function
limit
calculus
calculus computations

For every positive integer $n$, consider the function
\[f_{n}(x)=x^{2n}+x^{2n-1}+...+x^{2}+x+1(x\in\mathbb{R}). \]
a)Prove that $f_{n}$ attains its least value at a unique value $x_{n}$ of $x$.
b)Let $S_{n}$ be the least value of $f_{n}$. Prove that
i)$S_{n}>\frac{1}{2}\forall n$ and there does not exist a real number $a>\frac{1}{2}$ such that $S_{n}>a\forall n$.
ii)$(S_{n})$ is a decreasing sequence and $\lim_{n\to\infty}S_{n}=\frac{1}{2}$.
iii)$\lim_{n\to\infty}x_{n}=-1$.

a)Denote $m_{n}=\min_{-1\leq x\leq 1}f_{n}(x)$, because $\lim_{x\to\pm\infty}f_{n}(x)=+\infty$ we have $f_{n}(x)>m_{n}\forall |x|>a$ for some $a>1$, then $m_{n}'=\min_{-a\leq x\leq a}f_{n}(x)=\min_{x\in\mathbb{R}}f(x)$, so $f_{n}$ attains its least value .

Now, can you continuous?  :maybe:

First of all note that $f_{n}(-1)=f_{n}(0)=1$ and $f_{n}(x)>1$ if $x>0$ (obvious) and if $x<-1$ (this is because $x^{2k-1}+x^{2k}>0$ for all $k$ if $x<-1$). Therefore, the minimum of $f_{n}$ is attained at some point $x_{n}\in [-1,0]$. Now, $S_{n+1}\le f_{n+1}(x_{n})= f_{n}(x_{n})+x_{n}^{2n-1}+x_{n}^{2n}\le f_{n}(x_{n})=S_{n}$, so $S_{n}$ is non-increasing. To show that it is strictly decreasing, just observe that $S_{1}=\frac{3}{4}<1$ is attained at $x_{1}=-\frac{1}{2}$, so all $x_{n}$ must be strictly between $-1$ and $0$ to have $S_{n}<1$. Now write
\[f_{n}(x)=\frac{1}{2}+\frac12(x+1)^{2}(1+x^{2}+\dots+x^{2n-2})+\frac{1}{2}x^{2n}\]
whence $f_{n}(x)> \frac12$ for all $n,x$
Plug in $x=-1+n^{-2/3}$. You'll get
\[f_{n}(x)\le \frac12+\frac{1}{2}n^{-1/3}+\frac{1}{2}e^{-2n^{1/3}}\le \frac12+n^{-1/3}\,. \]
Thus $\lim S_{n}=\frac{1}{2}$. At last, $f_{n}(x)\ge \frac{1}{2}+\frac{1}{2}(x+1)^{2}$ so $\lim f_{n}(x_{n})=\frac{1}{2}$ implies $\lim x_{n}=-1$.

With some more work, one can find equivalents for $S_{n}-\frac12$ and $x_{n}+1$ but I'll leave it to someone else.

thanks!  :wink:

$x_{n}$ unique because $f'(x_{n})=0$ and equation $2nx^{2n+1}-(2n+1)x^{2n}+1=0$ has at most one real root in $(-1,0)$.


On CML contest 6 (which was the lamest CML I've ever taken, coincidentally), problem six mentioned that only four solutions were known to the Diophantine equation $x^3 + y^3 + z^3 = 3$, namely $(1, 1, 1)$ and $(4, 4, -5)$ and its permutations.  Has it been proven that there are no more solutions, or is this still an open problem?

This problem should be a difficult one.
In my book called [i]Unsolved Problems in Number Theory[/i], this problem was posted as an open question.  ( My book was first published in 1994.)
But the world is changing quickly, maybe someone has already solved this problem??
Go searching on the internet and good luck.  :)

See also :
http://www.mathlinks.ro/Forum/viewtopic.php?t=41825
and, more particularly (confirming that it's still open) :
http://www.mathlinks.ro/Forum/viewtopic.php?t=83307 .


The vertices of a convex pentagon have integer coordinates. Find the least possible area of the pentagon.

Picks formula says $A = I + B/2 - 1$, where there are I lattice points inside the polygon and B on the edges.

We cant get I = 0 because its convex(*).  The minimum of B is 5.  We can acheieve I = 1 and B = 5 easily.  Thus the area is 5/2.

(*): This requires some explanation: Basically, if there are no interior points, the polygon must fit on a horizontal or vertical strip of length 1. (If it doesn't, then it is atleast 2x2 which carries one point)  But then 3 points are collinear, contradiction.


algebra
polynomial
integration
real analysis
real analysis solved

Let $a_1,a_2,...,a_n$ be complex numbers. Prove that there is a point $x$ in $[0,1]$ such that
$|1-\displaystyle\sum_{k=1}^{n}a_ke^{2\pi i kx}|\geq1$

Or, in other words, given a complex polynomial $P$ s.t. $|P(0)|=1$, show that there is a $z$ with $|z|=1$ s.t. $|P(z)|\ge 1$.

We write $P(0)=\frac 1{2\pi i}\int _{\mathcal C}\frac{P(z)}zdz$, where $\mathcal C$ is the unit circle. This translates to $P(0)=\int_0^1 P(e^{2\pi i x})dx$, meaning that $\int_0^1 |P(e^{2\pi i x})|dx\ge \left|\int_0^1 P(e^{2\pi i x})dx\right|=|P(0)|=1$, and the conclusion follows.

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25126[/url] for a sharper estimate...


summer program
Mathcamp
geometry
HCSSiM
MathPath
AwesomeMath
PROMYS

Hello,

I'm student from Kazakhstan and I'd like to visit MathCamp. Who was there? What are your reviews?

Thanks.

Hello!  :) I went last year, and it was pretty awesome. What exactly do you want to know? Browsing the website (mathcamp.org) was very informative and answered all of my questions.

Thanks for replying. First, about classes: How many lessons per day? Who are teaching math? ; second, about camp: Where will we live? How much people in one room?.

For example: this year camp stays in Washington. Could I go myself to city?

P.S. If there are some mistakes in grammar, don't mention it. I've got some problems in it. :blush:

Best regrads.

Nah, your grammar's pretty good. :)

There's four class slots a day (five choices in each slot) as well as a lecture by a guest visitor, but you don't have to go to all of these (in fact they encourage you to take some breaks). The number of classes/lectures you attend is entirely up to you.

The teachers vary somewhat year to year, but they're all professors or graduate students. Of the more famous, John Conway visits just about every year and Dave Patrick came last year.

We live in the dorms of the college campus. I don't know if its always like that, but last year campers chose between zero and one roommates. There's also separate areas designated for guys/girls and early/late quiet hours.

The rule is that you can go wherever as long as you're with a group of at least three people and you have transportation plans for getting there and back.

I hope you decide to apply, the more the merrier. :D

Thank you very much. Are you going to apply this year?

Well, I don't have to apply again since I went last year, but I plan on going again this year. :)

Ok. Do you know smth interesting about this camp?

Umm let's see... Well, I really liked it because there's an incredible amount of freedom, especially compared to other camps I've been to. You can literally do *whatever* you want. Also, all the small quirky traditions and things make it really fun. :D

Thanks. One more thing. Are there some swimming pool or gym?

If there is a pool or a gym on campus, then we can probably use it. Indeed there is at UPS (see http://www.ups.edu/x3305.xml)

*whatever* we want??
lolll. whats that supposed to mean xP

i think im applying :)

[quote="math_king_kz"]Who knows the schedule of Mathcamp 2009?[/quote]

The schedule doesn't come out until the first day of Mathcamp. Check mathcamp.org if you'd like to see older schedules.

As for the "hall," we'll live in dorms at the University of Puget Sound.

@shelly32494: Apply; Mathcamp is awesome!

im applying!! :) hahaha just took a look at the problems... doable. i think XD

problems are very easy shely, so nice to meet you :D

I'm a little surprised that this year's quiz is shorter and seems easier than last year's, especially considering how the number of applicants has been rising. My perspective is colored considering I'm a lot better now (yay Mathcamp!), but I don't think I'd be able to solve 6.5 problems of last year's quiz in an hour.

[quote]the number of applicants has been rising[/quote]

How much people allowed?

There are generally 110-120 campers, of which 45-50 are returning.

what means returning? during the lesssons?

Returning means they have attended Mathcamp before. So each year there are around 60 to 85 campers that have never gone to Mathcamp before, and 45 to 50 who are returning for their second, third, or even fourth summer at Mathcamp.

Really, one can do just about whatever one wants at Mathcamp (including taking whichever classes you think sound interesting--the mentors just advise you on what classes you would probably like or are at your level, but they don't force you to take any particular class), as long as the Four Rules are obeyed:
1. Be excellent to each other.
2. No stupid stuff.
3. No fire.
4. Don't divide by zero (unless under staff supervision)
Along with what's usually called Rule 0, "Obey the rules of the college," and some assorted rules like checking into the dorms by 10 PM and not going off campus without a group of at least three people.

[quote="DPatrick"][quote="laughinghead505"]The teachers vary somewhat year to year, but they're all professors or graduate students. Of the more famous, John Conway visits just about every year and Dave Patrick came last year.[/quote]I don't know if I'm really that famous.

I am planning to come again this year.  (I don't know the details yet, but I've been invited back, and I've said "yes".)[/quote]

If I get into mathcamp, then I will be able to meet my calc teacher in person.  :)

Well, I think Mathcamp is the best summer math camp, but it really depends on what sort of camp you want. Rather than try to describe them, I'll just point out this thread: http://www.mathlinks.ro/viewtopic.php?t=247682 which is rather recent and has campers giving propaganda for both HCSSiM and Mathcamp :wink: .

And by searching, here's some other threads that compare the camps in various ways:
http://www.mathlinks.ro/viewtopic.php?t=247682
http://www.mathlinks.ro/viewtopic.php?t=183359
http://www.mathlinks.ro/viewtopic.php?t=30263
http://www.mathlinks.ro/viewtopic.php?t=199636

So who are going to visit Mathcamp this year say "YES"!! :lol:  :D

I am extremely interested in mathcamp.  I would consider myself a pretty good mathematician, because I usually score in the top few nationally.  It's just that I'm 13, so most of the problems on the qualifying quiz I can't solve.  I guess I'll wait a couple years, and then I'll definitely apply.  Mathcamp sounds like a really fun place!  I'm just scared I won't qualify in the future.

Don't give up just yet! The problems should take a long time. You need to keep trying new approaches; math doesn't come easy for anyone ;)

Also, we do have some 13 year-olds, so you are on the younger end, but you are certainly not too young.

First, I'll echo what David said: don't give up! It doesn't mean you're not ready for Mathcamp if you haven't gotten all of the problems already; the qualifying quiz problems are [i]supposed[/i] to take some pondering  :wink: .

Also, they don't expect people to get all of the problems on the qualifying quiz. I know I definitely didn't, and I got in. So you should send in whatever you have on the problems--they really want to see your work and how you approached the problems, not just right answers.

But if you don't get in, or decide to wait till you are older (although there have been 13 year olds both years I went to Mathcamp), there are lots of other camps, too, like MathPath, AwesomeMath, Promys, etc. I went to MathPath myself and it was [i]awesome[/i], and the oldest one can go to MathPath is 14 years old. Summer math camps are just generally lots of fun and interesting math people go to all of them, so I'd definitely encourage you to apply somewhere :-).

When I was 14, I got mail from Mathcamp after qualifying for AIME, and decided that it sounded interesting, but I'd wait a bit, get more used to spending an extended amount of time away from home, etc.

That was one of the worst decisions I've ever made.

question. we dont need to send in a transcript right?

I'm pretty sure on the MC application there was a place to put scores from contests, math classes one has taken, and stuff like that (at least I think that's what they wanted there--it has been two years since I applied...), but they don't need an official transcript beyond that.

actually, they only ask you to rate yourself in certain fields of mathematics (at least, thats what the online application asks for)

Scroll down on that page. What do you see?

they ask u to write all the contests/extra curricular events you are part of

[quote]
hmm.
can anyone give me some reasons why like mathcamp over like. promys/ross/hcssim??
[/quote]

Mathcamp is much more lenient than, say HCSSiM (which is also a good camp!), because you are allowed to pick your own classes, and aren't required to attend a rigorous schedule. Mathcamp also offers a bit more variety in the mathematical subjects that are introduced to the students. However, both are very good camps and I encourage you to apply to the other if you don't get into one (which happened to me! I still had a great time at the other one!  :lol: ). Also, if I remember correctly, Mathcamp has a much more diverse amount of students. I remember last year there were people from Turkey, China, all over the world!  :D


geometry
geometric transformation
reflection
circumcircle
geometry unsolved

Let $ ABCD$ be a quadrangle inscribed in circle and $ P$ be any one point in the plane.The orthogonal projection points of $ P$ on $ AB,BC,CD,DA,AC,BD$ are $ R,S,T,X,Y,Z$ respectively.Show that the three midpoints of the segments $ RT$、 $ SX$ and $ YZ$ are collinear.

[quote="mr.danh"]By Simson's theorem, we have four lines (RXZ), (RYS), (XRS), (YRZ), so the midpoints of RT,SX,YZ are collinear (Newton line)[/quote]

Wrong (this was the same flawed solution I gave long ago).

See
http://www.mathlinks.ro/Forum/viewtopic.php?t=4758
http://www.mathlinks.ro/Forum/viewtopic.php?t=169264

I am planning to write down my solution in detail and publish it in Math. Reflections, but this can take some months...

  darij

Sorry, I though that P is given on the circle. However, for every point P, this problem is easy, and we don't need links darij gave.

I have found a solution for the problrm,and even I regard it to be simple.My solution depends on the following lemma:

Lemma: Let $ a,b,c$ be three fixed lines in a certain plane and $ P$ be a point on $ c$.If $ P$ moves along $ c$,the midpoint of the segment linking the two  orthogonal projection ponts of $ P$ on $ a$ and $ c$ respectively moves along another fixed line proportionately.

[quote="Butterfly"][b][size=100]Lemma[/size][/b]: Let $ a,\ b,\ c$ be three fixed lines in a certain plane and $ P$ be a point on $ c$. If $ P$ moves along $ c$, the midpoint of the segment linking the two orthogonal projection points of $ P$ on $ a$ and $ b$ respectively, moves along another fixed line proportionately.[/quote]
[quote="77ant"]May I ask you a complete proof ? Butterfly .
I wonder it and the proof for the lemma :blush:[/quote]
This [b][size=100]Lemma[/size][/b] may be easy proved, as a direct application of the [b][size=100]ERIQ theorem[/size][/b] ( [b][size=100]E[/size][/b]qual [b][size=100]R[/size][/b]atios [b][size=100]i[/size][/b]n [b][size=100]Q[/size][/b]uadrilateral ), whose two elementary proofs, have already been posted in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=97493]Collinearity of orthocenters[/url], of [b][size=100]shobber[/size][/b].

