How do you prove that $ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a}\ge a\plus{}b\plus{}c$ for positive reals $ a, b, c$ with Cauchy-Schwarz?

Thanks.

[hide="Useful lemma"][b]Lemma[/b] (Titu Andreescu). $ \sum \frac {x_i^2}{y_i}\geq\frac {\left(\sum x_i\right)^2}{\sum y_i}$.

[i]Proof.[/i] This is a direct consequence of Cauchy-Schwarz. Multiply both sides by $ \sum y_i$ to get $ \left(\sum y_i\right)\left(\sum\frac {x_i^2}{y_i}\right)\stackrel{\text{\tiny{C\minus{}S}}}{\geq} \left(\sum\frac {x_i\sqrt {y_i}}{\sqrt {y_i}}\right)^2 \equal{} \left(\sum x_i\right)^2$. $ \Box$[/hide]

Applying that, $ \text{LHS}\geq\frac {(a \plus{} b \plus{} c)^2}{a \plus{} b \plus{} c} \equal{} a \plus{} b \plus{} c$.

[hide="Just a little-How to do it"]

You want Cauchy-Schwarz

1) Get rid of denominators, so the second multiple has to be like (xa+yb+zc), so the a,b,c in the denominators would disappear

2) You have already squares in numerators, so you can put x=y=z=1

Now we have $ (a\plus{}b\plus{}c)(\frac{c^2}a\plus{}\frac{a^2}b\plus{}\frac{b^2}c)\ge(a\plus{}b\plus{}c)^2$, from which the inequality follows...

[/hide]


integration
algebra
polynomial
real analysis
real analysis unsolved

${k=100}$
${\sum_{n=0}^{k}\frac{x^{n}}{n!}=\frac{e^{x}\Gamma (k+1,x)}{k!}}$

Simplifying your equation

${\int_{0}^{a}\frac{\Gamma (k+1,x)}{k!}\, dx=\frac{a \Gamma (k+1,a)+\Gamma (k+2,0)-\Gamma (k+2,a)}{k!}}$

this helps?!

i think no. h ehe

$f(a)=\int_{0}^{a}e^{-x}(1+\frac{x}{1!}+\frac{x^{2}}{2!}+...+\frac{x^{100}}{100!})dx$ is continuous, $f(50)<50$, and $f(51)>50$ so by the intermediate theorem there exists a root in the range $]50,51[$.

[quote="puuhikki"] $f(50)<50$, and $f(51)>50$ [/quote]

Why?  :maybe:

$f(50) < 50$ because the polynomial in that expression is bounded above by $e^{x}$ when $x \geq 0$.  The other one, I don't know :-)

[quote="N.T.TUAN"]
Why?  :maybe:[/quote]
Write $f$ as an enough accurate Taylor polynomial and compute its value. Then approximate its error term.

[quote="JBL"]$f(50) < 50$ because the polynomial in that expression is bounded above by $e^{x}$ when $x \geq 0$.  [/quote]

That also is all what I know on this problem  :D


What is the sum of the two least, distinct positive integers each of which has exactly eight positive factors?

The two are
$ 2 * 3 * 5 \equal{} 30$
and
$ 2^3 * 3 \equal{} 24$
The sum is $ 30 \plus{} 24 \equal{} \boxed{54}$.


This might be off-topic for this forum but from what I've read here there seems to be a few Exeter students. Just wanted to gauge the amount of Exeter people there are.

There are Exeterians, dont u worry. What's so special about them necessarily? :-P

[quote="tetrahedr0n"][quote]Exeterians...[/quote]

Ahem ... Exonians?[/quote]

LOL :lol:

[quote="MithsApprentice"][quote="tetrahedr0n"][quote]Exeterians...[/quote]

Ahem ... Exonians?[/quote]

-.- whatever[/quote]

hehe, exeterians. I like that better than exonians.  :D

[quote="guangweicao"]I'm not from Exeter,the other phillips school.
the better phillips school.
GO ANDOVER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!![/quote]

I also go to Philip's school...

All of you are from exeter?
Am I the only one from Andover?
[quote]filletwho said: I also go to Philip's school...[/quote]

And who is filletwho?
do i know u

No, my name is philip, thus I go to Philip's school

[quote]"Exonians"?????[/quote]

All along I thought you guys were called Exies
 :D


function
integration
calculus
calculus computations

Solve the differential equation $y'=x+y$.

The obvious solution is $y=Ce^x-x-1$. I'm just interested in the method in solving it.

Let me show you how to solve $y' + f(x)y = g(x)$. Find a function $F(x)$ which fulfills $F'(x) = f(x)$. I.e, it is a primitive to $f$. Now multiply the equation by $e^{F(x)}$ (this can be done without loss of information, since $e^x \ne 0$. Now, check that your equation now indeed is equivalent to $(ye^{F(x)})' = e^{F(x)}g(x)$. Now find a primitive to $e^{F(x)}g(x)$ to find all solutions
\[ y = e^{-F(x)}\int e^{F(x)}g(x)dx  \]

Try it on your equation, where $f(x) = -1$ and $g(x) = x$.

we can also write $y' - y =x$ and charasteristic equation $\lambda - 1 =0$ then general solution of homogeneous equation is $y_0 = C_1 e^{1x}$, and find partial solution for nonhomogeneous equation in form $y_1(x)=Ax+B$.


function
number theory unsolved
number theory

Im not sure if this is the right forum for this or not, but anyways:

Let $ f$ be a function such that $ f(n)$ is an integer whenever n is an integer. Prove that there is some integer $ n$ so that $ f(f(n))\not \equal{}n\plus{}3$.

Let's say $ f(f(n)) \equal{} n \plus{} 3$ for every integer $ n$.  Then $ f(n) \plus{} 3 \equal{} f(f(f(n))) \equal{} f(n\plus{}3)$.  From this, we can see that $ f$ preserves residue classes modulo $ 3$.  Combined with $ f(f(n)) \equal{} n \plus{} 3$, we see that $ f(0), f(1), f(2)$ are respectively $ 0, 1, 2$ modulo $ 3.$

From this, we see that $ f(3n) \equal{} 3(n \plus{} a)$ for some constant $ a$.  Now $ f(f(3n)) \equal{} 3n \plus{} 3 \equal{} 3(n\plus{}1)$ but $ f(f(3n)) \equal{} f(3(n\plus{}a)) \equal{} 3(n \plus{} 2a)$.  So $ 3(n \plus{} 2a) \equal{} 3(n\plus{}1)$ for all integers $ n$.  This means $ 2a \equal{} 1$ for some integer $ a$, impossible.

Let exist f, suth that $ f(f(n))\equiv n\plus{}3$. Then $ f(n\plus{}3)\equal{}f(f(f(n)))\equal{}f(n)\plus{}3$. Therefore $ f(n)\equal{}n\plus{}a_n$, were $ a_n\in \{a_0,a_1,a_2\}$ and $ a_n\equal{}a_m$ if $ n\equal{}m\mod 3$. $ n\plus{}3\equal{}f(n\plus{}a_n)\equal{}n\plus{}a_n\plus{}a_{n\plus{}a_n}$, therefore $ a_n\plus{}a_{n\plus{}a_n}\equal{}3$.
If $ 3|a_n$, then $ a_n\plus{}a_n\equal{}3$ - contradition. Therefore $ a_n\not \equal{}0\mod 3$ and exist only two means $ a, 3\minus{}a, 3\not |a$ and one of them meet twice in $ \{a_0,a_1,a_2\}$ Let $ g(n)\equiv f(n)\mod 3$, then we get $ g: \{0,1,2\}\to \{1,2\}$, suth that $ g(x)\plus{}g(x\plus{}g(x))\equal{}3$.
It give contradition.


Anybody has the answer key to 2007 and 2008 national mathcounts? Thank you.

Post this is the MATHCOUNTS resources topic.


abstract algebra
vector
Functional Analysis
real analysis
real analysis unsolved

Let $L$ be a Banach space. $M\subset L$ - closed linear subspace, and $x_0\notin M$. Does exist a point $y\in M$ such that $\rho(x_0,y)=\rho (x_0,M)$?

No, I don't think such a $y$ always exists.

First notice that if we restrict our attention to the subspace spanned by $M$ and $x_0$, this is a closed linear subspace of $L$, so it's also a Banach space. We can thus assume that within $L,\ M$ is actually a hyperplane: the kernel of a bounded linear functional. From now on, this is the setting we're working in.

Now, given a hyperplane $M$ of the Banach space $L$, there is essentially a unique continuous linear functional $f$ whose kernel is $M$. This is because every element of $L$ is of the form $ax_0+y$ for $a\in K$ (the scalar field, which we take to be either $\mathbb R$ or $\mathbb C$ for now), and given another functional $\tilde f$ with kernel $M$ such that $\tilde f(x_0)=\lambda f(x_0),\ \lambda\in K$, we have $\tilde f(ax_0+y)=\tilde f(ax_0)=a\tilde f(x_0)=\lambda af(x_0)=\lambda f(ax_0)=\lambda f(ax_0+y)$. In other words, $\tilde f=\lambda f$. Now take $K=\mathbb R$. Then a functional $\tilde f$ with the same kernel $M$ as $f$ (which we assume has norm $1$) and norm $1$ is either $f$ or $-f$. This means that the conditions $\ker f=M,\ \|f\|=1,\ f(x_0)>0$ determine $f$ uniquely.

A consequence of the above: if $f$ is a continuous linear functional on a real Banach space $L$ with norm $1$, kernel $M$, and $x_0$ such that $f(x_0)>0$, then $f(x_0)=d=\rho(x_0,M)$ (I'll denote this distance by $d$ from now on). To see this, notice that it's easy to construct, using the Hahn-Banach Theorem on the extension of functionals, a functional $\tilde f$ such that $\tilde f(y)=0,\ \forall y\in M,\ \|\tilde f\|=1$, and $\tilde f(x_0)=d$. The observations made in the previous paragraph imply that $\tilde f=f$, and we're done.

Now let $L=\ell^1$, the real Banach space of summable real sequences, and let the functional $f$ be defined by $f(x)=\sum_{n\ge 1}\left(1-\frac 1n\right)x^n,\ \forall x=(x^1,x^2,\ldots)\in\ell^1$. Clearly, $\|f\|=\sup_{n\ge 1}\left(1-\frac 1n\right)=1$. Let $M=\ker f$, and take $x_0\in\ell^1$ such that $f(x_0)>0$. By the above, $f(x_0)=d$. Now assume we can find $y\in M$ such that $\|x_0-y\|=d$. We then have $f(x_0-y)=f(x_0)=d=\|x_0-y\|$. We have thus found a vector $t=x_0-y$ such that $f(t)=\|t\|$. It's easy to see from the expression of $f$ though that there can be no such vectors; in other words, the norm of $f$ is bever reached: we always have $\|f(x)\|<\|x\|$ if $x\ne 0$. We have a contradiction.

I hope it's Ok. I'd hate to have written so much in vain :).


Compute $ \sum_{k \equal{} 1}^n \frac1{k(k \plus{} 1)(k \plus{} 2)(k \plus{} 3)}$

What result did you get the calculation of my hint?

Resume summation:

$ \equal{} \frac {1}{6}(\frac {2}{n \plus{} 2} \minus{} \frac {1}{n \plus{} 3})$

$ \equal{} \frac{1}{6}[\frac{n \plus{} 4}{(n \plus{} 2)(n \plus{} 3)}]$

[quote="stephencheng"]$ \sum_{k \equal{} 1}^{n}\frac {1}{k(k \plus{} 1)(k \plus{} 2)(k \plus{} 3)}$
$ \equal{} \frac {1}{2}(\sum_{k \equal{} 1}^{n}\frac {1}{k(k \plus{} 3)} \minus{} \sum_{k \equal{} 1}^{n}\frac {1}{(k \plus{} 1)(k \plus{} 2)})$
$ \equal{} \frac {1}{2}[\frac {1}{3}(\sum_{k \equal{} 1}^{n}(\frac {1}{k} \minus{} \frac {1}{k \plus{} 3}) \minus{} \sum_{k \equal{} 1}^{n}(\frac {1}{k \plus{} 1} \minus{} \frac {1}{k \plus{} 2})]$
$ \equal{} \frac {1}{2}[\frac {1}{3}(\frac {1}{1} \plus{} \frac {1}{2} \minus{} \frac {1}{n \plus{} 2} \minus{} \frac {1}{n \plus{} 3}) \minus{} (\frac {1}{2} \minus{} \frac {1}{n \plus{} 2})]$
$ \equal{} \frac {1}{6}(\frac {2}{n \plus{} 2} \minus{} \frac {1}{n \plus{} 3})$[/quote]

That's wrong in the 4th line.

Why do you use Telescoping Method that I gave it as hint? :wink:

and you can assure whether your answer is correct or not by plugging $ n \equal{} 1$ into the result.

[quote="stephencheng"]$ \sum_{k \equal{} 1}^{n}\frac {1}{k(k \plus{} 1)(k \plus{} 2)(k \plus{} 3)}$
$ \equal{} \frac {1}{3}[\sum_{k \equal{} 1}^{n}\frac {1}{k(k \plus{} 1)(k \plus{} 2)} \minus{} \sum_{k \equal{} 1}^{n}\frac {1}{(k \plus{} 1)(k \plus{} 2)(k \plus{} 3)}]$
$ \equal{} \frac {1}{3}(\frac {1}{6} \minus{} \frac {1}{(k \plus{} 1)(k \plus{} 2)(k \plus{} 3)} )$
$ \equal{} \frac {1}{18} \minus{} \frac {1}{3(k \plus{} 1)(k \plus{} 2)(k \plus{} 3)}$

P.S. Corrected, Kunny, using your hint.[/quote]

Incorrect :( , Take it easy! :lol:

hello, prove by induction that $ \sum_{k\equal{}1}^n\frac{1}{k(k\plus{}1)(k\plus{}2)(k\plus{}3)}\equal{}\frac{1}{18}\minus{}\frac{1}{3(n\plus{}1)(n\plus{}2)(n\plus{}3)}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, prove by induction that $ \sum_{k \equal{} 1}^n\frac {1}{k(k \plus{} 1)(k \plus{} 2)(k \plus{} 3)} \equal{} \frac {1}{18} \minus{} \frac {1}{3(n \plus{} 1)(n \plus{} 2)(n \plus{} 3)}$.
Sonnhard.[/quote]
Worning :D
Because we don't yet calculate an answer.


geometry
3D geometry
prism
conics
ellipse
congruent triangles
AMC

A cylindrical tank with radius $ 4$ feet and height $ 9$ feet is lying on its side. The tank is filled with water to a depth of $ 2$ feet. What is the volume of the water, in cubic feet?

