How can I determine whether the integer 9833 is prime? Thanks.

First, take the square root: about 100. Your best bet is to divide by every prime from 1 to 100. Sound fun? This is why RSA is so secure.

Clearly not 2, 3, 5, 7, or 11.

13?
17?
19?
23?
29?
31?
37?
41?
43?
47?
53?
61?
67?
71?
....

You divide each of those primes by the origial number, to test if it's prime, because for every factor under the square root, there is a factor which corresponds to it over the square root, so you only need to test the ones under.

[quote="MathStudent2"]SQRT(9833) = 99.16  If I am to round that it would give me 99.  Now listing all of the primes from 2 to 99 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

Well 3 goes into 99 even?  Or, should I round 99.16 to 99 at all before I check to see if the primes up to 99 divide evenly?  I know the answer is prime, just trying to figure it out.

thanks.[/quote]

what? i don't get your method? read Bill The Maniac's post, i think your mixing up something

the square root is 9833 is about 99. So divide 9833 by ALL PRIMES LESS THAN 99. so $9833 \div 2$, $9833 \div 3$, $9833 \div 5$, etc. until you get to 97. if any of them come out to be an integer then the original number (9833) is not prime

7h3.D3m0n.117, I did confuse what numbers to divide to check and see if it is prime.  I see it now.  Makes sense.  Thanks for everybody's help!


Why the commands \displaystyle and \limits don't work in my signature? I tried to type
[code]\left(\int\limits_{I}f(t)dt\right)^{k}= \int\limits_{I^{k}}f(t_{1})\cdot\ldots\cdot f(t_{k}) dt_{1}\ldots dt_{k}[/code]
and
[code]\displaystyle \left(\int_{I}f(t)dt\right)^{k}= \int_{I^{k}}f(t_{1})\cdot\ldots\cdot f(t_{k}) dt_{1}\ldots dt_{k}[/code]
but it is automatically replaced with
[code]\left(\int_{I}f(t)dt\right)^{k}= \int_{I^{k}}f(t_{1})\cdot\ldots\cdot f(t_{k}) dt_{1}\ldots dt_{k}[/code]
I don't know why, because I didn't exceed the 128 character limit.

Those commands are redundant on the forum; displaystyle is automatically on, and you just need superscripts and subscripts for the limits of your integral. The simplification of your code did not change the appearance of that line.

So why do $I$ and $I^{k}$ appear as subscripts and not as limits? I replaced $...$ with $\ldots$ today and before this change it worked fine.


Determine all positive integers whose squares end in $196$.

Here's a hint that should solve the problem.

The last digit of the number must be either $6$ or $4$ for it to end in $196$.

So these are the two cases:

Let the number be $100a+10b+c$ where $c=6,4$ (of course, all we need to know are the last three digits of the number since that determines the last three digits of the square)

We have $(100a+10b+c)^{2}=(100a+10b)^{2}+2c(100a+10b)+c^{2}=\cdots196$.

For the first case of $c=4$, we have:

$100^{2}a^{2}+1000ab+100b^{2}+800a+80b+16=\cdots196$

Looking at this $\mod{1000}$, we have:

$100b^{2}+800a+80b+16=196 \mod{1000}$

Now use trial and error and different conditions for $(a,b)$

Similarily, for the case of $c=6$, we have:

$100b^{2}+200a+120b+36=196 \mod{1000}$

and just use trial and error there for $(a,b)$

maybe not the most efficient method but eh...

This is the one method that works... the other is by factoring :)

If you  note that $196= \pm 14^{2}$ the answer is trivial,[hide]
 $(500k \pm 14)$where k is an integer..

 Or last 3digits has to be one of these.
(014), (1000-014=986  , 514, and (500-14 = 486)

[hide="because"]$\pm 14$ both mode 125 and 8 and CRT means $( \pm 14 \bmod 500$) [/hide][/hide]

lol wow i didn't realize $196$ was a square.... :oops:


analytic geometry
geometry
rhombus
geometric transformation
reflection
quadratics
symmetry

The graphs of the equations \[ y=k, \qquad y=\sqrt{3}x+2k, \qquad y=-\sqrt{3}x+2k, \] are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.$  These 63 lines cut part of the plane into equilateral triangles of side $2/\sqrt{3}.$  How many such triangles are formed?

[hide]There are 21 lines that are horizontal cutting the plane into 20 sections. Then the second kind of equation splits each of those sections into 20 rhombi. Finally the third kind of equation splits each rhombus along its shorter diagonal, to make 2 of each of the 400 rhombi, so $\boxed{800}$?[/hide]

[hide]
The equilateral triangles have two different orientations: point facing up or point facing down.

For the triangles with the point facing up:

The bases of all the equilateral triangles are the lines $y=k$, so the vertices opposite the bases of these triangles come from the intersection of the lines $y = \sqrt{3}x + 2a$ and $y = -\sqrt{3}x + 2b$ (the values of $k$ may be different in a given pair of intersecting lines).

Adding the two equations together and solving for $y$ to find the point of intersection, we get:

$y = a + b$

The value of $y$ must fall between $11$ and $-9$, since the horizontal lines only run between $10$ and $-10$, and an equilateral triangle of side $\frac{2}{\sqrt{3}}$ has a height of $1$.

The problem then becomes how many different pairs of numbers add up to $11, 10, 9, ... -9$

For $11$, $a$ can range from $10$ through $1$, so there are $10$ possible intersections. For $10$, $a$ can range from $10$ through $0$, so there are $11$ intersections. The number of intersections increases by $1$ each time, until the case where $y=0$, where there are $21$ intersections, and decreases by $1$ until the case $y=-9$, where there are $12$ cases. The sum of all these numbers is equal to $330$.

The case for where the triangles are point facing downward is almost identical, with $y$ ranging from $9$ through $-10$.

The total number of equilateral triangles is then equal to $2\times 330 = \boxed{660}$
[/hide]

If you notice, its just the number of intersection of the lines-1 (try for a triplet of 1 line, 3 lines-...it has to be odd since its going to be reflection) is the regions created

Now, the hard part is finding the number of intersections...

Again, by looking for a pattern, the number of intersections is (incredibly this works), {(nC2)-3[(n/3)C(2)]-1}/(2), where n is the number of total lines.  So its counting the number of ideal intersections of n lines, but it has to -3*(n/3)C(2) because they are parallel and will never interset.  Then divide by 2 and subtract 1 becuase out of the nC2-3((n/3)C(2)) points, there were (n+1)/2 that had three lines (like the point (0,0) which is concurrent) going through them and (n-1)/2 that had two lines going through them.

So we have a triplet of 21 lines so thats 63 lines.  [63C2-3(21C2)-1]/2=661.

Then from my first part, the number of intersections-1 is the number of regions, so 661-1=660...

[quote="Adunakhor"][hide]
The equilateral triangles have two different orientations: point facing up or point facing down.

For the triangles with the point facing up:

The bases of all the equilateral triangles are the lines $ y \equal{} k$, so the vertices opposite the bases of these triangles come from the intersection of the lines $ y \equal{} \sqrt {3}x \plus{} 2a$ and $ y \equal{} \minus{} \sqrt {3}x \plus{} 2b$ (the values of $ k$ may be different in a given pair of intersecting lines).

Adding the two equations together and solving for $ y$ to find the point of intersection, we get:

$ y \equal{} a \plus{} b$

The value of $ y$ must fall between $ 11$ and $ \minus{} 9$, since the horizontal lines only run between $ 10$ and $ \minus{} 10$, and an equilateral triangle of side $ \frac {2}{\sqrt {3}}$ has a height of $ 1$.

The problem then becomes how many different pairs of numbers add up to $ 11, 10, 9, ... \minus{} 9$

For $ 11$, $ a$ can range from $ 10$ through $ 1$, so there are $ 10$ possible intersections. For $ 10$, $ a$ can range from $ 10$ through $ 0$, so there are $ 11$ intersections. The number of intersections increases by $ 1$ each time, until the case where $ y \equal{} 0$, where there are $ 21$ intersections, and decreases by $ 1$ until the case $ y \equal{} \minus{} 9$, where there are $ 12$ cases. The sum of all these numbers is equal to $ 330$.

The case for where the triangles are point facing downward is almost identical, with $ y$ ranging from $ 9$ through $ \minus{} 10$.

The total number of equilateral triangles is then equal to $ 2\times 330 \equal{} \boxed{660}$
[/hide][/quote]

I didn't understand the part which you concluded:
The value of $ y$ must fall between $ 11$ and $ \minus{} 9$

[quote="Adunakhor"]
For $ 11$, $ a$ can range from $ 10$ through $ 1$, so there are $ 10$ possible intersections. For $ 10$, $ a$ can range from $ 10$ through $ 0$, so there are $ 11$ intersections. The number of intersections increases by $ 1$ each time, until the case where $ y \equal{} 0$, where there are $ 21$ intersections, and decreases by $ 1$ until the case $ y \equal{} \minus{} 9$, where there are $ 12$ cases. The sum of all these numbers is equal to $ 330$.
[/quote]
I think you over counted here, the diagonal lines don't always intersect on a horizontal line, and those intersections are not vertices of the correct triangles, so the number of triangles on each line actually increase and decrease by 2 each time and the total number of triangles facing one direction is actually $ 220$ so the total number of triangles is $ 440$.

If I am wrong, can someone tell me where I made a mistake please.

[quote="Differ"][hide]We have points (0, 0) (1, 12) (2, 36), where the x coordinate is the maximum value of x in the original equation. We graph it as f(x) and assume it is a quadratic, which becomes $ f(x) \equal{} 6x^2 \plus{} 6x$. Plugging in 10 gives us $ \boxed{660}$.[/hide][/quote]

How did you get (0, 0) (1, 12) (2, 36)?

In $ (u, v)$, $ v$ was the number of triangles I got when I had $ x \equal{} 2u \minus{} 1$ lines. I made the lines and counted the triangles.

Wow, this #6 is sooooo EVIL.. it took me almost an hour to get the ans correctly (didn't get it right in the first try b/c of dumb mistakes and poor insights)... but i finally got it right... yay!

Actually the answer is just $ 2\sum_{k \equal{} 0}^9 [2(2*10 \minus{} k) \plus{} 1] \plus{} 4(2*10 \minus{} 10)/2$
considering the symmetry in +ve and -ve y direction and considering each row by row... and of course not forgetting there's half of the usual no. of triangles in $ y \equal{} 11$... perhaps there's a much better solution that mine (which was basically brute-force)...

By the way could you come up with a formula for $ 2n \plus{} 1$ horizontal lines (in this case the y-intercepts range from $ \minus{} n$ to $ n$) and $ m \plus{} 1$ slanted lines for each sign (+ve or -ve)? It seems to me the formula is basically the same as the formula in my solution with special consideration for the row above the very topmost (and bottom-most)... in this problem the residue is $ 4[2n \minus{} n)]/2$ with $ n \equal{} 10$ which is even (the odd case will add one more 1 to it, but how do we combine the two separate odd and even case?). of course further modification must be added since the number of slanted and horizontal lines is not equal in this general case...  :idea:  :|

This problem falls under the study of hyperplane arrangements (see for example [url=http://ocw.mit.edu/OcwWeb/Mathematics/18-315Fall-2004/LectureNotes/index.htm]notes in MIT's OCW[/url], [url=http://www-math.mit.edu/~rstan/arrangements/arr.html]Richard Stanley's notes[/url] as published [url=http://www.ams.org/bookstore?fn=20&arg1=mathcomb&item=PCMS-13]in this book[/url], or [url=http://en.wikipedia.org/wiki/Arrangement_of_hyperplanes]wikipedia[/url])-- in particular, I think that the arrangement in this problem is closely related to the Shi arrangement, the Catalan arrangement, and other similar arrangements.

[quote="Adunakhor"]
For $11$, $a$ can range from $10$ through $1$, so there are $10$ possible intersections.[/quote]
Why is this true?


For a positive integer $K$, define a sequence, $\{a_n\}_n$, as following $a_1=K$, \[ a_{n+1} = \{ \begin{array} {cc} a_n-1 , & \mbox{ if } a_n \mbox{ is even} \\ \frac{a_n-1}2 , & \mbox{ if } a_n \mbox{ is odd} \end{array},  \] for all $n\geq 1$.

Find the smallest value of $K$, which makes $a_{2005}$ the first term equal to 0.

[quote="lightrhee"]This is the third problem of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=422487]the second round of 19th Korean Mathematics Olympiad[/url].

3.
For a positive integer K, define a sequence, {$a_n$}, as following:
$a_1=K$
$a_{n+1} =  \{ \begin{array} {cc} a_n-1 , & \mbox{ if } a_n \mbox{ is even} \\ \frac{a_n-1}2 , & \mbox{ if } a_n \mbox{ is odd} \end{array} (n \geq 1)$
Find the smallest value of K, which makes $a_{2005}$ the first term being 0.[/quote]

Let's redenote this sequence: $b_{1}=a_{2005}=0$, ...$b_{2005}=a_{1}$
The operation:  $b_{i+1}= \{ \begin{array} {cc}  [ \begin{array} {cc} 2b_{i}+1 \\ b_{i}+1 \end{array} ,& b_{i} \mbox{ is odd} \\ 2b_{i}+1 ,& b_{i} \mbox{ is even} \end{array}$
And the question is the smallest value of $b_{2005}$

Let's name this two operations: the operations with multiplying -the operations of the second type
The operations without multiplying - the ones of the first type.

Let's consider the sequence: $b_{1}=0 ,\ b_{2}=1 ,\ b_{3}=2,\ b_{4}=5,\ b_{2005}=2^{1003} -2$ (every odd operatoin ($2i \rightarrow 2i-1$) belongs to the second type; every even operation belongs to the first type)

We'll prove that $2^{1003} - 2$ is the minimal value of $K$.

Let's assume that min $K <2^{1003} - 2$ then

We can see $b_{2}=1$ then
either a) $b_{3}=2$
or      b) $b_{3}=3$

We can see that for any operation of the first type the next and the previous ones are of the second type.
So there will be $\geq [\frac{2002}{2}]=1001$ oprations of the second type   among the remaining 2002 operations.

