position?

Oxidizing the whole side chain to a -COOH would be a stronger possibility, but this only if we use hot, alkaline potassium permanganate.

How about this...(some of these names might be wrong, if you're confused just let me know)
Phenylacetylene + H2 with Lindlar Catalyst --> Phenylethylene
Phenylethylene +B2H6,diglyme then H2O2 and OH- --> 2-phenylethanol
2-phenylethanol + Na2Cr2O7 ---> 2-phenylethanoic acid
2-phenylethanoic acid + Methanol with acid ---> Methyl 2-phenylethanoate
Methyl 2-phenylethanoate + MeMgI followed by acid ---> 2-methyl-1-phenyl-2-propanol
2-methyl-1-phenyl-2-propanol + acid/heat ---> 2-methyl-1-phenylpropene


I'll be at Nationals this year.  I strongly encourage all you AoPSers who will be there to come introduce yourselves.  The easiest way to find me will be to look for the ESPN crew - I'll be doing the broadcast again this year and will be somewhere near them much of the time.

I hope to meet you all there!

Hmm...are you considering attending ARML (u of iowa) too? :-D

I'll be at San Jose State this year - hopefully we'll make it to other states in future years.


Solve the equation in rational numbers $x,y,z$:
\[x^2+y^2+z^2+3(x+y+z)+5=0\]

[hide]multiply both sides by 4 and rearrange the terms.[/hide]

[quote=chien than][hide]Let $ a\equal{}2x\plus{}3;b\equal{}2y\plus{}3;c\equal{}2z\plus{}3$
Then the given equation is equivalent to
$ a^2\plus{}b^2\plus{}c^2\equal{}7$(easy to continue :P )[/hide][/quote]

Isn't this part really easy ? Real problem is with rational number. can somebody solve it for all the rational numbers ?


AMC
AIME
AMC 10
AMC 10 B
AMC 12
AMC 12 B

If you did really good do you feel happy and excited? Conversely if you did bad are you pissed off, angry, and non-sociable?

I did really bad (about 20 pts less than last yr on AMC 12) and now I'm gonna be angry for about a week.

Eh, it wouldn't matter to me, though that's me in general :p

did well?

it affects me severely the 1st few days; then, it cools down

I'm not mad; I guess I'm just not particularly happy usually.

Also as a senior, I have to say yes.  This is my final year in high school math and I did better than I expected, so I thought it was a very appropriate end to this part of my math career.  So, yeah, it definetly brightened my day, but I'll probably forget about it by the end of this weekend.

It really does to me.

I did 14 and left 11 blank on the AMC 12 B, and if I got 1 wrong I will fail. I am so scared and dont even wanna look at the answers . I really expect to pass the AMC this time, so I would be very depressed if I dont : (

I would have to say yes.  I do well (very well) at math competitions, but I never seem to do well on the AMC.  The B test was the first time I even qualified.  So yes, I am happy.

Well, I suppose...but there hasn't been much change in my results, I've qualified for the AIME every time I've tried. But I would be very sad if I didn't next year.

It really doesn't matter as long as I get a high enough score to get like a 9+ or 10+ on AIME and qualify for USAMO.

Yes, quite a few things affect my mood.

For example, on my fantasy baseball draft yesterday, I was gonna steal aramis ramirez and miguel tejada at 40th and 41st pick (major steal), but somehow i got put on autopick and got billy wagner and jimmy rollins instead.

I was SO pissed... killed my mood for a day. Had that league in the bag with two major steals.

I was muy depressed after the AMC10B, and finding out I guessed 3 right didn't help that much either.

Wow, you guessed three right? That's PRO guessing skillz  :lol:

Rollins is pretty good for pick 41. That's all I have to say.

I'm really pissed because both my amc scores are low.

It affects my mood.  Bad tests affect my mood longer than good tests though.

[quote="mathnerd314"]I was a little depressed after I self-scored and found that I got a 99 on the AMC-12 :wallbash:[/quote]

Don't be. They're probably going to use the 5% rule again, and you'll be happy. But then if it's like 99.5, you can go and smack Mr. Dunbar  :D

[quote="ehehheehee"]Is it possible to get a 99.5?

Since 1.5*2=3 and you get 6 points for a correct answer, all scores that are whole numbers are divisible by 3.  Also, all scores that are not whole numbers must be 1.5 from a multiple of 3.

99.5+1.5=101 which is not divisible by 3.
99.5-1.5=98 which is not divisible by 3.

If I messed up somewhere up there, ignore this.[/quote]

You're right. And the cutoff really should be something like 100.5. But isn't it a lot nicer to call it 100?

definitely...I thought I did ok...I didn't think I qualified (10B)...but I thought I got at least a 90 or so...I got a 75.

[quote="kyyuanmathcount"]The A test got me nervous - I had never taken AMC12 before, so when I did better than most of my practices (138), I was pretty happy for the rest of the day.