$ \textbf{(A)}\ 24\pi \minus{} 36 \sqrt {2} \qquad \textbf{(B)}\ 24\pi \minus{} 24 \sqrt {3} \qquad \textbf{(C)}\ 36\pi \minus{} 36 \sqrt {3} \qquad \textbf{(D)}\ 36\pi \minus{} 24 \sqrt {2} \\ \textbf{(E)}\ 48\pi \minus{} 36 \sqrt {3}$

[hide]
The area of the cylinder's base = $ 16\pi$

and the area of that covered by the 2 ft of water = $ \frac{16\pi}{3} \minus{} 4\sqrt{3}$

and $ 9 (\frac{16\pi}{3} \minus{} 4\sqrt{3}) \equal{} \boxed{48\pi \minus{} 36\sqrt{3} \Rightarrow E}$
[/hide]

[quote="undefined117"][hide]
The area of the cylinder's base = $ 16\pi$

and the area of that covered by the 2 ft of water = $ \frac {16\pi}{3} \minus{} 4\sqrt {3}$

and $ 9 (\frac {16\pi}{3} \minus{} 4\sqrt {3}) \equal{} \boxed{48\pi \minus{} 36\sqrt {3} \Rightarrow E}$
[/hide][/quote]

Could you explain what you mean by the "area covered by the 2 feet of water"?
Pretty much everything that you said?  :huh:
Thanks

[hide]Gosh I really need to learn asymptote..

Since the cylinder is on its side, the water inside will form a prism with height 9 and the area of the base being however much of the original cylinder's base is covered by the water level (see image).

That area can be represented as being the area of sector AB minus the area of triangle ABC.

Note that triangle ABC, when the two congruent halves are combined, is an equilateral triangle with side length 4.

$ \frac{120^\circ}{360^\circ}\pi4^2 \minus{} 4\sqrt{3} \equal{} \frac{16}{3}\pi \minus{} 4\sqrt{3}$ is the area of the prism's base.

multiplying that by 9 will give us the volume, so

$ 9(\frac{16}{3}\pi \minus{} 4\sqrt{3}) \equal{} \boxed{48\pi \minus{} 36\sqrt{3}} \Rightarrow E$[/hide]

Just notice that it has to be a 120 degree sector.
Then 1/3*4^2pi=16/3pi
dont worry about the area of the triangle...
because 16pi/3*9=48pi, so only E is a logical choice (saves like 30 secs if you really needed the time at the end)

Sorry I think I'm missing something easy, but could you tell me how you figured out AB = 4 ?
Or how you know that it is 120 degrees?

AB does not equal 4, but $ AC \equal{} BC \equal{} 4$ because they are the radii (-us?), as was given in the problem.

AB (the water level) and the radius that is represented by the red line are perpendicular bisectors of each other.

It can be seen that the two congruent triangles formed have hypotenuse length 4, and a shorter leg of length 2. Because of the right angle adjacent to the side length 2, it can be seen that the triangles are 30-60-90 triangles (for more info, go [url=http://mathworld.wolfram.com/30-60-90Triangle.html]here[/url]).

$ 60^\circ \times 2 \equal{} 120^\circ$

In the end, how did you multiply by 9? Like, what type of volume formula did you use since this isn't a cylinder or half of one?

You multiply by 9 because the height is 9.

I know, but this isn't a cylinder and you didn't compute the volume of one of 9 things, you computed the area.

The volume we're trying to find is a sliced off part of a cylinder.
We multiply by nine just like we multiply by the height in a prism. (You find the area of the base, and then multiply by height.)
In other words, we first found the area of a slice of the "prism" one foot high, then multiplied it by nine.

Sorry, I'm confused now. You found the area of the slice, not the volume. And the water can't be half of a cylinder either, since it has an ellipse bases. Please help, thanks guys!

You currently have the area, correct? When you multiply it be $9$, you get the volume.

[quote="AruKasera"]I agree completely, but my question is, what type of a solid is this? If it's not half of a cylinder, what is it?[/quote]

It doesn't matter what it is (I don't think it has a name), we can calculate its area using Cavalieri's principle.


what happens when sym-tri nitro toluene is treated with $ \ce{(NH4)2S}$

The group in the para posn wrt to CH3 will get selectively reduced  :D
[hide]
This may be bcos the CH3 grp might be under steric strain due to the other 2 NO2 grps. So they mght not get reduced  :D

[/hide]

r u sure about ur answer....i am not sure of ur reasoning

im very sure of my product . See the NO2 grp adjacent to the CH3 grp is sterically hindered and hence the reducing agent cant reduce the NO2 grp easily (as it cant easily approach) so the NO2 grp which is free and not sterically hindered is easily reduced  :D . Think i have to make finetunings to my level of clarity  :D

There is no reasoning, because this is not a usefull reaction: several products can be formed, and this reaction can only be applied to about three compounds. Other synthetic routes are better.


function
LaTeX
search
real analysis
real analysis unsolved

Letting (x) denote the fractional part of the real number x, consider the function f(x)=sum of (nx)/n^2 (x real).
Find all discontinuities of f, and show that they form a countable dense set. Show that f is nevertheless Riemann-integrable on every bounded interval.

Learn to use latex:

$f(x) = \sum_{n = 1}^\infty \frac{\{nx\}}{n^{2}}$, where I have replaced your $(x)$ with $\{x\}$.

The first part was discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=discontinuous&t=113150]here[/url]. The second part follows from the fact that
[quote="Kent Merryfield"]The function $f$ is discontinuous at every rational number and continuous at every irrational number. [/quote]

Next time, you should mention the source. It's easier to search. Or, better yet, you could search topics on your own :D


limit
real analysis
real analysis unsolved

Sorry that this post doesn't answer your question but some tips to improve your TeX.  The command "\cdot" can be used for multiplication instead of ".", and to create an arrow, just type $ \rightarrow$.

Here is my solution (has other solution simple and I will post it later.
Let $ a_n$ is this sequence
${ \frac {1}{a_n} = \prod_{k = 0}^{n - 1}(1 + \frac {1}{2k + 1}} > \sum_{k = 0}^{n-1}\frac {1}{2k + 1}$
But $ \lim_{k=0}^{+\infty} \frac {1}{2k + 1} = + \infty$
So $ \lim_{n\to\infty}a_n = 0$

[quote="Hong Quy"]find lim :
$ \lim_{n \minus{} > \infty } \frac {1.3.5.7...(2n \minus{} 1)}{2.4.6.8...2n}$
sori, i edited it[/quote]
What if we wrote :
$ \lim_{n \minus{} > \infty } \frac {1.3.5.7...(2n \minus{} 1)\sqrt{n}}{2.4.6.8...2n}$ ?   :)
answer will be $ \frac{1}{\sqrt{\pi}}$

[quote="TTsphn"]Here is my solution (has other solution simple and I will post it later.
Let $ a_n$ is this sequence
${ \frac {1}{a_n} = \prod_{k = 0}^{n - 1}(1 + \frac {1}{2k + 1}} > \sum_{k = 0}^{n - 1}\frac {1}{2k + 1}$
But $ \lim_{k = 0}^{ + \infty} \frac {1}{2k + 1} = + \infty$
So $ \lim_{n\to\infty}a_n = 0$[/quote]
But $ \lim_{k = 0}^{ + \infty} \frac {1}{2k + 1} = + \infty$
who can solve this problem: $ \lim_{k = 0}^{ + \infty} \frac {1}{2k + 1} = + \infty$
 ? with a solution's  beatyful

$ a_n\equal{}\sum_{i\equal{}0}^n\frac{1}{2i\plus{}1}$
Then
$ a_{2n}\minus{}a_n\geq \frac{1}{4}$
$ a_{2^kn}\minus{}a_n\geq \frac{2^{k\minus{}1}}{4}$
$ a_n$ increasing .
So $ a_n\to\infty$


Is there an actual stateside contest for Wisconsin?
When i used to live in Ohio, there was like a OCTM...

The UW runs a talent search ([url]http://math.wisc.edu/~talent[/url]), and some schools participate in a statewide Math League (ask Mustafa, my school doesn't do that). The WI chapter of the MAA also administers a statewide test, which was in December.

Other than that, I think all our competitions are either regional or national.

Really? Our school takes MAA...um...I don't exactly know when but it's later

yea yea, i took the maa like last month. i got a perfect onit :) but they don't acutally give like a statewide ranking on it or anything like that right...

hey, anybody here went to last year's arml?

I didn't because my school doesn't do AMC even though I have tried...
For further ARML discussion you can go [url=http://www.artofproblemsolving.com/Forum/index.php?f=348]here[/url]

You should seriously find some way to do the AMC tests, they are very cool.

it's not as if I haven't tried...anyway Michael said that if in no way no how I could take it at my school, he would let me take it at Magellan, but I don't know if that would exactly work out :D Anyway, no laughing matter. I GOTTA TAKE THE AMC!!!!!!

All you have to do is sign up and then drive over to the right place for the right day and take the test. It would work.

I just found a nearby high school by asking around so I am taking both A and B on the test dates.

[quote="kstan013"]right place? say where?[/quote]

Magellan if you're going to do it there. Just wherever the test it being administered.

[quote="mustafa"]Sleep in a hotel that night :D . Or find somewhere closer. You made the drive plenty of times for Mathcounts Practice...[/quote]on weekends :wink:

Yeah, yeah, whatever. This is far more important than school :D .

Magellan does it? My MC teammates went to a high school to do it!

Michael did say "his school" and since he has no real affiliation with a school (e.g., a teacher), the high school and middle school are both "his school" from my understanding

[quote="ghjk"]Does anybody know Wisconsin can allow competitors to protest?[/quote]

What do you want to protest?

I want to protest my problem 1! Since I compared my solution to Wisonsin's solution and I see it's just different from the last part which is not a big cause to me. I just get 1 point for that problem 1 :( . Do you know how to protest, mustafa?

I [b]go to Magellan[/b], they don't offer the AMC. We go to Appleton West High School to take it, which is like a minute drive from Magellan middle school. They only offer AMC 12 A though. This information might not be accurate but its what I've been thinking this whole time.

[quote="Stargirl240"]*cough* The math teacher at Magellan didn't say anything about the AMC *cough*[/quote]
Mr. Blom's an idiot.


[b]Problema 2096:[/b]
Deputatii unui Parlament au format 2008 comisii, fiecare comisie are cel mult 10 membrii. Oricare 11 comisii exista un parlamentar care face parte din fiecare dintre ele. Demonstrati ca exista un parlamentar care face parte din fiecare comisie. ([i]Petrov[/i])

[b]Problema 2097:[/b]
Gasiti toate numerele prime $ p$ de forma $ a^2\plus{}b^2\plus{}c^2$, care au proprietatea ca: $ p|a^4\plus{}b^4\plus{}c^4.$ ([i]Senderov[/i])

2096. Este un rezultat de folclor c\u a, fiind dat\u a o familie (finit\u a sau infinit\u a) $ \mathcal{F} \equal{} (X_i)_{i \in I}$ de mul\c timi cu $ | X_i | \leq n$, astfel \^ inc\^ at intersec\c tia a oricare $ n\plus{}1$ dintre ele este nevid\u a, atunci intersec\c tia tuturor este nevid\u a. Fie $ A \neq B$ dou\u a dintre mul\c timi (dac\u a toate sunt egale, rezultatul este trivial); atunci $ 0 < | A \cap B | < n$, deci $ A \cap B \equal{} \{x_1, \ldots, x_k \}$, cu $ 1 \leq k \leq n\minus{}1$. Presupun\^ and rezultatul fals, pentru fiecare element $ x_i$ va exista (m\u acar) o mul\c time $ X_i$ astfel ca $ x_i \not \in X_i$ (mul\c timile $ X_i$ nu sunt neap\u arat distincte). Dar atunci, cele cel mult $ n\plus{}1$ mul\c timi $ A, B, X_1, \ldots, X_k$ vor avea intersec\c tia nevid\u a; fie $ x$ un element comun, vom avea evident $ x \neq x_i$ pentru orice $ i$, dar \c si $ x \in A \cap B$, absurd.

2097. Avem $ p \equal{} a^2\plus{}b^2\plus{}c^2 \ | \ a^4\plus{}b^4\plus{}c^4 \equal{} a^4\plus{}(b^2\plus{}c^2)^2\minus{}2b^2c^2 \equal{} a^4\plus{}(p\minus{}a^2)^2\minus{}2b^2c^2 \equal{}$ $ p^2 \minus{} 2pa^2 \plus{} 2(a^4\minus{}b^2c^2)$, deci $ p \ | \ 2(a^4\minus{}b^2c^2) \equal{} 2(a^2\plus{}|bc|)(a^2\minus{}|bc|)$. Atunci $ p\equal{}2$, pentru dou\u a dintre variabile egale cu $ \pm 1$ \c si a treia $ 0$; sau $ p \ | \ a^2\plus{}|bc| \leq a^2\plus{}2|bc| \leq a^2\plus{}b^2\plus{}c^2 \equal{} p$, f\u ar\u a solu\c tii; sau $ p \ | \ |a^2\minus{}|bc\parallel{} \leq a^2 \plus{} |bc| \leq a^2\plus{}b^2\plus{}c^2 \equal{} p$, conduc\^ and la $ a^2 \equal{} |bc|$, \c si similar, $ b^2 \equal{} |ca|$, $ c^2 \equal{} |ab|$, cu singura solu\c tie $ p \equal{} 3$, pentru toate cele trei variabile egale cu $ \pm 1$.
(Ideea vine mai u\c sor dac\u a lucr\u am \^ in $ \mathbb{Z}_p$, unde pentru $ c\neq 0$ se ob\c tine $ (ac^{\minus{}1})^2 \plus{} (bc^{\minus{}1})^2 \equiv \minus{}1 \pmod{p}$, $ (ac^{\minus{}1})^4 \plus{} (bc^{\minus{}1})^4 \equiv \minus{}1 \pmod{p}$, apoi $ (ac^{\minus{}1})^4 \plus{} (bc^{\minus{}1})^4 \plus{} 2(abc^{\minus{}2})^2 \equiv 1 \pmod{p}$, etc.)

moldovan, rezultatul t\u au este cel bun, dar solu\c tia este gre\c sit\u a, vezi factorizarea ...

:maybe:  am lasat doar raspunsurile - mi-am dat seama ce nu este bine


floor function
real analysis
real analysis unsolved

What is the  measure of the number $ 1<a<2$, where $ \{a^n\}(n\equal{}1,2,3,...)$ is dense in $ [0,1)$

Do you mean the measure of reals $ 1 < a < 2$ such that $ \left \{a^n - \lfloor a^n \rfloor\right \}_{n\in \mathbb{N}}$ is dense in $ [0,1)$?  There is a result by Hardy and Littlewood that $ \left \{a^n - \lfloor a^n \rfloor\right \}_{n\in \mathbb{N}}$ is equidistributed in $ [0,1)$ (and hence dense) for almost all real $ a > 1$.

Hardy, G. H. and Littlewood, J. E. "Some Problems of Diophantine Approximation." [i]Acta Mathematica[/i] 37, 193-239, 1914.

Actually I got the citation wrong, the page numbers should be 155-191 (193-239 is the second part of the same article, but it does not pertain to your question).