Also we can see that every operation of the second type increases the number of digits $b_{i}$ in the binary notation.

As both 2 and 3 have 2 digits in the binary notation
If there will be $>1001$ operation of the second type then the minimal value of K has $\geq 2+1002=1004$ digits (in BN),
but our example has 1003 digits(BN) $\Rightarrow$ K isn't min, thus there will be exactly 1001 such operations.

So the types of operations will alternate
In the case a) the previous($i=2 \rightarrow i=3$) operation was of the first type $\Rightarrow$
We have only one such sequence (our example)

In the case b) we have 2 such sequences. They give two values of $b_{2005}$ : $2^{1002}*3-3$ and $2^{1001}*5-1$
But they both are greater then $2^{2003}-2$ , from our example.
We have contradiction.

So the minimal value of $K$ equals to $2^{2003}-2$

My solution  bases on the lemma: the number of digits of a_n in binary system is at least 1 lesser than the $ a_{n+2}$ 's one (n>1)

[hide = Soln]

The answer is $2^{1003} -2.$ First, we will see that this number actually works. Indeed, we have that $a_{2i+1} = 2^{1003-i} - 2$ by induction. This immediately yields that $2^{1003} - 2$ works. Let's now show that it is the smallest one.

The key observation is that we cannot have two consecutive even numbers in the sequence. We will say that a positive integer $n$ is [b] cool [/b] if $a_{n+1} = a_n - 1$ and $a_n \neq 1.$ We've now shown that there cannot be two consecutive cool numbers. With this fact, if $a_1 = x < 2^{1003} - 2,$ then we can see that $a_{2i+1} < 2^{1003} - 2$ by induction on $i.$ We are now done.

$\square$

[/hide]


Let $s_n$ denote the sum of the first $n$ primes. Prove that for each $n$ there exists an integer whose square lies between $s_n$ and $s_{n+1}$.

show that there is a square between k and [k+2\sqrt{k}+2\].
so \[S_{n+1}>S_n +2 \sqrt{S_n}+2 \rightarrow (P_n -1)^2>4S_n\]
and we now \[(P_n -1)^2 > 4S_{n-1} \rightarrow  (P_n -1)^2+ 4P_n >4\rightarrow S_n\Leftrightarrow (p_{n}+1)^2 >4S_n\]


Mr. Smith commutes to the city regularly and invariable takes the same train home which arrives at his home station at 5 PM. At this time, his chauffeur always just arrives, promptly picks him up, and drives him home. One fine day, Mr. Smith takes an earlier train and arrives at his home station at 4 PM. Instead of calling or waiting for his chauffeur until 5 PM, he starts walking home. On his way he meets the chauffeur who picks him up promptly and returns home arriving 20 minutes earlier than usual. Some weeks later, on another fine day, Mr. Smith takes an earlier train and arrives at his home station at 4:30 PM. Again instead of waiting for his chauffeur, he starts walking home. On his way he meets the chauffeur who picks him up promptly and returns home. How many minutes earlier than usual did he arrive home this time?

[hide]
Regard $ 5: 00$ as $ 0$ in time line. Let Mr. S starts at time $ \minus{} x$ with speed $ v_1$, while the driver starts at time $ \minus{} t$ with speed $ v_2$. Then the distance between home and station is $ v_2t$. If they meet at time $ m$, then
$ (m \plus{} x)\cdot v_1 \plus{} (m \plus{} t)\cdot v_2 \equal{} v_2\cdot t$ (1)
Now, if $ T(x)$ is the arrived time when Mr.S gets off earlier $ x$ hours, then
$ T(0) \equal{} t, T(x) \equal{} 2(m \plus{} t) \minus{} t \equal{} 2m \plus{} t$
So the reduced time is $ \Delta T(x) \equal{} 2m$.
We see in (1) that $ m$ is a linear function in $ x$, with constants $ v_1, v_2, t$, so $ \Delta T(x)$ is also linear.
Since $ \Delta T(0) \equal{} 0$, $ \Delta T(1) \equal{} \minus{}\frac {1}{3}$, $ \Delta T(\frac {1}{2}) \equal{} \minus{}\frac {1}{6}$. Therefore, it's $ 10$ minutes earlier.[/hide]

[quote="mathgeniuse^ln(x)"]There is a way of doing this geometrically.[/quote]
How?

I think it involves drawing distance-time-diagrams and using the intercept theorem.

Yeah, and then you take note of the symmetry.


[url]http://www.funnytimes.com/lessercolumns/200306RL.html[/url]

Test your Patriotism and the threat you are to homeland security.

Just took the test, and the only thing I've got to say?

Oooops...

hmmmmmm, some of those questions are... well, mean and disturbing. I don't think that test is for real.
Ooooh, I get it! It's a joke! :D  :D  :D

On the job application, it said:
"Do you advocate the overthrow of the United States government by force or violence?"

I wrote "violence".

[color=green]Apparently, I am too. *high fives bob123*[/color]

Wow, I am too.  And the way it's looking, I'm definitely a terrorist.  (All but five of my answers were "d".)

Number 1 - I answered all four truthfully!

[quote="wereallgonnadie"]I don't think that test is for real.
Ooooh, I get it! It's a joke! :D  :D  :D[/quote]

Well it is from the funny times....

You know, the answers make it seem like getting 15 is bad, I mean I'd rather be a good friend of Bush and get a good government job with a high salary.

Hmm.....lets see... I got a score of....4020! Interesting to note my answer to question 1 was c. its fun to shoot things.... :D


[hide="My guess"]Since they can be represented as approximately $3.7 \cdot 10^{18}$ and $3.4 \cdot 10^{14}$, after multiplication yields approximately $1 \cdot 10^{33}$, so 34 digits?[/hide]

I've seen many math leagues put this type of problem as one of the last questions to try and make you waste a bunch of time.  Solution:  use scientific notation.

(3.659893456789325678*10^18)(3.42973489379256*10^14)=
12.552464296313423185890430413356*10^32=
1.2552464296313423185890430413356*10^33=
[b]34 digits[/b]

EDIT:
p.s. don't waste time multiplying it out; approximate like robinhe did.


Evaluate [2003!/(2002!+2001!+2000!+...+3!+2!+1!)] where brackets denote the greatest integer function.
I think the answer is 2001

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=143430

Please don't double-post.  :)


Find all n such that the sum of the divisor of n is 13 times the number of divisors of n

[hide]Let x be the number of divsors.
$13x=f_{1}+f_{2}+f_{3}+...+f_{x}$ or $13=\frac{f_{1}+f_{2}+f_{3}+...+f_{x}}{x}$  THis means that the average of the factors is 13.
1 factor would be 1, but 1 doesn't equal 13.
EVery other number has at least 2, but no prime has factors adding up to 26.
$39=x^{2}+x+1$ has no solution in integers.
Ditto for $52=x^{3}+x^{2}+x+1$ (3 is too small; 4 is too big.)
52=ab+a+b+1 or 51=ab+a+b, where a and b are distinct primes also has no solutions.  In mod 6, you get $3\equiv{ab+a+b}\mod{6}$, implying a and b are 1 mod 6.  7 and 13 are the smallest and 7*13>51
$64=x^{4}+x^{3}+x^{2}+x$ has no integer solution either.  (2 is too small; 3 is too large.)
$77=x^{5}+x^{4}+x^{3}+x^{2}+x$ has similar problems as 64 did.
$78=ab^{2}+b^{2}+a+ab+b+1$, again with distinct primes.  THis factors into $78=(b^{2}+b+1)(a+1)$  78's factors are 1,2,3,6,13,26,39,78.  Only $b^{2}+b+1=13$ $b^{2}+b+1=3$ yields positive integer roots, but for the latter equation b=1, which is not a prime.  Thus, b=3 and a=5, so 45 satisfies this condition.
$90=x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x$ has no integer solutions.  (1 is too small; 2 is too large.)  Because of that, no more power equations will be considered since 1 will always be too small, but after this power, 2 is too large.
$103=abc+ab+ac+bc$.  One of the 3 distinct primes must be 2 by parity.  Without loss of generality, say it's a.  $103=3bc+2b+2c$.  Now look at this mod 4.
$3\equiv{3bc}\mod{4}$.  THis means both b and c are either 1 mod 4 or 3 mod 4.  (3,7) yields 63+6+14=83, which is too small.  (3,11) yields 99+6+22, too large, meaning that there's no solution with 3 primes.
$104=ab^{3}+1+b^{3}+a+ab^{2}+b+b^{2}+ab=(a+1)(b^{3}+b^{2}+b+1)$
104's factors are 1,2,4,8,13,26,52,104.  None of these numbers satisfy the cubic equation for prime b.  Thus, there is no solution for 8 factors.
Now, somehow assume that all the numbers from 1-12 were somehow factors and only one factor (the number itself) was greater than 13.  To make the average thirteen, that number cannot exceed 13+12/2*13=91.  n cannot exceed 91.
Now, for 9 factors the minimum is 36, 36+18+12+9+6+4+3+2+1=91 which is less than 13*9=117.  The next would be 100, which is clearly too large.
For 10 factors the minimum is 48, 48+24+16+12+8+6+4+3+2+1=124 which is less than 130.  The next would be 80 which clearly is too large.
For 12 factors the minimum is 60,
60+30+20+15+12+10+6+5>12*13.
For 14 factors, the minimum is greater than 91, so 45 is the only solution.[/hide]

[hide]
Let $\tau$ be number of divisors and $\sigma$ be sum of divisors. Then we have
\[13\tau (n)=\sigma (n)\geq \tau (n)\sqrt{n}\]
where the inequality is AM-GM on the divisors of $n$. (Their geometric mean is $\sqrt{n}$ because we can pair any divisor $d$ with $\frac{n}{d}$, unless the divisor is $\sqrt{n}$ itself.) Therefore $n\leq 169$. Also, $\sigma(n)$ must be divisible by 13, so we must find a prime power dividing $n$ whose sum of divisors is divisible by 13. Let $p$ be a prime in such a prime power.

If $p$ appears once, then $p\equiv-1\pmod{13}\Rightarrow p\equiv-1\pmod{26}$. However, this implies $p\geq 103>\frac{169}{2}$. Therefore $n=p$ and $26=13\tau (p)=\sigma (p)\geq 104$ which is ridiculous.

If $p$ appears twice, then $p\leq 13$ and $0\equiv 1+p+p^{2}\equiv (p-3)(p-9)\pmod{13}$. This is only possible when $p=3$, so write $p=9m$ where $m$ is not divisible by 3, our equation then becomes $3\tau (m)=\sigma (m)$. Another divisor AM-GM gives $m\leq 9$, so $m$ is restricted to one of $1,2,4,5,7,8$. Only 5 satisfies the equation, which gives $n=45$.

If $p$ appears three times, then $p\leq 5$ and $0\equiv p^{2}+1\equiv (p-5)(p+5)$. Then $p=5$ and $n=125$, which clearly does not work because 125 only has 4 positive divisors.

If $p$ appears four times, then $p\leq 3$. $p^{5}\equiv 1\pmod{13}$, but raising both sides to the 5th power gives $p\equiv 1\pmod{13}$ which is not possible.

Otherwise, $p=2$. The order of 2 mod 13 divides 12 but not 6 or 4, so it must be 12. Then 2 must appear at least 11 times, which would make $n$ [b]WAY[/b] too big.

Therefore, the only solution is $\boxed{n=45}$.
[/hide]


inequalities
logarithms
inequalities unsolved

If $ a,b \in (0,1)$ then:

\[ \log _a \frac{2ab}{a\plus{}b} \cdot \log _b \frac{2ab}{a\plus{}b} \ge 1\]

Let $ a \equal{} e^{ \minus{} x}$ and $ b \equal{} e^{ \minus{} y}$. Then $ x,y> 0$ and the ineq is equivalent to $ \left(\ln\frac {2ab}{a \plus{} b}\right)^2\ge xy$ or, after some easy manipulations, $ e^x \plus{} e^y\ge 2 e^{\sqrt {xy}}$. (using the monotonicity of the $ exp$ function). But  $ (e^{\frac x2} \minus{} e^{\frac y2})^2\ge0$, thus $ e^x \plus{} e^y\ge 2e^{\frac {x \plus{} y}2}\ge 2e^{\sqrt {xy}}$. :)


function
integration
search
real analysis
real analysis unsolved

Let $ S$ be a linear subspace of $ C[0,1]$ that is closed as a subspace of $ L^{2}[0,1]$. Show that $ S$ is closed as a subspace of $ C[0,1]$.

Show that there is a constant $ M$ such that for all $ f \in S$, we have $ \parallel{}f\parallel{}_{2} \leq \parallel{}f\parallel{}_{\infty}$ and $ \parallel{}f\parallel{}_{\infty} \leq M\parallel{}f\parallel{}_{2}$.

Show that for each $ y \in [0,1]$ there is a function $ k_{y}$ in $ L^{2}$ such that for each $ f \in S$ we have $ f(y)\equal{}\int k_{y}(x)f(x) \; dx$.

The only examples of such $ S$ that I've been able to think of are finite dimensional. Is infinite-dimensional even possible?

According to [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1941621359&t=72292]another topic[/url] they're finite-dimensional. That proof works for the $ L^p$ norm with $ 1\le p\le 2$ but in fact, I think I can extend this to all $ 1\le p < \infty$. In other words: a linear space of continuous functions on the unit interval closed in the $ L^p$ norm for some $ 1\le p < \infty$ is finite dimensional. Equivalently, we can require that the $ L^p$ and $ L^\infty$ norms be equivalent on that particular space (that's how the problem was worded in that other topic).

Edit:

I posted the proof for the more general assertion in that other thread.


geometry
incenter
inradius
trigonometry
geometry proposed

Let $ ABC$ be a triangle and $ A'$ , $ B'$ and $ C'$ lie on $ BC$ , $ CA$ and $ AB$ respectively such that the incenter of $ A'B'C'$ and $ ABC$ are coincide and the inradius of $ A'B'C'$ is half of inradius of $ ABC$ . Prove that $ ABC$ is equilateral .

