inequalities
trigonometry
geometry
function
area of a triangle
Heron\u0027s formula
inequalities proposed

Prove that in any triangle the following inequality holds:

$\displaystyle \frac{a \sin A }{ \sqrt{(p-b)(p-c)}} + \frac{b \sin B }{ \sqrt{(p-a)(p-c)}} + \frac{c \sin C }{\sqrt{(p-a)(p-b)}} \leq 3 \sqrt{ 1+ \frac{R}{r}}.$

cheers! :D :D

It is clear that $\sum_{cyclic}{\frac{a\sin{A}}{\sqrt{(p-b)(p-c)}}}=\sum_{cyclic}{\frac{\sqrt{p(p-a)}a\sin{A}}{S}} (1)$ (using Heron's formula,where $S$ is the area of the triangle).On the other hand $1+\frac{R}{r}=1+\frac{abc\cdot{p}}{4S^{2}}=\frac{4S^{2}+abc\cdot{p}}{4S^{2}}=p(\frac{4(p-a)(p-b)(p-c)+abc}{4S^{2}})=p(\frac{a^{2}(p-a)+b^{2}(p-b)+c^{2}(p-c)}{4S^{2}}) (2).$
Now putting $(1)$ and $(2)$ in the original inequality we obtain  $\sum_{cyclic}{\frac{\sqrt{p(p-a)}a\sin{A}}{S}}\leq{\frac{3\cdot{\sqrt{p}}}{2S}\cdot{\sqrt{a^{2}(p-a)+b^{2}(p-b)+c^{2}(p-c))}}}$   or $\sum_{cyclic}{\sqrt{p-a}\cdot{a\sin{A}}}\leq{\frac{3}{2}\cdot{\sqrt{a^{2}(p-a)+b^{2}(p-b)+c^{2}(p-c)}}}.$
But after Cauchy we get that $\sum_{cyclic}{\sqrt{p-a}\cdot{a\sin{A}}}\leq{\sqrt{(a^{2}(p-a)+b^{2}(p-b)+c^{2}(p-c))(\sin^{2}{A}+\sin^{2}{B}+\sin^{2}{C})}}$, so we have to prove that  $\sin^{2}{A}+\sin^{2}{B}+\sin^{2}{C}\leq{\frac{9}{4}}$ and this is wellknown(I'll prove it if anybody don't know it).  :)
QED

[quote="Armo"]... we have to prove that  $\sin^{2}{A}+\sin^{2}{B}+\sin^{2}{C}\leq{\frac{9}{4}}$ ...[/quote]

it can be quickly proved using $\sin ^2x=\frac{1-\cos 2x}{2}$ , equation for cosine sum and known ineq $\cos A \cos B \cos C \leq \frac{1}{8}$
do you know other way ?

It's not as quick or as elegant, but here's another approach:

Use the fact that $\sin(x) = \sin(\pi - x)$ and take $A' = \mbox{min}(A,\pi-A)$, defining $B'$ and $C'$ similarly. We get $A', B', C' \in [0,\frac{\pi}{2}]$, and $0 \le A' + B' + C' \le \pi$. Let $f(x) = \sin^2(x)$. Since $\frac{d}{dx}(\sin^2(x)) = 2\sin(x)\cos(x) = \sin(2x)$, and $\frac{d^2}{dx^2}(\sin^2(x)) = 2\cos(2x)$, we have that $f(x)$ is convex over $[0,\frac{\pi}{4}]$ and concave over $[\frac{\pi}{4},\frac{\pi}{2}]$. We conjecture that the maximum of $f(A') + f(B') + f(C') = 3f(\frac{\pi}{3}) = \frac{9}{4}$. Note that $f$ is increasing over $[0,\frac{\pi}{2}]$, and that $f(x) \le 1$. Since, $f(\frac{\pi}{6}) = \frac{1}{4}$, it must be that if $f(A') + f(B') + f(C') > \frac{9}{4}$, each of $A', B', C'$ exceeds $\frac{\pi}{6}$. Now take

$g(x) := \frac{3}{4} + (x - \frac{\pi}{3})\frac{\sqrt{3}}{2}\, \forall x \in [\frac{\pi}{6},\frac{\pi}{3}];\, = f(x)\, \mbox{otherwise}$

It is clear that $g(x) \ge f(x)$ and $g''(x) \le 0\, \forall x \in [\frac{\pi}{6},\frac{\pi}{2}]$, but as $A',B', C' \in [\frac{\pi}{6},\frac{\pi}{2}]$, Jensen's inequality for concave functions establishes $f(A) + f(B) + f(C) \le f(A') + f(B') + f(C') \le g(A') + g(B') + g(C') \le \frac{9}{4}$.

Actually [b]Mequs[/b], my proof is the same as yours    :) .I'll try to find another one. :)


In a finite sequence of real numbers, the sum of any seven consecutive terms is negative, and the sum of any eleven consecutive terms is positive. Determine the maximum number of terms in the sequence?

1977 IMO, heres a link:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=367411#367411


Use vectors and vectors only to prove that the perperdicular bisectors of a given triangle are concurrent (as always).

Well I have solved this problem using lots of other methods, but I could not get through with this.

Can anyone help?

Call the intersection of the perpendicular bisectors of $BC,CA$ $O$ (:)). We have $(\overline {OC}+\overline {OA})\cdot \overline {CA}=0,\ (\overline {OA}+\overline {OB})\cdot\overline {AB}=0$ (the $\cdot$ is the dot product). This means that $|\overline{OC}|^2-|\overline{OA}|^2=0,\ |\overline{OA}|^2-|\overline{OB}|^2=0\Rightarrow |\overline{OB}|^2-|\overline{OC}|^2=0$, so $(\overline{OB}+\overline{OC})\cdot \overline{BC}$, which means that $O$ also lies on the third perpendicular bisector.


What did everyone think? The third was quite tricky.

Also, is the Open weighted more heavily than the other contests in determining camp selection?

Yeah it's weighed fairly heavily. I didn't do so great compared to other contests =/. I think $O(\sqrt(N))$ queries probably worked as well, at least that was my solution, though for some reason it is not working properly.

[hide="my results"]
  1  2  3  4  5  6  7  8  9 10
cheappal    *  *  *  *  *  *  *  *  *  *
connect     *  x  *  x  *  x  x  x  x
dining      *  x  *  x  x  x  x  x  x  x
[/hide]

I never do these contests properly, but I often open them up and have a look for an hour or so to see if I can solve anything. Hugely surprisingly, I managed to get cheappal and dining perfectly right in that time :S I'd like to see a solution to the other problem though; I'm never good at 'query' style problems.

I did rather well this time (silver)

[code]
            1  2  3  4  5  6  7  8  9 10 11 12 13
catchcow    *  *  x  *  x  *  *  *  *  x  *  *  x
fliptile    *  *  *  x  *  *  t  *  t  t  t  t  t
horizon     *  *  *  *  *  *  *  *  *  *
[/code]

Catchcow was just a silly mistake. When the farmer was at a higher number than the cow, I only initialized up to the cow, and my program made an empty file. The other x's were for similar array initialization problems.
Horizon was very easy, and I'm expecting it to be weighted around the same as catchcow, and much lower than fliptile. Although, a lot of people seem to have missed the priority queue usage, and did some other weird stuff.
Fliptile was really hard, and I got halfway, but missing the other half made my program look like a complete failure, so I just had the $O(2^{MN})$ solution, and I'm extremely surprised that they let me get away with 6 test cases (and I should've had test case 4 except that I think I was being stupid), since you can optimize it to $O(2^{\text{min}(M,N)}*other stuff)$.

[hide][code]cheappal    *  *  *  *  *  *  *  *  t  t
connect     *  x  x  *  *  *  *  x  t
dining      *  *  *  *  *  *  *  *  *  *
[/code][/hide]

Pretty reasonable. I liked this problem set a lot; it took me two hours to remember and apply maxflow right on dining, but I'm satisfied with the result. I spent another hour or so coming up with a slow (and apparantly partially wrong) solution to connect. cheappal looked really hard to me at first, but after spending the last hour on it I got a $O(M^{3})$ DP working - I actually like this problem a lot, too.

The most reasonable solution I can think of to connect is just to partition into intervals of size $100$ (or slightly less than $\sqrt(N)$ in general), and in each interval keep track of the connectivity from the left to the right-hand side of both the top and bottom boundary squares. Then for each query that modifies your graph, update the interval with the modified square. For each query that asks for whether two vertices are connected, you can use your precalculated information about intervals to quickly get to the interval you want, then just floodfill within the interval itself. This requires at most $100$ operations to get out of your starting interval, $150$ operations to get to the ending interval, and $100$ operations to floodfill to the right spot in the ending interval, for a total of $350$ operations. Multiplied by $50000$ queries, this is 17.5 million operations, which is fine, although slightly borderline, so it is a good idea to precalculate how your floodfill will propogate given information about the next column (whether top and bottom are connected, and which squares are connected to the next column as well). Working code is shown below. Here I have made use of the fact that $2^{x}= x+1$ if $x=0$ or $1$, which allowed me to do bitflicking a bit more cleanly:

[code]
/*
  ID: zealotp1
  LANG: C++
  PROG: connect
*/

#include <iostream>
#include <fstream>

using namespace std;

#define SPLIT (100)
#define MAX (15000)
#define ACC (0)
#define TOP (1)
#define BOT (2)

int checkpoints[MAX/SPLIT][4];
int arr[MAX];
int conn[4][8];
int N,r1,c1,r2,c2;
char c;

int _conn(int i,int p){
   int r = 0;
   if(i&TOP){
     if(p&(1<<TOP)) r |= TOP;
     if((p&(1<<ACC))&&(p&(1<<BOT))) r |= BOT;
   }
   if(i&BOT){
     if(p&(1<<BOT)) r |= BOT;
     if((p&(1<<ACC))&&(p&(1<<TOP))) r |= TOP;
   }
   return r;
}

int prop(int v1,int v2,int car){
   int cp = v1;
   while(cp < v2){
     car = conn[car][arr[cp]];
     cp++;
   }
   return car;
}

void update(){
  int col = min(c1,c2);
  if(r1==r2)
    arr[col]^=1<<(r1+1);
  else
    arr[c1]^=1;
  int pos = col/SPLIT;
  checkpoints[pos][0]=0;
  checkpoints[pos][1]=prop(pos*SPLIT,(pos+1)*SPLIT,1);
  checkpoints[pos][2]=prop(pos*SPLIT,(pos+1)*SPLIT,2);
  checkpoints[pos][3]=checkpoints[pos][1]|checkpoints[pos][2];
}

bool path(){
  if(c1==c2) return (r1==r2) || (arr[c1]&(1<<ACC));
  if(c1>c2) { int t = c1; c1 = c2; c2 = t; t = r1; r1 = r2; r2 = t; }
  int toprop = r1+1;
  while(c1/SPLIT != c2/SPLIT){
    toprop = (c1%SPLIT)?prop(c1,SPLIT*(c1/SPLIT+1),toprop):checkpoints[c1/SPLIT][toprop];
    c1 = SPLIT*(c1/SPLIT+1);
  }
  return ((1&arr[c2])?3:(r2+1))&prop(c1,c2,toprop);
}

int main(){
   //split the roads into intervals of size SPLIT
   //to add or remove an edge, simply update the interval -- O(4*SPLIT)
   //to determine connectivity, go through intervals until you are in the right
   //interval -- then search within the interval O(SPLIT+MAX/SPLIT)

   for(int i=0;i<4;i++)
     for(int j=0;j<8;j++)
       conn[i][j]=_conn(i,j);

   ifstream fin("connect.in");
   fin >> N;
   for(int i=0;i<N;i++){
     fin >> r1 >> c1 >> r2 >> c2;
     r1--;c1--;r2--;c2--;
     update();
   }
   while(1){
      cin >> c;
      while(!('A'<=c&&c<='Z')) cin >> c;
      if(c=='E') break;
      cin >> r1 >> c1 >> r2 >> c2;
      r1--;c1--;r2--;c2--;
      if(c=='O'||c=='C') update();
      if(c=='T') { cout << (path()?'Y':'N') << endl; cout << flush; }
   }

   return 0;
}
[/code]

I see. So it sounds like this problem is a buffed-up version of the range-minimum query problem. Funny since on that one I had the same $\sqrt{N}$ idea, but I didn't think of it this time :P.

I'm just wondering, who got invitations for the camp?

Darn...I got one test case away from advancing to gold :( (658/670 points)

Oh well, I'll just get to gold in october.

The task "connect" can also be solved using intervals of size 1, 2, 4, 8, 16, ...

That would result in all operations having a O(log(n)) time. Slightly better than the sqrt(n) solution in theory, but much more difficult to code (or rather, much more difficult to code and get it right!). :)

So how often are these contests held? I just signed up over the internet on the USACO website, so how long would it take before I can take one of these tests?

There's 6 contests a year, one a month, starting in November and a qualification contest for the different divisions in October.

There are also training pages that can help a LOT though, so you don't have to wait until October to do anything.

And one kid found out about USACO like two weeks or something before the Open, just worked on the training pages a lot, and ended up being on the team that year, or something crazy like that.  So they actually help a lot.


Hello,

To proof that RSA works I have to do the following:
(c = coded message, m is original message)

m = c^d mod N

c = m^e mod N

Replace c in first equation:

m = (m^e mod N)^d mod N

In a (short) proof of RSA that I have the next step is m^ed mod N.

I am wondering how they get from (m^e mod N)^d mod N to m^ed mod N.

I hope that someone of you can explain me this or give me a tip about where to start.

Thanks,
Mark

The mod N are not in the exponents... :wink:

You have $ c\equal{}m^e \mod[N]$ so that $ c \equal{} m^e \plus{} kN$ for some integer $ k$.
Now, $ c^d \equal{} (m^e\plus{}kN)^d$ then expanding, it leads to $ c^d \equal{} m^{ed} \plus{} k'N$ for some integer $ k'$, that is $ c^d \equal{} m^{ed} \mod [N]$.