Acta Mathematica is a fairly popular journal.  Judging by your location, I'm assuming you're at Babeş-Bolyai; I would think a research institution of that caliber would almost certainly have it available in their library.  Even if the library doesn't have it, it likely has some kind of sharing agreements with other universities and publishes, so they should be able to get it for you.

Failing all this, you can also purchase a (rather expensive) [url=http://www.springerlink.com/content/r6h60873n592p442/?p=999a6b285673498d8af9d39f14dfe59c&pi=5]copy from Springer[/url], though this particular article would cost you about $ \$$32.00 plus shipping.


analytic geometry
trigonometry
geometry solved
geometry

Assume South is +-0
Assume North is +-180
Assume East is -90
Assume West is +90
Known variables:
Current X, Y and Heading
Target X, Y

I need a formula that looks at your current heading, then your desired target, and calculates the shortest distance so I can instruct
to turn left or right to get there.

Any help would be great!

Your coordinate system is oriented so that the positive x-axis (0o) is in the south direction and the positive y-axis (90o) is in the west direction. Assume that heading means the oriented angle from -180o to 180o between the south direction (the positive x-axis) and the direction of travel.

Let $x_C, y_C$ be the current position and $\alpha_C$ the current heading. In your coordinate system, this line has an equation

$y - y_C = \tan{\alpha_C} \cdot (x - x_C)$

The line from the current position to the destination $x_D, y_D$ has an equation

$y - y_C = \frac{y_D - y_C}{x_D - x_C} \cdot (x - x_C) = \tan{\alpha_D} \cdot (x - x_C)$

where the destination heading $\alpha_D$ is

$\alpha_D = \arctan{\frac{y_D - y_C}{x_D - x_C}} + (x_D - x_C < 0) \cdot \frac{y_D - y_C}{|y_D - y_C|} \cdot 180^o$

To calculate the change of heading $\alpha_D - \alpha_C$,

$\tan{(\alpha_D - \alpha_C)} = \frac{\tan{\alpha_D} - \tan{\alpha_C}}{1 + \tan{\alpha_D} \cdot \tan{\alpha_C}} = \frac{\frac{y_D - y_C}{x_D - x_C} - \tan{\alpha_C}}{1 + \frac{y_D - y_C}{x_D - x_C} \cdot \tan{\alpha_C}} =$

$= \frac{y_D - y_C - \tan{\alpha_C} \cdot (x_D - x_C)}{x_D - x_C + \tan{\alpha_C} \cdot (y_D - y_C)}$

$\alpha_D - \alpha_C = \arctan{\left(\frac{y_D - y_C - \tan{\alpha_C} \cdot (x_D - x_C)}{x_D - x_C + \tan{\alpha_C} \cdot (y_D - y_C)}\right)}\ +$

$+ (x_D - x_C < 0) \cdot \frac{y_D - y_C}{|y_D - y_C|} \cdot 180^o$


inequalities
trigonometry
geometry
circumcircle
function
geometry solved

A+B+C=180 and A >= B >= C
Prove that a(C-B) + b(A-C) + c(B-A) >= 0

You didn't tell us what a, b, c are, but it's almost sure that a, b, c are the sidelengths of the triangle with the angles A, B, C, so that your question is:

[i]Let A, B, C be the angles and a, b, c the corresponding sidelengths of a triangle. Assume that $A\geq B\geq C$. Prove that $a\left(C-B\right)+b\left(A-C\right)+c\left(B-A\right)\geq 0$.[/i]

Indeed, after the Extended Sine Law, we have a = 2R sin A, b = 2R sin B and c = 2R sin C, where R is the circumradius of our triangle. In other words, a = f(A), b = f(B) and c = f(C), where the function f(x) is defined by f(x) = 2R sin x. Now, since the function sin x is concave on the interval $\left[0;\;\pi\right]$, and 2R is positive, the function f(x) = 2R sin x is also concave on the interval $\left[0;\;\pi\right]$. This interval is, of course, the interval where the angles A, B, C lie. Thus, our problem will immediately follow from the following more general problem:

[i][b]General problem.[/b] Let A, B, C be three arbitrary real numbers in an interval I, and let f(x) be a concave function on I. Assume that $A\geq B\geq C$. Then, $f\left(A\right)\left(C-B\right)+f\left(B\right)\left(A-C\right)+f\left(C\right)\left(B-A\right)\geq 0$.[/i]

The solution of this general problem is easy: Since $A\geq B\geq C$, we have $B-C\geq 0$, $A-C\geq 0$ and $A-B\geq 0$, so that $\frac{B-C}{A-C}\geq 0$ and $\frac{A-B}{A-C}\geq 0$. Also, clearly, $\frac{B-C}{A-C}+\frac{A-B}{A-C}=\frac{A-C}{A-C}=1$. Thus, since the function f(x) is concave on the interval I, we have by the definition of a concave function

$\frac{B-C}{A-C}f\left(A\right)+\frac{A-B}{A-C}f\left(C\right)\leq f\left(\frac{B-C}{A-C}A+\frac{A-B}{A-C}C\right)$;

since $\frac{B-C}{A-C}A+\frac{A-B}{A-C}C=\frac{\left(B-C\right)A+\left(A-B\right)C}{A-C}=\frac{B\left(A-C\right)}{A-C}=B$, this rewrites as

$\frac{B-C}{A-C}f\left(A\right)+\frac{A-B}{A-C}f\left(C\right)\leq f\left(B\right)$.

Multiplication by A - C (remembering that $A-C\geq 0$) yields

$\left(B-C\right)f\left(A\right)+\left(A-B\right)f\left(C\right)\leq \left(A-C\right)f\left(B\right)$.

Now, subtracting $\left(B-C\right)f\left(A\right)+\left(A-B\right)f\left(C\right)$, we get

$0\leq \left(A-C\right)f\left(B\right)-\left(\left(B-C\right)f\left(A\right)+\left(A-B\right)f\left(C\right)\right)$
$= \left(A-C\right)f\left(B\right)+\left(C-B\right)f\left(A\right)+\left(B-A\right)f\left(C\right)$
$=f\left(A\right)\left(C-B\right)+f\left(B\right)\left(A-C\right)+f\left(C\right)\left(B-A\right)$,

and the general problem is solved. $\blacksquare$

  darij

The  initial inequality is  very easy [tex] a(C-B)+b(A-C)+c(B-A)\geq c(C-B)+c(A-C)+c(B-A)=0  [/tex]

[b]EDIT:[/b] I am sorry, this is wrong.


Evaluate

$ \sum_{k\equal{}1}^{n}\sum_{r\equal{}0}^k r C_{r}^{n}$

$ \sum_{k\equal{}1}^{n} \sum_{r\equal{}1}^{k} rC_n^r \equal{}$ $ \sum_{k\equal{}1}^{n} \sum_{r\equal{}1}^{k} nC_{n\minus{}1}^{r\minus{}1} \equal{}$ $ n\sum_{k\equal{}0}^{n\minus{}1} (n\minus{}k)C_{n\minus{}1}^k \equal{}$ $ n\sum_{k\equal{}0}^{n\minus{}1} ((n\minus{}1)C_{n\minus{}2}^k \plus{} C_{n\minus{}1}^k) \equal{}$ $ n\left ( (n\minus{}1)2^{n\minus{}2} \plus{} 2^{n\minus{}1} \right )$.


For integers $ a, b$, define $ a*b\equal{}a^{b} \plus{} b^{a}$. If $ 2*x\equal{} 100$, find the value of $ x$.

$ 2 * x \equal{} x^2 \plus{} 2^x \equal{} 100$

Since the LHS increases very rapidly, we can just guess and check for *positive, even integers.

$ x \equal{} 6 \implies 6^2 \plus{} 2^6 \equal{} 36 \plus{} 64 \equal{} 100$

So $ \boxed{x \equal{} 6}$.

*Note: x must be positive, because if it's not, LHS is not an integer.
x must be even because if not $ x^2$ is odd so LHS is odd.


superior algebra
superior algebra solved

Let A \in Mn(C) such that there exist k \geq 2, integer
A^(k+1)+A^k=0. Let B=I+A
Prove that C=I+AB+(AB)^2+...+(AB)^(k-1) is invertible
---------------------------------------------------------------
I think we must prove that (AB) is nilpotent of order k
C^(-1)=I-AB
Can someone write a proof that (AB)^k=0
Thanks

Isn't it obvious?

We know that A k *(A+I_n) = 0_n and we want to prove that [A(A+I_n)] k =0_n, but [A(A+I_n)] k = A k *(A+I_n)*(A+I_n) k-1  = 0_n*(A+I_n) k-1  = 0_n.


Janice earns a monthly salary of $ \$600$ plus a $ 2\%$ commission on all her sales. How many dollars worth of sales must Janice make in a month to earn a total of $ \$1400$ that month?

That means Janice wants to earn $ 1400 \minus{} 600 \equal{} 800$ dollars commission. Thus, she must make $ 800 \div 0.02 \equal{} 40,000$ dollars worth of sales.

In an equation this would look like

Let $x$=the amount of money she earns in sales

$600+0.02(x)=1400$

$x=\boxed{40000}$


We have that $x,y,z,t$ are 4 real nimbers in the interval $[-1,1]$.
Show that $t^2(xy+yz+zx)+2t(x+y+z)+3\ge 0$. When does equality hold?

[quote="gnosis"]The inequality is the same as $(tx+1)(ty+1)+(ty+1)(tz+1)+(tz+1)(tx+1)\geq 0$, with equality for $\alpha=\beta=-t\in\{\pm 1\}$, where $\alpha,\beta$ are two of $x,y,z$.[/quote]

Great!
In the original solution we make the substitution $u=1/t$ and then we go on in standart way.


this is a romanian shortlist problem for nineth grade since 2002.

[quote="jli"][quote="Ponnamperuma"]Could you say what's the origin of this problem? 'cause I found it non-trivial, although quite easy (it took me only 5-10 minutes), but still remains the fact that until this year I wouldn't b able to attack it...

[hide="Anyway..."]
$2(x^{3}-xy+y^{3})=3(x^{2}+y^{2})-\to 2x^{3}+2y^{3}-3x^{2}-3y^{2}-2xy=0$
$(x^{2}+y^{2})(2x-3)=2xy$. Since the RHS is even, then $x^{2}+y^{2}|2xy$. But by the QM-GM inequality, $x^{2}+y^{2}\geq 2xy$.
Then it must be the case that $x^{2}+y^{2}=2xy$, which is $(x-y)^{2}=0$ and finally $x=y$.
Substituting, $2x^{2}(2x-3)=2x^{2}$ , $2x=4$ , $x=y=2$.
That's all folks... (at least, it should be!)  :wink:  [/hide][/quote]

nice solution.  shouldnt it be am-gm instead of qm-gm?[/quote]

Well, the qm-gm is the quadratic mean inequality, I think, so I think it is supposed to be there.

But nice solution.

And I hate to ruin your day, but $(0,0)$ is also a solution.

Quadratic Mean and Root Mean Square are the same thing, correct?

[quote="mathgeniuse^ln(x)"]

Well, the qm-gm is the quadratic mean inequality, I think, so I think it is supposed to be there.

 [/quote]

$x^{2}+y^{2}\geq 2xy$ follows from both am-gm or qm-gm :wink:

[quote="Ponnamperuma"]Since the RHS is even, then $x^{2}+y^{2}|2xy$. But by the QM-GM inequality, $x^{2}+y^{2}\geq 2xy$.
Then it must be the case that $x^{2}+y^{2}=2xy$[/quote]

I think you have a problem here, because if $2xy$ is negative, we can't say anything. Like $2^{2}+(-2)^{2}|(2 \cdot 2 \cdot-2)$.

EDIT: But you can say $|x^{2}+y^{2}| \ge |2xy|$ so either $x^{2}+y^{2}= 2xy$ or $x^{2}+y^{2}=-2xy$.


This is pure insanity. How you managed to obtain the last question to STEP II 2009 of the pure maths section BEFORE people in the UK had even gone to school for the eam (which started at 9 am) begs belief and it is rather appalling that you think what you have done here is acceptable.


Prove that any positive integer can be represented as an aggregate of different powers of $3$, the terms in the aggregate being combined by the signs $+$ and $-$ appropriately chosen.

Is it correct?

Let $ A\equal{}(a_0, a_1,\cdots,a_k)$ the digits of $ n$ in base $ 3$ $ (n\equal{}a_0\plus{}3a_1 \plus{}\cdots \plus{}3^k a_k)$ and $ a_i \in \{0,1,2\}$
Let $ g(A)\equal{}$ the amount of digits $ 2$ in $ A$
$ 1)$ If $ a_k\equal{}2$ then redefine $ A\equal{}(a_0,a_1,\cdots,a_{k\minus{}1},\minus{}1,1)$ notice that $ g(A)$ decrease in one unit from the previous one.
$ 2)$ If $ g(A)\equal{}0$ then we are done.
$ 3)$ Otherwise let $ a_j\equal{}$ the first element from the right of $ A$ equal to $ 2$ then
from $ 2$ $ 3^i\equal{}3^{i\plus{}1}\minus{}3^{i}$
redefine $ A\equal{}(a_0,a_1,\cdots,a_{j\minus{}1},\minus{}1,a_{j\plus{}1}\equal{}a_{j\plus{}1}\plus{}1,a_{j\plus{}2},\cdots,a_k)$
notice that the new $ a_{j\plus{}1} \in \{0,1,2\}$ thus $ g(A)$ is nonincreasing
Repeating the steps $ 2$ or $ 3$ and because the array $ A$ is finite, $ g(A)$ for same step reachs the rightmost element of $ A$ and then decrease by one.
At the end we get to $ g(A)\equal{}0$ and the result follows.

[hide="Seems too simple"]Consider the algorithm of taking every $ 2$ in the base $ 3$ expansion of the number, replacing it with $ \minus{}1$, and adding $ 1$ to the digit to the left, using standard rules for carrying in base $ 3$. Repeat this as much as needed until no $ 2$s are left. It's pretty obvious the algorithm works; if $ x$ is a digit for $ \minus{}1$ in base $ 3$, we can see that $ 2 \equal{} 1x$ in base $ 3$, justifying our procedure. The algorithm terminates because for there to be an infinite number of $ 2$s to replace, there has to be an infinite number of $ 1$s or $ 2$s to start out with, which is absurd. If there aren't infinite $ 2$s, we can only get more by adding $ 1$ (through the algorithm) to the $ 1$s. In the end, we are left with a number with $ 1,0,\minus{}1$ for digits in base $ 3$. Then expanding it solves the problem.[/hide]

In fact, that algorithm works for any base $ n$ and numbers $ a_1, a_2, ..., a_n$ that cover all residues $ \mod n$. It just may happen that you get an infinite representation (we get $ n$-adic numbers, thus for example $ 1\plus{}n\plus{}n^2\plus{}n^3\plus{}... \equal{}1\minus{}n$).