Using standard notation of $ \triangle A'B'C'$: $ a' = B'C', b' = C'A', c' = A'B',$ $ [A'B'C']$ is triangle area, $ p'$ semiperimeter, $ r' = \frac {_r}{^2}$ inradius.

$ A'I \ge 2r' \Longleftrightarrow \frac {b'c' (p' - a')}{p'} \ge 4r'^2 \Longleftrightarrow \frac {2[A'B'C'] r'}{\sin A' \tan \frac {A'}{2}} \ge 4r' [A'B'C] \Longleftrightarrow$

${ \sin A' \tan \frac {_{A'}}{^2}} \le \frac {_1}{^2} \Longleftrightarrow \sin^2 \frac {_{A'}}{^2} \le \frac {_1}{^4} \Longleftrightarrow \sin \frac {_{A'}}{^2} \le \frac {_1}{^2} \Longleftrightarrow A' \le 60^\circ$

All $ A', B', C'$ are outside the incircle $ (I)$ of $ \triangle ABC$ or some coincide with the incircle tangency points.

$ A'I, B'I, C'I \ge r = 2r' \Longrightarrow$ $ A' = B' = C' = 60^\circ \Longrightarrow$ $ A'I, B'I, C'I = 2r' = r \Longrightarrow$ $ \triangle A'B'C'$ is equilateral and coincides with the intouch triangle of $ \triangle ABC$ $ \Longrightarrow$ $ \triangle ABC$ is equilateral.

IMHO, it's easier to use $\frac{r'}{\overline{A'I}}=\sin\left(\frac{\alpha'}{2}\right)$.^^ Moreover, $\sin\left(\frac{\alpha'}{2}\right)\leqslant \frac{1}{2}$ doesn't necessarily yield $\alpha'\leqslant 60^{\circ}$, but the case $\alpha' \geqslant 150^{\circ}$ is easy to handle.

Suppose $\angle {C'A'B'}=A,\angle{A'B'C'}=B,\angle{A'C'B'}=C,A'B'=c,B'C'=a,C'A'=b$ and similarly define every thing adding $'$ to all of $a,b,c,S$ for $ABC$. Now $\frac {A'I}{c}=\frac {Sin(\frac {B}{2})}{Cos(\frac {C}{2})}$ so that implies basically $AI'=\sqrt{\frac {(S-a)bc}{S}}$. Now $AI'^2\geq 4r^2\implies bc\geq (S-b)(S-c)$ and now so multiplying those we get $abc\geq (S-a)(S-b)(S-c)\geq abc$ , and so we must have $a=b=c$. So we get $A'B'C'$ is equilateral.Now for instance, suppose $\angle{BA'C'}=x$ such that $x\leq 60^0$.Now easy application of sine rule gives $Sin(x+30^0)=2Sin(30^0)=1\implies x=60^0$. So we get in circle of $ABC$ meets $BC,CA,AB$ at $A',B',C'$.Now basically that implies $(S'-b')Sin(\frac {B'}{2})$ is fixed for all $a',b',c'$ and now suppose, $f(x)=x(S'-x)$. Clearly now we get $f(a')=f(b')=f(c')$ but $f(x)$ can't have at most two distinct roots, so wlog suppose $a'=b'$ now we have $a'(S'-a')=c(S'-c')\implies a'=c'$ , and so ultimately we get $a'=b'=c'$ and we are done.

Here's a synthetic solution:

Let $2R$ be the inradius of $ABC$ and $R$ be the inradius of $A'B'C'$ (and $\omega$, $\omega_1$ be the incircles). Then, since $B, D, A', C$ are collinear, we take its Pole in $\omega_1$, note that $ID = 2R$ so, $ID' = \frac{R}{2}$ where $D'$ is inverse of $D$ in $\omega$. But, the Polar of $A'$ in $\omega_1$ is the intouch chord, $E'F'$, so we have that $D'$ lies on $E'F'$. So distance of $I$ from $E'F'$ is atmost $R$ (since a point on $E'F'$ is at distance $R$ from $I$). So, we reflect $I$ over $E'F'$ (which is inside $\omega_1$, so $\angle E'IF' + \angle E'D'F' \geq 180^{\circ}$), and conclude that $\angle E'D'F' \geq 60^{\circ}$, and similarly for $D'F'E'$ and $F'E'D'$. So, due to angle sum property, each of them is $60^\circ$ and therefore, $A'B'C'$, $B'C'A'$ and $C'A'B'$ are all $60^{\circ}$. Now, since each angle is $60^\circ$, the distance from $I$ to the sides of $D'E'F'$ is $\frac{R}{2}$, so the inverse of $D$ in $\omega_1$, actually coincides with the foot from $I$ onto $E'F'$, which is $IA' \cap E'F'$, and $ID \cap E'F'$ is the inverse of $D$, so $D$ and $A'$ conicide and similarly, $E$ and $B'$, $F$ and $C'$ coincide thus giving us that $DEF$ is equilateral implying $ABC$ is equilateral.


inequalities
reflection
Cauchy Inequality
inequalities proposed

I prove this but my solution is very ugly . I use AM-GM for proving this problem . if u want I could send the solution .

[quote="hungkhtn"]Let $ a,b,c$ be positive. Prove that
\[ \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge (a+b+c)\sqrt{\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}}. \]
[/quote]
Yes, khashi70, you are right.
$ \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge (a+b+c)\sqrt{\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{a^{5}}{b}+\frac{a^{4}c}{b}+\frac{a^{5}c}{b^{2}}+\frac{2a^{3}b^{2}}{c}-a^{4}-2a^{3}c-2a^{2}bc\right)\geq0,$ which true because
$ \sum_{cyc}\left(\frac{a^{5}}{b}-a^{4}\right)\geq0$ is true by rearrangement,
$ \sum_{cyc}\left(\frac{a^{4}c}{b}-a^{3}c\right)\geq0\Leftrightarrow\sum_{cyc}(14a^{5}c^{2}+3b^{5}a^{2}+2c^{5}b^{2}-19a^{4}c^{2}b)\geq0$ is true  by Cauchy,
$ \sum_{cyc}\left(\frac{a^{5}c}{b^{2}}-a^{3}c\right)\geq0\Leftrightarrow\sum_{cyc}\left(23a^{7}c^{3}+8b^{7}a^{3}+6c^{7}b^{3}-37a^{5}c^{3}b^{2}\right)\geq0$ is true by Cauchy and
$ \sum_{cyc}\left(\frac{a^{3}b^{2}}{c}-a^{2}bc\right)\geq0\Leftrightarrow\sum_{cyc}\left(6a^{4}b^{3}+2b^{4}c^{3}+5c^{4}a^{3}-13a^{3}b^{2}c^{2}\right)\geq0$ is true by Cauchy.

Thank you, Arqady and khasi70. Your solution is very nice :D

Alternative solution: By Cauchy inequality,
$ \sum_{cycl}b(c+a)^{2}\sum_{cycl}\frac{a^{2}}{b}\ge(a^{2}+b^{2}+c^{2}+ab+bc+ca)^{2}$
It suffices to show that
$ (a^{2}+b^{2}+c^{2}+ab+bc+ca)^{2}\ge (a+b+c)\sqrt{\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}}\sum_{cycl}b(c+a)^{2}$
Since
$ a^{2}+b^{2}+c^{2}+ab+bc+ca\ge\frac{2}{3}(a+b+c)^{2}\ge\frac{a+b+c}{2(ab+bc+ca)}\succ b(c+a)^{2}$
it remains to prove
$ a^{2}+b^{2}+c^{2}+ab+bc+ca\ge 2(ab+bc+ca)\sqrt{\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}}=2\sqrt{(a^{2}+b^{2}+c^{2})(ab+bc+ca)}$
which is true by the AM-GM inequality.

Here is another solution i found .Hope to like it
By Cauchy Swartz $ \sum\frac{a^{2}}{b}\geq \frac{(a^{2}+b^{2}+c^{2})^{2}}{a^{2}b+b^{2}c+c^{2}a}$

So it suffices to prove that
$ (a^{2}+b^{2}+c^{2})^{3}(ab+bc+ca)\geq (a^{2}b+b^{2}c+c^{2}a)^{2}(a+b+c)^{2}$

By Vasc's inequality $ (a^{2}+b^{2}+c^{2})^{2}\geq 3(a^{3}b+b^{c}+c^{3}a)$
and $ 3(a^{2}+b^{2}+c^{2})\geq (a+b+c)^{2}$

So it remains to show that
$ (a^{3}b+b^{3}c+c^{3}a)(ab+bc+ca)\geq (a^{2}b+b^{2}c+c^{2}a)^{2}$

which is true by Cauchy -Swartz inequality .

A split-off of this topic ( http://www.mathlinks.ro/viewtopic.php?t=157543 ) containing a stronger version of the inequality was hidden until 17 March 2008 because of a running Mathematical Reflections problem.

  darij

[quote=hungkhtn]Let $ a,b,c$ be positive. Prove that
\[ \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge (a+b+c)\sqrt{\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}}.\][/quote]
The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$ \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge (a+b+c)\sqrt{\frac{26(a^{2}+b^{2}+c^{2})-19(ab+ac+bc)}{7(ab+bc+ca)}}.$$

[quote=qweDota]Notting that $(a+b+c)^2\geq ab+bc+ca$$$RHS=\sqrt{\frac{(a+b+c)^2}{ab+bc+ca}(a^2 +b^2 +c^2)}\leq \sqrt{3(a^2 +b^2 +c^2)}$$
And also
$$3(a^2 +b^2 +c^2) \leq 3(\frac{a^2 b}{c}+\frac{b^2 c}{a}+\frac{c^2 a}{b})\leq (\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a})^2$$
and the result follows[/quote]

[quote=luofangxiang][quote=hungkhtn]Let $ a,b,c$ be positive. Prove that
\[ \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge (a+b+c)\sqrt{\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}}.\][/quote]

[img]http://s5.sinaimg.cn/mw690/006ptkjAzy7qe2Gw27aa4&690[/img][/quote]

你的证明有很明显的错误

抱歉，不知道怎么把你也带上了，另一个有问题

[quote=KastavIvan]We also have  for \(a,b,c > 0\):
\[\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq \frac{(a+b+c)(a^2+b^2+c^2)}{ab+bc+ac}\][/quote]
https://artofproblemsolving.com/community/c6h326199p1748725

[quote=KastavIvan]We also have  for \(a,b,c > 0\):
\[\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq \frac{(a+b+c)(a^2+b^2+c^2)}{ab+bc+ac}\][/quote]
[url]https://artofproblemsolving.com/community/c6h157543p883191[/url]


function
algebra
polynomial
calculus
integration
system of equations
calculus computations

i can see that the answer $ f(x)\equal{}x$ works, and $ e^x$.. but i cant seem to prove that these are it?

[quote="mathxl"]if f a non constant polynomial and f(f '(x))=f΄(f(x)) for all real x, find f (maybe f has integer cofiecents)
This exercise has been taken from the post of a friend in romanian forum[/quote]

LET degree of polynomial=n.
then $ (n\minus{}1)n\equal{}n^2\minus{}1 \rightarrow n\equal{}1$

@kenn4000
$ e^x$ is not a polynomial.

ith_power i think you making a mistake about degree.
Both sides of equation have same degree [tex]$ n^2 \minus{} n$[tex]
 though you are right aout your comment for [tex]$ e^x$[tex]

Well $ f(x) \equal{} \frac {x^n}{n^{n \minus{} 1}}$ is a solution for every integer $ n$.  But there are others, for example:

$ f(x) \equal{} a_0 \plus{} \frac {x^2}{2}$ for any $ a_0\in \mathbb{R}$
$ f(x) \equal{} 9 \minus{} x^2 \plus{} \frac {x^3}{9}$

I suspect there are no other solutions, but that's an only slightly motivated guess.  Did the original thread get any further?

the original thread has no answers (is unsolved)
You are the first that make a nice approach at this exercise
How you found these solutions? the way of thinking i mean.
Thx!

For the first one just look for solutions of the form $ cx^n$, substitute into the equality, and solve for $ c$.  The same idea for the other two -- assume solutions of the form $ a_0 \plus{} a_1x \plus{} a_2x^2$ and $ a_0 \plus{} a_1x \plus{} a_2x^2 \plus{} a_3x^3$ and solve for the coefficients.  Finding the solutions comes down to solving a set of linear equations for the polynomial's coefficients.  I suspect there are no nontrivial solutions for order higher than $ 3$ only because the corresponding linear system becomes increasingly overdetermined as $ n$ grows (there are $ n(n \minus{} 1) \plus{} 1$ equations and $ n \plus{} 1$ degrees of freedom), and in particular I've verified directly that there are no nontrivial solutions for $ n \equal{} 4,5,6$; but this is of course not a proof, since it's not clear that there aren't certain $ n$ for which the linear system has so many redundant equations that solutions nevertheless exist.

Of course, if we follow the suggestion of the original post and consider only integral coefficients then $ f(x) \equal{} x$ is the unique solution: by looking at the leading coefficients we see that any solution has the form $ f(x) \equal{} \frac {x^n}{n^{n \minus{} 1}} \plus{} \ldots$.

