[quote="Valentin Vornicu"]Let $M$ be the set of the positive rational numbers less than 1, which can be expressed with a 10-distinct digits period in decimal representation.

a) Find the arithmetic mean of all the elements in $M$;

b) Prove that there exists a positive integer $n$, $1<n<10^{10}$, such that $n\cdot a - a$ is a non-negative integer, for all $a\in M$.[/quote]
a) we find the average of .1111111111 & .9999999999 and we are done
b) (im unfamilliar with a lot of notation) that means a can be any number inside M right? if so, then thats just a(n-1) is a non negative integer?

Why is the average of .1111111111 and .9999999999 equal to the arithmetic mean of set M?

I'm not sure how you do a) actually.

[hide="What I have"]I tried a) on my own and got .44444444445, which is 0.1111111111 below what Scrambled has.

But for b), that's easy because if $n=10^9$, then $na-a=[na]$ and is a positive integer.
[/hide]

Neither Zyloch nor Scrambled are correct for a). A periodic fraction with a period of 10 digits is something like 0,12345678901234567890123... (goes on forever) which we usually denote by 0,(1234567890).

Zyloch, $na-a$ is not an integer, because when you subtract the difference (there is a reason for setting $n<10^{10}$ ;) ) you do not overlap with the same digit (all the digits in the period are distinct).

For example take the number above: 0,(1234567890). Multiply it with $10^9$ we get 123456789,(0123456789) thus the difference between the latter and the number itself is not an integer.

[hide="Hint"]Try to find a "supplement" for each 10-distinct digit period fraction.  ;)[/hide]

Ah, Myth's hint helped quite a bit.

[hide]Since every element of $M$ has an unique complement that when the two are added together, they equal $0.9999999999$. Because of this, it's just sufficient to find the average of two of those complement numbers which is $\frac{0.9999999999}{2}=0.49999999995$. (I hope I didn't miss a digit in there lol)

As for b) I'm still a bit stuck. If $a$ ends in an even digit, it's easy to find such an integer $n$ such that $1<n<10^{10}$ because you can halve the resulting $a\cdot10^{10}$ so that we get $a\cdot5\cdot10^9$ as an integer and $n=5\cdot10^9+1$ (since $an-n=a(n-1)$).

If $a$ is odd however, I know that $a\cdot10^{10}$ will be divisible to 3 or 9, but $10^{10}$ will not be. Still have to work it out lol
[/hide]

[hide]Number (10^10-1)a is integer. What numerical properties does it have?[/hide]

Again Zyloch you are wrong :(
[hide]Any two complementary numbers add up to 1, whereas you (again) considered the fraction as having [b]exactly[/b] 10 digits over the decimal point. It is a periodic fraction, which means that it has an inifite number of digits after the decimal point ;)
So the average is $\dfrac 12$. [/hide]

Oh! Sorry lol I didn't read your previous post clearly enough.

I got 13 and i'm on the 22-th place and i need 3 more points...hopefully i will get them monday  :(


Let $ \alpha,$ $ \beta$ and $ \gamma$ are measureangle of triangle. Prove that
\[ \cos\frac {\alpha \minus{} \beta}{2} \plus{} \cos\frac {\beta \minus{} \gamma}{2} \plus{} \cos\frac {\gamma \minus{} \alpha}{2} > 1
\]

[quote="arqady"]Let $ \alpha,$ $ \beta$ and $ \gamma$ are measureangle of triangle. Prove that
\[ \cos\frac {\alpha \minus{} \beta}{2} \plus{} \frac {\beta \minus{} \gamma}{2} \plus{} \frac {\gamma \minus{} \alpha}{2} > 1
\]
[/quote]

Do you mean:
$ cos\frac {\alpha \minus{} \beta}{2} \plus{} cos\frac {\beta \minus{} \gamma}{2} \plus{} cos\frac {\gamma \minus{} \alpha}{2} > 1$ ?

Let $ \gamma\equal{}\min \{\alpha,\beta,\gamma\}$. Then we can reduce all variables by $ \gamma$, and not change the inequality, we get \[ \cos \frac{\alpha\minus{}\beta}{2}\plus{}\cos\frac{\alpha}{2}\plus{}\cos\frac{\beta}{2}>1\] where $ \alpha,\beta\ge 0$ and $ \alpha\plus{}\beta<\pi$. If we let $ 2x\equal{}\alpha$ and $ 2y\equal{}\beta$, then we have $ x,y\ge 0$ and $ x\plus{}y<\frac{\pi}{2}$. \[ \cos (x\minus{}y)\plus{}\cos x\plus{}\cos y>1\] Consider the statement: Given $ x,y\ge 0$, and $ x\plus{}y\le\frac{\pi}{2}$, then prove that \[ \cos(x\minus{}y)\plus{}\cos x\plus{}\cos y\ge 1\] with equality iff $ \{x,y\}\equal{}\left\{0,\frac{\pi}{2}\right\}$. This is easy to prove calculus...I don't know if there is a nicer proof for that statement (it should be easy I think).

[quote="Athinaios"]we know that
$ \cos{\frac {a}{2}} > \equal{} \minus{} 1$(1) and $ \cos{\frac {b}{2}} > \equal{} \minus{} 1$(2)
(1)*(2) = > $ \cos{\frac {a}{2}}*\cos{\frac {b}{2}} > \equal{} 1$
[/quote]
Are you sure? :wink:

Since $ \left[\frac{\pi}{2},0,\minus{}\frac{\pi}{2}\right] \succ \left[\frac{\alpha\minus{}\beta}{2},\frac{\beta\minus{}\gamma}{2},\frac{\gamma\minus{}\alpha}{2}\right]$ and $ \cos x$ is concave on $ \left[\minus{}\frac{\pi}{2}, \frac{\pi}{2}\right]$ the result follows from Karamata.


[quote="cezar lupu"]Let $x, y,z>0.$ Prove the following inequality:

   $\frac{x^{2}(y+z)}{(x+y)(x+z)}+\frac{y^{2}(z+x)}{(x+y)(y+z)}+\frac{z^{2}(x+y)}{(z+x)(y+z)}\leq\frac{x+y+z}{2}\blacksquare$ ;)[/quote]

if we set $x=a+b ; y = b+c ; z = a+c$ and $x+y+z =1$

So we must prove
$\large x,y,z>0 ; x+y+z=1$.then
$\large (1+x)(1+y)(1+z)\ge (1-x^2)^2+(1-y^2)^2+(1-z^2)^2$

this is an old problem

The  initial  inequality is  equivalent to  the  following geometric one :
In any triangle  ABC , prove  that  $a^4+b^4+c^4\leq2s^2(s^2-5r^2-7Rr)$

I just  put  $x+y=a , y+z=b , z+x=c$  so  $a,b,c$  are sides  of  triangle .
I also  used the  identities  $a+b+c=2s , a^2+b^2+c^2=2(s^2-r^2-4Rr) , a^3+b^3+c^3=2s(s^2-3r^2-6Rr)$

So here  goes my question :
How  can we  prove  that  geometric ineq , with no  use  of the initial  algebric one ?
Thank you for your time.


Let $ y_{n}\equal{}x_{n}\minus{}x_{n\minus{}1},y_{1}\equal{}1$, then $ y_{n\plus{}1}\equal{}\frac{1}{y_{n}\plus{}2y_{n\minus{}1}\plus{}...\plus{}ny_{1}}$, therefore $ \lim_{n}(ny_{n})\equal{}0$ and
$ \lim_{n}\frac{x_{n}}{log(n)}\equal{}0.$

Second prove.
If exist $ \lim_{x_{n}}{ln(n)}=c>0$, then for big n $ \sum_{k=1}^{n}x_{n}>\frac{cnln(n)}{2}$, therefore
${ x_{n+1}-x_{n}<\frac{2}{cnln(n)}\to x_{n}<const+\frac{2}{c}}ln(ln(n))$ - contradiction.


function
topology
real analysis
real analysis unsolved

What is the name of the Thm. that says that for any analytic $ f$ with $ f^{(n)}\equal{}0$ at $ x$, $ n \in \mathbb{N}$, there is an open ball/basis around $ x$ such that $ f \equal{} 0$?

No. The classic example: $ f(x)\equal{}\begin{cases}e^{\minus{}1/x^2}&x\neq 0\\0&x\equal{}0\end{cases}$

Ok so that was the wrong question to ask.  Is the function: $ f \equal{} 0$ for all $ x \leq 0$, $ f\equal{}e^{\minus{}1/{x^2}}$ for all $ x >0$ analytic?

I tend to phrase the "classic example" as $ g(x)\equal{}\begin{cases}e^{\minus{}1/x}&x>0\\0&x\le 0\end{cases}.$ Same idea in any case.

There is a distinct difference between "$ C^{\infty}$" and "analytic." An analytic function must be $ C^{\infty},$ but a $ C^{\infty}$ function need not be analytic. (Anywhere.)

Theorem: if $ f$ is analytic and $ f^{(k)}(0)\equal{}0$ for $ k\equal{}0,1,2,\dots,$ then there exists $ \delta>0$ such that for $ |x|<\delta,\ f(x)\equal{}0.$

Why so? The key here: what is the definition of analytic? A function is analytic if and only if it is locally given by a power series. That is, for each $ x_0,$ there is some neighborhood of $ x_0$ such that $ f(x)\equal{}\sum_{k\equal{}0}^{\infty}a_kx^k$ on that neighborhood. But then the coefficents $ a_k$ must be the Taylor coefficients $ \frac{f^{(k)}(x_0)}{k!}.$

Hence our theorem is true because the power series for $ f$ valid in a neighborhood of zero, then that series must be identically zero.

Since the examples given by jmerry and by me violate the conclusion of this theorem, then they must also violate the premise. They are not analytic functions, and as such provide concrete examples of the claim that $ C^{\infty}$ functions need not be analytic.

