show that there does not exist $ n \in \mathbb{N}$ such that

$ 2^n \equiv 1 \mod n$

[hide="Solution"]
By Fermat's Little Theorem, $ 2^{n}\equiv 2 \mod n$ if $ n$ is odd. If $ n$ is even, then $ 2^{n}\equiv x \mod n$
where $ x$ is even. So there exists no n such that $ 2^{n}\equiv 1 \mod n$.
[/hide]

I thought Fermat's Little Theorem only worked for prime $ n$ values? For example, $ 2^9 \equiv 8$ modulo $ 9$.

[hide="Solution"]$ n\equal{}1$ clearly does not work, so there are three cases.

Case 1: $ n$ is even. This is obvious; if $ 2^n \equiv 1$ modulo $ n$ with $ n$ even, then $ 2^n\minus{}1$ is odd and yet divisible by an even number: this is a contradiction.

Case 2: $ n$ is odd and prime. This is obvious; $ 2^{n\minus{}1} \equiv 1$ modulo $ n$ by Fermat's Little Theorem, so $ 2^n \equiv 2$ modulo $ n$.

Case 3: $ n$ is odd and composite. Therefore, $ 2^{\phi(n)} \equiv 1$ modulo $ n$. We would like to prove that $ \phi(n)$ does not divide $ n$. $ n \ge 3$, and it is well-known that $ \phi(n)$ is even for $ n > 2$ because there must exist an odd prime $ p$ to a certain power $ a$ that divides $ n$ since $ n$ is odd and composite, and $ \phi(p^a) \equal{} p^{a\minus{}1}(p\minus{}1)$, which is even since $ p\minus{}1$ is even, so since the totient function is multiplicative, $ \phi(n)$ is even for $ n > 2$, so $ \phi(n)$ cannot divide $ n$ because that would imply that $ n$ is even, which is a contradiction.

[/hide]

[quote="Ihatepie"][hide="Solution"]
By Fermat's Little Theorem, $ 2^{n}\equiv 2 \mod n$ if $ n$ is odd
[/hide][/quote]

FLT requires that $ n$ is prime

EDIT: okay its been proven, here is another [hide="solution"]

Clearly $ n$ cannot be even, so let $ n$ be odd. and let $ S$ be the set of integers, $ n$, such that $ 2^n \equiv 1 \mod n$. We will show that this set is empty by showing there does not exist a least element, $ n_0$.

Now, there must exist $ i, j \in \mathbb{N}$ such that $ 2^i \equiv 2^j \mod n \rightarrow 2^j(2^{i \minus{} j} \minus{} 1) \equiv 0 \mod n$

Since $ \gcd(2,n) \equal{} 1$ we have $ 2^{i \minus{} j} \equiv 1 \mod n$. So we know there must exist $ i \minus{} j \equal{} a$ such that $ 2^a \equiv 1 \mod n$. So assume $ a$ is the smallest such integer.

We know $ a < n$ because $ 2^k$ cycles through integers in $ \mod n$ in periods of $ a$, and $ 2^k \in \{1,2,3...,n \minus{} 1\}$ which has only $ n \minus{} 1$ elements. therefore if $ 2^n \equiv 1 \mod n$ then $ n$ is a mltiple of $ a$, or $ a|n$

Now, assume $ n_0$ is the least element in $ S$, we have $ 2^{n_0} \equiv 1 \mod n$. However $ a | n_0$ and $ 2^a \equiv 1 \mod n$, Therefore $ 2^a \equiv 1 \mod a$ and $ a \in S$
But $ a < n$ so this contradicts the assumption that $ n_0$ is the least element.

Hence $ S$ is empty and there are no $ n \in \mathbb{N}$ such that $ 2^n \equiv 1 \mod n$
[/hide]

Oops  :blush:

For some reason I just thought 2 had to be relatively prime with n.

Euler's Theorem/FLT holds true only if the base and modulus are relatively prime.

yup, you are right. the base and the moduli should be relatively prime.


counting
derangement
combinatorics unsolved
combinatorics

Here is the problem:
  Find the number of all permutations of n elements , such that:
  a) exactly m (0<=m<=n) elements  are on their starting places.
  b) there is not an element that is on its starting place

I hope i have written it clear

You're asking for the number of permutations of an $ n$-set with $ m$ fixed points.  It's easy to see that if we can solve this question for $ m = 0$ then we can solve it for all $ m$.  A permutation with 0 fixed points is a "derangement," and the problem of counting derangements has been discussed here several times in the past -- try a search.


find the ordered pair (x,y) that is the solution to the system:

(x+2y)/xy=11/12

(2x-3y)/xy=2/3

[hide="Hint"]Split the fractions and put some substitutions.[/hide]

[hide="solution"]We cross multiply each fraction:

11xy=12x+24y
2xy=6x-9y
4xy=12x-18y
42y=7xy
x=6
y=12/7[/hide]


Find all functions $ f: R\to R$ such that
$ f(x\plus{}y)\plus{}f(x)(f(y)\equal{}f(xy)\plus{}f(x)\plus{}f(y)$

[quote="sylvect"]Find all functions $ f: R\to R$ such that
$ f(f(x) \plus{} y^2) \equal{} f^2(x) \minus{} f(x)f(y) \plus{} xy \plus{} x$[/quote]
Sorry,but it from our test,Lock it mod http://www.mathlinks.ro/Forum/viewtopic.php?t=168836


[quote="gambs"]Find a particular solution to the given equation:

y"+y = sinx + x cosx[/quote]

If I did'nt misscalculated:
$y=\frac{x^{2}-Asin(2x+B)+C}{4}sinx-\frac{x+Acos(2x+B)}{4}cosx$

Where A,B,C are arbitrary.

[quote="gambs"]How did you arrive at that answer? What were your initial steps?
It isn't the answer given in the book but is similar. Also, the problem is to solve for the coefficients.

yp(x) = 1/4 (x^2 sinx - x cosx)[/quote]

If A=0 and C=0 u got the answer from the book I presume:))
Ofcourse the homogen part of equation is the wellknown equation $~y''+y=0~$ wich has its solutions: $~y=C_{1}sinx+C_{2}cosx$, $C_{1},C_{2}\in R$
So I searched your eqation's particular solutions in $~f(x)sinx+g(x)cosx$ form.


I got this from another forum.

How to play:
Each person makes a wish (Ex. I wish for a cookie) then the next person crushes that persons wish (Ex. But the cookie was made of pure bleach) and makes their own wish (Ex. I wish for a bleach free cookie).
I'll start.

I wish for a soda.

the soda was a carryon bomb

i wish for a element skateboard.

The skateboard was broken in half

I wish for 100 wishes

[quote="junggi"]btw there was another topic exactly like this(and 15+ pages too i think)[/quote]

Link Please?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=wish+corrupt&t=109351[/url]
haha nvm it's 36 pages

You get 100 wishes, but they expire in 1 nanosecond.

I wish I was able to memorize 100 digits of pi (I only know 57 digits).

You memorize 100 digits but they're the wrong ones.

I wish this was the original thread.

your bomb is a dud. i wish i had read a question on the nationwide test today correctly. (no mention of ___ whatsoever  :P )

[quote="footballrocks41237"][quote="junggi"]i pm the mod to delete this thread and you revive the original thread.

i wish the world exploded[/quote]

I revived the original.  Thanks.[/quote]
lollll i was jk in that post
sorry

LOL.  :blush:  Still, the original is back and running...

you read the question correctly, but all the other questions you read wrong  :lol:

i wish for a non-exploding-non-leaking-non-dud-and-100%-drinkable-with-nothing-happening-to-me soda

 :rotfl:

Soda is another name for sodium bicarbonate, or baking soda.  Your soda is dissolved in water.   You drink it.  It tastes terrible.  Also, you just drank lemon juice.  You vomit and burp a lot.  You feel miserable.

I wish money grew on trees (a classic).

wg, but there was a typhoon and your money tree got washed away before you got to pick the money
iwf, a friend forever

There is no reason to revive a 6-year-old thread, especially in the Fun Factory, where you are not posting something useful.


[b]5050(nr.1/2009):[/b] Demonstrati ca daca numerele $ a; b; c>0$ satisfac inegalitatea: $ a^3\plus{}b^3\plus{}c^3<5.abc$,
atunci numerele a, b si c sunt lungimile laturilor unui triunghi.([i]V.I. Kaskevici[/i]-Minsk)
[b]5051(nr.1/2009):[/b] Construiti un patrulater care este atat inscriptibil cat si circumscriptibil, cunoscand cazele celor doua cercuri
R si r si unghiul dintre diagonalele sale. ([i]A. Zaslavsky[/i])
[b]5055(nr.2/2009):[/b] In triunghiul ABC ducem mediana AM si inaltimea AH. Demonstrati ca: $ MH\ge\frac {|AB \minus{} AC|}{2}.$ ([i]S.I. Tokarev[/i])
[b]5063(nr.3/2009):[/b] Aratati ca in orice triunghi are loc inegalitatea: $ a^2\plus{}b^2\plus{}c^2\plus{}\frac{1}{16}.(|a\minus{}b|\plus{}|b\minus{}c|\plus{}|c\minus{}a|)\le 9.R^2.$([i]S.G. Abdullaev[/i])

Can anyone please translate the problem and give me the link to the website of this journal? I'm really curious to find out more about Romanian Math Journal, especially I only know the "Gazetta Matematica" journal. Does anyone know when will the contests hosted by this journal be in this year?


logarithms
inequalities
inequalities proposed

$x$ is a positive real number and $n \in Z$ and $n<0$.prove that:
$\frac{(n+1)^{n+1}}{n^{n}}>\frac{n^{n}}{(n-1)^{n-1}}$ :idea:

$[x\log x]''=\frac{1}{x},$ thus $x\log x$ is strictly convex. Now if we apply Jensen we get \[e^{x\log x+y\log y}> e^{(x+y)\log (x+y)/2}\] thus \[\frac{x^{x}}{\left(\frac{x+y}{2}\right)^{(x+y)/2}}>\frac{\left(\frac{x+y}{2}\right)^{(x+y)/2}}{y^{y}}\;\;\;\;\;(1)\] for positive reals $x,y.$ About negatives, $x\log x$ is concave, so we have \[(-x)^{-x}(-y)^{-y}< \left[\left(-\frac{x+y}{2}\right)^{-(x+y)/2}\right]^{2}\;\;\;\;\;\Leftrightarrow\;\;\;\;\;\frac{1}{x^{x}y^{y}}<\frac{1}{\left(\frac{x+y}{2}\right)^{x+y}}\] which is exactly (1). Thus the inequality (1) is true for all reals $x,y$ such that $xy>0.$

thank you.it seems that your solution is very hard,but nice.
is there any easy one?

can you explain your solution.plz.because may be it will be more usefull. :D


logarithms
number theory
prime numbers
number theory unsolved

Let $\sigma(n)$ be sum of all positive diviosrs of natural number $n$. Prove that if $\sigma(n)=5n$ than $n$ has more than 5 distinct prime divisors.

Let $\omega(n)$ be the number of distinct primes $p_1, p_2, \ldots, p_{\omega(n)}$ dividing $n$. Wlog, suppose $p_1 < p_2 < \ldots < p_{\omega(n)}$. You get $\frac{\sigma(n)}{n} < \prod_{k=1}^{\omega(n)} \frac{p_k}{p_k - 1} \leq 3\cdot \prod_{k=3}^{\omega(n)} \frac{p_k}{p_k - 1} \leq 3 \cdot (\frac{5}{4})^{\omega(n) - 2}$, as the map $]1,+\infty[\; \mapsto \mathbb{R}: x \mapsto \frac{x}{x-1}$ is strictly decreasing on its domain. Now consider (taking logarithms and making some calculations!) that $(\frac{5}{4})^{\omega(n) - 2} \leq \frac{5}{3}$, if $\omega(n) < 5$, and deduce that $5 = \frac{\sigma(n)}{n}$ only if $\omega(n) \ge 5$, q.e.d.

Nice, and by the way, good to see you back here.

Can somebody explain to me why
$\frac{\sigma(n)}{n} < \prod_{k=1}^{\omega(n)} \frac{p_k}{p_k - 1}$ ?

Note that $\sigma (p^k) = \frac{p^{k+1}-1}{p-1}<\frac{p^{k+1}}{p-1}$ which gives by multiplicativity the desired inequality.

[quote="hitlEULER"] Now consider (taking logarithms and making some calculations!) that $(\frac{5}{4})^{\omega(n) - 2} \leq \frac{5}{3}$, if $\omega(n) < 5$, and deduce that $5 = \frac{\sigma(n)}{n}$ only if $\omega(n) \ge 5$, q.e.d.[/quote]
You have a mistake in your solution!
Suppose $\omega(n)=4$ then we get from your solution $\frac{25}{4} \leq \frac{5}{3}$ - contradiction
You can solve it like this:
because as you said $\frac{x}{x-1}$ is strictly dicreasing so we need to show that with 5 smallest prime numbers except 1 that $\prod_{k=1}^{5} \frac{p_k}{p_k - 1} \leq 5$ so:
$\frac{2}{1}\cdot \frac{3}{2} \cdot \frac{5}{4} \cdot \frac{7}{6} \cdot \frac{11}{10}=4.8125$
So if $\sigma(n)=5n$ then there are more than 5 distinct prime divisors.


Last year, I became interested in: how many arrangements of an unrooted tree up to graph isomorphism are there for n vertices?

The sequence begins: 1, 1, 1, 1, 2, 3, 6, 11, 23, 47...

[url]http://www.research.att.com/~njas/sequences/A000055[/url]

I am really rather interested in this, but so far I have not been able to find much information on it.  Does a definitive recurrence or an explicit formula exist for this?

There's a formula on that page you linked to :)


Determine all pairs $(x, y)$ of positive integers with $y \vert x^{2}+1$ and $x \vert y^{3}+1$.

I think it may help to you :[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=25716[/url]

It may not help me :wink: Because there isn't a solution ,either.