Another way is that for a given positive integer $ n$, the decompositions with the form $ a_0 \plus{} \cdots a_n3^n$, where $ a_i \in \{\minus{}1,0,1\}$  are pairwise distincts.
Moreover any number with that form belongs to $ E_n \equal{} \{ \minus{}\frac {3^{n\plus{}1}\minus{}1}{2}, \cdots ,\frac {3^{n\plus{}1}\minus{}1}{2}\}$ and this set contains exactly $ 3^{n\plus{}1}$ integers. In another hand, there are exactly $ 3^{n\plus{}1}$ decompositions (three choices for each $ a_i$), thus each integer in $ E_n$ can be written as desired.
Since, this is true for any $ n$ we get the desired conclusion.

Pierre.


Do these problems fast. If you get two other people to do with you, that's even better as well.  :)

R1 - 1. The product of five consecutive positive integers is divisible by both 13 and 31. If that product is as small as possible, compute the smallest of the five integers.

R1 - 2. Let $ T \equal{} TNYWR$ and let $ K \equal{} \frac{T\minus{}1}{15}$.
Compute the value of $ (1 \minus{} i)^{2K}$.

R1 - 3. Three vertices of a cube are connected to form a triangle. Compute the maximum possible perimeter of this triangle if the edge of the cube is $ T$.


Determine $ a,b\in \mathbb{R}$ such that the function $ f: [0,2]\rightarrow [\minus{}1,3]\ ,\ f(x)\equal{}ax\plus{}b$ is bijective.

If $ f(x) \equal{} ax \plus{} b$ is bijective, then all the real numbers in the interval $ [0,2]$ must map to all the real numbers in the interval $ [\minus{}1,3]$. Note that the length of the first interval is $ 2$ and that the length of the second interval is $ 4$. Therefore, we let $ a \equal{} 2$ so we have $ [0,2] \mapsto [0,4]$. Now we let $ b \equal{} \minus{}1$ so that we have $ [0,4]\mapsto [\minus{}1,3]$. Thus, $ f(x) \equal{} \boxed{2x \minus{} 1}$.


number theory
Eulers function
system of equations
arithmetic sequence
IMO Shortlist
imo 1991
Laurentiu Panaitopol

I'll use Bertrand's Postulate: for $u\ge 2$ is a positive integer, there is a prime in the interval $(u,2u)$.

Clearly, $a_2$ must be equal to the smallest prime $p$ which does not divide $n$. If $p=2$, then $n$ is a prime since the common difference $a_{i+1}-a_i$ is equal to $1$, i.e. all positive integers less than $n$ are coprime to $n$. If, on the other hand, $p=3$, we find $n$ to be a power of $2$: the positive integers less than $n$ and coprime to it are precisely the odd ones. We may thus assume that $p\ge 5$. Furthermore, since $n>6$, the positive integers less than $n$ and coprime to it cannot be $a_1,a_2$ alone, i.e. $k=\varphi(n)\ge 3$.

By Bertrand's Postulate, the largest prime less than $p$ is strictly larger than $\frac{p-1}2$, so it cannot divide $p-1$. We will denote this prime by $q$. We know that $q|n$. It's easy to check that for $p\le 31$ there is a prime $r$ strictly between $a_2=p,\ a_3=2p-1$. $rq|n$, so, in particular, $pq<rq\le n$. On the other hand, if $p>32$, the two largest primes $r_1<r_2$ which are smaller than $q$ satisfy $r_2\ge\frac p4,\ r_1\ge\frac{r_2}2\ge\frac p8$ (again, Bertrand's Postulate), so $pq<\frac{p^2}{32}q\le r_1r_2q\le n$ so, once again, we have $pq<n$ (this is what I wanted to prove here).

Now, $q\not|p-1$, and all the numbers $1,1+p-1,\ldots,1+(q-1)(p-1)$ are smaller than $n$ (they are smaller than $pq$, which is smaller than $n$, according to the paragraph above). One of these numbers, however, must be divisible by $q|n$. Since our hypothesis tells us that all of them must be coprime to $n$, we have a contradiction.

The sequence is $ a_i \equal{} 1 \plus{} (p \minus{} 1)(i \minus{} 1), 1 \le 1 \le \phi(n), a_{\phi(n)} \equal{} 1 \plus{} (p \minus{} 1)(\phi(n) \minus{} 1) \equal{} n \minus{} 1$, where $ p$ is the smallest prime not dividing $ n$.
If $ p \equal{} 2$, then $ \phi(n) \equal{} n \minus{} 1$ so $ n$ is a prime.
If $ p \equal{} 3$, then $ n$ is coprime to all odd primes $ \le n$, so $ n$ is a power of $ 2$.
Otherwise, $ p \ge 5$. It implies that $ 6 | n$ and that if you take 4 (=5-1) consecutive integers ($ \le n$) then at most one of them is coprime to $ n$. Let $ t \equal{} \frac {n}{6}$.
Consider $ 3t \minus{} 1, 3t, 3t \plus{} 1$. At most one of $ 3t \minus{} 1$ and $ 3t \plus{} 1$ should be coprime to $ n$. But they have the same gcd with $ n$, so $ 3t \pm 1$ are not coprime with $ n$, which implies that $ t$ is odd, because $ (3t \minus{} 1, n) \equal{} (3t \minus{} 1,6t) \equal{} (3t \minus{} 1,2)$.

It implies that $ (3t \minus{} 2,6t) \equal{} (3t \minus{} 4,6t) \equal{} 1$:2 number with distance 2 that are coprime to $ n$, a contradiction.
(note: we need $ n > 6$ for $ 3t \minus{} 4 \ge 1$, because $ n \equal{} 6$ also satisfies the requirements)

here is my solution of this problem  :)

Let, $ a_2 \minus{} a_1 \equal{} a_3 \minus{} a_2 \equal{} ... \equal{} a_k \minus{} a_{k \plus{} 1} \equal{} s$ and $ a_k > a_{k \minus{} 1} > ... > a_2 > a_1 \equal{} 1$

[u]Case:1[/u]: $ s > 2$

if $ a_2$ is not a prime number it means that  there exist a divisor $ d$ of $ a_2$ such that $ 1 < d < a_2$. Hence $ \gcd(d,n) > 1\Rightarrow\gcd(a_{2},n) > 1$  which is a contradiction !!

Now let's suppose that $ a_i$ is not a prime for some $ i$. Hence there exist a prime divisor $ l$ of $ a_i$ such that $ 1 < l < a_i$ if $ l\neq a_j$ $ \forall j \le i$ then $ \gcd(l,n) > 1 \Rightarrow \gcd(a_i,n) > 1$ which is a contradiction !! and so there exist $ j > 1$ such that $ \forall i$ we have $ a_i \equal{} a_j^c$  and $ j < i$
By the last argument we have $ a_i \equal{} a_j^c \equal{} a_{j_2}^{c_2} \equal{} a_{j_3}^{c_3}... \equal{} a_2^d$
Usin our condition we get $ a_2^d \minus{} a_2^{d'} \equal{} a_2 \minus{} 1$ which is not true !!
we conclude that $ \forall i , a_i$ is a prime number . if $ a_2^2 < n$.Since $ a_i$ is a prime number it means that $ 2a_2 \minus{} 1$ is prime and $ 3a_3 \minus{} 2$ is prime and $ 4a_4 \minus{} 3$ is prime...$ (a_2 \plus{} 1)a_2 \minus{} a_2$ is prime $ \Leftrightarrow$ $ a_2^2$ is prime which is not true. Hence $ a_2 > \sqrt {n}$and it means that $ \forall k < \sqrt {n}$ we have $ \gcd(n,k) > 1$which implies that $ \gcd(k,n \minus{} k) > 1$,$ \forall k < \sqrt {n}$ . On the other hand there exist a $ l$ such that $ n \minus{} k \equal{} a_l$ and so $ \gcd(n,a_l) > 1$ which is not true !!
So the hypothesis of the case 1 is not true !! :D
Let's go now to the case $ s \equal{} 2$
we have
$ \gcd(n,1) \equal{} 1 \Rightarrow \gcd(n,3) \equal{} 1 \Rightarrow \gcd(n,5) \equal{} 1 ...\Rightarrow \gcd(n,2k \plus{} 1) \equal{} 1$ and so the power of two are a solution.
When $ s \equal{} 1$ we get easily the fact that n is prime

and we have done  :)  !!

if n is odd then the problem is trivial. if not then n must be square free because
if n=(2^k)((p_1)^ a_1)...((p_t)^a_t) (we suppose that n#2^m) then this two number are relatively prime to n:
2(p_1)...(p_t) + 1 and 2(p_1)...(p_t) - 1
if this two number are less than n then the problem woud be trivial. so n must be on this form
n=2(p_1)...(p_t) but u can easily see that in this case (n/2)+2 and (n/2)-2 are relatively prime to n. the rest is trivial(u can find that 3|n but (9,n)=1!!!!)

No. For any $n\geq 2$ we have $\gcd(n-1,n)=1$, so $a_k = n-1$. It is only when, as said, $\phi(n) = n-1$, that the conclusion $n$ prime was reached.

Nice problem. Really liked it. Sorry that I don´t know LaTex yet!
Ok, my solution is really simple:

First, solve the equalities two - by - two. Ater that, you get a general formula, from now on called G(n):
2a_i = a_i+1 + a_i-1

Hence:
1) 2a_2 = a_3 + a_1
2) 2a_3 = a_4 + a_2
          .
          .
          .
k-2) 2a_k-1 = a_k + a _k-2

Now, by definition, we know that 1 is relatively prime to every natural number > 1.
Therefore, a_i = 1 , i element of euler´s function of n (sorry, don´t know how to use latex).

Then, since 2a_2 = a_3 + a_1,  a_3 + a_1 = 0 (mod 2).
Now, since a_1 = 1, which is uneven, a_3 must be uneven.

Then, since 2a_4 = a_5 + a_3,     a_5 + a_3 = 0 (mod 2). Since we already established a_3 is uneven, so is a_5.

This process goes on, and we arrive to the conclusion that both terms on the right hand side of the equals are uneven, for the equalities on uneven places, starting with 2a_2 = a_3 + a_1.

Then, we see two cases: 1) a_2 Uneven. 2) a_2 Even.

[u]Case 1[/u]: a_2 Uneven.
Using the same analysis as before, since 2a_3 = a_4 + a_2,  a_4 + a_2 =0 (mod 4). Now, since a_2 is uneven, a_4 must also be uneven.

Then, using the same analysis in the remaining equalities we see that both terms to the right hand side of the equals will be uneven in the equalities on even places (starting in 2a_3 = a_4 + a_2).
 Earlier, we already established that both terms to the right hand side of the equals on uneven places were both uneven, independent of the parity of a_2.
Therefore, all terms are uneven.
Therefore, a_i is uneven, for i = 1,2,...,k.

Now, that means that n is not relatively prime to even numbers, so n must be even.

Now, lets assume n has at least one uneven factor.

That means that at least one of the uneven integers from 1 to n won´t be relatively prime to n.

Then, assume that p and p+2 are both relatively prime to n.
Since both are uneven, that means that a_i = p  ,  a_i+1 = p + 2.

Now, assume m-2, m and m+4 are relatively prime to n, but m+2  is not relatively prime to n.
That means a_i-1 = m-2   ,    a_i = m   ,   a_i+1 = m + 4
By G(n):, we get:
                    2m = m + 4 + m - 2
                 Therefore 2m = 2m + 2, so 0 = 2, which is obviously false, so this is a condradiction.
Therefore, n doesn´t have uneven factors, so [b]n = 2^t[/b].

[u]Case 2[/u]: a_2 Even.
Since 2a_3 = a_4 + a_2,   a_4 + a_2 = 0 (mod 2). Since a_2 is even, a_4 is also even.

Using the same analysis, we see that both terms to the right side of the equal sign will be even, for the equalities on even places.
We already established that both terms to the right hand side of the equals on uneven spaces were both uneven, independent of the parity of a_2.

Since this terms are even, that means n is relatively prime to some even numbers, so n is uneven.

Therefore, n is relatively prime to 2.

Since a_1 = 1, that means a_2=2.
Therefore, solving in G(n), a_3= 3.
Replacing in the next equality we get a_4 = 4.
In general, we get a_i = i , for i = 1,2,3,...,k.

Therefore, n is relatively prime to 1,2,3,...,k.

[b]That means n is prime[/b], thus finishing the demonstration.

Could somebody please check my solution? Thanks!!! :)

I found a really simple solution.
Let $d=a_{i+1}-a_i$. Then we have $a_i=1_1+d(i-1)=1+d(i-1)$. Since $\gcd(n-1,n)=1$, so $n-1$ is the largest positive integer smaller than $n$ and co-prime to $n$. Thus, $a_k=n-1$ and $1+d(k-1)=n-1$ which shows $d(k-1)=n-2$ and so $d$ divides $n-2$. We make two cases.
[b]Case 1:[/b] $n$ is odd. Then $\gcd(n-2,n)=1$. So, every divisor of $n-2$ is co-prime to $n$ too, so is $d$. Thus, there is a $j$ such that $d=a_j=1+d(j-1)$ which implies $d|1$. Therefore, in this case, $d=1$ and we easily find that all numbers less than $n$ are co-prime to $n$, so $n$ must be a prime.
[b]Case 2:[/b] $n$ is even. But then $\gcd(n-2,n)=2$ and also $d=1$ is ruled out. So $d>1$. Say, $n=2l$. Then, $d$divides $2(l-1)$. If $d$ is odd, $d|l$ and $d|l-1$ forcing $d|1$, contradiction! So, $d=2s$ with $s|l-1$. If $l=2r+1$ for some $r$, then $\gcd(r,n)=1$. So, again $r=1+d(j-1$ for a $j$. If $s$ is odd, then $s|r$ and thus, $s|1$. Similarly, we find that actually $d|2$ must occur. We  finally have, $d=2$. But then all odd numbers less than $n$ are co-prime to $n$. So, $n$ does not have any odd factor i.e. $n=2^k$ for some $k\in\mathbb N$.

And learning $\text{\LaTeX}{}$ is really easy. You just need to learn some symbols and put your formulas between two dollar signs. For more, you can see here:
http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About

Here is another solution. If $n$ is odd, then $a_{2}=2$, and the conclusion follows. ($n$ has no common divisor with all integers less than it). Now assume $n$ is even.
$i)$ $a_{2}=3$ , in this case n is pairwisely coprime with all odd integers less than itself, so that it should be a power of $2$.

$ii)$ $a_{2} > 3$. Observe that $3|n$. In this case, let $p$ be the smallest prime number not dividing $n$. We claim that $p \equiv1\; (mod\; 3)$. Because if  $p \equiv2\; (mod\; 3)$, then $a_{3}=2p-1$ will be divisible by $3$, contradicting with the fact that $3|n$. Now, we have an arithmetic sequence with common difference $p-1$. So, $(k-1)(p-1)=n-2$. But if we check both sides of this equation in $mod 3$, we see that $3|(k-1)(p-1)$; however, $3 \nmid n-2$, contradiction.