A slight extension of this theorem: if $ f$ and $ g$ are both analytic and there is a point $ x_0$ such that $ f^{(k)}(x_0)\equal{}g^{(k)}(x_0)$ for all $ k\ge 0,$ then there is some neighborhood of $ x_0$ on which $ f(x)\equal{}g(x).$

We even get this theorem:

If $ f$ and $ g$ are analytic on a [b]connected[/b] open set $ U$ and if there is some point $ x_0\in U$ such that $ f^{(k)}(x_0)\equal{}g^{(k)}(x_0)$ for all $ k\ge 0,$ then $ f(x)\equal{}g(x)$ for all $ x\in U.$

Ah, I guess not, then.  Thanks  !  Could you tell me the theorem's name?  Thanks again!


geometry
analytic geometry
summer program
Mathcamp
real analysis
real analysis unsolved

[quote="Kryant"]On the x, y coordinate plane, there is a finite set of line segments. Each line segment lies completely in the square bounded by (0, 0), (0, 1), (1, 0), and (1, 1), and the sum of the lengths of all the line segments is 18. Prove that there exists a straight line in the plane that crosses at least 10 of the segments.[/quote]

This problem is currently on the ongoing Mathcamp 2009 qualifying quiz ([url]http://mathcamp.org/quiz.php[/url]), so can a moderator please lock?


geometry
circumcircle
perpendicular bisector
geometry unsolved

Let quadrilateral $ABCD$ has angle $A$ = angle $C$ . Let $I$ is midpoint of $AC$ . Let $H,K,L$ in segment $BC,CA,AB$ respectively such that $DH,DK,DL$ are perpendicular to $BC,CA,AB$, respectively.
Prove that $I,H,K,L$ lie on a circle

Let M, N be the midpoints of CD, DA. $HM = \frac{CD}{2}= IN,$ because $\angle DHC$ is right and $IN \parallel CD$ is a midline of $\triangle ACD.$ Similarly, $LN = \frac{DA}{2}= IM,$ because $\angle DLA$ is right and $IM \parallel DA$ is another midline of $\triangle ACD.$ $\angle IMH = \angle IMC-\angle HMC = \angle INL-\angle LNA,$ because $\angle IMC = \angle INL = \angle D$ and the right angle triangles $\triangle DHC \sim \triangle DLA$ are similar, having equal angles $\angle C = \angle A.$ Consequently, the triangles $\triangle IHM \cong \triangle LIN$ are congruent by SAS and IH = IL, which means that I is the intersection of the perpendicular bisector k of HL with AC. Quadrilaterals DKHC, DKLA are both cyclic, because $DH \perp BC,\ DK \perp AC,\ DL \perp AB,$ hence $\angle CKH = \angle CDH = \angle ADL = \angle AKL.$ As $DK \perp AC,$ DK bisects the angle $\angle HKL$ internally and AC externally. Let (O) be the circumcircle of $\triangle HKL.$ DK, AC then meet (O) again at midpoints of the arcs HL opposite to K and containing K, respectively, and these midpoints are on the perpendicular bisector k of HL. Thus $I \equiv k \cap AC \cap (O)$, which means $I \in (O).$


What is the necessary and sufficient condition so that the angle bisector in the classical SL theorem can be replaced by two cevians?

[quote="jayme"]Dear Mathlinkers,
see for example :  http://forumgeom.fau.edu/FG2005volume5/FG200525index.html
or a analogue problem on this site proposed by Cosmin Pohoata, I think.
Sincerely
Jean-Louis[/quote]

Thanks but I have already seen that! I need a more general version of that! I am working on some aspects but have not succeeded!


let $x_i,y_i 1 \le i \le n$ be positive numbers  such that :

$\displaystyle            \sum_{i=1}^n x_i \ge \sum_{i=1}^n x_iy_i$

Prove :

$\displaystyle             \sum_{i=1}^n x_i \le \sum _{i=1}^n \frac{x_i}{y_i}$

Solution:

$\sum \frac{x_i}{y_i} = \sum \frac{{x_i}^2}{x_iy_i} \geq \frac{\left ( \sum x_i  \right)^2}{\sum x_iy_i}(*)$ and using ipoteses that $\sum x_iy_i \leq \sum x_i$ we get $(*) \geq \frac {\left ( \sum x_i  \right)^2}{\sum x_i} = \sum x_i$ .

q.e.d.

i used this inequality: If $a_i,b_i>0$ then we have: $\sum \frac{{a_i}^2}{b_i} \geq \frac { \left ( \sum a_i  \right )^2}{\sum b_i}$ which can be proved easily by using Cauchy-Buniakowski-Schwarz.

Another solution is to assume by contradiction that the inequality you have to show is false.Well,by adding this with the one from the hyphotessis and using the well known inequality $a+\frac{1}{a}\geq2$ we get very easily a contradiction. ;)


[quote="ckck"]Find the sum of the fractions:
$\frac{1}{2}$+$\frac{1}{6}$+$\frac{1}{12}$+$\frac{1}{20}$+...+$\frac{1}{9900}$

I'm sure there is a very quick way of doing this but i din't know it. Can someone plz explain.[/quote]
add a few of the terms and you should notice a pattern

btw, I think these are the numbers in a diagonal of the Leibniz triangle...

i don't have a clue how to do this either?

can someone explain?

-jorian

[hide="Hint for ckck"]
The 1st term is $1 \cdot 2 = 2$
The 2nd term is $2 \cdot 3 = 6$
The 3rd term is $3 \cdot 4 = 12$
The 4th term is $4 \cdot 5 = 20$

See the pattern?
[/hide]

[hide="Alternate Solution"]
We want to find the sum:
$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{9900}$

We can cleverly rewrite this as

$(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})...(\frac{1}{99}-\frac{1}{100})$

Most terms will cancel out, and we are left with $1-\frac{1}{100}=\boxed{\frac{99}{100}}$
[/hide]

[hide="Motivation"]
We have \[\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{99\cdot 100},\] or \[\sum_{n=1}^{99}\frac{1}{n(n+1)}.\] The denominator suggests that we split up the fraction using [b]partial fractions[/b], which you may or may not have learned yet. Basically, we split up the fraction as follows: \[\frac{1}{n(n+1)}= \frac{A}{n}+\frac{B}{n+1},\] for some constants $A$ and $B$ to be determined. Multiplying the equation by $n(n+1)$ we have \[1= A(n+1)+B(n).\] Since we are looking for $A,B$ such that this is true for [i]all[/i] values of $n$, we can substitute $-1\to n$ to get \[1 = A(-1+1)+B(-1) =-B\implies B=-1.\] Then we substitute $0\to n$ to get \[1 = A(0+1)+B(0) = A\implies A = 1.\] Therefore, the partial fraction decomposition is \[\frac{1}{n(n+1)}= \frac{1}{n}-\frac{1}{n+1}.\] The reason this is so useful is because the sum [b]telescopes[/b] as follows: \[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots+\left(\frac{1}{99}-\frac{1}{100}\right).\] Notice that every term cancels except for the first and the last terms. Our answer is \[\frac{1}{1}-\frac{1}{100}= \boxed{\frac{99}{100}}.\]
[/hide]

Telescoping is a pretty useful tool when we want to sum long, seemingly tedious, expressions. As chess64 did, basically the objective is to somehow manipulate the given expressions so that terms cancel. Try this one, it's harder:

$\frac{1}{1\cdot2\cdot3}+\frac{1}{3\cdot4\cdot5}+\cdots+\frac{1}{99\cdot100\cdot101}+\frac{1}{100\cdot101\cdot102}$.

[hide="The decomposition"]After some guess and check, I obtained:

$\frac{1}{x(x+1)(x+2)}= \frac{\frac{1}{2}}{x}-\frac{1}{x+1}+\frac{\frac{1}{2}}{x+2}$

(The fractions on top of the fractions make the partial fraction decomposition clear)[/hide]
13375P34K43V312,  your question is a little confusing. Which is the series that you want to compute,
$\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdots+\frac{1}{99\cdot100\cdot101}+\frac{1}{100\cdot101\cdot102}$

$\textrm{OR}$

$\frac{1}{1\cdot2\cdot3}+\frac{1}{3\cdot4\cdot5}+\cdots+\frac{1}{99\cdot100\cdot101}+\frac{1}{101\cdot102\cdot103}$

The original series has no pattern.

[quote="nutz_for2.718281828"][quote="ckck"]I kinda found a pattern. When you add the first two terms you get $\frac{2}{3}$. When you add the first three terms you get $\frac{3}{4}$. When you add the first four terms you get $\frac{4}{5}$. This continues until you reach 9900. I just need to find the number terms in this sequence. Could someone plz explain how to do that.[/quote]

[hide="hint"]When you add the first 2 terms, you get 2/3. If you add the first 3 terms, you get 3/4.[/hide]

Harder problem: Try to prove it for the sum of the first n terms.   :evilgrin:[/quote]
dude wtf you're just reiterating what he just said. your post wasn't a hint at all.


geometry
geometric transformation
rotation
projective geometry

Square ABCD has area 1 square unit. Point P is 5 units from its center. Set S is the set of all points that can be obtained by rotating P 90 degrees counterclockwise about some point on or inside the square. Find the area of set S.

Well, for this problem, I thought: hey, since each point on or in ABCD corresponds to one point in set S, the area of set S must be the same as that of ABCD, 1. However, the answer is supposedly 2. How can this be?

I think this is what explains that paradox I said ages ago about a rolling coin or something.
Think about this:
Take a line length 1, a line length 2, and pair up each point on the first line with double itself on the second line. Theres a 1-1 mapping, different length line.

Yeah, its a bunch of weird stuff to do with infinity. So...how can one do the above problem?

Should this be moved to Intermediate or Advanced? People would actually see the problem there...*sigh* :(

Dunno...maybe intermediate, but definitely not advanced. I found it grouped together with a bunch of easy-intermediate problems.

Select a point P that makes your life figuring out how to do it convenient. For example, let the line connecting P and the center be the perpendicular of AB. Consider the points on AB, BC, CD, and DA for your rotation to get the boundary. Do you see how it works?

Well, if P is perpendicular to AB, but closer to CD than AB, the rotated figure is a square of side length 2, whose image is rotated 45 degrees from that of the original ABCD. But why is this so, and how would I do this on an actual competition properly (without using a ruler/protractor and drawing out the figure)?

Can anyone give a clear explanation of the solution to this problem please? I still don't know how I would do it. Thanks.

hmm. Instead of thinking from the square's perspective and rotating the point, maybe you could think from the point's perspective and rotate the square. But I think you have to be careful about [i]how[/i] you rotate it.

But if I rotate the square, I get the same area of 1, and not the correct answer of 2.

This took me quite a while to work out what was going on.. but then I realised that

[hide]

Consider any point in the square.. when we rotate P about this point we get this little isosceles right angled triangle (connecting P to the centre and the new P to the centre and the Ps to each other ). In other words, the new set is made by rotating the square about P by 45 degrees, and then enlarging this by the square root of 2. (length of hypoteneuse is rt2 x length of side). So the squares new area will be rt2 squared = 2.

[/hide]

Ok. I think I understand now. Thanks. So that enlarging stuff has to do with projective geometry (which Im not too familiar with), right?


This forum is so dead. Nearly all the active people posting here are busy in Universities(esp UW :lol: ) so nobody is posting here. Let's keep this forum alive!!!

Of course it is dead.  Now that summer is over, people have lives again.

[quote="Sisyphus"]i hereby bring this forum to life[/quote]
Thank you. :omighty:  :rotfl:

Finally, I have access to the internet!!!!