JoeBlow, I'm not sure what you meant by a linear system of equations -- a system of equations, sure, but it's definitely not linear.  No other solutions for degree 7, either -- I suspect one could prove this by looking at the first $ n$ nonzero coefficients of the difference, but I'm not personally up for the experiement.  (These let you solve for all the coefficients below the $ (n \minus{} 1)$th in terms of the $ (n \minus{} 1)$th, and then hopefully the next coefficient just works out to something like $ c a_{n \minus{} 1}^k$, in which case you conclude.)

i thik problem must be with integer coefficients to be good, so JBL you are right f(x)=x
thx all  :D

I did miss the integer coefficients bit, I guess that simplifies things. :)

JBL -- you're right it's certainly not a linear system, I spoke hastily.  But it can be reduced to a system that's linear in every variable except one.  The $ n(n \minus{} 1) \minus{} 2$ coefficient depends only on $ a_{n \minus{} 1}$ and $ a_{n \minus{} 2}$, and is linear in $ a_{n \minus{} 2}$.  After substitution, the $ n(n \minus{} 1) \minus{} 3$ coefficient then depends only on $ a_{n \minus{} 1}$ and $ a_{n \minus{} 3}$, and is linear in $ a_{n \minus{} 3}$, etc.  So the set of equations, though nonlinear, can be solved by up to the last step simply by forward substitution, except that at the end you have to deal with some potentially high order polynomial in $ a_{n \minus{} 1}$.  And your suggestion seems right -- it's probably possible to prove that this last polynomial is just some power of $ a_{n\minus{}1}$.


No one in the mathcounts forum could solve this, so im posting it here.

Two cans containg coffee have the same shape and are made out of the
same tin. One can weighs 2 kg and is 12 cm high; the other weighs 1 kg
and is 9.5 cm high. Find the net weight of the coffee in both cans.

source:  Algebra Can be Fun

I figure it would be 3kg plus the weight of the coffee in both cans...

Am I missing something?

or are you supposed to find the weight of the cans themselves and subtract it from the total weight?

the whole question's kind of confusing. does Algebra Can Be Fun have an answer?

If the answer isn't 3 kg

Then it would be 3kg - the weight of the cans (not plus like someone said earlier) because 3kg is the net weight of the coffee + the can, and they just want the coffee. I'm not really sure how to do it, but that's how you would do it if you knew the wt. of the cans.

Oh, and we know the cans are similar, now that I think about it, you can probably set up some proportions that tie in the wt. of the can and the coffee, *shrugs*

kg is unit of mass, not weight. theyre asking for weight. in metrics, weight is measured in N (newtons)
2 kg * .12 m = .24 N
1 kg * .094 m = .094 N

net weight = .24 N + .094 N = .334 N

i think thats what theyre asking for. but if the cans are piled on top of each other:
net weight = (2 + 1) kg * .214 m = .642 N

If they'd asked for the net weight of the coffee on the moon, they probably would have wanted it in newtons. However being on earth kg do fine.

And a newton is defined as a $\frac {kg*m}{s^2}$, or the force required to accelerate a 1kg object by 1 meter per second squared. Weight in newtons is calculated by $F = ma$ or force (in newtons) equals mass times acceleration. The mass is the mass of the can. The acceleration is the acceleration of gravity or $\frac {9.8m}{s^2}$.

Then the weight of can #1 in newtons would be $2*9.8 = 19.6N$ and the weight of can #2 would be $1*9.8 = 9.8N$.

To clarify the problem, were looking for how many [b] grams [/b] of coffee  there are in each canister.

And I'm pretty sure you need to know either the radius of the cans or the weight of a unit area of the tin the cans are made of to solve the problem.


Στις πρωτες σελιδες του βιβλιου της Μαθηματικης Αναλυσης 1 υπαρχουν καποια στοιχεια απο τη Θεωρια των Συνολων, γινεται αναφορα στο ποτε δυο συνολα ειναι ισοδυναμα και στο ποτε ενα συνολο ειναι πεπερασμενο/απειρο/αριθμισημο.

Γνωριζουμε οτι ενα συνολο ειναι αριθμισημο οταν ειναι ισοδυναμο με το συνολο των φυσικων αριθμων, δηλαδη υπαρχει 1-1 και επι αντιστοιχεια απο το συνολο αυτο στο συνολο Ν των φυσικων αριθμων. Στις πρωτες σελιδες του βιβλιου της Αναλυσης 1 υπαρχουν καποιες προτασεις, μεταξυ των οποιων και οι παρακατω:

Το συνολο $ NxN$ ειναι αριθμισημο.
Το συνολο $ ZxZ$ ειναι αριθμισημο.
Το συνολο $ Zx\{Z \minus{} \{0\}\}$ ειναι αριθμισημο

Στο βιβλιο γραφει οτι οι αποδειξεις των παραπανω αφηνονται ως ασκησεις στον αναγνωστη. Πως μπορουμε ομως να τις αποδειξουμε; Μπορουμε να βρουμε μια 1-1 και επι αντιστοιχια απο το συνολο των φυσικων αριθμων σε ενα συνολο που εχει για στοιχεια του ολα τα διατεταγμενα ζευγη των φυσικων/ακεραιων αριθμων;

[quote="NickNafplio"]Το συνολο $ NxN$ ειναι αριθμισημο.
Το συνολο $ ZxZ$ ειναι αριθμισημο.
Το συνολο $ Zx\{Z \minus{} \{0\}\}$ ειναι αριθμισημο
[/quote]

Ένας πολύ απλός τρόπος να μετρήσεις το $ \mathbb{Z}^2 \equal{} \mathbb{Z}\times\mathbb{Z}$ είναι ο εξής: τα σημεία του, απλώς φτιάχνουν ένα grid στο επίπεδο. Ξεκίνα λοιπόν το μέτρημα από το κέντρο και άρχισε να διαγράφεις μία "τετραγωνική" "σπείρα". Δηλαδή:

$ 1\mapsto (0,0)$
$ 2\mapsto (1,0)$
$ 3\mapsto (1,1)$
$ 4\mapsto (0,1)$
$ 5\mapsto ( \minus{} 1,1)$
$ 6\mapsto ( \minus{} 1,0)$
$ 7\mapsto ( \minus{} 1, \minus{} 1)$
$ 8\mapsto (0, \minus{} 1)$
$ \ldots$

Έτσι, θα καλύψεις όλα τα σημεία του grid και θα έχεις την 1-1 αντιστοιχία σου. Tα υπόλοιπα σύνολα είναι υποσύνολα του $ \mathbb{Z}\times\mathbb{Z}$ οπότε...

Cheers,

Durandal 1707

Ευχαριστω πολυ.
Με καλυψατε πληρως :)

Να κάνω και εγώ κάποιες παρατηρήσεις με την σειρά μ γιατί τα έχω αρκετά φρέσκα.
Όπως γράφει ο κύριος Μπόρης, αριθμήσιμη ένωση αριθμήσιμων συνόλων είναι αριθμήσιμο σύνολο.
Αν γράψεις το  $ NxN$ σαν $ \bigcup ixN$, με i στους φυσικούς τότε προκύπτει άμεσα και τότε το $ NxNx...xN$ k φορές με k φυσικό είναι αριθμήσιμο. Και άν Α αριθμήσιμο σύνολο και B $ \subseteq$Α τότε το Β είναι αριθμήσιμο. Οπότε αφου το $ ZxZ$ είναι αριθμήσιμο, τότε και το $ Zx\{Z \minus{} \{0\}\}$ είναι αριθμήσιμο.

Κωστα, το προβλημα γινεται πολυ απλο αν το δουμε Γεωμετρικα (δηλαδη αν θεωρησουμε τα διατεταγμενα ζευγη σαν σημεια στο επιπεδο) οπως ο Durandal
Ο λογος που με ταλαιπορησε τοσο το προβλημα αυτο ειναι το οτι δεν πηγε ο νους μου σε καποια Γεωμετρικη λυση


I can't solve this problem. I think this is a good one.

 Let B and C be arbitrary points on sides AP and PD respectively of an
acute triangle APD. The diagonals of the quadrilateral ABCD meet at Q,
and H1,H2 are the orthocenters of triangles APD and BPC, respectively.
Prove that if the line H1H2 passes through the intersection point X (X and
Q are different.) of the circumcircles of triangles ABQ and CDQ, then it also passes
through the intersection point Y (Y and Q are different.) of the circumcircles of triangles
BCQ and ADQ.

Let $ \omega_1$ and $ \omega_2$ be two circles with diameters $ AC$ and $ BD$ respectivly.Let denote $ BB_1$, $ CC_1$, $ AA_1$, $ DD_1$ - an altitudes in triangles $ BPC$ and $ APD$ (points $ A_1$ and $ C_1$ lies on $ \omega_1$ $ B_1$ and $ D_1$ on $ \omega_2$). Points $ A, D_1, A_1, D$ lies on the same circle, thus $ H_1A*H_1A_1 \equal{} H_1D*H_1D_1$ , thus $ H_1$ lies on the radical axes of $ \omega_1$ and $ \omega_2$. Similar $ H_2$ lie on it, so this radical axe is $ H_1H_2$. Let $ M$ and $ N$ be a midpoints of $ AC$ and $ BD$ respectivly. By the condition $ X$ lie on the radical axe so $ XM^2 \minus{} CM^2 \equal{} XN^2 \minus{} DN^2$, but triangles $ XAC$ and $ XBD$ are similar(because $ \angle XAQ \equal{} \angle XBQ, \angle XCQ \equal{} \angle XDQ$), thus $ \frac {XM^2 \minus{} CM^2}{XN^2 \minus{} DN^2} \equal{} k^2$, where $ k$ is a similarity coefficient for triangles $ XAC$ and $ XBD$. So $ XAC$ and $ XBD$ are equal, thus $ AC \equal{} BD$. But analogically $ AYC$ and $ DYB$ are similar, thus the degree of point $ Y$ on circles $ \omega_1$ and $ \omega_2$ are equal(because $ YM \equal{} YN,MC \equal{} ND$), so it lie on $ H_1H_2$. Q.E.D
P.S. Sorry for my bad English :blush:


limit
logarithms
calculus
integration
derivative
function
calculus computations

Let $l_n$ be the length of the curve $y=x^n$ in $[0,1]$.

$\lim_{n\to +\infty}\frac{(2-l_n)n}{\ln n}=$ ?

i do not know how to do this because the integral is pretty difficult to evaluate

$\int_{0}^{1}\sqrt{1+n*x^{n-1}} \, dx$

but one idea that i had was that you could some how do l'hopital's rule take the derivative of the antiderivative to just get the function, but that does not work out that well

I don't have time to work on this for most of today. Here's a thought I had, which may or may not lead to anything useful:

Using the fact that $a+b-\sqrt{a^2+b^2}=\frac{2ab}{a+b+\sqrt{a^2+b^2}},$

$2-l_n=\int_0^11+nx^{n-1}-\sqrt{1+n^2x^{2n-2}}\,dx =\int_0^1\frac{2nx^{n-1}} {1+nx^{n-1}+\sqrt{1+n^2x^{2n-2}}}\,dx.$

Actually, Kent's formula leads to the answer fairly quickly. Indeed, just split the interval $[0,1]$ into 3 parts:
$\left[0, 1-\frac{(1+\delta)\ln n}{n}\right]$, $\left[1-\frac{(1+\delta)\ln n}{n}, 1-\frac{(1-\delta)\ln n}{n}\right]$, and  $\left[1-\frac{(1-\delta)\ln n}{n}, 1\right]$. It is not hard to see that the integrand is almost $0$ on the first interval, almost $1$ on the third and between $0$ and $1$ on the second for large $n$. Therefore, since the length of the second interval is much less than that of the third interval, the integral over the union of the second and the third intervals is about $\frac{\ln n}{n}$. As to the integral over the first interval, we can replace the denominator by $1$ and get the estimate $\left(1-\frac{(1+\delta)\ln n}{n}\right)^n\le e^{-(1+\delta)\ln n}=\frac 1{n^{1+\delta}}$ for it, which is much less than $\frac{\ln n}{n}$. So, $2-\ell_n\approx \frac{\ln n}n$ and the answer to the original problem is $1$.


trigonometry
real analysis
real analysis unsolved

Show that $ \sum_{1}^{\infty}\frac{\cos{nx}}{n}$ converges uniformly for any closed sub-interval, $ [a,b]$, of $ (0,2\pi)$, moreover show that the convergence is not absolute for any $ x$.

Dirichlet's test? We can even compute its sum.


[quote="hunter34"]No eyes. =/[/quote]

lol

[color=brown]my eye's r so dark brown u almost can't see my pupils  :D [/color]

[color=darkred]lots of dark brown... lol[/color]

Lots of Chinese people.

Proof:

All Chinese people are brown. There are many Chinese people in this AoPS forum. Thus it follows by the transitive property that there are many people with brown eyes in this AoPS forum. Q.E.D. :D

yeah... this poll is kinda random... but it is kinda subjective as to what color eye you have, for example sum1 might think havel is green or some ppl might think it is green or vice versa or whatever, although with brown its probably easy to see...yeah i know what your thinking this post is kinda random too...

[quote="1=2"]
[color=blue]so I'm the only light blue on AoPS.[/color][/quote]

[color=blue]This is false.[/color] Go light Blue!

[quote="1=2"][color=blue]so I'm the only light blue on AoPS.[/color]  :([/quote]

[color=blue]why???  I have the light blue eyes too( I'm Georgian but my Granny is russian ;) i have her eyes :) )  and here are many pupils wth light blue eyes, they just have not seen this poll yet.

Go light Blue![/color]

All the students of my classroom have brown eyes :o

A LOT OF PEOPLE HAVE BROWN EYES!!! GO BROOOOOOWWWNNN!!!


find
$ \lim_{x\rightarrow a} \frac{x^n sin (x)\minus{}a^n sin(a)}{x\minus{}a}$  with $ n \in N$* and $ x \in R$*

[quote="greatestmaths"]find
$ \lim_{x\rightarrow a} \frac {x^n sin (x) \minus{} a^n sin(a)}{x \minus{} a}$  with $ n \in N$* and $ x \in R$*[/quote]
We have:
$ \lim_{x\rightarrow a} \frac {x^n sin (x) \minus{} a^n sin(a)}{x \minus{} a}\equal{}lim_{x\to a} \frac{\left(x^n sin (x) \minus{} a^n sin(a)\right)'}{(x\minus{}a)'}$
$ \equal{}\lim_{x\to a} \dfrac{x^{n\minus{}1}(nsinx\plus{}xcosx)}{1}\equal{}a^{n\minus{}1}(nsin{a}\plus{}acosa)$


geometry
circumcircle
conics
ellipse
projective geometry
geometry unsolved

The circumcenter of ΔABC is O.
There exists ellipses which have O,P as two foci and touch segments AB,BC,CA.
Find locus of such point P (P is coplanar with ΔABC).
 :help: please~

A theorem say that the foci of an inscribed central conic are isogonal conjugates, so the locus is the orthocenter.