The B test was nice and relaxing, knowing I already had 138.  Thus, I did not make any dumb mistakes and got even better (only by 3 points, but w/e) and even happier (which did not have anything to do with eating free pizza afterwards).[/quote]

What a coincidence. My team had free pizza too.

My school only offers the A test, though :(

I think I'm going to barely make AIME, and then fail really miserably there (i.e. get 0 points)


In asymptote, what is the difference between "path" and "path[]"?

The same as between real and real[]. The first is a single object and the second is an array of objects. Note that Asymptote can automatically cast path into path[] (giving an array of length 1) but not vice versa. Another important thing to understand is that operator ^^ applied to 2 (or more) paths results not in a single path but in a path array of length 2 (or more). Some Asymptote commands (like intersectionpoints, for instance) that are documented as having paths as their arguments actually can take path arrays, but other cannot.


$ x_i , c \ge 0$
proof
$ \sqrt[n]{\prod^n_{i\equal{}1} (x_i\plus{}c)} \minus{} \sqrt[n]{\prod^n_{i\equal{}1} x_i} \ge c$

[quote="eeshift"]$ x_i , c \ge 0$
proof
$ \sqrt [n]{\prod^n_{i \equal{} 1} (x_i \plus{} c)} \minus{} \sqrt [n]{\prod^n_{i \equal{} 1} x_i} \ge c$[/quote]

Hint : Using Holder Inequality.

may you write it down, NextPeace

I can prove this inequality but it's not an elegant solution.

see this: http://www.mathlinks.ro/viewtopic.php?p=111563#111563

By Holder,

$ (x_1 \plus{} c)(x_2 \plus{} c)...(x_n \plus{} c)\ge(\sqrt [n]{x_1x_2...x_n} \plus{} c)^n$
$ \Leftrightarrow \sqrt [n]{\prod^n_{i \equal{} 1} (x_i \plus{} c)} \ge \sqrt [n]{\prod^n_{i \equal{} 1} x_i} \plus{} c$
Done.

thanks.

i haven't seen generalized Holder Inequality before
 :)


If the value of $ 2^x\plus{}2^{\minus{}x}$ is $ \sqrt[3]{3}$ then find the exact value of $ 8^x\plus{}8^{\minus{}x}$

Give the final result in its simplest form.

You already posted this here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1043089


floor function
trigonometry
geometry
geometric transformation
reflection
real analysis
real analysis unsolved

Remind the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=166042]formula[/url] $ \prod_{n \equal{} 1}^\infty \left[1 \minus{} \left(\frac {z}{n}\right)^m\right]^{ \minus{} 1} \equal{} \prod_{k \equal{} 0}^{m \minus{} 1} \Gamma\left(1 \minus{} \xi^k \, z\right)$ , whereas $ \xi \equal{} e^{\frac {2\pi i}{m}}$ .

Set $ m \equal{} 2p$ then $ \xi \equal{} e^{\frac {i\pi}{p}}$ and replace $ z$ by $ \sqrt {\xi} \, z$ , whereat $ \sqrt {\xi} \equal{} e^{\frac {i\pi}{2p}}$ .

$ \prod_{n \equal{} 1}^\infty \left[1 \plus{} \left(\frac {z}{n}\right)^{2p}\right]^{ \minus{} 1} \equal{} \prod_{k \equal{} 0}^{2p \minus{} 1} \Gamma\left(1 \minus{} \sqrt {\xi}^{2k \plus{} 1} z\right)$ which is $ \prod_{k \equal{} 0}^{p \minus{} 1} \Gamma\left(1 \minus{} \sqrt {\xi}^{2k \plus{} 1} z\right) \, \prod_{k \equal{} 0}^{p \minus{} 1} \Gamma\left(1 \minus{} \sqrt {\xi}^{2(k \plus{} p) \plus{} 1} z\right)$

$ \equal{} \prod_{k \equal{} 0}^{p \minus{} 1} \left[\Gamma\left(1 \minus{} \sqrt {\xi}^{2k \plus{} 1} z\right)\, \Gamma\left(1 \plus{} \sqrt {\xi}^{2k \plus{} 1} z\right)\right] \equal{} \prod_{k \equal{} 0}^{p \minus{} 1} \frac {\pi \sqrt {\xi}^{2k \plus{} 1} z}{\sin\left(\pi \sqrt {\xi}^{2k \plus{} 1} z\right)}$ after Euler's reflection formula.

