if (a[size=75]n[/size])	is an arithmetical progression of naturals with a[size=75]n[/size]<a[size=75]n+1[/size] and m>0
prove that there is k>0 with a[size=75]k+1[/size],a[size=75]k+2[/size],.....,a[size=75]k+n[/size]
composite numbers.

Let the progression be $a_n = a + dn$ with $d \geq 1$.
There are some different approaches now:

1.: In every $k=d(m+1)-1$ consecutive numbers there are at least $m$ numbers from the progression. But $(k+1)!+2 , (k+1)!+3 , .... , (k+1)!+k, (k+1)!+k+1$ are all composite (and when $(k+1)!+2 <a$, we just take a larger $k$).

2.: Lets choose $m$ different primes $p_1,p_2,...,p_m$ not dividing $d$.
By the Chinese Remainder Theorem, we can find (note that $d$ is invertible $\mod p_i$)) an integer $x$ such that $a+d(x+i) \equiv 0 \mod p_i^2$ for all $i$. But this gives us that $a+d(x+i)$ is composite for $i=1,2,3,...,m$. And when again $x$ isn't big enough, we just take a larger $x$, for example $x+\prod p_i^2$.

[3. Some assymptotical stuff (ugly).]


Find a number between 100 and 999 such that
\\
1. the sum of the digits is 12,
\\
2. the hundreds digit equals the sum of the tens and units digit, and
\\
3. one greater than the tens digit equals the hundreds digit.

let's consider the first and second conditions
if the hundreds digit equals the sum of the tens and units digit, and the sum of the digits is $ 12$, then the hundreds digit must be $ 6$

now we consider the third condition
if one greater than the tens digit equals the hundreds digit, and the hundreds digit is $ 6$, then the tens digit must be $ 5$

now we go back to the first condition
we know that the sum of the digits is $ 12$, and the hundreds digit is $ 6$, and the tens digit is $ 5$, so the ones digit must be $ 1$

therefore, our number is $ \boxed{651}$


About computer solutions, what count as valid languages? Does PHP count as one?

I used PHP and just turned it into pseudo code for the actual submission.

How about... don't use the computer to solve things :D

Then you won't have to ask these questions :wink:

[quote="#H34N1"]Also, if the graders do not know the program I used, will they hand it on to someone who does know?[/quote]
I can write BASIC, FORTRAN, Pascal, and C. I can read Perl, Python, Java, and C++. And I am only a mathematician; some of the graders are computer experts instead. We are bound to have someone nearby who can read Lisp or Ruby.

If you write in APL (see http://en.wikipedia.org/wiki/APL_(programming_language) ), which has been described as a "write once, read never" language, then I will be upset. Likewise for assembly language. (I have had to interpret uncommented assembly language while maintaining someone else's code. It was brutal.)

Quite simply, the mathematical parts of computer code frequently mimic mathematical notation, so they can be read by someone who does not know the language. Besides, since you will include an explanation of why you think your computer program actually answers the question, we graders will have some insight into your algorithm beforehand.

Erin Schram
USAMTS grader

If we're including code in our solutions, would you prefer pseudocode or the actual Java code?

Thanks.

I'll just convert it into pseudocode then.

[quote="elephantmaster"]If we're including code in our solutions, would you prefer pseudocode or the actual Java code?[/quote]

If you have written operational Java code already, skip translating it into pseudocode and send the Java.

If you routinely write pseudocode as preparation for writing operational code, send the pseudocode.

Don't go to a lot of trouble writing code for us graders. We grade your mathematics, not your programming.

Erin Schram
USAMTS grader


function
logarithms
real analysis
real analysis unsolved

Can someone solve the following problem?
    Let $E$ the space of continuous real functions from $[0,1]$ to $R$ that vanish in 0 and verify $|f(x)-f(y)|\leq |x-y|^a$ with $a\leq 1$ a constant. Let $N(x)$ the minimal number of balls of radius $x$ necessary to cover this set (with respect to the uniform norm). What is an equivalent for $lnlnN(x)$ when $x$ is close to 0?

Here is a small
[hide="Hint"]
Such functions can be determined up to $x$ from their values on the grid with step $x^{1/a}$ and you have quite a lot of freedom in assigning the values on this grid. Also $\ln\ln$ is a very slowly changing function, so don't be afraid to make [i]very[/i] rough estimates for $N(x)$.
[/hide]


Let G be the centroid of triangle ABC and (I) be the incircle of ABC.
Prove that G is inside (I) iff $\sqrt{p-a}$ , $\sqrt{p-b}$ , $\sqrt{p-c}$ are three side of a triangle :(

i think you should use that the area of the triangle is $A=p\sqrt{(p-a)(p-b)(p-c)}$
i think,i'm not sure :blush:

\[IG^2 = r^2 + \frac{1}{36}[5(a^2+b^2+c^2)-6(ab+bc+ca)].\]
So $G$ is inside $(I)$ if and only if
\[5(a^2+b^2+c^2)-6(ab+bc+ca) < 0 \Leftrightarrow\]
\[4(x^2+y^2+z^2) - 8(xy+yz+zx) < 0 \Leftrightarrow\]
\[-4(\sqrt{x}+\sqrt{y}-\sqrt{z})(\sqrt{y}+\sqrt{z}-\sqrt{x})(\sqrt{z}+\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}+\sqrt{z}) < 0\]
Since it is impossible that exactly two of $\sqrt{x}+\sqrt{y}-\sqrt{z}$, $\sqrt{y}+\sqrt{z}-\sqrt{x}$ and $\sqrt{z}+\sqrt{x}-\sqrt{y}$ are negative, so the above statement is equivlent to:

$\sqrt{x}+\sqrt{y}-\sqrt{z}$, $\sqrt{y}+\sqrt{z}-\sqrt{x}$ and $\sqrt{z}+\sqrt{x}-\sqrt{y}$ are both positive, or $\sqrt{x}, \sqrt{y}, \sqrt{z}$ form the lengths of sides of a triangle.


group theory
abstract algebra
search
modular arithmetic
superior algebra
superior algebra unsolved

prove that maximum cardinal of commutative subgroup of $S_n$ is $\leq 3^{\frac{n}{3}}$.

Wow! Is it that small? All I could do is to find a quite weak bound $6^{\frac{n+1}{2}}$ but my proof is already difficult for this bound. Sam-n, do you have a solution for this assertion?

actually I found it in a book (that is available on net :) ).but i search for new approach so I will be delighted if you share your idea. :)

Sam-n can you tell me where I can find that book on the Internet? It is for free? :)

Let $G\le S_n$ be a transitive abelian group, and let $\mathcal O_1,\ldots,\mathcal O_k$ be the orbits under the action of $G$ on $\{1,2,\ldots,n\}$. For each $i\in\overline{1,k}$, let $G_i$ be the group of permutations on $\mathcal O_i$ obtained by restricting the elements of $G$ to $\mathcal O_i$. Also, for every $g\in G$ and every $i\in\overline{1,k}$, let $g_i$ be the restriction of $g$ to $\mathcal O_i$. Now, the map $g\mapsto(g_1,\ldots,g_k)$ is clearly a monomorphism, so $G$ is isomorphic to a subgroup of $\prod_{i=1}^kG_i$. This, in turn, means that $G$ has at most $\prod_{i=1}^k|G_i|$ elements.

Each $G_i$ is an abelian transitive subgroup of $S_{|\mathcal O_i|}$, so it has $|\mathcal O_i|$ elements, according to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=67482]this problem[/url]. The conclusion is that an upper bound (which is easily seen to be attainable) for the number of elements of an abelian subgroup of $S_n$ is the solution to the following problem: what is the maximum value of the product of several positive integers $x_1,\ldots,x_k$ which sum up to $n$? When $n\equiv 0\pmod 3$, this number is exactly $3^{\frac n3}$, when $n\equiv 1\pmod 3$ it's $4\cdot 3^{\frac{n-4}3}$, and if $n\equiv 2\pmod 3$, it's $2\cdot 3^{\frac{n-2}3}$ (I think :)).

[quote="grobber"]Let $G\le S_{n}$ be a transitive abelian group :)).[/quote]
Hi,grobber,
I think here you mean $G\le S_{n}$ be a abelian group.
Not transitive. :) It make me hard to read it because of your such a little mistake.(Because I am a bigenner :) )
Anyway,Thanks for your proof.


No to your second question.  There were no middle schoolers from IL last year.

Though they attempted to change the USAMO selection process this year..? I think?

I think most of the attempts to change the USAMO selection process have been to give more of an advantage to AMC 12 takers. I didn't see anything about middle schoolers in the huge official report on the AMC site, and anyway, the AMC people have said the process will be the same this year as before.

Ben is right; the process, if anything, will be tweaked to favor those taking the more comprehensive and difficult 12 exams over those who take the 10 exams, just because a person with a score of $ n$ on the AMC 12, regardless of age, is more qualified to take the USAMO than someone with the same score on the AMC 10.

[quote="gauss1181"]How many Illinois middle schoolers do you predict will make USAMO this year?

And if there are no middle school qualifiers in a state after USAMO index is considered, will the top middle schooler be able to participate?[/quote]

not you for sure but maybe next year

I'm joking

gauss will make it this year
I'll make aime next year
.

saszs, did you take the AMC 10/12 this year? If so, did you not score high enough on it for AIME qualification?

I made so many mistakes on my AMC10.
I am a retard in 6th grade and got a 87.


geometry
trigonometry
Asymptote
vector
function
area of a triangle
Heron\u0027s formula

In a circle of radius 42, two chords of length 78 intersect at a point whose distance from the center is 18.  The two chords divide the interior of the circle into four regions.  Two of these regions are bordered by segments of unequal lenghts, and the area of either of them can be expressed uniquley in the form $m\pi-n\sqrt{d},$ where $m, n,$ and $d$ are positive integers and $d$ is not divisible by the square of any prime number.  Find $m+n+d.$

[hide]

$AC = 78$ $OC = 42$
$OY \perp AC$
$Y$ is the midpoint of AC.. $\longrightarrow YC=39$
By Pythagoras, $OY=9\sqrt{3}$
Since $IO=18$, $IY=9$, and thus $\angle IOY = 30$deg and $\angle YIB = 60$deg
$AI=IB=30$

So what we need to find is: (area of sector BOC) - [BOC] + [IBC]
Using law of cosines, $BC = 42$, so BOC is equilateral
(area of sector BOC) - [BOC] = $\frac{1}{6}\pi (42^2) - \frac{42^2\sqrt{3}}{4}$

Using Heron's formula, [IBC]=$\sqrt{(60)(12)(30)(18)}=360\sqrt{3}$

Our total is: $294\pi-81\sqrt{3}$
$m+n+d=294+81+3=\boxed{378}$

[/hide]

[quote="minsoens"][hide]

$ AC \equal{} 78$ $ OC \equal{} 42$
$ OY \perp AC$
$ Y$ is the midpoint of AC.. $ \longrightarrow YC \equal{} 39$
By Pythagoras, $ OY \equal{} 9\sqrt {3}$
Since $ IO \equal{} 18$, $ IY \equal{} 9$, and thus $ \angle IOY \equal{} 30$deg and $ \angle YIB \equal{} 60$deg
$ AI \equal{} IB \equal{} 30$

So what we need to find is: (area of sector BOC) - [BOC] + [IBC]
Using law of cosines, $ BC \equal{} 42$, so BOC is equilateral
(area of sector BOC) - [BOC] = $ \frac {1}{6}\pi (42^2) \minus{} \frac {42^2\sqrt {3}}{4}$

Using Heron's formula, [IBC]=$ \sqrt {(60)(12)(30)(18)} \equal{} 360\sqrt {3}$

Our total is: $ 294\pi \minus{} 81\sqrt {3}$
$ m \plus{} n \plus{} d \equal{} 294 \plus{} 81 \plus{} 3 \equal{} \boxed{378}$

[/hide][/quote]

I was wondering how exactly did you use the law of cosines?

[quote="minsoens"]\begin{eqnarray*}BC^2 & = & BI^2 + IC^2 - 2(BI)(IC)\cos(\angle BIC) \\
& = & 30^2 + 48^2 - 2(30)(48)\cos{60} \\
& = & 1764\end{eqnarray*}
$ \implies BC = 42$.[/quote]

Thank you, btw what kind of software do you use to attach a graph for the problem?

I use Asymptote: [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_(Vector_Graphics_Language)]tutorial[/url] and [url=http://asymptote.sourceforge.net/doc/Index.html]list of functions[/url].
You can also use Geogebra, which is much more interactive than Asymptote.

Actually instead of using the (computational-tend) Heron's formula to calculate the area of IBC, we could have just used $ 1/2*48*30*\sin60\equal{}12*30\sqrt{3}\equal{}360\sqrt{3}$ which is much cleaned...
but nice solution anyways.

Another solution although I assumed some things

[hide]
Using the image provided $\angle{AIB} = 120 ^{\circ}$ and $\angle{IOY} = 60 ^{\circ}$. Therefore if we draw in the other line from $O$ to line $CA$ with length $18$ we get an equilateral triangle.

Now draw $4$ more chords of length $78$ such that they intersect the same way as the first two. Construct the equilateral triangles using the same method as above. From there we can see the area we desire is one-sixth the area of the circle minus the 6 equilateral triangles (or the hexagon).