Thank you for your answer.
Seeing the theorem,I have searched and got it  :)
Can I ask you a favor ? I want to know the proof that orthocenter and circumcenter are isogonal conjugate.
Please give me the hint of the proof (with elementary euclidean geometry or your great projective geometry ^^)
 :)

[quote="77ant"]
Can I ask you a favor ? I want to know the proof that orthocenter and circumcenter are isogonal conjugate.[/quote]

It's simple: Let a triagle ABC with circumcenter O and orthocenter H. Then $ \angle OAC \equal{} \frac{180 \minus{} \angle AOC}2 \equal{} \frac{180 \minus{} 2 \angle ABC}2 \equal{} 90 \minus{} \angle ABC \equal{} \angle HAB$ and equaly for the others  :wink:

Oh ! It is very easy...
It is silly of me not to find it  :lol:


Official results of JBMO 2009 are attached. Medal cut-offs will be decided later.

Thanks a lot.

I guess the names of participants hidden under their codes will be announced as well?

Medal cut-offs:

13-23 = BRONZE

24-38 = SILVER

39-40 = GOLD

Attached is list with names of participants.

welldone Gledis Kallço (ALB4) ,he is only 5th grade.

[quote="sandu2508"][quote="math10"]here is result?Can delegat post problem JBMO?[/quote]
The problems have been posted here are the links to them:

Problem1: http://www.mathlinks.ro/viewtopic.php?search_id=748504771&t=285426
Problem2: http://www.mathlinks.ro/viewtopic.php?search_id=748504771&t=285428
Problem3: http://www.mathlinks.ro/viewtopic.php?search_id=748504771&t=285429
Problem4: http://www.mathlinks.ro/viewtopic.php?search_id=748504771&t=285431[/quote]
thank you very much :D .


graph theory
combinatorics theorems
combinatorics

Does anyone know a simple, nice and small   :D  proof of the well known theorem :

 A multigraph G is eulerian if and only if G is connected and every vertex of G is even.

Isn't any of the also well known proofes nice enough¿

E.g.:

Any eulerian cycle implies, when going along it in a given direction, the same numbers of enetring and leaving edges for any vertice, thus their number is even everywhere. Since you also reach every vertice, they are connected.

Converse:
Start at some random point $P$ and go from vertice to vertice without using any edge twice till you have to stop. Since, when doing this procedure, the number of still available edges neighbouring any point $A$ decreases by $1$ when entering $A$, you can leave that point $A$ again if that number was even before entering (since then it was $\geq 2$). Also this number will be even again after leaving. Thus the only vertice you can end up is $P$ since all others remain even everytime.
Now you have a path using some vertices and edges, and removing all edges (and vertices which's edges are all in use already) that are part of it from the graph leaves you with a smaller graph having the same properties again. Now start at any vertice that was part of the first cycle (such a vertice exists iff you are not done already cause of connectivity) for a new travel around the remaining graph and 'add' this new cycle to the first one. Do this over and over again till no vertices are left, you will be left with a eulerian cycle.


A bullet of mass $ m_1$ strikes a pendulum of mass $ m_2$ suspended from a pivot by a string of length $ L$ with a horizontal velocity $ v_0$. The collision is perfectly inelastic and the bullet sticks to the bob. Find the minimum velocity $ v_0$ such that the bob (with the bullet inside) completes a circular vertical loop.

What did I do wrong:
Using conservation of momentum, we have
\[ m_1 v_0 \equal{} \left(m_1\plus{}m_2\right) v_n.\]
Then
\[ v_n\equal{}\frac{m_1 v_0}{m_1\plus{}m_2}\]
and
\[ 0\equal{}v_n^2\plus{}2a dx.\]
Then
\[ v_n^2\equal{}\minus{}2 \int_0^\pi \minus{}g \sin \theta L d \theta?\]
I think that's where I screwed up?
If I follow my wrong steps, I end up with
\[ v_0\equal{}\left(m_1\plus{}m_2\right)2 \sqrt{Lg}/m_1,\]
but that's not the answer.

Do you know the answer? Then we can show you where you messed up.

The answer is
\[ \left(m_1\plus{}m_2\right) \sqrt{5Lg} / m_1.\]

[quote="123456789"]\[ 0 \equal{} v_n^2 \plus{} 2a dx.
\][/quote]

I'm not sure why you chose this formula... once the bob enters the vertical loop, there is nonconstant acceleration.

[hide="solution"]
I came up with the answer $ \left(m_1\plus{}m_2\right) \sqrt{5Lg} / m_1$ with a conservation of energy argument:

At the bottom of the loop, $ KE \equal{} \frac{1}{2}\left(m_1\plus{}m_2\right)v_n^2$.

At the top of the loop, if we consider the velocity, $ a_c \ge g$ since $ g$ is the acceleration downwards.  Thus $ v^2 \ge gL$, and $ KE \equal{} \frac{1}{2}\left(m_1\plus{}m_2\right)v^2 \ge  \frac{1}{2}\left(m_1\plus{}m_2\right)gL$.

The change in potential energy from the bottom to the top of the loop is $ U\equal{}\left(m_1\plus{}m_2\right)g \Delta h \equal{} 2\left(m_1\plus{}m_2\right)gL$.

Altogether, from conservation of energy, this means that:
$ \minus{}\Delta KE \equal{} \Delta U$
$ KE_{bottom}\minus{}KE_{top} \equal{} 2\left(m_1\plus{}m_2\right)gL$

Substituting $ KE_{top}\ge  \frac{1}{2}\left(m_1\plus{}m_2\right)gL$ and $ KE_{bottom} \equal{} \frac{1}{2}\left(m_1\plus{}m_2\right)v_n^2$ gives the desired solution.[/hide]

Thanks. I don't know what I was thinking with using that kinematic equation for non-constant acceleration.

I like this way of doing it, but is there any other way to do it too?

Yes, the place where most people are likely to mess up is assuming that at the top the desired velocity is 0; in fact, this velocity will just make the bob go down, and the resulting path will not be circular.

[quote="123456789"]In krsattack's post, can somebody explain why $ a_c \ge g$? I never really understood the concept of both forces going downwards.[/quote]

Sure.  For simplicity, in this case, I let $ m\equal{}m_1\plus{}m_2$.

Let $ F_t$ be the tension of the string, acting on the bob in a downward direction at the top of the loop.  Gravity points in the same downward direction with force $ mg$.  Therefore, the total force on the bob towards the center of the loop is

$ F_{total}\equal{}F_t\plus{}mg$

But the total force is just the force of centripetal acceleration, so $ F_{total}\equal{}ma_c$.  Since $ F_t \ge 0$, we have:

$ ma_c \equal{} F_t\plus{}mg \ge mg$

Divide by $ m$ to get the result.


Hi everyone,

I have got MikTeX and WinEdt on my machine, and am trying to figure out how I can tell them to look for a user defined file char.defs.tex.

When I say
\input {char.defs.tex}
it gives an error message saying that the file is not found. Do I have to set some sort of path?

It should be in a place that TeX can find it... either your 'texmf' folder or your project folder should work, I think.


More generally (for 2n) the sum gonna be equal to n+1  ????

@above the $\LaTeX$ is incorrect.

try using ^ for an exponent and {} after the exponent.
FOr example, if you wanted $i^{7i \cdot (I + 1)}$, you would do
[code] $i^{7i \cdot (I + 1)}$[/code]

Let $$a_k=\frac{k^3}{101^3-3\cdot 101^2k+3\cdot 101k^2}.$$

Claim: $$a_k+a_{101-k}=1.$$ Note that the denominator of $101^3-3\cdot 101^2k+3\cdot 101k^2$ is symmetric around $$k=\frac{3\cdot101^2}{2\cdot 3\cdot 101}=101/2,$$ so $a_k$ and $a_{101-k}$ have the same denominator. Furthermore, $k^3+(101-k)^3$ is equal to that common denominator, so their sum is 1.\\

Thus, the answer is 51.

remark: this is motivated by trying small numbers in place of $101$, you will see that the terms in the sum come in pairs that add to 1 and this easily generalizes

Nice! My first APMO problem. We see $\frac{x_i^3}{1-3x_i+3x_i^2}=\frac{x_i^3}{x_i^3+(1-x_i)^3}$; in particular, we are motivated to see that $1-x_i=x_{101-i}$. Combining, we get $$LHS=\sum_{i=0}^{50}\frac{x_i^3}{x_i^3+x_{101-i}^3}+\frac_{x_{101-i}^3}{x_i^3+x_{101-i}^3}=51.\blacksquare$$

im not sure whats wrong with my latex can someone fix

@above [quote=huashiliao2020]Nice! My first APMO problem. We see $\frac{x_i^3}{1-3x_i+3x_i^2}=\frac{x_i^3}{x_i^3+(1-x_i)^3}$; in particular, we are motivated to see that $1-x_i=x_{101-i}$. Combining, we get $$\text{LHS} =\sum_{i=0}^{50}\frac{x_i^3}{x_i^3+x_{101-i}^3}+\frac{x_{101-i}^3}{x_i^3+x_{101-i}^3}=51.\blacksquare$$

im not sure whats wrong with my latex can someone fix[/quote]


i have made a start... here it is:2
$ (a\plus{}b)^3(b\plus{}c)^3(c\plus{}d)^3(d\plus{}a)^3>\equal{}(2\sqrt{ab}2\sqrt{bc}2\sqrt{cd}\sqrt{da})^3\equal{}16^2abcd>\equal{}(a\plus{}b\plus{}c\plus{}d)^4$
Then what?

Help silouaaaaaaan

[quote="Γιώργος"] $ 16^2abcd \ge  (a \plus{} b \plus{} c \plus{} d)^4$[/quote]
it is true if and only if $ a\equal{}b\equal{}c\equal{}d$ .

[quote="Γιώργος"]does the inequality hold??????????[/quote]
by AM-GM we have for all $ a,b,c,d \ge 0$, $ (a\plus{}b\plus{}c\plus{}d)^4 \ge 4^4abcd$  .so for example if $ (a\equal{}b\equal{}c\equal{}1$ and $ d\equal{}2$) your inequality does'nt hold .

[quote="Γιώργος"]if a,b,c,d>0 prove that:
$ (a \plus{} b)^3(b \plus{} c)^3(c \plus{} d)^3(d \plus{} a)^3$ $ > \equal{}$ 16$ a^2b^2c^2d^2(a \plus{} b \plus{} c \plus{} d)^4$[/quote]
$ a\plus{}b\plus{}c\plus{}d\equal{}1$

$ (a\plus{}b)(b\plus{}c)(c\plus{}d)(d\plus{}a)$ $ \equal{}$ $ (ac\minus{}bd)^2\plus{}abc(a\plus{}b\plus{}c\plus{}d)\plus{}bcd(a\plus{}b\plus{}c\plus{}d)\plus{}cda(a\plus{}b\plus{}c\plus{}d)\plus{}dab(a\plus{}b\plus{}c\plus{}d)$ $ \geq abc\plus{}bcd\plus{}cda\plus{}dab\minus{}\minus{}\minus{}\minus{}\minus{}(1)$

by Newton Inequality $ (\frac{abc\plus{}bcd\plus{}cda\plus{}dab}{4})^2 \geq (\frac{ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd}{6})(abcd)$
$ \geq \sqrt{(\frac{a\plus{}b\plus{}c\plus{}d}{4})(\frac{abc\plus{}bcd\plus{}cda\plus{}dab}{4})}(abcd)$
$ \rightarrow (abc\plus{}bcd\plus{}cda\plus{}dab)^3 \geq 16a^2b^2c^2d^2(a\plus{}b\plus{}c\plus{}d)\minus{}\minus{}\minus{}\minus{}(2)$
by $ (1),(2) \rightarrow$ $ (a \plus{} b)^3(b \plus{} c)^3(c \plus{} d)^3(d \plus{} a)^3$ $ \geq$ 16$ a^2b^2c^2d^2(a \plus{} b \plus{} c \plus{} d)^4$

Here is my solution

First of all notice that $ (a\plus{}b)(b\plus{}c)(c\plus{}d)(d\plus{}a)\equal{}(a\plus{}b\plus{}c\plus{}d)(abc\plus{}bcd\plus{}cda\plus{}dab)$

So we have to prove that $ (abc\plus{}bcd\plus{}cda\plus{}dab)^3\geq 16a^2b^2c^2d^2(a\plus{}b\plus{}c\plus{}d)$

But from Maclaurin inequality we have
$ (\frac{abc\plus{}bcd\plus{}cda\plus{}dab}{4})^3\geq \frac{\sum abc\cdot bcd\cdot cda}{4}\equal{}\frac{a^2b^2c^2d^2(a\plus{}b\plus{}c\plus{}d)}{4}$

and we are done  :)

[quote="silouan"]Here is my solution

First of all notice that $ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} a) \equal{} (a \plus{} b \plus{} c \plus{} d)(abc \plus{} bcd \plus{} cda \plus{} dab)$

and we are done  :)[/quote]

Let $ b \equal{} 0$, then
$ ac(c \plus{} d)(d \plus{} a) \equal{} (a \plus{} c \plus{} d)(cda)$ or $ (c \plus{} d)(d \plus{} a) \equal{} (a \plus{} c \plus{} d)(d)$  :ninja:

Apollo's proof is nice :)

[quote="dien9c"][quote="silouan"]

But from [color=#FF4040]Maclaurin[/color] inequality we have
$ (\frac{abc\plus{}bcd\plus{}cda\plus{}dab}{4})^3\geq \frac{\sum abc\cdot bcd\cdot cda}{4}\equal{}\frac{a^2b^2c^2d^2(a\plus{}b\plus{}c\plus{}d)}{4}$
[/quote]
This inequality isn't Maclaurin inequality for 4 variables , it's must be
$\sqrt[3]{\frac{abc+bcd+cda+dab}{4}} \le \sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}$
Your inequlity is $(a+b+c+d)^3 \ge 16(abc+bcd+cda+dab)$. This is well known inequality and can be proved by AM-GM[/quote]
I don't agree with you.
Maclaurin's inequality for 4 variables is a following:
Let $a+b+c+d=4u$, $ab+ac+ad+bc+bd+cd=6v^2$,
$abc+abd+acd+bcd=4w^3$ and $abcd=t^4$,
where $a$, $b$, $c$, $d$, $v$ and $t$ are non-negatives. Prove that $u\geq v\geq w\geq t$. :wink:

[quote=sqing]Let $a,b,c,d$ be positive numbers such that $abcd(a+b+c+d)\geq4$ .prove that
$$(a+b)(b+c)(c+d)(d+a)\geq 16.$$[/quote]
It's much weaker than a starting inequality.


geometry
parallelogram
trapezoid
LaTeX
geometric transformation
reflection
inequalities

Let $(O)$ be a fixed circle with the radius $R$. Let $A$ and $B$ be fixed points in $(O)$ such that $A,B,O$ are not collinear. Consider a variable point $C$ lying on $(O)$ ($C\neq A,B$). Construct two circles $(O_1),(O_2)$ passing through $A,B$ and tangent to $BC,AC$ at $C$, respectively. The circle $(O_1)$ intersects the circle $(O_2)$ in $D$ ($D\neq C$). Prove that:
a) \[ CD\leq R  \]
b) The line $CD$ passes through a point independent of $C$ (i.e. there exists a fixed point on the line $CD$ when $C$ lies on $(O)$).