Since $ e$ and $ d$ have been chosen such that $ ed \equal{} 1 \mod[(p\minus{}1)(q\minus{}1)]$, where $ N \equal{} pq$ and $ p,q$ are distinct primes, Femat's little theorem ensures that $ m^{ed} \equal{} m \mod [N]$, as desired.

Pierre.

Hi Pierre.  My understanding was that $ p$ and $ q$ need not be distinct primes but only relatively prime.  Is that not the case?


calculus
derivative
function
real analysis
inequalities
integration
triangle inequality

Let $f$ be a differentiable function, defined on a compact interval.
What can be said about the Riemann / Lebesgue integrability of $f^\prime$?

I know that we can find $f$ such that $f^\prime$ is unbounded and not integrable.
But what if we suppose that $f^\prime$ is bounded? I can't find any counterexamples in this case.

Let's make it the interval $[0,1]$ just for convenience.

Now, define the function $f_{n}$ by letting $f_{n}(x)=2^{n}\left(f(2^{-n}(k+1))-f(2^{-n}k)\right)$ for $2^{-n}k\le x<2^{-n}(k+1),\ 0\le k<2^{n}.$ (Extend the furthest-right subinterval to include 1.)

Each $f_{n}$ is a step function, and $f_{n}(x)$ tends pointwise to $f'(x)$ everywhere. Furthermore, by the mean value theorem, every value that any $f_{n}(x)$ ever equals is taken from the range of $f'.$

Since $f'$ is a pointwise limit of measurable functions, it is measurable, and if we assume $f'$ is bounded, then it must be Lebesgue integrable as well.

I have to think a little more about Riemann integrability, but I'd guess that if $f'$ is bounded then it is Riemann integrable, and my sequence of step functions might have something to do with a proof.

[quote="Kent Merryfield"]$f_{n}(x)$ tends pointwise to $f'(x)$ everywhere.[/quote]
How would you prove this?
(If we have an incresing sequence $a_{n}\to x$ and a decreasing one $b_{n}\to x$, then the line through $\left( a_{n}, f(a_{n}) \right)$ and $\left( b_{n}, f(b_{n}) \right)$ should "approach" the tangent to the graph of $f$ at $(x,f(x))$ since both $(a_{n},f(a_{n}))$ and $(b_{n},f(b_{n}))$ tend to be on that tangent. Somehow I cannot come up with a rigourous argument from these ideas :oops: )

Make use of a convex combination. Suppose $a_{n}<x<b_{n},$ then

$\frac{f(b_{n})-f(a_{n})}{b_{n}-a_{n}}= \frac{x-a_{n}}{b_{n}-a_{n}}\cdot \frac{f(x)-f(a_{n})}{x-a_{n}}+\frac{b_{n}-x}{b_{n}-a_{n}}\cdot \frac{f(b_{n})-f(x)}{b_{n}-x}$

Assume the derivative exists, and then take the limit.

I do not understand something :blush: : how can we take the limit in the right side - the fractions containing f tend to f'(x) but what about the fractions with which they are multiplied? We write what the limit for the fractions containing f means with $\epsilon$ and then we multiply with the remaining terms from a rang above? Sry...I think my question is somehow nonsense :oops:  :maybe:  :(

[quote="Kent Merryfield"]Subtract $f'(x)$ from both sides. Since the right hand side is a convex combination (the coefficients are positive and add up to 1), we can rewrite that right hand side as

$\frac{x-a_{n}}{b_{n}-a_{n}}\left(\frac{f(x)-f(a_{n})}{x-a_{n}}-f'(x)\right)+\frac{b_{n}-x}{b_{n}-a_{n}}\left(\frac{f(b_{n})-f(x)}{b_{n}-x}-f'(x)\right)$

Now take absolute values, use the triangle inequality, and use that there is a $\delta$ such that if $0<|u-x|<\delta,$ then

$\left|\frac{f(u)-f(x)}{u-x}-f'(x)\right|<\epsilon$[/quote]
[quote="ciprian"] We write what the limit for the fractions containing f means with $\epsilon$ and then we multiply with the remaining terms from a rang above? [/quote]
Isn't this what I said? I was worried about the terms $\frac{x-a_{n}}{b_{n}-a_{n}}$ / other> I could not understand if we can write first with epsilon (without modifying the terms that add up to 1) and then add them up to 1... :maybe:

[quote="Kent Merryfield"]
I have to think a little more about Riemann integrability, but I'd guess that if $f'$ is bounded then it is Riemann integrable [/quote]
It is not so. The example is the same as the one of a function with zero derivative on a dense set. Just take $g(x)=x+\sum c_{k}\sqrt[3]{x-x_{k}}$ where $c_{k}>0$, $x_{k}$ are dense in $\mathbb R$ and the series converges and define $f=g^{-1}$. Then, on one hand, $f'=0$ on a dense set, so the Riemann integral of $f'$ over every interval $[a,b]$, if it exists, must be $0$, and, on the other hand, it must equal $f(b)-f(a)$. But $f$ is not a constant function.

@fedja:
How do you know that your series converges for all x? What if this does not hold?

If $c_{k}$ goes to zero rapidly enough (with some dependence on $x_{k}$), the series converges for all $x$. A simple estimate: $|x-x_{k}|^{\frac13}\le |x|^{\frac13}+|x_{k}|^{\frac13}$. If both $\sum c_{k}$ and $\sum c_{k}|x_{k}|^{\frac13}$ converge, the series converges absolutely for all $x$.


floor function
number theory open
number theory

Definition: Let us define a set of Manjil's primes as the triplet (p-6, p, p+6) such that they are consecutive primes and each differ from the succeding one by 6. The smallest set of Manjil's primes is (47, 53, 59) while there are 54 sets of Manjil's primes below 10,000.

Conjecture: There are infinite sets of Manjil's primes!

Please try to prove or disprove this conjecture.

In fact the number 6 can be replaced by other even numbers except 2

$ 6$ can only be replaced by other multiples of $ 6$.  See [url=http://en.wikipedia.org/wiki/Prime_triplet]prime triplet[/url] and [url=http://en.wikipedia.org/wiki/Sexy_prime#Sexy_prime_triplets]sexy prime triplets[/url].  In all likelihood, this is about as hard if not harder than the twin prime conjecture.

Thanks t0rajir0u for the links . I think this conjecture has no known solutions!

Maybe you can prove that there are no more consecutive primes like p and p+6
or p-6 and p+6...
but i think it's rather difficult too  :P

hey, try my conjecture  :D
i couldn't find any solution
"Let $ x$ be a positive number and $ M_x \equal{} \lfloor\pi^x\rfloor .10^x$
Prove that there are inifinite $ x$ so that $ M_x$ is a perfect square."

There are consecutive primes such that $ p$ and $ p\minus{}6$ are primes! There are approximately $ 58$ such pairs below $ 10,000$.

I will check ur conjecture! :lol:


Let ABC is a triangle
Prove that
$2(\frac{1}{sinA}+\frac{1}{sinB}+\frac{1}{sinC})-(cotA+cotB+cotC)\geq 3\sqrt{3}$

[quote="N.T.TUAN"][hide]1, $\frac{1}{\sin{A}}-\cot{A}=\frac{1-\cos{A}}{\sin{A}}=\tan{\frac{A}{2}}$

2, $\sum \sin{A}\leq \frac{3\sqrt{3}}{2}$

3, $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9 \forall a,b,c>0$[/hide][/quote]
It's easy. And here is a same problem that is interested.
$2(\frac{1}{sinB}+\frac{1}{sinC})-(cotB+cotC)\geq 2\sqrt{3}$

[hide="Filling in the gaps"]
\[\begin{enqarray*}2(\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C})-(\cot A+\cot B+\cot C) &\geq & 3\sqrt{3}\\ \text{With N.T.Tuan's hints, }\left(\sum \frac{1}{\sin A}\right)\frac{3\sqrt{3}}{2}&\geq & \left(\sum \sin A \right)\left(\sum \frac{1}{\sin A}\right)\\ \text{Also, }2(\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C})-(\cot A+\cot B+\cot C)&=& \sum \frac{1}{\sin A}+\sum \tan \frac{A}{2}\\ \text{Thus, }2(\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C})-(\cot A+\cot B+\cot C) &\geq & 3\sqrt{3}\\ \iff \sum \frac{1}{\sin A}+\sum \tan \frac{A}{2}&\geq & 3\sqrt{3}\\ \text{Applying a well known inequality in conjuction with the previously proven, }\\ \sum \frac{1}{\sin A}+\sum \tan \frac{A}{2}\geq 3\sqrt{3}\\ \iff 2(\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C})-(\cot A+\cot B+\cot C) \geq 3\sqrt{3}\end{eqnarray*}\]
[/hide]


inequalities
triangle inequality
inequalities unsolved

Let a,b,c be positive reals so that a+b+c=abc
Show that:

1/ \sqrt (a 2 +b 2 ) + 1/ \sqrt (b 2 +c 2 ) + 1/ \sqrt (c 2 +a 2 )  \leq  \sqrt (3/2)

Observing that a+b+c=abc is equivalent to saying that:
tanA+tanB+tanC=tanA*tanB*tanC, it becomes a triangle inequality.

More precisely, an acute-angled triangle inequality, since a, b, c are supposed to be positive.

But I'm not sure whether this really simplifies the problem...

  Darij

This may be one of the first inequalities I've ever seen where a+b+c=abc is given and a tangent substitution is useless.

sum(1/(a^2+b^2)^.5)<=sum(1/(2ab)^.5)=(a^.5+b^.5+c^.5)/(2abc)^.5=(a^.5+b^.5+c^.5)/(2(a+b+c))^.5<=(3/2)^.5, the first inequality following by AMGM and the second by Jensen's Inequality.

Wow! That's an interesting idea. My solution:
  We have  \sum  \sqrt 2/(a 2 +b 2 )  \leq  \sum  \sqrt 1/ab= (\sum  \sqrt a)/ \sqrt abc \leq  \sqrt 3(a+b+c)/ \sqrt abc= \sqrt 3

That's the way I thought the problem too, harazi


logarithms
inequalities
function
inequalities unsolved

[quote="dina013"]If $a, b, c > 0$, prove that:

                $(abc)^{\frac{a+b+c}{3}}\le a^{a}b^{b}c^{c}$.[/quote]
I have two solutions.

[b]Solution 1[/b]
Take $\ln$ of both sides. Then the inequality becomes \[\sum a\ln a\geqslant \frac{a+b+c}{3}\sum \ln a.\] Let's consider function $f(t)= t\ln t$. It is convex ($f''(t)=\frac1t$) and $\ln$ is concave, so from Jensen we have \[\sum a\ln a\geqslant 3\cdot\frac{a+b+c}{3}\cdot\ln\left(\frac{a+b+c}{3}\right)\geqslant (a+b+c)\cdot\frac{\sum\ln a}{3}\] which ends the proof.

[b]Solution 2[/b]
Also take $\ln$ of both sides. By rearrangement: \[\sum a\ln a\geqslant \sum a\ln a,\]  \[\sum a\ln a\geqslant \sum b\ln a,\]  \[\sum a\ln a\geqslant \sum c\ln a.\] Then summing up and dividing by $3$ ends the proof.


search
combinatorics unsolved
combinatorics

a)Prove that in any group of 6 people there are either 3 mutual friends either 3 mutual strangers.

b)Seventeen people correspond by mail with one another, each one with all the rest. In their letters only 3 topics are discussed. Each pair of correspondents deals with only 1 topic. Prove that there are at least 3 people who write to each other about the same topic.

Both have appeared before many times.  The first appeared most recently here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156926

Do a search.

i'm sorry..i'll search better next time :blush:


geometry
trigonometry
area of a triangle

A, B, C are on the circumference of a circle.  $AB=BC$, $\angle{BAC}=30^o$ and the length of minor arc $\widehat{BC}$ is $2\pi$.  If the area of the shaded angular region BAC is: $a+b\pi$, what is the value of $a+b$.

[hide]Call the center of the circle O and draw chord BC.  Then subdivide the desired area into 3 parts:
(i) sector BOC
(ii) triangle AOC
(iii) triangle AOB
We can find the area of each individually and the sum them up for the final answer.

Since $\angle{BAC}=30^o$, we know that $\angle{BOC}=60^o$.  We also know that $\angle{AOB}=\angle{AOC}$ and that they both equal $120^o$.

Now we can find the areas of each section:
(i) Sector BOC covers one sixth of the entire circle, and since it subtends and arc with length $2\pi$, the radius must be 6.  Sector BOC must also one sixth of the area of the entire circle, and since its radius is 6, sector BOC must have an area of $6\pi$.
(ii) A way to represent the area of a triangle is: $\frac{1}{2}ab\sin{C}$.  Using this, and the fact that $AO=BO=6$ and $\angle{AOB}=120^o$, the area of triangle AOB is 9.
(iii) Triangle AOC and triangle AOB are congruent, so triangle AOC has area 9.

(i)+(ii)+(iii)=total shaded region=$18+6\pi$.  This means that $a+b=\boxed{24}$[/hide]


[url]http://www.bored.com/games/media/worlds_hardest_game.swf[/url]

This is a really hard game.... I am stuck on level 23 with 689 deaths.

I didn't get past level 1, and I had 14 deaths. I quit after that.

SKILLZ

As of right now
Level 6
8 deaths

meh. never got past the first blue dot on level 1.

Level 6. deaths: 28. I will update again when I get to level 12.

UGH stuck on level 10 with 74 deaths!

EDIT: Checkpoint on level 12 with 101 deaths

EDIT2: I got level 13 on the first try. Not hard.

I can make a harder game... I'm stuck at level 10

I can make level one impossible

I remember this game...friend and i got past quite a few levels...but we didn't count deaths...they didn't really matter..the main achievement is getting to certian levels in the first place

[quote="Brut3Forc3"]Ah... reminds me of bounding on StarCraft... good ol' days.... :D[/quote]

DUDE  STARCRAFT PWNZ GO TEAM ZERG/PROTOSS FTW!!!!

Level 15 is no fun.
EDIT: Level 17 is no fun.
EDIT 2: Level 22 is no fun.
EDIT 3: Okay, level 23 is just absurd.

I made it to level 4 with only 1 death.