I don’t know how we can use this powerful theorem, for the solution of the proposed problem and so, I am looking forward in [b][size=100]Batterfly’s[/size][/b] approach.

Kostas Vittas.

I have a solution for this problem (but it's not mine), and in my opinion this problem is not easy as [b]mr.danh[/b] said, so I would like to see his solution first, and I'll present the solution I have in mind later.

Firstly,when $ P \equal{} O$(the circumcentre of Quadrangle $ ABCD$,the proposition is really true,with respect to $ O$, the three midponits of$ RT,SX$ and $ YZ$ coincide,and denote it as $ O'$

Secondly, when $ P$ lies on the circumcircle of  Quadrangle $ ABCD$,the proposition is true as well,.In this case,by Simson's theorem, the three point
groups$ RXZ, RYS, XRS$ and $ YRZ$ are collinear respectively,hence further by the theorem of Newton's line, with respect to complete quadrangle,the three midponits of$ RT,SX$ and $ YZ$ are collinear.Denote the three midpoints as $ L',M',N'$ respectively.

And then,with respect to a general $ P$, let $ OP$ meet the circle at $ P'$ and $ L,M,N$ be the three midponits of $ RT,SX$ and $ YZ$ with respect to general $ P$. By the lemma I posted above,we can easily conclude that $ O',L',L$are collinear ...etc,and $ \frac {O'L'}{O'L} \equal{} \frac {O'M'}{O'M} \equal{} \frac {O'N'}{O'N} \equal{} \frac {OP'}{OP}$.Because of $ L',M',N'$ being collinear,$ L,M,N$ are collinear.

Q.E.D


Let $X$ be the set of all lattice points in the plane (points $(x, y)$ with $x, y \in \mathbb{Z}$). A path of length $n$ is a chain $(P_0, P_1, ... , P_n)$ of points in $X$ such that $P_{i-1}P_i = 1$ for $i = 1, ... , n$. Let $F(n)$ be the number of distinct paths beginning in $P_0=(0,0)$ and ending in any point $P_n$ on line $y = 0$. Prove that $F(n) = \binom{2n}{n}$

Does anyone have a solution for this one?

Note that, given $P_i$, there are four possible places for $P_{i+1}$, which we may denote by up, down, left and right or $U, D, L, R$. So our chain becomes a sequence of $n$ letters $U, D, L, R$ with the same number of $U$s and $D$s.

We will find a bijection between these and the sequences of $2n$ letters, each $A$ or $B$, $n$ of each (which are obviously $\binom{2n}{n}$). To do this, just let $U=AA, D=BB, L=AB, R=BA$ and then do the obvious operations.

Severius was faster, but my solution is a little bit different:)
each move can be represented by complex number $1,-1,i,-i$ so every path can be represented by sequence of $n$ numbers $(a_1,..a_n)$ where $a_n \in \{i^n | n \in N\}$ but $\sum a_i$ must be real, so there are as many $i$ as $-i$
on $k$ places we can put $i$, next on $n-k$ places we can put $-i$ and then we can put real numbers in $2^{n-2k}$ ways so there are $\sum_{k=0}^{[n/2]} {n \choose k}{{n-k} \choose k}2^{n-2k}$ sequences, we change this sum and we have $\sum_{k=0}^{[n/2]} {n \choose 2k}{{2k} \choose k}2^{n-2k}$
and now we will prove that this sum is ${2n}\choose {n}$
${2n}\choose {n}$ is the number of ways we can choose $n$ poeple from $n$ girls and $n$ boys
assume that every girl has exactly one boyfriend and every boy has exactly one girlfriend, so we have $n$ pairs
let us take $2k$ boys and take $k$ boys from them, those chosen boys take their girlfriends and we add them to our $n$ group and those $k$ which were not chosen from $2k$ are not taken and their girlfriends neither
and finally we take $n-2k$ people from pairs which were not chosen at first time, we can do this in $2^{n-2k}$ ways couse we will take exactly one person from each pair, so we must sum it from $k=0$ to $k=[n/2]$ and we are done


use the following 4 numbers in any order to add, subtract, multiply or divide and come up with the number 16

3,2,8,5 = 16

You didn't provide enough information as if we had to use all of them only once, we couldn't use them as exponents, etc.

You do know that's 8 right?

EDIT: 007math just deleted his post.

can we repeat the no.s and signs?then it is easy.

8x3-2x5=24-10
               14.
   therfore,14+2=16.

Hey,

Nice solution [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=60030][u][b]Bellflower[/b][/u][/url].

Challenge: Use the following 4 numbers (only once) in any order to add, subtract, multiply, divide or use as exponents and come up with the number 16.

3,2,8,5 = 16
[hide="Hint."]
The ability to use the numbers as exponents introduces another solution apart from [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=60030][u][b]Bellflower[/b][/u][/url]'s.
[/hide]

Thanks,

-[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715][u][b]APSStudent[/b][/u][/url].

[hide="using exponents"](5-3)^(8/2)[/hide]
 :wink:

Hey,
[quote="Eulers_Apprentice"]
[hide="using exponents"](5-3)^(8/2)
[/hide]
:wink:
[/quote]
Another nice solution [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=46703][u][b]Eulers_Apprentice[/b][/u][/url].
[hide="My Solution."]
$ {{\left(\frac {8}{5 \minus{} 3}\right)}^{2}} \equal{} {16}$
[/hide]

Thanks,

-[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715][u][b]APSStudent[/b][/u][/url]


algebra
binomial theorem
combinatorics proposed
combinatorics

Evaluate $\sum_{k=0}^{(n-1)/2}2^{k}\cdot\binom{n}{2k}$ (if it's possible...).

More specifically, prove that it cannot be a perfect square.

[quote="me@home"]Hmm??? How do you get prove it can't be a perfect square from evaluate???[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=111392 :)

We have $\frac{n-1}{2}\in\mathbb{Z}$, so $2\nmid n$. After substitution $l=2k$ we can write the sum as
$\sum_{0\leqslant l\leqslant n\atop 2\mid l}\binom{n}{l}2^{\frac{l}{2}}$.
Let us denote it by $s_{0}$ and similar sum taken over $2\nmid l$ by $s_{1}$. Then by binomial theorem:
$s_{0}+s_{1}= \sum_{0\leqslant l\leqslant n}\binom{n}{l}2^{\frac{l}{2}}= \left(1+\sqrt{2}\right)^{n}$
and
$s_{0}-s_{1}= \sum_{0\leqslant l\leqslant n}\binom{n}{l}2^{\frac{l}{2}}(-1)^{l}=-\sum_{0\leqslant l\leqslant n}\binom{n}{l}2^{\frac{l}{2}}(-1)^{n-l}=-\left(\sqrt{2}-1\right)^{n}$.
Adding these two equations and dividing by $2$ we get
$s_{0}= \frac{\left(1+\sqrt{2}\right)^{n}+\left(1-\sqrt{2}\right)^{n}}{2}$,
as [b]chess64[/b] said :) .


1) Suppose an electron is released from rest in a uniform electric field whose magnitude is $ 5.90 \times 10^{3}\mathbf{V}/m$.

(a) Through what potential difference will it have passed after moving 1.00cm?
(b) How fast will the electron be moving after it has traveled 1.00cm?

2) An insulating rod having linear charge density $ \lambda = 40.0 \mu\mathbf{C}/m$ and linear mass density $ \mu = 0.100 kg/m$ is released from rest in a uniform electric field $ E = 100 \mathbf{V}/m$ directed perpendicular to the rod.
(a) Determine the speed of the rod after it has traveled 2.00m.
(b) How does your answer to part (a) change if the electric field is not perpendicular to the rod? (Ignore gravity)

1) (a) $ E = \frac{\Delta V}{d}\Rightarrow \Delta V = E \cdot d = 5.9 \times 10^{3}V/m \cdot 1.0 \times 10^{-2}m = 59\,V$.

(b) Newton's second law: $ F_{res}= ma$ and $ F_{res}= F_{el}= qE$, and so $ ma = qE \Rightarrow a = \frac{qE}{m}$.

Equation for constant acceleration motion: $ v_{f}^{2}= v_{0}^{2}+2ad \Rightarrow v_{f}= \sqrt{\frac{2qEd}{m}}$,

where $ q$ is the elementary charge ($ 1.6 \times 10^{-19}$ C) and $ m$ is the electron mass ($ 9.1 \times 10^{-31}$ kg).


Ο φερμά ισχυριστηκε οτι βρηκε μια απλη αποδειξη για τη μη επιλυσιμοτητα της εξισωσης
$x^{n}+y^{n}=z^{n}$ όπου $x,y,z \in \mathbb{N^{*}}$ και $n \in{N}$ , $n>2$.

Να λυθει τωρα η εξισωση $n^{x}+n^{y}=n^{z}$ οπου $x,y,z,n \in \mathbb{N}$.

H ιδεα δεν προηλθε απο το θεωρημα του φερμα ( :D ) αλλα απο μια αλλη εξισωση.

[i]Αφιερωμενη στον[b] Αλεξανδρο[/b][/i]

 :)

εννοεις οτι μονο n=0 ?

όχι, μια λύση έδωσα, θα το διευκρινήσω. Άλλωστε νομίζω ότι θα έπρεπε να είναι Ν*, σωστά socrates?

[quote="socrates"]
Να λυθει τωρα η εξισωση $n^{x}+n^{y}=n^{z}$ οπου $x,y,z,n \in \mathbb N$.
[/quote]
Νομίζω πως λύνεται εύκολα αν θεωρήσουμε την εξίσωση με βάση το $n$.
Τελικά προκύπτουν οι λύσεις:
1) $n=0$ και $x,y,z \in \mathbb N^{*}$
2) $n=2$ , $x=y$ και $z=x+1$ .

Είναι αρκετά πιο εύκολο από το τελευταίο θεώρημα του Fermat.  :P

Simfwnw apolita me ton stedes :)

Thanasi, euxaristw gia tin afierwsi...

Alexandros[/quote]

Σιγουρα καμια σχεση με φερμα και γουαιλς....
Απλα μοιαζουν...

Στεδες οκ!  :)


Is there a good way to make vectors underlined?

I am currently using
[code]\renewcommand{\v}[1]{\underbar{\(#1\)}}[/code]
which gives, for example, $ \underbar{\(v\)}$ and $ \underbar{\(y\)}$.  But this doesn't work well when I use a vector in a subscript (eg. $ \text{\Large \(D_{\underbar{\(v\)}} f\)}$ as opposed to $ \text{\Large \(D_vf\)}$ where the vector $ \underbar{\(v\)}$ in the subscript is "full-sized").  Is there an alternative?

Try \uline from the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=ulem]ulem package[/url]. The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=soul]soul package[/url] provides \ul which has a lot more adjustments but is trickier to use in math mode.


Mother is 21 years older than Little Joe. In six years, Mother will be five times as old as Little Joe. How old is Mother right now?

It's correct, but it'd make more sense if a conscientious moderator hadn't removed my extension. :P

[quote="Math Geek"]That's easy.
[hide]We set up a system of equations equations:
M=J+21
M+6=5J+30, so M=5J+24.  By the Transitive Property, J+21=5J+24, so it seems that we have hit a roadblock and this question can't be solved.  Thus the answer is:  The family, or at least Little Joe and his mother, do not exist. :D  However, Little Joe might still be in his mother's womb, meaning that he is according to the system of equations -3/4 years old.  Thus, his mother is 20.25 years old.[/hide]

I'm doing this on the top of my head, so I might have made a mistake.[/quote]

lol. i got the same answer, basically the same way, create a system of equation in x and y (where x and y are the ages of mom and little joe respectively), and we get that [hide]$ x=\frac{81}{4}$ or $ 20.25$[/hide]

wtf i just worked on it for like 10 mins and i didnt get it, and then i look at math geeks solution and apparently joe is negative years old now...so i guess i was doing it right..

what was the extension?

yeah..lol..i thought that was weird too...but then again, the problem never said little joe couldn't be un-born. :lol:
i don't know what the extension was, either...can you post it again, lazarus? :huh:

The extension read "Where is father?". I think it was intended more as a hint (to the fact that the baby was yet unborn) rather as an extension.


Let $ n,r \in \mathbb{N}_0$ and $ a_1,...,a_r,b \in \mathbb{Z}$. Let $ d \equal{} gcd(a_1,...,a_r,n)$. Show that the equation:

$ a_1 x_1 \plus{} a_2 x_2 \plus{} .... a_r x_r \equal{} b$ has exactly $ dn^{r \minus{} 1}$ different solutions in $ \mathbb{Z}_n^r$.

Yes, indeed. Forgot to mention that. That was the a) part of the exercise. I couldn't solve the b) part.

is it possible the book had a typo and they meant $ n^{r\minus{}1}/d$?

cf. $ 2x\plus{}4y\equal{}6$ in $ Z_8$


geometry
circumcircle
cyclic quadrilateral
power of a point
radical axis
geometry solved

Making a sketch it's seems that.. G is of no use ... so I was thinking that perhaps M is the midpoint of BG.. I don't know.. otherwise why would you need G?!

cheers!! :D

G4

Although it might not look like it, this is similar to problem 2 you posted.

For the cyclic quadrilateral BCEF M is the circumcenter and H is the intersection of the diagonals. GH is the angular polar of point A with respect to angle BGF, so it's also the polar of A with respect to the circumcircle of BCEF, so AM perpendicular to GH because M is the circumcenter of BCEF.

Hope it's correct!

I'm just looking through the old unsolved problems and I see they are not so unsolved after all...

[quote="hossein11652"]G4. Let AD, BE, CF be altitudes in traingle ABC. Let EF intersect BC in G. and let M the midpoint of BC, and H ABC's orthocenter. Prove that DH is perpendecular to AM.[/quote]

There's a typo here: Not the line DH is perpendicular to the line AM, but the line GH is perpendicular to the line AM.

For a rather elementary solution, see Lemma 2 of [url=http://de.geocities.com/darij_grinberg/Schroeder/Schroeder.html#6]the posting [Schroeder6] in my "Schröder points database"[/url].

  Darij

let the orth projection of H on AM is N.
Its easy to prove that BNEG are cyclic.
Because of AD is the (??) radical axis
of the two circles whith diameter BC and GM.
Consider the three circles and AD,BE,GN
they must intersect at one point---H
So it is proved that GH is perpendicular to AM.