Now write it as $ (\pi z)^p \, \sqrt {\xi}^{p^2} \, \prod_{k \equal{} 1}^p \left[\sin\left(\pi \sqrt {\xi}^{2k \minus{} 1} z\right)\right]^{ \minus{} 1}$ . It follows $ \prod_{n \equal{} 1}^\infty \left[1 \plus{} \left(\frac {z}{n}\right)^{2p}\right] \equal{} \frac {1}{(i\pi z)^p} \, \prod_{k \equal{} 1}^p \sin\left(\pi \sqrt {\xi}^{2k \minus{} 1} z\right)$

which can be written as $ \frac {1}{(i\pi z)^p} \, \prod_{k \equal{} 1}^{\lfloor\frac {p}{2}\rfloor} \sin\left(\pi \sqrt {\xi}^{2k \minus{} 1} z\right) \cdot \left[\sin\left(\pi \, \sqrt {\xi}^{\,p} z\right)\right]^{p \, \text{mod} \, 2} \cdot \prod_{k \equal{} 1}^{\lfloor\frac {p}{2}\rfloor} \sin\left(\pi \sqrt {\xi}^{2p \minus{} (2k \minus{} 1)} z\right)$

$ \equal{} \frac {\big[\sin(\pi i z)\big]^{p \, \text{mod} \, 2}}{(i\pi z)^p}\, ( \minus{} 1)^{\lfloor\frac {p}{2}\rfloor} \prod_{k \equal{} 1}^{\lfloor\frac {p}{2}\rfloor} \left[\sin\left(\pi \sqrt {\xi}^{2k \minus{} 1} z\right)\, \sin\left(\pi \sqrt {\xi}^{\, \minus{} (2k \minus{} 1)} z\right)\right]$

$ \equal{} \underbrace{\frac {i^{p \, \text{mod} \, 2} \, ( \minus{} 1)^{\lfloor\frac {p}{2}\rfloor}}{i^p}}_{ \equal{} 1}\, \frac {\big[\sinh(\pi z)\big]^{p \, \text{mod} \, 2}}{(\pi z)^p}\, \prod_{k \equal{} 1}^{\lfloor\frac {p}{2}\rfloor} \left[\sin\left(\pi \sqrt {\xi}^{2k \minus{} 1} z\right)\, \sin\left(\pi \overline{\sqrt {\xi}^{2k \minus{} 1}} z\right)\right]$

$ \boxed{ \equal{} \frac {\big[\sinh(\pi z)\big]^{p \, \text{mod} \, 2}}{(\pi z)^p}\, \prod_{k \equal{} 1}^{\lfloor\frac {p}{2}\rfloor} \left\{\cosh^2\left(\pi z\, \sin \frac {(2k \minus{} 1)\pi}{2p}\right) \minus{} \cos^2\left(\pi z\, \cos \frac {(2k \minus{} 1)\pi}{2p}\right) \right\}}$

For the last made step consider

$ \sin\big((a\pm ib)z\big) \equal{} \sin(az)\, \cos(ibz)\pm \cos(az)\,\sin(ibz)$ $ \equal{} \sin(az)\, \cosh(bz)\pm i\, \cos(az)\,\sinh(bz)$

$ \Rightarrow \, \sin\big((a \plus{} ib)z\big) \, \sin\big((a \minus{} ib)z\big)$ $ \equal{} \sin^2(az)\, \cosh^2(bz) \plus{} \cos^2(az)\, \underbrace{\sinh^2(bz)}_{\cosh^2(bz) \minus{} 1} \equal{} \cosh^2(bz) \minus{} \cos^2(az)$

In particular $ \sin\left(\pi \sqrt {\xi}^{2k \minus{} 1} z\right)\, \sin\left(\pi \overline{\sqrt {\xi}^{2k \minus{} 1}} z\right)$ $ \equal{} \cosh^2\left(\pi z\, \text{Im}\left(\sqrt {\xi}^{2k \minus{} 1}\right)\right) \minus{} \cos^2\left(\pi z\, \text{Re}\left(\sqrt {\xi}^{2k \minus{} 1}\right)\right)$

Seemingly from mathworld you copied [url=http://mathworld.wolfram.com/InfiniteProduct.html]formula 48[/url].

But the correct prefactor is $ \frac {\big[\sinh(\pi z)\big]^{p \, \text{mod} \, 2}}{(\pi z)^p}$ instead of $ \frac {1}{\pi^2\, z^2}\, \left(\frac {\sinh(\pi z)}{\pi z}\right)^{p \, \text{mod} \, 2}$ .


The mean of three number is 10 more than the least of the numbers and less than the greatest.  The median of the three numbers is 5.  What is their sum?

(A) 5
(B) 20
(C) 25
(D) 30
(E) 36

ANSWER: [hide]D[/hide]

Could you please give me the solution?

I don't know where you got this problem from, but it is certainly misstated. Both triples 0,5,25 and 2,5,29 satisfy the conditions with sums 30 and 36 respectively. My (wild) guess as to what the actual problem might be is that "less than the greatest" should really read "15 less than the greatest", in which case the problem reduces to a simple system of 2 linear equations with 2 unknowns.

[hide]okay I searched this probelm and it's 10 more than the least of the numbers and 15 less than the greatest.

So let the least number be x and the greatest be y.

(5+x+y)/3 = x+10 = y-15
5+x+y = 3x + 30 --> y = 2x + 25
5 + x + 2x + 25 = 3(2x+25-15)
30 + 3x = 6x + 30
3x = 0
x = 0
so y = 2(0) + 25 = 25.
The sum is then 0 + 5 + 25 = 30. D
[/hide]

Yes, that is the correct one that is if that problem is 2001 AMC 10, problem 16.