Timing is the key. Go one square at a time, then edge your way into the center and get back out of there fast. Then go for the green.

[quote="1=2"]Level 18 is SO hard! How do you do it!?![/quote]

yeah 18 took me a while.... which one is it?

[quote="1=2"]the one with all the yellow beacons.

I got to level 12 with 44 deaths!!![/quote]

yeah, that one was hard. have you beaten it yet?

[quote="1=2"]I made it to level 4 with only 1 death.[/quote]
Psh, I did it with 0. Usually I only start getting deaths in level 5.

For all those people asking how to get past level one, all you need to do is go forward when in comes away, then inch up when it comes towards you, and so on.

I accualy completed it with 150 kills  :lol:  :D

When I first looked at the high scores, there were exactly 299. There were 300 people on the high scores.

GAHH STILL CANT GET PAST LEVEL ONE!!!!!!!!!!! this is NOT good...

[quote="NapoleonXIV"]If you look the high score you'll see there's bunch of people with 0 kills
 :rotfl:[/quote]
Hackers.  That's all I have to say.

[quote="dreambold22"]ok i seriously need help getting past level 1.[/quote]
I'm not that great, and neither is this little video, but maybe this'll help:
[url]http://www.mediafire.com/?vkd9dddgwzt[/url]

(PS: Sorry that it is fat.)

EDIT: Found a "nicer" (cheating) way to beat it:
[url]http://www.mediafire.com/?n0d3zdjgwtv[/url]
Note: Go to the top row, then go right, and time it so you pass that approximate spot, then you should pass through. :D

i'll try that. tho level 2 will be a different story... :(  :(

Level 4, 17 deaths right now.
EDIT: Meh, level 7 with 60 deaths, then I felt bored.

I disagree with the hardest game thing. I say the hardest game in the world is IWBTG (i wanna be the guy, the movie, the game)

Look it up on the netz. It's horribly hard. It's also pretty hilarious though. Get the slo mo version cause you won't make it in the regular. I don't think anyone has
My friend beat the slo mo on hard. so...

18 was amazingly hard. I got there on my first try with 27 deaths, when I was done, I had 86. I finished the game on my 5th try with 213. (I personally think 18 was the hardest)

You have to start right after wing thats gonna turn around passes you and go into the direction where its gonna turn. Then try to get as close to middle as possible.
(Might take you few turns to collect all yellow)

The game is from Addictinggames.com and i beat it with only 294 deaths

umm... 28 deaths and i still can't make it out of level 1
i give up =[

im on level 2 now!!
w00t
but its still like impossible
for me at least

How about this becomes a competition to see who can get on the leaderboard?  If you suceed, put something like: "AoPS User:<insert username here>"


Find all 6-digit numbers $a_1a_2a_3a_4a_5a_6$ formed by using the digits $1,2,3,4,5,6$ once each such that the number $a_1a_2a_2\ldots a_k$ is divisible by $k$ for $1 \leq k \leq 6$.

a_1 is always divisible by 1.

a_1a_2 is divisible by 2 if a_2 is even, so it is 2,4,6.   (2)

a_1a_2a_3 is divisible by 3 if a_1 + a_2 + a_3 is 0 mod 3.   (3)

a_1a_2a_3a_4 is divisible by 4 if a_3a_4 is, in particular a_4 must be even.  (4)

a_1a_2a_3a_4a_5 is div. by 5 if a_5 = 0 or 5, so a_5 = 5.  (5)

a_1a_2a_3a_4a_5a_6 is div by 6 if a_6 is even and $\sum a_i$ is 0 mod 3, so a_4+a_5+a_6 must be.  (6)

-----

Try a_6 = 2.  From (6) we must have 5+2+a_4 0 mod 3, which cant happen.

Try a_6 = 4.  From (6)   a_4 is 0 mod 3, so it must be 3 or 6.
   From (4) a_4 is even, so it must be 6, and then a_3 must be 1.
   In (2) the only even number left is 2 = a_2.  Then a_1 = 3 which checks out.

Try a_6 = 6; from (6) we get a_4 = 1 or 4 ; must be 4 by (4).  Then from (2) we have a_2 = 2, and we can mix the choices.

In total, 321654, 123456, 321456

sorry for bringing this old post now.
I was searching for RMO problems.

SINGULAR, you should have checked your solution. Only 1 of your answers is correct - 321654
Also this one is a answer - 123654

a_5 = 5(fixed).

a_2, a_6,a_4 have to be 2, 4, 6 in some order.

that leaves us with 1 and 3 for a_1 and a_3.

if a_3 = 1, a_4 = 2 and a_2 = 4 and a_6 = 6

if a_3 = 3, a_4 = 6 and a_2 =  2 or 4  and a_6 = 4 or 2.

thus, the only numbers are

123654, 143652, 341256.  :clap2:

thus, the only numbers are

123654, 143652, 341256.  :clap2:[/quote]

ur 2nd answer is incorrect......143 is not divisible by 3
ooooo also the third one..... :P

oh, yes I was wrong  :blush:

if $ a_3$ is $ 1$, $ a_4$ can be $ 2$ or $ 6$.

Highly intuitive->

[hide="Click to see solution"]
[list=1]
[*] $ \because{a_1}{a_2}{a_3}{a_4}{a_5}$ is divisible by 5 $ \implies \boxed{{a_5} = 5}$

[*]$ {a_1}{a_2}{a_3}{a_4}{a_5}{a_6}$ will always be divisible by 3 as $ 1 + 2 + 3 + 4 + 5 + 6 = 21$ which is divisible by three.
$ \therefore$ if $ {a_6}$ is even then it will be divisible by $ 2$ also and hence by $ 6$.
$ \therefore{a_6} =$ [i]any even number[/i]

[*]also for $ {a_1}{a_2}$ to be divisible by 2, $ {a_2} =$ [i]any even number[/i]

[*][b]Now we can see that $ {a_1}$ and $ a_3$ belong to [i]either of the odd numbers 1 or 3[/i] since all the three even numbers are used up in $ a_2$, $ a_4$(which is also ovbiously even), $ a_6$. and $ {a_5} = 5$[/b]

[*]now $ {a_1}{a_2}{a_3}{a_4}$ is divisible by 4 $ \implies a_4$ is even. also its value will depend on wether the  preceding digit is even or odd. in this case, the [i]preceding digit $ a_3$ is odd[/i] by statement 4. hence $ a_4 \in \{2,6\}$

[*]this means that $ 4$ can come only in $ {a_2}$ or $ a_6$ using statement 4

[*]now $ {a_1}{a_2}{a_3}$ is divisible by 3 $ \implies {a_1} + {a_2} + {a_3} =$ [i]multiple of three[/i]. now $ a_1 + a_3 = 1 + 3 = 4$
and ${ a_2 \in \{2,4,6}\}$ using 3.

thus only if $ a_2 = 2$ we have $ {a_1} + {a_2} + {a_3} = 6 =$[i] multiple of 3[/i]
$ \boxed{\therefore a_2 = 2}$

[*] using statements 5,7 we have $ \boxed{a_4 = 6}$

[*] using statements 6,7 we have $ \boxed{a_6 = 4}$

[*]hence we are left only with {a_1} and {a_3} which can have values 1 or 3. hence we arrage 1,3 among $ a_1$ and $ a_3$ to get two answers $ \longrightarrow \boxed{123654,321654}$[/list]
[/hide]

[b]Edited: correction proposed courtesy saubhik[/b]

Statement 7: a little mistake; it should be $ a_1\plus{}a_3 \equal{} 1\plus{}3 \equal{} 4$    :oops:

I got 2 answers
[b]
123654
321654[/b]
Method:
1. a1 can be any digit and a6 is any even digit.
2.a5 has to be 5
3.a2 = 2,4 or 6
4.the 2 digit number formed by a3a4 can be only 12 , 16 , 24 , 32 , 36 or 64
case 1: a2 = 2
a3a4 = 16,36 or 64
subcase 1.1: a3a4 = 16
We now check if a1a2a3 is divisible by 3
here a2 + a3 = 3 and a1 = 3 or 4 ( the only digits not used up yet)
for a1 = 3 , conditions are satisfied so [b]321654[/b] is a solution.
subcase 1.2: a3a4 = 36
here a2 + a3 = 5 and a1 = 1 or 4 ( the only digits not used up yet)
for a1 = 1 and a1 = 4 , conditions are satisfied so we get 2 more solutions, [b]123654[/b] and [b]423651[/b]
but 423651 is rejected as it is odd(a6 must be even)
subcase 1.3: a3a4 = 64
proceed similarly as subcase 1.1 and subcase 1.2
to get no solution

case 2: a2 = 4
a3a4 = 12 , 16,32 or 36
similar steps as case 1
gives us no solution again

case 3: a2 = 6
a3a4 = 12 , 24 or 32
similar steps as case 1
and again no solution

Feel free to point out errors


who knows logarithms :lol:

       i just want to know

I do, but in the future please post constructive things, like math problems or actual conceptual questions you have. Don't just say "who knows logarithms". Thanks.

Any questions?


a can (cylindrical) is formed in such a way that a ball with a radius of 2 fits just inside. What is the area of the can?

How do you find the area of a can?
I assume you mean volume.

[hide]The radius and height of the can must be 2 and 4. So the answer is $16\pi$. [/hide]

whoops.. :blush: heh,heh, : P
well, ur answer is correctomundo!

[quote="mathmanman"]Well, in fact, you can calculate the area of a can if you want...  :)[/quote]
i thought you could only calculate areas of 2d figures.  You can calculate the surface area though. ;)

Bah, yeah, "surface area", that's what I meant  ;)

:rotfl:
Well, anyway, if you 'cut' the cylinder, you can make it 2D :P (but it's not a cylinder annymore haha)

If you cut the cylinder (assuming it has no base), lay it out, and measure its area, you are measuring its lateral area.

[hide]
If the ball just fits inside, then the radius has to be 2 as well.
$4(\pi(2)^2) = \boxed{16\pi}$[/hide]

[hide]since the ball just fits, the solution is $4\pi*2^2$=16 :idea: our answer :idea: [/hide]


geometry
circumcircle
incenter
geometry solved

Given a triangle $\triangle ABC$ , a point $K$ is chosen on the ray $AC$ and the point $L$ is chosen on the ray $BC$ such $AB=AK=BL$ . The point $M$ is chosen on the ray $AB$ and the point $N$ is chosen on the ray $CB$ such that $AC=AM=CN$ . The point $P$ is chosen on the ray $BA$ and the point $Q$ is chosen on the ray $CA$ such that $BC=BP=CQ$. Prove that the lines $KL,MN$ and $PQ$ are parallel.

In fact, all three lines KL, MN and PQ are perpendicular to the line OI, where O is the circumcenter and I the incenter of triangle ABC. See http://www.mathlinks.ro/Forum/viewtopic.php?t=14655 posts #3 and #9 for proofs of this.

  Darij


For problems like these, practice turning the words into math. For example, for the marbles question, you might let $ r$ equal the number of reds, $ b$ equal the number of blues, and $ g$ equal the number of greens. Then the problem is just saying:

$ r\plus{}g\plus{}b\equal{}140$
$ g\equal{}3r$
$ b\equal{}6r$
Find $ b$.

After you do this, the algebra is pretty straightforward.

[quote="jdrian10"][quote="panda1"][color=blue] Thank you so much. I know those problems sounded easy to you, but word problems are not my thing. Thanks again! :lol: [/color][/quote] Or if your not really good at algebra (thats me) you could use the model approach...
example: for number 1
[][][][][][] blue
[][][]        green
[]             red
B+G+R=140
In the drawing, there are 10 squares, so divide by 10
140/10=14
14= 1 box (red)

then count how many boxes there are in blue and multiply
14 x 6 = 84 blue marbles....

See? even a 9 year old could answer this type of question[/quote]

You are very rude and disrespectful, delete cyberbullying don't write it.
Also be mindful that some people are smarter than others, how would you like
it if someone made fun of your intelligence? Vaca el infierno, see what that means, loser.

[quote="panda1"][color=blue] How do I become better and more efficient at solving math story problems? Here are some examples:
A box of 140 marbles contains only red, green and blue marbles. there are three times as many gree marbles as red marbles. There are six times as many blue marbles as red marbles. How many blue marbles are there in the box?

Ann has 1$ \$$   in nickels, dimes and quarters with at least one of each. What is the difference between the largest number of coins that she could have and the smallest number of coins that she could have?

I am not looking for answers, but just tips and methods I could use to solve problems like these and other similar ones.

Thank you for your help!!! :D  :D
 [/color][/quote]

For the first problem, it screams ALGEBRA when I see three different quantities, and the relations between all three quantities, and some kind of equation. The variables would be represented by r, g, and b, and then you should write out all the relations that you know. Often, when solving for more than one variable, it would be a good idea to use the substitution method. Do u know what that is? You can find a lot of questions in this section of the thread that have systems of equations problems. Just click on the search option and type in "system", and you will get a lot of good examples.

For the second problem, this is a good example of how it is a good idea to write out what you know before you even attempt solving the problem. This is pretty much a logic problem because you can not solve it by any formula or algabraic way. It is smart to write out what you know, because once you have your solution, you can just look back and see if you followed all the rules stated in the problem. For example, it says that there is one of each coin. So you know that there is at least one penny, one nickel, one dime, and one quarter. This amount equals $ \$$0.41. Now, there is only 59 cents that we have to make out. The penny is the least amount of money, and therefore is least useful when trying to use the least amount of coins possible. So, if we use 59 pennies and the other four coins, then 63 coins would be the highest number of coins possible. Similarly, we would want to begin by using quarters, if we want to use the least amount of coins. You can have two quarters at the most. Then one nickel, and four pennies. This totals 11 coins.


:|  agar $r$ shoae dayere bashad va daroon an besoorate mohat $n$ zelie monazam gharar girad, masahate $n$ zelie ra bedast avarid.
(agar masahate $n$ zalieye monazam ra ta binahayat edame dahim, masahate hasel, masahate che shekle hendessist. toozih dahid.  :lol:

sin(180/n)*(r^2)*cos(180/n)*n ---->masahate n zelee
dayere chon :lim n mire be samte binahayate (sin(180/n)*(r^2)*cos(180/n)*n) o age bedast biarim mishe p(r^2) ,nemidoonam lazeme be dast biarim?