I just got my laptop yesterday. :D
I'll try to visit here more often.

Oh no kidding...I actually have a brief second to check what people are up to on this forum.

I'm sure I speak for all of us UW ppl when we say we're busy...there's a lot of stuff to do, lol.

[quote="lightrhee"][quote="J-I-M-B-O"]yeah, how u guys/girls find uw so far?[/quote]
Pretty good. I haven't got much work yet. Assignments are easy.[/quote]

hahaah.... thats good... enjoy! i loooveeddd first term hahaha...
took me 3 terms to learn to time manage haha... moral of the story, don't be like me!!!

[quote="lightrhee"][quote="J-I-M-B-O"]yeah, how u guys/girls find uw so far?[/quote]
Pretty good. I haven't got much work yet. Assignments are easy.[/quote]

 :!:    :o

 :(

I'm in the same camp as david rhee - assignments are easy.

how is it dead? there are still many posts


Find all $ n\in \mathbb{N}$ so that exist $ a,b,c$ such that $ a\plus{}b\plus{}c\equal{}0$ and $ a^n\plus{}b^n\plus{}c^n$ is a prime number.

Easy to check that if $ n\equiv 0 (\mod 2)$ then we chose $ a\equal{}1,b\equal{}1,c\equal{}\minus{}1$ then
$ a^n\plus{}b^n\plus{}c^n\equal{}3$ is a prime number.
Now consider $ n\equiv 1 (\mod 2)$
Suppose exist $ a,b,c$ satisfy condition .
Then we have :
$ a^n\plus{}b^n\plus{}c^n \equiv a\plus{}b\plus{}c \equal{}0 (\mod 2 )$
So $ a^n\plus{}b^n\plus{}c^n\equal{}2$
Exist two of three number have same sign .Suppose $ ab>0$
The equation equivalent :
$ a^n\plus{}b^n\minus{}(a\plus{}b)^n \equal{}2$
If $ a,b>0$ then easy to check  $ a^n\plus{}b^n\minus{}(a\plus{}b)^n<0$
If $ a,b<0$ then easy to check $ a^n\plus{}b^n\minus{}(a\plus{}b)^n>2$ because $ \minus{}na^{n\minus{}1}b<\minus{}2$
So the answer is $ n\equal{}2k$

Let $ p\equal{}a^n\plus{}b^n\plus{}(\minus{}a\minus{}b)^n$ (c=-a-b). If a,b=ad real, then $ a\equal{}\sqrt[n]{\frac{p}{1\plus{}d^n\plus{}(\minus{}1\minus{}d)^n}}, n\not \equal{}1$ solution for any d.
I think a,b,c are integer. If $ n\equal{}1$ $ a^n\plus{}b^n\plus{}c^n\equal{}0$.
Let $ n>1$,  then $ p\equal{}a^n\plus{}b^n\plus{}(\minus{}a\minus{}b)^n$ and $ 2|p$. Therefore $ p\equal{}2$. It give solution, when ab(a+b)=0 and
n even. For example $ a\equal{}1,b\equal{}\minus{}1,c\equal{}0$ n any even.

[quote="TTsphn"]Easy to check that if $ n\equiv 0 (\mod 2)$ then we chose $ a \equal{} 1,b \equal{} 1,c \equal{} \minus{} 1$ then
$ a^n \plus{} b^n \plus{} c^n \equal{} 3$ is a prime number.
Now consider $ n\equiv 1 (\mod 2)$
Suppose exist $ a,b,c$ satisfy condition .
Then we have :
$ a^n \plus{} b^n \plus{} c^n \equiv a \plus{} b \plus{} c \equal{} 0 (\mod 2 )$
So $ a^n \plus{} b^n \plus{} c^n \equal{} 2$
Exist two of three number have same sign .Suppose $ ab > 0$
The equation equivalent :
$ a^n \plus{} b^n \minus{} (a \plus{} b)^n \equal{} 2$
If $ a,b > 0$ then easy to check  $ a^n \plus{} b^n \minus{} (a \plus{} b)^n < 0$
If $ a,b < 0$ then easy to check $ a^n \plus{} b^n \minus{} (a \plus{} b)^n > 2$ because $ \minus{} na^{n \minus{} 1}b < \minus{} 2$
So the answer is $ n \equal{} 2k$[/quote]
 But $ a \equal{} 1,b \equal{} 1,c \equal{} \minus{} 1$ then $ a \plus{} b \plus{} c \equal{} 1$ not such that the condition $ a \plus{} b \plus{} c \equal{} 0$. We should choose $ a\equal{}0,b\equal{}1,c\equal{}\minus{}1$  :P

Do you means?
\[ n\equiv 1 (\mod 2)\to a^n \equiv a (\mod 2)
\]
My explaination:
\[ n\equiv 1 (\mod 2) \to n \equal{} 2k \plus{} 1,k\in \mathbb{N}
\]
Then if $ a \equal{} 2p$ we are done.
 If $ a \equal{} 2p \plus{} 1\to (2p \plus{} 1)^{2k} \equiv 1 (\mod 2)$ is always true. Because we have:
\[ a^n \equiv a (\mod 2) \iff a^{2k} \equiv 1 (\mod a)
\]
Is your explaination like me?
Can you explain clearly more?


all terms are positive and the last term reduces to $ n^k$ which is is greater than $ n\sqrt{n}\equal{}n^{1.5}$
since k>1.5

$ \frac{1^k}{n^{k\minus{}1}} \plus{} \frac{2^k}{(n\minus{}1)^{k\minus{}1}} \plus{} \frac{3^k}{(n\minus{}2)^{k\minus{}1}} \plus{}\cdots \plus{} \frac{n^k}{1^{k\minus{}1}} \ge n \sqrt[n]{\frac{(n!)^k}{(n!)^{k\minus{}1}}} \equal{} n\sqrt[n]{n!}$ by AM-GM.

The rest is showing that $ n\sqrt[n]{n!} \ge n\sqrt{n}$, which is true but I can't prove it yet.

oing from the last equation, we must prove $ \sqrt[n]{n!} \ge \sqrt(n) \implies n!^2 \ge n^n$

slightly rigorous solution:
$ n!^2\equal{} :$
n*n-1*n-2... 3*2 * 1
*
1*2 *  3...n-2*n-1*n
looking at vertical products, there are n of them and each is at least n
so the last equation of risingflame was correct


If $G$ is a graph, let $m$ be the minumum number of vertices such that removing them disconnects $G$; let $k$ be the minimum number of edges such that removing them disconnects $G$, and let $d$ be the minimum degree in $G$. Prove that:
(A) $m \leq k \leq d$
(B) If every vertex of $G$ has degree 3 then $m=k$

The first one is indeed obvious, and I believe the second one has been posted (by lightree, maybe? :?). I remember posting something on that thread. The problem was posed in terms of networks and computers.


quadratics
number theory unsolved
number theory

Prove that there are no integer solutions for x and y that satisfies the equation x^2+y^2=1986

$ 3\equal{}\minus{}1(mod4)$ occurs to an odd exponent in 1986

rhs is $ \equal{}0(mod3)$ and $ 0(mod4)$ hence $ x$ and $ y$ should be divisible by both 3 and 4 and therefore by 12.but rhs is not

Um, $ \text{RHS}\equiv 2\bmod4$, not 0.

Isn't there a solution by infinite descent? I should know... I just read this chapter in AoPS 2...

[quote="aadil"]rhs is $ \equal{} 0(mod3)$ and $ 0(mod4)$ hence $ x$ and $ y$ should be divisible by both 3 and 4 and therefore by 12.but rhs is not[/quote]
The argument concerning $ 4$ is completely wrong on two parts: rhs not $ 0 \mod 4$ and $ x^2 \plus{} y^2 \equiv 0 \mod 4$ doesn't imply that $ x,y \equiv 0 \mod 4$ (as $ 4$ is not prime/squarefree).
The part in $ 3$ is correct and renders the the part concerning $ 4$ redundant.

Abundant?
Anyway, as was said above, 1986 is equivalent to 6 mod 9, which cannot be the sum of 2 squares. (Quadratic residues mod 9 are 0,1,4,7.)


Determine all pairs of natural numbers $ (x,n)$ that satisfy the equation $ x^3 \plus{} 2x \plus{} 1 \equal{} 2^n$

$ (x\plus{}1)(x^3\minus{}x^2\plus{}x\plus{}1)\equal{}2^n$
But $ \gcd(x\plus{}1,x^3\minus{}x^2\plus{}x\plus{}1)\equal{}gcd(x\plus{}1,2)$
So $ x\plus{}1\equal{}2$ or $ x^3\minus{}x^2\plus{}x\plus{}1\equal{}2$
Solve this equation we have solution $ x\equal{}1$
Result : $ (x,y)\equal{}(1,2)$

Sorry,It's must be $ x^3\plus{}2x\plus{}1\equal{}2^n$  :blush:

[quote="apollo"]Sorry,It's must be $ x^3 \plus{} 2x \plus{} 1 \equal{} 2^n$  :blush:[/quote]

Considering mod 8 then mod 16 reveals that the only solutions could be $ <4$, 3,7

Therefore the only solutions can be $ (x,n)\equal{}(1,2)$


Prove that if $a_1,a_2,\dots,a_n>0$ have product $1$, then \[ \sum_{i=1}^n \sqrt{\frac{1+a_i^2}{2}} \le a_1 + a_2 + ... + a_n. \]

[quote="harazi"]The following problem has a one-line solution, but I spent one hour searching for it:
Prove that if a_1,...,a_n>0 have product 1, then:
sum sqrt[(1+(a_i)^2)/2]<=sum a_i.[/quote]
I am very curious to see your online solution, Harazi?

Thank you very much.

I am really fond of this problem.
We have $\sqrt x + \sqrt{\tfrac{x^2+1}{2}} \le x+1$. Thus, \[ n + \sum_{i=1}^n x_i \ge \sum_{i=1}^n \sqrt{x_i} + \sum_{i=1}^n \sqrt{\frac{x_i^2+1}{2}} \ge n+ \sum_{i=1}^n \sqrt{\frac{x_i^2+1}{2}} \] from AM-GM.
Unfortunately, it has two lines.

Wonderful solution for a very nice problem.