We know H. Brocard's theorem:
"The circumcenter of a cyclic quadrilateral is the orthocenter of its autopolar triangle".
Then $M$ is the circumcenter of $BCEF$ cyclic quadrilateral and $AHG$ is the autopolar triangle of $BCEF$, so $M$ is orthocenter of $AHG$. Therefore $H$ is the orthocenter of $AMG$ triangle, i.e. $GH$ is prependicular to $AM$. QED.


limit
real analysis
real analysis unsolved

Let {a_n}n≥1 be an infinite sequence with a_n ≥ 0 for all n. For n ≥ 1, let
b_n denote the geometric mean of a_1, . . . ,a_n. Suppose
$ \sum_{n\equal{}1}^{\infty}a_{n}$ is convergent. Prove that $ \sum_{n\equal{}1}^{\infty}b^{2}_{n}$
is also convergent.

[hide="Opinion"]If I understood correctly, $ b_{n}\equal{} (a_{1}\cdot a_{2}\cdot ...\cdot a_{n})^{1/n}$. As $ \sum a_{n}$ converges, then we know that $ \lim_{n\to\infty}a_{n}\equal{} 0$. Let's now compare both series through a limit:

$ \lim_{n\to\infty}\frac{b_{n}^{2}}{a_{n}}\equal{}\lim_{n\to\infty}\frac{\left[(a_{1}\cdot a_{2}\cdot ...\cdot a_{n})^{1/n}\right]^{2}}{a_{n}}\equal{}\lim_{n\to\infty}\frac{[a_{n}]^{2}}{a_{n}}\equal{}\lim_{n\to\infty}a_{n}\equal{} 0$,

and so, as $ \sum a_{n}$ converges, then also $ \sum b_{n}^{2}$ converges. Note that $ \sum b_{n}$ would also converge.

[/hide]

carcul could u please explain this  :blush:
[quote="Carcul"]
$ \lim_{n\to\infty}\frac{\left[(a_{1}\cdot a_{2}\cdot ...\cdot a_{n})^{1/n}\right]^{2}}{a_{n}}\equal{}\lim_{n\to\infty}\frac{[a_{n}]^{2}}{a_{n}}$
[/quote]

Let $ a_{n}$ be a sequence with $ a_{n}> 0$ for every n. Then

$ \lim_{n\to\infty}\left(a_{1}\cdot a_{2}\cdot ...\cdot a_{n}\right)^{1/n}\equal{}\lim_{n\to\infty}a_{n}$.

This is known as the geometric mean limit. You can show it by using the aritmetic mean limit which, in turn, can be proved from Cesaro-Stolz theorem.


function
number theory proposed
number theory

1. Suppose $ f(d)\equal{}\sum_{d|n}g(n)$ where the summation is over all multiples of $ d$. Show that
\[ g(d)\equal{}\sum_{d|n}\mu\left(\frac{n}{d}\right)f(n)\]
and conversely, assuming that the series are absolutely convergent.

2. Show that if $ f(n)\equal{}\sum_{d^k|n}g\left(\frac{n}{d^k}\right)$ then
\[ g(n)\equal{}\sum_{d^k|n}\mu(d)f\left(\frac{n}{d^k}\right)\]
and conversely.

1.  This follows from the usual statement of Mobius inversion by taking the dual poset.  One has the relationship $ \mu(x, y) \equal{} \mu'(y, x)$ between the Mobius function of a poset and its dual.


trigonometry
analytic geometry
geometry
calculus
calculus computations

$ \frac{\partial}{\partial r} \equal{} \sin{\varphi} \frac{\partial}{\partial \rho}\plus{}\frac{\cos{\varphi}}{\rho}\frac{\partial}{\partial \varphi}$ is a Laplacian Operator in spherical coordinates.

How do I show that $ \frac{\partial^{2}}{\partial x^{2}}\plus{}\frac{\partial^{2}}{\partial y^{2}}\plus{}\frac{\partial^{2}}{\partial z^{2}}\equal{}\frac{\partial^{2}}{\partial \rho^{2}}\plus{}\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial \varphi^{2}}\plus{}\frac{1}{\rho^{2}\sin^{2}{\varphi}}\frac{\partial^{2}}{\partial \theta^{2}}\plus{}\frac{2}{\rho}\frac{\partial}{\partial \rho}\plus{}\frac{\cot{\varphi}}{\rho^{2}}\frac{\partial}{\partial \varphi}$?

I am using spherical coordinates to be $ (\rho, \varphi, \theta)$.

With the multivariable chain rule and a lot of pain. You can split the computation into two steps: $ \Delta f\equal{}\nabla\cdot \nabla f$ (Laplacian = divergence of the gradient). Both divergence and gradient are first-order differential operators, which makes the chain rule more manageable. You can check your work [url=http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates]here[/url].

I know I have to use the chain rule.  It looks like it will take hours.

Could somebody please help me get started?  If someone could show me $ \frac{\partial^{2}}{\partial x^{2}}$, that would be great and then I can do the rest.

[quote="JRav"]Could somebody please help me get started?  If someone could show me $ \frac {\partial^{2}}{\partial x^{2}}$, that would be great and then I can do the rest.[/quote]
let $ x\equal{}\rho\;\sin\varphi\;\cos\theta$
$ y\equal{}\rho\;\sin\varphi\;\sin\theta$
$ z\equal{}\rho\;\cos\varphi$

so, $ \rho\equal{}\sqrt{x^2\plus{}y^2\plus{}z^2}$
$ \varphi\equal{}\cos^{\minus{}1}\left(\dfrac{z}{\sqrt{x^2\plus{}y^2\plus{}z^2}}\right)$
$ \theta\equal{}\tan^{\minus{}1}\left(\frac{y}{x}\right)$

then,
$ \dfrac{\partial}{\partial x}\equal{}\dfrac{\partial \rho}{\partial x}\;\dfrac{\partial}{\partial \rho}\plus{}\dfrac{\partial \varphi}{\partial x}\;\dfrac{\partial}{\partial \varphi}\plus{}\dfrac{\partial \theta}{\partial x}\;\dfrac{\partial}{\partial \theta}$
$ \equal{}(\sin\varphi\;\cos\theta)\;\dfrac{\partial}{\partial \rho}\plus{}\left(\dfrac{\cos\varphi\;\cos\theta}{\rho}\right)\;\dfrac{\partial}{\partial \varphi}\minus{}\left(\dfrac{\sin\theta}{\rho\;\sin\varphi}\right)\;\dfrac{\partial}{\partial \theta}$

and so,
$ \dfrac{\partial^2}{\partial x^2}\equal{}(\sin\varphi\;\cos\theta)\;\dfrac{\partial\left((\sin\varphi\;\cos\theta)\;\dfrac{\partial}{\partial \rho}\plus{}\left(\dfrac{\cos\varphi\;\cos\theta}{\rho}\right)\;\dfrac{\partial}{\partial \varphi}\minus{}\left(\dfrac{\sin\theta}{\rho\;\sin\varphi}\right)\;\dfrac{\partial}{\partial \theta}\right)}{\partial \rho}$
$ \plus{}\left(\dfrac{\cos\varphi\;\cos\theta}{\rho}\right)\;\dfrac{\partial\left((\sin\varphi\;\cos\theta)\;\dfrac{\partial}{\partial \rho}\plus{}\left(\dfrac{\cos\varphi\;\cos\theta}{\rho}\right)\;\dfrac{\partial}{\partial \varphi}\minus{}\left(\dfrac{\sin\theta}{\rho\;\sin\varphi}\right)\;\dfrac{\partial}{\partial \theta}\right)}{\partial \varphi}$
$ \minus{}\left(\dfrac{\sin\theta}{\rho\;\sin\varphi}\right)\;\dfrac{\partial\left((\sin\varphi\;\cos\theta)\;\dfrac{\partial}{\partial \rho}\plus{}\left(\dfrac{\cos\varphi\;\cos\theta}{\rho}\right)\;\dfrac{\partial}{\partial \varphi}\minus{}\left(\dfrac{\sin\theta}{\rho\;\sin\varphi}\right)\;\dfrac{\partial}{\partial \theta}\right)}{\partial \theta}$

expand to get,
$ \equal{}(\sin\varphi\;\cos\theta)\;(\sin\varphi\;\cos\theta)\;\left(\dfrac{\partial}{\partial \rho^2}\right)$
$ \plus{}(\sin\varphi\;\cos\theta)\;\left(\minus{}\left(\dfrac{\cos\varphi\;\cos\theta}{\rho^2}\right)\dfrac{\partial}{\partial\varphi}\plus{}\left(\dfrac{\cos\varphi\;\cos\theta}{\rho}\right)\dfrac{\partial^2}{\partial\varphi\;\partial\rho}\right)$
$ \minus{}(\sin\varphi\;\cos\theta)\;\left(\minus{}\left(\dfrac{\sin\theta}{\rho^2\;\sin\varphi}\right)\;\dfrac{\partial}{\partial\theta} \plus{}\left(\dfrac{\sin\theta}{\rho\;\sin\varphi}\right)\;\dfrac{\partial^2}{\partial\theta\;\partial\rho}\right)$
$ \plus{}\left(\dfrac{\cos\varphi\;\cos\theta}{\rho}\right)\left((\cos\varphi\;\cos\theta)\;\dfrac{\partial}{\partial\rho}\plus{}(\sin\varphi\;\cos\theta)\;\dfrac{\partial^2}{\partial\rho\partial\varphi}\right)$
$ \plus{}\left(\dfrac{\cos\varphi\;\cos\theta}{\rho}\right)\left(\minus{}\left(\dfrac{\sin\varphi\;\cos\theta}{\rho}\right)\dfrac{\partial}{\partial\varphi}\plus{}\left(\dfrac{\cos\varphi\;\cos\theta}{\rho}\right)\dfrac{\partial^2}{\partial\varphi^2}\right)$
$ \plus{}\left(\dfrac{\cos\varphi\;\cos\theta}{\rho}\right)\left(\left(\dfrac{\sin\theta}{\rho\;\sin^2\varphi}\right)\;\dfrac{\partial}{\partial\theta} \minus{}\left(\dfrac{\sin\theta}{\rho\;\sin\varphi}\right)\;\dfrac{\partial^2}{\partial\theta\;\partial\varphi}\right)$
$ \minus{}\left(\dfrac{\sin\theta}{\rho\;\sin\varphi}\right)\;\left((\sin\varphi\;\sin\theta)\;\dfrac{\partial}{\partial\rho}\plus{}(\sin\varphi\;\cos\theta)\;\dfrac{\partial^2}{\partial\rho\partial\theta}\right)$
$ \minus{}\left(\dfrac{\sin\theta}{\rho\;\sin\varphi}\right)\;\left(\minus{}\left(\dfrac{\cos\varphi\;\sin\theta}{\rho}\right)\dfrac{\partial}{\partial\varphi}\plus{}\left(\dfrac{\cos\varphi\;\cos\theta}{\rho}\right)\dfrac{\partial^2}{\partial\varphi\partial\theta}\right)$
$ \minus{}\left(\dfrac{\sin\theta}{\rho\;\sin\varphi}\right)\;\left(\minus{}\left(\dfrac{\cos\theta}{\rho\;\sin\varphi}\right)\;\dfrac{\partial}{\partial\theta} \minus{}\left(\dfrac{\sin\theta}{\rho\;\sin\varphi}\right)\;\dfrac{\partial^2}{\partial\theta^2}\right)$

likewise, from $ \dfrac{\partial^2}{\partial y^2}$, you will get $ 9$ terms, and from $ \dfrac{\partial^2}{\partial z^2}$, you will get $ 4$ terms --- collectively, they will involve the co-efficients only those of $ \left(\dfrac{\partial^2}{\partial \rho^2},\;\dfrac{\partial^2}{\partial \varphi^2},\;\dfrac{\partial^2}{\partial \rho\;\partial\varphi},\;\dfrac{\partial}{\partial \rho},\;\dfrac{\partial}{\partial \varphi}\right)$. you then need to collect all of the like terms and add them up. i will show the cases below for $ \dfrac{\partial^2}{\partial \rho^2}$ and $ \dfrac{\partial}{\partial \rho}$

the co-efficient of $ \dfrac{\partial^2}{\partial \rho^2}$ is $ \left(\cos^2\varphi\plus{}\sin^2\varphi\;\sin^2\theta\plus{}\sin^2\varphi\;\cos^2\theta\right)$ which becomes $ 1$ upon simplification, and, the co-efficient of $ \dfrac{\partial}{\partial \rho}$ is $ \left(\frac{\sin^2\varphi}{\rho}\plus{}\frac{\cos^2\varphi\;\sin^2\theta}{\rho}\plus{}\frac{\cos^2\theta}{\rho}\plus{}\frac{\cos^2\varphi\;\cos^2\theta}{\rho}\minus{}\frac{\sin^2\theta}{\rho}\right)$, which after simplification is $ \frac{2}{\rho}$

similarly, get the co-efficients for the rest $ 7$, which are $ \left(\dfrac{\partial}{\partial \varphi},\;\dfrac{\partial^2}{\partial\varphi^2},\;\dfrac{\partial}{\partial \theta},\;\dfrac{\partial^2}{\partial\theta^2},\;\dfrac{\partial^2}{\partial \rho\;\partial\theta},\;\dfrac{\partial^2}{\partial \rho\;\partial\varphi},\;\dfrac{\partial^2}{\partial \varphi\;\partial\theta}\right)$, and you will see that the co-efficients for $ 4$ of those are just plain $ 0$ (upon simplification), they are those of $ \left(\dfrac{\partial}{\partial \theta},\;\dfrac{\partial^2}{\partial \rho\;\partial\theta},\;\dfrac{\partial^2}{\partial \rho\;\partial\varphi},\;\dfrac{\partial^2}{\partial \varphi\;\partial\theta}\right)$

collecting the terms for $ \dfrac{\partial^2}{\partial \rho^2}$ and $ \dfrac{\partial}{\partial \rho}$ together gives $ \left(\dfrac{\partial^2}{\partial \rho^2}\cdot 1\;\plus{}\;\dfrac{\partial}{\partial \rho}\cdot \left(\frac{2}{\rho}\right)\right)$, which can be compactly written as $ \dfrac{1}{\rho^2}\;\dfrac{\partial}{\partial \rho}\;\left(\rho^2\cdot \dfrac{\partial}{\partial \rho}\right)$. i have given you here $ 2$ of the $ 5$ terms of the overall laplacian, i.e. those of $ \dfrac{\partial^2}{\partial \rho^2}$ and $ \dfrac{\partial}{\partial \rho}$. you then only need to find the other $ 3$, those of $ \left(\dfrac{\partial^2}{\partial \varphi^2},\;\dfrac{\partial}{\partial \varphi},\;\dfrac{\partial^2}{\partial \theta^2}\right)$, and write them in a similar compact form.

at the end, you should eventually get the answer
$ \nabla^2\equal{}\boxed{\dfrac{1}{\rho^2}\;\dfrac{\partial}{\partial \rho}\;\left(\rho^2\cdot \dfrac{\partial}{\partial \rho}\right)\;\plus{}\;\dfrac{1}{\rho^2\;\sin^2\varphi}\left(\dfrac{\partial^2}{\partial \theta^2}\plus{}\sin\varphi\;\frac{\partial\left(\sin\varphi\;\dfrac{\partial}{\partial \varphi}\right)}{\partial\varphi}\right)}$

There ought to be a geometric derivation of this formula. Consider a spherical box with opposite vertices $ (\rho,\theta,\phi)$ and $ (\rho\plus{}d\rho,\theta\plus{}d\theta,\phi\plus{}d\phi)$. By the divergence theorem the flux of $ \nabla f$ out of the box is approximately $ \Delta f dV$, where $ dV\equal{}\rho^2 \sin\phi\,d\rho\,d\theta\,d\phi$. Now, the surface area of the inner spherical face is $ \rho\sin\phi\,d\theta\,d\phi$, so the flux through it can be estimated by $ \minus{}\frac{\partial f}{\partial \rho}\rho^2\sin\phi\,d\theta\,d\phi$. Combining this with the contribution of the outer spherical face, we get approximately $ \frac{1}{\rho^2}\frac{\partial}{\partial \rho}\left(\rho^2 \frac{\partial f}{\partial \rho}\right)dV$, hence the first term of the Laplacian. And so on...