:)  be nazaram miresse eshtebast.
man be yek formoole digeyi residam $s(n)=\frac{1}{2}nr^{2}\sin\frac{2\pi}{n}$ va kamelan doroste.  :maybe:

bebinid man masahate chand zel ee ro goftama,
nemidoonam in chizi ke shoma migin masahate chie,age masahate n zel'eeas chera pi dare formool?

Har dotashoon dorostan.  $2\sin \theta\cos\theta =\sin 2\theta$. Azadeh, $2\pi$ radian.

man ebarate [b]boxedexe[/b] ro dar nazar nagerefte boodam. dar in soorat formoli ke bedast ovordid kamelan doroste.  :oops: [/b]


Does anyone know how to draw graphs (as in a data structure nodes and edges, not something with an xy-plane) in Latex? I need something where it can all be done in Latex, i.e., I don't want to import a picture or something like that. The only thing I could find was something called XY-Pic, but it looked extremely complicated...
Is there something else, or could someone give a quick explanation of how to use XY-Pic to draw graphs?
Thanks!

Have a look at [url=http://www.dpmms.cam.ac.uk/~al366/xyintroduction/]xy-tutorial[/url]. The author does say that "[i]There is also the authoritative reference of xy-pic diagrams xy-refer, but I have found some parts a bit difficult to understand.[/i]" which is what you found, so hopefully his explanations will be simpler. If you can get hold of a copy, [url=http://www.amazon.com/LaTeX-Graphics-Companion-Illustrating-Postscript/dp/0201854694/sr=8-1/qid=1162682272/ref=pd_bbs_sr_1/102-6787070-3364907?ie=UTF8&s=books]The LaTeX Graphics Companion[/url] devotes a whole chapter to xy-pic.

You could also look at [url=http://home.in.tum.de/~baueran/techdoc/latex/]Introduction: Drawing proper Graphs, Trees and related Data Structures using LaTeX[/url] which uses graphs.sty (there's a link to where to download that package).

You could use [url=http://tug.org/PSTricks/main.cgi]PSTricks[/url] (lots of examples on the site) or [url=http://jpicedt.sourceforge.net/en/]jPicEdt[/url] which is a visual editor which will spit out LaTeX code for you. I think [url=http://www.matf.bg.ac.yu/~janicic/gclc/]GCLC/WinGCLC[/url], although meant for geometry, is another visual editor which could be used for graphs.

If you find that these or anything else works for you, do report back so others can benefit from what you have discovered.

The link for jPicEdt doesn't work. http://sourceforge.net/project/showfiles.php?group_id=50377&package_id=43951&release_id=104705 is a link that works.

[quote="stevem"]The link worked yesterday; I hope it comes back soon as it had information about jPicEdt not available on the download site.[/quote]
How long have you been a mod in this forum for? I never noticed you were until like a week ago (though that probably was me being careless)...

[quote="stevem"]Only a few days ago, which is why you didn't see it before.[/quote]
I didn t see it too, infact a very good TeXer like you must be the moderator since the creation of this forum. :)


Prove, for any natural number n, there exists a number such that its decimal expression consists of only 1 and 2, and that it is a multiple of 2^n.

It is an easy problems . Prove by induction.

The approach is indeed by induction. We have a base case for $ n\equal{}1$ of $ 2$. Now:

Assume that there exists an integer $ s$ such that $ s$'s decimal expansion consists of only $ 1$s and $ 2$s and it is divisible by $ 2^n$, and furthermore, that it has exactly $ n$ digits. We will show that one of the two numbers

$ 10^n \plus{} s$
$ 2*10^n \plus{} s$

has exactly $ n\plus{}1$ digits, all of which are $ 1$ or $ 2$, and is divisible by $ 2^{n\plus{}1}$. The first part of the claim is clear. Now, consider the second part. If $ 2^{n\plus{}1} | s$, we have that clearly $ 2^{n\plus{}1}  | 2*10^n$ as well, and we pick the second number.

If this is not the case, then $ s/2^n$ is $ 1 (\text{mod } 2)$. Also, $ 10^n/2^n \equal{} 5^n$ is $ 1 (\text{mod }2)$. Then we have that $ 10^n/2^n \plus{} s/2^n$ is congruent to $ 0 (\text{mod }2)$, or $ 2^{n\plus{}1}$ divides $ 10^n\plus{}s$, and we are done.


geometry
3D geometry
trigonometry
conics
ellipse
ratio
analytic geometry

A cylindrical hole of radius 1 is drilled along one of the diagonals of a cube of side length 3.  Find the area of one of the six congruent faces of the resulting solid.  (Express your answe in simplest radical form in terms of pi.)

Anyone who can get this one deserves recognition!

Let $A_1B_1C_1D_1$ be the bottom face and $A_2B_2C_2D_2$ the top face of the cube, the edges $A_1A_2, B_1B_2, C_1C_2, D_1D_2$ being vertical. Assume that the hole is drilled along the body diagonal $A_1C_2$. The face and body diagonals have lengths $A_1C_1 = A_1B_1 \sqrt 2 = 3 \sqrt 3$ and $A_1C_2 = A_1B_1\sqrt 3 = 3 \sqrt 3$. Let $\phi = \measuredangle C_1A_1C_2$. From the right angle triangle $\triangle A_1C_1C_2$, $\cos \phi = \sqrt{\frac 2 3} = \frac{2\sqrt 3}{3}$, $\sin \phi = \frac{1}{\sqrt 3} = \frac{\sqrt 3}{3}$. The drill bit axis forms the angle $90^o - \phi$ with the outside normal to the top face plane $A_2B_2C_2D_2$ and the circular drill bit cross-section projects to this plane as an ellipse centered at the vertex $C_2$, with the major and minor axes $p = \frac{r}{\sin \phi} = \sqrt 3$, $q = r = 1$, where $r = 1$ is the drill bit radius and the ellipse major axis is along the top face diagonal $A_2C_2$. Let $M, N$ be the intersections of the ellipse with the edges $B_2C_2$ and $C_2D_2$. We have to calculate the the area of the elliptic sector $MC_2N$, where the angle $\measuredangle MC_2N = 90^o$ is right and the lines $C_2M, C_2N$ symmetrical with respect to the face diagonal $A_2C_2$, i.e., with respect to the ellipse major axis. To do this, we use a parallel projection to project the ellipse to a circle $(C_2)$ with radius, for example, equal to the ellipse major axis $p = \sqrt 3$. The points $M, N$ are projected into points $M', N'$ on the circle $(C_2)$ and the angle $\measuredangle MC_2A_2 = \frac{\measuredangle MC_2N}{2} = 45^o$ into the angle $\psi = \frac{\measuredangle M'C_2N'}{2}$, the tangent of which is equal to the ratio of the ellipse major and minor axes, $\tan \psi = \frac p q = \sqrt 3 = 60^o$. The area of the circular sector $M'C_2N'$ is

$|M'C_2N'| = \pi p^2 \cdot \frac{2\psi}{360^o} = \frac{\pi p^2}{3} = \pi$

Since area ratios are preserved in a parallel projection, the area ratio of the circular and elliptic sectors is equal to the area ratio of the circle $(C_2)$ and the ellipse:

$\frac{|M'C_2N'|}{|MC_2N|} = \frac{\pi p^2}{\pi pq} = \frac p q = \sqrt 3$

Hence,

$|MC_2N| = |M'C_2N'| \cdot \frac q p = \frac{\pi}{\sqrt 3} = \frac{\pi \sqrt 3}{3}$

and the area of the remaining part of the cube face $A_2B_2C_2D_2$ is

$|A_2B_2C_2D_2| - |MC_2N| = 9 -  \frac{\pi \sqrt 3}{3} \doteq 7.1862$

Correct, though it should be expressed as 9 - [(pi)(root3)(3)]

(Mandelbrot 3)

[hide]I got the same answer as yetti but with a slightly different method.

If we let the axis of the cylinder be the line $(a,a,a)$ in x,y,z coordinates, then we wish to find the set of points in the x-y plane that have a perpendicular distance of 1 from that line (this describes the intersection of a cylinder with a plane). The translation of this condition is that:

1) $(x-a)^2 + (y-a)^2 + (a)^2 = 1$, and
2) $(x-a, y-a, -a) \cdot (a,a,a) = 0$ (where this is the dot product).

From 2) we have $(x-a) + (y-a) = a$, or $x + y = 3a$. Solving for $a$ and substituting into 1) we get:

$\Big(\frac{2x - y}{3}\Big)^2 + \Big(\frac{2y - x}{3}\Big)^2 + \Big(\frac{x + y}{3}\Big)^2 = 1$, or $2x^2 -2xy + 2y^2 = 3$.

This is the equation of an ellipse centered at the origin with semi-major axis $\sqrt{3}$ along the line $y = x$, and semi-minor axis of $1$ along the line and $y = -x$.

The area that's cut from each side of the cube will be the area of this ellipse in the first quadrant. I found this area using calculus, but Yetti's method works just as well. The area, as he said, turns out to be $\frac{\pi \sqrt{3}}{3}$. Therefore the area of the remaining square will be $9- \frac{\pi \sqrt{3}}{3}$.[/hide]


LaTeX
calculus
probability
calculus computations

1)
Lets say that we have a tree. The first day the height of the tree is 1 meter. The second day is 50cm higher. (1.50m). The third day is 25cm higher(1.75m)The fourth day is 12.5cm higher. The next day is 6.25cm higher, next day is 3.125cm higher and so on. Make a Sigma notation that will show the height of the tree after n days...
2)
The leaves of the tree are:
first day=1
second day=3
third day=7
fourth day=15
fifth day=31
sixth day=63
seventh day=127
day8= 255 leaves and so on......
A sigma notation that will give the number of leaves after n days....
Thank you in advance

Τhanx man...even though i think the first answer is just
1/2^n-1

Theres 3 things you need to know.
1) $2^{1-n}=\frac{1}{2^{n-1}}\longrightarrow$ check by yourself
2) Use $\LaTeX$, things look nicer and its more likely you get a response ;)
3) This problem is high-school, not calculus :P

I am not going to argue...But Mathematica gave me a very strange number when i put that 2-.....Believe me the problem is calculus.If not then i am not studying in a university but in elementary school.....i wrote just a part of it...the easier part...As for latex...the original problem was a word problem...i dont think it needed latex...

$a^{-b}=\frac{1}{a^{b}}$ sounds familliar?

Why do you think that this problem is a calculus problem?

[quote="ypsonas"]If not then i am not studying in a university but in elementary school.....[/quote]
Alas, nowadays there exist university courses that actually belong if not to elementary, then to middle school :lol: :(. Also, even a decent university course can start with the school material. For instance, the introduction to probability course usually starts with some elementary combinatorics that would belong to the Olympiad or even Pre-Olympiad or Intermediate subfora here. On the other hand, don't worry too much about where to post when in doubt: if you post in a completely wrong place, some moderator will move your post eventually :).

As to $\LaTeX$, your first post in this thread does not require it, indeed, but the second one would look much nicer $\LaTeX$ed (or, at least, with properly put parentheses).

[quote="ozolexil"]
$\sum_{i=1}^{n}2^{1-n}=2-\frac{1}{2^{n-1}}$
[/quote]
It should rather be $\sum_{i=1}^{n}2^{1-i}=2-\frac{1}{2^{n-1}}$

I still dont understand why are we "fighting" here...We are all in the same side...I repeat my self....I am Greek and i study in the university of Athens..Pythagora, Eukleidis,  Thalis...were greeks as you know....and you should know that we are not playing in our university...The problem i have to solve is 3 pages long..it just begins with this...i found the answers but i had to be sure to continue...Sorry about Latex but with all these "elementary" projects in a poor greek university i didnt have time to learn it...When i will finish this one may be i will find the time to write the whole problem for the REAL students....A....and i forgot to mention that the name of the lesson i am having the semister is CALCULUS I....calclulus calculus....hm....i have to write a letter to the university.....they are lying to us...this is not calculus.......
"I KNOW ONLY ONE THING.....THAT I KNOW NOTHING"
SOCRATES....GRATE GREEK FILOSOFER.......
"TAKE THE EGG AND GIVE A GOOD HAIR CUT"
GREEK RHYME
"I WANTED TO FALL...THANK YOU YOU PUSHED ME"
SOTIRIS...A GREEK THAT THINKS HE IS STUDYING IN A UNIVERSITY

[quote="fedja"]
[quote="ozolexil"]
$\sum_{i=1}^{n}2^{1-n}=2-\frac{1}{2^{n-1}}$
[/quote]
It should rather be $\sum_{i=1}^{n}2^{1-i}=2-\frac{1}{2^{n-1}}$[/quote]
Thanks fedja  :blush:

Could a moderator please delete ypsonas's previous post in this thread, along with this message?  Thanks.

[quote="ypsonas"]I still dont understand why are we "fighting" here...
[/quote]
Nobody's fighting :) We merely tried to explain to you that a particular problem (and, especially, a part of a problem) given in the Calculus university course may belong not to Calculus subforum but to some other subforum here. There is absolutely nothing wrong with it! I often start my Introduction to Analysis courses with a few exercises about understanding statements with quantifiers. It is not Analysis, of course, just elementary logic, but if I don't do it this way, many students have a lot of trouble struggling with various $\varepsilon-\delta$ definitions later on. Also, you should understand that a calculus problem can include "non-calculus" parts and that when you choose to post a part of the problem, the right subforum for it is determined by the content of this part only, not by the absent context or by the name of the course it comes from. That's all! :lol:

As to learning $\LaTeX$, it takes half an hour to learn how to type simple formulae and you can easily do it by looking at other people's posts and experimenting with TeXer on this site.

To JBL: I can delete it but I do not see a compelling reason for doing so. It is a bit off topic, of course, but it is just his 18th post here and I don't think we should be really hard with newcomers. On the other hand, some other mod may think differently. Then this reply of mine should be removed too.


[quote="mihail911"]Harold tosses a nickel four times. The probability that he gets at least as many heads as tails is..[/quote]
That's from AMC 8..