Here's a cute proof for the first point in this problem:

Perform an inversion of pole $C$ and power $4R^2$. For a point $X$, let $X'$ be its image through the inversion. $CD'$ is the diagonal of the parallelogram $DA'D'B'$, which is at least twice the altitude of $CA'B'$, which is $2R$. this means that $CD'\ge 4R$, and, since $CD'\cdot CD=4R^2$, we get $CD\le R$.

We can use the same inversion to show that the fixed point we have to find for the second point is the intersection of $CD$ with the circle $(AOB)$ (the point different fro $D$; we can show that $A,O,B,D$ are concyclic; see below).

First of all, $A'O'D'B'$ is an isosceles trapezoid (the same notations as in my previous post), meaning that $A',O',D',B'$ are concyclic, meaning that so are $A,O,B,D$ (call this assertion $(*)$). It is also easy to see that $\angle CO'D'=\frac\pi 2$, so $\angle CDO=\frac \pi 2\ (**)$. From $(*),(**)$ we derive the conclusion that $CD$ intersects the circle $(AOB)$ in the antipode of $O$.

a) In my exercise , I showed  <ODC = 90 .
 b)  In my exercise , I proved $ ADOB $ is a cyclic quadrilateral

a) Prove that [tex] OO_1CO_2 [/tex] is a parallelogram. [tex] O_1O_2 [/tex]  intersects $ OC $ and $ CD $ at $ M $ and $ H $.
    We have [tex] CH \leq CM \Leftrightarrow CD \leq R [/tex] :)
b) I proved that [tex] AEFB [/tex]  is a cyclic quadrilateral where [tex] AE,BF [/tex] are two tangents of [tex] (O),E \in (O_1);F \in (O_2) [/tex].
Then [tex] CD,AE,BF [/tex] converge.   :)
To nttu: Làm bài tốt chứ ? :)

in my exercise,i show that D lies on the symmedian through C of triangle ABC

As for a) part of the question, my solution is as follows.

First of all, by simple angle-chasing one observes that $\Delta ACD$ is similar to $\Delta CBD$ and hence $CD^2=AD\cdot BD$. Let $E=(O)\cap DB\backslash B$, then $\angle AEB = \pi - \angle DCA - \angle DCB = \pi - \angle DBC - \angle DAC = \angle DAE$ and therefore $\Delta AED$ is isosceles. Then $AD=ED$ and from the definition of the power of the point we obtain $CD^2=AD\cdot BD= DE \cdot DB = R^2 - DO^2 \leq R^2$, because point D lies inside $(O)$ ($R$ - radius of $(O)$).

The only weak place in my proof, in my opinion, is that one, where I say, that $D$ lies inside $(O)$. It must be obvious, but I just can't prove it ingeniously :blush:. It is obvious, that $D$ lies inside $\angle ACB$, but why can't it lie outside $(O)$. Please help me! How to prove this fact without writing tales, with pure geometry?

P.S. How to write "\" and "{}" in LaTeX?

[color=red][Moderator edit: For "\", use \backslash; for "{}", use \{ and \} or, better, \left\{ and \right\}.][/color]

GROBBER    can  you  explain why we obtain the parallelogam  you  mention in your solution?

well  im  thinking its not that  obvious ,  seeing that  were  probably talking about  CA'D'B'  and  not DA'D'B'.
  my other  question is  why is the altitude from C  to  A'B'  equal  to  2R?  IM  not sure if its obvious or not.

Well, this problem dies with inversion. Invert through $C$ with arbitrary power, and the image of $X$ will be denoted as $X$ for all points for simplicity. We get the following: $C$, and 2 points $A, B$. Then $O$ is the reflection of $C$ over $AB$. Then $O_1, O_2$ map to the lines parallel to $BC, AC$ through $A, B$ respectively, so $ACBD$ is a parallelogram. So $D$ is the reflection of $C$ over the midpoint of $AB$. Since perpendicular give shortest distance, $CO \le CD$, but we inverted, so the inequality should be flipped, so (a) is done. For (b), clearly $CD$ is a median, so in the original, $CD$ is a symmedian, so it passes through the intersection of the tangents from $A, B$ (famous lemma). So (b) is finished.


Consider a quarter of a circle of radius 16. Let r be the radius of the circle inscribed in this quarter of a circle. Find r. Round r to the nearest whole number.

If we were to draw the figure with the inscribed circle inside the quarter circle, let the radius of the inscribed circle be r. If we drew r from the center of the inscribed circle to the three tangency points, we notice that a square with side length r is formed with the two radii and the segments of the quarter circle. Therefore we can draw an entire radius of the quarter circle by using the square's diagonal and the radius r of teh inscribed circle. By doing this, we can get $ r$ $ (1\plus{}\sqrt{2})$ $ \equal{}$ $ 16$, simplifying to $ r$ $ \equal{}$ $ \frac{16}{1\plus{}\sqrt{2}}$. This simplifies to $ r$ $ \equal{}$ $ 16$ $ \sqrt{2}$ $ \minus{}$ $ 16$, or $ \boxed{7}$, to the nearest whole number.

So there is a circle inscribed within a quarter of a circle.
The quarter circle has a radius of $ 16$. Instead of this image, let us take a full circle with $ 4$ circles inscribed within it.
Using the radii of these $ 4$ small circles, we can make a square.
For the time being, let us consider the radius of the smaller circles (what we want) to be $ r$.
Thus, the side of the square is $ 2r$.
With this being said, the diagonal of the square is $ 2r\sqrt{2}$.
Then, to get the diameter of the whole big circle, we must add $ 2r$ to the diagonal of the square.
Now, we have $ 2r\sqrt{2}\plus{}2r \equal{} 32$.
Simplify this to get $ 2r \times (\sqrt{2}\plus{}1) \equal{} 32$.
Now $ r \times (\sqrt{2}\plus{}1) \equal{} 16$.
So $ r \equal{} \frac{16}{\sqrt{2}\plus{}1}$.
By solving this equation, we get $ r  \approx \fbox{7}$
I hope you understood this; it might have been difficult without pictures.

[quote="akhil0422"]So there is a circle inscribed within a quarter of a circle.
The quarter circle has a radius of $ 16$. Instead of this image, let us take a full circle with $ 4$ circles inscribed within it.
Using the radii of these $ 4$ small circles, we can make a square.
For the time being, let us consider the radius of the smaller circles (what we want) to be $ r$.
Thus, the side of the square is $ 2r$.
With this being said, the diagonal of the square is $ 2r\sqrt {2}$.
Then, to get the diameter of the whole big circle, we must add $ 2r$ to the diagonal of the square.
Now, we have $ 2r\sqrt {2} \plus{} 2r \equal{} 32$.
Simplify this to get $ 2r \times (\sqrt {2} \plus{} 1) \equal{} 32$.
Now $ r \times (\sqrt {2} \plus{} 1) \equal{} 16$.
So $ r \equal{} \frac {16}{\sqrt {2} \plus{} 1}$.
By solving this equation, we get $ r \approx \fbox{7}$
I hope you understood this; it might have been difficult without pictures.[/quote]
Eh, aren't you basically repeating what I just said? There's no need for this.

But then the second poster would be notified of new post. Leave this to PM-s.

[quote="isabella2296"]Look at the times. You submitted while the second poster was typing.[/quote]
 :(  :blush:

errr didn't notice that...sry


What is the sum of the first 10 terms of the arithmetic sequence 1, $ \minus{}1$, $ \minus{}3$, $ \ldots$?

The first term is 1.
The next nine terms are -1,-3,-5,-7,...,-17.
Added up the next nine is -9^2=-81
So, our answer is -81+1=-80

We know that the sum of the first $n$ odd numbers is $n^{2}$. Hence we find the the sum of the first $9$ negative odd numbers and add it onto $1$.

$1+(-1)(9^{2})=\boxed{-80}$


Prove that:  $ (a\plus{}2b\plus{}3c)^2\leqslant2(a^2\plus{}4b^2\plus{}9c^2)\plus{}8ab$ :?:

[quote="Pirkuliyev Rovsen"]Prove that:  $ (a \plus{} 2b \plus{} 3c)^2\leqslant2(a^2 \plus{} 4b^2 \plus{} 9c^2) \plus{} 8ab$ :?:[/quote]
It's equivalent to $ (a\plus{}2b\minus{}3c)^2\geq0.$


function
inequalities
inequalities proposed

Given positive integer $ n,$ real numbers $ x_{1}\leq x_{2}\leq \cdots\leq x_{n},$ and $ y_{1}\geq y_{2}\geq\cdots\geq y_{n}$ satisfying $ \sum_{i \equal{} 1}^{n}ix_{i} \equal{} \sum_{i \equal{} 1}^{n}iy_{i}.$ Prove that for arbitrary real number $ \alpha, \sum_{i \equal{} 1}^{n}x_{i}[i\alpha]\geq \sum_{i \equal{} 1}^{n}y_{i}[i\alpha].$ (where $ [x]$ denotes the greatest integer less than or equal to $ x$)


function
limit
algebra
functional equation
complex analysis

I need help in finding all the functions satisfying:
$ f(t\plus{}s)\equal{}f(t)f(s)$

Do you know anything else about the function? Like is it continuous or differentiable?

if its differentiable...

$ f'(x) \equal{}\lim_{h \to 0} \frac {f(x\plus{}h)\minus{}f(x)}{h}\equal{} \lim_{h \to 0} \frac {f(x)(f(h)\minus{}1)}{h}\equal{}f(x) \lim_{h \to 0} \frac {f(h)\minus{}1}{h}$

letting $ \beta \equal{} \lim_{h \to 0} \frac {f(h)\minus{}1}{h}$ we see $ f'(x)\equal{}\beta f(x)$

and so we get $ f(x)\equal{}Ae^{\beta x}$

as for A, we can see that $ 0\equal{}f(s)\minus{}f(s)\equal{}f(s)\minus{}f(s\plus{}0)\equal{}f(s)\minus{}f(s)f(0)\equal{}f(s)(1\minus{}f(0)), \forall s$

so either $ f(s)\equal{}0, \forall s \to A\equal{}0$ or $ f(0)\equal{}1 \to A\equal{}1$

ignoring the boring identically zero solution, let $ A\equal{}1, \to f(x)\equal{}e^{\beta x}$ for your favorite $ \beta$

$ f(x) \equal{} e^{g(x)}$ where $ g(x)$ is a solution to the [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy functional equation[/url].  I discuss this briefly [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=207609]here[/url].


[quote="jhcreinhardt"]ok try this one

$\sin 10^\circ \sin 50^\circ \sin 70^\circ$[/quote]

[hide]$X = \sin 10^\circ \sin 50^\circ \sin 70^\circ$
or, $2\cos 10^\circ X = 2\cos 10^\circ \sin 10^\circ \sin 50^\circ \sin 70^\circ$
or, $2\cos 10^\circ X = \sin 20^\circ \sin 50^\circ \sin 70^\circ$

since $\sin 70^\circ = \cos 20^\circ$, we have
$4\cos 10^\circ X = 2\sin 20^\circ \sin 50^\circ \cos 20^\circ$
or, $4\cos 10^\circ X = \sin 40^\circ \sin 50^\circ$

since $\sin 50^\circ = \cos 40^\circ$, we have
$8\cos 10^\circ X = 2\sin 40^\circ \cos 40^\circ$
or, $8\cos 10^\circ X = \sin 80^\circ$

since $\sin 80^\circ = \cos 10^\circ$, we have
$8\cos 10^\circ X = \cos 10^\circ$

or, $X = \frac{1}{8}$[/hide]


Heh...
That's probably my fault. I had meant to make the test around the state level, but obviously, I got carried away. Just a tad though.  :lol:

I may make another, but that would have to be after I get back from vacation.

Thanks, you have no idea how much I appreciate that. Its my last year and I really want to do well. I truly need the practice!

                                        -Thanks Again!

[quote="AGeneson"]I have been thinking about the most recent mathcounts mocks and they are all about national and up. We should probably start doing some other levels too, because not everyone :blush: is good enough to do national++ in the time limit without bombing it lol. If anyone has any ideas for one or could make one etc..... please reply

                                                            -Thanks[/quote]

And somebody doing problems in the Pre-Olympiad Forum isn't good enough for national level problems?  :?  ;)

Eh, I haven't seen her do problems in Pre-Olympiad Forum ;).