Doing the math, we get

$\frac{42^2\pi-6\cdot\frac{18^2}{4}\sqrt{3}}{6} = 7\cdot42\pi -  \frac{18^2}{4}\sqrt{3} = 294\pi - 81\sqrt{3}$

[/hide]

stole wiki's diagram

[asy]
size(200); pathpen = black + linewidth(0.7); pen d = dashed+linewidth(0.7);
pair O = (0,0), E=(0,18), B=E+48*expi(11*pi/6), D=E+48*expi(7*pi/6), A=E+30*expi(5*pi/6), C=E+30*expi(pi/6), F=foot(O,B,A);
D(CR(D(MP("O",O)),42)); D(MP("A",A,NW)--MP("B",B,SE)); D(MP("C",C,NE)--MP("D",D,SW)); D(MP("E",E,N)); D(C--B--O--E,d);D(O--D(MP("F",F,NE)),d); MP("39",(B+F)/2,NE);MP("30",(C+E)/2,NW);MP("42",(B+O)/2);
[/asy]
If we find how much is covered in the other two regions, we can take the complement and divide by 2 to find the area of the region bounded by the two unequal segments. Notice that the area of regions AEC and DEB is equal to sector(DOB) + sector(AOC) + area(DOBE) - area(AECO). We can find this. First, we must figure out the angle of each of the sectors. Notice that triangle AOE has sides 18,30,42. Applying Law of Cosines, we find that $\cos(\angle AOE) = \frac{11}{14}$. So the angle of sector AOC is $2\cdot \arccos(\frac{11}{14}).$ To find the angle of DOB, we can again apply Law of Cosines on triangle DOE. This has sides 42, 18, 48. We get $\cos(\angle DOE) = -\frac 17$. So the angle measure of $DOB$ is $360-2 \cdot \arccos( - \frac 17)$. Adding these two sector angles, we get $360 + 2(\arccos (\frac{11}{14}) - \arccos(-\frac 17)).$ Now let $x = \arccos (\frac{11}{14}) - \arccos(-\frac 17)$. We can take the cosine of both sides to get $\cos(x) = \cos(\arccos (\frac{11}{14}) - \arccos(-\frac 17)).$ Using the subtraction formula, we get $\cos(x) = -\frac{11}{14} \cdot  \frac{1}{7} + \frac{5\sqrt3}{14} \cdot \frac{4\sqrt3}{7} = \frac{1}{2}$. Then $x= \pm 60$ degrees. But it has to be negative, since if it's positive, the sum of the angles is more than $360$. Then we get $360 - 2(60) = 240$. Since it covers 240 degrees, this is $\frac 23 \cdot 42^2 \cdot \pi = 1176 \pi$.

Now we need to find area(DOBE) - area(AECO). DOBE is two 18, 42, 48 triangles. The altitude to the 48 side is $9\sqrt 3$. Then the area of both is $432 \sqrt 3$. For AECO, we have two 18, 30, 42 triangles. Each has area $135 \sqrt3$ (herons) so both have area $270 \sqrt 3$. This gives $432 \sqrt 3 - 270 \sqrt3 = 162 \sqrt 3$. So the area of the two regions not bounded by unequal segments is $1176 \pi + 162\sqrt3$. Taking the complement, we get $1764\pi - (1176 \pi + 162\sqrt3) = 588\pi - 162\sqrt3$, and dividing by 2, we get $\boxed{294 \pi - 81\sqrt3} \implies \boxed{378}$.


A picture is attached to a vertical wall by means of string AC with length l forming an angle α with the wall. The height of the picture BC = d. The bottom of the picture is not fastened. At what coefficient of friction between the picture and the wall will the picture be in equilibrium?

For diagram,
http://www.flickr.com/photos/30014811@N06/3439916273/

I keep getting the value of μ in terms of weight of the body but that is not a known variable. How to find μ in terms of the lengths given?

The friction will be:

     cot(alfa)+2(d^2-l^2sin(alfa)^2)^.5/(lsin(alfa))


Consider $A = \{ x \in \mathbb N_{0}| x = 2 ( p^{2}-q ) ( q^{2}-p ), \, p,q \in \mathbb N_{0}\}$.

Prove that:

(a) $2006 \in A$;

(b) $A$ contains an infinity of perfect squares;

(c) a set $B$ such that $|B| = p^{2}-q, \, | \mathcal P (B) | = q^{2}-p$, where $p,q \in \mathbb N_{0}$, does not exist.

[i]Note.[/i] $\mathcal P(B)$ ought to denote the [url=http://en.wikipedia.org/wiki/Power_set]power set[/url] of $ B$.

[i]Cristian Alexandrescu, Adrian Stoica[/i]


Let $ABCD$ be a bicentric quadrilateral with incircle $(I)$. If $K,M,L,N$ are the touch points of $(I)$ with $AB,BC,CD,DA$ respectively and from $I$ draw the altitude $IQ$ to $BD$. Prove that: $[ABD].[QML]=[BCD].[KQN]$

Let $(O,R)$ be the circumcircle of $ABCD.$ $ \triangle KQN$ is pedal triangle of $I$ WRT $\triangle ABD$ and $ \triangle QML$ is pedal triangle of $ I$ WRT  $\triangle BDC.$ By Euler's theorem we have:

$ \frac {|\triangle KQN|}{| \triangle ABD|} \equal{} \frac {p(I,(O))}{4R^2} \ \ , \ \ \frac {|\triangle QML|}{|\triangle BCD|} \equal{} \frac {p(I,(O))}{4R^2}$

$\Longrightarrow \ |\triangle ABD| \cdot |\triangle QML| \equal{} |\triangle BCD|\cdot |\triangle KQN|$


inequalities
function
inequalities unsolved

for every positive integer n,find the largest subset M(n) of the real numbers such that
n +  \sum (x_i)^(n+1)  \geq  n \pi (x_i) +  \sum x_i
for all x_1,x_2,x_3...............x_n belonging to M(n)
when do we have equality?

Some inequalities given in this contest are extremely difficult. And I find this one very difficult too. I think I solved the case n odd ( the whole R is the answer), but the other case is strange. I cannot figure what set could it be. Help!!!

Oh, my God! It's exactly what I was thinking (and praying not to be so): it seems that for even n the set is [-1, infinity).

your having the solution will be of help only if u post it,hxtung :D

Ok, I've finally found a solution to this terrible problem. Please, tell me, am I the only one who finds it very intriguing? Now let's kill it:
    First I prove that for odd n we have M=R. Take x_1,...,x_n reals. We have |nx_1*...*x_n+x_1+...+x_n|<=n|x_1|*...*|x_n|+|x_1|...+|x_n|. So, it is enough to prove that for any a_1,...,a_n>=0 we have  \sum (a_I)^(n+1)+n \geq n a_1*...*a_n+ \sum a_i. Take  \sum a_i=nx. Then  \sum (a_i)^(n+1) \geq nx^(n+1) and na_1*...*a_n \leq nx^n. So, it is enough to prove that x^(n+1)+1 \geq x^n+x, which reduces to (x^n-1)(x-1) \geq 0, true.
    Now comes the difficult part. I will prove that for even n the set M is the set of those numbers greater than or equal to -1. From the condition we have that x^(n+1)+1 \geq x+x^n for any x in M.Suppose x<-1 and write x=-a, with a>1. Then 1+a>a^n+a^(n+1), false! Thus one inclusion is proved. Now, take n even and x_1,...,x_n \geq -1. We have seen that if they are all positive, then it's true. So, we may assume at least one is negative. If they are all negative, then x_i=-a_i, with 0<a_i<1 and it's trivial. So, assume that x_1,...,x_k are negative and the rest are positive. Write x_1=-a_1,....,x_k=-a_k. The inequality becomes  \sum (a_i-(a_i)^(n+1))+ \sum (x_i)^(n+1) \geq  \sum x_i+n(-1)^k*a_1*...*a_k*x_1*...*x_k-n. Obviously, it suffices to prove that  \sum (x_i)^(n+1)+n \geq nx_(k+1)*...*x_n+x_(k+1)+...+x_n. Again write  \sum x_i=(n-k)x. Then with the same argument it suffices to show that (n-k)x^(n+1)+n \geq (n-k)x+nx^(n-k). This can be written (x^n-1)/n*x \geq (x^(n-k)-1)/(n-k). And this is easy since the function f(x)=(t^x-1)/x is increasing for x>=1 and decreasing for x<1.


calculus
MIT
college
topology
linear algebra
number theory
real analysis

I'm interested in learning all of the math I can, however, the magnitude of material is daunting. Because of this I am interested in finding what the best order to learn each subject is. By best I mean: how easily one can move from one topic to another without missing concepts. I'm very interested in your opinions.

Please note: I can say with confidence that I have already learned all high school level mathematics usually offered. Therefore I am talking about college level topics and beyond.

Are you interested in taking accredited high school/university classes?

I will be attending college in the near future, however I wish to teach myself for the time being.

At the undergraduate level, there isn't really an "order":  it's more like a poset of prerequisites.

It would help if you were more specific about both what you already know and what your interests are.  If you intend to be serious about learning pure mathematics, important prerequisites include having a strong grounding in linear algebra and formal proof.  Most people will expect you to be familiar with elementary differential equations and multivariable calculus as well.  After that, I would say that it is important, if you are going for a "classical" mathematical education, to have a strong grounding in abstract algebra, analysis, and topology.  It wouldn't hurt to learn some number theory to supplement the algebra.  If you're more interested in computer science or combinatorics, you might want to emphasize discrete mathematics instead, and if you're going to do that then the sky's the limit:  the nice thing about discrete math (as the state of the art currently stands) is that most of its subfields have relatively few prerequisites.

Thank you for the posts.

t0rajir0u, to be more specific, I am interested in pure mathematics. As for my background, I am well acquainted with most linear algebra and multivariate calculus. To revise my question, I would like to see what order you believe undergrad and graduate level math should be learned. By undergrad and graduate math, I mean anything from real analysis to algebraic topology.

And the answer is that that's a silly question: even among as homogeneous a group as (for example) MIT math graduate students, the courses that people have taken (not even to speak of the order they've taken them in) vary dramatically.  t0r already mentioned the "keystone" classes that almost everyone has taken, and that also often helped them develop a sense of what direction they wanted to send their studies in: abstract algebra, analysis, and intro topology.  (I actually think relatively fewer people have taken a differential equations course -- the mathematical physics people, sure, but among the pure math and discrete math people, it seems to me to be less common.  Maybe I just haven't asked the right people, though.)  I think everyone should probably take a basic discrete math class at some point, too.  But after that basic level, it's really a matter of taste -- I've never taken algebraic topology because I didn't find topology very enthralling, but plenty of my fellow-students have taken three semesters or more of it (one or two undergrad, one or two grad).


Determine how many integers of 2-digit that are divisible by their numbers.

I don't think I understand the question.  By "numbers", do you mean "digits"?  In that case, do you mean divisible by [b]both[/b] digits?

yes Ravi. In that case, do you mean divisible by both digits? --> That's correct.

Okay, then 14 seems right.  Call the two digits $ a$ and $ b$.  Then we want $ 10a \plus{} b$ to be divisible by $ a$ and $ b$.  Thus $ b$ must be divisible by $ a$ and $ 10a$ must be divisible by $ b$.  Hence $ b \equal{} ka$ where $ k$ is a factor of 10.  If $ k \equal{} 1$, we get 9 solutions.  If $ k \equal{} 2$, we get 4 solutions.  If $ k \equal{} 5$, we get 1 solution.  If $ k \equal{} 10$, we get 0 solutions.

I'm an idiot  :D , i'm sorry luiseduardo

No Lousin.
14 = number of integers that are divisible by their numbers.

Fourteen isn't a number in this case, but "how many"

understand ?


If    $ a,b,c,x,y,z\neq 0$      and     $ \dfrac{x^2\minus{}yz}{a}\equal{}\dfrac{y^2\minus{}xz}{b}\equal{}\dfrac{z^2\minus{}yx}{c} \neq 0$.

Prove that :

$ \dfrac{a^2\minus{}bc}{x}\equal{}\dfrac{b^2\minus{}ac}{y}\equal{}\dfrac{c^2\minus{}ab}{z}$

[hide]Let the first three fractions be equal to $ k$.

Then $ a={x^2-yz\over k},b={y^2-xz\over k},c={z^2-xy\over k}$, hence

\begin{eqnarray*}{a^2-bc\over x}=\frac{x^4-2x^2yz+y^2z^2-y^2z^2+xy^3+xz^3-x^2yz}{k^2x}={x^3+y^3+z^3-3xyz\over k^2}\end{eqnarray*}

Symmetric expression is obtained in all three cases.[/hide]


Assume that $ a_i,b_i$ are positive integers with $ \gcd(a_i,b_i)\equal{}1$ and $ 1 \le i \le n$.
Prove that we can write all positive rational number $ \frac{l}{m}$ in the form

$ \frac{l}{m}\equal{}\frac{\sum_{i\equal{}1}^{n} x_i^{a_i}}{\sum_{i\equal{}1}^{n} y_i^{b_i}}$

where $ x_i,y_i$ are natural numbers.

$ n \equal{} 1$,  is  easy  to  prove.

then  use  the  fact  that
if$ \frac {p_1}{q_1} \equal{} \frac {p_2}{q_2} \equal{} .... \equal{} \frac {p_n}{q_n} \equal{} q$,  then
$ q \equal{} \frac {p_1 \plus{} p_2 \plus{} .. \plus{} p_n}{q_1 \plus{} q_2 \plus{} .. \plus{} q_n}$


Do you know any good websites with problems and stuff?

Google goodly maths. It has a lot of past competitions.

try the one right here.

http://www.artofproblemsolving.com

In case you don't know, TZ=Totally Zealous.

what happened to those amc 8 practice tests on unl.edu? they disappeared...  :(

where else can you find them?

They took it down.

Now, you have to pay $1 per year, which is 25 problems. They mail it to you with a solution pamphlet. If you want to write on it, then you need to buy it bulk. (Its a violation of copyright law if you photocopy it if you don't run a school).