Try Complimentary Counting.

There are 2^4=16 different outcomes. Of these, there would be 6 outcomes with two heads and two tails, 1 with all heads, and 4 with 3 heads. Probability: 11/16

[hide]the probablility that he gets heads more times than tails is equal to the probablity that he gets more tails than heads. there are 6 ways he can get the same number of each.  therefore the probabliity that he gets more heads than tails is $ \frac{16\minus{}6} {16*2}\equal{}\frac{5} {16}$ the problem says atleast. so we add the $ \frac{6} {16}$, $ \frac {11} {16}$[/hide]


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For $n\geq 1$ let $M$ be an $n\times n$ complex array with distinct eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_k$, with multiplicities $m_1,m_2,\ldots,m_k$ respectively. Consider the linear operator $L_M$ defined by $L_MX=MX+XM^T$, for any complex $n\times n$ array $X$. Find its eigenvalues and their multiplicities. ($M^T$ denotes the transpose matrix of $M$).

I try for Moubi's generalization.

Let a and x an eigen value and eigen vector for A
Let b and y an eigen value and eigen vector for B

Then the matrix X=x*y^t is not nul and L(X)=(a+b)X (easy checks).

If all the (ai+bi) are different, P(X) = :prod: (X-ai-bi).

Otherwise, it is know that the caracteristic polynomial depends continuously of the matrix (all coefficients are polynomials of the coefficients), so if one change slightly A to A' such as all (ai'+bi) become different (the ai' depends continuously of the A' also). Then one can use a limit argument.

I am sure of the correctness of last argment but it seems sensible. Moubi, your opinion ?

$a$, $b$ endomorphism associate to the matrices $A$, $B$
The caracteristic polynomial of the linear map $u\rightarrow au+ub$ of $L(E)$
is  $\prod_{ij}(x-a_i-b_j)$ where $a_i,b_j$ eigenvalues of $A,B$

Let $e_i$ a basis s.t. mat(b,e) is triangular.

$mat(a;e) = (a_{ij})$, $mat(b;e) = (b_{ij})$ with $b_{ij}= 0$ if i>j

Consider $g_{ij}$ basis of $L(E)$ asociate to $e_i$
$g_{ij}(e_k) = \delta_{jk}e_i$

$F= au$ and $G=ub$
$F(g_{ij})=\sum_{l}a_{li}g_{lj}$
$G(g_{ij}) = \sum_{m}b_{jm}g_{im}$

We write the basis $g_{ij}$ with a good order like
$g=(g_{11},...,g_{il},...,g_{nl},...,g_{nn})$

$mat(F;g),mat(G,g) \in M_{n^2}(C)$

$mat(F,g) = (A_{kj}) = A $ if k=j and 0 elsewhere
$mat(G,g) = (b_{jk}I_n)$
$mat(F+G;g)$ is triangular matrix with diagonal $A+b_{jj}I_n$

The caracteristic polynomial of $F+G$ is
$\prod_{j=1}^{n}det(xI_n- A - b_{jj}I_n) = \prod det((x-b_{jj})I_n-A) $
Use $det(yI_n-A) = \prod(y-a_i)$
we get $\prod_{i,j}(x- a_i - b_j) $

[quote=Moubinool]More generally if $A,B \in Mn(C)$ and consider the linear operator
$L(X)=AX+XB$ on $Mn(C)$ the caracteristic polynomial of L is given by

$P(X)=\prod_{i,j}(X-a_i - b_j)$ where $a_i,b_j$ are the eigenvalues of $A$

and  $B$[/quote]

What is the proof of this statement any help .
so above is proof..but the word like endomorphism confuse me i will see.
also what is mat in above proof?


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Rules for Mock AMC E

1.	This is a 25 question, multiple choice test.  Each question has 5 answer choices marked A, B, C, D and E. Only ONE answer choice is correct.
2.	SCORING: Each correct answer is worth 6 points, each problem left blank is worth 2.5 points, and each incorrect answer is worth 0 points.
3.	Only aids permitted are scratch paper, graph paper, a ruler, a compass, a protractor, erasers, and calculators that are acceptable for the SAT.
4.	You have 75 MINUTES to work on this test.  At the end of the time you will PM your answers to joml88 as promptly as possibly.

(Note: if for some strange reason you are not able to PM me the answers I will accept answers through email if they are sent to jlaurendi@gmail.com)

[b]Mock AMC E[/b]

                                      [size=200]Joseph Laurendi [/size]

                               [size=100]Proofread by tokenadult[/size]

1.  The sum of the digits of the year $1999$ is $28$.  Let abcd be the next smallest (i.e. next earliest) year where the sum of the digits equals $28$ ($a, b, c, \ \mathrm{and} \ d$ are digits, not necessarily distinct).  Compute $a\times b\times c\times d$

(A) $1296$       (B) $28$       (C) $729$       (D) $56$        (E) $216$

2.  The sum of the interior angles of a triangle is $180^\circ$.  How many sides does a convex polygon have if the sum of its interior angles equals $2700^\circ$?

(A) $13$       (B) $14$       (C) $15$      (D) $16$      (E) $17$

3. John wants to take three classes at the AoPS.  He prefers that one of them  be an intro class, another be an intermediate class, and the other to be olympiad.  If there are intro classes on Monday, Wednesday, and Friday, intermediate classes Tuesday, Thursday, and Friday, and olympiad classes Monday, Tuesday, and Wednesday how many ways can he choose the three classes according to what he prefers?(He can only take one class per day)

(A)  $9$      (B) $15$        (C)  $14$      (D) $27$     (E) $10$

4.  Let $x$ equal the sum of the $6$ distinct prime factors of $2^{24}-1$.  What is $x$?

(A) 286    (B) 289   (C) 4125   (D)  4128   (E) 12319

5.  If three successive rolls of a die are all prime numbers, what is the probability that they are all the same?

(A) $\frac{1}{3}$  (B) $\frac{4}{27}$  (C) $\frac{2}{27}$  (D) $\frac{1}{9}$  (E) none of these

6.  What is the sum of the coefficients of the expanded form of $(2x-3y+3)^4$?

 (A) 0    (B) 1   (C) 16   (D) 625  (E) 1000

7.  How many $7$ step paths are there from $(3,2)$ to $(7,5)$ in the coordinate plane?  (One can only take one unit steps paralled to the x and y axes e.g.  from (3,2) one can only go to (3,3) or (4,2))

(A)  35  (B) 12  (C)  128   (D)  20  (E)  none of these

8. If the product of the ages of a group of teenagers is $10584000$, then the sum of their ages is

(A)  85   (B) 86  (C) 87  (D) 88  (E) 89

9.  What is the binary (base 2) representation of the base 10 fraction 1/5?

(A)  $.\overline{1001}$  (B) $.\overline{0011}$  (C) $.\overline{0101}$  (D)  $.\overline{00111}$  (E) $.\overline{01101}$

10.  Team A and Team B play a best of 7 series. Team A has a $\frac{3}{4}$ probability of beating Team B in each game. What is the probability that the series goes to the 7th game?

(A) $\frac{405}{2048}$  (B) $\frac{135}{512}$   (C) $\frac{135}{1024}$  (D) $\frac{1}{7}$  (E) $\frac{135}{2048}$

11. If $\sin{x}=3\cos{x}$, then what is the value of $\sin{x}\cos{x}$?

(A)  1/6  (B) 1/5  (C) 2/9  (D) 1/4  (E) 3/10

12.  What is the number of distinct solutions (x,y,z) to the equation $x^2+yz=y^2+xz$ if x, y and z are integers between 1 and 5 inclusive?

(A) 31    (B)  8    (C) 6  (D)  25  (E)  37

13.  What is $x^3$ if  $\displaystyle \frac{2}{x}=\frac{y}{3}=\frac{x}{y}$?

(A) 6   (B) 12  (C) 216   (D) 24  (E) none of these

14.  A circular lighthouse floor has a circular rug in the center. The lighthouse keeper observes that if he places a 10-foot pole on the floor so that each end touches the wall, then it is tangent to the rug. Find the area of the floor that is not covered by the rug.

(A) $5\pi$   (B) $15\pi$  (C) $20\pi$  (D) $25\pi$  (E) $100\pi$

15.  If $\cos{\theta}+\cos{2\theta}=0$ and $0\leq \theta \leq \frac{\pi}{2}$, what is $\cos{\theta}$ equal to?

(A) 0  (B) 1  (C) $\frac{1}{2}$  (D) $\frac{\sqrt{3}}{2}$ (E) $\frac{\sqrt{2}}{2}$

16.  Two parallel chords in a circle have lengths 6 and 8.  The distance between them is 1.  What is the diameter of this circle?

(A)  12  (B) 14  (C) 10  (D) 9  (E) $10\sqrt{3}$

17.  Let m be the number of digits of $2^{2005}$, and let n be the number of digits of $5^{2005}$.  What is $m+n$?

(A) 2003    (B) 2007    (C) 2004    (D) 2005   (E) 2006

18. In $\triangle ABC$, AB=AC=1, BC=k, and $\displaystyle \frac{\tan{B}}{\tan{A}}=2001$. What is $k^2$?

(A)  $\frac{1}{1001}$  (B) 1  (C) $\frac{1}{10}$  (D)$\frac{4}{81}$  (E) $\frac{1}{100}$

19.  A circle is inscribed in a triangle whose sides are 10, 10 and 12 units in measure.  A second, smaller circle is inscribed tangent to the first circle and to the two equal sides of the triangle.  What is the measure of the radius of the second, smaller circle?

(A)  3   (B) $\frac{3}{4}$  (C) 2  (D) 1  (E) $\frac{3}{5}$

20.  If $f(x)$ is a monic (leading coefficient of 1) quartic polynomial such that $f(-1)=-1$, $f(2)=-4$, $f(-3)=-9$, and $f(4)=-16$, find $f(1)$.

(A) 1   (B) 23  (C) -1  (D) 25 (E) 27

21. Let a be the integer such that $\displaystyle{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{22}+\frac{1}{23}=\frac{a}{23!}}$ What is the remainder when a is divided by 13?

(A) 0  (B) 2  (C) 5  (D) 7  (E) 10

22.  Let distinct x and y be drawn from $\{ 1,2,3,\ldots,99\}$ such that each ordered pair (x,y) is equally likely.  If the probability that the sum of the units digits of x and y is less than 10 is $\frac{a}{b}$,where a and b are relatively prime, what is a+b?

(A)3  (B) 4851  (C) 7477 (D) 2627  (E) 55

23.  $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=$

(A) $\sqrt{5}-1$   (B) 1  (C) $\sqrt[3]{2}$  (D) $\sqrt{5}-\sqrt[3]{2}$   (E) 6/5

24.  What is the smallest number d such that fewer than half of the positive integers with d-digits have all distinct digits?

(A) 4  (B) 5  (C) 6  (D) 7  (E) d is bigger than 7

25. In trapezoid ABCD, $AB\parallel DC$ and $AB < DC$, the lines AC and DB intersect at P.  The sum of the areas of $\triangle APD$ and $\triangle BPC$ is $1996$.  If the lengths of the bases $\overline{AB}$ and $\overline{CD}$ are integers and the distance between them is an integer, what is the minimum area of ABCD?

(A) 5988  (B) 2994  (C) 4500  (D) 3992  (E) 4491

Perhaps I missed this, but is someone working on a mock AIME? It would be fun to go back to the old system where there are qualifying rounds for the mock AIME and stuff, although based on what I got, I don't think I will make it.

Do you plan on reposting the contest, because I would like to finish it as I didn't have time since I got home late

[quote="RC-7th"]About how long will it take for you to grade our answers? Several minutes? Hours?Days?Weeks?Months?Years?[/quote]

Nah... Just a few millenia.

I got mine back in about a minute, but I sent mine in 5 minutes early.

I also have mine back, and I didn't send in mine early (at least, I didn't by my clock).

Quote:8. If the product of the ages of a group of teenagers is ,
then the sum of their ages is:
[hide]
Since 10,584,000 = (26)(33)(53)(72)

the only possible "teen" factoring is: (142)(153)(16)

Therefore, the sum is 14 + 14 + 15 + 15 + 15 + 16 = 89 ... answer (E)[/hide]
Quote:14.  A circular lighthouse floor has a circular rug in the center.
The lighthouse keeper observes that if he places a 10-foot pole on the floor
so that each end touches the wall, then it is tangent to the rug.
Find the area of the floor that is not covered by the rug.
[hide]

Let R = radius of the floor. r = radius of the rug.

Let the ends of the pole be A and B, tangent to the rug at C.

Let O be the center of the room (and rug).

In right triangle OCB: R2 - r2 = 52

Multiple through by [pi]: [pi]R2 - [pi]r2 = 25[pi]

The left side is the area of the floor that is not covered by the rug.

Therefore: 25[pi] ... answer (D)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I dealy love this problem!

Note that the uncovered area is a function of the length of the pole only.

The size of the circular room and the size of the rug have no bearing.

This problem is analogous to the famous "Hole in a Sphere" problem.[/hide]

Soroban, there is a page here: [url]http://artofproblemsolving.com/Forum/viewtopic.php?t=18066[/url] for the solutions.  I don't we have all of them up so you can fill in the spots on that thread.

Quote:13.  What is  if  ?
[hide]

From 2/x = y/3, we have: [1] xy = 6

From 2/x = x/y, we have: [2] x2 = 2y

Multiply [1] and [2]: (xy)(x2) = (6)(2y)

And we have: x3y = 12y ---> x3 = 12 ... answer (B)[/hide]

[quote="joml88"]Hello!??

Did you listen to a word I said?  If you had you would not have posted that on this page but instead on the solutions page.  That problem was already solved on there anyways.[/quote]

Apparently not. Though, you shouldn't waste your breath.


[quote="zhaobin"]$(a+b)(a^n+b^n) \ge (a^{n+1}+b^{n+1})$
and similar[/quote]

Would you mind posting the full solution? I do not get it :(

Yassin,  $(a^{k+1}+b^{k+1})(a^{n-k}+b^{n-k}) \ge (a^{n+1}+b^{n+1})$ then multiply everything


510 digits were printed in numbering the pages of a book.  If the first page of the book is numbered "1", how many pages does the book have?