I think you're confusing AGeneson with JGeneson, becuase all of AGeneson's posts have been in the MATHCOUNTS forum.

Are you two related or something?


Find {x: (x-2)^2 > 9}

and

Find {(x,y): x = 7 - 4y and y = 7 - 4x}

Thanks!

Please post your own progress first. We will not do your homework for you, but we will give help. And someone move this to HSB.

Assuming you mean:

1.) Solve for $ x$: $ (x\minus{}2)^{2}>9$

2.) Solve for $ (x,y)$: $ x\equal{}7\minus{}4y$ and $ y\equal{}7\minus{}4x$

Then:

1.) Take the square root of each side: $ \sqrt{(x\minus{}2)^{2}}>\sqrt{9}\rightarrow (x\minus{}2)>3$. Then add $ 2$ to each side: $ x>5$.

2.) First put these equations into standard form: $ x\plus{}4y\equal{}7$ and $ 4x\plus{}y\equal{}7$. Now solve using whatever method you like. Because this is a simple enough system I'll do it algebraically: First I multiply the first equation by $ \minus{}4$: $ \minus{}4(x\plus{}4y\equal{}7)\rightarrow \minus{}4x\minus{}16y\equal{}\minus{}28$. Now I'll add this to the second equation which yeilds: $ \minus{}15y\equal{}\minus{}21$. Now Ill solve this: $ y\equal{}\displaystyle\frac{7}{5}$. Now I'll substitue the $ y$ value into one of the original equations: $ x\plus{}4(\frac{7}{5})\equal{}7$. Now I'll solve for $ x$: $ x\plus{}\frac{28}{5}\equal{}7$. Now subtract to get: $ x\equal{}\frac{7}{5}$. So the answer is $ (\frac{7}{5},\frac{7}{5})$.

On Number 1 I got:
[hide="1"]
Take square roots and get $ \sqrt{(x\minus{}2)^2}>\sqrt{9}$, so [u]either[/u] $ x\minus{}2>3$ [u]or[/u] $ x\minus{}2<\minus{}3$. Solving these we get either $ x>5$ or $ x<\minus{}1$[/hide]

Oh okay, I only managed to find out x < 5 but not the other solution

Cheers!

I'll put my progress on next time


:huh: Why is a topic from a math olympiad in the intermediate topics forum? :huh:

In the UK, math olympiads start at a [b]much[/b] easier level than they do in the USA. (There is an olympiad specifically for 11-13 year olds, though that is of course very easy by olympiad standards).

This is question 1 from round 1 of the main British Olympiad, but that is really at an equivalent level to the AIME, although it does require proofs. Whenever I or someone else have posted a question from this olympiad (even the questions 4 and 5 on the paper) in Intermediate, it has been answered, although the harder ones are perhaps Pre-Olympiad.

Olympiad does[b] not[/b] imply USAMO difficulty at all.

I don't think that it's a particularly elegant question. (I had hoped that there was a nicer way to do it than I did, but that seems unlikely.)

Basically, I think the easiest way is to (by an exhaustive search or otherwise) find all numbers that can expressed as the sum of two squares (where the squares are of numbers $\leq 17$).

This, in regard to the equation $(a+1)^{2}+(d+1)^{2}=(b+1)^{2}+(c+1)^{2}$, gives us about $8$ different cases to check, which we just put into the other equation to see which ones work.

Long, but not too long.

[quote="roadnottaken"][quote="tim1234133"]I have four children. The age in years of each child is a positive integer between $2$ and $16$ inclusive and all four ages are distinct. A year ago, the square of the age of the oldest was equal to the sum of the squares of the other three. In one year's time the sum of the squares of the ages of the oldest and the youngest will be equal to the sum of the squares of the other two children.

Decide whether this information is sufficent to determine their ages uniquely, and find all possibilities for their ages.[/quote]

Here goes...  Label the children's ages $a$, $b$, $c$, and $d$ such that $2\le a<b<c<d\le16$.  We now have
\begin{align*}(a-1)^{2}+(b-1)^{2}+(c-1)^{2}&=(d-1)^{2}\\ a^{2}-2a+1+b^{2}-2b+1+c^{2}-2c+1&=d^{2}-2d+1\\ a^{2}-2a+b^{2}-2b+c^{2}-2c-d^{2}+2d&=-2\\ \\ (a+1)^{2}+(d+1)^{2}&=(b+1)^{2}+(c+1)^{2}\\ a^{2}+2a+1+d^{2}+2d+1&=b^{2}+2b+1+c^{2}+2c+1\\ a^{2}+2a-b^{2}-2b-c^{2}-2c+d^{2}+2d&=0\\ \\ 4a-2b^{2}-2c^{2}+2d^{2}&=2\\ 2a-b^{2}-c^{2}+d^{2}&=1\\ \end{align*}
The seventh equation is simply the sixth equation minus the third equation.

This does not seem to be enough information to determine their ages uniquely.  Either I'm missing something :(  , or this is about to devolve into a [b]lot[/b] of case-work :wallbash:  :help: .[/quote]

Instead of substracting the sixth equation and third equation, you could also add them to get $a^{2}+1 = 2(b+c-d)$. This implies that $a$ is odd. Furthermore, we know that $b+c \leq 14+15$ and $d\geq 5$, so $a^{2}+1 = 2(b+c-d) \leq 2(29-5) = 48$, so $a < 7$. The only possibilities for $a$ are 1, 3 and 5.
For $a=1$, we get $1 = b+c-d$, so $d+1 = b+c$, so $(b+c)^{2}= (d+1)^{2}= (b+1)^{2}+(c+1)^{2}-(a+1)^{2}= b^{2}+2b+c^{2}+2c-2$, so $bc-b-c+1 = 0$, which is only possible if $b=1$ or $c=1$. Impossible.
For $a=3$, we get $5 = b+c-d$, so $d+1 = b+c-4$, so \[b^{2}+c^{2}+16-8b-8c+2bc = (b+c-4)^{2}= (d+1)^{2}= (b+1)^{2}+(c+1)^{2}-(a+1)^{2}= b^{2}+2b+c^{2}+2c-14,\] so $bc-5b-5c-15 = 0$, so $(b-5)(c-5) = 10$. If $b-5 < 0$, then $b-5 \in \{-10,-5,-2,-1\}$, i.e. $b\in\{-5,0,3,4\}$. This is only possible when $b\in\{3,4\}$, but then $c-5\in\{-5,-10\}$, which is impossible. So $b-5 \geq 0$, so $(b-5,c-5)\in\{(1,10),(2,5),(5,2),(10,1)\}$, but $b-5 < c-5$, so $(b-5,c-5)\in\{(1,10),(2,5)\}$, i.e. $(b,c)\in\{(6,15),(7,10)\}$. Then $d = b+c-5 \in\{16,12\}$. So we have two possibilities here: $(a,b,c,d)\in\{(3,6,15,16),(3,7,10,12)\}$. And those two quadruples seem to be solutions to our system \[\left\{\begin{array}{l}(a-1)^{2}+(b-1)^{2}+(c-1)^{2}= (d-1)^{2}\\ (a+1)^{2}+(d+1)^{2}= (b+1)^{2}+(c+1)^{2}\end{array}\right.\]
So, no unique solution...


If $ x,y,z$ are positive numbers, then

$ p\sum{\left(x^3\plus{}xyz\right)^p}\geq p\sum{\left(x^2y\plus{}x^2z\right)^p}$

holds for any real number $ p.$

As a special case $ p\equal{}1$, we have

$ \sum{x(x\minus{}y)(x\minus{}z)}\geq 0.$

when $p=2$,
$ \sum{\left(x^3+xyz\right)^2}\geq
\sum{\left(x^2y+x^2z\right)^2}$
$LHS-RHS=\sum{(x^6+x^2y^2z^2-x^2y^4-x^4y^2)}$
which $\geq 0$ due to Schur inequality in which $x$ becomes $x^2$, $y$ becomes $y^2$, $z$ becomes $z^2$

when $p=\frac{m}{n}$ in which $p\geq1$ and $m,n$ are both odd numbers, from Power-Mean Inequality and schur inequality we can get:
$\frac{(LHS-RHS)^\frac{1}{p}}{6}=\frac{(\sum{\left(x^3+xyz\right)^p}+\sum{\left(-x^2y-xy^2\right)})^\frac{1}{p}}{6}$$\geq$$\frac{\sum{\left(x^3+xyz\right)}-\sum{\left(x^2y+x^2z\right)}}{6}$$\geq0$


The graphs of $ \minus{}2x \plus{} 5y \equal{} 7$ and $ 10x \plus{} ky \equal{} \minus{}6$ are perpendicular. What is the value of $ k$?

$ \minus{}2 \cdot 10 \plus{} 5 \cdot k \equal{} 0 \iff k \equal{} \boxed{4}$.


A trapezoid $ ABCD$ with $ AB \parallel CD$ is inscribed in a circle $ k$. Points $ P$ and $ Q$ are chosen on the arc $ ADCB$ in the order $ A\minus{}P\minus{}Q\minus{}B$. Lines $ CP$ and $ AQ$ meet at $ X$, and lines $ BP$ and $ DQ$ meet at $ Y$. Show that points $ P,Q,X,Y$ lie on a circle.

Dear mathlinkers,
1. XY is the pascal's line of the inscribed hexagone PCDQABP and is parallel to AB
2. According to a converse of Reim's theorem, P, Q, X and Y are concyclic.
Sincerely
Jean-Louis

$ \angle PXQ\equal{}\frac{1}{2}(arc AC\plus{}arc PQ)\equal{}\frac{1}{2}(arc BD\plus{} arc PQ)\equal{}\angle PYQ$
So $ PYXQ$ is cyclic quadrilateral.

$ m(\angle XPY)\equal{}180^{\circ}\minus{}m(\angle CPB)\stackrel{AB\parallel CD}{\ \ \equal{}\ \ }180^{\circ}\minus{}m(\angle DQA)\equal{}m(\angle XQY)$ a.s.o.


limit
calculus
derivative
calculus computations

Evaluate:  $\lim_{x\to\infty} (\sqrt{x^2+x}-x)$.

[hide]1/2[/hide]

$\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {x^2  + x}  - x} \right) = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x^2  + x - x^2 }}{{\sqrt {x^2  + x}  + x}} = $
$ = \mathop {\lim }\limits_{x \to  + \infty } \frac{1}{{\sqrt {1 + \frac{1}{x}}  + 1}} = \frac{1}{2}$

Let's change the question:

$\lim_{x\to\infty}\left(x^3+x^2+1\right)^{\frac13}-x$

Can't use the old "rationalize the numerator" trick on that one, can you?

---

For a class I'm teaching, I promised a handout introducing big-O and little-o notation. When I get it written next week, some parts of it will wind up here - and yes, this problem will be an example.

[quote="Peter VDD"][quote="Jimmy"]Thanks, I can't belive I forgot about L'Hopital. :lol:

I'll be interested in seeing the big-O and little-o sheet.[/quote]

where do you see de l'hopital? :?[/quote]

My mistake, I looked at it quickly and thought he took the derivative of the numerator and denominator.


prove that for all non negative integers m and n $ (2m)!(2n)!/m!n!(m!\plus{}n!) is an integer$

[hide="Well-known"]This is IMO 1972 no. 3. See http://www.mathlinks.ro/viewtopic.php?p=139664#139664.[/hide]


Okay in this competition the signups begin now, and I will wait until there are 20 people that have signed up. In the competition I will randomly divide the 20 people into 2 equal teams. Then the two teams will have a competition. The team that loses will vote somebody off of their team. This person is eliminated. This process will continue until there are only twelve people left. Then they will all be against each other. They will still do challenges, but the winner is the only person that gets the immunity idol, which means that nobody else can vote for that person. At the end of each round the person with the immunity idol must return it, and then they play for it again. This will happen until there are 3 people left. Once they are down to the top three contestants, they will do a challenge for the immunity idol, and the winner autimatically moves on to the last round. The winner also picks the person that he wants to compete against in the last round. In the last round the people who made it to the top twelve other than the top two will vote for the person who they want to win the competition.  Then they will also do a FTW competition. (If you don't know what that is please ask.) The winner gets a free vote added to there current total of votes. Most of the challenges will take place on this post so please keep checking out this post to see what is going on and to see the challenges. Good Luck!!! Signups start now.

Okay budi713 is the only person that has joined.

err i don't really understand the rules, but i'll get it once we start. Sure, I'll join.

Okay so far we have three people that have joined;

budi713
ernie
jjx1

I join! Votes will be confidential right?

budi713
ernie
jjx1
mewto55555
pacman2812

Votes will be confidential, although you can tell others who you are voting for.

um is this a old thread now or are we still continuing sign up.

someone with a username starting with dragon96 and and ends in dragon96 joins. That person thinks the rules are a bit unclear, so that person asks for an explanation about how a team "loses". FTW? longest word?

Basically the two teams compete in some kind of random challenge and the team that loses will be the one that has the vote. At the vote each person private messages me their vote. (I will keep the votes secret. The person you vote for is the person that you want to get rid of in order to win more challenges. Remember, you want to have as good a team as possible because you don't want to get voted off, and you are guaranteed to make it to the next round if your team wins.) I have already decided the challenges, but if you guys have any really good ideas, please pm them to me, and I might consider using them.

budi713
ernie
jjx1
mewto55555
pacman2812
dragon96
kl2836

no one is signing up on this thread. I better keep it on the top

Please WIll more people sign up. I want this to start

I don't get this mafia game. Explain the rules when you make challenges . I didn't know what mafia was.

OK, every night, someone gets killed by the mafia.

Every day, everyone (including the mafia) votes off a person (called lynch).

This continues until all the people are gone except for one.

Alright, this is annoying. I have only gotten two votes. When should the deadline be?