ARML email Support MATHCOUNTS HCSSiM geometry probability

AoPS hosts a number of local communities including those for Alabama, Florida, Texas, Upstate New York, and Southern California.  These communities are a great way to create social networks between high caliber students and also facilitate ARML teams and other local events.  These communities are hidden until you are added as a member by a moderator.  If there are active members of our forum who would like to take on the responsibility of moderating other such local communities, please contact me at crawford@artofproblemsolving.com.

If you think you can round up enough AoPSers for a local community not listed above, PM me and we will create a community for your city or state area.

[b]Edit: Much of this post no longer applies as we have revamped the entire local/national community forum area.[/b]

The Southern California group (one of the ones MCrawford refers to above) is run as a "hidden forum", meaning that you would only see it as an option if the moderator (me) has admitted you to the group. I would assume that the other groups referred to are run the same way.

If you are a high school or middle school student in Southern California and are interested in joining an ARML team or otherwise in finding out what this is about, send me a private message.

[quote]Alabama, Florida, Texas, Upstate New York, and Southern California.[/quote]

Is there one for the midwest? I find that a lot of communities revolve around bigger cities where interest is great. For those of us stuck in states that don't have numerous programs, it is a little bit more difficult. If AoPS decides to create a community for the midwest - or specifically for my home state, MI, I volunteer to moderate it.

Likewise, students, teachers, and math enthusiasts in Alabama who wish to join the Alabama community can email me at crawford@artofproblemsolving.com.

We may soon have a way to see which groups are available and have an automated email sent to these community moderators when students wish to join these hidden communities.

[quote="tetrahedr0n"]
Is there one for the midwest? I find that a lot of communities revolve around bigger cities where interest is great. For those of us stuck in states that don't have numerous programs, it is a little bit more difficult. If AoPS decides to create a community for the midwest - or specifically for my home state, MI, I volunteer to moderate it.[/quote]

I think Michigan is certainly large enough to have its own community.  If you would be interested in moderating it or know the ARML coach (to see if they would be interested), I could set that up soon.  Let me know.

[quote="joml88"]Hmm... sounds very cool.

I would love to have one for South Carolina but honestly I don't really think there are too many South Carolinians on the forum....I don't know of any active ones right now.  I'm sure that Marcus (gauss202) would be probably be able to get in touch with more schools and such to get people on here.  But the last time he was on here was a little after ARML.

<sigh>[/quote]

Perhaps the Carolinas together would make a good sized community?  Marcus would certainly make a good moderator if he is interested.  Otherwise, either you or someone else could moderate?  Let me know.

[quote]I think Michigan is certainly large enough to have its own community. If you would be interested in moderating it or know the ARML coach (to see if they would be interested), I could set that up soon. Let me know.[/quote]

I am interested in moderating. The issue with ARML is that it is hard to say what is going to happen this year. Up to last year, every year MI has had an ARML team known as the "MI All Star Team" determined by finish in the MMPC competition. MMPC director David Redman and former director Robert Messer were coaches. Last year, there was a lack of funding/interest, so the All Star Team system stopped. ICAE - a local math club/circle offered to send a team, altough it was difficult to go due to disorganization in travel plans. This year might be different - either Mr. Messer will use the old system, or ICAE might continue the trend started last year. I'll try to find some more information.

Hi, I've just emailed to indicate an interest in forming and maintaining a Minnesota AoPS community too. I just met someone this weekend who would be good to recruit into the group.

[quote]Hey Tetrahedr0n, did you get my PM?[/quote]

I did. I'm sorry I didn't respond a little earlier, but I've been very busy with finals the last two days. I'm going to post here soon to attract more members as well.

[quote="Peter VDD"]Is this only for the usa? I wouldn't mind to do Belgium/Flanders if you're Europe is okay too :)[/quote]

We plan to build some national communities in the relatively near future.  They may even have some different features, though we are in discussion of the ideas.  We will keep you posted.

ok, if you ever need me just drop me a pm or a mail. I'll unsubscribe from this topic for now. :)

Do you suppose the Massachusetts group would be interested in expanding to a New England group?  I don't suppose Vermont could support a forum of its own ;-)

ondrob, yes we could do that. But please remember local and national communities are just for accomodation and for other (local) discussions you may want to have.

we expect all members to use the main forums just as before, and even more.

please present to me on email a possible structure of the forums the Czech-Slowack community should posses.

I'd moderate or just join Connecticut if there were one. It it's too small, we can probably combine with Rhode Island to make Rhodenecticut. (Or just southern new england, to be bland.  :) )

[quote="SnowStorm"]I could do Colorado although I doubt there is anyone from here on these forums.[/quote]

There is already a Colorado community.  The moderator is rcv.  It is one of the new ones and is just getting going.

[quote="SnowStorm"]I could do Colorado although I doubt there is anyone from here on these forums.[/quote]
Hi, SnowStorm!  I have added you to the Colorado group.  Send me a PM or introduce yourself over on the Colorado Forum.  See you there!

Everyone can request to join the usergroup by going to usergroups -> selecting the group and pressing the join group button (on the second row).

[quote="yif man12"]I don't any of the admin to feel rushed by this, but I'm just curious. Is there going to be one made for Connecticut? If so, there's no rush. I'm just curious about it.[/quote]

Are you volunteering to moderate?

[quote="yif man12"]Sure, I'd be glad to![/quote]Done.

just wondering... if even hungarian is in now, why not dutch? :) I know some people of this board are from belgium/holland..

[quote="aznphatso"]Is there one for Northern California, specifically the SFBA?[/quote]

Yes, and you're a member.  Do you not see it in the North American communities area?

So there it is, staring at me right in the face. Thanks!

[quote="Valentin Vornicu"][quote="Peter VDD"]Is this only for the usa? I wouldn't mind to do Belgium/Flanders if you're Europe is okay too :)[/quote]Well, now you have your own Dutch/Flanders community :)[/quote] :clap:  :clap2:  :omighty:


Putnam
linear algebra
linear algebra unsolved

Suppose $P$ and $Q$ are $n\times n$ matrices over some field. Suppose that $P^2=P,\,Q^2=Q,$ and $I-P-Q$ is invertible.

Prove that $P$ and $Q$ have the same rank.

$P$ has the same rank as $P(I-P-Q)=-PQ$, and $Q$ has the same rank as $(I-P-Q)Q=-PQ$, so the two ranks are equal.

I gave that problem to my "Putnam preparation" class, and one of my students came up with precisely that proof. Once you've found the argument that grobber just gave, there's no point looking for any alternatives.


I have no idea on how to approach this problem...
A square array of dots with $10$ rows and $10$ columns is given. Each dot is coloured either blue or red. Whenever two dots of the same colour are adjacent in the same row or column, they are joined by a line segment of the same color as the dots. If they are adjacent but of different colors, they are joined by a green line segment. In total, there are $52$ red dots. There are $2$ red dots at the corners with an additional $16$ red dots on the edges of the array. The remainder of the red dots are inside the array. There are $98$ green line segments. The number of blue line segments is:

$(A) 36$          $(B) 37$         $(C) 38$          $(D) 39$           $(E) 40$

Source: CMC-Fermat 2001- Question #25

[hide="some really obvious stuff I came up with"]
There are a total of $48$ blue dots.
$16$ blue dots are on the edges.
$2$ blue dots are on corners.
The maximum number of green line segments is $180$[/hide]

Hi Hellstar. Below is hint but reply for solution if you want it :)

[hide="hint"]Idea is simple: Look at each dot. Each segment contains 2 dots. Count the number of segments incident from blue dots (note that some dots have different number of segments incident from it) But there are two types of segments... oops did I say too much?[/hide]

[hide="so basically..."]
If you focus on segments the total number of segments will be $180$. So the total number of blue and red segments will be $82$ since the number of green segments is $98$. So following your hint, since there are a total of $48$ blue dots ($2$ on corners, $16$ on edges, and $30$ inner), there will be a total of $2*2+16*3+30*4=172$ segments issuing from blue dots.  These segments can only be blue or green, because at least one dot is blue.  Subtracting the $98$ green segments, we are left with $74$ blue segments. These are segments that connect blue to blue so we overcounted by a factor of $2$ and we divide $2$ to get $37$. Is that right?[/hide]

Yup  :thumbup: That's exactly what I had in mind


geometry
circumcircle
perpendicular bisector
geometry proposed

In triangle ABC, bisectors of angles B and C intersect ar point E. The area of the circumcirle of triangle BCE equals [i]q[/i]. Find the area of the circumcircle of triangle ABC if BC=[i]d[/i]

Perpendicular bisector of BC meets BC at its midpoint M and the circumcircle (O) at P, Q. The intersection on the opposite side of BC than A is circumcenter of the $ \triangle EBC.$ Let $ R,\ \rho$ be radii of the circles (O), (P), OB = OC = R, PQ = 2R. The line BC is inversion of (O) in (P), $ PM \cdot PQ \equal{} \rho^2,$
$ OM \equal{} OP \minus{} PM \equal{} R \minus{} \frac {\rho^2}{2R},\ \ \ OM^2 \equal{} R^2 \minus{} \rho^2 \plus{} \frac {\rho^4}{4R^2}$
$ \frac {d^2}{4} \equal{} BM^2 \equal{} OB^2 \minus{} OM^2 \equal{} \rho^2 \minus{} \frac {\rho^4}{4R^2}$
$ R^2 \equal{} \frac {\rho^4}{4\rho^2 \minus{} d^2},\ \ \ \pi R^2 \equal{} \frac {(\pi \rho^2)^2}{4 (\pi \rho^2) \minus{} \pi d^2}$


Oh, yeah, nemo. I may not be able to understand much about your polynomial problems, but ateast I had the satisfaction of seeing you needing help on a concept. that's almost unheard of for you  :rotfl:  btw, out of curiosity, can you give me a brief explanation of what you're learning so far?

Umm... dude this is a forum, not a discussion page. If you really need to talk email or pm me. Anyways, can someone look at my other post.


A circle in the 1st quadrant touches the x-axis at point A and touchest the y-axis at point B. It touches the line $x+2y-3-\sqrt5 = 0$ at point C(a,b). Find a+b

Be more specific...By touches, do you mean that it is tangent to???

From my pics. :D
[hide="My answer"]
$OB=\frac{3+\sqrt5}2\\
OA=3+\sqrt5\\
AB=\frac{5+3\sqrt5}2\ \ (from\  Pythagoras)$
\begin{eqnarray*}
&p+r&=\frac{3+\sqrt5}2&...(1)\\
&r+q&=3+\sqrt5&...(2)\\
&p+q&=\frac{5+3\sqrt5}2&...(3)\\
\frac{(1)+(2)+(3)}2;&p+q+r&=\frac{7+3\sqrt5}2&...(4)\\
(4)-(3);&r&=1
\end{eqnarray*}
radias of this circle is 1 and center is (1,1)
and equation of this circle is $(x-1)^2+(y-1)^2=1$
for (a,b) is common point in 2 equation
$(a-1)^2+(b-1)^2=1\ \ and\ \ a+2b-3-\sqrt5=0$
I got $a=\frac{5+\sqrt5}5\ \ and\ \ b=\frac{5+2\sqrt5}5$
Then $a+b=\frac{10+3\sqrt5}5$  :blush: [/hide]


trigonometry
limit
inequalities
algebra unsolved
algebra

Determine the number of real solutions of the system \[\left\{ \begin{aligned}\cos x_{1}&= x_{2}\\ &\cdots \\ \cos x_{n-1}&= x_{n}\\ \cos x_{n}&= x_{1}\\ \end{aligned}\right.\]

Only one solution $x_{i}=x_{0}$, were $cos(x_{0})=x_{0}$, because $|cos'x_{0}|=sinx_{0}<1$.

Note $ x_{0}=\cos x_{0},$ $\cos^{(n)}x= \underbrace{\cos\dots\cos}_n x $
$ x_{i} \in (0,1),$ $ i =1, \dots,n $
for $ n=1 $,it’s obvious.
$ n\geq2,x_{n}=\cos x_{n-1} \in (-1,1),$ $ x_{1}=\cos x_{n} \in (\cos1,1) \subset (0,1),$ $\dots,$ $ x_{n}= \cos x_{n-1} \in (0,1).$ so $ x_{i} \in (0,1),i =1,\dots,n $
When $ x_{1}=x_{2},x_{1}= \cos x_{1},$ $ x_{1}=x_{0}. $ Hence $ x_{i}= x_{0},i =1,\dots,n $ is the solution.
When $ x_{1} > x_{2},n\neq2.n $ must be an even number.
If $ n=2k+1,k \in N^+, x_{1} > x_{2} \Rightarrow \cos x_{1} > \cos x_{2}\Rightarrow  x_{2} > x_{3}\dots$
We get $  x_{1} > x_{2}, x_{2} < x_{3},\dots, x_{2k} > x_{2k+1}, x_{2k+1} > x_{1}, x_{1} < x_{2} $.contradiction.
When $ n=2 $,$ \cos^{(2)} x_{1} = x_{1}. $ let $ g(x)=\cos(\cos x) - x ,g $'$ (x)=\sin(\cos x)\sin x -1<0 $,and $ g(x_{0})=0, x_{0} $ is the only solution. but $ x_{1} > x_{2} $ contradiction.
For $ n\geq4,$ let $ n=2k,$ suppose $ x_{1} > x_{3} $ or $ x_{1} < x_{3},( x_{1}\neq x_{3}, $ see $ n=2) $,$ x_{1} > x_{3} \Rightarrow \cos^{(2)} x_{1} > \cos^{(2)} x_{3} \Rightarrow x_{3} > x_{5},\dots,$ have $  x_{1} > x_{3}>$ $\dots $ $ > x_{2k-1} > x_{1} $contradiction. when $  x_{1} < x_{3}\dots $
In similar way,thereis no solution when $ x_{1} < x_{2} $
as Rust said $ x_{i}=x_{0}$ is the only solution.
[hide]For $0<x<1$
$\Rightarrow \cos^{(2)} 0< \cos^{(2)}x<\cos^{(2)}1$

$\Rightarrow \lim_{n\rightarrow\infty} \cos^{(2n)}0<\lim_{n\rightarrow\infty}\cos^{(2n)}x<\lim_{n\rightarrow\infty}\cos^{(2n)}1$