Hint: [hide]Break it down into cases. For example, consider the case of the number printed being only one digit. What about if there were some two-digiters  printed? Three? So on.[/hide]

Mmm...close (well, not really , but I didn't want to sound too mean.). Another hint:[hide] start with how many digit total all the one-digit page numbers make. Then two digits, then three. That should give you a rough estimate. Then you could actually calculate the number of digit total all the one-digit page numbers added together makes, the number of digit total all the two-digit page numbers added together makes, and so on.[/hide]

I believe that is correct. Now try it if we were living in a world where everybody counted in base 5. (no, jk ;))

Why are you just kidding? There's nothing wrong with that problem if it has an answer (which it may not; I haven't checked). Well, you have to change the number of digits so that it doesn't have a 5 in it, but other than that...

yes, I realize the problem can be done; but it's not really necessary to do it since it's the same calculations as before, except using powers of 5. So what's the point in doing it? Never do a problem you already know how to do.

206.  just look at the ones, tens, and hundreds individually.


There are two questions I want to ask:

First: If a letter may be used twice, how many ways can the word "RADAR" be formed using the following diagram, always moving one space horizontally or vertically from one letter to the next?

                             R
                          R A R
                       R A D A R
                          R A R
                             R

The answer for this question is: 144 ways

Second: Starting at the M in the center, you may move left, right, up, or down to an adjoining letter. How many distinct paths can be followed to spell the word MATH?

                                          H
                                       H T H
                                    H T A T H
                                 H T A M A T H
                                    H T A T H
                                       H T H
                                          H

The answer for this question is: 28 paths..

I tried to do same method I used for upper question to do the bottom question but I don't get the same answer... Why is that?
Please help!!

Hmm.. are you sure you stated the question right? In the first one, if you can only move horizontally or vertically, theres only 4 ways. If you allow diagonally, I make it 196 ways. Maybe it said you can't use the same letter twice or something?

There are supposed to be spaces -- they are supposed to be diamond-type shapes, I believe.  Like this:

[hide]R A [/hide]R

[hide]A [/hide]R A R

R A D A R

[hide]A [/hide]R A R

[hide]R A [/hide]R

The second problem is supposed to look like problem 26 from

http://www.mathcounts.org/Problems/04sprintC.pdf

Sorry if I posted this in two forums, but I did it to help the people solving it.

By the way, the questions are different because in one case you move in and out, while in the other you move only in one direction.  This should affect the answer, but in general if you apply the same theory, it should work roughly the same.

I think the first question is Question #10 from Warmup-14..
Second question is the Question #26 from the site that is availble in this forum... The one with mathcounts.org,etc....
Hopefully, this helps and I can get some help..

do you understand the solution the mathcounts site gives?

um, just wondering, why did you post this on two forums?

I don't know why I put in two forums, or not that I remember..
Um.. And I don't get it the solution in MATHCOUNTS site..
So, that's why I'm asking great Mathletes in this site... :D


Obviosly $ 5$, because for all odd k $ k*5\equal{}5\mod 10$.

[quote="ellahoo"]Hi,

I am new here and sorry if I am not following some rules correctly.

can anyone help me with this question?

What is the smallest digit in the result of this product:

1*3*5* ..... *2001*2003

I know the smallest digit must be less than 4 but i can't go any further[/quote]

Stirling equivalence shows that required number begins with $ 308\ldots$

So the smallest digit in the required number is 0.

thank you [b]Patrick[/b], but is there any theoretical way to prove that? many people have solved it by using a calculator or computer program, but i needed to solve it in a step by step method.


After giving away registration codes last month, now Opera has removed the advertising from the free version, so now Opera is FREE!

What do ya'll think of this?

Great. I will be trying it out very soon!

Using Opera...it's ridiculously faster than IE.

Ugh. Anyone know why all of my webpages are being displayed in Arabic or Greek or w/e that is?

[quote="stupidkid"]Does anyone know why they made it free? Was it because of the competition from Firefox?[/quote]
[url]http://www.opera.com/free/[/url] this is what they say.

I downloaded it yesterday and am using it right now !:)
[size=59]
EDIT: I fixed the url and quote.[/size]

[quote="MithsApprentice"]So I'm the only one getting pages displayed in a weird language?[/quote]Maybe you're just the only one finding it a weird language? :P

[I think you should check your encoding... probably that one is misconfigured]

[quote="math92"]what is opera?[/quote]
A web browser like Internet Explorer.

i got it. it's pretty cool. don't know if i like it more than firefox or not. as for the language thing, when i went to install or download (can't remember which) i was asked what language i wanted. is it possible that you changed this without realizing?

I haven't downloaded Opera yet, but I'm probably now willing to make the switch from Firefox to Opera.

Can someone convince me how Opera is better (if it is)?

Firefox seems perfect for me currently.

Another question, does Google Web Accelerator work on Opera?

[quote="Peter VDD"]
[I think you should check your encoding... probably that one is misconfigured][/quote]

Honestly, I'm not completely technologically challenged. :P That's configured just fine. In fact, it was the first item I checked.

[quote="MithsApprentice"]So I'm the only one getting pages displayed in a weird language?[/quote]Opera 8.5 has some issues I think with UTF-8 encoding (which is what we use on the site BTW) ... anyway it was obvious they are going to make it free ... they weren't earning any money from it for a long time now, and in plus Firefox was taking away little of what they had as customers before :)

Wow, it seems Opera is a lot faster now than when it was in the 7.0 versions. Pretty nice.

[quote="stupidkid"]For some reason Opera seems to load pages faster than Firefox in Linux. But Opera takes up more resources so if you have a slow computer I would recomment FF over Opera.[/quote]

Not on my slow comp, firefox doesnt work but opera does  :rotfl:

Yes, in fact it's well know that it's quite the opposite: Firefox eats the most resources, and Opera can deal with little RAM and processor power :)


find the product of all numbers of the form (1-1/n)^-1 for 2 :le: n :le: 15


algebra
system of equations
algebra unsolved

guys seriously, answer at least one pleaseeee

this is the only questions I have left and I need this im begging you

please at least answer one

I need this please if anyone's listening

[quote="Croyt"]Please answer or give advice on the following...

Solve the non-linear system of equations for real solutions

1. x^2 + 3y^2 = 3

x+y = 8

2. y = 4x^2 +5

x+y = -8

3. y = x^2 + 2

y = -x^2 + 4

Please! Thank you so much![/quote]
 all's easy.
  solution 3:
$ 2y \equal{} 6$ hence  $ y \equal{} 3$ and $ x^2 \equal{} 1$
answer : $ ( \minus{} 1,0) ; (1;0)$
solution 1 ;  $ x \plus{} y \equal{} 8 \to x \equal{} 8 \minus{} y$(*)
 from (*)  with  (1)   we  have ;  $ 4 (y \minus{} 2)^2 \plus{} 45 \equal{} 0$
thus  system of equations hasn't root
solution 2 :prove  to like  problem 1


[quote="cezar lupu"]Let ABC a triangle where I is incenter,G is the intersection of medians and M,N are the midpoints AB and AC.
 Prove that the lines IG and BC are parallel if and only if M,I,N are collinear.[/quote]

IG is parallel to BC iff IG is parallel to MN since MN // BC

If M,I,N are collinear, since G is never on MN, how can IG // BC?

Did I misunderstand something here?

I'm sorry I have made a stupid mistake. I'll post the correct problem. :oops:

[quote]
Here it is: Let ABCbe a triangle and I is the incenter and G is the centroid.Prove that IG//BC then b+c=2a.
[/quote]
It's still the same.
At least to my understanding, you mean IG // BC in any triangle.  However it's obvious that it's not. We can prolong AI intersects BC at H, and prolong AG intersects BC at J. Since AJ is a median, AG/GJ = 1/2. So, only when AI / IH = 1/2 = AG/GJ, IG//BC.


geometry
circumcircle
incenter
projective geometry
power of a point
radical axis
geometry proposed

O,I are respectively the circumcenter and the incenter of a non-isosceles triangle ABC. The incircle (I) touches sides BC,CA,AB at D,E,F respectively. $ P\equal{}FD\cap AC, Q\equal{}DE\cap AB$. M,N are the midpoints of PE,QF. Prove that $ OI\perp MN$.

$ K \in AI$ is midpoint of $ EF,$ $ I_a$ is excenter in the angle $ \angle A.$ The external bisectors $ I_aB, I_aC$ of the angles $ \angle B, \angle C$ cut $ CA, AB$ at $ X, Y.$

$ KM \parallel FP$ is a midline of the $ \triangle PEF$ $ KN \parallel EQ$ is a midline of the $ \triangle QEF$. But $ FP \parallel I_aX$ (both $ \perp BI$) and $ EQ \parallel I_aY$ (both $ \perp CI$). The $ \triangle KMN, \triangle I_aXY$ are perspective from $ A$, by Desargues theorem, the intersections of their corresponding sides are collinear. As $ KM \parallel I_aX, KN \parallel I_aY,$ their perspective axis is the line at infinity and $ MN \parallel XY.$ $ XY \perp OI$ has been proved multiple times, for example, at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=202692[/url], etc.

[size=134][color=darkred][b]Nice and easy problem ![/b][/color][/size]

[quote="mr.danh"] [color=darkred]Let $ w = C(O,R)$ , $ C(I,r)$ be the circumcircle and the incircle respectively of a non-isosceles $ \triangle\ ABC$ . The incircle  touches $ BC$ , $ CA$ , $ AB$

at $ D$ ,$ E$ ,$ F$ respectively. Denote $ P = FD\cap AC$ , $ Q = DE\cap AB$ and $ M$ , $ N$ are the midpoints of $ PE$ ,$ QF$ . Prove that $ OI\perp MN$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Since $ OI\perp MN$ $ \Longleftrightarrow$ $ IM^2 - IN^2 = OM^2 - ON^2$ , I"ll prove $ \mathrm {R.H.S.}$  Since the divisions $ (B,F,A,Q)$ , $ (C,E,A,P)$

are harmonically, from an well-known property obtain $ \|\begin{array}{c} ME^2 = MA\cdot MC \\
\ NF^2 = NA\cdot NB\end{array}$ . Thus, $ \|\begin{array}{c} IM^2 = ME^2 + r^2 \\
\ IN^2 = NF^2 + r^2\end{array}\|$ $ \implies$

$ IM^2 - IN^2 = ME^2 - NF^2 =$ $ MA\cdot MC - NA\cdot NB =$ $ p_w(M) - p_w(N) =$ $ (OM^2 - R^2) - (ON^2 - R^2) =$ $ OM^2 - ON^2$ .[/color]

Already been solved here:
http://www.mathlinks.ro/viewtopic.php?t=207163

It is easy to show that the points $ M,\ N,$ lie on the radical axis of the circles $ (O)$ and $ (I).$

The line segments $ AD,\ BE,\ CF,$ are concurrent at [b][size=100]Gergone point[/size][/b] $ G_{e},$ as well.

From the complete quadrilateral $ CDG_{e}EAB,$ we have that the points $ B,\ F,\ A,\ Q,$ are in harmonic conjugation and because of $ NF \equal{} NQ,$ we conclude that $ (NF)^{2} \equal{} (NA)\cdot (NB)$ $ ,(1)$

From $ (1),$ we conclude that the point $ N,$ lies on the radical axis of $ (O),\ (I)$ and similarly for the point $ M.$

Hence, we conclude that $ OI\perp MN$ and the proof is completed.

Kostas Vittas.


[quote="zhangqiang"]???
let a=b=c=1
is there anything wrong?[/quote]
sorry,i edited.thank you :)

[quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that:$ abc \equal{} 1$.Prove that:
$ \frac {1 \plus{} a \plus{} ab}{a(a \plus{} b)^4} \plus{} \frac {1 \plus{} b \plus{} bc}{b(b \plus{} c)^4} \plus{} \frac {1 \plus{} c \plus{} ca}{c(c \plus{} a)^4}\geq\ \frac {81}{16(ab \plus{} bc \plus{} ca)^2}$[/quote]
As with $ abc\equal{}1$, we have
\[ \sum \frac{a}{1\plus{}a\plus{}ab} \equal{}1\]
Therefore, by Cauchy Schwarz and Iran 96, we have
\[ LHS \equal{}\left( \sum \frac{1\plus{}a\plus{}ab}{a(a\plus{}b)^4}\right) \left( \sum \frac{a}{1\plus{}a\plus{}ab}\right) \ge \left( \sum \frac{1}{(a\plus{}b)^2}\right)^2 \ge \frac{81}{16(ab\plus{}bc\plus{}ca)^2}.\]
;) :)

[quote="can_hang2007"][quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that:$ abc \equal{} 1$.Prove that:
$ \frac {1 \plus{} a \plus{} ab}{a(a \plus{} b)^4} \plus{} \frac {1 \plus{} b \plus{} bc}{b(b \plus{} c)^4} \plus{} \frac {1 \plus{} c \plus{} ca}{c(c \plus{} a)^4}\geq\ \frac {81}{16(ab \plus{} bc \plus{} ca)^2}$[/quote]
As with $ abc \equal{} 1$, we have
\[ \sum \frac {a}{1 \plus{} a \plus{} ab} \equal{} 1
\]
Therefore, by Cauchy Schwarz and Iran 96, we have
\[ LHS \equal{} \left( \sum \frac {1 \plus{} a \plus{} ab}{a(a \plus{} b)^4}\right) \left( \sum \frac {a}{1 \plus{} a \plus{} ab}\right) \ge \left( \sum \frac {1}{(a \plus{} b)^2}\right)^2 \ge \frac {81}{16(ab \plus{} bc \plus{} ca)^2}.
\]
;) :)[/quote]
your proof very nice :)


Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let $ B$ be the total area of the blue triangles, $ W$ the total area of the white squares, and $ R$ the area of the red square. Which of the following is correct?