Hence the problem is done.

[quote="grupyorum"][quote="AFdrolf"][quote="grupyorum"]Here is another solution. If $n$ is odd, then $a_{2}=2$, and the conclusion follows. ($n$ has no common divisor with all integers less than it). Now assume $n$ is even.
$i)$ $a_{2}=3$ , in this case n is pairwisely coprime with all odd integers less than itself, so that it should be a power of $2$.
[/quote]

The fact that a_2 = 3 doesn´t imply that it is comprime with all odd integers less than itself... Take 20 for example:
a_2=3, but it is not coprime with 5[/quote]

Actually, if $a_{2}=3$, then $a_{3}=5$, $a_{4}=7$, .. $a_{k}=n-1$ (remember $n$ is even). So, n is pairwisely coprime with all odd integers less than itself ;)

By the way, $20$ does not satisfy the conditions of the problem, so that it's wrong ;)[/quote]

Oh, sorry about that, I thought you meant that if ANY integer n is coprime with 3, n was coprime with all pairwise integers less than itslef.. I actually used a similar approach as you in my solution. Well thought solution that of yours!

Let $p$ be the smallest prime number such that $p$ doesn't divide $n$.It is trivial that $a_1=1$ , $a_2=p$ , $a_k=n-1$ so all the equalities are equal to $p-1$.Now for cases $p=2$ and $p=3$ we trivialy obtain that prime $n$ and every $n=2^k$ are solutions.Now we consider the case $p>3$.By summing the equalities we get $(p-1)(k-1)=n-2$. Now because $p>3$ we have $3|n$ so $3$ doesn't divide $n-2$ so we obtain $p \equiv 2 $ modulo $3$.That means $p-1 \equiv 1$ which immediately implies $3|a_3$ which is a contradiction so we are finished.

If $n$ is odd then $a_1 = 1, a_2 = 2 \rightarrow a_k = k$ for all $k$, so if $n$ is not prime then $n = pq$ for some $p, q > 1$, which means $p$ should not be the value of $a_i$; but it is, contradiction.

Suppose $n$ is even. If $n$ is not divisible by $3$ but has an odd factor greater than 1, then $a_1 = 1, a_2 = 3 \rightarrow a_k = 2k - 1$ and we are done in a similar manner to above. Thus assume $n$ is divisible by $3$. Because $n > 6$, it is clear that $\phi(n) > 2$ (by considering the prime factorization of the largest odd factor of $n$), so there exists $p \neq 1, n-1$ relatively prime to $n$. Thus $a_2 \neq n-1$. Because if $p$ is relatively prime to $n$ then $n - p$ is also, we have $a_2 < \frac{n}{2}$. Now, because $1$ and $n-1$ are values of some $a_k$, the common difference, $p-1$, must divide $(n-1) - 1 = n-2$, so since $3 | n$, we have $3$ relatively prime to $n-2$ and thus $p-1$. Therefore, $a_1, a_2, a_3$ form a complete set of residues modulo $3$, which means one of these numbers is not relatively prime to $3$ and thus $n$, contradiction. Thus $n$ is a power of $2$ if $n$ is even, completing the proof.

I'll use Bonse's Inequality: $\[p_{m+1}^2 < p_1p_2\cdots p_m\]$, for all $m \ge 4$, $p_1 = 2$. \\

Let $p$ denote the smallest prime which does not divide $n$. \\

Case 1) $p = 2$. This implies $n$ is a prime.
Case 2) $p = 3$. This implies $n = 2^k$ for some $k \in \mathbb{N}$. \\
Case 3) $p = 5$. Since $n > 6$, we have that $n > 9$. Therefore $a_3 = 9$, but $\gcd (n, 9) > 1$.
Case 4) $p \ge 7$. Write $n = \[p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}\]$. We have $(p - 2)^2 < p_1p_2\cdots p_k \le n = \[p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}\]$. Hence $(p-2)^2 = a_i$, a contradiction since $\gcd(p-2, n) > 1$.

I think I have carefully avoided advanced NT, so here is my solution:

Clearly, $a_1 = 1$. Now let $p$ be the smallest prime number which is not a divisor of $n$. Hence, $a_2 = p$, and the common difference $d = p-1$. Observe that since $a_k = n-1$, we have that \[ d \mid a_k - a_1 \], or \[ p - 1 \mid a_k - a_1 = n-2 \]. This means that all prime factors of $p-1$ are prime factors of $n-2$. However, due to the minimality of $p$, all primes factors of $p-1$ are prime factors of $n$. By the Euclidean algorithm, the only prime factors of $p-1$ are $2$, so $p = 2^{2^k} + 1$ or $p = 2$, by the theory of Fermat primes. We will now do casework on $p$.

Case 1: $p = 2$
This case is relatively easy. $a_2 = 2$, so all numbers less than $n$ must be relatively prime to $n$. However, if $n = pq$, where $p > 1$ and $q > 1$, observe that $p$ is not relatively prime to $n$, so $n$ must be prime for this case to work.

Case 2: $k = 0$ and $p = 3$.
This case is somewhat tricky. Observe that $n$ is even. Since $a_2 = 3$, we have that all numbers less than $n$ which are odd are relatively prime to $n$. However, if $n$ has a odd divisor $p > 1$, clearly $p$ is not relatively prime to $n$, so $n$ must be a power of two.

Case 3: $k \ge 1$.
Note that $3$ must be a divisor of $n$, so $a_k$ cannot be a multiple of $3$ for any $k$. I will contradict this statement. First, I claim that $k = \phi(n) > 2$ for all integers $n > 6$. $\phi{n} = 1$ is trivial, so let me show that $\phi(n) = 2$ has only solutions less than or equal to $6$. Since $\phi$ is a multiplicative function, observe that $n$ cannot be a multiple of a prime greater than $3$, because for such primes $p$ we have that $\phi(p^k) = (p-1)p^{k-1} \ge p-1 > 2$, which is contradiction. Hence, $n$ is just composed of factors of $2$ and $3$. If it divides both, and is of the form $2^m 3^n$ for $m$ and $n$ positive, we get that we need $2^{m} 3^{n-1} = 2$, so $n = 1$ and $m = 1$ in this case. If it is of the form $n = 2^k$, we get $k = 2$. And if it is of the form $3^k$, we get $k = 1$. $3$, $4$, and $6$ are all less than or equal to $6$, so $\phi(n) > 2$. Now we're almost done. Since $k > 2$, the number $a_3 = 2p-1$ must be in this sequence. However, $a_3 = 2p-1$ is of the form $2^{2^n+1} + 1$, which is a multiple of $3$, and we have contradiction, so we are done.

Here is my solution.

Clearly that $a_1=1$. Let $a_{i+1}-a_i = d \; (1 \le i \le \varphi(n)-1)$ then $$a_{i}=(i-1)d+1 \; \; (1 \le i \le \varphi(n)). \qquad (1)$$
We can see that if $n$ is a prime then $n$ satisfies the condition. We consider the case when $n$ is a composite number. Let $p$ be the smallest prime divisor of $n$ then $p \le \sqrt{n}$. Note that $\varphi (n) > \sqrt{n} \ge p$ for all prime $n>6$  since $p_i^{k_i}-p_i^{k_i-1} \ge p_i^{k_i/2}$ for all $p \ge 3$. Therefore, if $p \nmid d$ then there will exists $1 \le i \le \varphi(n)$ such that $(i-1)d \equiv -1 \pmod{p}$ or $p \mid a_i$, a contradiction since $\gcd (a_i,n)=1$. Thus, $p \mid d$. We consider two cases:

If $d \ge 2p$ then $p+1 \ne a_i \; (1 \le i \le \varphi(n))$. Hence, $\gcd (p+1,n)>1$. This means there exists a prime $q>p \ge 2$ such that $q \mid p+1$. Since $q>p$ so we get $q=p+1$ or $p=2,q=3$, a contradiction since $n>6$. Thus, $d <2p$ or $d=p$.

If $p=d=2$ then all odd numbers $a_i \le n$ are relatively prime to $n$. This can only happen when $n=2^x \; (x \in \mathbb{N})$.

If $p=d >3$ then $p+3 \ne a_i \; \forall 1 \le i \le \varphi(n)$ since $a_i \equiv 1 \pmod{p} \; \forall 1 \le i \le \varphi(n)$. Thus, $\gcd (p+3,n)>1$ or there exist a prime $q>p>3$ such that $q \mid p+3$. Note that $2 \mid p+3$ so $q \le \tfrac{p+3}{2}<p$, a contradiction.

Thus, $n$ must be either a prime number or a power of $2$.    $\blacksquare$

Let $ \,n > 6\,$ be an integer and $ \,a_{1},a_{2},\cdots ,a_{k}\,$ $(k\geq5)$  be all the natural numbers less than $ n$ and relatively prime to $ n$ . If $k$ is even ,and $a_{2} \minus{} a_{1} \equal{} a_{4} \minus{} a_{3} \equal{}a_{6} \minus{} a_{5}  \cdots \equal{} a_{k} \minus{} a_{k \minus{} 1} ,$ find   all such positive integers $n$.

[quote=Rewritten Problem Statement]Let $n\in\mathbb Z_{\geq 2}$ be an integer and $ \,a_{1},a_{2},\cdots ,a_{k}\,$  be all the natural numbers less than $ n$ and relatively prime to $ n$. If
\[ a_{2} \minus{} 1 \equal{} a_{3} \minus{} a_{2} \equal{} \cdots \equal{} \left(n-1\right) \minus{} a_{\varphi(n) \minus{} 1}=\Delta> 0,
\]
prove that $ \,n\,$ must be either a prime number or a power of $ \,2$.[/quote]
It is clear that we have $\Delta\mid n-2$. Hence $\gcd(n,\Delta)\in\{1,2\}$. Let $n=2^{\alpha}p^{\beta}\prod q_i^{\gamma_i}$ where $p$ is the lowest odd prime divisor of $n$. If $p$ does not exist then we necessarily have $n=2^{\alpha}$, a case that is easily checked to give $\Delta=1$. If $n\not \in\left\{p,2p\right\}$ then we have $\varphi(n)\geq p$, meaning that our arithmetic sequence contains a multiple of $p$, a clear contradiction, since $p\nmid \Delta$. The case $n=p$ is also easily checked, giving $\Delta=2$. For $n=2p$, if $p\geq 5$ we have $\Delta=2$ by considering $a_1=1,a_2=3$, but then $p$ is on the arithmetic sequence which is ludicrous. Note that $n=6$ also works. We conclude that the solutions are $\boxed{n\in\left\{p,6,2^{\alpha}\right\}}$

[quote=shinichiman][quote=sqing][url=http://www.artofproblemsolving.com/community/c6h1343797p7308398]China Mathematical Olympiad 2017 Q5[/url][/quote]

Hi sqing, what is your reason behind linking this topic to Q5 China MO 2017 ? I don't see any connection between these two problems.[/quote]
I judged wrong .
Sorry .

Let $p$ be the smallest prime that does not divide $n$. Let $p \ge 5$ (the other cases were treated in the other solutions). $a_{1}=1$, $a_{2}=p$, so $a_{k}-a_{k-1}=p-1$. We easily get by induction that $a_{i}=(i-1)p - i + 2$, so $a_{p-2}=p(p-3) - p+ 4=(p-2)^2$, so $(n,p-2)=1$. This contradicts the minimality of $p$.

[quote=pavel kozlov]There is simple solution without using Bertrand's Postulate...

We have a[1]=1, a[2]=p, where p is the least prime, which is not a divisor of n; a[k]=n-1, r=p-1 is the difference of the progression a[i].
   If n is odd, then a[2]=2  :arrow:  1, 2, ..., n-1 is our progression  :arrow:  n is a prime.
   If i is even, then p>2.
   1.  If p=3, then 1, 3, ..., n-1 is our progression  :arrow:  n is a power of 2.
   2.  If p>3, then n is divisible by 3. we have n-1=a[k]=1+(p-1)(k-1)  :arrow:  p-1 is divisor of n-2.  :idea:  Let q be a random prime divisor of p-1. q | p-1 | n-2, but q|n, because q<p  :arrow:  q|2 and q=2. It means p-1=2^i  :arrow:  p=2^i + 1. So as p is prime >3,  i  is even.
   But a[3]=2p-1=2^(i+1) + 1 is divisible by 3, but n is divisible by 3 too  :arrow:  gcd(n, a[3])>1. This contradiction ends our proof.  :lol:[/quote]

This is exactly what I did.

Let the smallest prime not dividing $n$ be $p$. Then the sequence has to be  $1, p, 2p-1, 3p-2, 4p-3, \dots, n-1$.

Since the common difference in this arithmetic sequence is $p-1$, we have that $p-1$ must divide $(n-1)-1=n-2$. But by the assumption of $p$ being the smallest prime, we also have that $p-1$ divides $n$. Therefore, $p-1$ divides $n-(n-2)=2$.

So $p$ is either $2$ or $3$. If $p$ is $2$, then the sequence is $1, 2, 3, \dots. n-1$, so $n$ is prime. If $p$ is $3$, then the sequence is $1,3,5,\dots n-1$ so $n$ is relatively prime to all odd numbers less than it and must be a power of $2$.