$\Rightarrow \lim_{n\rightarrow\infty} \cos^{(2n-1)}1<\lim_{n\rightarrow\infty}\cos^{(2n)}x<\lim_{n\rightarrow\infty}\cos^{(2n)}1$

Suppose$\lim_{n\rightarrow\infty}\cos^{(n)}1=a$,so$\lim_{n\rightarrow\infty}\cos^{(n)}x=a $ , take $ x=x_{0},\lim_{n\rightarrow\infty}\cos^{(n)}x=x_{0} $
It seems Samuel R. Kaplan had published something in [i]Mathematics Magazine [/i]in 2007.
The price just kill me, although I want to have a look. :wink:
[url]http://www.ingentaconnect.com/search/article?option2=author&value2=Samuel+R.+Kaplan&sortDescending=true&sortField=default&pageSize=10&index=1[/url][/hide]

Solution:It is easy to prove that $|\cos{z}-\cos{t}|{\leq}|z-t|$, $(1)$.Subtracting the first equation from the second we obtain $\cos{x_1}-\cos{x_2}=x_2-x_3$. Hence, by the inequality $(1)$ we have $|x_1-x_2|{\ge}|x_2-x_3|$.Similarly, we have $|x_2-x_3|{\ge}|x_3-x_4|{\ge}...{\ge}|x_{n-1}-x_n|{\ge}|x_{n}-x_1|{\ge}|x_1-x_2|$.So everywhere should stand equal signs $x_1=x_2=...x_n$Equation $\cos{x}=x$ has a unique solution.So the system has a unique solution.Ok. :lol:

To [b]szj[/b]. This [url]http://arxiv.org/pdf/1212.1027.pdf[/url] is free, and might be as interesting to you as Kaplan's.
By the way, this (also free) seems to be the very Kaplan's article [url]http://www.maa.org/sites/default/files/Kaplan2007-131105.pdf[/url] (it has 2 pages, like the ingenta link mentions).


function
real analysis
real analysis unsolved

Let $f: R\to R$ be a strictly monotonous function over $Q$.Show that it can be prolonged into a unique monotonous function over $R$.Show that this new function is continuous.
I hope it wasnt already posted.

what do you mean ? you just suppose at the beginning that if $x<y \in \mathbb{Q}$ then $f(x) \leq f(y)$ ? then there is a unique prolongement. It is of course false. There is a statement problem ..

alekk is right.. you need one more hypothesis on $f$, or ask for a unique class of functions, each couple of having a countable number of points of difference (which is not much useful, though :p)

Is it ok right now with the hypothesis of strict monotonicity?I hope so.

hm... no, not enough, yet:
just consider the following function: $f(x) = x$ if $x<\sqrt{2}$, and $f(x) = x+1$ otherwise...
the value for $f(\sqrt{2})$ can be chosen in the whole interval $\left[\sqrt{2}, 1+\sqrt{2}\right]$, and for none of these values the resulting function is continuous..


Since the central angle is $ 45$ and $ \frac{360}{45} \equal{} 8$, we have an octagon. If its perimeter is $ 96$, then each side is of length $ \frac{96}{8} \equal{} 12$.

[quote="luiseduardo"]The central angle of a regular polygon measures 45 degrees and the perimeter is equal to 96 cm. Calculate the far side of that polygon.

(My answer was 12 cm)[/quote]

All solutions aside, because the problem *already* establishes that it is a regular polygon you
wouldn't bother to type "far side" as all of the sides have the same length, and you can't speak
logically of "calculat[ing] the far side" of a polygon, but you could calculate the *length* of the
side of a regular polygon.   So, then the problem should just directly state:

"Calculate the length of a side of that polygon."


linear algebra
matrix
inequalities
inequalities proposed

Let $a,b,c$ be three positive reals. Prove that
$\frac{a+\sqrt{ab}+\sqrt[3]{abc}}{3}\le{\sqrt[3]{a\cdot{\frac{a+b}{2}}\cdot{\frac{a+b+c}{3}}}}$.

It is proven that A(G(m)) is less or equal to G(A(m)). Below is from the source.
-----------------------------------------------------------------------------------------------------------
If A is a matrix with m rows, n columns and nonnegative elements, let $G_1,G_2,...,G_m$ denote the geometris means of the rows of A and let $A_1, A_2,...,A_n$ be the arithmetic means of the columns of A. Then,
geometric mean of $A_1, A_2,...,A_n \geq$ arithmetic mean of $G_1,G_2,...,G_m$.
proof)
Let A be  $\left(\begin{array}{cccc}a_{11}&a_{12}&...&a_{1n}\\a_{21}&a_{22}&...&a_{2n}\\...&...&...&...\\a_{m1}&a_{m2}&...&a_{mn}\end{array}\right) $.
Consider one of the rows of A, let's say the first row. After dividing $a_{1k}$ by $A_k$, we get a list of n numbers: $\frac{a_{11}}{A_1},\frac{a_{12}}{A_2},...,\frac{a_{1n}}{A_n}$.
Applying AM-GM yields
\[\frac{\frac{a_{11}}{A_1}+\frac{a_{12}}{A_2}+...+\frac{a_{1n}}{A_n}}{n}\geq \sqrt[n]{\frac{a_{11}a_{12}...a_{1n}}{A_1A_2...A_n}}=\frac{G_1}{G(A_1, A_2,...,A_n)}\].
After repeating this procedure for all rows of A, we may add all m inequalities.
This will give
\[\sum \frac{\frac{a_{k1}}{A_1}+\frac{a_{k2}}{A_2}+...+\frac{a_{kn}}{A_n}}{n}\geq \sum \frac{G_k}{G(A_1, A_2,...,A_n)}\].
Rearranging this inequality gives us,
geometric mean of $A_1, A_2,...,A_n \geq$ arithmetic mean of $G_1,G_2,...,G_m$.


function
algebra
polynomial
functional equation

Find all the increasing functions $f:{1,2,...,10}\longrightarrow\{1,2,...,100\}$ satisfying $x+y|xf(x)+yf(y)$ for all $1\leq x,y\leq 10$.

[quote="JBL"][hide="Solution modulo details and hard work"]So, for any pair (x, y), let $d = gcd(x, y)$ so $x = md, y = nd, gcd(m, n) = 1$.  Then we have:
$md + nd\mid md f(x) + nd f(y)$ and so
$m + n \mid m f(x) + n f(y)$
$m + n \mid n (f(y) - f(x))$
$m + n \mid f(y) - f(x)$
and thus $f(x) \equiv f(y) \mod \frac{x+y}{gcd(x,y)}$.

So, in particular, we have
[hide="The key step"]
$f(1) \equiv f(10) \mod 11$
$f(1) \equiv f(8) \mod 9$
$f(8) \equiv f(10) \mod 9$
so by the Chinese Remainder Theorem,
$f(1) \equiv f(10) \mod 99$.[/hide]
[hide="All the hard work"]
This gives us three cases:
Case 1) f(1) = f(10).
$f(1) \equiv f(9) \mod 10$
$f(1) = f(10) \equiv f(9) \mod 19$
So $f(1) \equiv f(9) \mod 190, f(1) = f(9) = f(10)$.  Then
$f(1) = f(9) \equiv f(8) \mod 17$, $f(1) \equiv f(8) \mod 9$ so $f(1) \equiv f(8) \mod 153$ and f(1) = f(8) = f(9) = f(10).  Similar comparisons (7 with 1 and 8, 5 with 8 and 9, 6 with 5 and 10 and 1, 4 with 9 and 7, 3 with 10 and 7, 2 with 9 and 7 and 3) give us by the CRT and size of the range that the only functions here are the 100 constant functions.

Case 2) f(1) = 1, f(10) = 100.
$1 = f(1) \equiv f(9) \mod 10$ and $100 = f(10) \equiv f(9) \mod 19$, so we are left with no choice but f(9) = 81.  We can continue this process, which I admit I am loathe to do because it's not very interesting and we should be able to do it just by repeated applications of CRT and the fact that our range is pretty small.  In fact, in this case, the unique solution is $f(x) = x^2$.
Case 3) f(1) = 100, f(10) = 1.  By the uniqueness of the solution to the previous problem, it seems clear that the only solution here is $f(x) = 101 - x^2$, although to be perfectly honest I haven't bothered to check if that's true.[/hide][/hide]

If anyone comes up with something more elegant, I'd be obliged.[/quote]

There exist another short way.Try it yourself:[hide] $x+y|xf(x)+yf(y)\Rightarrow x+y|(f(x)(x+y)-xf(x)-yf(y))\Rightarrow x+y|y(f(x)-f(y))$.Put $x=y+1\Rightarrow 2y+1|f(y+1)-f(y)\Rightarrow f(y+1)-f(y)\ge 2y+1\Rightarrow f(10)\ge100$ with equality iff $f(x)=x^2$.[/hide]

The generic solutions to the unrestricted functional equation on $\mathbb{Z}$ are the even polynomials with integer coefficients. As it happens, only the constants, $n^2$, and $101-n^2$ can fit in the restricted range.

The unrestricted solutions certainly don't have to be monotone.

[quote="JBL"]That's very nice, but does it necessarily work?  Because we could have $f(y+1) - f(y) = 2y +1$ and $f(y + 2) - f(y+1) = -(2(y+1) + 1)$ -- if the function needed to be monotonic, that's very nice, but what if it needn't be?  I should be able to bounce around, in theory.  Or have some of the differences be 0.  How can you deal with those circumstances?[/quote]

Hmmm,I apologies,I forgot to say that $f$ is increasing. :blush: .But you solved a more difficult problem. :lol:


Let $ \{X_i\}$ be a sequence of independent identically distributed random variables such that
\[ E|X_i|<\infty \textrm{ and } EX_i\equal{}v>0.\]
Define $ f(y)$ to be the smallest $ n$ such that
\[ X_1\plus{}X_2\plus{}\cdots\plus{}X_n\geq y.\]
Find the limit
\[ \lim_{y\rightarrow \infty}\frac{f(y)}{y}.\]

Compare [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20848]this old topic.[/url]

One difference: in that other topic, we were assuming that $ X_i$ was nonnegative. bravado does not seem to be making that assumption here. Does that make a difference? Maybe not, but I haven't checked to be sure.

Fix $ y > 0$.  We have $ S_{f(y) \minus{} 1} < y \leq S_{f(y)}$ by definition.  Hence

Divide through by $ f(y)$, so that
\[ \frac {S_{f(y) \minus{} 1}}{f(y) \minus{} 1}\cdot \frac {f(y) \minus{} 1}{f(y)} < \frac {y}{f(y)} \leq \frac {S_{f(y)}}{f(y)}
\]
Since $ f(y)\to\infty$ as $ y\to\infty$ (why?), using the SLLN on both sides of the above shows that a.s. $ \frac {y}{f(y)}\to v$, or $ \frac {f(y)}{y}\to \frac {1}{v}$.


Here are some problems in Math Zoom Year-round classes.  In the following exercises, find real solutions unless indicated otherwise:

1. Solve for $ x,y,z$:
\[ \left\{
\begin{array}{rcl}
x^2 \plus{} xy \plus{} xz \minus{} x &\equal{}& 2 \\
y^2 \plus{} yz \plus{} yx \minus{} y &\equal{}& 4 \\
z^2 \plus{} zx \plus{} zy \minus{} z &\equal{}& 6
\end{array}
\right.\]

2. Solve for $ x$ and $ y$:
\[ x^2 \minus{} 2x\sin(xy) \plus{}1 \equal{} 0\]

3. Let $ x$, $ y$, and $ z$ be positive real numbers, and
\[ (x^2\plus{}1)(y^2\plus{}2)(z^2\plus{}8) \equal{} 32xyz\]
Find $ x$, $ y$, and $ z$.

4. Given that $ a^2\plus{}2a\minus{}5\equal{}0$, and $ b^2\plus{}2b\minus{}5\equal{}0$, find all possible values of $ \frac{b^2}{a}\plus{}\frac{a^2}{b}$.

5. Given that $ a$ and $ b$ are distinct positive integers, and the two equations $ (a\minus{}1)x^2 \minus{}(a^2\plus{}2)x \plus{} (a^2\plus{}2a)\equal{}0$ and $ (b\minus{}1)x^2 \minus{}(b^2\plus{}2)x \plus{} (b^2\plus{}2b)\equal{}0$ have a common root, find all possible values of $ \frac{1}{2a^2}\plus{}\frac{4}{3b^2}$.

6. Solve for $ x$, $ y$, and $ z$:
\[ \left\{
\begin{array}{ccl}
x\plus{}y &\equal{}& 4 \\
xy\minus{}z^2 &\equal{}& 4
\end{array}
\right.\]

7. In $ \triangle ABC$, $ \angle C\equal{}90^\circ$, and the area of the triangle is $ 30$.  What is the minimum possible value for the hypotenuse $ c$?

8. Solve for $ x$:
\[ \frac{1}{x^2\plus{}2x\minus{}3} \plus{} \frac{18}{x^2\plus{}2x\plus{}2} \minus{} \frac{18}{x^2\plus{}2x\plus{}1} \equal{} 0\]

9. Solve for $ x$:
\[ \frac{4x}{x^2\plus{}x\plus{}3} \plus{} \frac{5x}{x^2\minus{}5x\plus{}3} \equal{} \minus{}\frac{3}{2}\]

10. Solve for $ x$:
\[ \sqrt{4x\minus{}1} \plus{} \sqrt{4x\plus{}1} \minus{} \sqrt{16x^2\minus{}1} \equal{} 4x\]

[hide]1. Let $ s \equal{} x \plus{} y \plus{} z$, Add the equations to get $ s^2 \minus{} s \equal{} 12\implies s \equal{} 4, \minus{} 3$ Each equation is $ x(s \minus{} 1) \equal{} 2$, etc answer follows.

2. Rearrange the equation $ (x \minus{} \sin xy)^2 \equal{} \minus{} cos^2xy$. So $ x \equal{} \sin xy$ and $ \cos xy \equal{} 0$. Clearly, $ \sin xy \equal{} \pm 1 \equal{} x$, and we easily solve for $ y$.