Congratulation, Harazi.


calculus
derivative
real analysis
real analysis unsolved

Given $ y\equal{}\frac{(\sqrt{x}\plus{}1)(x^2\minus{}\sqrt{x})}{x\sqrt{x}\plus{}x\plus{}\sqrt{x}}\plus{}\frac{1}{15}(3cos^2x\minus{}5)cos^3x$, then find $ \frac{dy}{dx}$.

hello
[hide="hint"]
for $ \frac{(\sqrt{x}\plus{}1)(x^2\minus{}\sqrt{x})}{x\sqrt{x}\plus{}x\plus{}\sqrt{x}}$ use log
and then differentiate it.
$ [\frac{1}{15}(3cos^2x\minus{}5)cos^3x]'$=$ cos^2xsinx\minus{}cos^4xsinx$
[/hide]
thank u

The options are:
(A) $ 1\minus{}sin^3xcos^3x$
(B) $ 1\minus{}sin^2xcos^3x$
(C) $ 1\plus{}sin^2xcos^2x$
(D) $ 1\plus{}sin^3xcos^2x$
(E) None of these

[quote="thomaskurian89"]The options are:
(A) $ 1 \minus{} sin^3xcos^3x$
(B) $ 1 \minus{} sin^2xcos^3x$
(C) $ 1 \plus{} sin^2xcos^2x$
(D) $ 1 \plus{} sin^3xcos^2x$
(E) None of these[/quote]
ans is D.after simplification of course :D  :D

[quote="thomaskurian89"]Given $ y \equal{} \frac {(\sqrt {x} \plus{} 1)(x^2 \minus{} \sqrt {x})}{x\sqrt {x} \plus{} x \plus{} \sqrt {x}} \plus{} \frac {1}{15}(3cos^2x \minus{} 5)cos^3x$, then find $ \frac {dy}{dx}$.[/quote]
infact  $ \frac {(\sqrt {x} \plus{} 1)(x^2 \minus{} \sqrt {x})}{x\sqrt {x} \plus{} x \plus{} \sqrt {x}} \equal{} x$ :wallbash:  :wallbash:  :rotfl:

No, it doesn't.  (Although $ x \minus{} 1$ isn't [i]that[/i] different.)

hello, we have $ \frac{(1\plus{}\sqrt{x})(\sqrt{x}\plus{}x^2)}{\sqrt{x}\plus{}x\plus{}x^{\frac{3}{2}}}\equal{}x\minus{}1$ and $ \frac{1}{15}(3\cos(x)^3\minus{}5)\cos(x)^3\equal{}\frac{1}{240}\left(\minus{}30\cos(x)\minus{}5\cos(3x)\plus{}3\cos(5x)\right)$ so we have the first derivative with respect to $ x$ as $ 1\plus{}\frac{1}{240}(30\sin(x)\plus{}15\sin(3x)\minus{}15\sin(5x))$
Sonnhard.


2 hydroxy 1,2,3 propanetricarboxylic acid + h2so4->

wrong. maybe that would be one of minor products. but thats not the major product

Perhaps if you could be more specific about the reaction conditions then we could give better answers. Is the H2SO4 diluted or concentrated? Is there any heating? You have to say all those things. Of course, if this is another problem from your famous book then I am not surprised.

sorry forgot to mention that
temp:45-50
and its conc.h2so4
p.s which famous book were you referring to?

Given those conditions, I think two things are to be expected: dehydration and decarboxylation.

yeah but in this case it does take place in a way you would expect it to
well the product is 3 oxo pentandioic acid
this is mechanism( i think):
dehydration of OH does not take place as it would lead to a carbocation adjacent to a carbonyl group.
so dehydration of hydroxyl of cooh takes place also acylium ion is quite stable.
then co is eliminated along with hydrogen alpha to oh group
whichh gives an enol tautomerises to keto


Let $ x_1, x_2, ..., x_n$ are all reals that greater than 1.

Prove that

$ \sum^{n}_{i\equal{}1} \frac{\sqrt{x_i^2(x_{i\plus{}1}\minus{}1)^2\plus{}x_{i\plus{}1}^2}}{x_i x_{i\plus{}1}} \geq \frac{\sqrt{2}}{2}n$,

where $ x_{n\plus{}1}\equal{}x_{1}$.


function
integration
real analysis
real analysis unsolved

Let $f$ be a continuous function with real values, defined on $[0,1]$ and let $ 0<g<1$. Assume that the average of $f$ on any interval of $(0,1)$ of length $g$ is smaller than 1. Can the average of $f$ over $[0,1]$ be greater than 1?

I think the average of $f$ can exceed $1$ iff $g$ is not of the form $\frac 1n$ (one implication is obvious: if $g=\frac 1n$, then it's impossible for $f$ to have an average greater than $1$ on $[0,1]$). Is this correct, harazi?

I'd post what I did, but I worked on a slightly different problem: given a continuous function $F:[0,1],\ F(0)=0$ s.t. $F(x+g)<F(x)+g,\ \forall x\in[0,1-g]$, can we have $F(1)>1$?

We can work with $F(x)=\int^x_0 f\mbox{d}t$, and we get a somewhat weaker problem (the one in the paragraph above), but when we construct counterexamples for $g\ne\frac 1n$, we can take our $F$ to be smooth enough (like continuously differentiable, for instance), so that it can be interpreted as an integral. That's why I think it suffices to solve the problem I stated.

Sorry if it's all nonsense :).

Yes, the similarity with the universal chord problem is pretty clear.


Find all positive integers $ n$ such that both $ n \plus{} 2008$ divides $ n^2 \plus{} 2008$ and $ n \plus{} 2009$ divides $ n^2 \plus{} 2009$.

$ n \plus{} 2008|n^2 \plus{} 2008 \minus{} (n \plus{} 2008) \equal{} n^2 \minus{} n$ and $ n \plus{} 2009|n^2 \plus{} 2009 \minus{} (n \plus{} 2009) \equal{} n^2 \minus{} n$ and since $ gcd(n \plus{} 2008,n \plus{} 2009) \equal{} 1$ we have :
$ (n \plus{} 2008)(n \plus{} 2009)|n^2 \minus{} n$ , but for $ n > 0 , n^2 \minus{} n \le (n \plus{} 2008)(n \plus{} 2009)$, so no solutions exist in this case.

[quote="tchebytchev"][....] but for $ n > 0 , n^2 \minus{} n \le (n \plus{} 2008)(n \plus{} 2009)$, so no solutions exist in this case.[/quote]

The unique solution is $ n\equal{}1$.

[quote="test20"][quote="tchebytchev"][....] but for $ n > 0 , n^2 \minus{} n \le (n \plus{} 2008)(n \plus{} 2009)$, so no solutions exist in this case.[/quote]

The unique solution is $ n \equal{} 1$.[/quote]
yes $ 1$ is the only solution to the inequality  $ n^2 \minus{} n \le (n \plus{} 2008)(n \plus{} 2009)$


f : N->R , $ f(1)\equal{}3$and $ f(m\plus{}n)\plus{}f(m\minus{}n)\minus{}m\plus{}n\minus{}1\equal{} \frac{f(2m)\plus{}f(2n)}{2}$ ,$ m,n \geq 0$ __Find f !!

[quote="KapioPulsar"]f : N->R , $ f(1) \equal{} 3$and $ f(m \plus{} n) \plus{} f(m \minus{} n) \minus{} m \plus{} n \minus{} 1 \equal{} \frac {f(2m) \plus{} f(2n)}{2}$ ,$ m,n \geq 0$ __Find f !![/quote]

$ f(n)\equal{}n^2\plus{}n\plus{}1\qquad n\in N$.

Later I will post the complete solution... :roll:

@KapioPulsar
$ n \in N \implies f(n) \in N$ while you said that $ f: N \to \mathbb{R}$. You are wrong or distract or...maybe I am wrong  :oops:

[quote="KapioPulsar"][quote="prester"][quote="KapioPulsar"]f : N->R , $ f(1) \equal{} 3$and $ f(m \plus{} n) \plus{} f(m \minus{} n) \minus{} m \plus{} n \minus{} 1 \equal{} \frac {f(2m) \plus{} f(2n)}{2}$ ,$ m,n \geq 0$ __Find f !![/quote]

$ f(n) \equal{} n^2 \plus{} n \plus{} 1\qquad n\in N$.

Later I will post the complete solution... :roll:

@KapioPulsar
$ n \in N \implies f(n) \in N$ while you said that $ f: N \to \mathbb{R}$. You are wrong or distract or...maybe I am wrong  :oops:[/quote] no it is $ f: N \to \mathbb{R}$ !![/quote]

My solution
[hide]
Let $ P(m,n)$ the assertion $ f(m \plus{} n) \plus{} f(m \minus{} n) \minus{} m \plus{} n \minus{} 1 \equal{} \frac {f(2m) \plus{} f(2n)}{2}$

In the statement $ 0 \in N$

$ P(0,0) \iff 2f(0) \equal{} f(0) \plus{} 1 \iff f(0) \equal{} 1$

$ P(m,0) \iff 2f(m) \minus{} m \minus{} 1 \equal{} \frac {f(2m) \plus{} 1 }{2} \iff$
$ \iff f(2m) \equal{} 4f(m) \minus{} 2m \minus{} 3 \qquad m \ge 0, m \in N \qquad (*)$

Substituting $ (*)$ in the original assertion we have

$ f(m\plus{}n) \plus{} f(m\minus{}n) \equal{} 2f(m)\plus{}2f(n)\minus{}2n\minus{}2\qquad m \ge 0, m \in N$

Putting $ n\equal{}1$ we have

$ f(m \plus{} 1) \equal{} 2f(m) \plus{} f(m \minus{} 1) \plus{} 2 \qquad m \ge 0, m \in N$

By induction we get

$ f(m \plus{} n) \equal{} (n \plus{} 1)f(m) \minus{} nf(m \minus{} 1) \plus{} n(n \plus{} 1) \qquad m \ge 1, n \ge 1$

Putting $ m \equal{} 1$ we have $ f(n \plus{} 1) \equal{} (n \plus{} 1)f(1) \minus{} nf(0) \plus{} n(n \plus{} 1) \equal{} 3(n \plus{} 1) \minus{} n \plus{} n(n \plus{} 1)$

$ m \equal{} n \minus{} 1 \implies f(n) \equal{} n^2 \plus{} n \plus{} 1 \qquad n \ge 0, n \in N$

It's simple to check that this $ f(n)$ satisfies the original statement

So the solution is $ f(n) \equal{} n^2 \plus{} n \plus{} 1\qquad n \ge 0, n \in N$
[/hide]


geometry
analytic geometry
vector
pigeonhole principle
modular arithmetic
combinatorics unsolved

This was a caseworky problem. However, I think the following method makes it *relatively* nice. First of all, notice that all numbers with prime factors at most 3 can be expressed as $2^{a}3^{b}$. Notice that the product of three such numbers is a cube if and only if the exponents on the 2 and the 3 are both divisible by 3. Therefore the problem is equivalent to finding three elements out of a set X of nine (not necessarily distinct) in $\mathbb{Z}_{3}^{2}$ whose sum is $(0, 0)$, where addition is defined by adding the two corresponding coordinates in $\mathbb{Z}_{3}^{2}$ (i.e. vector addition).