I solved the problem with the condition that $n\in \mathbb{N}.$

Obviously $n=1$ is a solution$,$ as well as $n=2, 3, 4,$ so assume that $n\geq 5.$ Let the common difference be $d.$ The condition implies $$a_{\varphi(n)}=a_1+d(\varphi(n)-1)\iff n-2=(\varphi(n)-1)d\hspace{1cm}(*)$$ as it's easy to notice that $a_{\varphi(n)}=n-1.$

If $d=1,$ we get that $n$ is prime$,$ which is easily seen to work$.$
If $d=2,$ we get that $n$ is a power of $2,$ as $$2=\frac{n}{\varphi(n)}=\prod_{p\mid n}\bigg(\frac{p}{p-1}\bigg)\geq 2$$ with equality only when $\text{rad}(n)=2.$ It is again not that hard to verify that all such $n$ work$.$
Suppose $d\geq 3$ and notice that $\varphi(n)$ is even for $n>2,$ so if $d$ is odd$,$ by equation $(*)$ we obtain $n$ must also be odd$,$ but then $\gcd(2, n)=1,$ meaning $a_2=2$ and $d=1,$ contradiction$.$ Therefore $d\geq 4$ and $n$ is even$.$
Let $p$ be the smallest prime not dividing $n.$ Now all the numbers $2, \cdots , p-1$ are not relatively prime to $n$ and so $a_2=p \Rightarrow d=p-1.$

Suppose that $r$ is a prime factor of $p-1.$ Because $p$ is minimal$,$ $r$ must divide $n.$ By equation $(*)$ we see that $r\mid 2,$ therefore $p-1=2^k$ for some $k\geq 2$ (recall that $4\leq d=p-1$)$.$ But since $p$ is a prime$,$ $k$ cannot have any odd factor$,$ from which we get $k=2^t$ and $p=1+2^{2^t}.$

Finally$,$ suppose $\varphi(n)\geq 3$ so that $a_3=2p-1$ exists$.$ Note that $p\equiv 2\pmod 3 ,$ meaning $a_3\equiv 0\pmod 3$ and as $p>3,$ we see that $3$ also divides $n,$ a clear contradiction$.$ This means that $\varphi(n)\leq 2$ and the only possibility is $n=6,$ which is also easily verified$.$

Firstly $n \text{ odd} \iff n \text{ prime }$:
$2 \nmid n \implies a_2=2 \implies a_i \text{ are } 1, 2, 3, \ldots \implies n \text{ prime}$
So, let $n$ even:
Let $p$ be the smallest prime with $p \nmid n$
Case 1: $p=3$
$\implies a_i \text{ are } 1, 3, 5, \ldots \implies n \text{ is a power of 2 }$
Case 2: $p>3 \implies 3 \mid n$
$\implies$ none of $m \in \{ 2, 3, \ldots, p-1 \}$ can be an $a_i$ else: $\gcd(m, n) = 1 \implies$
for any prime $q \mid m \text{ we must have } q \nmid n \text{ but } q \leq m <p$ which is a contradiction.
So, $a_2=p \implies a_r=1+(p-1)(r-1) \implies p \equiv 1 \pmod{3} ( \text{ else, } p \equiv 2 \pmod{3} \implies 3 \mid a_{r \equiv 0 \pmod{3}} ) \implies p-1 \equiv 0 \pmod{3}$ which is a contradiction. $\blacksquare$
Q.E.D

No idea if this is correct :clown:
First note that the problem obviously works for primes. So henceforth assume $n$ is composite.
[b]Case 1[/b]. $2 \nmid n$
Let $n=p_1^{a_1}p_2^{a_2} \dots p_k^{a_k}, p_i \ne 2 \forall i \in \{1,2, \dots, k\}$. Note that $\frac{p_1p_2 \dots p_k-1}{2}, \frac{p_1p_2 \dots p_k+1}{2}$ are two consecutive naturals, less than $n$ and coprime to each $p_i$, hence coprime to $n$. Therefore we deduce that if this were a part of an AP, all $p_i$s sure to occur in the sequence, but they not coprime to $n$. This is a contradiction.
[b]Case 2.[/b] $2 \mid n$
Let $n= 2^{a_0}p_1^{a_1}p_2^{a_2} \dots p_k^{a_k}, p_i \ne 2 \forall i \in \{1,2, \dots, k\}, a_0 \geq 1$. Note that $p_1p_2\dots p_k+2, p_1p_2\dots p_k+4$ are two [b]odd[/b] natural numbers, which are coprime to each of $p_1,p_2,\dots p_k$, hence coprime to $n$. Also, note that $n \geq 2p_1p_2\dots p_k \geq p_1p_2\dots p_k+4 \iff p_1p_2 \dots p_k \geq 4$ which is true since $n>6$. (The only case where it is violated is where $p_1=3$ so $n = 6$). Therefore the common difference of the AP at maximum is $2$, so it would cover all odd primes in the factorisation, a contradiction. The only case where it does not, is when there is no odd prime in the prime factorisation ! This means that $n$ is a power of $2$ which is easily seen to work $\blacksquare$

Does this work?

The condition says that the positive integers less than and relatively prime to $n$ form an arithmetic sequence.

Case 1: $n$ is odd. Then $1$ and $2$ are relatively prime to $n$, so the common difference of the arithmetic sequence is $1$. But since $n-1$ is in the sequence, this means all of the positive integers less than $n$ are relatively prime to $n$, implying $n$ is prime.

Case 2: $4 | n$. Then $\frac{n}{2}-1$ and $\frac{n}{2}+1$ are relatively prime to $n$, so the common difference of the arithmetic sequence is $2$. But since $n-1$ is in the sequence, this means all the odd positive integers less than $n$ are relatively prime to $n$, so since $n$ is even $n$ must be a power of $2$.

Case 3: $2 || n$. Then $\frac{n}{2}-2$ and $\frac{n}{2}+2$ are relatively prime to $n$, so the common difference of the arithmetic sequence is $4$.  Since the sequence contains $1$ and $n-1$, it contains all the positive integers less $n$ which are $1$ mod $4$. In particular, $3 | n$. But if $n = 9$, $7$ is not in the sequence but relatively prime to $n$, a contradiction. If $n > 9$, $9$ is in the sequence but not relatively prime, also a contradiction.

[hide=solution(MONT)][quote=IMO 1991#2]Let $ \,n > 6\,$ be an integer and $ \,a_{1},a_{2},\cdots ,a_{k}\,$  be all the natural numbers less than $ n$ and relatively prime to $ n$. If
\[ a_{2} \minus{} a_{1} \equal{} a_{3} \minus{} a_{2} \equal{} \cdots \equal{} a_{k} \minus{} a_{k \minus{} 1} > 0,
\]
prove that $ \,n\,$ must be either a prime number or a power of $ \,2$.[/quote]
Clearly $a_1=1,a_k=n-1$.
Now the question asks for an AP of coprime integers.
Say the common difference be $d$.
$1+(k-1)d=n-1 \implies d=\frac{n-2}{k-1}$.
Now say $p$ be a prime factor of $n$ and assume $p<k$.
Now $(p,cd+1)=1$ for all $c$ s.t $0<c<k$.
Since $p<k$ there exist $c_1,c_2$ such that,
$c_1d+1 \equiv c_2d+1\pmod p$.
Since we are working modulo $p$,we can safely assume,
$p\not |c_1-c_2 \implies d \equiv 0 \pmod p$.
This gives,
 $\frac{n-2}{k-1} \equiv \frac{-2}{k-1}\equiv 0 \pmod p \implies p|2$
Or $n=2^a$.
For $p>k,k=p-1$ or $n$ is a prime.[/hide]

My soln not seen above solutions yet  when n is a prime it is trivial, now let n be prime power then it's easy to see n is power of 2, So now let n be non prime power composite number , then let it,s smallest divisor greater than 1 be p , then 1,2...p-1 are co prime to n so common difference would be 1,we can easily check p =2 gives the previous soln
[color=#f00]Claim 1:[/color] : No such n is there
Proof : $2-1...= a_p-(p-1) $
But this mean a_p is p and is coprime to n CONTRADICTION!!!
Hence the claim and result.

It is easy to check that primes and powers of $2$ satisfy the condition. Also, the condition readily gives $a_1<a_2<\ldots<a_k$. Assume that $n$ is not prime and odd. Hence, $a_2=2$ and $a_2-a_1=2-1=1$. As $n>6$, $n-1=a_k\neq a_2=2$, and to keep the condition satisfied we must have $\gcd(n,k)=1$ for all $1\le k<n$, implying that $n$ is prime, violating our assumption.

	Now assume that $n$ is not prime even, and not a power of $2$. Let $p$ be the smallest prime not dividing $n$. Now we must have $a_2=p$, because if there was otherwise another number $q>1$ coprime to $n$ smaller then $p$ then $q$ would have a prime factor $r$ smaller than $p$, which must in turn be also relatively prime to $n$, which is non-sense.  Hence, in order for the condition to be fulfilled, we must have
	$$a_1=1,a_2=p,a_3=2p-1,a_4=3p-2,\ldots,a_k=n-1$$
	Hence, $p-1|(n-1)-1$. But clearly $p-1|n$, so $p-1|2$, i.e. $p=2$ or $p=3$. In these cases, we get that $n$ must be prime or a power of $2$, respectively, so we are done.

Really good problem!
We do casework on $a_2$. If $a_2=2,3$, we can easily obtain $n$ is a prime or a power of two, respectively. Now let $a_2=p>3$, so that we obtain $a_3=2p-1$ (the difference is apparently $p-1$).
Note that by summing, $(k-1)(p-1)=n-2$. Now we finish by modulo $3$: if $p \equiv 2 \pmod 3$, then $3|2p-1$, but $3|n$, absurd; otherwise $3|n-2$, but we have again $3|n$ due to the fact that $a_2>3$, contradiction.

Note that there are $\phi(n)$ terms, with the first term being $1$ and the last term being $n-1$, so the common difference is $$d=\frac{n-2}{\phi(n)-1}.$$ Now, consider any prime $p$ dividing $n$.\\

If $p>2$, then $d$ is relatively prime to $p$ since $p$ does not divide $n-2$ if it divides $n$. However, if $d$ is relatively prime to $p$, there exists some $0\leq s\leq p-1$ for which $$1+ds\equiv 0\pmod{p}.$$ However, since we need the entire arithmetic progression to be relatively prime to $n$ (and thus to $p$), we must have $$\phi(n)-1\leq p-2\rightarrow \phi(n)\leq p-1.$$ Since $\phi(p)$ is already $p-1$, and $\phi$ is multiplicative, we also have $\phi(n)\geq p-1$, with equality at $n=p$ and $n=2p.$ We can rule out $n=2p$ since then $$d=\frac{2p-2}{p-2}=2+\frac{2}{p-2},$$ which is only an integer for $p=3$ but we require $n>6$, so this case is resolved.\\

Otherwise, there is no prime $p>2$ dividing $n$, so $n$ is a power of 2 and we are done.

$a_2-a_1=a_3-a_2=...a_k-a_{k-1}>0 \implies a_1<a_2.....<a_k \implies (a_2-1)(k-1)+2=n$\\
For odd $n, (a_2-1)$ has to be odd and since $a_2$ is the least prime the doesnt divide $n$ (trivial),$a_2=2 \implies $ every positive integer less than $n$ is coprime to $n \implies $ $n$ is a odd prime.\\
If $n$ even, write $n=2^{\alpha_1}.{p_2}^{\alpha_2}......{p_j}^{\alpha_j}.$
Now, $(p_c-1)(k-1)+2==2^{\alpha_1}.{p_2}^{\alpha_2}......{p_j}^{\alpha_j}$ where $p_c=a_2$ ; taking $p_c-1=2f$, we get $f(k-1)+1=2^{\alpha_1-1}.{p_2}^{\alpha_2}......{p_j}^{\alpha_j} \cdots (i)$.\\
Note that $a_2=3$ makes $n$ a power of $2$. So suppose $a_2>3$. Now take the least prime $p_i$, from the canonical factorization of $n$, such that $p_i$ and $p_{i+1}$ has at least one prime between them. The least of these 'in-between' primes $=p_c$. (If theres no prime between any $p_i$ and $p_{i+1}$ for $1 \leq i \leq k-1$ then the same argument we give below holds because then there would be no primebetween $p_k$ and $p_c$.)Thus the prime factors of $p_c-1$ will also be the factor of $n$. Now if any prime besides $2$ divides $p_c-1$ we get a contradiction in $(i) \implies p_c-1=2^z$. Now as $z$ has to be odd since otherwise $3|2^z-1=p_c$. But then$3|a_3=2^{z+1}-1$ which is a contradiction since $(3,n)=3$. So we are done.
The existence of $a_3$ is assured by the fact that for all $n>6, k \geq4$ by the formula for eulers totient function.

Forgot that I did this problem a year ago already so here is another solution (which I like but which is more complicated):

If $n$ is odd then $a_{i+1}-a_i=1$, so in fact $a_i=i$ as $a_1=1$. Thus $n$ is relatively prime to all numbers smaller than $n$, so $n$ is prime. Otherwise $n$ is even. If now assume that $n$ has at least one odd prime divisor, and let $q$ be the smallest. Also, let the smallest odd prime not dividing $n$ be $p$, so that $a_1=1, a_2=p, a_3=2p-1, a_4=3p-2,\ldots$ so that in general, $a_i=(i-1)p-(i-2)\iff a_i=1+(i-1)(p-1)$.

	The idea is now to seek a contradiction by showing that some $a_i$ is  $0\mod q$ at least once. Consider $i-1\in I=\{0,1,\ldots,q-1\}$. We now notice that in total, there are $\phi(n)$ numbers $a_k$ relatively prime to $n$, but as $\phi(n)\ge\sqrt{n}\ge q$ for $n>6$ the, index set $I$ delivers $a_i$ that are all relatively prime to $n$. However note that as $\gcd(p-1,q)=1$, the set $(p-1)I+1$ is still a complete residue system modulo $q$, so there exists one $i\in I$ such that $a_i\equiv 0\mod q$, a contradiction.


Let a,b,c>0 such that   a+b+c=1                                Prove that
$ 1\leq\sum\frac{(a\plus{}b)b}{c\plus{}a}$

[quote="tranquoc"]Let $ a,b,c>0$ such that   $ a\plus{}b\plus{}c\equal{}1$                                Prove that
$ 1\leq\sum\frac {(a \plus{} b)b}{c \plus{} a}$[/quote]
$ 1\leq\sum_{cyc}\frac {(a \plus{} b)b}{c \plus{} a}\Leftrightarrow\sum_{cyc}(a^4\minus{}a^2b^2\plus{}a^3c\minus{}a^2bc)\geq0,$ which obviously true.

$ \sum\frac{b(a\plus{}b)}{c\plus{}a} \geq \sum a$
<-> $ \sum\frac{a(a\plus{}b\plus{}c)\plus{}b(a\plus{}b\plus{}c\minus{}a\minus{}c)}{c\plus{}a} \geq 2\sum a$
<->$ \sum\frac{a\plus{}b}{a\plus{}c} \geq 3$
done :rotfl:


propozitia matematica "80 este mai mare sau egal cu 70" este adevarata?

:D $ 80 \ge 70$ este adevarat din cate stiu eu


projective geometry
geometry proposed
geometry

Let ABC be a triangle. The A-excircle (Ia) touches BC at D. AD meets the excircle (Ia) at D and K. E,F are on AK such that CF=CD, BE=BD. Show that F lies on the polar of E with respect to the excircle (Ia).

It is enough to prove that the point $ F$ is the harmonic conjugate of $ E,$ with respect to the points $ D,\ K.$

From the similar isosceles triangles $ \bigtriangleup BDE$ and $ \bigtriangleup CDF,$ we have that $ \frac {ED}{FD} \equal{} \frac {BE}{CF}$ $ ,(1)$

Let be the point $ L\equiv BE\cap CK$ and based on the below [b][size=100]Lemma[/size][/b], we conclude that $ BE \equal{} EL$ $ ,(2)$

So, because of $ (2)$ and $ CF\parallel EL,$ we have that $ \frac {EL}{CF} \equal{} \frac {EK}{FK}$ $ \Longrightarrow$ $ \frac {BE}{CF} \equal{} \frac {EK}{FK}$ $ ,(3)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {ED}{FD} \equal{} \frac {EK}{FK}$ $ \Longrightarrow$ $ \frac {ED}{EK} \equal{} \frac {FD}{FK}$ $ ,(4)$

Hence, from $ (4)$ we conclude that the point $ F,$ as the harmonic conjugate of $ E,$ with respect to the points $ D,\ K,$ lies on the polar of $ E$ with respect to the $ A$-excircle $ (I_{a})$ of $ \bigtriangleup ABC$ and the proof is completed.