3. $ x^2 \plus{} 1\ge 2x$, $ y^2 \plus{} 2\ge 2\sqrt {2} x$, $ z^2 \plus{} 8\ge 4\sqrt {2}z$ by AM-GM, multiply them to get that $ (x^2 \plus{} 1)(y^2 \plus{} 2)(z^2 \plus{} 8)\ge 32xyz$ with equality iff $ x \equal{} 1$, $ y \equal{} \sqrt {2}$, $ z \equal{} 2\sqrt {2}$.

4. If $ a \equal{} b$, then we get $ 2( \minus{} 1\pm\sqrt {6})$. If $ a\ne b$, then they are the distinct roots of the quadratic $ y^2 \plus{} 2y \minus{} 5 \equal{} 0$, we want $ \frac {a^3 \plus{} b^3}{ab} \equal{} \frac {(a \plus{} b)^3 \minus{} 3ab(a \plus{} b)}{ab}$ and the answer follows.

5. Factor the equations to get $ (x \minus{} a) ( \minus{} 2 \minus{} a \minus{} x \plus{} a x) \equal{} 0$. Let $ r$ be the common root, then we have that $ a \equal{} r$, $ b \equal{} \frac {r \plus{} 2}{r \minus{} 1} \equal{} 1 \plus{} \frac {3}{r \minus{} 1}$. Clearly $ r > 0$ is an integer so $ r \minus{} 1 \equal{} 1,3$. Check if $ a,b$ are distinct, etc and compute.

6. $ (t \minus{} x)(t \minus{} y) \equal{} t^2 \minus{} t(x \plus{} y) \plus{} xy \equal{} t^2 \minus{} 4t \plus{} 4 \plus{} z^2 \equal{} 0\implies (t \minus{} 2)^2 \equal{} \minus{} z^2$ so $ z \equal{} 0$, $ x,y \equal{} 2$.

7. $ ab \equal{} 60$, $ c \equal{} \sqrt {a^2 \plus{} b^2}\ge \sqrt {2}ab \equal{} 60\sqrt {2}$; equality occurs when $ a \equal{} b$.

8. Let $ u \equal{} x^2 \plus{} 2x \plus{} 1$. The equation is then $ \frac {1}{u \minus{} 4} \plus{} \frac {18}{u \plus{} 1} \minus{} \frac {18}{u} \equal{} 0\implies u \equal{} 8,9$; the solutions for $ x$ follow.

9. Let $ u \equal{} x^2 \plus{} x \plus{} 3$, then we have $ \frac {4x}{u} \plus{} \frac {5x}{u \minus{} 6x} \plus{} \frac {3}{2}$ factors as $ (u \minus{} 4x)(u \plus{} 4x)$ and the answer follows.

10. Let $ u \equal{} \sqrt {4x \minus{} 1}$ and $ v \equal{} \sqrt {4x \plus{} 1}$. Then $ u \plus{} v \minus{} uv \equal{} \frac {1}{2}\cdot (u^2 \plus{} v^2)\iff 2(u \plus{} v) \equal{} (u \plus{} v)^2$. So either $ u \plus{} v \equal{} 0$ or $ u \plus{} v \equal{} 2$. In the first case, $ u,v\ge 0$, so $ u \equal{} 0;v \equal{} 0$ is inconsistent. In the second case, $ v^2 \minus{} u^2 \equal{} 2$, so $ v \minus{} u \equal{} 1$, $ (u,v) \equal{} (1.5,0.5)$ which gives $ x \equal{} \frac {5}{16}$ which works.[/hide]


calculus factorial algebra polynomial logarithms limit

$\frac{0}{0}=\frac{a}{b}$
Multiplying by $b$ to clear the denomenator, we have
$\frac{0 \cdot b}{0}=a$
$\frac{0}{0}=a$
Multiply by $0$ to get
$0=0$
So it doesn't matter what $a$ or $b$ are; $\frac{0}{0}$ can equal anything, so it is undefined.

[quote="riddler"]sorry if this is dumb but what is $0^0$? and $0/0$? practically I even know it will be $0$ but...
and one more.... you have this equation....
$x+2=0$
Multiply by $x-2$ both sides.... you get a different solution to the question?huh :wallbash_red:  this is weird or what?
I will appreciate if you explain your answer. This question is really annoying :mad:[/quote]

It's strange because anything to the 0th power is supposed to be 1, but 0 to any power is 0. These two contradict, and I think that it is undefined.

0/0 also is strange because it has three possible solutions.
-anything divided by itself is 1
-0 divided by anything is 0
-anything divided by 0 is undefined.

I don't know what to tell you about that. I think it's undefined.

When you multiply both sides by x-2, the left side turns into $(x+2)(x-2)$ which turns into $x^2 - 4$. The right side is 0, and multiplying it by x-2 still makes it 0. (do you see why?)

Solving, $x^2 - 4 = 0$
$x^2 = 4$
so, $x = 2$, [i]or[/i] $x=-2$

And the solution to x + 2 = 0 is $x = -2$.

i understand but this should happen in math??!!!! :mad:  its really annoying!

[quote="riddler"]hey, chess64 how can it be indetermined?eh? :rotfl:  because $0^0$ should be either $0$ or $1$.
But what????????????????????????? :(  :?
I need the exact answer.  :([/quote]

Indeterminant form. Do you really want me to get into calculus?  :P

[quote="236factorial"]Oh things can be worse than this.
A moderator by the name of jili posted this one:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38781[/url][/quote]

that ones easier than mine besides mines impossible. :P

Of course its impossible. Its hard to find the value of an indeterminant number  :D

[b]What does "0/0" mean?[/b]0/0 means the the number c such that 0×c = 0. What value of c will make this equation true? How about 1? or 2, or -26/31? Yes! c can be any number and still satisfy 0×c = 0. Therefore, 0/0 does not mean any particular number - or even anything until we give it some new meaning.

[b]What does "00" mean?[/b]
In trying to answer this, we could first ask what "0n" means for n ≠ 0, n is an integer.
For instance, "03" = 0×0×0 = 0.
01 = 0.
So 0n is the value of the string where 0 occurs n times. Then, 00 is the value of the string where 0 occurs 0 times. But if 0 occurs 0 times, the string has no value and meaning

All of this was gotten from [url=http://www.mathpath.org/concepts/division.by.zero.htm]http://www.mathpath.org/concepts/division.by.zero.htm[/url]

I just found this out: :blush:

[hide]0^0= (0^1)/(0^1)
0^0= 0/0

So, the values of 0^0 and 0/0, whatever they may be, are the same.[/hide]

It doesn't. [b]0^1/0^1[/b] should be 0^(1-1)=0^0.

0^1/0^1 = 0/0.

[hide]The weird part is...

If you cancel out 0^1/0^1, you get 1. But 0/0 can be any number.

 :arrow: Nothing's perfect, even math. There are little aspects in the subject that simply can't be solved. [/hide]

Well, it has been established that : 0^0 = 1
It's juste like : 0! = 1 (where "!" = "factorial")

0^0 is NOT equal to 1. By the quotient of powers, we can consider

$\frac{0^n}{0^n}=0^0$. This would work for any base other than 0. But the problem with $\frac{0^n}{0^n}$ is because you have 0's in both the numerator and the denominator and that's $\frac{0}{0}$. Which of course is undefined. Therefore 0^0 is undefined.

For x+2=0, when you multiply both sides by x-2, you get $x^2-4=0, x^2=4, x=2 or -2$. But one of them is an extraneous solution.

All right? 0^0 is NOT equal to 1 or 0 because it's undefined, 0/0 is NOT equal to 0 because it's undefined too, and for the equation, when you introduce a square term you also introduce an extraneous solution most of the time.

Hope that helps.

OK, my mistake!
0^0 is undefined, but a^0 = 1 (except when a = 0, duh)

[quote="riddler"]sorry if this is dumb but what is $0^0$? and $0/0$? practically I even know it will be $0$ but...
and one more.... you have this equation....
$x+2=0$
Multiply by $x-2$ both sides.... you get a different solution to the question?huh :wallbash_red:  this is weird or what?
I will appreciate if you explain your answer. This question is really annoying :mad:[/quote]

$\frac{0}{0}$ is $0$.

$0^0$ is 1

$x+2=0$'s answer is -2. x=-2

$0^0$ is considered undefined because you can't tell if it is 0 or 1.  It can never be defined, and if someone says it can, they are wrong.

Same goes with $\frac{0}{0}$. It has three answers: 1, 0, and undefined.

But it can be proved that it is not equal to 0 or 1, so it can only be undefined.

[quote="chess64"]Depending on the context, $0^0$ is considered to be of interdeterminant form, 1, or undefined.[/quote]

is there a word called interdeterminant?

I believe he means [url=http://mathworld.wolfram.com/Indeterminate.html]Indeterminate[/url].

thanks. also, i didn't mean to double post.  :)

Riddler, just to try to answer the question about $x+2=0$...
Suppose you take a polynomial $(x-a_1)(x-a_2)(x-a_3)...(x-a_n)=0$.  That's a polynomial of degree n.  Now multiply both sides by $(x-2)$.  Notice that when $x=2$, $(x-2)=0$.  Thus, you've just increased the degree to $n+1$, and you've added the root $2$ by putting that in!  Multiplying both sides by something doesn't always give the same equation: you just won't remove any roots.  Realize that when you multiply by $(x-2)$, you're implicitly assuming that $x\neq2$, because  if x=2, then you're just multiplying both sides of the equation by 0, and getting no information.  Same way, if you multiply both sides by $\frac{1}{x-2}$, you're implicitly assuming that $x\neq2$.
Another way to look at it: whenever you get roots which you had to dramatically change the original equation to get, you MUST check them, because many may not be valid, some may yield identitities orthings that aren't possible.

$0^0$ and $\frac{0}{0}$  are both undefined!!!

and for x+2=0
the ans. is -2 ,

since you times (x-2) to both sides
the (x+2)(x-2)=0,
x= 2 or -2

But the reality is that x+2=0
so the ans is -2 only. :P

Indeterminate forms can only be evaluated through the process of limit.

0/0 is not undefined, it's indeterminate form. I'm not sure about 0^0 though, I've never run across it before. If 0/0 = 0^0, then I guess it's indeterminate form too, but I'm not sure if they're equal.

Ah, didn't see that last post. It seems this was already cleared up, sorry.

Tip: In the future if you find that your post is bad, use the delete button if nobody's replied to it. If you find there's something wrong with it, use the edit button. Don't double post. I realized you didn't know, but now you do.

[quote="mathnerd314"][quote="boxedexe"]Well, $x^0=1$ according to exponential rules.  I have come up with the follwoing:

$x^0=x^{a-a}=\frac{x^a}{x^a}$, and since anything divided by itself equals $1$, $x^0=1$.

Somethings in life are meant to stay undefined for the intelligibility of humans.

Masoud Zargar[/quote]

Objection.

Everything you've said is true EXCEPT when x=0. 0 divided by 0 is not 1. Then you'd find yourself trying to calculate $\frac{0^a}{0^a}=\frac{0}{0}$. On a side note, we'd better hope that $a \neq 0$.

0 divided by 0 is not 1. You can make it equal anything with calculus. Or by these means:

We see that
0*0=0
0*1=0
0*2=0
0*3=0
0*4=0
0*5=0
0*6=0
..........

Dividing both sides by zero (with a fire extinguisher ready in case your paper bursts into flames), we get

$\frac{0}{0}=0$
$\frac{0}{0}=1$
$\frac{0}{0}=2$
$\frac{0}{0}=3$
$\frac{0}{0}=4$
$\frac{0}{0}=5$
$\frac{0}{0}=6$
.........

So $\frac{0}{0}$ is indeterminate, and so is $0^0$ because of the arguments earlier in this thread.[/quote]

Yes, you're correct.  :oops: I forgot to give the conditions.

Masoud Zargar

[quote="G-UNIT"]With all due respect, boxedexe, a great deal of the posts I have seen you make end in something along the lines of "...hence, the result is incomprehensible to the human mind".  Althought this is true in many cases, just because an idea dosent fit our human intuition dosen't mean it is impossible or undefinable or that it will remained unsolved among future generations.  Instead of pulling out the phrase like a rifle give it some thought,  as in this case it back-fired.  :D[/quote]

Well, most of my posts are not among those lines.  However, I still agree with you. :) I actually believe that even if something is not the way the human mind understands it, it could still be intelligible.  An example would be Quantum physics and its myteries.  Next time, I'll try to word my ideas in a way that it is both clear and that my ideas are not misunderstood.:)  In this case, I don't believe it backfired; it was just a simple mistake of not giving the conditions.

Masoud Zargar

hey riddler , as for the x-2=0 , let me tell you another way similar to k81o7

look, you know $ x-2=0 $

when you multiply  by x+2 you get $ x^2-4 = 0 $

note that these 2 expressions are different !when you have multiplied by x+2 , you have actually CHANGED the equation . its no longer the same equation as x-2! hence the answers will obviusly be different

i think you mean to say that on multiplying both sides of the equation by the same ummmm thing , (here  x+2) the equation shouldnt change because the 2 things get cancelled . WHICH IS WRONG because they do change. It will ONLY not change if the 'thing ' is a CONSTANT like 2,3 ,4 , ok ??

like if you have n=m+1 doesnt mean that this is same as n^2 = (m+1)n !!

[quote="eminem"]like if you have n=m+1 doesnt mean that this is same as n^2 = (m+1)n !![/quote]
You can do this, you just have to make sure you eliminate the extraneous solution $n=0$ that you gain by multiplying both sides by $n$. (Of course, it is possible that $n=0$ really is a solution, so you have to test for this before eliminating it.)

Since you guys have spent two pages on this, I think I'll step in.

If by $\dfrac00$ you mean literally zero divided by literally zero divided by literally zero, then the fraction must remain undefined. There's no way to give it a unique meaning. As someone pointed out, the equation $0\cdot x=0$ has many solutions, so you can't single out just one.

On the other hand, if you mean $\dfrac00$ as a shorthand for "the limit of an expression in which both the numerator and denominator tend to zero," or less formally, as "very small divided by very small", then you have what is referred to in calculus as an indeterminate case - anything can still happen, and we have to do more work to decide the case. We have to ask HOW small over HOW small.