Please show your neat steps. Thank you.

1. One angle is 6* (degrees) more than its complement. Find the measure of both angles.
2. The measures of the angles of a given triangles are x*, (2x)*, and (3x-12). Find the measure of all 3 angles.
3. L = a + (n-1)d, for n
4. 1/3(x-10)-4x>5/6(2x+1)-1 Note: show in set notation, graph, and internal notation
5. Joe wants to buy a $20,000 car. If the dealership offers 4.4% interest on a 4-year loan, how much will he actually pay for the car if he makes monthly payments? How much will 1 monthly payment be?

[hide="#2"]
$ x\plus{}2x\plus{}3x\minus{}12 \equal{} 180$
$ 6x\minus{}12 \equal{} 180$
$ 6x \equal{} 162$
$ x \equal{} 27$
So our angles are $ \boxed{27}$, $ \boxed{54}$ and $ 3(27)\minus{}12$ or $ \boxed{69}$
[/hide]

[hide="#5"]

Just use the formula $ A \equal{} P\left(1\plus{}\frac{R}{N}\right)^{NT}$

Where A is total amount earned, P is base amount, R is the "interest rate", N is number of times money is compounded per year and T is the time, in years.

[/hide]

[hide="#1"]Let the angle be $ x^{\circ}$. Then $ x \equal{} 90\minus{}x\plus{}6$ or $ x \equal{}48^{\circ}$, and $ 90\minus{}x \equal{} 42^{\circ}$.[/hide]

[quote="Anonymous52111"]Please show your neat steps. Thank you.
2. The measures of the angles of a given triangles are x*, (2x)*, and (3x-12). Find the measure of all 3 angles.
?[/quote]
angle sum of a triangle is $ 180^{o}$. So $ x\plus{}2x\plus{}3x\minus{}12 \equal{} 180^{o}$
$ 6x\minus{}12 \equal{} 180$
$ 6x \equal{} 192$
$ x \equal{} 32$

sub in for x for each angle and obtain the other 2 angles

-------
[quote="Prime factorization"][hide="#2"]
$ x\plus{}2x\plus{}3x\minus{}12 \equal{} 180$
$ 6x\minus{}12 \equal{} 180$
[/hide][/quote]
qn 2 6x=192

1. Let x be the angle so 90-x+6=x so 2x=96 and the angles measure 48 and 42 degrees.
2. 6x-12=180 so 6x=192. The angles are 32, 64, and 84 degrees.

[quote="Anonymous52111"]Please show your neat steps. Thank you.

1. One angle is 6* (degrees) more than its complement. Find the measure of both angles.
2. The measures of the angles of a given triangles are x*, (2x)*, and (3x-12). Find the measure of all 3 angles.
3. L = a + (n-1)d, for n
4. 1/3(x-10)-4x>5/6(2x+1)-1 Note: show in set notation, graph, and internal notation
5. Joe wants to buy a $ \$$20,000 car. If the dealership offers 4.4% interest on a 4-year loan, how much will he actually pay for the car if he makes monthly payments? How much will 1 monthly payment be?[/quote]
set notation im not very good at but ill do invterval notation and try set. im too lazy to graph

multiply both sides by 6 to get rid of he fractions and then distribute.
2x-20-4x>10x+5
-2x-20>10x+5
-25>12x
-25/12>x
interval:$ (\minus{}\infty,\minus{}25/12)$
set:$ {x|\minus{}25/12>x}$ i think thats it.
graph: once agin, too lazy

#3
L = a + (n-1)d for n
L - a = (n-1)d
(L-a)/d = n-1
n = [(L-a)/d]+1
By the way, this happens to be the formula for finding the "n"th term (L) of an arithmetic sequence with first term "a" and the difference "d".


$ F_0 \equal{} 1, F_1 \equal{} 1, F_n \equal{} F_{n \minus{} 1} \plus{} F_{n \minus{} 2} \text{ } \forall \text{ } n \ge 2$.

Prove that the polynomial given by $ \displaystyle \sum_{i \equal{} 0}^{n} F_ix^i$ has exactly one real root when $ n$ is even and has no real roots when $ n$ is odd.

Edit: Changed $ n$ to $ i$.

Interestingly enough, that's not what people usually mean by [url=http://mathworld.wolfram.com/FibonacciPolynomial.html]Fibonacci polynomial[/url].  Also, you probably want $ F_i x^i$.

[hide="Large hint"] Multiply by $ 1 \minus{} x \minus{} x^2$. [/hide]

[hide="Inspiration"]
The generation function of the Fibonacci sequence is $ G(x)\equal{}\frac 1{1\minus{}x\minus{}x^2}$.
[/hide]


I feel teamwork is imperative. Thus, I feel that I shud learn about team rounds. Can some people help me with these problems:

#2. A recipe that makes 48 cookies calls or 2 1/4 cups of flour. A 5 pound bag of flour contains 20 cups. How many pounds of flour are needed for 960 cookies.

[hide]The answer is 11.25 pounds, how do u do it?[/hide]

#3. An 8 x 8 checkerboard has alternating black and white squares. How many distinct squares, with sides on the grid lines of the checkerboard (horizontal and vertical) and containing atleast 4 black squares, can be drawn on the checkerboard?

[hide]The answer is 91 squares, how do u do it?[/hide]

2. Ok so it says you can make 48 cookies with 2 1/4 cups. How many are needed for 960 cookies. Since 960/48 is 20, you multiply 2 1/4 by 20 to get 45

Since a 5 pound flower bag equals 20 cups you use a proportion

45/20=x/5

By cross multiplying you get

225=20x

x=11.25

3. Well if you think about it, it has to be at least 3 squares wide to carry at least 4 black squares.  So find all 3 squares wide.

It's 36

Find all 4 squares wide

It's 25

Find all 5 squares wide

It's 16

Find all 6 squares wide

It's 9

Find all 7 squares wide

It's 4

Find all 8 sqaures wide

It's 1

So

36+25+16+9+4+1=91

A faster way is recognizing that the amount of sqaures for each size is 6^2, 5^2...

This is the kind of kid that I think deserves [color=red]a lot[/color] of credit. Thank you very much.
One more thing.  There were more questions I have wrong. Do u happen to have that test, so u can explain to me the answers to all the other questions?

well I don't have those tests, and in fact I've never seen those problems. So if you post more problems here, or pm them to me I can answer them and give solutions.


topology
analytic geometry
logarithms
real analysis
real analysis unsolved

In Rn diffeomorphisms can be locally composed into a composition of finitely many elementary diffeomorphisms (diffeomorphisms which alter at most one coordinate). Is there a counterexample of where this cannot be done globally (on a domain). Is there a counterexample of where the analogous decomposition cannot be done for a homeomorphism?

The map given in polar coordinates by $ (r,\theta)\mapsto (r,\theta \plus{} \log r)$ is a bi-Lipschitz homeomorphism, but I would guess that it cannot be factored into finitely many elementary homeomorphisms... (I have no proof, just the vague idea that you can only create a finite amount of "winding" with an elementary homeomorphism).

Similarly, one can twist the open unit disk by the diffeomorphism $ (r,\theta)\mapsto (r,\theta \plus{} \log (1 \minus{} r))$, possibly creating the same obstruction to global decomposition.


linear algebra
matrix
quadratics
linear algebra unsolved

There are two possibilities for the next problem: either it is very challenging and nice ( which I doubt), either it is wrong (as usual):
   Prove or disprove: for any symmetric positive matrix $A\in SL_n(Z)$ with $n\leq 4$ there is a matrix $ B\in GL_n(Z)$ such that $ A=B^{t}B$. What about if $n>4$?  For $n=2$ this implies: if $ ab=c^2+1$ and $a,b,c$ are positive integers, then there are numbers $ x,y,z,t$ such that $ a=x^2+y^2, b=z^2+t^2, 1=|xt-yz|, c=xz+yt$, which, as far as I remember, was in a very very old shortlist for IMO.

This is equivalent to asking whether any strictly positive quadratic form on $Z^n$ with determinant $+1$ is equivalent to the standard Euclidean form.  It is true for $n \le 4$: see "Symmetric bilinear forms" by Milnor and Husemoller (Springer, 1973) for example, and the proof uses Minkowski's theorem.

Yes, I also proved it like this, but unfortunately I did not know whether it is classical result. I apologize. Yet, could you give a counter-example for $n=5$?

The result is stated, without proof, to be true for $n \le 7$ and a counterexample is given for $n=8$.

And yet as far as I know, Mordell constructed in 1937 or 1938 a counter-example for $n=6$, but I cannot understand it, since it says smply that it is the linear form associated with the Dynkine lattice $E_6$. I have no idea of what this means.  :?

No, the Cartan matrix for $E_6$ has determinant 3 and that for $E_7$ has determinant 2.  The matrix for $E_8$ has determinant 1 and is indeed not equivalent to the usual inner product: indeed it is of "type II", i.e. all its values are even.

There's a [url=http://en.wikipedia.org/wiki/Root_system]good Wikipedia article[/url] on the subject.


geometry
rectangle
perimeter
probability
ratio
circumcircle
3D geometry

1. The diagonals of a rectangle intersect at point P. Point P is 5 centimeters further from the shorter side than the longer side. The perimeter of the rectangle is 44 centimeters. What is the number of square centimeters in the area of the rectangle?(98 Nats#27)
[hide="Answer"]96 square centimeters[/hide]
2. Great Aunt Minnie's age the square of Great Uncle Paul's age is 7308. Great Uncle Paul's age plus the square of Great Aunt Minnie's age is 6974. What is the sum of Great Aunt Minnie's age and Great Uncle Paul's Age?(98 Nats#29)
[hide="Answer"]168. I tried the nearest square number, but got 169..[/hide]
3. In $ \triangle ABC,\angle C$ is a right angle. Point $ M$ is the midpoint of $ \overline{AB},$ point N is the midpoint of $ \overline{AC},$ and point $ O$ is the midpoint of $ \overline{AM},$. The perimeter of $ \triangle ABC$ is 112cm, and $ ON\equal{}12.5$cm. What is the number of square centimeters in the area of quadrilateral $ MNCB?$(00 Nats#17)
[hide="Answer"]252 square centimeters [/hide]
4. The probability of drawing a red marble from a jar of marbles is $ \frac15$. If three marbles are drawn with replacement, what is the probability that none of the marbles drawn is red? Express your answer as a common fraction.(00 Nats#20)
[hide="Answer"]$ \frac{64}{125}$[/hide]
5. A radio station in central Pennsylvania is having a birthday contest. One of the 366 days of the year is randomly selected with equal probability. If the month of a contestant's birthday matches the month selected- or the day of the month of the contestant's birthday mathces the day of the month selected-or both- the contestant wins a compact disc. Otherwise, the contestant wins nothing. What is the probability that a contestant with a December 31 birthday will win? Express your answer as a common fraction.(02 Nats#9)
[hide="Answer"]$ \frac{37}{366}$[/hide]
6. How many points $ (a,b)$ satisfy both $ a^2\plus{}b^2\equal{}25$ and the area of the triangle with vertices $ (\minus{}5,0),(5,0),(a,b)$ is 10?(02 Nats#20)
[hide="Answer"]4 points[/hide]
7. What is the ratio of the area of a quare inscribed in a semicircle with radius $ r$ to the area of a square inscribed in circle with radius $ r$? Express your answer as a common fraction.(02 Nats #24)
[hide="Answer"]$ \frac25$[/hide]
8. The radius of the inscribed circle is 6cm. What is the number of centimeter in the length of $ \overline{AB}?$ Express your answer in simplest radical form.*Picture attached below.(01 Nats #22)
[hide="Answer"]$ 12\plus{}12\sqrt{3}$[/hide]

[hide="1"]Let $ x$ be the length of the shorter side.  Then the length of the longer side is $ x \plus{} 10$.

So $ 4x \plus{} 20 \equal{} 44$
$ x \equal{} 6$

And the area is 96.[/hide]
[hide="2"]
Let  $ A$ be the Aunt's age and $ U$ be the uncle's age.

$ A \plus{} U^2 \equal{} 7308$
$ A^2 \plus{} U \equal{} 6974$
Subtracting gives $ U^2 \minus{} A^2 \plus{} A \minus{} U \equal{} 334$
$ (U \minus{} A)(U \plus{} A \minus{} 1) \equal{} 334$
$ 334 \equal{} 2*167$

Thus $ U \minus{} A \equal{} 2$ and $ U \plus{} A \equal{} 168$.  Our desired answer is then $ 168$[/hide]

[hide="3"]It is easy to see that $ \bigtriangleup ANM \sim \bigtriangleup ACB$.

The ratio is $ \frac {1}{2}$.  Thus, the area of quadrilateral $ MNCB$ is $ \frac {3}{4}$ of the area of triangle ABC.
Since $ ON \equal{} OM \equal{} OA$ because $ O$ is the center of the circumcircle of triangle ANM, we can see that $ AM \equal{} 25\Longrightarrow AB \equal{} 50$.