[quote="jhredsox"][quote="kyyuanmathcount"][quote="anirudh"]TZ's website.[/quote]

And he knows EXACTLY where that is.  Here's a link:

	http://www.freewebs.com/totallyzealous/Home.html[/quote]

and he knows [size=200][color=red]EXACTLY [/color][/size]how to make it clickable!

[url]http://www.freewebs.com/totallyzealous/Home.html[/url]

 :rotfl:

-jorian[/quote]

 :blush:

tz used to have much more on there, he took stuff down for some reason.

Buying the tests may end up being the best method.

are there any amc 8 sites, to practice before tuesday?

[quote="13375P34K43V312"]Or you can build a time machine and steal them from test sites.[/quote]
Heh heh heh.
I know of an internet time machine...

[quote="anirudh"][quote="bpms"][quote="star99"]cached links on google?[/quote]
I am sworn to secrecy, but that doesn't mean I can't go back, download them, and post them. I'll do them in a minute...
Can someone post the links? (You can find the links in your browsers history with ctrl h)...[/quote]

When are you? :huh:[/quote]
When someone posts the links...

[quote="anirudh"]The links to what?[/quote]
The amc 8 practice tests.
You can find them by hitting ctrl h

[quote="anirudh"]http://www.unl.edu/amc/e-exams/e4-amc08/archive8.shtml

Is that what you want?[/quote]
I meant the practice tests themselves.

[quote="anirudh"]I don't have the links to those... :([/quote]
If you or anyone else has visited them in the past three weeks then they do.


I'm afraid I don't quite follow the pattern here.

Edit:  Aha, I misread some of those multiplications as additions.  I should get glasses or something  :blush:

$\sum_{k=1}^{\infty}\frac{1}{k(k+1)2^{k}}$.

Masoud Zargar

Let $S = \sum_{k=1}^{\infty}\frac{1}{k \cdot 2^{k}}$.  Then

$\sum_{k=1}^{\infty}\frac{1}{k(k+1) 2^{k}}= \sum_{k=1}^{\infty}\left( \frac{1}{k \cdot 2^{k}}-\frac{1}{(k+1) 2^{k}}\right) = S-2(S-\frac{1}{2}) = 1-S$

... I seem to remember knowing how to evaluate $S$ at some point, but I've forgotten  :(  Could someone enlighten me?  Or am I on the wrong track here?


Given a natural number $ n\geq 2$, find the smallest integer $ k$ with the following property:
Every set consisting of $ k$ cells of an $ n\times n$ table contains a nonempty subset $ S$ such that in every row and in every column of the table, there is an even number of cells $ \in S$.


group theory
abstract algebra
least common multiple
superior algebra
superior algebra unsolved

Let $ p$ be a prime, and $ H$ a subgroup of the group of permutations of order $ p$. How many morphisms are there from $ H$ to $ \mathbb{Z}\slash\left(p\mathbb{Z}\right)$ ?

some ideas...

if $|H|=m, (m,p)=1$ then $\sigma\in H$ has order prime with $p$ but because $\mathbb Z_p$ and $p$ prime results that the only morphism from $H$ to $\mathbb Z_p$ is the null morphism...

now if $|H|=pm, (m,p)=1$. $\phi\in Hom(H,\mathbb Z_p)$ a morphism.

if $\sigma\in H, \sigma=c_1c_2\ldots c_k$ the decomposition in distinct cycles then the order of $\sigma$ is the least common multiple of the lengths of its cycles, number which is prime with $p$. So $\sigma$ has order $p$ (that means it's a cycle) or has order $d$ with $(d,p)=1$.

if ord$(\sigma)=d, (d,p)=1$ then $\phi(\sigma)=\widehat 0$.

now we have to look only at those $\sigma\in H$ for which ord$(\sigma)=p$ => only at cycles of length $p$...

and that's kinda all what I managed until now...

back..

I want to show that if $H$ is not a cyclic group of order $p$ then the only morphism from $H$ to $\mathbb Z_p$ is the null one.

If $H$ has no elements of order $p$ from the post above we conclude that the unique morphism is the null one.

If $H$ has both elements of order $p$ and of order prime with $p$ then let $\sigma, \tau\in H$, so that ord$(\sigma)=p$ and ord$(\tau)=d, (d,p)=1$.
Let $f: H\to \mathbb Z_p$ be a morphism. Suppose $\widehat a = f(\sigma)\neq \widehat 0$.Then the order of $f(\sigma)$ in $\mathbb Z_p$ is $p$.
Write $\sigma=(i_0,i_1,\ldots,i_{p-1})$. For some $r,s, r\neq s\in \overline{0,p-1}$ we have that $\tau(i_s)=i_r$.
$\sigma^k=(i_0,i_k,i_{2k},\ldots )$ where the indices are taken modulo $p$. We can suppose that $r=0$ (else just permute circulary the therms of $\sigma$ and you get the same permutation with $i_r$ on the first position).
Now $\sigma^{s}(i_0)=i_s$. But $\sigma^s \tau (i_s) = \sigma^s(\tau(i_s)) = \sigma^s(i_0)=i_s$, so $\sigma^s\tau$ does not have the order $p$ anymore and it is not $e$ (because ord$(\tau)=d, (d,p)=1$ and ord$(\sigma^s)=p$ ).

So $\widehat 0 = f(\sigma^s\tau) = f(\sigma)^s+f(\tau)=\widehat{sa}+\widehat0\Rightarrow \widehat a=0$, contradiction.
So we can have in our group $H$ only permutations of order $p$, hence $|H|=p^u, u\ge 1$. $|S_p|=p!$, $p$ prime and due to Lagrange theorem it follows that $|H|=p$, so $H$ is cyclic, generated by a cycle of length $p$. In this situation there are $p$ morphisms from $H$ to $\mathbb Z_p$.

In all the other cases there is only one morphism , the null one.

hope it's correct...


Βρείτε όλες τις συναρτήσεις  $f: \mathbb{Z} \mapsto \mathbb{Z}$
έτσι ώστε $f(x^3 +y^3 +z^3 )=(f(x))^3+(f(y))^3+(f(z))^3$

Τι σας λέει ?Beautiful e?

The main lemma is : for each integer number n>=4 we have x,y,z,t,w integer numbers s.t their absolute values less than n and:
n^3=x^3+y^3+z^3+t^3+w^3

To prove this ,we may use this identity:
(2n+1)^3=(2n-1)^3 + (n+4)^3 - (n-4)^3 - 5^3 - 1^3

http://www.mathlinks.ro/Forum/topic-26140.html


calculus
integration
algebra
polynomial
sum of roots
product of roots

find the product of the solution(s) to the following equation: $2^\left(2x-3)$ $-129 \times2^{x}+2^{4}=0$

[quote="ashwinrk_jain"]let replace $2^{x}$ by y the equation becomes
$\frac{y^{2}}{8}-129y+2^{4}= 0$
the sum of roots of this equation is $129*8$ that is $2^{x_{1}}+2^{x_{2}}= 129*8$
we can write the equation as
$2^{x_{1}-3}+2^{x_{2}-3}= 129$
now from above we can say that one of the power of the 2 must be 1 because R.H.S is odd.
hence one of the root is 4. and from product of roots of the above equation is we get the other root is 3. hence the product of roots are 12.[/quote]
[hide]$129=128+1=2^{7}+2^{0}$
$2^{7}+2^{0}=2^{x_{1}-3}+2^{x_{2}-3}$
$x-3=7,0 \ x=3,10$
Then the product is $\boxed{30}$[/hide]

[quote="xxxxx"][quote="ashwinrk_jain"]let replace $2^{x}$ by y the equation becomes
$\frac{y^{2}}{8}-129y+2^{4}= 0$
the sum of roots of this equation is $129*8$ that is $2^{x_{1}}+2^{x_{2}}= 129*8$
we can write the equation as
$2^{x_{1}-3}+2^{x_{2}-3}= 129$
now from above we can say that one of the power of the 2 must be 1 because R.H.S is odd.
hence one of the root is 4. and from product of roots of the above equation is we get the other root is 3. hence the product of roots are 12.[/quote]

how did u get the $\frac{y^{2}}{8}$???[/quote]
$2^{2x-3}=\frac{2^{2x}}{2^{-3}}$
$2^{x}=y$
$\therefore 2^{2x-3}=\frac{y^{2}}{8}$

oh i m so sory i made a mistake.....
yes the roots are 3 and 10......

[quote="stergiu"][quote="ashwinrk_jain"] $2^{x_{1}-3}+2^{x_{2}-3}= 129$
now from above we can say that one of the power of the 2 must be 1 because R.H.S is odd.
 [/quote]

[b]Just a question on your  nice solution :[/b]

[hide]Does this hold  for any real values of $x$ or you consider olny natural or integral values of $x_{1}, x_{2}$ ?.Probably in your school you solve equations of this kind only in the set of integers.[/hide][u]
Babis[/u][/quote]
Are there any other numbers?


geometry
3D geometry
sphere
trigonometry
Pythagorean Theorem
trig identities
Law of Cosines

Two persons are at the same point of departure.They go opposite direction. After having gone for 3 metres from the point of departure, they turn left and continue going for 4 metres.
 How far have they gone from the point of departure?

 Collect,[/img][/tex]

By applying theorem phytagoras,

the distance from the point of departure for each person is 5m.

by adding it together it is 10m total.

Am I right?

I'm assuming that when you turn, it's 90 degrees.

So, they walk 7 meters each and end up 10 meters from each other.

Look at the picture.

You draw the diagram, and you should have two right triangles with side lengths 3 and 4. You apply pythagorean theorem and a person is 5 units from the point of departure each.

I'm pretty sure it is a plane becuase people can't walk up (towards the sky) or down, at least none of the people I know :D .

Sphere of radius... 2? Heh. That would make things a bit more interesting.

There's also the one where this guy walks 10km south, 10km east, and 10 km north, and winds up in the same place. Of course, that's slightly different. Easy too, I think.

Anywhere from 1 to 7 meters, except it cannot be exactly 7 meters.

1 meter if you turn 180 degrees
The distance approaches 7 meters as the turn angle approaches 0 degrees

Each person is 5m from the starting point.  This can be determined using the pythagorean theorem.  Furthermore, the two people are 10m apart since they both waled opposite directions but turned the same direction, left, relative to themselves.  Thus the distance apart would be 5m*2=10m.

by using theory of pythgoras..
                          (or what ever that guy's name is..)
his theory

A^2 + B^2=C^2

3^2 + 4^2 = 25
25=5^2
so the length of the diagonal is 5.

u know these problems come out at the IQ tests.
and it's like 2nd grade math

Here's a trickier question.

How much would each person have to turn after three meters so four meters after turning they are twelve meters apart? Assume they both turn left, and express in degrees rounded to the nearest hundredth.

Look below for a hint:

[hide]

Law of Cosines

[/hide]

Look below for the answer

[hide]

62.72 degrees

[/hide]

Look below for the solution

[hide]

Draw a triangle ABC with sides 3, 4, and 6.

Use the Law of Cosines to find the angle opposite the side with length 6.

The angle is about 117.28 degrees.

To find out how much each person turned, just subtract 117.28 degrees from 180 degrees.

The turn size is about 62.72 degrees

[/hide]

Since they will be 12 meters apart, both are 6m apart
now we have triangle with sides of 3,4 and 6 and we want to find the angle opposite 6 so by unsing the law of cosines we get

$ a^2=b^2+c^2-2bc \cosh $
$ 6^2=4^2+3^2-2*4*3 \cosh $
$ 11=-24 \ cosA $
$ \ cosA =-11/24 $
$ A=117.28^\circ $
$ 180-117.28=62.72 $
therefore they turned 62.72 degrees to the left


There are numbers from 1 to 7. Find the number of ways that you can arrange them in a manner such that no two adjacent numbers sum to multiple of 2 or 3.

Basically, to find the answer, we know that there are 7 numbers, 4 even, and 3 odd. We know that two even numbers or odd numbers cannot be next to each other. So, we can only pair odds with evens. So kind of like: odd, even, odd, even, odd, ... . Then, the pairs that sum up to three are 1,2 ; 2,7; 3,6 ; 4,5. So, we just do some casework and find the answer with the restrictions mentioned above. I [i]think[/i] we can solve it like that. Now I'll try it...and post it when I get it :)

It's definitely not casework.

I believe it involves recursion.

Yeah...now I've come to realize that. Casework get's pretty nasty.

Well, if this is in a row...

[hide="hint"] The sum of any two adjacent numbers must be congruent to 1 or 5 (mod 6).[/hide]

[hide="guess?"] I'm guessing that only sequences of the form 1.......7 or 7........1 work, which gives 4.[/hide]

[hide="okay here is some justification"] Each number has two unique numbers that it must be surrounded by (so the sum is 1 or 5 (mod 6)), but 1 and 7 are special as they're both congruent to 1 (mod 6). This way, a value 1 (mod 6) may be used twice, one to begin and one to end the sequence.

In other words, the numbers 2 through 5 may not be used to either begin or end the sequence, as they will be needed somewhere else.[/hide]

Wait, are they arranged in a row or in a circle?

Finally, ...

[hide="Solution"]
I think it is well established that it must be odd, even, odd, even, ....

First, we think about how we can arrange the evens. We can have six arrangements.

_ 2 _ 4 _ 6 _
After going throught the numbers, we see there are 2 possible arrangements(It isn't hard, I just have to go soon). This also takes care of 642, since it is just this flipped. We have a total of 4 so far.

_ 2 _ 6 _ 4 _
Again, this had two possible arrangements. Similar to the last one, we can flip it and get 4 different ways, bringing the total up to 8.

_ 4 _ 2 _ 6 _
Yet again, two more arrangements. Doubling, we get 4. So the final total is 12.