All indeterminate cases result from the conflict of tendencies that pull in opposite directions. A numerator that tends to zero would tend to make the fraction go to zero; a denominator that tends to zero would tend to make the fraction go to infinity. Those are conflicting tendencies, and we must do more work to gauge their relative strengths.

It's much the same for $0^0.$

If we mean literally zero raised to the literally zero power, than we must give up and leave the quantity undefined. If we're faced with a troublesome question of how to define $a^b$ we use $a^b=e^{b\ln a}.$ Since $\ln 0$ is not defined, that doesn't help us.

HOWEVER, there is one exception to the above paragraph, which has to do with representing a polynomial or power series in summation notation. By agreed upon convention, the expression $P(x)=\sum\limits_{k=0}^na_kx^k$ has the same identical meaning as $P(x)=a_0+a_1x+a_2x^2+\cdots a_nx^n.$ Included in this is the notion that $P(0)=a_0.$ But this isn't a definition of $0^0$; it's a notational convention having to do with summation notation.

If by $0^0$, we mean a shorthand for a limit in which a base tends to zero, raised to a power that tends to zero, then we have another indeterminate case. Anything can still happen, and we have work to do. Again, there are conflicting tendencies. The power tending to zero is trying to pull this limit in to 1. The base tending to zero is trying to push the limit outwards to either zero or infinity. A good first step for the work needed to sort out the conflicting tendencies is usually to take the logarithm of it.

Some examples:

$\lim\limits_{x\to0^+}x^x=1.$

$\lim\limits_{x\to0^+}x^{\left(\dfrac1{\ln x}\right)}=e.$

$\lim\limits_{x\to0^+}\left(e^{-\dfrac1x}\right)^{\dfrac1{\ln x}}=\infty$

Discussion of examples like this would belong in "Calculus Computations and Tutorials."


geometry
3D geometry
modular arithmetic
number theory
relatively prime
number theory unsolved

find all primes $ p$ and $ q$ such that $ p \plus{} q \equal{} (p \minus{} q)^3$
 :D

It posted before .You can solve the general
$ m\plus{}n\equal{}(m\minus{}n)^3$
With your problem solution is
$ (p,q)\equal{}(5,3)$

[hide]
$ p\plus{}q$ divides $ (p\minus{}q)^3$ and $ (p\plus{}q)^3$, so it divides
\[ (p\plus{}q)^3\minus{}(p\minus{}q)^3\equal{}2q(3p^2\plus{}q^2)\]
So $ p\plus{}q$ divides $ 2(3p^2\plus{}q^2)$ as $ p\plus{}q$ and $ q$ are relatively prime ($ p\equal{}q$ clearly doesn't work).

Similarly, $ p\plus{}q$ divides $ (p\plus{}q)^3\plus{}(p\minus{}q)^3$ from which we can conclude $ p\plus{}q$ divides $ 2(p^2\plus{}3q^2)$.

Now we get that $ p\plus{}q$ divides $ 2(p^2\plus{}3q^2)\plus{}2(3p^2\plus{}q^2)\equal{}8(p^2\plus{}q^2)$ and $ 8((p\plus{}q)^2\minus{}p^2\minus{}q^2)\equal{}16pq$. Because $ p\plus{}q$ is relatively prime to $ p$ and $ q$, $ p\plus{}q$ divides 16. But also $ p\plus{}q$ is a cube greater than 1. We get $ p\plus{}q\equal{}8$. For $ p>q$ this occurs only when $ p\equal{}5,q\equal{}3$.
[/hide]

[u]Solution.[/u]  Take the equation modulo $ p\plus{}q$ to obtain $ 0\equiv (p\minus{}q)^3\equiv 8p^3\pmod{p\plus{}q}$.  Clearly, $ \gcd(p,q)\equal{}1\iff \gcd(p,p\plus{}q)\equal{}1$.  Hence, $ p\plus{}q|8$.  Take the equation modulo $ p\minus{}q$ to obtain $ 0\equiv p\plus{}q\equiv 2p\pmod{p\minus{}q}$.  Since $ \gcd(p,p\minus{}q)\equal{}1$, $ p\minus{}q|2$.  It is obvious that $ (p,q)\equal{}(5,3)$. $ \Box$


Η ιδεα της κατασκευης-παραλλαγης προεκυψε απο λυκειακη ασκηση
Εστω $ C_1,C_2,C_3$ τρείς ομόκεντροι κύκλοι κεντρου $ O$ και $ AA',BB',CC'$ τρείς διάμετροι των $ C_1,C_2,C_3$ αντίστοιχα και τα παραλληλογραμμα $ OAKB,OALC,OBMC$.
Αν $ OMHA',OLIB',OKUC'$ είναι επίσης παραλληλόγραμμα να αποδειχθεί οτι
$ (OH)+(OU)+(OI)\geq{(OA)+(OB)+(OC)}$

(G.Bas. easy)

Δεν συγκινεί κανέναν μάλλον
Αναγεται ευκολα στην ανισωση $ |a\plus{}b\minus{}c|\plus{}|a\minus{}b\plus{}c|\plus{}|\minus{}a\plus{}b\plus{}c|\geq{|a|\plus{}|b|\plus{}|c|}$ οπου a,b,c τα διανυσματα ΟΑ,ΟΒ,ΟC ή εικόνες των μιγαδικων.........

Γειάς σας


Give $a,b \in Z^{+}$ such as $ab(a+b)\vdots a^{2}+ab+b^{2}$
Prove that :
$\mid a-b\mid >\sqrt[3]{ab}$

[quote="Redriver"] $ab(a+b)\vdots a^{2}+ab+b^{2}$[/quote]

Is that a "divides" sign?


1--Sydsaeter, Hammond - Mathematical Methods of Economics
2-- RGD Allen, Mathematical Analysis for Economics

3--R.Dornbusch, S.Fischer - Macroeconomics
4--5Macroeconomics by Makiw
5--Mankiw - Macro + Micro - Study Guide
6--Macroeconomic Theory and Policy by William H. Branson

7--microeconomics by Mankiw
8--Intermediate Microeconomics by Hal Varian
9--workouts in intermediate microeconomics by hal varian
10-- Microeconomics Analysis by varian
11--Microeconomics by Pindyck and Rubinfeld

12--Basic Econometrics by Damodar N Gujarati
13--Essentials of Econometrics by Damodar N Gujarati
14--applied statistics for business and economics: an essential version by allen L.webster
15--Statistics for Business and Economics (with CD-ROM) by David R. Anderson, Dennis J. Sweeney, and Thomas A. Williams

16--Krugman and Obstfeld (KO), 8th edition (Pearson Low Price Edition)

I would be thankful if someone can provide me with the ebooks and solution manuals of books mentioned above.I am really in need of these books for my exams and therefore one of my friend referred me to this site...i am really hopeful.
most of solution manual are available in authors or publisher website..but it wont let me download as they require some code or something.
any torrents or link to these books would be highly appreciated.
thank you

please someone help me...i urgently need solution manuals..and ebooks..please

thanks


i know how to compute $ \pmod{n}$, where $ n$ is a large number, but how does one compute $ 4^{10}\plus{}5^9\plus{}6^8\plus{}7^7\pmod{13}$?

Thanks in advance

[u]Strategy:[/u]
Take it term by term. Each term:
- convert smaller bases into larger bases while reducing the exponent.
- convert the larger bases into smaller bases via mod13, which does not raise the exponent
Thus, in effect, repeating the above pair of steps over and over will just reduce your exponents until they are very reasonable.

[hide="Specific Solution"]

$ 4^2 \equal{} 16 \equiv 3 \pmod{13}$
$ \implies 4^{10} \equiv 3^5 \equiv 3 \cdot 9^2 \equiv 3 \cdot 16 \equiv 3 \cdot 3 \equiv 9 \pmod{13}$

$ 5^9 \equiv 5 \cdot (25)^4 \equiv 4 \cdot ( \minus{} 1)^4 \equiv 5 \pmod{13}$

$ 6^8 \equal{} 36^4 \equiv ( \minus{} 3)^4 \equiv 3^4 \equiv ( \minus{} 4)^2 \equiv 16 \equiv 3 \pmod{13}$

$ 7^7 \equiv ( \minus{} 6)^7 \equiv \minus{} 6 \cdot (36)^3 \equiv \minus{} 6 \cdot ( \minus{} 3)^3 \equiv 6 \cdot (27) \equiv 6 \pmod{13}$

Adding the above, we get $ 4^{10} \plus{} 5^9 \plus{} 6^8 \plus{} 7^7 \equiv 9 \plus{} 5 \plus{} 3 \plus{} 6 \equiv 10 \pmod{13}$[/hide]


A boy has four red marbles and eight blue marbles.  He arranges his twelve marbles randomly, in a ring.  What is the probability that no two red marbles are adjacent?

[hide]Suppose he first puts the 8 blue marbles in a ring.  Then, he puts the four red marbles one at a time.  The probabilities that the first red marble is not adjacent to any other red marbles in the ring is $1$.  The probability for the second is $\frac{11}{13}$, for the third, $\frac{10}{14}$, and for the fourth, $\frac{9}{15}$.  So then, the total probability in question is $\frac{11*10*9}{13*14*15}=\frac\boxed{{33}{91}}$[/hide]


modular arithmetic
geometry
3D geometry
Diophantine Equations

Prove that if $n$ is a positive integer such that the equation \[x^{3}-3xy^{2}+y^{3}=n.\] has a solution in integers $(x,y),$ then it has at least three such solutions.  Show that the equation has no solutions in integers when $n=2891$.

Note that if $ (x,y)$ is a solution in inetegers to $ x^3 \minus{} 3xy^2 \plus{} y^3 \equal{} n$ then $ (y \minus{} x, \minus{} x)$ and $ ( \minus{} y,x \minus{} y)$ are also solutions in integers to $ 3 \minus{} 3xy^2 \plus{} y^3 \equal{} n$, because:
$ (y \minus{} x)^3 \minus{} 3(y \minus{} x)( \minus{} x)^2 \plus{} ( \minus{} x)^3 \equal{} y^3 \minus{} 3y^2x \plus{} 3x^y \minus{} x^3 \minus{} 3(x^2y \plus{} x^3) \minus{} x^3 \equal{} x^3 \minus{} 3xy^2 \plus{} y^3 \equal{} n$;
$ ( \minus{} y)^3 \minus{} 3( \minus{} y)(x \minus{} y)^2 \plus{} (x \minus{} y)^3 \equal{} \minus{} y^3 \plus{} 3(x^2y \minus{} 6y^2x \plus{} y^3) \plus{} x^3 \minus{} 3x^2y \plus{} 3y^2x \minus{} y^3 \equal{} x^3 \minus{} 3xy^2 \plus{} y^3 \equal{} n$.
Now, we can never have $ (x,y) \equal{} (y \minus{} x, \minus{} x)$ because this yields $ x \equal{} y \equal{} 0$ and then $ n \equal{} 0$ which is impossible. The same happens when $ (x,y) \equal{} (y,x \minus{} y)$ and $ (y \minus{} x, \minus{} x) \equal{} (y,x \minus{} y)$. Therefore, if $ (x,y)$ is an integer solution to $ x^3 \minus{} 3xy^2 \plus{} y^3 \equal{} n$ then $ (y \minus{} x, \minus{} x)$ and $ (\minus{}y,x \minus{} y)$ are two other different solutions in integers to $ 3 \minus{} 3xy^2 \plus{} y^3 \equal{} n$, this proves part one.

Assume there do exist integers $ x,y$ such that $ x^3 \minus{} 3xy^2 \plus{} y^3 \equal{} 2891$. Now, $ 2891\equiv 2\pmod 9$; because any cube of an integer is $ \minus{} 1,0,$ or $ 1\pmod 9$ then if $ x\equiv 0\pmod 3$ or $ y\equiv 0\pmod 3$ then $ x^3 \minus{} 3xy^2\equiv 0\pmod 9$ or $ \minus{} 3xy^2 \plus{} y^3\equiv 0\pmod 9$, respectively, then $ y^3\equiv 2\pmod 9$ or $ x^3\equiv 2\pmod 9$, respectively which is impossible. Therefore, we cannot have $ x$ or $ y$ congruent to $ 0\pmod 3$. Therefore, $ x^3,y^3$ can be conruent to $ \minus{} 1$ or $ 1\pmod 9$.
Now, if $ x^3\equiv \minus{} 1\pmod 9, y^3\equiv 1\pmod 9$ or $ y^3\equiv \minus{} 1\pmod 9, x^3\equiv 1\pmod 9$ then we must have $ \minus{} 3xy^2\equiv 2\pmod 9$ which is impossible as $ 3|( \minus{} 3xy^2)$.
Therefore, we must have $ x^3\equiv y^3\pmod 9$ then $ x^3 \plus{} y^3\equiv 2$ or $ \minus{} 2\pmod 9$ then $ \minus{} 3xy^2\equiv 0$ or $ 4\pmod 9$. However, $ \minus{} 3xy^2$ cannot be $ 4\pmod 9$ as $ 3|( \minus{} 3xy^2)$. On the other hand, $ \minus{} 3xy^2$ cannot be $ 0\pmod 9$ because then $ 3|(xy^2)$ so $ x$ or $ y$ is congruent $ 0\pmod 3$ which we proved before cannot be possible. Therefore, we get a contradiction and the equation $ x^3 \minus{} 3xy^2 \plus{} y^3 \equal{} 2891$ has no integer solutions QED.