So $ AC \plus{} BC \equal{} 62$ while $ AC^2 \plus{} BC^2 \equal{} 2500$
It is easy to see that $ AC \equal{} 14$ and $ BC \equal{} 48$, as $ 62$ is a multiple of $ 31 \equal{} 24 \plus{} 7$.

Therefore, our desired area if $ \frac {3}{4}*24*14 \equal{} 252$.[/hide]

[hide="4"]
The probability that one marble is not red is $ \frac {4}{5}$.  Since it is without replacement, we cube this to get the desired answer.[/hide]

[hide="5"]We simply need to find the number of days that satisfy the conditions.

All days in December work, as well as the 31st day of every month (that has 31 days).

Those in december: 31
January, March, May, July, August, October: 6 more.

Thus, $ \frac {37}{366}$[/hide]

[hide="6"]It is easy to see that $ b\equal{}\pm 2$.

Since there are 2 values of a that correspond to those two values of b, there are 4 points.[/hide]

[hide="7"]draw the radius from the center of the semicircle to the two points of the square that lie on the semi circle.

Obviously, the center of the circle is the midpoint of one of the sides.

It is then easy to see that the area of the square can be expressed as $ \frac{4}{5}r^2$.

The area of the other square is $ 2r^2$ (diagonals are both 2r).

Thus, our ratio is $ \frac{2}{5}$[/hide]
[hide="8"]
Let $ X$ be the tangent point to the circle on $ BC$, where $ C$ is the right angle.

Then, $ BX\equal{}BY$, where $ Y$ is the tangent point on $ AB$.

Draw $ BO$, where $ O$ is the center of the circle to see that  $ BY\equal{}6\sqrt{3}\equal{}BX$

Thus the total length of $ BC$ is $ 6\plus{}6\sqrt{3}$.

Then our desired answer is twice of this, $ 12\sqrt{3}\plus{}12$[/hide]

sorry mathcrazed, but I still dont get what you are saying for number 6.
BTW: I just switched my browser to Mozilla Firefox!

For #6, $ (a, b)$ is on a circle with radius five, and there are exactly four points on that circle such that the triangle with the given vertices has area 10.


I tried to solve this equation but.......... well, you try

Solve for the variables

z=5(1.02)^x+7(1.05)^y
z=40(.97)^x+50(.95)^y
z=12x+15y

I recant what I said. I misread the problem at first glance, apologies.

White horse king,  why shouldn't this be on Pre-Olympiad??  Away, go ahead and try substituing and solving.   If you can, I would like an answer.

This is not really an olympiad problem because an answer cannot really be found under olympiad conditions (that is, with only pencil and paper, and on a time limit).  Hence, it isn't suitable for Pre-Olympiad.

This problem does not have an algebraic solution.  X.x


Prove $ (\sin {A}\plus{}\cos {A})(1\minus{}\sin {A}\cos {A}\equal{}\cos^{3} {A}\plus{}\sin^{3} {A}$

$ x^3\plus{}y^3\equal{}(x\plus{}y)(x^2\minus{}xy\plus{}y^2)$

$ \cos^3A\plus{}\sin^3A\equal{}(\cos A\plus{}\sin A)(\cos^2A\minus{}\cos A\sin A\plus{}\sin^2A)$

$ \cos^2A\plus{}\sin^2A\equal{}1$

therefore, $ \cos^3A\plus{}\sin^3A\equal{}(\cos A\plus{}\sin A)(1\minus{}\cos A\sin A)$


geometry
ratio
trigonometry
articles
AoPSwiki
geometric transformation
reflection

In $\triangle ABC$, $AB = 360$, $BC = 507$, and $CA = 780$.  Let $M$ be the midpoint of $\overline{CA}$, and let $D$ be the point on $\overline{CA}$ such that $\overline{BD}$ bisects angle $ABC$.  Let $F$ be the point on $\overline{BC}$ such that $\overline{DF} \perp \overline{BD}$.  Suppose that $\overline{DF}$ meets $\overline{BM}$ at $E$.  The ratio $DE: EF$ can be written in the form $m/n$, where $m$ and $n$ are relatively prime positive integers.  Find $m + n$.

Make use of Menelaus' theorem...............

[hide="hint:"]make use of the fact that $DE : EF = \frac{\tan{\angle MBF}}{\tan{\angle DBM}} - 1$[/hide]

Why do i keep getting a different answer from kalva..??

well.... u use Menaleus Theorem twice in this pbm. right?

I'll try one more time...

[hint]I believe you apply it to $\triangle GBF$ and $\triangle DBC$[/hint] should I just post Kalva's solution here?

[quote="GoKorea"]I got 240/49,,

Does anyone know the answer or where the answer is?[/quote]

you are right.

Just one more step, since its m/n, m+n=240+49=289

http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2003-aa/03answersaime1.html

number 15.

[quote="nsato"]In $ \triangle ABC$, $ AB = 360$, $ BC = 507$, and $ CA = 780$.  Let $ M$ be the midpoint of $ \overline{CA}$, and let $ D$ be the point on $ \overline{CA}$ such that $ \overline{BD}$ bisects angle $ ABC$.  Let $ F$ be the point on $ \overline{BC}$ such that $ \overline{DF} \perp \overline{BD}$.  Suppose that $ \overline{DF}$ meets $ \overline{BM}$ at $ E$.  The ratio $ DE: EF$ can be written in the form $ m/n$, where $ m$ and $ n$ are relatively prime positive integers.  Find $ m + n$.[/quote]
For computation, instead consider the triangle as above except $ AB=120,BC=169,CA=260$. In the following, let the name of a point represent the mass located there.
By the angle bisector theorem, we can place mass points on $ C,D,A$ of $ 120,289,169$ resp. Thus, a mass of $ \frac{289}{2}$ belongs at $ F$ (seen by reflecting $ F$ across $ BD$, to an image which lies on $ AB$).
Having determined $ CB/CF$, we reassign mass points to determine $ FE/FD$. This setup involves $ \tri CFD$ and transversal $ MEB$. For simplicity, put a masses of $ 240,289$ at $ C,F$. To find the mass we should put at $ D$, we compute $ CM/MD$: applying the angle bisector theorem again and using the fact $ M$ is a midpoint, we find
\[ \frac{CM}{MD}= \frac{169\cdot\frac{260}{289}-130}{130}=\frac{49}{289}\]
At this point we could find the mass at $ D$ but it's unnecessary.\[ \frac{DE}{EF}=\frac{F}{D}=\frac{F}{C}\frac{C}{D}=\frac{289}{240}\frac{49}{289}=\boxed{\frac{49}{240}}\]

Where can i learn about assigning mass points for geometry problems? (like how it was used in me=home's solution?)

Here are some great articles on the topic:
http://sanjosemathcircle.org/handouts/2007-2008/20071114.pdf
http://mathcircle.berkeley.edu/archivedocs/1999_2000/lectures/9900lecturespdf/mpgeo.pdf

Hope it helped.

You can also check out the AoPSWiki:
http://www.artofproblemsolving.com/Wiki/index.php/Mass_points

[hide="Elementary Solution"]
We will solve this problem in general, letting $a = BC$, $b = AC$, and $c = AB$. The diagram uploaded by [b]me@home[/b] will suffice for my solution.

Because we want to find $DE:EF$, it is intuitive to draw a line through $F$ parallel to $\overline{BM}$ that intersects $\overline{AC}$ at $K$. From the similar triangles created, we find that the problem is easily solved once we compute $BF:FC$.

To reduce the clutter in the diagram, remove the median temporarily. Note that $\triangle GBF$ is isosceles because $\angle GBD = \angle FBD$ and $\overline{BD} \perp \overline{GF}$. Consider the point $H$ on $\overline{BF}$ such that $BH = AB = c$. It can easily be shown that $\triangle DAG \cong \triangle DHF$ as a consequence of $\triangle GBF$ being isosceles and the definition of $H$. It follows that $AD = DH$ and $\angle ADG = \angle HDF$. By vertical angles, $\angle ADG = \angle FDC$. Thus, $\overline{DF}$ is an angle bisector of $\triangle HDC$.

From the angle bisector theorem on $\triangle ABC$, we have $AD = \frac{bc}{a + c}$. Furthermore, we have $DH = AD = \frac{bc}{a + c}$. Applying the angle bisector theorem on $\triangle HDC$, we find $HF = \frac{c(a - c)}{a + c}$ and $FC = \frac{a(a - c)}{(a + c)}$. Moreover, $BF = BH + HF = \frac{2ac}{a + c}$. Thus, $\frac{BF}{FC} = \frac{2c}{a + c}$.

Going back to the similar triangles we created, we have $\triangle CFK \sim \triangle CBM$. So we have $\frac{CK}{MK} = \frac{FC}{BF} = \frac{a + c}{2c}$. Because $CM = \frac b2$, it follows that $MK = \frac{2bc}{2(a + c)}$. We also have $DM = CD - CM = \frac{ab}{a + c} - \frac b2 = \frac{b(a - c)}{2(a + b)}$.

Finally, we have $\triangle DEM \sim \triangle DFK$. So $\frac{DE}{EF} = \frac{DM}{MK} = \frac{a - c}{2c}$. Interestingly, the answer is independent of the length of $AC$.

We are given that $a = 507$ and $c = 360$ from the problem so the ratio is $\frac{507 - 360}{2\cdot 360} = \frac{49}{240}$ so $m + n = \boxed{289}$.
[/hide]

[hide="Let's pwn this problem"]
Let $Y$ be a point on line segment $\overline{BC}$ such that $AY\perp BD$, and let $X=AY\cap BD$, $Z=BM\cap AY$.  Note that $Y$ is the reflection of $A$ across $\overline{BD}$, so $BY=BA=360$ and $YC=507-360=147$.

Note that $AY\parallel DF$, and furthermore $AX=XY$.  Therefore $\triangle AXM\sim\triangle AYC$, so $XM=\tfrac12YC=\tfrac{147}2$.  Additionally, $XM\parallel BC$ so $\triangle BYZ\sim\triangle MXZ$.  Now \[\dfrac{DE}{EF}=\dfrac{XZ}{ZY}=\dfrac{XM}{YB}=\dfrac{147/2}{360}=\dfrac{49}{240}\] and the requested answer is $\boxed{289}$.[/hide]

[hide="Thank you USAMTS/Yufei"]
Let $D'$ be the projection of $C$ onto $BD$, and select $M'\equiv BM\cap CD'$. Now, it is well-known that $D'$ lies on the $C$-midline. Thus, we have that $$\frac{DE}{EF}=\frac{D'M'}{M'C'}$$ Letting $a=BC$ and cyclic variations, in barycentric coordinates we can derive that $D'\equiv (a:c-a:c)$ and $M'\equiv (a:c-a:a)$. Then we get $$\frac{CD'}{CM'}=\frac{1-\frac{1}{2}}{1-\frac{a}{a+c}}=\frac{a+c}{2c}=\frac{289}{240}\Longrightarrow \frac{D'M'}{M'C'}=\frac{D'C}{MC'}-1=\frac{49}{240}$$ whence $49+240=\boxed{289.}$
[/hide]

[quote=shiningsunnyday][hide="Even quicker"]
From cross-ratios \[(A, M; D, C) \stackrel{B}{=} (A', E; D, F)\] where$A' = AB \cap DF.$
[/hide][/quote]

Please how can DA' , DE, FA' & FE be determined?


Question 2:

In the convex quadrilateral $ABCD$, points $M$ and $N$ lie on the side $AB$ such $AM=MN=NB$, and points $P,Q$ lie on the side $CD$ such that $CP=PQ=QD$. Prove that

Area of $AMCP$ $=$ Area of $MNPQ$ $=$ $\frac{1}{3}$ Area of $ABCD$.

[hide]$[ACM]=(1/3)[ACB]$, since AMC's base is 1/3 of ABC's, and they share the same height.
Similarly, $[ACP]=(1/3)[ACD]$.
$[AMCP]=[AMC]+[ACP]=(1/3)([ACB]+[ACD])=(1/3)[ABCD]$.

$[APM]=[MPN]$, since they share the same base length and height.
Similarly, $[CMP]=[PMQ]$.
$[APM]+[CMP]=[MPN]+[PMQ]\implies [AMCP]=[MNPQ]$.
[/hide]


trigonometry
inequalities unsolved
inequalities

an easy problem but i can't make it :(  :( :
Prove:$(\cos^{n}A+\sin^{n}A)^{n+1}>(\cos^{n+1}A+\sin^{n+1}A)^{n} (0<A<\frac{\pi}{2})$


superior algebra
superior algebra unsolved

If a and b are elements of a group satisfying $a^-1.b^2.a^1$  =$b^3$ and

$b^-1.a^2.b^1$   =$a^3$       Prove that a=1=b.

The authors claim that this is a very hard problem .Do you agree?

Hello Michael!

There is a rather short proof: I found one that basically consists of three equation chains with 16 terms all in all. But this tells you nothing about how easy it is to find these three lines of text. ;)
Many of the basic properties like this have tricky proofs (one never sees where to use which axiom), and although I was guided by an automatically generated proof (using Prover9), it took me three hours to find my short proof.

LiMa

Playing further with the prover, I found out this:

If G is a left-cancellable semigroup (the operation is associative and $\forall x \forall y \forall z\ (xy = xz \Rightarrow y = z)$) and a, b are elements of G with $ba^{3}=a^{2}b$ and $ab^{3}=b^{2}a$, then for all g in G we have ga = g. EDIT: the correct conclusion is "ag = g" (see below).

Obviously this is equivalent to a = e for a cancellable monoid (or a group) G.

LiMa

Hello LiMa
Thanks for the details concerning semigroups and monoids in your second post-this adds an interesting dimension to the problem.
Also thanks for the time you have put into finding the proof.It looks as if the authors claim is indeed correct

Michael N.

[quote="LiMa"]
If G is a left-cancellable semigroup (the operation is associative and $\forall x \forall y \forall z\ (xy = xz \Rightarrow y = z)$) and a, b are elements of G with $ba^{3}=a^{2}b$ and $ab^{3}=b^{2}a$, then for all g in G we have ga = g.

LiMa[/quote]
Hi,can you tell me why it is ture?

Hm... somewhere between the prover and the forum I had messed up something. The proposition I posted 08 Jun 2006 is wrong, as can be seen from the example $G = \{0, 1\}, x y : = y, a = b = 0, g = 1$.
Correct is this proposition, which has a slightly different conclusion:

If G is a left-cancellable semigroup (the operation is associative and $\forall x \forall y \forall z\ (xy = xz \Rightarrow y = z)$) and a, b are elements of G with $ba^{3}=a^{2}b$ and $ab^{3}=b^{2}a$, then for all g in G we have $ag = g$.