$ Answer\equal{}12$
[/hide]

[hide="Cases Solution"]I got [b]12 [/b]as well. Of course, I didn't see recursion right off the bat. I used cases like so:

First, notice that the seven spaces must be alternating from even to odd (to keep from a multiple of two). That is:

Odd Even Odd Even Odd Even Odd

Then, write down the numbers we can pair with each even number.

2: 3 or 5

4: 1, 3, or 7

6: 1, 5, or 7

And using cases from there isn't too hard. 12 will be the answer.[/hide]

I am not sure how to use recursion to try and solve this problem, although using the cases method kind of does hint at a recursive solution.

7 is probably small enough that the recursive approach isn't worth doing.  Here's one recursive attack, but maybe it can be made more elegant:

For $ 0 \leq i \leq 5$ and nonnegative integers $ b, c, d, e, f, g$, let $ a_i(b, c, d, e, f, g)$ be the number of words consisting of $ b$ copies of 0, $ c$ copies of 1, $ d$ copies of 2, ..., $ g$ copies of 5, ending in $ i$, so that the sum of any two consecutive letters is 1, 5, or 7.  Then we have a ridiculous system of equations that begins $ a_0(b, c, d, e, f, g) \equal{} a_1(b \minus{} 1, c, d, e, f, g) \plus{} a_5(b \minus{} 1, c, d, e, f, g)$ and has five other similar equations.  We have a bunch of initial values like $ a_0(0, ...) \equal{} 0$ and $ a_0(1, 0, 0, 0, 0 ,0) \equal{} 1$, and then we're trying to compute $ 2\sum_i a_i(1, 2, 1, 1, 1, 1)$.  Unless there are some significant improvements to be made in this ridiculously complicated system of equations, though, this is really equivalent to a brute force counting method.  (Literally the only simplification built-in so far is that 1 and 7 are interchangeable.)

I think that probably a casework argument like those above is the nicest way to approach the particular problem given.

Here are some related questions to think about (with at least one but no more than four guaranteed to be easy ;)):

What is the number of permutations of the numbers 1 through $ n$ such that the sum of no two consecutive elements is even?

What is the number of permutations of the numbers 1 through $ n$ such that the sum of no two consecutive elements is divisible by 3?

What is the number of permutations of the numbers 1 through $ n$ such that the sum of no three consecutive elements is divisible by 2?

What is the number of permutations of the numbers 1 through $ n$ such that the sum of no three consecutive elements is divisible by 3?

What is the number of permutations of of the numbers 1 through $ n$ such that the sum of no two consecutive elements is divisible by either 2 or 3?

JBL do you have the solutions to the problems you proposed? Thanks.

Well, let me give you some meta-comments, from which you can deduce which ones I have a solution to ;)

The first is an elementary counting exercise that is completely appropriate for this forum, or maybe even HSB.

One answer to the third is "not very many."  Think about the parities of the elements, and imagine trying to build such a permutation of length longer than four.

For the second, one can write down a system of recursive equations as I did in my previous post.  I'm not sure about getting an exact formula, but from this system one can acquire generating functions in three variables and then see about extracting coefficients.

For the fourth and fifth, the same techniques as in the the previous comment should work in principle.  In practice may be another matter.


superior algebra
superior algebra unsolved

Let $k \subset F$ - be an extension of finite transcendence degree, $\bar{F}=F$.Prove that any field homomorphism $\phi : F \to F$ such that $\phi = id$ on $k$ is an isomorphism

I guess I have something but I'm not completely sure it's correct.
Take $x$ such that $\phi(x)=0$, the transcendance degree is finite so $\exists n$ such that $x^{n}+a_{n-1}x^{n-1}+\ldots+a_{0}=0$ with $a_{i}\in k$. Then $\phi(x)^{n}+a_{n-1}\phi(x)^{n-1}+\ldots+a_{0}=0$. But $\phi(x)=0$ so $a_{0}=0$ and we see that if we choose $n$ minimal we have that  $x$ is in $k$. Then $x=\phi(x)=0$.

[quote="Rodolphe2005"]correct me if I make errors with my english please  :)
Merci ![/quote]

How about maths?

You tried to prove that $\phi$ is injective. First of all, $\phi$ is a morphism of fields, so it is immediately injective.
Let's look at what you did:
You took $x\in F$ and said that it is algebraic over $k$. This is not a consequence of the hypothesis. $F$ is of finite trancendence degree over $k$, not algebraic over $k$.

Sorry, shame on me ... I write too quickly.
But I don't understand why finite transcendance degree doesn't imply algebraic, since if $n$ is the transcendance degree and if I take $x$ in $F$ then $x,x^{2},\ldots,x^{n+1}$ is not free ?
Could you explain me this point since I apparently misunderstand something. Thanks.

PS: I don't write in my signature that you could correct maths errors since it's obvious.

[quote="Rodolphe2005"]Sorry, shame on me ... I write too quickly.
But I don't understand why finite transcendance degree doesn't imply algebraic, since if $n$ is the transcendance degree and if I take $x$ in $F$ then $x,x^{2},\ldots,x^{n+1}$ is not free ?[/quote]

Just because of the definition of a transcendence element: we call an element transcendent if it is not algebraic. The finitness of a transcendence degree just means that we can choose finite number of transcendence elements $t_{1},t_{2},...,t_{n}$ in $F$ over $k$ and $F$ is algebraic over $k(t_{1},t_{2},...,t_{n})$, but surely not over $k$.

You may have a look at http://en.wikipedia.org/wiki/Transcendence_basis.

[quote="eugene"] $F$ is algebraic over $F(t_{1},t_{2},...,t_{n})$, but surely not over $k$.[/quote]

algebraic over $k(t_{1},\ldots,t_{n})$ you mean, of course

Let $t_{1}, \ldots, t_{n}$ be a $k$-transcendence basis of $F$. Because transcendence bases are mapped to transcendence bases and transcendence bases have the same cardinality, $F/\phi(F)/\phi(k(t_{1},\ldots,t_{n}))$ is algebraic, but $F$ is algebraically closed, hence also $\phi(F)$, so $\phi(F) = F$.


Is there a lazy way to do

EDWARD x R = MURROW[/u]

[quote="jhredsox"]Didn't you already ask this and people said no?

Lazy person+asking for a lazy solution=lazy

-jorian[/quote]
No,  I've never posted this before.

EDIT: I realize others have posted this before, but nobody gave a sufficient answer, and I don't want to get yelled at for awakening old topics.
.

[quote="machack"]where did u get this question im REALLY curious (as u can see by my posting in the 2nd link, i actaully tried this... )[/quote]
Harker Math Invitational 2004 Individual Contest
It's a Bay Area competition.

[quote="rcv"]I have found a solution to the Edward R. Murrow problem.  I don't think the problem is bogus, at all.[/quote]
Can you tell me what i did wrong? or maybe repeats are allowed?

it wasn't that hard. process of elimination

i actually did try process of elimination and thought i found out that r cannot be any of 0-9
THOUGH... Why does this belong in ths forum i thought it was for mathcounts only..

muhahaha i found my mistake! :blush:  :blush:
[hide]the answer is 486128*2=972256
sry for the wrong answer in the other post[/hide]
[size=75]
[color=red]<Please use hide tags.  Please don't multiple-post.    -- moderator rcv>[/color][/size]


function
search
algebra proposed
algebra

Find all functions $f : \mathbb{N}\rightarrow\mathbb{N}$ such that :
$f(f(n)) \leq \frac{n+f(n)}{2}$   for all natural $n$.

This has been posted before. Search for it.

By the way, this problem was also used in one of the South African selection tests for IMO 2005 :) It's pretty easy though.


Three real numbers $a, b, c \geq 0$ satisfy the inequalities $a^2 \leq b^2 + c^2$, $b^2 \leq c^2 + a^2$, and $c^2 \leq a^2 + b^2$. Prove that

$(a+b+c)(a^2+b^2+c^2)(a^3+b^3+c^3) \geq 4(a^6+b^6+c^6)$

Here's my solution, which actually proves a stronger result.  Note that

\[(a^2+b^2+c^2)^3 - 4(a^6+b^6+c^6) = 3\sum_{\textrm{cyc}} a^4(b^2+c^2-a^2) + 6a^2b^2c^2 \geq 6a^2b^2c^2,\]
From Cauchy, we have

\[(a+b+c)(a^3+b^3+c^3) \geq (a^2+b^2+c^2)^2,\]

and thus

\[(a+b+c)(a^2+b^2+c^2)(a^3+b^3+c^3) \geq (a^2+b^2+c^2)^3 \geq 4(a^6+b^6+c^6)+6a^2b^2c^2.\]

However, there's a much much simpler solution.  Note that

\[(a+b)^2 \geq a^2+b^2 \geq c^2 \Rightarrow a+b \geq c,\]

and analogous inequalities.  So assume WLOG that $a \geq b \geq c$ and we have

\[(a+b+c)(a^2+b^2+c^2)(a^3+b^3+c^3) \geq (2a)(2a^2)(a^3+b^3+c^3) \geq 4(a^6+b^6+c^6).\]


Determine the value of the infinite series:
$ \sum_{n\equal{}0}^{\infty} \frac{3^{n}\plus{}5^{n}}{8^{n}}$.

$ 1990$ MANDELBROT

What a cake.

$ \sum_{n\equal{}0}^{\infty} \frac{3^{n}\plus{}5^{n}}{8^{n}}$
$ \equal{} \left (\sum_{n\equal{}0}^{\infty} \left ( \frac{3}{8} \right )^n \right ) \plus{} \left (\sum_{n\equal{}0}^{\infty} \left ( \frac{5}{8} \right )^n \right )$
$ \equal{} \frac{1}{1 \minus{} \frac{3}{8}} \plus{} \frac{1}{1 \minus{} \frac{5}{8}}$
$ \equal{} \frac{8}{5} \plus{} \frac{8}{3}$
$ \equal{} \frac{64}{15}$.

Hope I didn't make any silly mistakes.


algebra
polynomial
calculus
derivative
inequalities unsolved
inequalities

Prove that:$a+b+c+d \geq \frac{2}{3}(ab+bc+ca+ad+ac+bd)$
where $a;b;c;d$ are positive real numbers satisfying $2(ab+bc+cd+da+ac+bd)+abc+bcd+cda+dab=16$

[quote="Erken"]Could you please post a complete proof here?
Thank you.[/quote]
  Here is the essentials: Consider the polynom $ f(x) \equal{} (x \minus{} a)(x \minus{} b)(x \minus{} c)(x \minus{} d)$.Using Rolle's theorem to obtain that :$ f'(x) \equal{} (x \minus{} u)(x \minus{} v)(x \minus{} w)$ for some $ u,v,w$.Make a substituting between the monomials of $ (a,b,c,d)$ and $ (u,v,w)$.And we will reach a much  easier problem.

[hide="following the essentials"]
Let $ P(t)\equal{}(t\minus{}a)(t\minus{}b)(t\minus{}c)(t\minus{}d)$, then the problem is equivalent to proving $ P^{(3)}(0)\geq 2P^{(2)}(0)$, where $ P$ is a polynomial with four nonnegative roots satisfying $ P^{(2)}(0)\plus{}P^{\prime}(0)\equal{}16$. Let $ Q(t)\equal{}\frac{P^\prime (t)}{4}$, so $ Q$ has leading coefficient 1. Also, all of $ Q$'s roots are nonnegative, since by Rolle's Theorem and the fact that a root with multiplicity $ m$ is a root of multiplicity $ m\minus{}1$ of the derivative, $ P^\prime$ has at least three nonnegative roots.

We then get $ Q^\prime (0)\plus{}Q(0)\equal{}4$ and need to prove $ Q^{(2)}(0)\geq 2Q^\prime (0)$. Letting $ Q$'s roots be $ x,y,z$ with $ x\geq y\geq z$, we must prove
\[ xy\plus{}yz\plus{}zx\plus{}xyz\equal{}4\Rightarrow x\plus{}y\plus{}z\geq xy\plus{}yz\plus{}zx\]
To do this, let $ s\equal{}x\plus{}z,p\equal{}xz$ so that we have
\[ y(s\plus{}p)\equal{}4\minus{}p\]
\[ (s\plus{}p)(s\plus{}y\minus{}sy\minus{}p)\equal{}s^2\plus{}ps\plus{}4\minus{}p\minus{}4s\plus{}sp\minus{}sp\minus{}p^2\]
\[ \equal{}s^2\plus{}sp\minus{}p^2\minus{}4s\minus{}p\plus{}4\]
\[ \equal{}(s\minus{}2)^2\plus{}p(s\minus{}p\minus{}1)\]
Showing that this is nonnegative reduces to proving $ (x\minus{}1)(z\minus{}1)$ nonpositive. This is true since $ x\geq 1$ and $ z\leq 1$. (If $ x<1$ then $ x,y,z<1$ so $ xy\plus{}yz\plus{}zx\plus{}xyz<4$, a contradiction. Similarly, $ z$ can't be greater than 1.)
We then get $ s\plus{}y\geq sy\plus{}p\Rightarrow x\plus{}y\plus{}z\geq xy\plus{}yz\plus{}zx$, as desired.
[/hide]


The half-life of radium is 1690 years.  If 20 grams are present now, how much will be present in 100?

One hundread, what? Years?
[hide]
$ x = x_{i}\left( \frac{1}{2}\right)^{\frac{n}{t}}$
$ x_{i}= 20$
$ n= 100$
$ t= 1690$
Substituting $ x \approx 19.1963018$
[/hide]


Let a_1,...,a_n reals such that sum (a_i)^2=1. Prove that:
   n+1>=(sum a_i)*( sum a_i+sum (a_i)^3).