Let $\omega$ be a ninth root of unity. Note that $x^3-3xy^2+y^3=(x-y(\omega+\frac{1}{\omega}))(x-y(\omega^2+\frac{1}{\omega^2}))(x-y(\omega^4+\frac{1}{\omega^4}))$.
Now note that $(x-y(\omega^2+\frac{1}{\omega^2}))=(x+2y-(\omega+\frac{1}{\omega})^2y)$. Now note that $(\omega+\frac{1}{\omega})(\omega^2+\frac{1}{\omega^2})(\omega^4+\frac{1}{\omega^4})=-1$ and $(\omega+\frac{1}{\omega})(x-y(\omega^2+\frac{1}{\omega^2}))=(-y+(\omega+\frac{1}{\omega})(x-y))$. So using $x^3-3xy^2+y^3=(x-y(\omega+\frac{1}{\omega}))(x-y(\omega^2+\frac{1}{\omega^2}))(x-y(\omega^4+\frac{1}{\omega^4}))$ we get that if $f(x,y)=x^3-3xy^2+y^3$ then $f(x,y)=f(-y,x-y)$ and also $f(x,y)=f(-y,x-y)=f(y-x,-x)$. Hence one solution induces at least three solutions. Now $2891=7^2*59$. Using the $x=t+\frac{1}{t}$ substitution which makes $x^3-3x+1=t^3+1+\frac{1}{t^3}$, then cyclotomic polynomials and field theory implies that for any prime divisor $p$ of $x^3-3x+1$, we have $p$ is $9k \pm 1$ for some $k$. Thus if $7|x^3-3xy^2+y^3$, we see that $7|x,y$. Thus, a solution to $x^3-3xy^2+y^3=2891$ does not exist.


function
LaTeX
quadratics
algebra
quadratic formula

Let $ f(x) \equal{} x^3 \plus{} 3x \plus{} 1$, where $ x$ is a real number. Given that the inverse function of $ f$ exists and is given by
\[ f^{\minus{}1}(x) \equal{} \left( \frac {x \minus{} a \plus{} \sqrt {x^2 \minus{} bx \plus{} c}}{2} \right)^{1/3} \plus{} \left( \frac {x \minus{} a \minus{} \sqrt {x^2 \minus{} bx \plus{} c}}{2} \right)^{1/3}
\] where $ a$, $ b$ and $ c$ are positive constants, find the value of $ a \plus{} 10b \plus{} 100c$.

Eh, I used a really lame brute-force method that is probably wrong...

[hide]
Since $ f(f^{ \minus{} 1}(x)) \equal{} x$ for all x, let's substitute some convenient values!
First, $ x \equal{} 0$ gives $ f^{ \minus{} 1}(0) \equal{} (\frac { \minus{} a \plus{} \sqrt {c}}{2})^{1/3} \plus{} (\frac { \minus{} a \minus{} \sqrt {c}}{2})^{1/3}$.
Also, solving for $ x$ in $ x^3 \plus{} 3x \plus{} 1 \equal{} 0$ using the cubic formula, the above value is also equal to $ (\frac {( \minus{} 1) \plus{} \sqrt {5}}{2})^{1/3} \plus{} ...$(messy LaTeX)
Which gives $ a \equal{} 1$ and $ c \equal{} 5$ since they are both positive.

Similarly, I found $ b \equal{} 2$ using the cubic formula with $ f(f^{ \minus{} 1}( \minus{} 1)) \equal{} \minus{} 1$.
So the answer is $ \boxed{521}$.[/hide]

[hide]
For the sake of convenience, let $ A \equal{} x \minus{} a$, $ B \equal{} x^2 \minus{} bx \plus{} c$, then $ f^{ \minus{} 1}(x) \equal{} \left(\dfrac{A \plus{} \sqrt {B}}{2}\right)^{1/3} \plus{} \left(\dfrac{A \minus{} \sqrt {B}}{2}\right)^{1/3}$.

Let $ y \equal{} f(x) \equal{} x^3 \plus{} 3x \plus{} 1$, then $ f^{ \minus{} 1}(y) \equal{} x$, so
\[ \left(\dfrac{A \plus{} \sqrt {B}}{2}\right)^{1/3} \plus{} \left(\dfrac{A \minus{} \sqrt {B}}{2}\right)^{1/3} \minus{} x \equal{} 0
\]
Therefore
\[ \dfrac{A \plus{} \sqrt {B}}{2} \plus{} \dfrac{A \minus{} \sqrt {B}}{2} \minus{} x^3 \equal{} \minus{} 3x\cdot \left(\dfrac{(A \plus{} \sqrt {B})(A \minus{} \sqrt {B})}{4}\right)^{1/3}
\]
Simplifying
\[ A \minus{} x^3 \equal{} \minus{}3x\cdot \left(\dfrac{A^2\minus{}B}{4}\right)^{1/3}
\]
Substituting $ A \equal{} y \minus{} a, B \equal{} y^2 \minus{} by \plus{} c, y \equal{} x^3 \plus{} x \plus{} 1$
\[ 3x \plus{} 1 \minus{} a \equal{} \minus{} 3x\cdot \left(\dfrac{(b \minus{} 2a)(x^3 \plus{} x \plus{} 1) \plus{} a^2 \minus{} c}{4}\right)^{1/3}
\]
Plugging in $ x \equal{} 0$, we get $ a \equal{} 1$, so
\[ x \equal{} \minus{} x \cdot \left(\dfrac{(b \minus{} 2)(x^3 \plus{} x \plus{} 1) \plus{} 1 \minus{} c}{4}\right)^{1/3}
\]
Dividing both sides by $ x$
\[ \dfrac{(b \minus{} 2)(x^3 \plus{} x \plus{} 1) \plus{} 1 \minus{} c}{4} \equal{} \minus{} 1
\]
Rearranging
\[ (b \minus{} 2)x^3 \plus{} 3(b \minus{} 2)x \plus{} b \minus{} c \plus{} 3 \equal{} 0
\]
Hence, $ b \equal{} 2, c \equal{} 5$, and our answer is $ 521$.
[/hide]

Solving the equation isn't really brute force.

Subsitute $ x \equal{} y \minus{} \frac{1}{y}$

to get $ f(y) \equal{} y^{6} \plus{}y^{3} \minus{} 1$, which can easily be solved with the quadratic formula

[quote="iniquitus"]Solving the equation isn't really brute force.

Subsitute $ x \equal{} y \minus{} \frac {1}{y}$

to get $ f(y) \equal{} y^{6} \plus{} y^{3} \minus{} 1$, which can easily be solved with the quadratic formula[/quote]

Is there any motivation behind this substitution?  It clearly gets the job done, but I don't see how you would come up with that directly from the given cubic.

Apparently it is the same idea used in the derivation of cubic formula:
http://mathworld.wolfram.com/VietasSubstitution.html

yes. The cubic equation is actually two simple substitutions.

For

$ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} 0$ (divide by the leading coefficient),

substitute $ x \equal{} y \minus{} \frac {a}{3}$ to obtain

$ y^3 \plus{} dy \plus{} e \equal{} 0$ (the depressed cubic), and then substitute $ y \equal{} z \minus{} \frac {d}{3z}$ to obtain a quadratic.

The substitution is apparently a common technique when seling with cubics, as it gets rid of the $ x^2$ term.
For example,
Let $ P(x) \equal{} x^3\minus{}3x^2\plus{}5x$.  If $ P(a)\equal{}1$ and $ P(b)\equal{}5$, then find $ a\plus{}b$.
Try a substitution here.


Ross Mathematics Program
geometry
geometric transformation
reflection
email

I have been thinking about what to do this summer, and based on my interests, I have narrowed down my choices to these:

1. Attend Ross Mathematic Program for the second year as a junior counselor
2. Internship related to mathematics (working with professors, etc.)
3. Work (tutoring, teaching, etc.)
4. Any other suggestions?

I am also applying to RSI, but I probably won't get in... :)
The reason this is in the college forum is because I want to know what colleges will think of each of these options. I know I should put my interests and my passion in front of what colleges think, but at this point, all of my options reflect my passion.

My opinion is that you should do what you think is the most unique and fruitful opportunity (the best combination of the two).

- For example, you can always teach or tutor -- however, if it's a particularly special teaching opportunity, then you might consider it.
- If you've already been to Ross, it might not be as fruitful to do that again as it would be to do an internship under a professor, especially if the latter involves research.
- At the same time, if interning for the professor actually just means making his/her research paper look presentable or stapling copies, then it's not so hot.

My suggestion is that, to judge the fruitfulness aspect, write down a list of all the skills and experience you hope to gain from each one, and then rank them the set of skills/experience that seems most attractive. Then, take into account which ones you will opportunities you will be able to put off (most likely tutoring/teaching) and decide which one of them seems most attractive in the short run.

Note that it is [i]always[/i] a good idea to think about the skills and experience you hope to gain from something you plan to engage yourself in, and also to reflect on them seriously when you're done. This way, you'll internalize their value, and be able to express yourself most effectively when college applications come around, and again later when interviews for jobs/internships come around.

I'd say, do whatever you think you'll enjoy most... likely, that's the thing you'll be able to talk about most truthfully when it comes time for college apps.

So I guess working with a professor would be better than going to Ross again. Yeah, I'll put in plenty of work if I work with a professor, it won't be mainly the professor's work.

My question is how did u find a professor to work with? inside connections?

[quote="buzzer11"]Just send emails to professors in the field you are interested in... But a LOT of them say no. Maybe about 90% haha.[/quote]

I had emailed about 10 professors, and only 2 had even responded. One rejected me and one welcomed me, which was pretty awesome. I mean, I had expected a lot of rejections, but not a lot of ignorals.

Yeah, same with me. About half of my contacts didn't respond.

That's nothing to get surprised or demoralized about -- professors are busy people! Some of them are particularly bad with e-mail, and occasionally forget to respond to e-mails from their own students.

Keep in mind that there's absolutely nothing rude in shooting a follow-up e-mail if you haven't heard back in a week (of course, keep it polite, respectful, and sensitive to their busy schedule).


algebra
polynomial
complex numbers
calculus
calculus computations

If we define $ P_m(n) \equal{} \sum_{k \equal{} 1}^n \sec^{2m} \left(\frac {k\pi}{2n \plus{} 1}\right)$ then

$ P_1(n) \equal{} 2\, n(n \plus{} 1)$

$ P_2(n) \equal{} \frac {2^3}{3} \, n(n \plus{} 1)(n^2 \plus{} n \plus{} 1)$

$ P_3(n) \equal{} \frac {2^3}{3\cdot 5} \, n(n \plus{} 1)(8n^4 \plus{} 16n^3 \plus{} 19n^2 \plus{} 11n \plus{} 6)$

$ P_4(n) \equal{} \frac {2^7}{3^2\cdot 5\cdot 7}\, n(n \plus{} 1)(17n^6 \plus{} 51n^5 \plus{} 82n^4 \plus{} 79n^3 \plus{} 54n^2 \plus{} 23n \plus{} 9)$

It looks like $ P_m$ is always a polynomial with $ \deg(P_m) \equal{} 2m$ and $ n(n \plus{} 1)|P_m(n)$ .

In particular $ P_3(6) \equal{} 326592$ .

Similarly we can define $ Q_m(n) \equal{} \sum_{k \equal{} 1}^n \cot^{2m} \left(\frac {k\pi}{2n \plus{} 1}\right)$ .

$ Q_1(n) \equal{} \frac13 \, n(2n \minus{} 1)$

$ Q_2(n) \equal{} \frac {1}{3^2\cdot 5}\, n(2n \minus{} 1)(4n^2 \plus{} 10n \minus{} 9)$

$ Q_3(n) \equal{} \frac {1}{3^3\cdot 5\cdot 7}\, n(2n \minus{} 1)(32n^4 \plus{} 112n^3 \plus{} 8n^2 \minus{} 252n \plus{} 135)$

$ Q_4(n) \equal{} \frac {1}{3^4\cdot 5^2\cdot 7}\, n$ $ (2n \minus{} 1)(192n^6 \plus{} 864n^5 \plus{} 496n^4 \minus{} 2248n^3 \minus{} 1388n^2 \plus{} 3834n \minus{} 1575)$ .

Here $ Q_m$ seems to be always a polynomial with $ \deg(Q_m) \equal{} 2m$ and $ n(2n \minus{} 1)|Q_m(n)$ .

Does somebody see a general formula for $ P_m$ and $ Q_m$ ?

I also have been looking for the proving of
(k=1..n),  SUM[cosec(k*pi/(2n+1))^2)] = 2/3*n*(n+1)
(it can be deduced from Q1 (n) mentioned above)
I have found a proving for this special case [Q1(n)] in the Wikipedia at Basel-Problem ( http://en.wikipedia.org/wiki/Basel_problem )
 but I think it's too complicated and hope there exists another way of proving, in which there is no need for using Complex-numbers,  Moivre's-, Newton's- (binomial)-, and Viete's- formulas, so there can be a proving which uses only equivalent trigonometric and algebric transformation.


What is the sum of the first six positive multiples of 3?

The first six positive multiples of 3 are 3,6,9,12,15,18
Noticing that 3+18=6+15=9+12=21, the answer is 3*21=$ \boxed {63}$


[quote="chess64"]Given three non-collinear points $A,B,C$, construct a circle with centre $C$ such that the tangents from $A$ and $B$ are parallel.[/quote]

Let $D$ and $E$ be the tangency points, since the tangents are parallel $D,C,E$ would be collinear, $C$ is the centre so $C$ is the middlepoint of $DE$, also $DE$ is perpendicular to both tangents.

$DC=CE$ would be the circle's radium, now we have the radium, so by compass we can construct the circle.

 :D

[hide]Find $M$ the midpoint of $AB$, draw $l_1$ tru $A$, $l_1 \|l$.  Find the distance from $M$ to $l_1=d$. Draw a circle with center $C$ and radius $d$.
if $C=M$ draw circle with radius $AM$
 [/hide]
Are u interested in constuction problems? wud u join such a thread? Pls reply!

[quote="delta"][hide]Find $M$ the midpoint of $AB$, draw $l_{1}$ tru $A$, $l_{1}\|l$.  Find the distance from $M$ to $l_{1}=d$. Draw a circle with center $C$ and radius $d$.
if $C=M$ draw circle with radius $AM$
 [/hide][/quote]
Of course you mean that $l$ is the line that joins $M$ and $C$

[quote]Are u interested in constuction problems? wud u join such a thread? Pls reply![/quote]
I love geometry!
Most of all, I love construction problems!!  :D  :)

I have a solution, could someone please tell me if it works pls

[hide = Solution]
Let $M$ be the midpoint of segment $AB$. Let the line $MC$ be $\ell$. Now let $\ell_A$ and $\ell_B$ be the lines parallel to $\ell$ passing through points $A$ and $B$, respectively. Let points $P$ and $Q$ be the projections of $C$ in the direction perpendicular to $\ell$ onto $\ell_A$ and $\ell_B$, respectively. The the circle passing through points $P$ and $Q$ with centre $C$ works, as $CP \perp \ell_A$ and $CQ \perp \ell_B$.
[/hide]


Let $n$ points be given on a circle, and let $nk + 1$ chords between these points be drawn, where $2k+1 < n$. Show that it is possible to select $k+1$ of the chords so that no two of them intersect.