Note that if there were four or fewer distinct elements in X, then some three elements are identical, and these three will obviously sum to $(0, 0)$. Thus it would suffice to prove that in any 5-element subset Y (this time with distinct elements) of $\mathbb{Z}_{3}^{2}$, there are some three which sum to $(0, 0)$.

Now let $f(a, b) =-a-b$, and note that $f(a, b) \not\in \{a,b\}$ and $f(a, b)$ is the unique solution to $x \in \mathbb{Z}_{3}^{2}$ such that $a+b+x = (0, 0)$. There are ${5 \choose 2}= 10$ pairs of elements in Y, and for any $a, b \in Y$, we would be done if $f(a, b) \in Y$. Hence we will assume the contrary (f(a, b) is never in Y), in which case the 10 pairs of elements map to 4 elements not in Y.

By pigeonhole, this means that there are some three distinct pairs (but we don't know if their components are distinct) $(x_{1}, x_{2}), (x_{3}, x_{4}), (x_{5}, x_{6}) \in Y^{2}$ for which $f(x_{1}, x_{2}) = f(x_{3}, x_{4}) = f(x_{5}, x_{6}) = t$ for some $t \not\in Y$.

Of course, $x_{1}\ne x_{2}, x_{3}\ne x_{4}, x_{5}\ne x_{6}$, since we only considered pairs. Then suppose some two of the $x_{i}$ are equal; WLOG $x_{1}= x_{3}$. But since $-x_{1}-x_{2}= t =-x_{3}-x_{4}$, we have that $x_{2}= x_{4}$. This means that the pairs $(x_{1}, x_{2})$ and $(x_{3}, x_{4})$ are identical, which is contradiction. The alternative, however, is that all the $x_{i}$ are distinct, implying that Y contains six elements, which is also contradiction.

Therefore, there must be three elements of Y whose sum is zero, as desired.

I agree that this problem has many solutions, but I was very happy when I came up with this one, so I have to post it here :)

The beginning is the same. So we have these possibilities for exponents $(0,0),(0,1),(1,0),(1,1),(1,2),(20),(2,1),(2,2)$. Now put them in the fields of $3\times 3$ square like this
\[(0,1)...(2,1)...(1,1) \]

\[(1,2)...(0,2)...(2,2) \]

\[(2,0)...(1,0)...(0,0) \]
Now, notice that if we take any three numbers from the sam row, column or diagonal, then their sum is divisible with three. I don't think big diagonals only, if you look more cearfull, you can se that sum of $a[1,2],a[2,1],a[3,3]$ is also divisible with 3, and for others "broken" diagonal too.

We want to choose 9 numbers from there such that: we don't choose one number three times, we don't take three numbers form the same row or column, and the additional condition, we don't take three numbers from any "diagonal". Now, just to fulfill first two conditions, we see that we must take $5$ or $6$ different fields (if we take less then $5$ fields then we must take one field three times, on the other hand, if we take more than $6$ fields, we will surtanly have one whole row or column). Now it's not hard to deduce that third condition can never be fulfilled if the first two are.
Nice isn't it :)

Bye

Nice, but how do you prove it neatly? I thought about it that way during the contest, but it seemed very hard/annoying to justify on paper.

I have totaly forgotten on this problem. It's really isn't so hard to write all cases. Just look at corner fields that we picked. If we pick only one corner field, we must pick 4 no-corner fields, and this case isn't hard to deduce. Now, if we pick 2 corner fields, then there are two sub-cases, that really isn't hard and long to write. Three and four corner fields are also easy to write all down. Maybe it isn't so neatly, but I really liked the idea of that $3\times 3$ table :)

Bye

[hide]
Note that all the numbers $n_i$ can be expressed as $2^{a_i}3^{b_i}$. Let $x_i,y_i$ be the remainders when $a_i,b_i$ are divided by 3. 3 numbers $n_i,n_j,n_k$ multiply to a perfect cube iff $x_i+x_j+x_k \equiv y_i+y_j+y_k \equiv a_i+a_j+a_k \equiv b_i+b_j+b_k \equiv 0 \pmod 3$. In particular, we can replace all $2^{a_i}3^{b_i}$ with $2^{x_i}3^{y_i}$. A number $n_i$ is [i]equivalent[/i] to $2^{x_i}3^{y_i}$.

Order these 9 numbers in a 3x3 grid as shown:

$2^03^0, 2^13^0, 2^23^2$
$2^23^0, 2^03^1, 2^13^2$
$2^13^0, 2^23^1, 2^03^2$.

Note that the product of each row, column, and set of 3 squares such that none are in the same row or column (henceforth 'diagonal') is a perfect cube.

Assume the contrary.

Define a number to be [i]shaded[/i] if at least one $n_i$ is equivalent to it. Note that there can be at most 2 $n_i$ equivalent to the same number. Therefore, by the pigeonhole principle, there are at least 5 shaded squares. Let a shaded square have a value of 1 and a non-shaded square to have a value of 0. The value of a row/column/diagonal is the sum of the values in that row/column/diagonal. Then the sum of the values over all the rows, columns, and diagonals is at least 20, as each square is counted 4 times.

Assume WLOG that $2^03^0$ is shaded. This leaves 8 rows, columns, and diagonals not containing $2^03^0$, with at least a total sum of 16. As the value of any row, column, or diagonal cannot be 3, the value of each row/column/diagonal not containing $2^03^0$ must be $2$.

Consider the center square. If it is not shaded, then all the numbers except $2^03^1, 2^03^2$ must be shaded, absurd. Therefore the center square is shaded, thus $2^03^2$ is not shaded, thus $2^13^0, 2^03^1, 2^23^2$ are shaded, contradiction.

Therefore the statement must be true.[/hide]

I thought the whole 3x3 grid thing was really nice when I first found it, but it took quite a while to finish after that.

This result (that on a 3-by-3 grid, any five points determine a generalized line) is well-known to players of the card game SET.  If we'd like to phrase it in terms suitable for possible generalization, we're looking at the problem of choosing the maximal number of points in the vector space $\mathbf{F}_3^2$ over the field $\mathbf{F}_3$ with 3 elements that does not contain an entire (affine) line.  I believe that when we replace "2" with "5" in the statement already the corresponding answer is unknown.

As mentioned previously, the problem quickly solves if it can be shown that a set, $X$, of five distinct elements in $\mathbb{Z}_{3}^{2}$ must have three distinct elements that sum to zero.  Let $S_0$ be the subset of elements of the form $(0,a)$, $S_1,S_2$ defined similarly.  Then if any of $S_1,S_2,S_3$ have three elements, we are clearly finished.  If not, one $S_i$ has one element, the other two each.  So  WLOG assume the elements of $X$ are $(0,a),(1,b),(1,c),(2,d),(2,e)$.  The equations $a+b+x \not\equiv 0, a+c+x \not\equiv 0$ have only one solution in $\mathbb{Z}_{3}$, so we are done.

[quote="mathbuzz"]by using a well known result from additive combinatorics[/quote]
In other words, "the result of the problem follows immediately if we take the statement of the problem as known."  :roll:

I have a solution that borders on pigeonhole and blind casework bashing:

Let $(p_i, q_i)$ denote the exponents of 2 and 3 in the prime factorization of the $i^{th}$ number respectively. Let $(a_i, b_i)$ be the residues of $p_i$ and $q_i$ in modulo three respectively. Also let $r_0, r_1, r_2$ be partitions of A = {$a_i$} and B = {$b_i$} that divide the elements of A and B into the residues corresponding to their indices in modulo three.

It suffices to prove the existence of a triple (i, j, k) for which $a_i + a_j + a_k \equiv b_i + b_j + b_k \equiv 0$ (mod 3).

We split the problem into two cases:

[b]Case 1: $\exists r_i$ that contains at least 5 elements of A.[/b]
If there exists three $b_i$ corresponding to A's elements in $r_i$ in the same residue group, then the sum of those three $b_i$ is congruent to 0 mod 3. Multiplying their corresponding numbers yields a power of three.

Else, by pigeonhole, each residue must contain at least one $b_i$. Summing any three $b_i$, taking one from each residue group, gives us a multiple of three. Hence, we can multiply their corresponding numbers to get a power of three.

[b]Case 2: There do not exist 5 elements of A that belong to a single residue group.[/b]
By pigeonhole, there are at least two residue groups that have at least three elements, and each residue group must have at least one element from A. Suppose on the contrary that there do not exist three $b_i$ in such a configuration that sum to a multiple of three.

WLOG, let $a_1, a_2, a_3$ belong to residue group $r_0$, $a_4, a_5, a_6$ belong to residue group $r_1$, and $a_7$ belong to residue group $r_2$.

We let D = {$b_1, b_2, b_3$}, E = {$b_4, b_5, b_6$} and F = {$b_7$}.
Note that $b_1, b_4, b_7$ cannot all have different residues or the same residues. Hence (wlog), it follows that $b_7 \equiv X$ (mod 3), $b_1 \equiv X$ (mod 3), and $b_4 \equiv Y$ (mod 3). Also note that group Ds and E cannot all have the same residues.

It follows that group E must have the residues Y and Z among its elements. Similarly, group D must have the residues X and Y among its elements. Since $b_7 \equiv X$, we can choose three elements among D, E, F that have three distinct residues. Summing these elements gives us a multiple of three, and multiplying their corresponding numbers gives a power of three; contradiction.

Hence, we can multiply three numbers to obtain a power of three.

[quote=hmida99][quote=N.T.TUAN]Let $S$ be a set of $9$ distinct integers all of whose prime factors are at most $3.$ Prove that $S$ contains $3$ distinct integers such that their product is a perfect cube.[/quote]

that reminds me with some hwo to
[url]https://artofproblemsolving.com/community/c6h60788p366595[/url][/quote]

Wait I was going to refer to that problem after I posted my solution... What a coincidence.

It remind me somewhere?? This type is fimiliar

[quote=mathbuzz]let the 9 numbers be $a_1,a_2,....,a_9$
Let $a_n=2^{p_n}.3^{q_n}$ [$p_n ,q_n \ge 0$]
So, let $X={(p_n,q_n):n=1,2,...,9}$ . as $n(X)=4.3-3$ , hence by using a well known result from additive combinatorics, we deduce that , there exists $i,j,k$ such that $(p_i+p_j+p_k,q_i+q_j+q_k)=(0,0) (mod3)$
Hence $a_{i}a_{j}a_{k}$ is a perfect cube  :lol:[/quote]
My solution was the same, I think this one is really elegant.

Can someone pls check my solution?