   Harazi

Here is my sol,for n \geq 3
(Note: \sum a_i mean sum a_i; i in {1,...,n}
  \sum  a_i*a_k  mean sum a_i*a_k;  (i,k) in{1,...,n} 2 ,i#k
  \sum  a_i*a_k*a_l mean ... i#k,i#l,l#k)
a_i=a_i*( \sum a_i 2 )
So
 RHS= (\sum a_i )*( 2*\sum a_i 3 + \sum a_i*a_k 2 )
       = 2 *\sum a_i 4 +3* \sum a_i*a_k 3 + \sum a_i 2 *a_k 2 + \sum a_i*a_k*a_l 2
LHS=(n+1)*( \sum a_i 2 ) 2
     =(n+1)* \sum a_i 4  +(n+1)* \sum a_i 2 *a_k 2
We need to prove that
    (n-1) \sum a_i 4 +n \sum a_i 2 *a_k 2  \geq 3* \sum a_i*a_k 3 + \sum a_i*a_k*a_l 2
We have
   1)  (a_1-a_2) 4  \geq 0
 so   a_1 4 +a_2 4 +6*a_1 2 *a_2 2  \geq 4*(a_1 3 *a_2+a_2 3*a_1
so     3/4 (n-1) \sum a_i 4 +9/4  \sum a_i 2 *a_k 2  \geq 3*  \sum a_i*a_k 3
   2)   1/4 (n-1) \sum a_i 4  \geq 1/4  \sum a_i 2 *a_k 2
   3)   a_1 2 *a_2 2 +a_1 2 *a_3 2  \geq 2a_1 2 *a_2*a_3
so    (n-2) \sum a_i 2 *a_k 2  \geq  \sum a_i*a_k*a_l 2
 1),2) & 3) qed
Sorry for my bad symbols(and English)
By this method,i think you can make it stronger


How many different ways can the letters of the word THREE be scrambled?

The permutations of this is 5!, but this counts each name twice, since the different 'E's are indistinguishable. So, your final answer is $ \frac{5!}{2!} \equal{} \boxed{60}$


[quote="archimedes1"][hide="a"]If $ f(x)=f(2a)^{x}$ and $ f(x+2)=27f(x)$ for all $ x$, then what is $ a$?[/hide]
[hide="b"]Consider a family of functions $ f_{b}(x)$ such that $ f_{b}(0)=b$ and $ f_{b}(x)=2^{a}f_{b}(x-a)$.  What is $ f_{c}(2x)$ in terms of $ f_{b}(x)$?[/hide][/quote]

[hide="a"]a. plug in $ x=0$ now you get $ f(0)=1$ so now $ f(0+2)=27(1)$   so with that we get to know that every even number will just increase be powers of 27 for example $ f(4)=f(2+2))=27f(2)=27^{2}$. So now $ f(2)=f(2a)^{2}$ which is basically $ 27=f(2a)^{2}$ so $ f(2a)=3\sqrt{3}$.....since every even increases by a power of 27 and f(2a) is the sqrt of it $ a=\frac{1}{2}$. to check you can do $ f(6)=f(1)^{6}$  basically $ f(6)=(3\sqrt{3})^{6}=3^{9}$  and also $ f(6)=f(4+2)=27\cdot f(4)=27^{3}=3^{9}$ so $ a=\boxed{\frac{1}{2}}$[/hide]

[quote="mindboggling"][hide="Alternate solution?"]
$ f(x) = f(2a)^{x}$
$ f(2a) = f(2a)^{2a}$
$ a = 1/2$

is this a valid method?
[/hide][/quote]

What if $ f(2a) = 0$ or $ f(2a) = 1$?  ;)

He needed to add in these considerations:

$ f(2a)=0 \implies f(0) = f(2a)^{0}= 0^{0}$, which is indeterminate, violating the "for all x".

$ f(2a)=1 \implies f(x)=1 \implies 1 = 27(1)$, contradiction.

Even still, the solution is shorter than the first one.

There's actually one tiny last case, which is that $ f(2a) =-1$ and $ a$ is half an odd integer.   :P

That depends: if the domain is meant to be the real numbers, then $ f(2a)$ must be positive (since $ a^{x}$ is only defined as a function from $ \bf R \to R$ if $ a > 0$) and in fact we only have to eliminate the case $ f(2a) = 1$.

well can someone try the second problem?


In any MCQ test that you take you guess atleast 5-6 answers. For those who say they have never done guess work so far.
take a look at
 [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=197467[/url] :rotfl:
and for an  small read on guess work
[url]http://www.goiit.com/posts/list/community-shelf-the-mathematics-behind-guessing-answers-in-objective-12590.htm[/url]
I know this may sound crazy,but how many answers do u guess ,how exactly do u guess?
My guesses always tend to bacfire even though i eliminate 1 or 2 ans. also at times. :rotfl:
so i guess at the max only 2 -3 and at times 4.
What abt you? :D

Depends on the subject really :D I take abt 2 guesses in chem, abt 3 guesses in math and [u][b]dont guess 2 to 3 in physics [/b][/u]:rotfl: :D

Just 5-6??????????

i've never guessed less than 75% of answers overall and 95% in bio :lol:  :D

cha 2 bad :maybe:
evry1s acting like deyv never guessd an ans.
now thats hypocrisy again :rotfl:  :rotfl:

You are talking abt guessing....I see..

In XIth, we had mock NTSE exam for 2nd round...
I was forced to write it and usually I don't read even the questions if I am bored...

I took it as a forum to check my luck factor...
So I finsihed the paper within 15-20 mins and guess how much I got....47/100
(without reading the questions...)

-ve marking doesn't affect me at all.....
there hasn't been a MCQ exam where I haven't attempted all questions and more often than not..it helps...(in my case)

You guys are awake at 2:23 and 3:06??? :o Arent you all tuning to the biological clock? :rotfl:

Well im serious man elasti, i gave my truthful opinion and case. Otw i might have as well said that i guess full of math n chem as well :rotfl:

Ahhh..Guessing is the game, that I always play. I do guess a lot, with a very decent success ratio. But very few of these guesses are the stupid guesses, based on Inky Pinky Ponky or 1.2.3... you go free.  :rotfl: Its usually an intelligent guess, after eliminating some options, using haphazard formulas to get an idea of the answer or approximating super fast to get the closest, or probably because I've seen this problem before. Oh! It may also be because of the raghavan instinct.

I guess:
2-3 Physics, 3-6 Chemistry, 2-5 Maths.

P.S: @Soumya: Better tune your biological clock man.

~~no hypocrisy~~

i really guess most of my answers as it saves time --but i have methods to guess out correctly :D  :D

in maths, the best way to get answers is to try out the options, for known values!!
in chemistry, my guesses almost always go wrong, but that doesn't deter me. i keep guessing!!
in physics i abstain from guessing most of the time, (as either i solve it, or i don't have a clue about it!!)
so my shot will be,
physics 0-1  chemistry 4-5   maths 5-6

a guessing hint--(from some book)
if all options fail go 4 "c" it has >25% chance of being correct ♥♥♥ cool!!!! :D  :D


hi,
we consider the equation
\[ a_{0}x^{n} \plus{} a_{1}x^{n \minus{} 1} \plus{} ... \plus{} a_{n} \equal{} 0, a_{i} \in \mathbb{R}, \forall i \equal{} 0,...,n
\]
Prove that the inequality $ (n \minus{} 1)a_{1}^{2} \geq 2na_{0}a_{2}$ is necessar that the equation has real roots.

Suppose that $ x_1,x_2,\cdots,x_n$ are the roots (maybe not distinct) of the equation.

From $ \displaystyle\sum_{1\leq i<j\leq n}(x_i-x_j)^2\geq 0$ (Since all th roots are real.)

$ \displaystyle(n-1)\sum_{i=1}^{n}x_i^2-2\sum_{1\leq i<j\leq n}x_ix_j\geq 0$

$ \displaystyle(n-1)\left(\sum_{i=1}^{n}x_i\right)^2-2n\sum_{1\leq i<j\leq n}x_ix_j\geq 0$

$ \displaystyle(n-1)\left(\frac{a_1}{a_0}\right)^2-2n\left(\frac{a_2}{a_0}\right)\geq 0$

$ \displaystyle a_{0}^2\left((n-1)\left(\frac{a_1}{a_0}\right)^2-2n\left(\frac{a_2}{a_0}\right)\right)\geq 0$

$ \therefore (n-1)a_{1}^2\geq 2na_0a_2$


A bouncy ball is dropped from a helicopter flying at a height of 1,500 feet. Each time the ball hits the ground it bounces straight back up 2/25 of the height from which it just fell. How many feet will the ball have traveled when it hits the ground for the fourth time? Express your answer as a decimal rounded to the nearest thousandth.

$ 1500\plus{}2\cdot\frac{2}{25}(1500)\plus{}2\cdot\frac{2}{25}\cdot\frac{2}{25}(1500)\plus{}2\cdot\frac{2}{25}\cdot\frac{2}{25}\cdot\frac{2}{25}(1500)$.  It's the first few terms of a geometric series.


But Texas History is probably an elemental part of your curriculum, so it follows that it should be part of the Academic Decathlon.

Academic decathlon etc is supposed to cover topics ranging from Economics to History to Math...so it makes sense for them to put Texan history questions in there...no, I don't do decathlon although I'm trying (vainly) to convince my school that we should try for it...


I think (it's just a suppostion) that I found a way of proving inequalities ( a very bug majority of them).
It is really nice. Till now, almost (not all) the IMO ineqs i've seen are easily proved. I think that after that, ineqs dont become chalenging.
Just A Supposition.
:) :D

and what is this miraculous method you found ? :?

Yeah I'd like to know that method too :D
Or is it a secret :? ?

So, Dust what is it? :) Now you've been to the IMO, had an inequality to solve, got 7p on it (but you've fought for them :D), so what's the miraculous method? :P


inequalities
geometry
inequalities proposed

Hi all member of ML! this is my first post anh i hope you like it
        Let ABC is a triangle has BC=a,CA=b,AB=c and m,n,p are the positive number.Prove that
  $mab+nbc+pca\geq\frac{12mnp\sqrt{3}S}{m^{2}+n^{2}+p^{2}}$
 with S is the area of ABC

I can prove the stronger
$mab+nbc+pca\geq\frac{36\sqrt{3}mnp}{(m+n+p)^{2}}S$
:D

I can prove the stronger with $\frac{12\sqrt{3}mnpS}{mn+np+pm}$ :D  :D

okie that's nice and I can prove more
$mab+nbc+pca\le\frac{(mn+np+pm)^{2}}{mnp}R^{2}$
:D :D :D

Ok that you must post your solutions of your stronger :D thanks
  i hope this is true
        $mab+nbc+pca\geq 4\sqrt{mn+np+pm}S$

I don't know your last inequality true (I think it can't ) but I can prove the stronger than Albanian Eagle (and weaker than yours) :D
$mbc+nca+pab\ge\frac{4\sqrt{3}(mn+np+pm)}{m+n+p}S$
now I don't have time, i will promise to post its solution later.

gemath post your solutionplease i can't wait more


Let $ ABC$ be a triangle, $ D$ midpoint of $ AB$ and $ E$ on $ BC$ such that $ BE\equal{}2.CE$. If $ \angle CDA\equal{}\angle BAE$, find $ \angle BAC$.

[hide]
call the intersection of CD and AE, P.
draw the line from D parallel to AE, which intersects CB at a point F.
CP=PD by similarity
PD=AP because triangle APD is isosceles
because the median from A to CD is half the length of CD, CAD is right

[/hide]


mathleague.org
Mandelbrot Competition
National Assessment
\/closed

I think that a Mandelbrot forum would be a nice addition to the AoPS forum.
What do you guys think of this?

First of all: What's Mandelbrot?
Second of all: Would it be really useful? why?
Third of all: Would people have reasons to post productively there?

Most discussion of Mandelbrot in the past has taken part in the "Other US Contests and Programs" forum http://www.artofproblemsolving.com/Forum/index.php?f=136 which is a perfectly good place for it.  Other problems have been posted in Intermediate or Pre-Olympiad (or, more inappropriately, in some of the middle-school fora).

[quote="isabella2296"]Pacman: [url=http://www.mandelbrot.org/]The Mandelbrot Competition[/url] was founded by Sam Vandervelde, Richard Rusczyk, and Sandor Lehoczky. It's a high school competition where participants can compete either individually or in teams. Currently, Sam Vandervelde runs the competition, Richard Rusczyk manages AoPS(as you should know :P ), and Sandor Lehoczky has a career on Wall Street.[/quote]
Aah *facepalm* those are the disadvantages of being non-american.
mm... JBL is right; there are many math competitions, and if you did create a forum for every math competition, that would be overexcessive.

I really agree- There should be a separate thread for Mandelbrot. I actually wrote about this one time too. Should/Can we pm the administrators about the possibility?

You can use the Contests forum for that: http://www.mathlinks.ro/index.php?f=136 (scroll down under the subforums to find the general threads.) There is no need to create a forum for just a handful of topics that might pop up.

The Mandelbrot is a major competition that many people are interested in.  Currently there are fora for Rocket City Math League (four posts since October), National Assessment and Testing (one topic in the last year), mathleague.org (no posts since June), and NOML, which hasn't even existed for years.  I would recommend deleting some of those and adding a Mandelbrot forum, which will probably attract a fair amount of interest during the competition season.

You've made your point. I've created the forum.