The automatic proof I have is 29 to 94 lines long (depending on the level of detail you want), and I have no intention to refactor it into a human readable form, but I post it here in almost original form.

LiMa

[hide="automatic proof"]
This proof differs from the one Prover9 produced by removing the operation sign and parentheses and the lines in which the law of associativity is used. "para" stands for paramodulation, which is a kind of substitution.

1 x y z = x y z.  [input] % associativity
2 b a a a = a a b.  [input] % axiom 1
3 a b b b = b b a.  [input] % axiom 2
4 a c != c.  [input] % negated conclusion: there is a constant 'c' such that...
5 x y != x z | y = z.  [clausify] % left-cancellation, '|' is 'or'

6A a a b x = b a a a x.  [para (2 (a 1) 1 (a 1 1))]
6 b a a a x = a a b x.  [copy 6A flip a]
7A b b a x = a b b b x.  [para (3 (a 1) 1 (a 1 1))]
7 a b b b x = b b a x.  [copy 7A flip a]
8 x a c != x c.  [resolve (5 b 4 a)]
9A x y z != x y u | z = u.  [para (1 (a 1) 5 (a 1))]
9 x y z != x y u | z = u.  [para (1 (a 1) 9A (a 2))]
10A b a a b b a = a a b b b b.  [para (3 (a 1) 6 (a 1 2 2 2))]
10 b a a b b a = a b b a b.  [para (7 (a 1) 10A (a 2 2))]
11 x y a c != x y c.  [resolve (5 b 8 a)]
12 x y z a c != x y z c.  [resolve (5 b 11 a)]
13A x y z u != x y z v | u = v.  [para (1 (a 1) 9 (a 1 2))]
13 x y z u != x y z v | u = v.  [para (1 (a 1) 13A (a 2 2))]
14A a b b a a b = b b a a a a.  [para (2 (a 1) 7 (a 1 2 2 2))]
14 a b b a a b = b a a b a.  [para (6 (a 1) 14A (a 2 2))]
15A b a a b b a x = a a b b b b x.  [para (7 (a 1) 6 (a 1 2 2 2))]
15 b a a b b a x = a b b a b x.  [para (7 (a 1) 15A (a 2 2))]
16A a b b a a b x = b b a a a a x.  [para (6 (a 1) 7 (a 1 2 2 2))]
16 a b b a a b x = b a a b a x.  [para (6 (a 1) 16A (a 2 2))]
17 x y z u v a c != x y z u v c.  [resolve (9 b 12 a)]
18 a a b x != b a a y | a x = y.  [para (6 (a 1) 13 (a 1))]
19A b a a b a a b x = a b b a b a a x.  [para (6 (a 1) 15 (a 1 2 2 2 2))]
19 a b b a b a a x = b a a b a a b x.  [copy 19A flip a]
20A a b b a b b a b x = b b a a a b b a x.  [para (15 (a 1) 7 (a 1 2 2 2))]
20B a b b a b b a b x = b a a b b b a x.  [para (6 (a 1) 20A (a 2 2))]
20 a b b a b b a b x = b a b b a a x.  [para (7 (a 1) 20B (a 2 2 2))]
21 a b b a b c != b a a b b c.  [para (15 (a 1) 17 (a 1))]
22A b a a b b b a a b a = a b b a b b b a a b.  [para (14 (a 1) 15 (a 1 2 2 2 2 2))]
22B b a b b a a a b a = a b b a b b b a a b.  [para (7 (a 1) 22A (a 1 2 2))]
22C b a b a a b b a = a b b a b b b a a b.  [para (6 (a 1) 22B (a 1 2 2 2))]
22D b a a b b a b = a b b a b b b a a b.  [para (10 (a 1) 22C (a 1 2 2))]
22E a b b a b b = a b b a b b b a a b.  [para (15 (a 1) 22D (a 1))]
22F a b b a b b = a b b b b a a a b.  [para (7 (a 1) 22E (a 2 2 2 2))]
22G a b b a b b = a b b b a a b b.  [para (6 (a 1) 22F (a 2 2 2 2 2))]
22H a b b a b b = b b a a a b b.  [para (7 (a 1) 22G (a 2))]
22I a b b a b b = b a a b b b.  [para (6 (a 1) 22H (a 2 2))]
22 a b b a b b = b a b b a.  [para (3 (a 1) 22I (a 2 2 2))]
23A a a b b b a b c != b a a b a a b b c.  [resolve (18 b 21 a)]
23B a b b a a b c != b a a b a a b b c.  [para (7 (a 1) 23A (a 1 2))]
23C b a a b a c != b a a b a a b b c.  [para (16 (a 1) 23B (a 1))]
23 b a a b a a b b c != b a a b a c.  [copy 23C flip a]
24A b a b b a x = a b b a b b x.  [para (22 (a 1) 1 (a 1 1))]
24 a b b a b b x = b a b b a x.  [copy 24A flip a]
25A b a b b a a b x = b a b b a a x.  [para (24 (a 1) 20 (a 1))]
25B b b a a b a x = b a b b a a x.  [para (16 (a 1) 25A (a 1 2))]
25 b a b b a a x = b b a a b a x.  [copy 25B flip a]
26A a b b a b b a x = b a a b a b b x.  [para (7 (a 1) 16 (a 1 2 2 2 2))]
26B b a b b a a x = b a a b a b b x.  [para (24 (a 1) 26A (a 1))]
26C b b a a b a x = b a a b a b b x.  [para (25 (a 1) 26B (a 1))]
26 b a a b a b b x = b b a a b a x.  [copy 26C flip a]
27A a b b a a a b b a b x = b a a b a a a b b a x.  [para (15 (a 1) 16 (a 1 2 2 2 2 2))]
27B a b a a b b b a b x = b a a b a a a b b a x.  [para (6 (a 1) 27A (a 1 2 2))]
27C a b a b b a a b x = b a a b a a a b b a x.  [para (7 (a 1) 27B (a 1 2 2 2))]
27D a b b a a b a x = b a a b a a a b b a x.  [para (16 (a 1) 27C (a 1 2 2))]
27E b a a b a a x = b a a b a a a b b a x.  [para (16 (a 1) 27D (a 1))]
27F b a a b a a x = b a a a a b b b a x.  [para (6 (a 1) 27E (a 2 2 2 2))]
27G b a a b a a x = b a a a b b a a x.  [para (7 (a 1) 27F (a 2 2 2 2 2))]
27H b a a b a a x = a a b b b a a x.  [para (6 (a 1) 27G (a 2))]
27I b a a b a a x = a b b a a a x.  [para (7 (a 1) 27H (a 2 2))]
27 b a a b a a x = a b a a b x.  [para (6 (a 1) 27I (a 2 2 2))]
28A a b b a a b a b b a x = b a a a b b a b x.  [para (15 (a 1) 16 (a 2 2 2 2))]
28B a b b b a a b a a x = b a a a b b a b x.  [para (26 (a 1) 28A (a 1 2 2))]
28C a b b a b a a b x = b a a a b b a b x.  [para (27 (a 1) 28B (a 1 2 2 2))]
28D b a a b a a b b x = b a a a b b a b x.  [para (19 (a 1) 28C (a 1))]
28E a b a a b b b x = b a a a b b a b x.  [para (27 (a 1) 28D (a 1))]
28F a b a b b a x = b a a a b b a b x.  [para (7 (a 1) 28E (a 1 2 2 2))]
28G a b a b b a x = a a b b b a b x.  [para (6 (a 1) 28F (a 2))]
28H a b a b b a x = a b b a a b x.  [para (7 (a 1) 28G (a 2 2))]
28 a b a b b a x = b a a b a x.  [para (16 (a 1) 28H (a 2))]
29A a b a a b b b c != b a a b a c.  [para (27 (a 1) 23 (a 1))]
29B a b a b b a c != b a a b a c.  [para (7 (a 1) 29A (a 1 2 2 2))]
29C b a a b a c != b a a b a c.  [para (28 (a 1) 29B (a 1))]
29 false.  [copy 29C xx a] % we came to a contraction
[/hide]

Using associativity and 11 applications of the two rules $baaa=aab$ and $abbb=bba$ one can directly prove: abbabbaabbc == abbabbaabbac.
Left cancellation then proves $c = ac$ for all c.

LiMa


Hint: Work with the easier cases first.

First do $x_{1}\leq x_{2}$ only.

Then do $x_{1}\leq x_{2}\leq x_{3}$ only.

And so on.

[hide="Hint 2"]
Find sequences:
$a<b<c<d<e<f$
$a<b<c<d<e$
$a<b<c<d$
$...$
$a$

Then stick in numbers that are equal to ones already there to make them length 6.[/hide]


function
integration
real analysis
real analysis unsolved

Let $f: [0,1]\rightarrow\mathbb{R}$ be a continuous function, differentiable on $(0,1)$, with the property $\exists a\in (0,1]$ such that $\int_0^a f(x)\ \mathrm{d}x=0$.  Prove that\[ \left|\int_0^1f(x)\ \mathrm{d}x\right|\leq\frac{1-a}{2}\sup_{x\in (0,1)}|f'(x)| \]
Masoud Zargar

There exists $x_0 \in (0,a)$ such that $f(x_0)=0$.Let $x_{max}$ be the point from $[a,1]$ where $f(x)$ attains it's maximum, then $|f(x_{max})|=|f(x_{max})-f(x_0)| \leq (x-x_0)sup_{[0,1]}|f'(x)| \leq xsup_{[0,1]}|f'(x)|$   Now $|\int_a^1 f(x)dx | \leq \int_a^1 |f(x)|dx \leq \int_a^1 x sup_{[0,1]}|f'(x)| dx= \frac{1-a}{2} sup_{[0,1]}|f'(x)|$


I have installed texniccenter and MikTex but after building sourcefiles I cannot view the output.  When I try I get a box that reads:
[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]
Cannot execute command

Can anyone help?

This was recently discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42360]here[/url].


calculus
derivative
vector
calculus computations

This problem shouldn't be hard, I just don't understand something: "let $ f(x,y,z) = x^{2}+y^{2}-z^{2}$.  Find the directional derivative of $ f$ at $ (3,4,5)$ along the curve of intersection of the two surfaces $ 2x^{2}+2y^{2}-z^{2}= 25$ and $ x^{2}+y^{2}=z^{2}$."  In particular, what direction is being specified by the phrase "along the curve of .... " ?

The tangent direction to the curve of intersection. You actually don't need to know the equation of this curve. Indeed, the tangent line is the intersection of the tangent planes to the surfaces, which means that you can find its direction vector by taking the cross-product of two normals.


What is the division algorithm for polynomials? Its used in one of the aime solutions for #9 of the 2003 alternate aime. Can someone please explain to me how the division algorithm works.

http://www.unl.edu/amc/a-activities/a7-problems/a7-1-archives/2003/AIMEpractice/03AIMEaltq.html

The division algorithm goes something like if you have polynomials F(x) and G(x)
Then
F(x)=G(x)*H(x)+R(X) for some polynomials H(x) and R(x)

Consider our division: \[ \frac{a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0}{x^m + b_{m-1} x^{m-1} + ... + b_1 x + b_0}  \]

In case we have a leading coefficient for the denominator, just divide through and add that in at the end.

Now add $a_n x^{n-m}$ to the problem and subtract that value times the lower polynomial in the numerator. We have \[ a_n x^{n-m} + \frac{a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 - (a_n x^{n-m})(x^m + b_{m-1} x^{m-1} + ... + b_1 x + b_0)}{x^m + b_{m-1} x^{m-1} + ... + b_1 x + b_0} = a_n x^{n-m} + \frac{a_{n-1} x^{n-1} + ... + a_1 x + a_0 - (a_n x^{n-m})(b_{m-1} x^{m-1} + ... + b_1 x + b_0)}{x^m + b_{m-1} x^{m-1} + ... + b_1 x + b_0}  \]

Hey...we've reduced the degree of the numerator. Remember that $n \ge m$ for this to work. Now, reduce the degree of the new numerator in the same method, and continue until the numerator has a lower degree than the denominator.

Also note that this is the general case and therefore looks [i]very[/i] messy, especially the numerator of $a_{n-1} x^{n-1} + ... + a_1 x + a_0 - (a_0 x^{n-m})(b_{m-1} x^{m-1} + ... + b_1 x + b_0)$. Normally this will just be a few simply numbers to multiply and add, so when doing this in practice it won't be so evil.
Also note that much of the time the original denominator will be linear $\rightarrow$ much easier than this.

It works just like long division.  Also, you can use synthetic division.

Yeah, you can use synthetic iff the denominator is $x-c$ for a constant $c$. Otherwise, use long division, which always works, it's basically the same as normal long division. Look an algebra book for examples.

Look at this page.

http://homepage.smc.edu/kennedy_john/POLYTHEOREMS.PDF

Hope it is useful.


For how many integers $ x$ is $ \frac{x^3\plus{}2x^2\plus{}3x\plus{}4}{x\plus{}5}$ an integer?

[hide="Answer"]We can write $ x^3\plus{}2 x^2\plus{}3 x\plus{}4\equal{}P(x) \cdot (x\plus{}5)\plus{}Q$, for some integer $ Q$ and some integer polynomial $ P(x)$.

Plugging in $ x\equal{}\minus{}5$ into this form gives $ P(\minus{}5) \cdot (\minus{}5\plus{}5)\plus{}Q\equal{}Q\equal{}\minus{}86$.

Thus, $ \frac{x^3\plus{}2x^2\plus{}3x\plus{}4}{x\plus{}5} \equal{} \frac{P(x) \cdot (x\plus{}5) \minus{} 86}{x\plus{}5} \equal{} P(x) \minus{} \frac{86}{x\plus{}5}$

Clearly, $ P(x)$ is an integer, so we need $ x\plus{}5|86$.

86 has 8 factors: $ \pm 1, \pm 2, \pm 43, \pm 86$, so there are 8 possible integer values for x+5, and thus 8 possible values for x.[/hide]

I used the same approach as zuton. $ 86\equal{}43^1\cdot 3^1\Rightarrow d\equal{}4$ [i]positive[/i] divisors. Multiplying by two to account for $ f: \pm x\to\mathbb{Z}$ we see that there are eight such integers as listed by zuton.

How did you arrive at $ Q \equal{} \minus{}86$, Zuton? I'm missing something here.

[b]Remainder Theorem:[/b]  The remainder when $ P(x)$ is divided by $ x \minus{} a$ is $ P(a)$.

The Zuton Force essentially re-proved that theorem.  Do you understand the first line?