[b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ D$ be, the tangency point of the $ A$-excircle $ (I_{a}),$ to its side-segments $ BC.$The line segment $ AD$ intersects $ (I_{a})$ at one point so be it $ K$ and let be the point $ L\equiv BE\cap CK,$ where $ E$ is the point on $ AD,$ such that $ BE \equal{} BD.$ Prove that $ BE \equal{} EL.$[/color][/size][/b]

Kostas Vittas.

PS. I will post here later the proof of the above [b][size=100]Lemma[/size][/b], I have in mind.

[quote="vittasko"]
[b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ D$ be, the tangency point of the $ A$-excircle $ (I_{a}),$ to its side-segments $ BC.$The line segment $ AD$ intersects $ (I_{a})$ at one point so be it $ K$ and let be the point $ L\equiv BE\cap CK,$ where $ E$ is the point on $ AD,$ such that $ BE \equal{} BD.$ Prove that $ BE \equal{} EL.$[/color][/size][/b][/quote]
[b][size=100]PROOF.[/size][/b] - Let $ M,\ N$ be, the tangency points of $ (I_{a})$ to the sidelines $ AB,\ AC$ respectively and we denote the point $ Q\equiv BC\cap MN.$

Based on the [b][size=100]Ceva theorem[/size][/b], we conclude that $ P\equiv AD\cap BN\cap CM$ and so, fron the complete quadrilateral $ BCNMAQ,$
we have that the points $ B,\ D,\ C,\ Q,$ are in harmonic conjugation.

Hence, the pencil $ K.BDCQ,$ is also in harmonic conjugation and so, it is enough to prove that $ BL\parallel KQ.$

$ \bullet$ The line segment $ MN,$ is the polar of $ Q$ with respect to $ (I_{a})$ and because it passes through the point $ Q,$
we conclude that the polar of $ Q$ wrt $ (I_{a}),$ passes through the point $ A.$

So, the line segment $ AD,$ is the polar of $ Q$ with respect to $ (I_{a}),$ because it connects the point $ A$ with the point $ D,$
as the tangency point of $ (I_{a})$ with the line through the point $ Q.$

Hence, the line segment $ QK,$ tangents to $ (I_{a})$ at point $ K$ and then, we have $ I_{a}K\perp KQ$ and it is enough to prove that $ I_{a}K\perp BL.$

$ \bullet$ Let be the point $ T\equiv BL\cap I_{a}K$ and it is easy to prove that the quadrilateral $ BFKT,$ where $ F$ is the orthogonal projection of $ B$ on $ AD,$

is cyclic, because of $ \angle DKT \equal{} \angle KDI_{a} \equal{} 90^{o} \minus{} \angle KDC \equal{} 90^{o} \minus{} \angle FDB \equal{} \angle FBD \equal{} \angle FBE$ $ ,(1)$

So, we conclude that $ \angle BTK \equal{} \angle BFK \equal{} 90^{o}$ $ \Longrightarrow$ $ BL\perp I_{a}K$ $ ,(2)$

From $ (2),$ we conclude that $ BL\parallel KQ$ $ ,(3)$

From $ (3)$ and because of the pencil $ K.BDCQ$ is in harmonic conjugation, we conclude that $ BE \equal{} EL$ and the proof is completed.

Kostas Vittas.

The A-excircle $ (I_a)$ touches $ AB, AC$ at $ U, V.$ $ (B)$ is a circle with center $ B$ and radius $ BD \equal{} BE.$ $ (C)$ is a circle with radius $ CD \equal{} CF.$ $ (B), (C)$ are externally tangent at $ D,$ $ I_aD$ is their single common internal tangent.  Points $ A, D, E$ are collinear, their polars $ UV, BC, e$ WRT $ (I_a)$ are concurrent at $ P.$ By Menelaus theorem for the $ \triangle ABC$ cut by the line $ PUV,$ $ \frac {\overline{PB}}{\overline{PC}} \equal{} \frac {\overline{UB}}{\overline{UA}} \cdot \frac {\overline{VA}}{\overline{VC}} \equal{} \frac {\overline{UB}}{\overline{VC}} \equal{} \minus{}\frac {\overline{DB}}{\overline{DC}}$ $ \Longrightarrow$ $ P$ is the external similarity center of the circles $ (B), (C).$ Line $ I_aE$ cuts $ (B)$ again at $ X.$ Power of $ I_a$ to $ (B)$ is $ I_aD^2 \equal{} I_aE \cdot I_aX$ $ \Longrightarrow$ $ e \equiv PX \perp I_aX$ is the polar of $ E$ WRT $ (I_a),$ cutting $ (B)$ again at a point $ Y$ diametrically opposite point to $ E.$ From similarity of the isosceles $ \triangle DBE \sim \triangle DCF,$ $ BY \equiv EB \parallel CF$ $ \Longrightarrow$ $ FY$ goes through the external similarity center $ P$ of $ (B), (C),$ which means $ F \in PY \equiv PX \equiv e.$


A 7.00 kg object starting from rest falls through a viscous medium and experiences a resistive force  = -b , where  is the velocity of the object. The object reaches one half its terminal speed in 4.94 s.

How far has the object traveled in the first 4.94 s of motion?

[quote="kl2836"]...a resistive force = -b , where is the velocity of the object.[/quote]
I think you're missing something.  From Newton's second law, $ \sum\mathbf{F}\equal{}m\mathbf{a}$.  It looks like the two forces are $ \mathbf{g}$ and the viscous force, which at the moment, I can't quantify.

if acceleration is time varying, and i want to find the distance s= (1/2)at^2,

how do i write the integral form of s

The resistive force seems to be = -bv.
If so, castigioni's answer is OK, but to came to it you can't put $ \;s \equal{} (1/2)at^2\;$ since (you alone said it) [b]acceleration is time varying[/b].

Do you manage differential equation?

yes, you need a differential equation

[hide]
so you make the equation $ ma \equal{} mg \minus{} bv$
set $ a \equal{} \frac {dv}{dt}$

$ m\frac {dv}{dt} \equal{} mg \minus{} bv$
solve...

get to $ m*\frac {dv}{mg \minus{} bv} \equal{} dt$
integrate $ \minus{} \frac {1}{b}ln(mg \minus{} bv)*m \equal{} t$
then, solve and get $ v \equal{} \frac {m(g \minus{} e^{\frac { \minus{} bt}{m}})}{b}$
and then, you can integrate that from 0 to 4.94

[/hide]

wondering if there's any other way?

note:  $ \int{(s)}dt \equal{} v$

if you want me to, i can go into a little more detail

There may be another way, but I can't think of one at the moment.  The presence of the drag force $ bv(t)$ sets this up as a nice equation to use integrating factors on.


calculus
integration
logarithms
complex analysis
complex analysis unsolved

Hi, could anyone help me on these integrals please?

1. Verify the following:
$ \int_{0}^{\infty}\frac{\log{x}}{(1\plus{}x^2)^2}dx\equal{}\minus{}\frac{\pi}{4}$

2. Compute
$ I\equal{}P.V. \int_{\minus{}\infty}^{\infty}\frac{xe^{i{\lambda}x}}{x^2\minus{}b^2}dx$
where $ \lambda>0,b>0$.

Thanks in advance,

For the first one, note that the numerator is a multi-valued function.  So first, specify a branch cut.  The one running along the positive real axis is a good choice (why?).  Call the integral we want $ I$.  When we cross the branch, we get $ \minus{}2\pi i I$ (why?).  The poles occur at $ x\equal{}\pm i$, each with multiplicity 2.  So we need the residues.  After that,
$ \oint_{\Gamma}f(z)\ dz\equal{}(1\minus{}2\pi i)I\equal{}2\pi i(\text{Res}(i)\plus{}\text{Res}(\minus{}i))\Rightarrow I\equal{}\frac{2\pi i(\text{Res}(i)\plus{}\text{Res}(\minus{}i))}{1\minus{}2\pi i}$.
Now you need to calculate the residues, which I'll leave to you as an exercise.

No problem.  Now the second problem seems a bit simpler.  First note there are two poles: $ x\equal{}\pm b$ so we know where we need to take residues.  Construct a contour that runs along the real axis, with indentations around the two poles, since they both lie on the real axis.  Close the contour with an upper-semicircle.  Now we constructed our contour so that there are no poles inside so
$ \oint_{\Gamma}f(z)\ dz\equal{}0\equal{}I\plus{}\int_{S_{\delta_{1}^{\minus{}}}}f(z)\ dz\plus{}\int_{S_{\delta_{2}^{\minus{}}}}f(z)\ dz\plus{}\int_{C_{R}^{\plus{}}}f(z)\ dz$
The last integral is the upper-semicircle, which goes to 0 by Jordan's Lemma.  The second and third integral are given by $ \minus{}\pi i\text{Res}(f,\minus{}b)$ and $ \minus{}\pi i\text{Res}(f,b)$, respectively.  Now just take the residues and you're done.


Como ya sabran hoy se cumple la segunda semana de venir celebrando el [b]Reto de la Semana[/b].

Este conciste en una lista de problemas cuyas reglas son:
- No colgar las soluciones y/o comentarios en los respectivos topics durante la semana. Solo comentarlos luego de que se conlcuya el tiempo.

- Enviar sus soluciones a Email : jibarotimbalero@hotmail.com (manuel) , pascual2k1@hotmail.com (pascual)

- Duracion: 1 semana

- Luego de la semana, se publicaran las soluciones mas sobresalientes.

Asi que la dinamica es simple: Colgamos problemas, los resuelven, nos mandan las brillantes soluciones en cualquier formato (preferiblemente Latex, pero no es requisito), Las revisamos, Colgamos las mas lindas con elk respectivo nombre del autor y finalmente las discutimos.

Para que esto sea grande deben participar todos y enviar soluciones todos! Piensen que les sirve como un entrenamiento ya que nosotros hacemos las veces de jurado!

Suerte y espero participen.

Finalmente una encuesta. Espero sugerencias.

[b]que tal le parece la iniciativa? Alguna sugerencia? Aproveche y opine aqui:[/b]

hola!! :) Bueno a mi parecer esto que realizan esta muy bien pero hasta ahora no entiendo en que semana se esta...además el tipo de problemas que proponen en algunos casos son complicados( para mi son un poco dificiles ¡¡¡¡pero no imposibles!!! ;) ), no se si puedan insertar sus soluciones oficiales (es decir de los organizadores)eso es mi parecer...
Hasta otra oportunidad :P

lo que pasa es que estamos como en un receso durante vacaciones y mientras se acababa el año escolar, ya vamos a iniciar otra tempoorada de problemas esta semana. Acerca de las soluciones la idea es que los usuarios compartan las suyas, claro esta que si hay una que no han podido resolver entonces nos toca a los moderadores.

hola!!!  yo tambien participare en el reto de la semana

      por ahora no tengo ningun pregunta  :D ...
      pero  me parece muy buena la iniciativa

      fabiola-peru :ninja:


geometry
analytic geometry
trigonometry
search

In triangle ABC, BD is a median.  CF intersects BD at E so that BE=ED.  Point F is on AB.  THen if BF=5, what is BA?

Does anyone know how to do this?

THis was an old ASHME problem and the solution was a little sketchy..but taht was the problem, with all the information that I gave..

Note that if you stretch or skew the plane in any direction, if (generically) A is the midpoint of BC then A will always be the midpoint of BC. That won't change. There are certain constructions that will not change under stretch and skew transformations.

Notice that through just stretching and skewing, one can transform any triangle into any other triangle. Since the construction presented in this problem does not change with stretching and skewing, it applies to all triangles.

You can therefore pick whichever triangle is most convenient for you.

A good pick would be a 45-45-90 right triangle with the right angle at B or an equilateral triangle.

But I dont see how taht helps..I swear this problem is mistyped or something in my book

Using that method

[hide]Ok, so let's assume for a moment that the triangle is a 45-45-90 triangle formed by $A(4k,0), \; B(0,0), \; C(0,4k)$

Since $BD$ is a median, $D\equiv (2k,2k)$, and the line $\overline{BD}$ is the line that connects $(0,0)$ and $(2k,2k)$, namely $y=x$

$E$ is the midpoint of $\overline{BD}$, so $E \equiv (k,k)$

Since $\overline{CF}\equiv \overline{CE}$, $\overline{CF}$ is the line passing through $(0,4k)$ and $(k,k)$. This line is clearly $y=4k-3x$, and intercepts $\overline{AB}$ when $y=0 \implies 3x=4k \implies x=\frac{4}{3}k$. Thus, $F \equiv \left( \frac{4}{3}k , 0 \right)$ Hence, $BF=\frac{4}{3}k$ and $BA=4k=3\frac{4}{3}k = 3 BF = 3(5)=15$[/hide]

A more natural solution would involve mass points and/or using area ratios. You may consider looking for a good resource on mass points (also called barycentric coordinates) - I'm not the right person to ask for a clear explanation on that. Although there are plenty of other people on this forum that can.

[hide="solution, there is probably an easier one"]Using a 45-45-90 triangle with $\angle B$ as the vertex as TZF suggested, we must first note that since the triangle is isosceles the median from the vertex is the altitude and thus $\overline{CD}\perp\overline{AC}$. Now label $ED$ as $x$. Obviously, $EB=x$ as well. Since the median from the right angle is half the side it meets in a right triangle, $DC=2x$. Because $BDC$ is also a 45-45-90, $BC=2\sqrt{2}x$. Now, obviously, $\tan{DCE}=\frac{1}{2}$, $\tan{BCF}=\tan{BCE}=\frac{5}{2\sqrt{2}x}$. To combine these, we have that \begin{eqnarray*}\tan{BCD}&=&1\\ &=&\tan{(DCE+BCE)}\\ &=&\frac{\tan{DCE}+\tan{BCE}}{1-\tan{DCE}\tan{BCE}}\end{eqnarray*}
I don't feel like typing out all the simplification here to find $x$, so you can do that yourself and then multiply by $2\sqrt{2}$. I may have messed it up, but the answer I got was $\boxed{15}$.[/hide]

The problem has a unique solution...

Lets use mass points. [use the aops search if you don't know what that is] Let $w_{A}$ denote the weight at $A$. WLOG $w_{a}=1$, since $AM=MC$, $w_{C}=1$ and $w_{M}=1+1=2$. Since $BE=EM$, $w_{B}=2$. Then $w_{A}*AF=w_{B}*BF\implies AF=10$, so $AB=5+10=\boxed{15}$

[hide="umm yeaaa ok"]
Since $BE=EC,$ $[BEC]=[EDC]$ and since $AD=DC,$ $[AED]=[EDC]$

$\frac{BF}{AF}=\frac{[BEC]}{[AEC]}=\frac{[BEC]}{[AED]+[EDC]}=\frac{[BEC]}{2[BEC]}=\frac{1}{2}$

so
$AF=10$ and $BA=15$
[/hide]

[hide]Yay mass points!
Put a mass of $1$ on $A$ and $C$, so there's a mass $2$ on $D$ and thus $B$, so $2*BF=AF \implies AF=10 \implies AB=\boxed{15}$[/hide]

[quote="cincodemayo5590"][quote="Altheman"]Lets use mass points. [/quote]

Does anyone know a good introductory text for mass points? (I know what they are. :P )[/quote]

An introductory text?