Prove that if $ x,y,z>0$ then \[ \frac{1}{3}(x\plus{}y\plus{}z) \left(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z} \right) \geq \frac{x}{y}\plus{}\frac{y}{z}\plus{}\frac{z}{x}.\]

[quote="Mîndrilă Claudiu"]Prove that if $ x,y,z > 0$ then
\[ \frac {1}{3}(x \plus{} y \plus{} z) \left(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \right) \geq \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x}.
\]
[/quote]

$ F(x,y,z)\equal{} \frac {1}{3}(x \plus{} y \plus{} z) \left(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \right) \minus{} \left(\frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \right).$
I think there must be other constrain to this inequality because it may not be true
For example:
$ x\equal{}\frac{5}{28} , y \equal{} \frac{89}{28} , z\equal{}\frac{33}{28}$ then $ F \equal{} \frac{38416}{44055}$
But
 $ x\equal{}\frac{13}{14} , y \equal{} \frac{55}{14} , z\equal{} \frac{27}{14}$  then $ F \equal{}\frac{\minus{}9604}{57915}$

Indeed, it should be reversed. I`m sorry, I guess I was a bit tired  :blush:

Actually, I think [b]Gibbenergy[/b] showed that it doesn't work even if it is reversed.

This is believable. Multiply the thing by 3 and subtract what was originally the RHS from both sides. You get:

$ 3 \plus{} \left( \frac{x}{z} \plus{} \frac{y}{z} \plus{} \frac{z}{y} \right) \leq 2 \left( \frac {x}{y} \plus{} \frac{y}{z} \plus{} \frac{z}{x} \right)$

Note that $ 3 \leq \left( \frac {x}{y} \plus{} \frac{y}{z} \plus{} \frac{z}{x} \right)$ by AM-GM, but either $ \left( \frac {x}{y} \plus{} \frac{y}{z} \plus{} \frac{z}{x} \right)$ or $ \left( \frac{x}{z} \plus{} \frac{y}{z} \plus{} \frac{z}{y} \right)$ can be bigger by symmetry.

What is the flaw in stating that:$ \frac{1}{3}(x+y+z)(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \geq \frac{1}{3}.9=3(*)
\leftrightarrow \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq \frac{1}{3}(x+y+z)(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) \geq 3
\leftrightarrow \frac{x}{y} + \frac{y}{z} + \frac{z}{x}\geq 3 (*)$

Where both of the (*)'s are true by AM-GM inequality as $ x, y, z$ are positive reals??  :huh:

[NB: Sorry but I am unable to operate LaTeX correctly and so am fearful that I might make a casting mistake. Hence I refuse to use it until I am certain that I can use it flawlessly!]

Polta: the fact that both sides are greater than or equal to $ 3$ does not proove that one is greater that the other

BTW, Latex is very easy to use, just put $ \$$ signs around everything...[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php[/url] this will show you the basic commands

Also, perhaps there should be a restriction that $ abc\equal{}1$, in which case the equality is easily proven after substituting $ x\equal{}\frac{a}{b}, y\equal{}\frac{b}{c}, z\equal{}\frac{c}{a}$...

If you mean $ xyz\equal{}1$, then since the inequality is homogeneous you can assume ir wlog. So, show us your solution with it :P


algorithm
Princeton
college
geometry
3D geometry

1. What does not work as well as you would like it to work?
2. What job(s) would you like to see solved?
3. What problem(s) would you like to see solved?
4. If you could invent something to make your life easier, what would you invent?
5. What is the most annoying problem
at home?_________________
at school?________________
at work?__________________
at the airport?____________
on the road?_______________
at the supermarket?____________
at the bank?_______________
at the mall?____________
at the________________

Serious answers please...

1. computer programs
2. the U.S. presidency
3. the middle east
4. computers that turned on faster and froze less
5. home: homework
school: people who take dating casually
work: ?
airport: long lines
road: traffic jams
supermarket: people who dress in revealing clothes
bank: the hostile atmosphere
mall: girls who flaunt their bodies because they have nothing better to show, and so they take pride in being viewed as objects

Is this only for US residents? If not:

1. Kyoto protocol
2.
3. religuous fundamentalists / global warming
4. universal file format
5. school: too many subjects (14)
    road: rude and careless drivers
    supermarket: unpolite personel
    bank: conceited personel
    mall: females that dy their hair to be blonde and put to much make-up  and then think that   everybody looks at them

Deimos, when it wasn't mentioned what nationalities were required, it is by default not necessary to be from the USA, not the other way around :wink:

1.
2. job solved? I'm sorry:  I don't understand the question.
3. on an international level : the Israeli-Palestinian issue (with a right of return for Palestinians)
  on a national level (to be blunt) : Belgium
 on a local level : the fact :mad:  that anyone can steal a bike here and victims are considered "losers"
4. something that would allow me to track down stolen bikes  :mad: without the perpetrator being able to remove it
5.
school : professors who get a poor evaluation every year and don't mind
work : ?
airport : that a USA officer in Toronto got mad at me when I manipulated his thumbprint device, even though it was obvious the thing was unstable
road : that I am going on foot because my bike has just been stolen... and that the city of Ghent doesn't have enough bikepaths
supermarket : noisy, crying children
bank : no problems, but they do mail me "fun youngster magazines" :|
mall :? that would be about the same as supermarket here :huh:

1. My printer
2. I'm not sure what you mean by jobs
3. War.
4. Can you invent more interesting teachers.
5. Home: Boring homework
School: Teachers that assign boring homework
Work: N/A
Airport: Taking off your shoes for security
Road: Pesky police officers that don't let 13 year olds drive
Supermarket: A disturbing lack of free samples
Bank: I love my bank, they give me free popcorn on fridays, no complaints there.
Mall: Condescending salespeople

1. Maple's simplification algorithm
2. Preparation for Princeton University Math Contest [gah so much work]
3. The closed-mindedness and apathy of the general populace
4. A teleportation device [long range, hopefully]
5.
home: [dorm] Right now, no AC; [home] fighting with sister over computer use
school: I dunno, I like Princeton
work: n/a
airport: airplanes are too small. I am always uncomfortable in the seats, and sometimes in the tiny planes I have to bend over in the aisle and if I straighten a bit my shoulders will touch the ceiling...
road: other drivers
supermarket: on coupon tripling day they won't triple our internet coupons so what we expected was $\$$10 of groceries turned out to be $\$$40. OWNED
bank: I can't remember my PIN so I can't use an ATM
mall: rank commercialism

[quote="fredbel6"]Deimos, when it wasn't mentioned what nationalities were required, it is by default not necessary to be from the USA, not the other way around :wink:
[/quote]
you are right, this is an international forum.  :)

[quote]

 on a local level : the fact :mad:  that anyone can steal a bike here and victims are considered "losers"
[/quote][quote]
4. something that would allow me to track down stolen bikes  :mad: without the perpetrator being able to remove it
[/quote][quote]
road : that I am going on foot because my bike has just been stolen... and that the city of Ghent doesn't have enough bikepaths
[/quote]

Somebody stole your bike?   :o  Police think that you've lost it?    Well, I thought that croatian police is the only one that do nothing. :mad:

Yes, I started as a Mathlinker, I still log in through [url]http://www.mathlinks.ro[/url], if it's not an international forum...it's Romanian.  :)

Uhm, first of all you need to know that here only rich or posh students go to high school or the university in their own car (btw you have to be 18 anyway).  University cities wouldn't survive every city driving up and down in his own car anyway.  The fact that most of Flanders looks about as flat as this encourages that (no wonder we used to be part of the "United Netherlands"):wink:
[img]http://www.rudidebruyne.be/Rudis%20fotos/Landschappen/Landschap%20Polder%20Vlaanderen/Polder%20Vlaanderen%20Landschap%20(4).jpg[/img]
I'm 21 and I've already lost three bikes due to theft.  You can avoid it, if you are willing to buy a really expensive lock, more expensive than the bike itself.  I just went to the police station and they wrote it down and that was it (I was lucky, in other places they ENCOURAGE you to go steal one yourself, I guess you see where the problem comes from).  But I don't believe in that.  I wasn't raised that way.  I believe in sweet, ice cold revenge, and I might get it once he figures out how bad the brakes are. :diablo:

[quote="Xevarion"]You can get several 9V batteries and wire them to a step-up transformer and every time you leave the bike, wrap the live wire from the high voltage end on a metal part. Anyone who touches the bike w/o first pulling off the wire gets zapped! :D :D[/quote]
Not only would the police consider me a violent criminal if I did that.... I would even agree with the police in that case:
You must understand that Flanders is a biking region, and bike parkings are often crowded like : [img]http://www.times-up.org/images/ed2004-bike-parking.jpg[/img]
It's impossible to get a bike in or out without touching other bikes. :)

[quote="Xevarion"][quote="fredbel6"]
Not only would the police consider me a violent criminal if I did that.... I would even agree with the police in that case:

It's impossible to get a bike in or out without touching other bikes. :)[/quote]

Sorry. :( My thought was more like NYC: [url=http://static.flickr.com/42/86236769_f09ee7d224_o.jpg]example[/url]. People chain them to poles. Usually there aren't many together.[/quote]

I log in from mathlinks page too.

xeravion, why is that bike full of paper-wasted garbagge

[quote="deimos"]
xeravion, why is that bike full of paper-wasted garbagge[/quote]
Because people have this weird urge to drop their garbage on the ground, in rivers, alongside the roads, busses, a bike's bag.... anything is better than a plain old litter bin.

1. my computer screen
2. what do you mean?
3. Hodge conjecture
4. a macintosh that can operate windows programmes.
5. my brother
backround noise
I don't work.
I never go to the airport.
boredom
I almost never go to the mall.
I need a better question.

1. my musical instrument.......sigh, I totally need a new one.
2. Jobs in underdeveloped countries...
3. THe world needs to be cleaner...devdloped countries try to pollute less...and groups should be sent to underveloped countries to bring cleaner water and stuff.
4. Something that gives me more energy, something that will keep me awake and alert even on 6 hours a day...
5. at home--grr, my home itself
    at school--being so tired throughtout the day since half the week I can get 5-7 hours sleep at most...
    at work? well, I don't have a professional career yet..
    at the airport--when it's organized, when flights are delayed etc
    on the road--acciednts.
    supermarket--um, I don't know...
    bank--i rarely step inside banks...
    mall--I totally agree with wobster... and also when people spend WAY too much time at the mall.

OH, and if you want serious answers, this should probably be moved to Round Table

1. My mind
2. Ted Stevens and Condoleezza Rice. George Bush is fine.
3. More abortion rights, legal gay marriage, basic health care for everybody, a balanced budget, more competent teachers, corruption, etc.
4. A compact backpack that could roll since my backpack is 30+ pounds and rapidly growing
5.
My mother
My Spanish teacher.
They don't allow magnets on the airplane
Snow is shovelled too quickly
Incompetent Cashiers
Incompetent Tellers
Overpricing and lack of availability

Nutz... The new macbook can use Windows if you buy something extra I think :)

1. What does not work as well as you would like it to work?
my rubiks cube i hate the stickers
2. What job(s) would you like to see solved?
people gone forever
3. What problem(s) would you like to see solved?
how to see into the 4d world
4. If you could invent something to make your life easier, what would you invent?
 the super nice rubik's cube that stays forever and i'd sell it for the price of a grand shimmel
5. What is the most annoying problem
at home? the fact i have to pack up all my math after i'm done
at school? the fact that everyone doesn't celebrate Moebius Strip Independence Day or Klein Bottle Recognition Day..
at work(swimming)? People oh and the fact you can't talk to a person without them saying, "shut up you're freaking people out"
at the airport? people staring
on the road? people who bike and school buses that... oh and people with "w" stickers
at the supermarket? people
at the bank? lollipops
at the mall? people
at the library? librarians

[quote="funcia"]
on the road? people who bike and school buses that... [/quote]
 :mad:  Care to elaborate?
You don't seem to be that tolerant :|

[quote="funcia"]I wasn't trying to say anything against you(i hope.) Bikers who bike on the road in one of the US's most dangerous biking spots... they ruin traffic more for the ATL. Now in Savannah at SCAD-I'd understand since it's all bikes almost near the college from what I saw...[/quote]
It's not their fault.  There should a bikepath.

There is. I bike there. I'm not completely against bikes....they just break every 5 weeks.

I am completely against bikes, partly because they are dangerous (seen that way, the bike-thieves are philanthropists), but mainly because I am the only one among  ALL my friends who can't ride a bike.  :(

Hot news: A policeman - a fat good-for-nothing policeman - brought back fred's bike!  :D Howzzat?

I learnt a valuable lesson : never tell Bubka something before you wrote it on Mathlinks... :maybe:

Yes, this afternoon, a policeman (who wasn't that fat and certainly not good-for-nothing) drove by our house and gave me back my bike.  He found it (not locked) in a town nearby.  He said it was the third time he found one there.

 :D So thank you, police!! :10:

(My back tire is flat, I suspect that that villain vandalized it on purpose  :mad:  :| )


i think there must be some another people other than US citizen must take USAMO and can go to MOSP. Now they're not taking us because we're not american. Then why we have taken AMC and AIME, though my AIME score is 4. My point is not that. If we'll not take USAMO and MOSP, why we're taking AMC and AIME?

By non American citizen do you mean not residing in the US or residing in the US and not possesing citizenship?

From what I hear you only need to be a legal resident of the US.

[quote="beta"][quote="paladin8"]Well, it is the [b]USA[/b] Math Olympiad. And MOP is for training the [b]USA[/b] IMO Team. Why would they take non-US citizens if they aren't allowed to be part of the team? I don't see any logic in that except to train other countries' teams/people.[/quote]

I think denizuzun's asking why did US even let international students to take AMC/AIME, which I'm not sure why's that either.[/quote]

I can only speculate that there is a small profit margin involved. A better reason is that many people like taking contests - why else would people on the east coast of the U.S. take the Asian-Pacific Math Olympiad? Open environments are usually good ideas, and not just in math. Of course, at some point they cannot help but focus on developing the U.S. IMO team, so it becomes gradually more slanted towards the U.S.