@t0rajir0u,
you mean its possible to find the answer 8 in one step ?
is it possible to get the answer in one step ?

Define "one step."  There are two calculations being made here:  one of the remainder term (which determines what values $ x$ can take), and one of the number of values $ x$ can take given the remainder term.  If you want a shorthand, the answer is $ 2 \sigma_0(P(\minus{}5))$, but I don't consider a single formula very satisfactory since it doesn't really address the concepts involved.

Thanks, [b]t0rajir0u[/b] — I forgot all about polynomial division and the remainder theorem. XD

f(-5) = -86 which is the remainder when dividing by x+5 by the Remainder theorem.
that means that we need 86 / x+5 to be an integer. 86 = 43 x 2 so their are four factors x2 because it is + or - 5 = 8

Slightly off subject but what exactly is contained in a precalc course? And doesn't Introduction to algebra of AoPs cover most of Algebra 2(except maybe something like sigma notatation)?

:P
Well I didn't receive an introduction to sigma notation in precalculus. Nor was I introduced to DeMoivre's/Euler's Formula. I don't know what the algebra books cover exactly but I'm sure they're quite comprehensive. A standard precalc course should clarify the concepts learned in algebra II as well as familiarize you with trigonometry & its many identities.

No wonder, i am taking this course at the local univ and the subjects in precalc and college algebra are almost the same! :D

They should be...

I'm taking Alg II Honors at my high school and we've covered summation (sigma) notation, all of Alg II, and trig. DeMoivre's and Euler's Formulas were also in the textbook (we didn't cover them though).  :P

im taking algebra 2 in middle school now. but all these problems are making my head hurt. u guys are all really smart, man.

math hurts my head... -me!!!


what is ninhydrin reaction???? explain with an example!!!! is it related to proteins or amino acids ??? explain with a example!!!! :D

Ninhydrin (triketohydrindene hydrate) is a chemical used to detect ammonia or primary and secondary amines. When reacting with these free amines, a deep blue or purple color known as Ruhemann's purple is evolved. Ninhydrin is most commonly used to detect fingerprints, as amines left over from peptides and proteins (terminal amines or lysine residues) sloughed off in fingerprints react with ninhydrin.

click on the picture to view it.


For how many $n$ in{1, 2, 3, . . . , 100} is the tens digit of $n^2$ odd?

[hide="Solution!!!"]Since we need $n^2 \equiv k (\mod{20})$ cuz it's the tens digit and so we use $20$ and we need $k$ to be an element of ${10,11,12,13,...19}$, since $n^2$ will be in the form $(n+20)^2$, we just need to find how many of the first $20$ numbers will have the tens digit in it's square odd and then multiply it by $5$ since there are $100$ numbers. There are a total of $4$ of these numbers so the answer is $\boxed{20}$.[/hide]

I believe this is an old AMC problem from the 80s or something...


function
trigonometry
complex analysis
complex analysis unsolved

[quote="moloy_kb"]Find the Complex Fourier Coefficients of the Function

f(@)=A*(1+B*Sin@+C*Cos2@), defined in the limit of 0 to 2Pi, where A, B, C are the constants[/quote] Is that meant to be $ f(\theta) \equal{} A ( 1 \plus{} B\sin {\theta} \plus{} C \cos{2\theta})$ or $ f(\theta) \equal{} A ( (1 \plus{} B)\sin {\theta} \plus{} C \cos{2\theta})$?


calculus
integration
function
inequalities
invariant
real analysis
real analysis unsolved

Let $f:[0,1]\to\mathbb{R}$ a derivable function on $(0,1)$,continuous on $[0,1]$ and $f(0)=0$ and $0\leq f'(x)\leq 1$,for all $x\in (0,1)$.Prove that: $(\int_{0}^{1} f(x)dx)^2\geq\int_{0}^{1} f^3 (x)dx$ ;)

Note that this inequality is scale-invariant- we could replace the interval $[0,1]$ with $[0,a]$ for any $a$.

Let $L(t)=\left(\int_0^tf(x)\,dx\right)^2$ and $R'(t)=\int_0^t(f(x))^3\,dx$
Using the Fundamental Theorem, $L'(t)=2f(t)\int_0^tf(x)\,dx$ and $R(t)=(f(t))^3$. We want to show that $L(t)\ge R(t)$, so we want $L'(t)\ge R'(t)$ and $2\int_0^tf(x)\, dx\ge (f(t))^2$. This is clearly true at $t=0$; now we differentiate again.
$\left(\frac{L'(t)}{f(t)}\right)'=2f(t)$
$\left(\frac{R'(t)}{f(t)}\right)'=2f(t)f'(t)$

The inequality is clear. Integrating, $\left(\frac{L'(t)}{f(t)}\right)\ge\left(\frac{R'(t)}{f(t)}\right)$. Multiplying by $f(t)$ and integrating again, $L(t)\ge R(t)$ for all $t\ge0$, and we are done.
We have equality if and only if $f'$ is identically 1 and $f(x)=x$.

Very nice solution.Mine is a a little bit different.In fact I have,bonus another solution using Cauchy mean value theorem. ;)


We have 6 different natural numbers and the numbers are <108.To prove that we always can choose three of them-a,b and c ,and a<bc,b<ac and c<ab.Thanks in advance.

Let a<b<c

If c<ab then b<ac also a<bc

We'll be finished if we prove that always exist such number which is bigger than other two and is smaller than the product of the same two numbers.

Let $ k_1 < k_2 < k_3 < k_4 < k_5 < k_6$ and $ k_3 > k_1k_2; k_4 > k_2k_3; k_5 > k_3k_4; k_6 > k_4k_5$

The smallest possibl value of $ k_3$ is 1*2+1=3
The smallest possible value of $ k_4$ is 2*3+1=7
The smallest possible value of $ k_5$ is 3*7+1=22
The smallest possible value of $ k_6$ is 7*22+1=155>108 - contradiction  :)

Wait,
I did mistake:

Let $ k_1<k_2<k_3<k_4<k_5<k_6$ and $ k_3\ge{k_1k_2}$ ; $ k_4\ge{k_2k_3}$ ; $ k_5\ge{k_3k_4}$ ; $ k_6\ge{k_4k_5}$

And smallest possible values of terms would be:

1;2;3;6;18;108

The last terms is 108 - contradiction.


I have an ugly solution therefore if you have an easy solution, why don't you post it before, I will post mine later! Thanks :)

[quote="N.T.TUAN"]Denote $a=y'+z'\\b=z'+x'\\c=x'+y'$,

We have $a,b,c$ are sides of the triangle and $x'y'z'(x'+y'+z')=S(a,b,c)^{2}$.

Final, we have to find minimum of $ma^{4}+nb^{4}+pc^{4}$, this is quite easy!  :wink:[/quote]
Really you have an "easy" proof or no proof  :D

That's right it is my ugly solution ,too!
$\sum (y^{2}+z^{2})a^{4}\ge\sum\frac{(y+z)^{2}}{2}a^{4}\ge \frac{1}{6}(\sum (y+z)a^{2})^{2}\ge \frac{16}{6}(\sum(x+y)(x+z))S^{2}\ge 32\frac{xy+yz+zx}{3}S^{2}$
But really I am finding a nice solution for this too weak inequality! :)

Happy hunting!  :wink:

P/S : In my solution I used $(x^{2}+y^{2})(y^{2}+z^{2})\geq (y^{2}+zx)^{2}$,...


If $ z=x+iy$, $ \alpha = a+ib$ are complex numbers and $ \bar{z},\bar{\alpha}$ their conjugates, show that if c is real, then

(iii)$ \bar{\alpha}(\bar{z}-z)+\alpha(\bar{z}+z)+c$=0 [b]REPRESENTS TWO STRAIGHT LINES[/b]

After simplifying I obtain this equation

$ 2ax-2by+c-i (2ay-2bx)$=0

What statement do I write now? or is there more working to do??

That problem, as you have written it, doesn't make sense.

sry missed the most important part.

thanks bye


linear algebra
matrix
inequalities
linear algebra unsolved

Let $A$ a matrix of order $n$ with $det(A)=0$.Prove that:

               $rank(A^*)={0,1} $ ;)

Since $*$ is an overused notation, we should clarify that this problem refers to the classical adjoint. This problem is also independent of field.

It is immediate consequence of Sylvester inequality and the definition of adjoint.


Let $ x_1, x_2, ..., x_n$ be positive real numbers such that $ \sum\limits_{i \equal{} 1}^n {ix_i }  \equal{} 1$
Prove that
\[ \sum\limits_{i\equal{}1}^n{\frac{{ix_i }}{{x_i  \plus{} 1}}} \ge \frac{{n\left( {n \plus{} 1} \right)}}{{n^2  \plus{} n \plus{} 2}}\]


[quote="fuzzylogic"]Let $a, b, c$ be the sides of a triangle, $m_a,m_b,m_c$ be the corresponding medians. Prove that for any reals $x,y,z$:

$(x+y+z)(xa^2+yb^2+zc^2) \geq 4(yzm_a^2 + zxm_b^2 + xy m_c^2)$.

This is a reformulation of a problem posted before (I will add a link to that post after someone post a solution). Try to find a different solution if you knew that post. :-)[/quote]

Replace $ m_a^2 = \frac{2b^2+2c^2-a^2}{4} $ fuzzylogic's inequality becomes

$ a^2(x-y)(z-x) + b^2(x-y)(y-z) + c^2(z-x)(y-z) \leq 0 $

$ mna^2 + npb^2 + pmc^2 \leq 0 $   if  $ m + n + p=0 $

This one comes from fuzzylogic general inequality. It also can be proved by replace $ c^2 = a^2 + b^2 -2abcosC $

Thanks Hoang.

It's time to reveal the story. In replying a PM to a ML/AoPS member. I wrote:

[quote]Dear fuzzylogic,

you have posted two nice inequalities about medians

[quote="fuzzylogic"]Let $a, b, c$ be the sides of a triangle, $m_a,m_b,m_c$ be the corresponding medians. Prove that

$m_abc+m_bca + m_cab \geq 4m_am_bm_c$.[/quote]

[quote="fuzzylogic"]Let $a, b, c$ be the sides of a triangle, $m_a,m_b,m_c$ be the corresponding medians. Prove that

$m_aa^2+m_bb^2 + m_cc^2 \geq 4m_am_bm_c$.[/quote]

I have solved them in this way. I hope it is correct.

As Darij said  m_a,m_b,m_c are the sides of a triangle.

 I have computed the medians of this triangle and I have found they

are \frac{3a}{4},\frac{3b}{4},\frac{3c}{4}.

So we can treat m_a,m_b,m_c as sides of a triangle and a,b,c as

\frac{4}{3} of its medians and our inequalities become

\sum{ \frac{m_am_b}{ab}} \geq \frac{9}{4}

\sum {\frac{m_a^2}{bc}} \geq \frac{9}{4}

already solved by you using a general inequality.
[/quote]

Yes, this is exactly the way I derived these inequalites. I called the "duality".

I have been playing with the inequalities:

$(x+y+z)(xa^2+yb^2+zc^2) \geq 4(yz m_a^2 + zx m_b^2 + xy m_c^2)$ ------ (1)

$(x+y+z)(xm_a^2+ym_b^2+zm_c^2) \geq \frac{9}{4}(yz a^2 + zx b^2 + xy c^2)$ ------ (2)

(1) is equivalent to the inequality at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2896

(2) came directly from the general inequality at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15558

Although I posted different proofs for (1), I found out that (1) can also be proved by "duality" from (2).

After proving (1) by "duality", I realized that I can create some interesting ones from the known inequalities. That's the story.

I tried to apply the "duality" to the inequality at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15600

Guess what happened? It turned into itself. :-)

_______________________________________________________

PS. I just applied "duality" to solve the problem at:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15979


Tet dau tien xa nha nhug cong nhan  vui wa ko co time de noi nho nha xen vao
  That hanh fuc   ...................

Em này làm gì mà la Tết um sùm vậy. Lâu ngày trở lại, chỗ này thê lương quá !
Tết qua rồi, nhưng cũng chúc mấy con vịt ở đây năm mới trí tuệ mới !!! :)

$ (\sqrt {2} \plus{} 1)^{\frac {6x \minus{} 6}{x \plus{} 1}} \equal{} (\sqrt {2} \minus{} 1)^{ \minus{} x}$
  ta co nhan xet :
$ (\sqrt {2} \plus{} 1)(\sqrt {2} \minus{} 1) \equal{} 1$ do do :
$ (\sqrt {2} \minus{} 1)^{ \minus{} x} \equal{} (\sqrt {2} \plus{} 1)^x$
  phuong trinh tro nen don gian : $ \frac {6x \minus{} 6}{x \plus{} 1} \equal{} x$   ( $ x$  khac$ \minus{} 1$ )
 $ x^2 \minus{} 5x \plus{} 6 \equal{} 0$hay  $ x \equal{} 2$ va$ x \equal{} 3$

[quote="Hong Quy"][quote="nkht-tk14"][quote="quangchuc"]Anh Quy bay gio hoc o truong nao ay nhi?[/quote]
Anh Quý bị BKHN đuổi học rồi :D đang lang thang vất vưởng giờ lại sang ĐH HN học =))[/quote]
cac chu dung hu a the, thiet tinh la`  fai xa BK ..........[/quote]

He he, anh Quý có phải là anh lehoan trên diendantoanhoc.net không nhỉ?

Exactly.
Ah` co 1 diendan moi mong ai  ham toan thi` vo go'p vui:
[url]http://mathvn.org[/url]


Below are the front view and side view of some arrangement of unit cubes.  Let $N$ denote the largest number of possible unit cubes in an arrangement wiht the two views shown below, and let $n$ denote the least number of possible unit cubes.  What is the value of $N-n$?

Also, does anyone have anytips about solving front view and side view problems?
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=633[/img]

I got that:

[hide="Click to reveal hidden content"]

$n=10$ and $N=26$

So my answer is $16$, but I don't know if that's right. I'm never good with visual problems like this[/hide]

I[hide] was thinking N=26 and n=8 (from what I saw, every cube in the front view could be a cube in the side view. [/hide]

I just realized I'm wrong. $n$ is less than 10, I donno the exact value.

Can little cubes float without a base? Just wondering  :P

how can you get 8...

just count everyone on the front, and you still have 4 side way ones... not mentioning the one on second floor

sorry for this "double post"
[hide]
n=12 and N=26

14 is the answer[/hide]

Yes I agree, 14 is the answer. How did you guys get 10 or 8? :?

I am pretty sure I am right. I took a weird 9 week class at school where all we did was solve logic puzzles and front-side cube problems.