Number the points clockwise from $1$ to $n$. Then each chord is a pair $(a, b)$ with $1 \leq a < b \leq n$.

We'll show the problem for $k=1$ first: in this case, we want to prove that there are no $n+1$ chords such that any two of them intersect. Let $(a, b), (c, d)$ be any two chords with $a \leq c$. Then they intersect iff $c \leq b \leq d$. Now, take a $2 \times n+1$ board; write a chord on each column with the smaller number on top. Order the chords by smaller first element, from left to right (if $a \leq c$, then $(a, b)$ goes first), and order those with equal first element by smaller second element. Now note that no number on the second row can be smaller than a number on the first row (as otherwise the corresponding chords would be disjoint). Also, for two chords with different 1st element, the one on the right must have greater or equal 2nd element (as otherwise the second would be contained in the first, and they wouldn't intersect). So the second row is ordered as well. Let the first row be $a_1, \ldots, a_{n+1}$ and the second be $b_1, \ldots, b_{n+1}$. We know that $a_1 \leq \ldots \leq a_{n+1} \leq b_1 \leq \ldots \leq b_{n+1}$. Consider $b_{n+1}-a_1$. Clearly this number is $>0$ and $\leq n-1$. On the other hand, each chord has greater sum of elements than the one to the left, so that $a_{i+1}+b_{i+1}-a_i-b_i \geq 1$ for $i=1, \ldots, n$.

Now consider the sum of the terms $a_{i+1}+b_{i+1}-a_i-b_i$ for $i=1, \ldots, n$. It is at least $n$, but it telescopes to $a_{n+1}+b_{n+1}-a_1-b_1 \leq b_{n+1}-a_1<n$, absurd.

The problem is done for $k=1$. But now it's easy to do it for any $k$: for any set of intersecting chords, we can order them by first element from smaller to larger. If we order those with equal first element by the second element, then it is clear that the ordering contains no cycles, so we're looking at a partially ordered set. Since it's a poset, there's either a chain of length $n+1$ (that is, $n+1$ pairwise intersecting chords) or an antichain of length $k+1$ (disjoint chords). But we've already proved that there were no $n+1$ pairwise intersecting chords. Thus we're done.

See also this [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2782715&sid=0b5d42e0500b3977698e32a4e49c7df8#p2782715[/url] for the very case $k=1$.


Jim has a perfect number of puzzle pieces that fit into a rectangle. Each one has a one-digit positive integer on each of them. The sum of the numbers is 41, and none of the numbers are composite. What is the lowest even possible product of all of the numbers?

(No cookies for anybody who guesses what song I'm listening to right now.)

Sorry, wrong forum. I thought I was posting in the MathCounts form - somebody move this, as it's too easy for Getting Started in my opinion.

[quote="Treething"]Jim has a perfect number of puzzle pieces that fit into a rectangle. Each one has a one-digit positive integer on each of them. The sum of the numbers is 41, and none of the numbers are composite. What is the product of all of the numbers?

(No cookies for anybody who guesses what song I'm listening to right now.)

Sorry, wrong forum. I thought I was posting in the MathCounts form - somebody move this, as it's too easy for Getting Started in my opinion.[/quote]

Well, we can solve it too, why not?

[quote="Treething"]Jim has a perfect number of puzzle pieces that fit into a rectangle. Each one has a one-digit positive integer on each of them. The sum of the numbers is 41, and none of the numbers are composite. What is the product of all of the numbers?[/quote]
Does that mean he has 6 or 28 pieces? Since any number of pieces can fit into a rectangle. And can the numbers be used more than once?

[hide] ok so 6 obviously can't work (6x7=42, 5x7+5 =40) so we know its 28. and lets just assume its 1s, 2s, and 3s.

x + y +z = 28
x+ 2y +3z =41

so y + 2z =13

then I just figured there had to be about 5  3s so i tested it and ti was too high and I found that 4 3s work. so its 19 1s, 5 2s and 4 3s.

3^4 x 2^5 = 2592[/hide]

[quote="Treething"][quote="José"][quote="Treething"]Sorry, I forgot to add a couple words that slightly change the meaning of the problem.[/quote]

Yes, and the result... huge difference! ejeje lol it's all OK....[/quote]

i can laugh in two languages as well =o[/quote]
I can sneeze in two languages.  :P

[quote="ch1n353ch3s54a1l"]The only perfect numbers between 1 and 41 are 6 and 28. You can't have six non composites wtih one digit add up to 41. For 28, you can have $1\times25+7\times2+2=41$ and multiplying $1^{25}\times7^2\times2=98$?[/quote]

That sounds about right.

Now solve it if you can have two-digit numbers as well :) .


quadratics
function
inequalities proposed
inequalities

Let $A = \sqrt{1-2a^{2}}$, etc.  Then we wish to minimize

$\sum_{cyc}\frac{1+2A-A^{2}}{2}= \sum_{cyc}\frac{ 2-(A-1)^{2}}{2}\ge \boxed{6}$


I enjoyed solving this problem very much:

Prove that \[\tan{\frac{3\pi}{11}} + 4\sin{\frac{2\pi}{11}} = \sqrt{11}.\]

I am posting it here since my solution is very algebraical.

hmm... I saw this equality many times, I wonder if solution mustn't be simpler when I see your hint. I think squaring and... (maybe I saw a proof like that, somewhere).

There were extensive discussion of some trig problems [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=few+trig&t=12775]here[/url] including this problem.

[quote]$\displaystyle(\tan(3s) + 4\sin(2s))^2 = 11 - \frac{(\tan(8s) + \tan(3s))\cos(8s)}{\sin(s)\cos(3s)}$[/quote]

was proved [url=http://www.artofproblemsolving.com/Forum/files/trig-pf2.pdf]here[/url].

Something similar:
Prove that    $tan\frac{\pi}{19}-4sin\frac{6\pi}{19}+4sin\frac{7\pi}{19}+4sin\frac{8\pi}{19}=\sqrt{19}.$

[quote="Arne"]Stil no solution?

Hint: let $\zeta = e^{\frac{i\pi}{11}}$...[/quote]

Ok, I followed your hint and solved it. ;)

What we want to prove is :

$\frac{\zeta^4 - \zeta^{-4}}{i \cdot (\zeta^4 + \zeta^{-4})} + \frac{\zeta - \zeta}i = \sqrt{11}$

We simplify, square, we use the fact that : $\zeta^{11} = -1$
Then, with some boring computations, we get :
$\zeta^{20} + \zeta^{18} + \ldots + 1 = 0$

But $\zeta^{22} = 1$ so $(\zeta-1)(\zeta+1)(\zeta^{20} + \zeta^{18} + \ldots + 1) = 0$.
Since $\zeta \ne -1$ and $\zeta \ne 1$, we're done.

[quote="Arne"]I guess we need a LOT of calculations to prove that?[/quote]
Not so it is a lot of calculations. But also not a little.

See the application of the [u]Altheman's method[/u] for $n=11$ from
the my last reply to the post http://www.mathlinks.ro/Forum/viewtopic.php?t=84400

[quote]$\displaystyle(\tan 3x + 4\sin 2x)^2 = 11 - \frac{\cos 8x(\tan 8x + \tan 3x)}{\sinh\cos 3x}\ \ \ \ \ (1)\ .$[/quote][b] Proof.[/b] I shall use only  the following formulas :

[quote]\[ 2\sin x\cos y=\sin (x+y)+\sin (x-y)\ .  \]\[ 2\cos x\cos y=\cos (x+y)+\cos (x-y)\ .  \]\[ 2\sin x\sin y=\cos (x-y)-\cos (x+y)\ .  \]\[ 2\sin^2x=1-\cos 2x\ .  \]\[ 2\cos^2x=1+\cos 2x\ .  \]\[ 2\sin x\cos x=\sin 2x\ .  \][/quote] Prove easily that the identity $(1)$ is equivalently with the identity :

$\sin x(\sin 3x+4\sin 2x\cos 3x)^2=11\sin x\cos^23x-\sin 11x\ \ \ \ \ (2)\ .$

$1.\blacktriangleright 4\cdot\math {LHS}=4\sin x(\sin 3x+2\sin 5x-2\sin x)^2=$
$4\sin x(\sin^2 3x+4\sin^25x+4\sin^2x+4\sin 3x\sin 5x-4\sin 3x\sin x-$
$-8\sin 5x\sin x)=2\sin x[(1-\cos 6x)+4(1-\cos 10x)+4(1-\cos 2x)+$
$+4(\cos 2x-\cos 8x)-4(\cos 2x-\cos 4x)-8(\cos 4x-\cos 6x)]=$
$2\sin x(9-4\cos 2x-4\cos 4x+7\cos 6x-4\cos 8x-4\cos 10x)=$
$18\sin x-4(\sin 3x-\sin x)-4(\sin 5x-\sin 3x)+7(\sin 7x-\sin 5x)-$
$-4(\sin 9x-\sin 7x)-4(\sin 11x-\sin 9x)\Longrightarrow$
$\boxed {\ \mathrm {LHS}=\frac 14\cdot (22\sin x-11\sin 5x+11\sin 7x-4\sin 11x)\ }\ .$

$2.\blacktriangleright 4\cdot \mathrm {RHS}=22\sin x(1+\cos 6x)-4\sin 11x=$
$22\sin x+11(\sin 7x-\sin 5x)-4\sin 11x\Longrightarrow$
$\boxed {\ \math {RHS}=\frac 14\cdot (22\sin x-11\sin 5x+11\sin 7x-4\sin 11x)\ }\ .$

Therefore , $\math {LHS}=\math {RHS}\Longleftrightarrow (2)\Longleftrightarrow (1)\ .$

[b]Remark.[/b] In the particular case $x: =\frac{\pi}{11}\ ,$ i.e. $\tan\frac{8\pi}{11}+\tan \frac{3\pi}{11}=0$
we will obtain the identity $\boxed {\ \tan \frac{3\pi}{11}+4\sin \frac{2\pi}{11}=\sqrt{11}\ }\ .$


It is a very minor issue but it will, probably, take less than a minute to fix it, so I decided to bring it up: when you view user's profile, the space after the word contact is missing (so it appears as "contactfedja" instead of "contact fedja", at least, on Firefox)


inequalities
Sequence
algebra
Recurrence
IMO Shortlist

Let $ a > 2$ be given, and starting $ a_0 \equal{} 1, a_1 \equal{} a$ define recursively:

\[ a_{n\plus{}1} \equal{} \left(\frac{a^2_n}{a^2_{n\minus{}1}} \minus{} 2 \right) \cdot a_n.\]

Show that for all integers $ k > 0,$ we have: $ \sum^k_{i \equal{} 0} \frac{1}{a_i} < \frac12 \cdot (2 \plus{} a \minus{} \sqrt{a^2\minus{}4}).$

[quote=""ISL questions and solutions from 1995-2001" by A.Ajorloo"]
note that $ a > 2$,so there exists a positive real $ b$ such that $ a \equal{} b \plus{} \frac 1b$ hence:

$ a^2 \equal{} b^2 \plus{} \frac 1{b^2} \plus{} 2\Rightarrow a^2 \minus{} 2 \equal{} b^2 \plus{} \frac 1{b^2}$

so we have:

$ a_2 \equal{} a(a^2 \minus{} 2) \equal{} \left(b^2 \plus{} \frac 1{b^2}\right)\left(b \plus{} \frac 1b\right)$

$ a_3 \equal{} \left((\frac {a_2}{a_1})^2 \minus{} 2\right)a_2 \equal{} \left((b^2 \plus{} \frac 1{b^2})^2 \minus{} 2\right)a_2 \equal{} \left(b^4 \plus{} \frac 1{b^4}\right)\left(b^2 \plus{} \frac 1{b^2}\right)\left(b \plus{} \frac 1b\right)$

continuing this,we get that:

$ a_n \equal{} \left(b^{2^n \minus{} 1} \plus{} \frac 1{b^{2^n \minus{} 1}}\right)\cdots \left(b^2 \plus{} \frac 1{b^2}\right)\left(b \plus{} \frac 1b\right)$

so we have:

$ \sum_{i \equal{} 0}^n\frac 1{a_i} \equal{} 1 \plus{} \frac {b}{b^2 \plus{} 1} \plus{} \frac {b^3}{(b^2 \plus{} 1)(b^4 \plus{} 1)} \plus{} \ldots \plus{} \frac {b^{2^n \minus{} 1}}{(b^2 \plus{} 1)(b^4 \plus{} 1)\cdots (b^{2^n} \plus{} 1)}$

on the other hand we have:

$ \frac 12\left(a \plus{} 2 \minus{} \sqrt {a^2 \minus{} 4}\right) \equal{} \frac 12\left(b \plus{} \frac 1b \plus{} 2 \minus{} (b \minus{} \frac 1b)\right) \equal{} 1 \plus{} \frac 1b$

so we have to show that:

$ \frac {b^2}{1 \plus{} b^2} \plus{} \frac {b^4}{(1 \plus{} b^2)(1 \plus{} b^4)} \plus{} \ldots \plus{} \frac {b^{2^n}}{(1 \plus{} b^2)(1 \plus{} b^4)\cdots (1 \plus{} b^{2^n})} < 1$

now note that:

$ \frac {b^{2^k}}{(1 \plus{} b^2)\cdots (1 \plus{} b^{2^k})} \equal{} \frac 1{(1 \plus{} b^2)\cdots (1 \plus{} b^{2^k \minus{} 1})} \minus{} \frac 1{(1 \plus{} b^2)\cdots (1 \plus{} b^{2^k})}$

now summing up this equation for $ 1\leq k\leq n$ we get that:

$ \frac {b^2}{1 \plus{} b^2} \plus{} \ldots \plus{} \frac {b^{2^n}}{(1 \plus{} b^2)\cdots (1 \plus{} b^{2^n})} \equal{} 1 \minus{} \frac 1{(1 \plus{} b^2)\cdots (1 \plus{} b^{2^n})} < 1$

QED[/quote]

Let $S(a)$ be the sum of the LHS of the inequaltiy when $a_1=a$ when $k$ goes to infinity.