Each number here can be represented with prime factors of 2 & 3.So, we have to find powers of 2 as well as 3, which add up to a multiple of 3 separately.Let the powers of 2 of those 3 requested  numbers be a,b & c.So,if we take a+b+c modulo 3, one of the following four patterns must hold-
0+0+0
1+1+1
2+2+2
1+2+0
pigeonhole principle says that  one of 0,1 or 2 must occur at least thrice among the powers of 2 of those 9 numbers.So,some casework suggests that there are minimum 14 ways to choose 3 numbers out of the nine numbers such that the sum of the powers of 2 of them is 0 mod3.now we need to ensure that at least one of those ways also satisfies the same for powers of 3.In case of the minimum number of ways, there are at least 2 ways satisfying this,using pigeonhole.The others also satisfy in the same way.

Is the solution OK?I am a beginner.so,pls consider if there's any silly mistake.


trigonometry
complex numbers
geometry solved
geometry

In the three sides of a triangle ABC we construct three equilateral triangles, ABX, BCY and CAZ. Now we construct a new triangle joinning the three centers os the triangles. Prove that this new triangle is equil...

I found a very ugly proof using complex numbers and lots of trigonometry, I was hoping someone could show me another one...

it's the famous napolean triangle, i bet you can find a proof on the internet easily by searching napolean triangle?

See e. g.

http://www.cut-the-knot.org/proofs/napoleon_intro.shtml

http://agutie.homestead.com/files/Geoproblem_B.htm#Napoleon

  Darij


function
logarithms
linear algebra
matrix
calculus
derivative
algebra

Let $0 < a_1 < a_2 < \ldots < a_n$. If $b_1,\ldots,b_n$ are real numbers s.t. for all $x \in \mathbb{R}$ we have $b_1 (a_1)^x + \ldots+ b_n (a_n)^x=0$ then $b_1=\ldots=b_n=0$.

This is in my analysis homework at limits of functions chapter. How do we use them to solve this problem?

On the other hand, this problem is trivial by linear algebra (vandermonde).

[quote="Ravi B"]How about dividing both sides by ${a_n}^x$, and then taking the limit as $x$ approaches infinity?[/quote]

well, I thought of dividing by $\left( a_1 \right)^x$. but i couldn't conclude it...

The theorem we want to prove: the functions $a_1^x,a_2^x,\dots$ are linearly independent. Two ideas:

Suppose $b_n\neq0$. As $x\to\infty$, $\frac{b_ka_k^x}{b_na_n^x}\to0$ when $k<n$, so $\frac{\sum_k b_ka_k^x}{b_na_n^x}\to1$ and the sum cannot be identically zero. Reduce $n$ by one and repeat.

Consider the differential equation $f^{(n)}(x)+c_{n-1}f^{(n-1)}(x)+\dots+c_1f'(x)+c_0f(x)=0,$ where $t^n+c_{n-1}t^{n-1}+\dots+c_1t+c_0=(t-\ln(a_1))(t-\ln(a_2))\cdots(t-\ln(a_n)).$ Set $r_i=\ln a_i.$ The functions $a_1^x,a_2^x,\dots,a_n^x$ are solutions of this differential equation. Their Wronskian is $\det\begin{bmatrix}e^{r_1x}&e^{r_2x}&\cdots&e^{r_nx}\\ r_1e^{r_1x}&r_2e^{r_2x}&\cdots&r_ne^{r_nx}\\ \vdots&\vdots&\ddots&\vdots\\ r_1^{n-1}e^{r_1x}&r_2^{n-1}e^{r_2x}&\cdots&r_n^{n-1}e^{r_nx}\end{bmatrix}= e^{(r_1+r_2+\dots+r_n)x}\cdot\det \begin{bmatrix}1&1&\cdots&1\\ r_1&r_2&\cdots&r_n\\ \vdots&\vdots&\ddots&\vdots\\ r_1^{n-1}&r_2^{n-1}&\cdots&r_n^{n-1}\end{bmatrix}$ $=e^{(r_1+r_2+\dots+r_n)x}\cdot\prod_{i<j}(r_j-r_i)\neq0$ and we are done.

thanks, jmerry. the first solution is nice.

the second is also nice, but i don't know some terms (e.g. wronskian), hence I can't understand it. anyway, we can apply vandermonde after taking $x=0,1,\ldots,n-1$.

That's a standard term in the study of ordinary differential equations (a class I never had). The $i,j$ term is the $(i-1)$st derivative of the $j$th solution.

The Wronskian is useful because it satisfies a first-order homogeneous linear equation (which we can always solve) when the $n$ solutions it is built from satisfy an $n$th order homogeneous linear equation, and it is nonzero exactly when the solutions are linearly independent. I'm using the second property here, in the very easy direction.
There is also a version for use with linear systems.

Of course that's a Vandermonde determinant- I was pretty sure you knew what to do with one.


Billy Bob, Killy Kob, Hilly Hob, and Jilly Job were returing from Dilly Dob's house. It was midnight and pure pitch black. They reach a bridge. They have only 1 flashlight. The bridge was made in the USA so it was very weak. Only up to 2 people can cross it at a time. They must use the flashlight for EVERY crossing. At their fastest, Billy Bob can cross it in 1 minute, Killy Kob in 2, Hilly Hob in 5, and Jilly Job in 10. A big bad wolf is waiting in the dark and if they don't cross it in 17 minutes the big bad wolf will eat them. Can you save Billy Bob, Killy Kob, Hilly Hob, and Jilly Job!?!?!?!?

Please provide a detailed solution and save Billy Bob, Killy Kob, Hilly Hob, and Jilly Job.
[hide="OR ELSE"]The Big Bad Wolf will come after you next!!![/hide]

First you send billy bob and killy bob. It will take 2 min to cross. Then the flashlight on the right side of the bridge. Now billy bod return to the left side taking flash light in 1 min. So the total time is 2+1= 3min completed. Noe the Hilly Bob and Jilly Bob Goes. They take 10 min to cross the bridge. So the total is now 10 + 3 = 13min
Now the flashlight is again on the right side. Now killy bob take it and return to the left side of the bridge where his brother billy bob is waiting. His journey take 2 min. So the Total time taken is now
13 + 2 = 15 min. Now Both will Cross the bridge safely in 2 min. The Total time elpased is 15+2 = 17 min. And now they are Safe.


I didn't do state, but I took it for practice just nwo and I got a 40. Does anyone know the NJ cutoff?

[url=http://www.artofproblemsolving.com/Forum/index.php?f=241]this would probably be a better place to post it.[/url]


geometry
Triangle
similar triangles
area of a triangle
IMO
IMO 1967

$A_0B_0C_0$ and $A_1B_1C_1$ are acute-angled triangles. Describe, and prove, how to construct the triangle $ABC$ with the largest possible area which is circumscribed about $A_0B_0C_0$ (so $BC$ contains $B_0, CA$ contains $B_0$, and $AB$ contains $C_0$) and similar to $A_1B_1C_1.$

We construct a point $P$ inside $A_0B_0C_0$ s.t. $\angle X_0PY_0=\pi-\angle X_1Z_1Y_1$, where $X,Y,Z$ are a permutation of $A,B,C$. Now construct the three circles $\mathcal C_A=(B_0PC_0),\mathcal C_B=(C_0PA_0),\mathcal C_C=(A_0PB_0)$. We obtain any of the triangles $ABC$ circumscribed to $A_0B_0C_0$ and similar to $A_1B_1C_1$ by selecting $A$ on $\mathcal C_A$, then taking $B= AB_0\cap \mathcal C_C$, and then $B=CA_0\cap\mathcal C_B$ (a quick angle chase shows that $B,C_0,A$ are also colinear).

We now want to maximize $BC$. Clearly, $PBC$ always has the same shape (i.e. all triangles $PBC$ are similar), so we actually want to maximize $PB$. This happens when $PB$ is the diameter of $\mathcal C_B$. Then $PA_0\perp BC$, so $PC$ will also be the diameter of $\mathcal C_C$. In the same way we show that $PA$ is the diameter of $\mathcal C_A$, so everything is maximized, as we wanted.

The acuteness of the triangles ensured that $P$ is inside $A_0B_0C_0$. I'm sure I used it in there somewhere :D.

also of this problem, I remember, I already posted a solution some time ago.

missing too?  :(

PS
i have to check the grobber's solution: at first insigth, it seems equal to mine. If this is the case, I don't resend it.

ciao

buon anno!!!

About lost solutions in the Geometry subforum, see http://www.mathlinks.ro/Forum/viewtopic.php?t=20496 .

Happy New Year!

  Darij


Does anyone know wheter this is or isn't true?

are there infinitely many primes $ p$ and $ q$ such that $ p\equal{}q\plus{}2$

It is conjecture, see "The prime pages".


Hello,

I was just wondering (for all you MikTeX users out there), does anyone know about a program called ActivePerl? I recently downloaded MikTeX 2.4 with WinEdt 5.4 and learned that a package in MikTeX called ConTeXt usese PerlScripts, so I should download the recent version of ActivePerl, which is what I did. However, I am not sure what do do now. Does anyone else use ActivePerl here or know what I am talking about? I tried looking @ some documentation, but that did not tell me anything :(

-joe d. crawford
P.S-nice forum you've got here-whoever maintains this site :D

that would be me :) .. and nope sorry I don't use Perl, just php, and hopefully soon, latex :D:D

[quote=barry.yyc]16 year revival o.O[/quote]

plz do not bump 16 year old threads


for how many integer values $n \ge 1$ , is $\sqrt{n+\sqrt{n}}$ integer ?


probability
conditional probability
IMO
IMO 2004

This might seem ridiculous but it is not a joke, as the Belgian team is discussing it now and is very confused about it.

You are visiting a family with two children. You are drinking tea with the mother and the father. At a given moment one of the children enters the room. The other one does not show up. What is the probability that the second child is a boy if the father says
a) "this is paul, our eldest child";
b) "this is paul"?

I take it that by second child you are referring to the younger child, not the "other" child?

we look at the following sample space:

    Oldest child    Youngest child
1.  Girl            Girl
2.  Girl            Boy
3.  Boy             Girl
4.  Boy             Boy

if Paul is the eldest child we eliminate 1 ans 2, and since (boy, girl) and (boy, boy) are equally likely the probability is 1/2.

If we only know that one child is a boy we eliminate 1, and each other pair is equally likely so the probability is 1/3.

No, that's wrong, I made the same mistake :blush:
The answer is 1/2 in both cases! Believe it or not!

It looks pretty easy to me... :?

a) Obviously the second child can be either a boy or a girl so the probability is half.

b) If Paul is the elder child then the probability is half.  Same if Paul is the younger child.  Since it's equally likely that Paul is either the elder or younger child, our total probability is half.