I think that U.S. resident thingy is good, because people dont immigrate to U.S. and possess a residency card unless they intend to go back to their country, which i dont see it as possiblility for anybody. And, U.S. is one of the countries in the world with the best secondary school education and opportunities to receive merits through scoring well on the tests such as USAMO. It is a great opportunity for them, as well as for United States as a country, because then they will have a wide range of capable students to choose from. Recent results of non-born U.S. representatives have shown that non-US born are capable of competing at the same level.

[quote]And, U.S. is one of the countries in the world with the best secondary school education[/quote]

i don't know if i'd agree with that ;)

As in the number of students to which education is available to, it is one of the best, but as far as how good the education itself is....i don't agree

Well, Mr. Dimitar, the word "best" generally refers to quality. Yes? :P

[quote="seungjun"]I think you only need to be a US permanent resident (green card) to take the USAMO.[/quote]

Thank god. Only got that thing last year...

[quote="DPopov"]Hey, some of us aren't as smart as you Miss Apprentice  :P  :P[/quote]

You're just asking to be pwned at ARML. Just asking for it...

Where's Samson when you need him...

Hey Samson, when you read this go pwn Dimitar.

hahaha, and how do u propose he do that?

I prophesize that in five minutes, either an admin or a powerhungry moderator will say "This is spam. Go to your room, sit in the corner, and never do this again."

About the title: It's not true that non-US citizens "must" take the AMC. This applies for everybody else, too :p .

A conjecture:

A problem with the Asian Pacific Mathematical Olympiad is, that due to the time-zones, it is unfeasible to have all students write the contest at the same time.  This seriously jeopradizes the validity of the results.

It is well known that students in some Asian countries well exceed most of the top American and Canadian students.  If the United States of America Mathematical Olympiad allowed students in Asian countries to participate, they will be writing the contest after students in Canada and the United States of America.  The Mathematical Association of America and/or its affiliated sponsors will question the awarding of cash prizes to the students in those countries.  Furthermore, said affiliates have no interest in international students with regards to "handing out money", and would rather keep the money in the area.

In short: Asian students cannot write the USAMO because they can't write it at the same time.

[quote="xxreddevilzxx"]And, U.S. is one of the countries in the world with the best secondary school education[/quote]
I'm sure most math teachers in US would really disagree with this statement :)

[quote="xxreddevilzxx"]Recent results of non-born U.S. representatives have shown that non-US born are capable of competing at the same level.[/quote]The most recent result shows that the best contestant in the US IMO Team last year was a .... Russian guy, a [b]non-US citizen[/b]. :)

Of course this is not a spam. And none of the admins or moderators told me something like that.

I just want to show something unfair.

Also about English: english is not my native lang. So as if I'm not a langage prof; I cannot understand all the grammar kinda stupid things in 1,5 year. So I'm not a linguistic man and I'm a mathematician; and sorry for those Englishman.

[quote="Valentin Vornicu"][quote="xxreddevilzxx"]And, U.S. is one of the countries in the world with the best secondary school education[/quote]
I'm sure most math teachers in US would really disagree with this statement :)
[/quote]
i meant to say that in terms of opportunity, not necessarily as of quality....United States' education is very available and open. And we as students can explore many educational options and easily access other extracurricular options.

I used to live in Korea. The education there, in my mind, is not the best for students. Sure, they push students to work hard in middle and secondary school, and that has shown in years, of how smarter they are compared to foreign countries' students of same ages, noticeably that to United States. However, since they study so hard in their youth, and barely enjoy any extracurricular and community services and such, by the time they go to college, they quit studying. My mom told me, in the recent research, 90% of the college students have studied just to get accepted. Students picked to the Olympiad are mostly first-year high schoolers; by their second yr, they are too busy prepping for college.

back to the original question: should Non-US citizens be allowed to take USAMO? I think yes, because it is one of the few opportunities given to them. But can they be on the US TEAM? no, i dotn think so, because if you do well here on U.S. tests, your country will know and they sure will scout you back to their team.

[quote="h_s_potter2002"]Just a question, Isn't the USAMO required to be taken if you want to be on the Canadian team?  Because then, you don't even need to live in the US to take the USAMO.[/quote]

Well, plenty of Canadians take the USAMO.  In fact, some of them have even made Winner and HM.

Just to clarify on the issue of USAMO for Canadian team selection. USAMO is one of three exams that the Canadian IMO committee uses to select the team. If a Canadian student writes the USAMO, then the Canadian jury will need to look at the script (as opposed to just taking the scores from the American judges). If the Canadian jury feels that a student has the potential in making the team, they may let the student write USAMO unofficially (that is, only the Canadian jury marks it).

[quote="white_horse_king88"]Cool! I'm canadian![/quote]

Wait, why is it that I have a Canadien citizenship???

A U.S. Green Card.

And a Chinese passport?


function
inequalities
inequalities proposed

Find the smallest real $ K$ such that for each $ x,y,z\in\mathbb R^{ \plus{} }$:
\[ x\sqrt y \plus{} y\sqrt z \plus{} z\sqrt x\leq K\sqrt {(x \plus{} y)(y \plus{} z)(z \plus{} x)}
\]

We want to find the [b]smallest[/b] K.

I claim $ K\equal{}3/(2\sqrt{2})$

$ 8(x\sqrt{y}\plus{}y\sqrt{z}\plus{}z\sqrt{x})^2 \le  9(x\plus{}y)(y\plus{}z)(z\plus{}x)$

$ 8x^2y\plus{}8y^2z\plus{}8z^2x\plus{}16xy\sqrt{yz}\plus{}16yz\sqrt{zx}\plus{}16xz\sqrt{xy} \le 9\sum_{sym}x^2y \plus{}18xyz$

$ 16xy\sqrt{yz}\plus{}16yz\sqrt{zx}\plus{}16xz\sqrt{xy} \le x^2y\plus{}y^2z\plus{}z^2x\plus{}9y^2x\plus{}9z^2y\plus{}9x^2z\plus{}18xyz$

By AM-GM $ z^2x\plus{}(9)y^2x\plus{}(6)xyz\ge 16\sqrt[16]{z^2x*y^{18}x^9*x^6y^6z^6}\equal{}16xy\sqrt{xz}$

Sum up cyclically. We can get equality when x=y=z=1, so we know that K cannot be any smaller.

Or you could do simple AM-GM like I did... lol.

My solution, also with AM-GM:  :lol:

We want to find the smallest K such that

$ (x\sqrt {y} \plus{} y\sqrt {z} \plus{} z\sqrt {x})^2 \leq K^2(x \plus{} y)(y \plus{} z)(z \plus{} x)$

But
$ (x\sqrt {y} \plus{} y\sqrt {z} \plus{} z\sqrt {x})^2 \equal{} \sum_{cyc}x^2y \plus{} 2(\sum_{cyc} xy\sqrt {yz}) \leq \sum_{cyc}x^2y \plus{} 2(\sum_{cyc} \frac {xyz \plus{} xy^2}{2} ) \equal{} (x \plus{} y)(y \plus{} z)(z \plus{} x) \plus{} xyz \leq (x \plus{} y)(y \plus{} z)(z \plus{} x) \plus{} \frac {1}{8}(x \plus{} y)(y \plus{} z)(z \plus{} x) \equal{} \frac {9}{8}(x \plus{} y)(y \plus{} z)(z \plus{} x)$

Therefore, $ K^2\geq \frac {9}{8} \rightarrow K \geq \frac {3}{2\sqrt {2}}$ with equaltiy if and only if $ x \equal{} y \equal{} z$.

[quote="Omid Hatami"]Find the smallest real $ K$ such that for each $ x,y,z\in\mathbb R^{ + }$:
\[ x\sqrt y + y\sqrt z + z\sqrt x\leq K\sqrt {(x + y)(y + z)(z + x)}
\]
[/quote]

My solution using Cauchy-Schwarz:

We have
\begin{align*}LHS & = \sqrt x\sqrt {xy} + \sqrt y\sqrt {yz} + \sqrt z\sqrt {zx} \\
& \le\sqrt {\left(x + y + z\right)\left(xy + yz + zx\right)} \\
& \le\frac {3}{2\sqrt 2}\sqrt {(x + y)(y + z)(z + x)}\end{align*}
where the last inequality follows from $ 8(x + y + z)(xy + yz + zx)\le 9(x + y)(y + z)(z + x)$, which is well known.

[hide=storage] I claim $min\{K\} = \frac{3}{2\sqrt{2}}$ .
[color=#006D77]equality case[/color]: $x=y=z=1$
[color=#E29578]Why It works:[/color] Square both sides to get
$(\sum_{cyc} x\sqrt{y} )^2 \leq  \frac{9}{8} \prod_{} (x+y) $
we have $\frac{9}{8} \prod_{}(x+y) \geq (\sum x)(\sum xy) \geq_{\text{C-S}} (\sum_{cyc} x\sqrt{y} )^2 $   $\blacksquare$

[quote=Allnames][quote="rofler"]We want to find the [b]smallest[/b] K.

I claim $ K \equal{} 3/(2\sqrt {2})$

$ 8(x\sqrt {y} \plus{} y\sqrt {z} \plus{} z\sqrt {x})^2 \le 9(x \plus{} y)(y \plus{} z)(z \plus{} x)$

$ 8x^2y \plus{} 8y^2z \plus{} 8z^2x \plus{} 16xy\sqrt {yz} \plus{} 16yz\sqrt {zx} \plus{} 16xz\sqrt {xy} \le 9\sum_{sym}x^2y \plus{} 18xyz$

$ 16xy\sqrt {yz} \plus{} 16yz\sqrt {zx} \plus{} 16xz\sqrt {xy} \le x^2y \plus{} y^2z \plus{} z^2x \plus{} 9y^2x \plus{} 9z^2y \plus{} 9x^2z \plus{} 18xyz$

By AM-GM $ z^2x \plus{} (9)y^2x \plus{} (6)xyz\ge 16\sqrt [16]{z^2x*y^{18}x^9*x^6y^6z^6} \equal{} 16xy\sqrt {xz}$

Sum up cyclically. We can get equality when x=y=z=1, so we know that K cannot be any smaller.[/quote]
There are many way too kill it
Use pqr
Or Use ABC theorem :Fix $ x\plus{}y\plus{}z,\sum xy$ then it is a crease function with $ xyz$ so we only need check with $ x\equal{}y$after some small caculator we will find $ K$ followed  above[/quote]

I am new to inequalities and would like to know what is that treorem you called, the ABC theorem

I searched "ABC theorem" and didn't find anything about it. I got something about the ABC conjecture (which is an interesting one) but I know that's not what's used here since that remains unproven.

Search uvw method instead. You’ll find similar things.
(But as someone mentioned in some other Iran forum, you cannot use UVW,ABC method in Iran.)

Oh, right. I've heard of the uvw method being mentioned before, and I felt like that was what was being mentioned, but I forgot which 3 letters they were.

Using Cauchy Schwarz Inequality,
$$
\begin{aligned}
x\sqrt y + y\sqrt z + z\sqrt x
&\leqslant\sqrt{(xy+yz+zx)(x+y+z)}\\
&=\sqrt{(x+y)(y+z)(z+x)+xyz}\\
&\leqslant\sqrt{\frac98(x+y)(y+z)(z+x)}\\
&=\boxed{\frac{3\sqrt 2}4}\sqrt {(x + y)(y + z)(z + x)}.\blacksquare
\end{aligned}$$

A simple argument. Taking $x=y=z$, we get $K\ge K^* = 3/(2\sqrt{2})$. We now show the inequality with $K^*$. Without loss of generality, let $x+y+z=1$. Using concavity of $x\mapsto \sqrt{x}$, we have $x\sqrt{y}+y\sqrt{z}+z\sqrt{z}\le \sqrt{xy+yz+zx} = \sqrt{(xy+yz+zx)(x+y+z)}$. From here, the result follows using the well-known $9(x+y)(y+z)(z+x)\ge 8(x+y+z)(xy+yz+zx)$.


Has anyone heard back from the HCSSiM program yet? If yes, when did you guys send your Interesting Test, and in what form did you hear back in (i.e. e-mail/snail mail)?

Thanks!

Mel, there are no patterns to be discerned from the data you desire...

HCSSiM does rolling admissions, but not in a regular sort of rolling way.  The order in which responses are sent out has little-to-no relation to the order in which we receive Interesting Tests.  Frequently we receive an IT weeks before a sponsor recommendation arrives, for example.  Additionally, there are irregularities in response time introduced by the state of one of the Co-Directors' desks.

As for email vs. snail mail, all responses are sent by snail mail but some are sent by email as well.  We use email when there's a deadline to be met, and more frequently as the summer approaches.

All that said,
(a) I think we've responded to approximately half of the applicants whose ITs we've received.
(b) If there's a question about a particular applicant/application, please PM me and I'll look into it.
(c) We are still taking applications, and welcome their submission at http://www.hcssim.org .


Given

$ x\plus{}y\plus{}z\equal{}1$
$ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}2$
$ x^{3}\plus{}y^{3}\plus{}z^{3}\equal{}3$,

show that

$ xy\plus{}yz\plus{}zx\equal{} \frac{\minus{}1}{2}$

and that

$ x^{2}y\plus{} x^{2}z\plus{} y^{2}x \plus{} y^{2}z \plus{} z^{2}x\plus{}z^{2}y\equal{} \minus{}1$

and deduce that

$ xyz\equal{}\frac{1}{6}$.