[asy]unitsize(3mm);
fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue);
fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,red);
path onewhite=(-3,3)--(-2,4)--(-1,3)--(-2,2)--(-3,3)--(-1,3)--(0,4)--(1,3)--(0,2)--(-1,3)--(1,3)--(2,4)--(3,3)--(2,2)--(1,3)--cycle;
path divider=(-2,2)--(-3,3)--cycle;
fill(onewhite,white);
fill(rotate(90)*onewhite,white);
fill(rotate(180)*onewhite,white);
fill(rotate(270)*onewhite,white);[/asy]

$ \textbf{(A)}\ B \equal{} W \qquad \textbf{(B)}\ W \equal{} R \qquad \textbf{(C)}\ B \equal{} R \qquad \textbf{(D)}\ 3B \equal{} 2R \qquad \textbf{(E)}\ 2R \equal{} W$

[hide]We can divide it into triangles and find there are 24 (if I counted right) blue and 24 white. A.[/hide]

[quote=OhhhYouKnow]Time is precious. Learn to recognize FASTER ways to do problems you can already do.

[hide]Don't count ALL of the triangles. Note that the diagram can be divided into 4 identical quadrants. Now, just count the triangles in the upper left quadrant: Red 4, White 6, Blue 6 Answer: A[/hide][/quote]

you could actually split it into eigths and do it even faster!


If $ 5x \minus{} 3 \equal{} 12$, what is the value of $ 5x \plus{} 3$?

You could also note that if $ 5x \minus{}3 \equal{} 12$, then $ 5x \equal{} 15$, so our answer is just $ 15 \plus{} 3 \equal{} 18$.


Let $ A_0\equal{}(0,0)$. Distinct points $ A_1,A_2,\ldots$ lie on the $ x$-axis, and distinct points $ B_1,B_2,\ldots$ lie on the graph of $ y\equal{}\sqrt{x}$. For every positive integer $ n$, $ A_{n\minus{}1}B_nA_n$ is an equilateral triangle. What is the least $ n$ for which the length $ A_0A_n\ge100$?

$ \textbf{(A)}\ 13\qquad
\textbf{(B)}\ 15\qquad
\textbf{(C)}\ 17\qquad
\textbf{(D)}\ 19\qquad
\textbf{(E)}\ 21$

[hide]Consider the sequence of integers, $ \{1,1,2,2,3,3,4,4,5,5,\ldots\}.$  We can sum adjacent pairs of elements of the sequence in two different ways:  $ \{1,3,5,7,9,\ldots\}$ and $ \{2,4,6,8,10,\ldots\}.$

The sum of the first $ n$ odd integers is equal to $ n^2.$  The attached figure shows, in red, the sequence of the sums of the odd integers, along the $ x\text{ \minus{} axis}$, and the points $ (n^2,\sqrt {n^2}).$

The sum of the first $ n$ even integers is equal to $ n(n \plus{} 1) \equal{} 2T_n,$ where $ T_n$ is the $ n^{\text{th}}$ triangular number.  The attached figure also shows, in black, the sequence of the sums of the even integers along the $ x\text{ \minus{} axis}.$

The diagram demonstrates how a sequence of similar isosceles triangles, increasing arithmetically in size, can be fit in the graph of $ y \equal{} \sqrt {x}.$

But the problem specifies that the sequence of triangles must be equilateral.  If we scale our drawing, using a horizontal scale factor of $ f$, and a vertical scale factor of $ \sqrt {f},$ we will have a different sequence of similar isosceles triangles, which still fits in the curve $ y \equal{} \sqrt {x}.$

So, let's compute the scale factor, $ f,$ so that we have an equilateral triangle.  After scaling, the three points of the leftmost triangle will become $ A_0 \equal{} (0,0),$ $ A_1 \equal{} (2f,0),$ $ B_1 \equal{} (f,\sqrt {f}).$  We apply the distance formula to two sides and solve:

$ |A_0 A_1|^2 \equal{} |A_0 B_1|^2$
$ (2f)^2 \equal{} f^2 \plus{} f$
$ 3f^2 \minus{} f \equal{} 0$
$ (f)(3f \minus{} 1) \equal{} 0$
$ f \equal{} 0$ or $ f \equal{} \frac {1}{3}$

The base points are thus $ A_n \equal{} (f\times 2T_n,0) \equal{} \Big(\frac {2}{3}T_n,0\Big).$

Now, find $ n,$ such that $ \frac {2}{3}T_n\ge 100\longrightarrow T_n\ge 150.$

Calculate $ T_{16} \equal{} 136,$ $ T_{17} \equal{} 153.$  So $ n \equal{} 17$ is the least $ n$ at which the distance from $ A_0$ to $ A_{17}$ is greater than or equal to $ 100.$[/hide]

That's a tremendous observation to make, and hence a nice solution  :)

I went for the more thoughtless and slightly more time-consuming algebra-crank:

[hide="Solution"]
Consider $ A_n$ and $ A_{n \plus{} 1}$. Then, we know that the $ x$ coordinate of $ B_{n \plus{} 1}$ is $ \frac {A_n \plus{} A_{n \plus{} 1}}{2}$, and thus the y coordinate was $ \sqrt {\frac {A_n \plus{} A_{n \plus{} 1}}{2}}$. We also know that this is $ \frac {\sqrt {3}}{2} (A_{n \plus{} 1} \minus{} A_{n})$ by equilateral triangle side ratios. Squaring and multiplying by 4, and bringing everything to one side, we get: $ 3A_{n \plus{} 1}^2 \plus{} 3A_n ^2 \minus{} 6 A_{n \plus{} 1} A_n \minus{} 2A_{n \plus{} 1} \minus{} 2A_n \equal{} 0$, an invariant.

Setting the case of n equal to the case of n+1, we get:

$ 3(A_{n \plus{} 2}^2 \minus{} A_n ^2) \minus{} 6 A_{n \plus{} 1} (A_{n \plus{} 2} \minus{} A_n) \minus{} 2 (A_{n \plus{} 2} \minus{} A_n) \equal{} 0$
$ 3(A_{n \plus{} 2} \plus{} A_n) \minus{} 6 A_{n \plus{} 1} \minus{} 2 \equal{} 0$
$ A_{n \plus{} 2} \minus{} 2A_{n \plus{} 1} \plus{} A_{n} \equal{} \frac {2}{3}$
$ (A_{n \plus{} 2} \minus{} A_{n \plus{} 1}) \minus{} (A_{n \plus{} 1} \minus{} A_n) \equal{} \frac {2}{3}$

Thus we have a quadratic equation. For convenience, we can multiply by 3 and aim over 300, and then we can just crank it out with finite differences to get $ 3A_{17} \equal{} 306 > 300 \implies \boxed{C. \; 17}$[/hide]

This was, by the way, my favorite problem on the test

Let $ x_n$ denote the $ x$-coordinate of point $ A_n$. Consider the system: \[ x\equal{}y^2\qquad y\equal{}\sqrt{3}\cdot (x\minus{}x_n)\]
\[ x\equal{}3\cdot (x\minus{}x_n)^2\iff x^2\plus{}\left(\minus{}2x_n\minus{}\frac{1}{3}\right)\cdot x\plus{}x_n^2\equal{}0\] The two solutions of this system correspond to the vertex of the $ n$th triangle reflected over the $ x$-axis and the vertex of the $ n\plus{}1$th triangle. The $ x$-coordinate of the $ n$th triangle is the average of the $ x$-coordinates of $ A_{n\minus{}1}$ and $ A_n$. Noting that the sum of the roots is equal to the opposite of the $ x$ coefficient, \[ \frac{x_{n\minus{}1}\plus{}x_n}{2}\plus{}\frac{x_n\plus{}x_{n\plus{}1}}{2}\equal{}2x_n\plus{}\frac{1}{3}\iff (x_{n\plus{}1}\minus{}x_n)\minus{}(x_n\minus{}x_{n\minus{}1})\equal{}\frac{2}{3}\]

Hence $ y_n: \equal{}x_n\minus{}x_{n\minus{}1}$ is an arithmetic series. We can set $ x_{\minus{}1}\equal{}x_0\equal{}0$, it follows that \[ x_n\minus{}x_{n\minus{}1}\equal{}\frac{2}{3}\cdot n\] Then \[ x_n\equal{}(x_n\minus{}x_{n\minus{}1})\plus{}(x_{n\minus{}1}\minus{}x_{n\minus{}2})\plus{}...\plus{}(x_1\minus{}x_0)\plus{}x_0\equal{}\frac{1}{3}\cdot (n)(n\plus{}1)\] Now we need to find the minimal $ n$ such that $ x_n\ge 100$. That is, we must solve $ x_n\ge 100$ and $ x_{n\minus{}1}< 100$. It is easy to see that $ n\equal{}17$ is the correct $ n$, so $ \boxed{C}$ is the answer.

Once we find this: $(x_{n+1}-x_n)-(x_n-x_{n-1})=\frac{2}{3}$, how do we find out that each x-value is just $\frac{2}{3}$ times a triangular number if we don't notice it on the spot?

In other words, what method can we use to find the general formula if we are given something like this: $(x_{n+1}-x_n)-(x_n-x_{n-1})=\frac{2}{3}$

The finite difference minus the next one equals $\frac{2}{3}$. Thus each finite difference is $\frac{2}{3}$ more than the next. This means that the sequence of finite difference is increasing by $\frac{2}{3}$ each term. We see the first finite difference is $0$. So the sequence of finite differences is $\frac{2}{3}n$. We can add and telescope to find any arbitrary term.

I understand what you mean by [quote=always_correct]

This means that the sequence of finite difference is increasing by $\frac{2}{3}$ each term.[/quote]

but I don't understand how [quote=always_correct]We can add and telescope to find any arbitrary term.[/quote]

Look at the last piece of latex Altheman shows. See what's happening?


Let $k\geq2$ be an integer, what is the value of
\[\sum_{n=2}^{n=+\infty}\frac1{n^k-1}\ ?\]

I have strictly no idea whether this is well-known, hard, etc. except of course for the case $k=2$, which is trivial...

I think the following link will be interesting for you http://www.mathlinks.ro/Forum/viewtopic.php?t=5101 (Goldenbach's theorem). Though it doesn't give the answer for your question.

I'm not sure but I think it's SRMO (Silk Road Mathematical Olympiad)2003


The sum of the base- 10 logarithms of the divisors of $ 10^n$  is $ 792$ . What is $ n$?

I didn't like the first solution on the wiki because it resorted to basically plugging into the multiple choice. If someone could explain expand on the second solution or provide an entirely different method all together, then I would greatly appreciate it.

you mean this solution?

http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=2008&p=1050540

well, once you get the equation $ n(n \plus{} 1)^2 \equal{} 1584$, the quickest way to solve it (with or without a calculator) is to look for a cube near 1584 (since the LHS of the equation is slightly larger than $ n^3$. $ 12^3 > 1584$, so you know $ n < 12$. you then try 11, and 11 works. but, in this case, you *know* 11 has to work since it's the only choice $ < 12$. and that's it.

another way to get it is to use this fact: for every natural number $ k$, the product of the positive divisors of $ k$ equals $ k^{\frac {d(k)}{2}}$. then since "sum of $ \log$'s equals $ \log$ of the product," you get $ 10^{792} \equal{} (10^n)^{\frac {d(10^n)}{2}}$, i.e. $ 1584 \equal{} nd(10^n) \equal{} n(n \plus{} 1)^2$.

[hide]
There's a rule of logarithms that states:
$ \log{a_1}\plus{}\log{a_2}\plus{}\log{a_3}\plus{}\cdots\plus{}\log{a_n}\equal{}\log{a_1a_2a_3\cdots a_n}$
Therefore, the base-10 logarithm of the product of the divisors of $ 10^n$ equals $ 792$.
Note that $ 10^n\equal{}2^n\cdot5^n$, and by counting divisors, there are $ (n\plus{}1)^2$ divisors of $ 10^n$.
Also note that for every divisor $ k$ of $ 10^n$, there is a unique divisor $ l$ of $ 10^n$ such that $ kl\equal{}10^n$.*
Therefore, the product of the divisors is $ (10^n)^{\frac{(n\plus{}1)^2}{2}}$.* The base-10 logarithm of this is $ n\cdot\frac{(n\plus{}1)^2}{2}\equal{}792$ or $ n(n\plus{}1)^2\equal{}1584$. Note that $ 1584\equal{}2^4\cdot3^2\cdot11\equal{}11\cdot12^2$. Therefore, $ n\equal{}11$.
*If $ n$ is even, then $ 10^n$ would have an odd number of divisors. The odd one out would be $ k\equal{}\sqrt{10^n}$, and $ k\equal{}l$. The formula still holds because $ \sqrt{10^n}\equal{}10^{\frac{n}{2}}$.
[/hide]

[hide]
Using log properties, 792= log(product of the factors of 10^n)

The formula for the product of the factors of a number n is n^((number of divisors of n)/2). For example the product of the factors of 12 is 12^6/2=12^3 because 12 has 6 factors.

Now, 10^n=2^n * 5^n, so it has (n+1)^2 factors. Thus, the product of the factors of 10^n is 10^(n(n+1)^2/2). The log of this quantity is just the exponent, so now, n(n+1)^2=792*2=1584

1584 is divisible by 11 because 1-5+8-4=0, which is a multiple of 11. 1584=11*144. It works! n=11.[/hide]


let ABC be a triangle. and (C) be a circle which passes through B and C. and let (C') be a circle which is tangent to (C) ,AB,AC at D,E,F respectively and I is the incenter of ABC. prove that ID is the bisector of (< BDC).

Maybe you meant "prove that ID is the bisector of (< BDC)" rather than "prove that ID is the bisector of (< EDF)"? If this is the case, you can find your problem discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=6086 .

  darij


calculus
derivative
function
limit
logarithms
calculus computations

Function $f(x)=x^{x}$ , and we can use $x^{x}=e^{x lnx}$ to easily get it's derivative $f'(x)=x^{x}(ln{x}+1)$

but why can't we do it by directly using the definition of derivative ?

$f'(x)=\lim_{\Delta x\to 0}\frac{\Delta f(x)}{\Delta x}=\lim_{\Delta x\to 0}\frac{(x+\Delta x)^{x+\Delta x}-x^{x}}{\Delta x}$

by this method , I can't continue to do .