Thus, $S(a)=\frac{1}{1}+\frac{1}{a}+\frac{1}{ (a^2-2)a }+\frac{1}{((a^2-2)^2-2)(a^2-2)a}+\cdots$. We have that $a(S(a)-1)=S(a^2-2)$.

Let $S(a)=f(a)+\frac{2+a-\sqrt{a^2-4}}{2}$.  Subsituting and simplifying: $af(a)=f(a^2-2)$.

It is clear to see that $1\le S(a)\le 2$ (since $a>2$). Thus, if $f(a)=\epsilon \neq 0$, then there exists $x$ such that $|f(x)|>2$, but that would imply that $S(a)=f(a)+\frac{2+a-\sqrt{a^2-4}}{2}$ is outside of the range of $[1,2]$, contradiction.  Thus, $f(a)\equiv 0$ and $S(a)=\frac{2+a-\sqrt{a^2-4}}{2}$.

Thus, all the partial sums are less than $S(a)=\frac{2+a-\sqrt{a^2-4}}{2}$, as desired.

The solution as given in Bhabak Ghalebi's post is hard to realize(probably too hard....its not an usual transormation).Are we to have no easier solutions???


How can an international applicant applying for MIT, Princeton, Harvard etc. stand out? Australia doesn't really have a 'good' high school qualification to be honest. Will things like an Olympiad medal/certificate help? I have read about the requirements of some University but don't quite get them.

International applicants stand out the same way domestic applicants do.

That having been said, international admissions at selective colleges is [b]very[/b] competitive; at MIT, for example, the admission rate is about 5%.  Even if you are extremely well-qualified, it is not easy to get accepted.

In my mind, instead of applying to anywhere he weakly hopes to be accepted one is at the state to choose applying to more logical place to be able to enter.
For example, Harvard is the exact strongest university all over the world for its intellectual capacities and international qualities, but i think number of who choose MIT is bigger than number of who choose Harvard.

What does the [url=http://www.amt.canberra.edu.au/]Australian Mathematics Trust[/url] recommend to strong high school students in Australia for getting ready for challenging university courses?


A graph has 120 points. A weak quartet is a set of four points with just one edge. What is the maximum possible number of weak quartets?

Are you finding a solution? It is here http://www.mathlinks.ro/Forum/viewtopic.php?t=15588 .

I had that document but the solution is too long, I don't understand because my mothertongue is English. So, who can solve it in a shorter solution? Thanks a lot :)

It seems that your mothertongue is [i]not[/i] English, but before you tell us what exactly you don't understand we cannot help you. Please ask any questions in the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=17341 .

Note that this is one of the hardest combinatorics problems of IMO Shortlists, so there shouldn't be a shorter solution.

  darij


geometry
geometric transformation
homothety
geometry unsolved

1.In an acute angled $ \triangle ABC$ points $ D,E,F$ are located on the sides $ BC,CA,AB$ respectively,such that
$ \frac{CD}{CE}\equal{}\frac{CA}{CB}$,

$ \frac{AE}{AF}\equal{}\frac{AB}{AC}$,

$ \frac{BF}{BD}\equal{}\frac{BC}{BA}$.

Prove that $ AD,BE,CF$ are altitudes of $ \triangle ABC$.

2.The circumference of a circle is divided into eight areas by a convex quadrilateral $ ABCD$ with four arcs lying inside the quadrilateral and the remaining four lying outside it.The lengths of the arcs lying inside the quadrilateral are denoted by $ p,q,r,s$ in anticlockwise direction starting from some arc.
Suppose $ p\plus{}r\equal{}q\plus{}s$.
Prove that $ ABCD$ is a cyclic quadrilateral.

3.Let $ BE$ and $ CF$ be the altitudes of an acute angled $ \triangle ABC$,with $ E$ on $ AC$ and $ F$ on $ AB$.Let $ O$ be the point of intersection of $ BE$ and $ CF$.Take any line $ KL$ through $ O$ with $ K$ on $ AB$ and $ L$ on $ AC$.Suppose $ M$ and $ N$ are located on $ BE$ and $ CF$ respectively such that $ KM$ is perpendicular $ BE$  and $ LN$ is perpendicular to $ CF$.Prove that $ FM$ is parallel to $ EN$.

1. Let $ AD',BE',CF'$ be the altitudes of $ \triangle ABC$. Then $ D'E'\parallel DE,E'F'\parallel EF,F'D'\parallel FD$. This is only possible if $ D'\equal{}D,E'\equal{}E,F'\equal{}F$. Q.E.D.

[b]2[/b] [geogebra]92cc838cc92b6cc6fbe776f238cae2f6e53965a3[/geogebra]
For the denomination of points see the picture. Then, by the assumption $ 2K_DK_B\plus{}2L_BL_D\equal{}L_CK_A\plus{}K_DK_B\plus{}L_AK_C\plus{}L_BL_D$
(by $ XY$  we denote the half of the degree mesaure of arc $ \smallsmile XY$). So $ \angle ABC\plus{}\angle CDA\equal{}(K_DK_B\plus{}K_BL_A\plus{}L_AK_C\plus{}K_CL_B\plus{}L_BL_D\minus{}K_AL_C)\plus{}(L_BL_D\plus{}L_DL_C\plus{}L_CK_A\plus{}K_AK_D\plus{}K_DK_B\minus{}L_AK_C)\equal{}2K_DK_B\plus{}2L_BL_D\plus{}K_BL_A\plus{}K_CL_B\plus{}L_DL_C\plus{}L_CK_A\plus{}K_AK_D\equal{}K_AK_D\plus{}K_DK_B\plus{}K_BL_A\plus{}L_AK_C\plus{}K_CL_B\plus{}L_BL_D\plus{}L_DL_C\plus{}L_CK_A\equal{}\pi.$ Thus $ ABCD$ is cyclic.

[b]3[/b]
$ KM \bot BE, BE \bot AC$ so $ KM \| EL$. Thus $ \triangle KMO \sim \triangle LEO$ with the coefficient $ \frac{KO}{OL}$. Analogiously $ \triangle OFK \sim \triangle ONL$ with the same coefficient. So there exists a homothety with the center $ O$, which transforms  $ \triangle MKF$  into $ \triangle NLE$. Thus $ FM \| EN$ Q.E.D.

1
The first given equality is equivalent to $CD\cdot BC=CE\cdot AC$ $\implies$ $\triangle ACD \sim \triangle BCE$, hence $\widehat {ADC} = \widehat {BEC}$ $(1)$.
Similarly, from $BD\cdot BC=BF\cdot BA$ we get $\widehat{ADB}=\widehat{BFC}$ $(2)$.
As $\widehat{ADB}+\widehat{ADC}=180^\circ$ , we get $\widehat{BFC}+\widehat{BEC}=180^\circ$ $(3)$.
From $AF\cdot AB=AE\cdot AC$ the quad $BCEF $ is cyclic, hence $\widehat{BFC}=\widehat{BEC}$ $(4)$, from the relations $(3)$ and $(4)$ getting $BE\perp AC$ and $CF\perp AB$, done

Best regards,
sunken rock

3
From $AFOE$, $FKMO$ and $ONLE$ cyclic, we get $\widehat{BFM}+\widehat{CEN}=\hat {A}$, done.

Best regards,
sunken rock


geometry
cyclic quadrilateral
geometry unsolved

Let $ABCD$ be a cyclic quadrilateral of area 8. If there exists a point $O$ in the plane of the quadrilateral such that $OA+OB+OC+OD = 8$, prove that $ABCD$ is an isosceles trapezoid.

We have $OA+OB+OC+OD \geq AC+BD \geq 2 \sqrt{AC.BD} \geq 2 \sqrt{2S} = 8$.

The equality happens when $AC = BD$ and $AC$ is perpendicular to $BD$. Hence $ABCD$ is an isosceles trapezoid.

Quite a contrived problem - any such convex quadrilateral will have equal-length, perpendicular diagonals, with only possible positioning for point $O$ at their meeting point ... whence the trivial conclusion - fortunately the other problems in the set were slightly more challenging  :)

OA+OB+OC+OD>=AC+BD>=2√(AC*BD)>=2*√(2*S)=2*√(2*8)=2*4=8 equality for AC=BD and AC⟘BD and how ABCD is quadrilateral resulting conclusion.


AoPSwiki
number theory
modular arithmetic

Apparently, most schools do not teach modular arithmetic, but it is required for most math competitions.

Can somebody please post a link to a website that explains (in-depth) modular arithmetic, from the basics to the more advanced?

Thanks.

Beginning and intermediate number theory from AoPSWiki:
[url]http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction[/url]
[url]http://www.artofproblemsolving.com/Wiki/index.php/Intermediate_modular_arithmetic[/url]
This pdf by Naoki Sato has a section (#4) on modular arithmetic:
[url]http://www.artofproblemsolving.com/Resources/Papers/SatoNT.pdf[/url]


factorial
number theory
least common multiple
prime numbers
prime factorization

well im new here and im kind of lost.  :? . well anyway someone have told me that i can ask a question throught here so here it is:

Find a number that can be devided by 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 :D

It was really hard so......i thought you could help me. thank you. ;)

[hide]umm all u really have to do is find a number with factors of 1,2,3,4,5,6,7,8,9,  and 10. So you can just multiply them all together and you get 3628800. :D [/hide]

If instead, you wish to find the [i]smallest[/i] such number, you would first prime factorize all the numbers:

[hide](don't need 1 because everything's divisible by 1)
$2=2$
$3=3$
$4=2^2$
$5=5$
$6=2\cdot3$
$7=7$
$8=2^3$
$9=3^2$
$10=2\cdot5$

Then, we see that the only prime numbers used was 2, 3, 5, and 7.

We find the number that has the greatest exponent for each of them, namely, $2^3,3^2,5,7$

Multiply them together to get $2520$ as the smallest possible such number.[/hide]

Well...... danfvh i think im finding for the LEAST NUMBER that can be.......u know. Well anyway i think Zyloch gave the best information for me to do the problem. :lol: But i still dont much get it. :( Sorry im just a 7th grader. Could somebody explain it better??
 :D

well i think i get it now my father explained it to me a while ago ( after i read Zyloch's message) :lol: He's good in math you know :first:

It's a matter of finding the least common multiple. So, I'll give a simple example.

What is the smallest integer that can be divided evenly by 6 and 8?

The answer is 24. That shouldn't be too hard, just test it yourself.

Now, notice 24 = 2 * 2 * 2 * 3. And 6 = 2 * 3. And 8 = 2 * 2 * 2. If we multiple 6 * 8, we are multiplying (2 * 3) * (2 * 2 * 2), by substitution. Follow?

But we don't need that 2 in the prime factorization of 6 - factoring a number into the primes which when multiplied together gives that number - since the 2's in the 8 supply plenty of 2 factors. So instead, we have 3 * 2 * 2 * 2 as the smallest integer that evenly divides 6 and 8.

Same with 4 and 6. 4 = 2 * 2. 6 = 2 * 3. The number only needs to be divisible by 2 * 3 and 2 * 2, so how about 2 * 2 * 3? That works. Your answer is 2 * 2 * 3 = 12.

So with 10!, which is 10 factorial, equal to 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1, we factor each of the numbers involved.

Now follow Zyloch's reasoning about the prime factorization of the LCM, least common multiple of those numbers, and you get 2520.

um.......uh........i think i get it know. :wacko:  thanks! in anyway! :)

If you're a 7th grader, you might want to go take a look in the middle school forums.

actually JBL here in the philippines im just 1st years highschool! begginer in highschool stuffs like that! in our country 1st year highschools are BEGGINER!!!!! :mad:

No offense was intended, certainly.  I think there must be confusion about our grade numbering systems.  In the United States, 7th grade is the next-to-last year of middle school, typically serving students 11 to 13 years old.  High school is from grades 9 to 12 (ages approximately 14 to 18).  Is the numbering system different in the Phillipines?

yes!!!! very much different highschool levels starts with first year approximately 12-14 years old and ends at forth year 15-17 years old

im just 13 years old so im just a 1st year highschool here in the philippines and a 7th grader in the united states

ohhhhhhhhhhh!!!

So confusing :wacko:


geometry
circumcircle
incenter
cyclic quadrilateral
geometry proposed

Given cyclic quadrilateral $ ABCD$. Four circles each touching its diagonals and the circumcircle internally are equal. Is $ ABCD$ a square?

[i]Author: C.Pohoata, A.Zaslavsky[/i]

Let $ I_a, I_b, I_c, I_d$ be the incenters of triangles $ BCD, ACD, ABD, ABC$. Applying [b]Sawayama-Thebault's[/b] theorem we have $ I_a$ lies on $ O_2O_3$, i.e $ I_b, I_c, I_d$. But $ R_{O_1}\equal{}R_{O_2}\equal{}R_{O_3}\equal{}R_{O_4}$ then $ R_{I_a} \equal{} R_{I_b} \equal{} R_{I_c} \equal{} R_{I_d}$
This result is well-known: $ I_aI_bI_cI_d$ is a rectangle. But $ I_aI_b//CD$, i.e then $ ABCD$ is a rectangle. Moreover, 4 radius of $ (O_1), (O_2), (O_3), (O_4)$ are equal thus we can easy show that $ ABCD$ is a square.


Let $ ABCD$ be a trapezoid inscribed in a circle C of diameter AB, with AB and CD parallel, AB=2CD. Let Q  be the point where AD meets BC and P the intersection of the tangents to C at B,D. Calculate the area of ABPQ in terms of the area of PDQ.

You've got 2002 and 2006 in Resources. http://www.mathlinks.ro/viewtopic.php?p=1313573#1313573