Hence find $ a,$ and $ c$, independent of $ n$, such that

$ S_{n\plus{}1} \equal{} aS_{n} \plus{} bS_{n\minus{}1}\plus{}cS_{n\minus{}2}$,

where $ S_{n} \equal{} x^{n} \plus{}y^{n}\plus{}z^{n}.$

Andrew Tom, we can see the same post.[url=http://www.mathlinks.ro/viewtopic.php?t=313241]here[/url].

Yes, I didn't realise that this one had been sent as my computer seems to be slow and when I got back, it said "webpage not available". Sorry.


$ Question 1$
Use the four numbers $ 1, 2 ,3, 4$ each only once without clubbing the numbers together and with any mathematical signs/notation you like and arrive at an expression to make a $ 6$ digit prime number.

$ Question 2$
Now make a $ 7$ digit prime number.

$ Question 3$
Now try above without using $ \plus{}$ and $ \minus{}$ signs.

$ Question 4$
Only symbols you can use are $ ($ and $ )$ and $ !$ and can concatenate digits.

[quote="CircleSquared"][hide="1 unconfirmed"]
$ (3!!)^2 \plus{} 4! \minus{} 1$

[/hide]

[hide="2 unconfirmed"]
$ (3!!\cdot4)^2 \plus{} 1$
[/hide][/quote]
The question asks for 7-digit numbers; both of those are 6-digit.

The notation $ n!!$ is known as the double factorial, which is not the same as $ (n!)!$. I won't go into details, but the way you wrote it would be $ 3!!\equal{}3\cdot1\equal{}3$.

[quote="i_like_pie"][quote="CircleSquared"][hide="1 unconfirmed"]
$ (3!!)^2 \plus{} 4! \minus{} 1$

[/hide]

[hide="2 unconfirmed"]
$ (3!!\cdot4)^2 \plus{} 1$
[/hide][/quote]
The question asks for 7-digit numbers; both of those are 6-digit.

The notation $ n!!$ is known as the double factorial, which is not the same as $ (n!)!$. I won't go into details, but the way you wrote it would be $ 3!! \equal{} 3\cdot1 \equal{} 3$.[/quote]

May be the mistake there, but while I calculated $ n!!$ i took it to mean $ (n!)!$ which I am sure is the same thing meant by the earlier post.

Yes, I interpreted it that way also (double factorials are rarely used), but the numbers aren't prime anyway.

Wait wait wait how is $ (3!!\cdot4)^2 \plus{} 1 \equal{} 8294401$ not 7 digit?
Yes I meant $ (3!)!$, I didn't know $ 3!!$ was something different.

EDIT: New unconfirmed: $ 4!^{3 \plus{} 2} \minus{} 1 \equal{} 7962623$

CONFIRMED # 2: $ 4\cdot\left(\left(3!\right)!\right)^2\plus{} 1 \equal{} 2073601$

[quote="CircleSquared"]Wait wait wait how is $ (3!!\cdot4)^2 \plus{} 1 \equal{} 8294401$ not 7 digit?[/quote]
I said they weren't prime; I wasn't thinking straight when I said that was 6-digit. :D

[quote="CircleSquared"]New unconfirmed: $ 4!^{3 \plus{} 2} \minus{} 1 \equal{} 7962623$[/quote]
$ 7962623 \equal{} 23\cdot346201$

Someone confirm or unconfirm $ (3! \minus{} 2)^4 \plus{} 1 \equal{} 331777$ for me, please.

w00t

Number three is impossible
Number four might be $ (3!)!\equal{}720$ concatenated with $ 4!2\equal{}48$ and $ 1$, or $ 720481$.

I know, I'm kind of cheating here by not using a multiplication sign.

Why do you say # 3 is impossible? Not that I can find a way to do it... :D

720481 is prime, and I think 2(4!) suffices the "(, ), and ! only". Nice job, again. :)

Btw, $ !n$ is called the subfactorial of $ n$, so once again your notation could've been mistaken for something else. :D

[quote="xxxyyyy"]$ Question 1$
Use the four numbers $ 1, 2 ,3, 4$ each only once without clubbing the numbers together and with any mathematical signs/notation you like and arrive at an expression to make a $ 6$ digit prime number.

$ Question 2$
Now make a $ 7$ digit prime number.

[/quote]

For the two first questions, note that, without restriction on the notations which can be used, any integer may be obtained with $ 1,2,3,4$ :

$ n\equal{}\frac{\ln(\frac{\ln(1\plus{}3)}{\ln(\sqrt{\sqrt{\sqrt{...\sqrt{4}}}})})}{\ln(2)}$ with exacty $ n$ square root signs.


abstract algebra
group theory
superior algebra
superior algebra solved

what is an example of an infinite Abelian group that is not cyclic,
and an example of a finite Abelian group that is not cyclic

$\mathbb{Z}^2$ and $\mathbb{Z}_2^2$ with the elementwise addition


find all positive integers x,y,z such that

xyz=5(x+y+z)

(My method might be too complex. Thanks for any helps.)

Since $x$, $y$ and $z$ are symmetric in the equation, we may without lost of generality assume $x \leq y \leq z$. Solving the equation for $z$, the result is

$(1) \;\; z \;=\; \frac{5(x \:+\: y)}{xy \:-\: 5}.$

Hence $xy \: \geq \: 6$. Moreover $z \: \geq \frac{x+y}{2},$ hence $xy \: \leq \: 15$ by (1). Therefore we have the following possibilities:

$\begin{tabular}{|c|c|c|c|c|}\hline xy & x & y & z & \\ \hline 6 & 1 & 6 & 35 & * \\ \hline & 2 & 3 & 25 & * \\ \hline 7 & 1 & 7 & 20 & *\\ \hline 8 & 1 & 8 & 15 & * \\ \hline & 2 & 4 & 10 & *\\ \hline 9 & 1 & 9 & 12,5 & \\ \hline & 3 & 3 & 7,5 & \\ \hline 10 & 1 & 10 & 11 & * \\ \hline & 2 & 5 & 7 & * \\ \hline 11 & 1 & 11 & 10 & \\ \hline 12 & 1 & 12 & 9 2/7 & \\ \hline & 2 & 6 & 5 5/7 & \\ \hline & 3 & 4 & 5 & * \\ \hline 13 & 1 & 13 & 8,75 & \\ \hline 14 & 1 & 14 & 8 1/3 & \\ \hline & 2 & 7 & 5 & \\ \hline 15 & 1 & 15 & 8 & \\ \hline & 3 & 5 & 4 & \\ \hline \end{tabular}$

The 8 "*"-marked solutions in the table above and their permutations give all positive integer solutions of equation (1)

Thanks, it looks nice. But is there an even simpler one?

I don't know, maybe this will be easier....
One of this numbers must be $\leq3$ - otherwise $1=5 \left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\leq 5\cdot \frac{3}{16}=\frac{15}{16}<1$.
So we can assume that $x\in\{1,2,3\}$. We have also \[xyz=5(x+y+z)\]  \[(xy)(xz)=5x^{2}+5xy+5yz\]  \[(xy-5)(xz-5)=5x^{2}+25\] For $x=1,2,3$ it's equivalent: $(y-5)(z-5)=30$, $(2y-5)(2z-5)=45$, $(3y-5)(3z-5)=70$. Now we can easily examine these cases and find all the solutions.


1)Solve  the following equiation!

[(2x-1)/3]+[(4x+1)/6]=(5x-4)/3

where [x]=x-{x}.

2)Solve the system :

x+[y]=20/3
y-[x]=21/5

thank you!!! bye bye  :)

these problems are not that hard!!!

here is a sol for the 1st:

[(2x-1)/3]+[(4x+1)/6]=(5x-4)/3  we can see that (2x-1)/3+1/2=(4x+1)/6
we denote (2x-1)/3=y and we get that x=(3y+1)/2

then we get to solve the following eqs:
[y]+[y+1/2]=(5y-1)/2. But we know that:
[y]+[y+1/2]=[2y] (HERMITE IDENTITY) and the eqs is eqiuvalent to:
[2y]=(5y-1)/2.

Lets denote (5y-1)/2=k integer. we get that y=(2k+1)/5 and x=(3k+4)/5.

[(4k+2)/5]=k (integer) and k<=(4k+2)/5<k+1
so we get that k is in {-2,-1,0,1,2} and x is in {-2/5,2,1/5,7/5,4/5}

cheers! :D :D :D

And now for the 2nd one:
we can write that:

x=[x]+a, a in [0,1) and y=[y]+b, b in [0,1)
we can write the sistem like so:
[x]+a+[y]=6+2/3
[y]+b-[x]=4+1/5 so we get that [x]+[y]=6 and [y]-[x]=4 and a=2/3 and b=1/5 and solving the system we get that: x=5/3 and y=26/5.


Engineers use the formula $ L\equal{}\frac{25T^4}{H^2}$ to find the crushing
load for square pillars. If $ T \equal{} 4$ and $ H \equal{} 8$, what is the value
of $ L$?


inequalities
triangle inequality
inequalities proposed

Let $ABC$ be a triangle $(BC=a,\ CA=b,\ AB=c)$ and let $x,\ y,\ z$ be any real numbers. Prove that:

$\sum a(x+y-2z)(x+z-2y)\le 0.$ (with equality iff $x=y=z$). A particular case.....$15Rr\le p^2+3r^2.$

This is really nice, since the proof is absolutely easy, but the inequality has some nice consequences.

Here is the proof: First denote p = y + z - 2x, q = z + x - 2y and r = x + y - 2z; then, we have p + q + r = (y + z - 2x) + (z + x - 2y) + (x + y - 2z) = 0, while the inequality in question becomes $aqr+brp+cpq\leq 0$. So we have to show the following fact:

[color=blue]If a, b, c are the sidelengths of a triangle, and if p, q, r are three reals satisfying p + q + r = 0, then $aqr+brp+cpq\leq 0$.[/color]

But this is actually easy: Since a, b, c are the sidelengths of a triangle, we have b + c > a by the triangle inequality; thus, $\left(\sqrt{b}+\sqrt{c}\right)^2=b+c+2\sqrt{bc}>b+c>a=\left(\sqrt{a}\right)^2$, so that $\sqrt{b}+\sqrt{c}>\sqrt{a}$, and similarly, $\sqrt{c}+\sqrt{a}>\sqrt{b}$ and $\sqrt{a}+\sqrt{b}>\sqrt{c}$. Hence, $\sqrt{a}$, $\sqrt{b}$, $\sqrt{c}$ are also the sidelengths of a triangle. Also, q, r, p are reals satisfying q + r + p = 0. Hence, by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=4357]the Kvant problem M1166[/url], we have $\left(\sqrt{a}\right)^2qr+\left(\sqrt{b}\right)^2rp+\left(\sqrt{c}\right)^2pq\leq 0$. In other words, $aqr+brp+cpq\leq 0$. And this is exactly what we wanted to prove. $\blacksquare$

  darij

The inequality becomes the relation:\[\sum (b+c-a)(y+z-2x)^2\ge 0.\]

[i]Darij[/i], your inequality is true generally:
If $x,y,z\in R$, then $a^2yz+b^2zx+c^2xy\le (x+y+z)^2.R^2.\ \ \ (*)$
The power of the point $M(x,y,z)$ with respect to the triangle $ABC$ is:
$p(M)=-a^2yz+b^2zx+c^2xy=OM^2-R^2\Longrightarrow \ (*)$. [i]With friendship, Levi[/i].


search
inequalities unsolved
inequalities

let $ x,y,z \in R$ such that $ x^2\plus{}y^2\plus{}z^2\equal{}2$. prove that
$ x\plus{}y\plus{}z \leq 2 \plus{} xyz$

there is a hint here: [url]http://www.mathlinks.ro/viewtopic.php?search_id=1641323056&t=155[/url]

[quote="reza1370"]let $ x,y,z \in R$ such that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2$. prove that
$ x \plus{} y \plus{} z \leq 2 \plus{} xyz$[/quote]
It 's here http://www.mathlinks.ro/viewtopic.php?t=32399


can you tell us the conditions of a, b, c satisfied?

Positive reals, then:

a(b+c) is always positive, and it is trivial

i think ou have lost one of the statements

May be $\sum a^5(a+b) \ge 0$ . But....it's very easy.


Could someone explain to me how to draw a right triangle using cosine, sine, tan and a compass? As well as a pencil and all that other stuff... I am quite confused and can't figure how to draw one... Also if you need denominators, use 100... Or, try and make an angle using cosine, tan, sine and a compass. I'm really confused, so also draw it on the computer if possible... Thanks

You don't necessarily need trig to draw right triangles. Instead, draw a circle with center $ O$ along with its diameter that has endpoints $ A$ and $ B$. Then pick a point $ P$ on the circle which is not a point on the diameter of the circle. Connect $ \overline{AP}$ and $ \overline{BP}$. You have a right triangle now with hypotenuse length $ AB$.


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Does there exist a bijective map $f:\mathbb{N} \rightarrow \mathbb{N}$ so that $\sum^{\infty}_{n=1}\frac{f(n)}{n^2}$ is finite?

This is very well-known. You can use the rearrangement inequality to show that the $n$'th partial sum is at least as large as $\sum_{i=1}^n\frac 1i$, so the answer is "no".

Yes, that was also what I had. I just wondered if it could be that easy... compared to some other number 2 questions :?

We only need to prove that: for $n\in\mathbb{N}^+$, $\sum^{3n}_{k=n+1}\frac{f(k)}{k^2}>\frac 19$.
There are at least $n$ numbers bigger than $n$ in $f(n+1),\cdots ,f(3n)$, so $\sum^{3n}_{k=n+1}\frac{f(k)}{k^2}>\sum^{3n}_{k=n+1}\frac{f(k)}{(3n)^2}>\frac{1}{9n^2}\cdot n\times n=\frac 19.$