I don't really think it warrants an entire book, it's just a clever trick derived from Menelaus...

Here's an explanation, though: [url]http://mathcircle.berkeley.edu/mpgeo/mpgeo.html[/url]


function
geometry
3D geometry
sphere
analytic geometry
integration
real analysis

$f: \mathbb{R} \to  \mathbb{R}$ a $C^2$ function .
Suppose there exists a solution $u$ to the equation $(E)$, where $u$ is a function fom the unit ball to $\mathbb{R}$:

$(E):$
1/$\Delta{u}+f(u)=0$ on the open unit ball
2/ $u(x)=0$ on the unit sphere
3/ $u(x) \geq 0$ everywhere on the unit ball

Does it follows that $u(r)$ is decreasing (decreases on every radius) ?

Do you mean $f(x)$, a $C^2$ function on the unit ball, or $f(u)$, a $C^2$ function on $\mathbb R$?

The answer is YES.  The problem seems to assume that the solution $u$ depends only on $|x| = r$, i.e. $u(x) = U(r)$. That's correct but nontrivial (afaik).

In polar coordinates the problem becomes $U''(r) + \frac{n-1}{r}U'(r) + f(U(r)) = 0$, with $n$ = space dimension. Also, $U'(0) = 0$ and $U(1) = 0$. It's clear that the solution is in $C^4$. We must show that $U'(r) \le 0$ for $0 < r < 1$.

Suppose not, then there exist $0 \le a < c < b \le 1$ such that $U(a) = U(b) < U(c)$. We may assume that $U'(a) = 0$, i.e. $U$ has a local  minimum at $a$.

Multiply the equation with $U'(r)$ and integrate from $r=a$ to $r=b$. Then
$\frac{1}{2}\left((U'(b))^2 - (U'(a))^2 \right) + \int_a^b \frac{n-1}{r}(U'(r))^2dr + F(U(b))-F(U(a)) = 0$
where $F'=f$. That is,
$\frac{1}{2}(U'(b))^2 + \int_a^b \frac{n-1}{r}(U'(r))^2 dr = 0$
implying $U' = 0$ on $[a,b]$. This contradicts the assumption that $U$ is non-constant on this interval and proves the assertion. :)


Linarite is a mineral containing copper and lead. Its chemical formula is $ PbCuSO_{4}(OH)_{2}$. This mineral is soluble in both nitric acid and sulfuric acids, but its non reactive in HCL. A student was given a sample and asked to dissolve it to produce a solution of lead, copper and sulfate ions.Which acid should the student use? and why?

[hide="Opinion"]Perhaps nitric acid should be more appropriate, since by using sulphuric acid the sulphate concentration could be high enough to precipitate some of the Pb2+ ions.[/hide]


calculus
integration
function
real analysis
real analysis unsolved

Here is a nice integral. It can be calculated exactly.

Let $x>0$. Calculate

$\int_{0}^{\infty}\frac{t^{x}}{(e^{t}+1)^{2}}dt=$

Very nice benorin. There is another approach of this integral involving Mellin's transform.

Let $f(t)=\frac{1}{e^{t}+1}$, for $t>0$.

Then $f'(t)=-f(t)+\frac{1}{(e^{t}+1)^{2}}$.

Now take the Mellin transform of the two functions and we get that:

$M[f,\alpha]=g(\alpha)\xi(\alpha)$ and

$M[f', \alpha]=-g(\alpha)\xi(\alpha)+M[\frac{1}{(e^{t}+1)^{2}},\alpha]$
where $g(x)=\Gamma(x)(1-2^{1-x})$.

The two Mellin transforms are related by the formula:

$M[f',\alpha]=-(\alpha-1)M[f,\alpha-1]$, provided that $t^{\alpha-1}f(t)$ vanishes at 0 and infinity.

It follows that:

$M[\frac{1}{(e^{t}+1)^{2}},\alpha]=g(\alpha)\xi(\alpha)-(\alpha-1)g(\alpha-1)\xi(\alpha-1)$.

Replacing $\alpha$ by $x+1$ we recover your result.


Hello !!

lets $ x,y,z >0$ and $ x\plus{}y\plus{}z\equal{}1$

prove that :

$ \frac{12}{xyz} \plus{} (\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z})^3 \geq (\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z})^2 \plus{}\frac{4}{xyz}(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z} )$

[quote="*210*"]Hello !!

lets $ x,y,z > 0$ and $ x \plus{} y \plus{} z \equal{} 1$

prove that :

$ \frac {12}{xyz} \plus{} (\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})^3 \geq (\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})^2 \plus{} \frac {4}{xyz}(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} )$[/quote]
Well,that's my inequality that I posted before in a moroccan math forum.

Here is my generalization:

If $ x,y,z,k > 0$ and $ x \plus{} y \plus{} z \equal{} k$ then:

$ \frac {9k \plus{} 3k^2}{xyz} \plus{} k.(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})^3 \geq (\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})^2 \plus{} \frac {4k^2}{xyz}.(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} )$

$ *$ The case $ k \equal{} 1$ gives the inequality you've posted.


function
algebra
functional equation
algebra unsolved

Find all functions $ \ f: R^\plus{}\rightarrow\ R^\plus{}$ such that

$ f(x \plus{} \frac{1}{y}) \plus{} f(y \plus{} \frac{1}{x}) \equal{} f(x \plus{} y \plus{} \frac{1}{x} \plus{} \frac{1}{y})$   for all $ x,y > 0$

[hide]$ f(x)\equal{}0$ is the obvious solution. The equation satisfies Cauchy's Functional Equation, so its continous solution is $ f(x)\equal{}xf(1)$[/hide]

[quote="basted"][hide]$ f(x) \equal{} 0$ is the obvious solution. The equation satisfies Cauchy's Functional Equation, so its continous solution is $ f(x) \equal{} xf(1)$[/hide][/quote]

It not satisfies Cauchy's Functional Equation because the equation not holds for every positive real numbers a ,b
that $ f(a\plus{}b)\equal{}f(a)\plus{}f(b)$

For any $ a,b>0$ with $ ab\ge4$, we can solve
$ x\plus{}\frac1y\equal{}a$ and $ y\plus{}\frac1x\equal{}b$.
Hence, we have
$ f(a\plus{}b)\equal{}f(a)\plus{}f(b)$ when $ ab\ge4,a>0,b>0$.
Now we assume $ x\ge2$. Let $ n\ge m>0$ be integers. Then we have
$ f(nx)\equal{}nf(x)\equal{}mf(\frac{nx}m)$.
In other words, $ f(rx)\equal{}rf(x)$ whenever $ r\ge1$ is rational and $ x\ge2$.
Now for any $ q>r$, we can find integer $ n$ such that $ n(q\minus{}r)>1$.
Then we have for $ x\ge 2$,
$ nf(qx)\minus{}nf(rx)\equal{}f(nqx)\minus{}f(nrx)\equal{}f(nqx\minus{}nrx)>0$, or $ f(qx)>f(rx)$.
Using $ r_n$ increasing converging to $ q$ and another $ r_m$ decreasing converging to $ q$, monotonicity will gives us $ f(qx)\equal{}qf(x)$ for any $ q\ge1$ and $ x\ge2$.
Hence, we have $ f(x)\equal{}\frac x2f(2)$ for $ x\ge2$.

For any $ x<2$, we have
$ f(x)\equal{}f(x\plus{}\frac4x)\minus{}f(\frac4x)\equal{}\frac{f(2)}2(x\plus{}\frac4x\minus{}\frac4x)\equal{}\frac x2f(2)$.
So we prove $ f(x)$ is linear. Note we do not need continuity condition.
Q.E.D.


What is $f^{-1}{(x)}$ given $f(x)=x+\lfloor{x} \rfloor$?

[hide]When x is an integer, this is trivial.  Inverse is y=x/2
When x is a positive non-integer, we have x+a=y where a is x's integer part as the function.  The inverse is x-a=y.
When x is a negative non-integer, we have x+a-1=y where a is x's integer part as the funtion.  The inverse is x+1-a=y.
Now, when x is an integer, basementfuntion(x)=x
When x is a positive non-integer, basementfunction(x)=int. part of x.
When x is a negative non-integer, basementfunction(x)=int. part of x-1.
Therefore, when y=x+basementfunction(x) is the function, x-basementfunction(y)=y is the inverse.[/hide]

But how can you write $f^{-1}{(x)}$ as some function of x explicitly?

[hide]For all y, y=x-basementfunction(x)/2.
This works because whatever that's added to the ys in the original equation, there was at most one non-integer and one integer.  This causes the non-integer part of x to be the same as y's corresponding part.  All I had to do was to isolate that part and remove the excess.  If we set y as a(Int. Part)+b(Not Int. Part) and basementfunction(y) as a for positives and integers in the original function, we add these to equal x.  We get 2a+b.  To get y back, we need to get a+b with only functions involving x, which is 2a+b.  Since basementfunction(x) is 2a and x is 2a+b, I can synthesize y=a+b if I divide by two.  But the basementfunction for negatives works not as the int. part but as the int. part-1.  In other words, when y is negative non-integer, a+b=y and basement function(y)=a-1.  Adding these up, we get 2a+b-1 as x but the basement function of this is 2a-2, so dividing by two works here also, yielding 2a+b-1-(a-1) or a+b=y.[/hide]


there are 3 Roses the white the red and the black rose
each is a symbol for a wizard clan
each has 2 or 3 wizards and the rest a rest warriors
it works like this each rose will secretly discuss and decide to try to kill a person on one of the other roses by use of a tournament. one rose challenges another and the remaining rose does a vote on who they want to win and pm me then i or someone i tell to post the event will tell who won and who died
signups are now. o and wizards are more powerful and can only be killed in a war of the roses or a special challenge.
[b]
PM ME IF YOU WANT TO JOIN I WILL GIVE MORE RULES AND WHICH ROSES PEOPLE ARE IN SOON.[/b]

BATTLES:

tournament challenge:
one rose puts a warrior up and challenges another rose then challenged rose places a warrior or wizard for battle. if it is a warrior remaining rose puts it to a vote to decide the winner, but if it is a wizard i will discuss with the wizard(s) in the remaining rose to decide.

special challenge: two warriors from a rose challenge a person from another. if that person is a wizard the wizard dies but if he isnt the team losesa warrior.

War of the roses:
one wizard goes out and declares war with another rose its kinda like a wizards duel except each wizard brings 2 warriors and there are 3 battles which the remaining rose and i decide the outcome of each one. the rose that wins 2 of them win the "war" and the mage and warrior survive but the other wizard does not, but the other warriors do. Can only be declared if team declaring have only 1 wizard.

more roses will be added as necessary to make the game intresting. i will announce the time when we have enough roses to create alliances of dragons

I especially like the lack of clear sentences. Also, explain later? No thanks.

Mathblitz, TALK IN COMPLETE SENTENCES.

Anyway, I join :P

[quote="eggylv999"][quote="Dojo"][quote="mathblitz"]ok ive showed whats happening if you want to join you can i might add sorcerers and enchanters to the lot. sorcerers make talismans that enhance winning chance and enchanters enchant warriors so they switch to enchanters rose.[/quote]

yeah... eh... no

Just use the first rules.
/in[/quote]
agree with dojo why did dojo make two posts[/quote]

It's when you click submit, and the internet lags, and you click submit again to try to  make it go faster.

Mathblitz, could you post everything you posted in the first post in COMPLETE SENTECES?

Thank you.  That is all.

Proper capitilization would be nice too. And punctuation. no one wants to read rules when they are written like this if they are like this you cant tell where they end and begin so it becomes confusing.

YEA SURE. There is one problem though. i can only edit once for some reason and im trying to contact Klebian about this

The deadline for editing posts is no more than 48 hours.  (It may be less, but I know it's definitely not more than that.)

my games tend to nt get past the sign up stages because not enough people join.
Do not revive this thread. I am working on a better game for less people.


Have the entry forms been processed yet? Because I faxed mine in like two weeks ago, and approval hasn't showed up yet. Or do we actually need to fax it at the same time in which we e-mail the solutions?

Well it said to mail the registration forms with the solutions but I'm not sure what to do with emailed solutions...

It says that mine has been received; I faxed mine in the same day I emailed my solutions in, though no confirmation of the solutions yet. Not fussed, though.

[quote="13375P34K43V312"]Have the entry forms been processed yet? Because I faxed mine in like two weeks ago, and approval hasn't showed up yet.[/quote]

Maybe you should resend your entry form.  I mailed mine weeks before sending in my solutions, and they processed it about five days later.


The quadratic equation $ x^2 \plus{} mx \plus{} n \equal{} 0$ has roots that are twice those of $ x^2 \plus{} px \plus{} m \equal{} 0$, and none of $ m,n,$ and $ p$ is zero. What is the value of $ n/p$?

$ \textbf{(A)}\ 1\qquad
\textbf{(B)}\ 2\qquad
\textbf{(C)}\ 4\qquad
\textbf{(D)}\ 8\qquad
\textbf{(E)}\ 16$

[hide="Attempt"]Roots of first equation be $2r,2s$, so $2r+2s=-m,4rs=n$. From second equation, roots are $r,s$, so $r+s=-p,rs=m$. Now,
\[ \frac{n}{p}=\frac{4rs}{-(r+s)}=\frac{8rs}{-2(r+s)}=\frac{8m}{m}=8 \]
My answer is [b]D[/b][/hide]

Quite a fun problem. My solution is what I would do during a contest, although it does make a foolish assumption at the start...

[hide = "Solution"]
Note from the start that I assume the roots are integers  :blush: Let the roots of the first equation be $2a$  and $2b$. Then the roots of the first equation are $a$ and $b$. Vieta's on both equations gives $ab=m, 4ab=n, a+b=-p, 2a+2b=-m$. Seeing a $-m$ and $m$, we add those two equations to get $2a+2b+ab=0$. SFFT on this gives $(a+2)(b+2)=4$. Note that $a+2$ and $b+2$ cannot be equal to $2$ since then that would make them zero, and the question wouldn't work out. Instead let $a+2=4$ and $b+2=1$. Then we get $a=2$ and $b=-1$. Since $n/p=4ab/(a+b)$ and $a=$2 and $b=-1$, we plug in the values to get a final answer of $8$, or [b]D[/b].

[hide]Let the roots of $x^2+px+m$ be $r_1,r_2$. Then $r_1+r_2=-p$, so $2r_1+2r_2=-m=-2p$. Similarly, $r_1r_2=m\Rightarrow 4r_1r_2=n=4m$. So $\frac np=\frac{4m}{m/2}=\boxed{8}$.[/hide]