So... how are you going to evaluate those limits? It's possible, but that's the hard way. The method you mentioned (or equivalently logarithmic differentiation) is the easy way.

[quote="Love U"]Function $f(x)=x^{x}$ , and we can use $x^{x}=e^{x lnx}$ to easily get it's derivative $f'(x)=x^{x}ln(x+1)$

but why can't we do it by directly using the definition of derivative ?

$f'(x)=\lim_{\Delta x\to 0}\frac{\Delta f(x)}{\Delta x}=\lim_{\Delta x\to 0}\frac{(x+\Delta x)^{x+\Delta x}-x^{x}}{\Delta x}$

by this method , I can't continue .[/quote]

[quote="Love U"]  $f'(x)=x^{x}ln(x+1)$  [/quote]it's not true.
I show you answer as you want:
1) we need to find
$\lim_{\epsilon\to 0}\frac{(x+\epsilon)^{x+\epsilon}-x^{x}}{\epsilon}=\lim_{\epsilon\to 0}\frac{ e^{(x+\epsilon)\ln{(x+\epsilon)}}-e^{x\ln{x}}}{ \epsilon }=lim_{\epsilon\to 0}\frac{ e^{(x+\epsilon)(\ln{x}+\ln{(1+\frac{\epsilon}{x})})}-e^{x\ln{x}}}{\epsilon}$

by Taylor formula you can show that
$\ln{(1+\frac{\epsilon}{x})}=\frac{\epsilon}{x}+O(\epsilon^{2})$
therefore we have

$lim_{\epsilon\to 0}\frac{e^{(x+\epsilon)(\ln{x}+\ln{(1+\frac{\epsilon}{x})})}-e^{x\ln{x}}}{\epsilon}= lim_{\epsilon\to 0}\frac{e^{(x+\epsilon)(\ln{x}+(1+\frac{\epsilon}{x}+O(\epsilon^{2}))}-e^{x\ln{x}}}{\epsilon}= lim_{\epsilon\to 0}\frac{e^{x\ln{x}}(e^{\epsilon\ln{x}+\epsilon+\frac{\epsilon^{2}}{x}+O(\epsilon^{2})}-1)}{\epsilon}$
also by Taylor you can show that
$e^{\epsilon\ln{x}+\epsilon+\frac{\epsilon^{2}}{x}+O(\epsilon^{2})}-1=\epsilon\ln{x}+\epsilon+\frac{\epsilon^{2}}{x}+O(\epsilon^{2})$
therefore we will be have
$lim_{\epsilon\to 0}\frac{e^{x\ln{x}}(e^{\epsilon\ln{x}+\epsilon+\frac{\epsilon^{2}}{x}+O(\epsilon^{2})}-1)}{\epsilon}=lim_{\epsilon\to 0}\frac{e^{x\ln{x}}(\epsilon\ln{x}+\epsilon+\frac{\epsilon^{2}}{x}+O(\epsilon^{2}))}{\epsilon}$=
$e^{x\ln{x}}(\ln{x}+1)$
Also note that by analogysly method you can find derivetive of this $x^{x^{x}}$ good exercises  :wink:

hey, very cool ! :thumbup: i made one correction...

[quote="Extremal"]1) we need to find
$\lim_{\epsilon\to 0}\frac{(x+\epsilon)^{x+\epsilon}-x^{x}}{\epsilon}=\lim_{\epsilon\to 0}\frac{ e^{(x+\epsilon)\ln{(x+\epsilon)}}-e^{x\ln{x}}}{ \epsilon }=\lim_{\epsilon\to 0}\frac{ e^{(x+\epsilon)(\ln{x}+\ln{(1+\frac{\epsilon}{x})})}-e^{x\ln{x}}}{\epsilon}$

by Taylor formula you can show that
$\ln{(1+\frac{\epsilon}{x})}=\frac{\epsilon}{x}-O(\epsilon^{2})$
therefore we have

$\lim_{\epsilon\to 0}\frac{e^{(x+\epsilon)(\ln{x}+\ln{(1+\frac{\epsilon}{x})})}-e^{x\ln{x}}}{\epsilon}= \lim_{\epsilon\to 0}\frac{e^{(x+\epsilon)(\ln{x}+\frac{\epsilon}{x}-O(\epsilon^{2}))}-e^{x\ln{x}}}{\epsilon}=e^{x\ln{x}}\left( \lim_{\epsilon\to 0}\frac{e^{\epsilon(\ln{x}+1)+O(\epsilon^{2})}-1}{\epsilon}\right)$

the last step follows from
$(x+\epsilon)(\ln{x}+\frac{\epsilon}{x}-O(\epsilon^{2}))\;=\;x\ln x+\epsilon-x\; O(\epsilon^{2})+\epsilon \ln x+\frac{\epsilon^{2}}{x}-O(\epsilon^{3})\\ \\ =\;x\ln x+\epsilon(\ln x+1)+O(\epsilon^{2})\\ \\ \left[ \;\;\;\quad since \;\;\; O(\epsilon^{2})\gg O(\epsilon^{3}) \;\;\;\quad and \quad\;\;\;-x\; O(\epsilon^{2})+\frac{\epsilon^{2}}{x}\;\sim \; O(\epsilon^{2}) \;\;\;\quad \right]$

now, by Taylor you can show that
$e^{\epsilon(\ln{x}+1)+O(\epsilon^{2})}-1=\epsilon(\ln{x}+1)+O(\epsilon^{2})$
therefore, we will have
$e^{x\ln{x}}\left( \lim_{\epsilon\to 0}\frac{e^{\epsilon(\ln{x}+1)+O(\epsilon^{2})}-1}{\epsilon}\right) \;=\;e^{x\ln{x}}\left( \lim_{\epsilon\to 0}\frac{\epsilon(\ln{x}+1)+O(\epsilon^{2})}{\epsilon}\right) \\ \\ =\;e^{x\ln{x}}\left( \lim_{\epsilon\to 0}\left( (\ln{x}+1)+O(\epsilon)\right) \right) \;=\;e^{x\ln{x}}(\ln{x}+1)=\boxed{x^{x}(\ln x+1)}$[/quote]

[quote="Love U"]Function $f(x)=x^{x}$ , and we can use $x^{x}=e^{x lnx}$ to easily get it's derivative $f'(x)=x^{x}ln(x+1)$[/quote]
Just to make sure you know:

$f'(x) = x^{x}(\ln(x)+1)$


inequalities
geometry
circumcircle
inradius
inequalities proposed

Let  ABC  is  an  acute trianlge. Prove  that
max(ha,hb,hc)>=R+r
Thanks!

[i]unknown source[/i]

This is what I call beautiful inequalities:

Well here is a solution:
We use Carnot's relation R+r=d_a+d_b+d_c where d_a is the distance between the center of the circumscriebed triangle to BC.
And the inequality comes from the fact that:
min{a,b,c}*(d_a+d_b+d_c)<=2S=min{a,b,c} *max{h_a,h_b,h_c}

cheers! :D :D

Well the inequality could be extended to any triangle!
In case that it is a right angled traingle with m(A)=90 we rewrite the inequality like so:
max{b,c}>=a/2+p-a which is equivalent to max{b,c}>=(b+c)/2 which is obious!

cheers!

Thanks  for  the solution!!!
I  hope you  know  that  Carnot  relation  holds  for any  point  in  the  interior  of  a  triangle.
nickolas.

I just accidentally came across this topic and wanted to make a little comment, namely that both of the remarks

[quote="Lagrangia"]Well the inequality could be extended to any triangle![/quote]

and

[quote="nickolas"]Carnot  relation  holds  for any  point  in  the  interior  of  a  triangle.[/quote]

are wrong.

The problem itself, namely to show the inequality $\max\left\{h_a;\;h_b;\;h_c\right\}\geq R+r$ for an acute triangle ABC with altitudes $h_a$, $h_b$, $h_c$, circumradius R and inradius r, seems to be quite a classical problem. I saw it mentioned at http://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=29627 .

  darij


modular arithmetic
Ring Theory
polynomial
number theory proposed
number theory

(a) Show that the product of two numbers of the form $ a^2\plus{}3b^2$ is again of that form.

(b) If an integer $ n$ is such that $ 7n$ is of the form $ a^2\plus{}3b^2$, prove that $ n$ is also of that form.

Oh, how trivial!

a) Let $ x \equal{} a^2 \plus{} 3b^2$, $ y\equal{}c^2 \plus{} 3d^2$

$ xy \equal{} a^2c^2 \plus{} 3a^2d^2 \plus{} 3b^2c^2 \plus{} 9b^2d^2 \equal{} (ac \plus{} 3bd)^2 \minus{} 6abcd \plus{} 3a^2d^2 \plus{} 3b^2c^2 \equal{} (ac \plus{} 3bd)^2 \plus{} 3(ad \minus{} bc)^2$

Yeah, part a is quite trivial. :D  What about part b?

I've got no idea other than dumbassing (checking all possible value for $ a,b$ modulo 7). :oops:

assume that $ 7n\equal{}a^2\plus{}3b^2$
then $ n\equal{}x^2\plus{}3y^2$, where: $ x\equal{}\frac{2a\minus{}3b}{7}, y\equal{}\frac{2b\plus{}a}{7}$ or $ x\equal{}\frac{2a\plus{}3b}{7}, y\equal{}\frac{2b\minus{}a}{7}$

First case: $ 7|b$. Then also $ 7|a$, and so:  $ n/7\equal{}(a/7)^2\plus{}3(b/7)^2$, and because $ 7\equal{}2^2\plus{}3(1)^2$, by part (a) we get that n is of the requested form.

Second case: $ b,a$ are not divisible by 7.
$ a^2\plus{}3b^2 \equiv 0 \pmod 7 \implies (ba^{\minus{}1})^2 \equiv \minus{}1*3^{\minus{}1} \equiv 2 \pmod 7$
$ \implies ba^\minus{}1 \equiv 3,4 \pmod 7$
Again we have 2 cases - 1st case - $ ba^{\minus{}1} \equiv 3 \pmod 7$. It's easy to verify that $ x\equal{}\frac{2a\minus{}3b}{7}, y\equal{}\frac{2b\plus{}a}{7}$ are integers, so $ n\equal{}x^2\plus{}3y^2$ and we're done. 2nd case - $ ba^{\minus{}1} \equiv 4 \pmod 7$. It's easy to verify that $ x\equal{}\frac{2a\plus{}3b}{7}, y\equal{}\frac{2b\minus{}a}{7}$ are integers, so $ n\equal{}x^2\plus{}3y^2$ and we're done (now for real...).

Note that $ 7 \equal{} 2^2 \plus{} 3 \cdot 1^2$.  What you are really doing in part b) is performing division in $ \mathbb{Z}[ \sqrt { \minus{} 3} ]$, a subring of the [url=http://en.wikipedia.org/wiki/Eisenstein_integers]Eisenstein integers[/url].  We are given that $ 7 | a^2 \plus{} 3b^2$, so either $ 2 \plus{} \sqrt { \minus{} 3} | a \plus{} b \sqrt { \minus{} 3}$ or $ 2 \minus{} \sqrt { \minus{} 3} | a^2 \plus{} b \sqrt { \minus{} 3}$, hence either

$ \frac {a \plus{} b \sqrt { \minus{} 3} }{ 2 \plus{} \sqrt { \minus{} 3} } \equal{} \frac {(2a \plus{} 3b) \plus{} (2b \minus{} a) \sqrt { \minus{} 3}}{7}$

or

$ \frac {a \plus{} b \sqrt { \minus{} 3} }{2 \minus{} \sqrt { \minus{} 3} } \equal{} \frac {(2a \minus{} 3b) \plus{} (2b \plus{} a) \sqrt { \minus{} 3}}{7}$.

A simpler version of this computation is as follows:  if $ 2m$ is a sum of two squares, so is $ m$.

[quote="bambaman"]
Second case:
$ a^2 \plus{} 3b^2 \equiv 0 \pmod 7 \implies (ba^{ \minus{} 1})^2 \equiv \minus{} 1*3^{ \minus{} 1} \equiv 2 \pmod 7$
$ \implies ba^ \minus{} 1 \equiv 3,4 \pmod 7$
.[/quote]
I don't understand it.Can you explain clearly?

[hide="Solution"]
For the first part, we can notice that $ a^2 \plus{} 3b^2 \equal{} (a \plus{} bi\sqrt {3})(a \minus{} bi\sqrt {3})$.  Then, we get\[ (a^2 \plus{} 3b^2)(c^2 \plus{} 3d^2) \equal{} (a \plus{} bi\sqrt {3})(c \plus{} di\sqrt {3})(a \minus{} bi\sqrt {3})(c \minus{} di\sqrt {3})\]which is\[ [(ac \minus{} 3db) \plus{} (ad \plus{} bc)i\sqrt {3})][(ac \minus{} 3db) \minus{} (ad \plus{} bc)i\sqrt {3}]\]This is $ k^2 \plus{} 3l^2$ when $ k \equal{} ac \minus{} 3db$ and $ l \equal{} ad \plus{} bc$.

For the second part, we notice that\[ 7(a^2 \plus{} 3b^2) \equal{} 7a^2 \plus{} 21b^2 \equal{} (2a \minus{} 3b)^2 \plus{} 3(2b \minus{} a)^2\][/hide]

EDIT: Oh, sorry!  I misread part (b).

[quote="The QuattoMaster 6000"]For the second part, we notice that
\[ 7(a^2 \plus{} 3b^2) \equal{} 7a^2 \plus{} 21b^2 \equal{} (2a \minus{} 3b)^2 \plus{} 3(2b \minus{} a)^2
\][/quote]
This is the converse of what is being asked.


superior algebra
superior algebra solved

I guess it's not a hard problem, but it took me some time to solve it:

Let (G,.) be a group s.t.

a)x^2=0 => x=0
b)(xy)^2=(yx)^2 for all x and y \in G.

Show that G is commutative.
[/i]

It is enough to show that x2 is injective.  :)

For this take x2=y 2
Then xy-1=x-1y
So (xy-1)2=(x-1y)2=(yx-1)2=(xy-1)-2
This gives (xy-1)4=1 => (xy-1)2=1 => xy-1=1 => x=y.:)