The thing to note is that knowing that there is already a boy means that it's not an even spread among (girl, girl), (girl, boy), (boy, boy), (boy, girl) anymore.

[quote="zscool"]I still believe that the answer to b) is 1/3, saying that the other child is a boy or a girl with 1/2 probability is not enough since this problem is a conditional probability problem and new information was added.[/quote] you mean that if one couple has a boy (or a girl) then chances are that the next kid is gonna be the same gender as the first? :)
Almost all my friends (boys) have only sisters, so statistically speaking for me, it's quite the opposite of 33% chances to be a girl :P :)
Anyway, I belive that in both cases it's 1/2. Applying the classical definition of probability, the second child can either be a boy or a girl. That's it. The fact that the first, or the second one's name is Paul, does not influence that fact.

The answer is in deed 1/2. The problem with zscool's analysis of the second problem is that Paul could be either the younger or the older of two sons. Those possibilities have to be counted separately.

A correct sample space for the second problem is:

Older Child - Younger Child
Girl - Paul
Boy - Paul
Paul - Girl
Paul - Boy

So the probability that the other child is a boy is 1/2.

Another way to think of it, is that there is an equal chance that Paul has either an older sister, a younger sister, an older brother, or a younger brother. In exactly half of these possibilities Paul has a brother.

These types of problems are often quite confusing. It helps to sketch a sample space when possible.

I agree that the answer is 1/2, but there is another similar problem on a robotics-based forum  that I also believe is 1/2.  However, most people are saying that the answer is 1/3.

This is the problem:
Mr and Mrs Smith have two children. You ring at their door, and one of their two children, a girl opens the door. What is the probability that the other child is also a girl? (Warning: medium difficulty)

And here is the link:
[url]http://www.chiefdelphi.com/forums/showthread.php?threadid=29393[/url]

Thanks!

P.S. Those who do not know what Chief Delphi is or FIRST Robotics in general please contact me or other people on the Chief Delphi forums - Robotics is great for those who are in or have graduated from school and you can become involved in a local team or start your own - it will be the time of your life - it never hurts to ask!

That problem is equivalent to the one given here. The people in that other Forum are wrong. You should copy and paste the explainations given here to that thread.


geometry
trigonometry
complex numbers
power of a point
radical axis
geometry theorems

This is a really interesting result:

Let $\Gamma_i$, $0\leq i\leq 5$ , be a sequence of circles, all tangent to the circle $\Gamma$ in $A_i$.
The circle $\Gamma_i$ is tangent to the circles $\Gamma_{i-1}$ and $\Gamma_{i+1}$ (take mod 6 in the subscripts).
Then the lines $A_iA_{i+3}$ are concurrent.

The proof is easy with inversion. In a next mssage I post my demonstration.

Ass.:Johann

This problem interested deeply me about five years ago. I take a long trip to find the original book "The Seven Circles Theorem and other new theorems" by Evelyn,Coutts,Tyrrell in an University Library very far from my house.
The original proof is by inversion, but I was informed that in the Mathematical Gazzette there is a very simple proof by complex numbers and in the Pi Mu Epsilon Journal there is a beautiful proof by simple theorems of euclidean geometry.
However I think the proof by inversion is the most neat, simple and natural.
I am very interested also in Japanese Temple's Geometry problems and it is very strange that this theorem about circles was not found in that period.

Johann and Manlio: I don't know any of the proofs you are speaking about! I have always been proving the theorem using Ceva in trigonometric form, and this does work, although one has to use the $\sin\frac{A}{2}=\sqrt{\frac{\left(c+a-b\right)\left(a+b-c\right)}{4bc}}$ formula for triangles and this is a bit of calculation.

  Darij Grinberg

And finally :D my proof:

LEMMA:
"If two circumferences are tangent and your diameters have lenghts 2a and 2b, then your common tangent have a lenght 2*$\sqrt{ab}$"

PROOF: Diect consequence of Pythagoras' theorem.

And now the theorem!

Invert all figure in $A_6$ . Let it $i(P)$ the inverse of set P of points in this plane.
Then $i(\Gamma_6)$ and $i(\Gamma')$ are parallel lines, and then the diameters of $i(\Gamma_1)$ and $i(\Gamma_5)$ are equal.
Let y,x,1,z,y the semidiameters (there are "rays" in English?) of the circles
$i(\Gamma_1)$,$i(\Gamma_2)$,$i(\Gamma_3)$,$i(\Gamma_4)$,$i(\Gamma_5)$,
respectively.
$i(A_1 A_4)$ is a circle passing through $A_6$, $i(A_1)$ ,$i(A_4)$;
$i(A_2 A_5)$ is a circle passing through $A_6$, $i(A_2)$ ,$i(A_5)$;
and clearly $i(A_3 A_6)$=$A_3 A_6$.

The original statement is equivalemt to show that these three last pictures are concurrent in exactly two points, or to show that the line is radical axis of the two circles.
$A_6$ are obviously common to the three pictures,  then it suffices we prove that

$\overline{i(A_3)i(A_2)}*\overline{i(A_3)i(A_5)}$=$\overline{i(A_3)i(A_1)}*\overline{i(A_3)i(A_4)}$.

Now, use the lemma, do the algebraic manipulations and be happy! :lol: !
 :D
Ass.: Johann

[color=red][Moderator edit: "Semidiameter" is "radius" in English. The plural form is "radii".][/color]


And the minimum is $-\frac{5+2\sqrt{7}}{12}$?


calculus
derivative
function
inequalities unsolved
inequalities

Let $ a,b>0$ and $ 0\le v\le 1$. Show $ (a^vb^{1\minus{}v}\minus{}a^{1\minus{}v}b^v)^2\le (1\minus{}2r)(a\minus{}b)^2$, where $ r\equal{}min\{v,1\minus{}v\}$.  :maybe:

The proposer has got a proof, but it is lengthy. :blush:

[quote="miwalin"]Let $ a,b > 0$ and $ 0\le v\le 1$. Show $ (a^vb^{1 \minus{} v} \minus{} a^{1 \minus{} v}b^v)^2\le (1 \minus{} 2r)(a \minus{} b)^2$, where $ r \equal{} min\{v,1 \minus{} v\}$.  :maybe:[/quote]
Why does it need a lengthy proof? Wlog $ v\le\frac12$. Putting $ x: \equal{}\sqrt{\frac{a}{b}}$ and $ k: \equal{}1\minus{}2v\in[0,1]$, it becomes
$ (x^k\minus{}x^{\minus{}k})^2\le k\left(x\minus{}\frac1x\right)^2$. Wlog assume $ x\ge1$. Replacing $ k$ by $ k^2$, we have thus to prove for $ k\in[0,1]$
\[ x^{k^2}\minus{}x^{\minus{}k^2}\le k\left(x\minus{}\frac1x\right)\] which is easy with derivatives,:blush: fixing $ x\ge1$ and considering it as a function of $ k$. :)


a) $a,b,c>0$.prove that
$a^3+b^3+c^3+abc \geq \frac{4}{3}(a^2b+b^2c+c^2a)$
b) find k max satisfy:
$a^3+b^3+c^3+kabc \geq (1+\frac{k}{3})(a^2b+b^2c+c^2a)$ $\forall a,b,c>0$

[quote="hungkhtn"]Use old way. It has been posted.[/quote]
please solve it !!!!!!!!  :D    :blush:


I am sorry, but I didn't know where to post this, so...

I am looking for a program that draws graphs of functions. Can somebody tell me some names?

Thank you very much!  :D

I think [url=http://www.artofproblemsolving.com/Forum/index.php?f=144]this[/url] is a more appropriate forum, but before you post there, use the [url=http://www.artofproblemsolving.com/Forum/search.php]search function[/url]; I'm sure this question has come up before.

Mathematica works very nicely (actually, I think it does pretty much everything well).

Mathematica is ridiculously expensive if you're just going to use it as a graphing application.
I prefer [url=http://www.graphcalc.com/]GraphCalc[/url].

I like NuCalc, but you have to pay money for it.


algebra
polynomial
system of equations
national olympiad

VietnamMO1994
Day1
1.Solve the system of equations in $x,y,z$ :
                                  $x^3+3x-3+ln(x^2-x+1)=y$
                                   $y^3+3y-3+ln(y^2-y+1)=z$
                                     $z^3+3z-3+ln(z^2-z+1)=x$
2.Let $ABC$ be a triangle and let $A',B',C'$ be points symmetric to $A,B,C$ with respect to $BC,CA,AB$ respectively .Find the necessary and sufficient condition for $ABC$ so that $A'B'C'$ is equilateral.
Day2
1.Let be given a convex polygon $A_{0}A_{1}...A_{n}$ ($n>2$).Initially ,there are $n$ stones at $A_0$.In each move we can choose two vertices $A_i$ and $A_j$(not necessarily different),take a stone from each of them ad put a stone onto a vertex adjacent to$A_i$and onto a vertex adjacent to$A_i$.Find all values of$n$such that after finitely many such moves we can achieve that each vertex except$A_0$contains exactly one stone . 2. 3.Do there exist polynomials$P(x),Q(x),T(x)$with positive integer coefficient such that$T(x)=(x^2-3x+3)P(x)$,
                $P(x)=(\frac{1}{20}.x^2-\frac{1}{15}x+\fra{1}{12})Q(x)$?

\[\text{Day 1}\]
1. http://www.mathlinks.ro/Forum/viewtopic.php?t=135399
2. http://www.mathlinks.ro/Forum/viewtopic.php?p=765790#765790
3. http://www.mathlinks.ro/Forum/viewtopic.php?p=765798#765798
\[\text{Day 2}\]
4. http://www.mathlinks.ro/Forum/viewtopic.php?p=765802#765802
5. http://www.mathlinks.ro/Forum/viewtopic.php?p=765803#765803
6. http://www.mathlinks.ro/Forum/viewtopic.php?p=765804#765804


What is the average of integers a, b and c if the average of the two numbers a and 2b is 7 and the average of the two numbers a and 2c is 8?

[hide] We have

$ \frac{a\plus{}2b}{2}\plus{}\frac{a\plus{}2c}{2}$

$ \implies \frac{a}{2}\plus{}\frac{a}{2}\plus{}\frac{2b}{2}\plus{}\frac{2c}{2}$

$ \implies a\plus{}b\plus{}c$.

Therefore, $ a\plus{}b\plus{}c\equal{}7\plus{}8\equal{}15$, so $ \frac{a\plus{}b\plus{}c}{3}\equal{}\boxed{5}$.[/hide]


floor function
combinatorics unsolved
combinatorics

Let  A = {$ n \in N$ | 5 divides n}
      B = {$ n \in N$| both n and $ \frac {n}{5}$ contain only odd digits}
and C = {$ n \in N$ | both n and $ \frac {n}{5}$ contain exactly 2006 digits} where N
      denotes the set of all natural numbers.
      Find the number of elements in $ A \cap B \cap C$

Let $ D(m,k)$ denote the $ k$-th digit of $ m$ from the right.