Actually, [hide]236! is right[/hide]

[hide="Solution"]In the front view, each block can correspond to one block in the side, so $n=8$.

The max is if each of the 4 blocks in the side view base (4) has 5 blocks in the Front View base (5), so that's 20.  Also, the 2 on the top of the side view can be filled in with the 3 blocks in the front view, so that's 6. $N=26$.

$N-n=18$.

Sorry for the bad explanation.  :D [/hide]

Wow, I remember this problem it was the only one I could do, basically because I wasted three years of math on it
[hide]
Basically for $n$ you can get 8 and for $N$ you can get 26. So $26-8=18$
If you spend three years on this, you will see that its easy[/hide]


Let $ a,b$ be integers greater than $ 1$. What is the largest $ n$ which cannot be written in the form $ n \equal{} 7a \plus{} 5b$ ?

$\textbf{(A)}\ 82  \qquad\textbf{(B)}\ 47  \qquad\textbf{(C)}\ 45  \qquad\textbf{(D)}\ 42  \qquad\textbf{(E)}\ \text{None}$

So a=2+x, b=2+y, x,y $ \ge$ 0. Then 7a+5b=7x+5y+24. The largest number not expressable as 7x+5y is $ (7 \cdot 5) \minus{} (7 \plus{} 5) \equal{} 23$. Thus, the answer is 23+24=47.


Let p,q,r be three different primes, $ a \ge 2$ integer

Find all solutions if any so that both of these equations hold simultaneously:

$ q \plus{} r \equal{} p^{a \minus{} 1} \plus{} 1$
$ qr \equal{} p^a \plus{} 2$

System is symmetrik for $ q<\minus{}\minus{}>r$. We can desired $ q\le r$.
If $ p\equal{}2$, then $ qr\equal{}p^a\plus{}2$ even, so $ q\equal{}2$ and $ r\equal{}2^{a\minus{}1}\minus{}1\equal{}2^{a\minus{}1}\plus{}1$.
Therefore $ p>2$. $ q,r$ are solutions of $ x^2\minus{}(p^{a\minus{}1}\plus{}1)x\plus{}p^a\plus{}2\equal{}0$.
Diskriminant
$ D\equal{}p^{2a\minus{}2}\minus{}4p^a\plus{}2p^{a\minus{}1}\minus{}7\equal{}(p^{a\minus{}1}\minus{}2p)^2\plus{}2p^{a\minus{}1}\minus{}7\minus{}4p^2\equal{}(p^{a\minus{}1}\minus{}2p\plus{}1)^2\minus{}4p^2\plus{}4p\minus{}8$.
If $ a>3$ diskriminant can not be square ($ p\ge 3$). If $ a\equal{}3$ $ D\equal{}(p^2\minus{}2p\minus{}1)^2\minus{}2p\minus{}8$ and $ (p^2\minus{}2p\minus{}2)^2<D<(p@\minus{}2p\minus{}1)^2$ for p>3.
If p=3=a, then $ D<0$. Therefore $ a\equal{}2$. If $ p\equal{}3$, then $ q\plus{}r\equal{}4,qr\equal{}11$.
Therefore $ a\equal{}2,p>3$. Because $ 3|p^2\plus{}2$ we find $ q\equal{}3$ and $ r\equal{}p\minus{}2\equal{}\frac{p^2\plus{}2}{3}$.
Therefore system had not solutions.

[quote="Rust"]System is symmetrik for $ q < \minus{} \minus{} > r$. We can desired $ q\le r$.
If $ p \equal{} 2$, then $ qr \equal{} p^a \plus{} 2$ even, so $ q \equal{} 2$ and $ r \equal{} 2^{a \minus{} 1} \minus{} 1 \equal{} 2^{a \minus{} 1} \plus{} 1$.
Therefore $ p > 2$. $ q,r$ are solutions of $ x^2 \minus{} (p^{a \minus{} 1} \plus{} 1)x \plus{} p^a \plus{} 2 \equal{} 0$.
Diskriminant
$ D \equal{} p^{2a \minus{} 2} \minus{} 4p^a \plus{} 2p^{a \minus{} 1} \minus{} 7 \equal{} (p^{a \minus{} 1} \minus{} 2p)^2 \plus{} 2p^{a \minus{} 1} \minus{} 7 \minus{} 4p^2 \equal{} (p^{a \minus{} 1} \minus{} 2p \plus{} 1)^2 \minus{} 4p^2 \plus{} 4p \minus{} 8$.
If $ a > 3$ diskriminant can not be square ($ p\ge 3$).

[b]I cannot see why????[/b]

 If $ a \equal{} 3$ $ D \equal{} (p^2 \minus{} 2p \minus{} 1)^2 \minus{} 2p \minus{} 8$ and $ (p^2 \minus{} 2p \minus{} 2)^2 < D < (p@ \minus{} 2p \minus{} 1)^2$ for p>3.
If p=3=a, then $ D < 0$. Therefore $ a \equal{} 2$. If $ p \equal{} 3$, then $ q \plus{} r \equal{} 4,qr \equal{} 11$.
Therefore $ a \equal{} 2,p > 3$. Because $ 3|p^2 \plus{} 2$ we find $ q \equal{} 3$ and $ r \equal{} p \minus{} 2 \equal{} \frac {p^2 \plus{} 2}{3}$.
Therefore system had not solutions.[/quote]

[b] I cannot see why diskriminant cannot be square for $ a>3$ [/b]

if $ a>3$ then the expression would be between two squares.
$ (p^{a\minus{}1}\minus{}2p\plus{}1)^{2}>D>(p^{a\minus{}1}\minus{}2p)^2$

[quote="shoki"]if $ a > 3$ then the expression would be between two squares.
$ (p^{a \minus{} 1} \minus{} 2p \plus{} 1)^{2} > D > (p^{a \minus{} 1} \minus{} 2p)^2$[/quote]


linear algebra
matrix
calculus
integration

let M be the set of matrices with determinant prime to p. How do we show that the set M is a group under the matrix multiplication.

thanks.

In general, to show that a set is a group under a binary operation, you need to show four things:

1) There is a right and left identity element in the set (in your case a matrix $ A$ such that $ AB \equal{} BA \equal{} B$ for all $ B \in M$).

2) The binary operation is associative (in your case $ (AB)C \equal{} A(BC)$ for all $ A,B,C \in M$).

3) The set is closed under the binary operation (in your case if $ A,B \in M$ then $ AB \in M$)

4) Every element in the set has an inverse that is in the set (in your case for each $ A \in M$ there exists $ B \in M$ such that $ AB$ is the identity element from 1).

Its not difficult to prove 1 through 3, but 4 isn't really true...if a matrix has a determinant that is an integer greater than 1, then its inverse has a determinant that is a fraction. From the way you stated the problem, it seems like only matrices with integral determinant are in $ M$.

You probably want to take everything $ \bmod p$ to make this work.  Then it's a corollary of the more general fact that a matrix over a commutative ring with unit determinant is invertible.


Hello,

I've just watched a Russian movie about the secrets of water. More of the information was rather far-fetched and hard to prove on practice, but there were also introduced some possible and very interesting properties of the water. For instance, there were mentioned that the water has a "memory" and due to some clusters in its micro-molecular level it can remember the influence of the surrounding  background :| There was said that the water molecules modify their structure under the expose to sound or some other energetic field, and thus change their properties.
In a result, for example, some rice sunk in glass with water which was placed in a room where were played classical music (Mozart and etc.) fermented, while other rice sunk in glass with water "listening" to hard rock music putrefied.

Have you ever heart something about this question and can someone give more information?

Just "Вода" ("Water") but probably the movie can be found as "The Greatest Secrets of Water" or something like that. I watched the movie on the Россия телеканал by the way if this can help. The scenarist is Михайл Вайгар.


Bob and Jill are trying to find a day that they both have off so they can go to the zoo they both started their jobs on the same day. Jill works for 6 days gets off the 7th day. Bob works for 12 days and then gets the 13th and 14th days off what is the first day they both have off?
[hide="Solution"]First we need to list the congruences
n== 0 mod 7
n== 1 mod 14
n== 0 mod 14
Quite obvisbily the first day they both have off is the 14th day of work  [/hide]

You wouldn't really need to use modular arithmatic for this problem. It is easy to see that they both have off on the 14th day...

[quote="Complex_Ninja"][quote="Math Champion"]Why do you post a question and then answer it yourself? The point of this forum is to have other people help you.[/quote]
Maybe because she wanted other people to solve it in a different way..  I don't know :([/quote]Or perhaps she would like to express her generosity and share a math problem & solution with other individuals with similar interests.

I started this thread to show how you would solve a problem like this.

Oh...Well, as the previous posts have said, you didn't really need modular arithmetic to solve this problem, as it was obvious. Thanks for the clarification though.


Prove that if a,b,c>0 verify a^2+b^2+c^2=4-abc then
  sum sqrt(a(2+a))<=sqrt(prod(1+2/a)).

I think it can be proven that a=2cosA, b=2cosB, c=2cosC, with ABC an acute triangle. Maybe this works; I don't have the time to check on that right now.

Grobber was right when said that we can put  a = 2cosA, b = 2cosB, c= 2cosC where ACB acute triangle. From that we can easily get a+b+c<=3. But from this point, we will go by other way:
We have, by C-B-S:  Sum \sqrt(a(a+2)) <= \sqrt(3(a^2+b^2+c^2+2a+2b+2c)).
Is sufices to prove that
      3(a^2+b^2+c^2+2a+2b+2c) <= (a+2)(b+2)(c+2)/abc
<=> 3(4-abc+2a+2b+2c)abc <= (a+2)(b+2)(c+2)
<=> 11abc + 6(a+b+c)abc <= 3a^2b^2c^2 + 2(ab+bc+ca) + 4(a+b+c) + 8  (*)
From a+b+c<=3 we have ab+bc+ca<=3, abc<=3
So
6abc <= 3a^2.b^2.c^2 + 3 (AM-GM)  (1)
5abc <= 5 (2)
4(a+b+c)abc <= 4(a+b+c) (3)
2(a+b+c)abc = 2(ab.bc+bc.ca+ca.ab) <= (2/3)(ab+bc+ca)^2 <=2(ab+bc+ca) (4)
Sum (1), (2), (3), (4) we have (*).

I leave proving a+b+c <=3 (algebraically) for you. This is also an interesting problem (not so hard)

Namdung

I have a little bit different solution. First of all, we have c^2+abc=4-a^2-b^2<=2(2-ab), so ab(c+2)<=(2-c)(2+c). So ab<=2-c. If we set a=2cosx, b=2cosy, c=2cosz, we will find that sum sqrt([(2-a)(2-b)]/[(2+a)(2+b)])=1. So, 1>=sum csqrt(ab)/sqrt([(2+a)(2+b)]. Now multiply this one with sqrt(prod(1+2/a)).


Prove that the equation $a^{3}=b^{2}+1$ has no solutions in positive integers.

i made this problem this morning and it took me not more than fifteen minutes to solve it. but now it has been more than an hour but no solutions. it's a nice problem. so have a try.. :P

noone interested?? :(
i have a beautiful but long solution.

I hope this is a good solution ( i'm not very good at Number Theory, so...).
$a^{3}=b^{2}+1$
$(a-1)(a^{2}+a+1)=b^{2}$ so from here we have:
$(a-1)k=b^{2}$ and $(a^{2}+a+1)t=b^{2}$ with $tk=1$
but we must solve the equations in positive integers so
$a-1\leq b^{2}$ and $a^{2}+a+1\leq b^{2}$ so $t=k=1$
We will have to solve the system
$a-1=b^{2}$
$a^{2}+a+1=b^{2}$, which have no solutions.

look at my post again. then you'll understand.

so soon? i'm a bit busy right now so i'll do it later. besides let the others think about it. is it okey with you? :wink:

doesn't anybody have a solution? or should i post mine?

here's my solution:
$a^{3}=b^{2}+1\Rightarrow (a-1)(a^{2}+a+1)=b^{2}$
let $\gcd(a-1, a^{2}+a+1)=g$
note that $(a^{2}+a+1)-(a+2)(a-1)=3$ so $g|3\Rightarrow g=3,1$
but if $g=1$ then we have $a^{2}+a+1$ as a perfect sqaure, which is impossible.
so $g=3$. Let $a-1=3k$
so $9k(3k^{2}+7k+1)=b^{2}\Rightarrow b=3p$
$\Rightarrow k(3k^{2}+7k+1)=p^{2}$
but we have $(3k^{2}+7k+1)-k(3k+7)=1$
so $\gcd (3k^{2}+7k+1, k)=1\Rightarrow k=x^{2}, 3k^{2}+7k+1=y^{2}$  $(1)$
so $x^{2}(3x^{2}+7)=(y+1)(y-1)\Rightarrow x^{2}\le y-1< y+1\le 3x^{2}+7$
because other cases result to contradiction.
so either $x^{2}|y-1$ or $x^{2}|y+1$

case1: if $x^{2}|y-1$ let $y=kx^{2}+1$
$(1)$ becomes $3x^{2}+7=k(kx^{2}+2)\Rightarrow x^{2}(k^{2}-3)=7-2k$
note that $k^{2}\ge 3\Rightarrow 2k\le 7$, contradiction.
and $k^{2}\le 3\Rightarrow 2k\ge 7$, contradiction.
so no solution in this case.

case2: if $x^{2}|y+1$ let $y=dx^{2}-1$
which implies $3x^{2}+7=d(dx^{2}-2)\Rightarrow x^{2}(d^{2}-3)=2d+7$
so $x$ is odd, $d$ is even.
and the last equation implies $d^{2}-3\le 2d+7\Rightarrow (d-1)^{2}\le 11\Rightarrow d\le 4$
so $d=2,4$
but these produce no solutions.$\blacksquare$  :D

[quote="RDeepMath91"]
a) $q = 3k+1 \Rightarrow k(3k+1) = p^{2}(p^{2}+1)$
[/quote]
Shouldn't it be $k(3k+2)$?

[quote="nayel"]
so either $x^{2}|y-1$ or $x^{2}|y+1$
[/quote]why?can you explain this step?

Yes, I think that both of the posted solutions are wrong and the mistake is the same:
$a|bc$ and $(b,c)=1$ DOES NOT imply that $a|b$ or $a|c$.

@Tomici0:yes,you are absolutely right.
@all:who can solve the problem?

I think that's quite well-known problem and the most popular solution uses $Z[i]$. Here: [url]http://www.mathlinks.ro/Forum/contest/2nd%20Edition/2nd%20edition%20-%20round%207%20solutions.pdf[/url] (problem 2) you can find this solution, but there is completely elementary one as well (which is much more complicated).