Everybody except you and AIMEman has quit, so I suppose we should just vote between the two of you to declare a winner. You may post reasons why you should be voted for on this thread.

ahem when did I quit. i never did. I jsut pmed if this game was dead and you said it was. Doesn't mean I quit. I'm still in

Oh, when I said "you", I meant to say jjx1. Mods please change that to jjx1.

jjx1 and AIMEman please give everybody reasons to vote for you to win.

wait. I think AIMEman is afk. Anway if everyone is inactive on AOPS who would vote?

Hmm.... Me I guess.

Votecount;
jjx1
AIMEman

jjx1, you have an advantage. You can post why you think you should be voted for, and you can vote for yourself.

i think I should be voted because lots of people know me more and less of AIMEman. I think. Also because i'm a mod :lol:. Oh, and I vote myself

Votecount;
jjx1 - 1 vote
AIMEman

Two weeks from  now, whoever has the most votes will be the winner. If there is a tie, then I will pick a person based on what they have to say.

Alright, it has been long enough. jjx1, you are the winner of AoPS's worst survivor competition ever!!! Even though it was a bad competition, jjx1 deserved to win it, because he was always eager to get the game going, and he was always telling me to continue. jjx1, you are the winner, sorry the competition didn't go as well.

Thank you Wick. I don't get why people sign up for a game and don't play it.

Yeah, me neither. Thanks people that actually participated, and sorry about those who didn't.

Hm... I don't remember quitting.  Oh well...


What is the sum of the roots of $ 2x^2 \plus{} 1 \equal{} 3x$ ? Express your
answer as a common fraction.

$ 2x^2 \plus{} 1 \equal{} 3x$

$ 2x^2 \minus{} 3x \plus{} 1 \equal{} 0$

By Vieta's, the sum of the roots is $ \frac{3}{2}$.

Alternatively, factor:

$ (2x \minus{} 1)(x \minus{} 1) \equal{} 0$

$ x \equal{} \frac{1}{2}, 1$

This sum yields the same answer.


logarithms
function
Euler
real analysis
real analysis unsolved

Denote:

$ f(s) \equal{} \sum^{\infty}_{n \equal{} 1}( \minus{} 1)^n\frac {\ln(n)^2}{n^s}$

Now, denote:

$ f(s) \equal{} \frac {d^2 F(s)}{ds^2}$

From this we can get:

$ F(s) \equal{} \sum^{\infty}_{n \equal{} 1}\frac {( \minus{} 1)^n}{n^s} \equal{} \minus{} \eta(s)$

where $ \eta(s)$ is the Dirichlet eta function that can be represented as:

$ \eta(s) \equal{} (2^{1 \minus{} s} \minus{} 1)\zeta(s)$

where $ \zeta(s)$ is the Riemann's zeta function. From this we get:

$ F(s) \equal{} (1 \minus{} 2^{1 \minus{} s})\zeta(s)$

and

$ f(s) \equal{} \ln(2)^2\,2^{1 \minus{} s}\zeta(s) \minus{} \ln(2)\,2^{2 \minus{} s} \zeta^{'}(s) \plus{} (2^{1 \minus{} s} \minus{} 1)\zeta^{''}(s)$

Plugging in $ s \equal{} 2$ we get:

$ f(2) \equal{} \frac {(\ln(2)\pi)^2}{12} \minus{} \ln(2)\,\zeta^{'}(2) \minus{} \frac {\zeta^{''}(2)}{2}$

Also, this may help:

$ \zeta^{'}(2) \equal{} \frac {\pi^2}{6}\left(\gamma \plus{} \ln(2\pi) \minus{} 12\ln(A)\right)$

where $ \gamma$ is the Euler-Mascheroni constant and $ A$ is the Glaisher-Kinkelin constant:

$ A \approx 1.28242712$

yes. but what about $ \zeta''(2)$? or how to find $ \sum_{n\equal{}1}^{\plus{}oo}(\minus{}1)^{n}\frac{\ln^{k}n}{n^{k}}$ where $ k \in N$

Well, we could use the identity:

$ \frac{d^n}{ds^n}\eta(s)\equal{}(\minus{}1)^n\sum^{\infty}_{k\equal{}1}(\minus{}1)^k\frac{\ln(k)^n}{k^s}$

and since

$ \eta(s)\equal{}2^{1\minus{}s}\zeta(s)\minus{}\zeta(s)$

we get:

$ \sum^{\infty}_{k\equal{}1}(\minus{}1)^k\frac{\ln(k)^n}{k^s}\equal{}\frac{1}{2^{s\minus{}1}}\sum^{n}_{i\equal{}0}(\minus{}1)^i\binom{n}{i}\ln(2)^{n\minus{}i}\frac{d^i}{ds^i}\zeta(s)\minus{}(\minus{}1)^n\frac{d^n}{ds^n}\zeta(s)$

now, we can put $ s \rightarrow n$ and get

$ \sum^{\infty}_{k\equal{}1}(\minus{}1)^k\frac{\ln(k)^n}{k^n}\equal{}\frac{1}{2^{n\minus{}1}}\sum^{n}_{i\equal{}0}(\minus{}1)^i\binom{n}{i}\ln(2)^{n\minus{}i}\frac{d^i}{ds^i}\zeta(s)\Big|_{s\equal{}n}\minus{}(\minus{}1)^n\frac{d^n}{ds^n}\zeta(s)\Big|_{s\equal{}n}$

It's easy to check for $ n\equal{}2$:

$ \sum^{\infty}_{k\equal{}1}(\minus{}1)^k\frac{\ln(k)^2}{k^2}\equal{}\frac{1}{2}\sum^{n}_{i\equal{}0}(\minus{}1)^i\binom{2}{i}\ln(2)^{2\minus{}i}\frac{d^i}{ds^i}\zeta(s)\Big|_{s\equal{}2}\minus{}\frac{d^2}{ds^2}\zeta(s)\Big|_{s\equal{}2}\equal{}$
$ \equal{}\frac{1}{2}\left(\ln(2)^2\zeta(2)\minus{}2\ln(2)\zeta^{'}(2)\plus{}\zeta^{''}(2)\right)\minus{}\zeta^{''}(2)\equal{}$
$ \equal{}\frac{1}{2}\left(\ln(2)^2\zeta(2)\minus{}2\ln(2)\zeta^{'}(2)\minus{}\zeta^{''}(2)\right)$

and that's OK! And what about $ \zeta^{''}(2)$? Well, what about $ \pi$? Or $ \gamma$?

Same method what you use was posted by jmerry here http://www.mathlinks.ro/Forum/viewtopic.php?t=143417  where after some disscusing we say that we can calculate this sum but it has not good form. now I represent problem like this form $ \sum ( \minus{} 1)^{n}\frac {\ln^{2} n}{n^{2}}$ and as we seen no one don't post other soultion. thanks anyway milin.

Well, the approach I gave is quite understandable because the derivation of $ \zeta(s)$ is well defined and in close connection with your problem. I must say that I see no reason why you couldn't rewrite the sum you gave like an expression with known functions like $ \zeta(s)$!?


function
superior algebra
superior algebra unsolved

Let G be a finite nonempty set with an operation * such that:
1. G is closed under *.
2. * is associative.
3. Given $ a,b,c \in G$ with a*b=a*c, then b=c
4. Given $ a,b,c \in G$ with b*a=c*a, then b=c
Prove that G must be a group under *.

Need help to find a condition where G is [b]no longer [/b]a group if G is an infinite set?

An infinite counterexample is $ (\mathbb{N} \minus{} \{1 \}, \times)$.  The problem here is that inverses don't exist even though the cancellation law holds, so we can take out the identity.  Another example is $ (\mathbb{N}, \plus{} )$.  The key word here is [url=http://en.wikipedia.org/wiki/Semigroup]semigroup[/url]; the second example is the free semigroup on one letter.

As for the problem, fix an element $ a \in G$.  The two cancellation laws imply that $ aba \equal{} aca \Leftrightarrow b \equal{} c$ for any $ a, b, c \in G$.  The function $ f(g) \equal{} aga$ is therefore injective, which on a finite set (and this is the step that fails when $ G$ is infinite!) implies that it is surjective, hence bijective, so there exists $ g$ such that $ f(g) \equal{} aga \equal{} a$.  Then $ ag \equal{} ga$ is the identity (and $ g$ the inverse of $ a$): $ agab \equal{} ab \implies gab \equal{} b$ and $ baga \equal{} ba \implies bag \equal{} b$.  Additionally, there exists $ h$ such that $ f(h) \equal{} aha \equal{} b \implies ahaga \equal{} bga \equal{} aha \equal{} b$ and $ agaha \equal{} agb \equal{} aha \equal{} b$.

Thank you very much for the detailed response.  :lol:  I'm still trying to absorb all of it.   I think I'm missing some of your points related to the infinite set.

 By the (N-{1}, x) this is all natural numbers minus the set of {1}.

 Doesn't a finite set of natural numbers also not have an inverse...because the inverse would be rational and not a natural number?

 For N-{1}, does the -{1} give a zero when N = 1, again no inverse?

 I'm struggling with how there can be a difference between the inverses for a finite vs infinite set.

 I think I'm missing some of your points related to the infinite set.

 Could you elaborate on the comment "`the function f(g) = aga is therefore injective on a finite set, but fails on an infinite set"'.

Could you recommend anything to read on the finite vs infinite set topic?

[quote="smurf"]Doesn't a finite set of natural numbers also not have an inverse...because the inverse would be rational and not a natural number?[/quote]
A finite set of natural numbers also isn't closed.  (Unless we are talking about the singleton $ \{ 1 \}$, which is the trivial group with one element and doesn't constitute a counterexample to the problem.)

[quote="smurf"]For N-{1}, does the -{1} give a zero when N = 1, again no inverse?[/quote]
That's not what the notation means.  I mean the set $ \{ 2, 3, 4, 5, ... \}$ under the group operation of multiplication.

[quote="smurf"]Could you elaborate on the comment "`the function f(g) = aga is therefore injective on a finite set, but fails on an infinite set"'.[/quote]
An injective function associates distinct outputs with distinct inputs.  An injective function from a set $ S$ of $ n$ elements to itself outputs $ n$ distinct values in $ S$, which must therefore be all of $ S$.  In other words, injective implies surjective.

Infinite sets behave quite differently.  It's easy to construct injective functions from an infinite set to itself that are not surjective.  For example, if we tried to apply the same argument as above to $ \mathbb{N} \minus{} \{1\}$, we would run into functions like $ f(g) \equal{} 4g$ which are clearly not surjective.  The problem here is that infinite sets contain proper subsets with the same [url=http://en.wikipedia.org/wiki/Cardinality]cardinality[/url] as themselves (and that is the word you should look up for more information).  This might be a little nonintuitive, and indeed this notion gives rise to such paradoxical results as [url=http://mathworld.wolfram.com/Banach-TarskiParadox.html]Banach-Tarski decomposition[/url].

Again, thank you very much for your comments.
I understand that (N-{1}, X) and (N, +) would not work because their inverses do not exist.

Would a infinite counterexample be:

G is as a set of Integers (Z, +) and the function f(g) = 4a where a = {...-2, -1, 0, 1, 2,...} and b = {...-8, -4, 0, 4, 8, ...}  is a group since the inverse  (-4g) and identitiy (0) exist.

But is not a group for the infinite case for (Z, +)  since a = {...-2, -1, 0, 1, 2,... to Infiniti} ,  b= {...-8,-7, -6, -5  -4, 0,1,2,3,  4, 8, ...}
This counterexample would not be a group because it is not surgective (onto), but would be injective.  Need both to be bijective and thereby countable?

I'm still not sure i understand the difference between an infinite group and finite group other than for a infinite group to be countable it has to be bijective = surgective and injective (one-one)

[quote="smurf"]G is as a set of Integers (Z, +) and the function f(g) = 4a where a = {...-2, -1, 0, 1, 2,...} and b = {...-8, -4, 0, 4, 8, ...}  is a group since the inverse  (-4g) and identitiy (0) exist.[/quote]

You're confusing notation here and we're talking about two statements.

[b]Statement 1:[/b]  The original problem.  A finite semigroup with the cancellation property is a group.

[b]Statement 2:[/b]  An injective function from a finite set to itself is surjective, hence bijective.  This is not true for infinite sets.

I have given you a proof of the first statement and mentioned that the first statement doesn't hold if we remove the "finite" condition because of the second statement.  $ (\mathbb{Z}, \plus{})$ and the function $ f(g) \equal{} 4g$ is an example of the second statement but has nothing to do with the first.  You also do not have its inverse [i]function[/i] (which is [b]not[/b] the same thing as a group inverse) correct.  The inverse function of $ f(g) \equal{} 4g$ is $ h(g) \equal{} \frac{g}{4}$, which is clearly not defined on all of $ (\mathbb{Z}, \plus{})$.  That's why $ f(g)$ is not a bijection on $ \mathbb{Z}$ even though it is injective (again, statement 2, not statement 1).

[quote="smurf"]But is not a group for the infinite case for (Z, +)  since a = {...-2, -1, 0, 1, 2,... to Infiniti} ,  b= {...-8,-7, -6, -5  -4, 0,1,2,3,  4, 8, ...}
This counterexample would not be a group because it is not surgective (onto), but would be injective.  Need both to be bijective and thereby countable? [/quote]

Again, you're confusing Statement 1 and Statement 2.  $ (\mathbb{Z}, \plus{})$ is an infinite group, not a finite semigroup, so it has nothing to do with Statement 1.

[quote="smurf"]I'm still not sure i understand the difference between an infinite group and finite group other than for a infinite group to be countable it has to be bijective = surgective and injective (one-one)[/quote]

Infinite [i]semigroup[/i].  As far as this proof goes, Statement 2 constitutes the only difference (but it's a very important one).

Thanks for the response.

Sorry, another question...

When testing for closure (G is closed under *) , does this include the group inverse or the function inverse or both?

Neither.  Closure only implies that the result of the $ *$ operation on elements of $ G$ is in $ G$.