If $ D(m,k)$ is odd: $ D(5m,k) \equal{} 5 \plus{} \lfloor \frac {5D(m,k \minus{} 1)}{10}\rfloor$
because for all digits $ d$, $ 5d < 50$, we have $ \lfloor \frac {5d}{10}\rfloor < 5$
Hence $ 5 \plus{} \lfloor \frac {5D(m,k \minus{} 1)}{10}\rfloor < 10$

Lemma: If all digits of $ m$ are odd, then all digits of $ 5m$ are odd iff all but the left-most digit of $ m$ are $ 1,5,9$.

Proof: Suppose $ m$ has $ l$ digits.
For all $ k\le l$ we have $ D(5m,k) \equal{} 5 \plus{} \lfloor \frac {5D(m,k \minus{} 1)}{10}\rfloor$
And since for all those $ k$, all digits of $ 5m$ are odd if $ D(5m,k)$ is odd, $ 5 \plus{} \lfloor \frac {5D(m,k \minus{} 1)}{10}\rfloor$ is odd or $ \lfloor \frac {5D(m,k \minus{} 1)}{10}\rfloor$ is even.
But the only odd digits $ d$ with the property $ \lfloor \frac {5d}{10}\rfloor$ even are $ 1,5,9$.
Hence $ D(m,k \minus{} 1) \equal{} 1,5,9$ for all $ k\le l$

Lemma proved.

by the condition of $ C$, $ \frac {n}{5}$ must be between $ \left[10^{2006},2\cdot 10^{2006}\right )$
Hence the first digit of $ \frac {n}{5}$ is $ 1$
And by the condition of $ B$ and by our Lemma, we have all other digits are $ 1,5,9$ and that is a sufficient condition.

Hence there are other $ 2005$ digits of $ \frac {n}{5}$ and we have $ 3$ choices for all those digits. Thus the toatal number of choices for $ \frac {n}{5}$ satisfying the conditions of $ B,C$ is $ 3^{2005}$. For all those $ \frac {n}{5}$, there is a corresponding $ n$ with all the properties of $ A,B,C$.

$ \boxed{|A\cap B\cap C| \equal{} 3^{2005}}$


there are 3 altitudes in a triangle.... (obviously)

if the lengths of two of the altitudes are $h$ and $k$ with $h$ does not equal $k$, find the MAXIMUM and MINIMUM length of the thrid altitude, $p$, in terms of $h$ and $k$...

hint: use the traingle inequality, if a,b,c are sides of a tirangle, a < b+c

You have the triangle inequality backwards.  It should be $a<b+c$.

c'mon plz i need help

i got no ideas   :blush:

any ideas flying around in ur head??      :(

We have $\frac{2S}{h}=a$, $\frac{2S}{k}=b$ and $\frac{2S}{p}=c$, where $S$ is the area of the triangle. It follows that $p$ satisfies $\frac{1}{h}+\frac{1}{k}>\frac{1}{p}>\left|\frac{1}{h}-\frac{1}{k}\right|$ and it is the only condition.


inequalities
trigonometry
calculus
derivative
inequalities unsolved

Show that in any triagle we have $\sum\frac{1}{1+\sin\frac{A}{2}}\geq 2$

Jensen's does it, since $\sin(x/2)>0$ for $x\in (0,\pi)$

Take the second derivative of $\frac {1}{1+\sin(x/2)}$.  You should see that it's positive.  From there, we need only apply Jensen's inequality.

using $\sum \frac{a_i^2}{b_i} \geq \frac{(\sum a_i)^2}{\sum b_i} $
then
$\sum \frac{1}{1+\sin \frac{A}{2}} \geq \frac{(1+1+1)^2}{3+\sum \sin \frac{A}{2}} $
It is well known that.
$\sum \sin \frac{A}{2} \leq \frac{3}{2} $
so the inequality holds.


The sequence $ (a_n)$ is defined by the recursion $ a_0\equal{}a_1\equal{}1, (n\plus{}1)a_{n\plus{}1}\equal{}(2n\plus{}1)a_n\plus{}3na_{n\minus{}1}$. Prove that the sequence consists of integer numbers.

My hint:You can prove by induction that  $ a_{n\plus{}2}\equal{} a_{n\plus{}1}\plus{} a_1.a_n\plus{}a_2.a_{n\minus{}1}\plus{}...\plus{}a_n.a_1$ .

It's wrong!
$ a_0\equal{}a_1\equal{}1,a_2\equal{}3,a_3\equal{}7: a_3\neq a_2\plus{}a_1a_1$

$ f(x)\equal{}\sum a(n)x^n$ find a differential equation with $ f'$ and $ f$

solve it

expand the solution you'll get $ a(n)$ in term of $ n$


calculus
integration
limit
trigonometry
logarithms
real analysis
calculus computations

Find $ \lim_{n\to\infty} \frac{\pi}{n}\left\{\frac{1}{\sin \frac{\pi (n\plus{}1)}{4n}}\plus{}\frac{1}{\sin \frac{\pi (n\plus{}2)}{4n}}\plus{}\cdots \plus{}\frac{1}{\sin \frac{\pi (n\plus{}n)}{4n}}\right\}$

The given Riemann sum can be written as

$ \pi \int_{0}^{1}\csc \frac{\pi }{4}\left( x\plus{}1\right) dx\equal{}\allowbreak 4\ln \left( \sqrt{2}\plus{}1\right)$


integration
logarithms
probability and stats

Let $ X_{1},X_{2},...$ be the sequence of independent random variables uniformly distributed on interval $ [1,2]$. Prove that sequence:
\[ \frac{1}{n}(\frac{X_{1}}{X_{2}}\plus{}\frac{X_{2}}{X_{3}}...\plus{}\frac{X_{n}}{X_{n\plus{}1}})\] converges almost surely and find its limit.

Results of a simulation: the value of that sequence for $ n\equal{}2500$ for ten successive runs.

1.03962001
1.04032509
1.03794365
1.03943993
1.04064367
1.03824317
1.04109146
1.04067459
1.03910050
1.03863660

Let $ Y_j\equal{}\frac{X_j}{X_{j\plus{}1}}.$ Then each $ Y_j$ is a random variable with a particular distribution (details to be worked out later) that is confined to the interval $ [\tfrac12,2]$ and thus has a mean (details to be worked out later) and a finite variance. (We don't even know about the variance.)

The sequence $ (Y_j)$ is not independent; $ Y_j$ and $ Y_{j\plus{}1}$ should be negatively correlated. But the sequence $ (Y_1,Y_3,Y_5,\dots)$ is i.i.d. as is also the sequence $ (Y_2,Y_4,Y_6,\dots).$ So break the average into two separate sums, one for the odd terms and one for the even terms. By the strong law, each half-average should converge [i]a.s.[/i] to the mean of $ Y,$ so combining the two should give us that the whole average converges [i]a.s.[/i] to the mean of $ Y.$ What is that mean?

$ E(Y)\equal{}\int_1^2\int_1^2\frac{x_1}{x_2}\,dx_1\,dx_2\equal{}\frac32\,\ln 2\approxeq 1.03972077$

That's our answer. Note that in comparison to the comment I made on your other problem that the question of possible [i]a.s.[/i] convergence does make sense here with the approach to a constant limit.


agha man ye seri masaleye kheili kheili ba hal dar tamaame zavaayaaye nazarie adaad peyda kardam ke age kasi oono nadare mitoone download kone.
mokhlesiiiiiiiiiiiiiiim.

khoobe dar morede hamin masale ha ye zare topico betterkoonim az bahs dar moredeshoon...khoobe?
khoob nis?bashe har joor mayelin.

taghdim be hameye bozorgane khafan az jomle babak khan va mohammadhosein khan ke kheili mokhleseshoonam va mokhlese hameye barobachze dige ham hastam(baraye in esm ovordam ke felan hamin doostan on hastan).

agha felan ba ina haal konid ta bazam befrestam :blush:  :D
be mahze inke ino ferstaadam badi ro ham mifressam. ;)

agha inam do ta dige(hamash hamin biid...tamoom shod! :( )

agha ooni ke goftam momkene nayam aghab oftaad...
let's drink some polynomials...enjoyyyyyyyyy


geometry
3D geometry
sphere
linear algebra
matrix
number theory proposed
number theory

Find all integers $n$ for which the rational points on the sphere of radius $n$ in $R^3$ are dense on the sphere.

This sounds really weird :?. Is this sphere of radius $n$ you're talking about supposed to be centered at the origin? Because if it is, it's clear that we only need to check whether the rational points are dense on the unit sphere, since if $S$ is a set of rational points dense on the unit sphere, $nS$ is also a set of rational points, dense on the sphere with radius $n$, so we only need to check whether the rational points are dense on the unit sphere (and they are, which is not at all hard to prove, but this is not the point here :)).

Is it possible that you want not radius $n$ but $\sqrt n$¿
Then the problem in http://www.mathlinks.ro/Forum/viewtopic.php?t=56576 would give that only the $n$ that can be represented as sum of three squares can fulfill this property, and I think they all do then.

Any orthogonal matrix with rational entries maps rational points of the sphere to rational points of the sphere. $SO_3(\mathbb{Q})$ is dense in $SO_3(\mathbb{R})$, which acts transitively on the sphere. (We prove this looking at the unit sphere)

If there is one rational point, they are dense.


limit
function
inequalities
calculus
derivative
algebra
binomial theorem

figre it out!!!
$\displaystyle\lim_{x\to1}\frac{(1-\sqrt{x})(1-\sqrt[3]{x})...(1-\sqrt[n]{x})} {(1-x)^{n-1}}$
thanks.

Wow, you were able to get the answer without writing a solution!

Could you spread the knowledge?

Let $t=1-x$ and $t\to 0$ as $x\to 1$, the limit becomes
$\lim_{x\to 1}\frac{(1-\sqrt{1-t})(1-\sqrt[3]{1-t})...(1-\sqrt[n]{1-t})} {t^{n-1}}$
It is known that $(1+x)^\alpha \sim 1+\alpha x$ as $x\to 0$, so $1-(1-t)^\alpha \sim \alpha t$
Hence
$\quad \lim_{x\to 1}\frac{(1-\sqrt{1-t})(1-\sqrt[3]{1-t})...(1-\sqrt[n]{1-t})} {t^{n-1}}$
$= \lim_{x\to 1}\frac{\frac 12 t\cdot \frac 13 t\cdots \frac 1n t}{t^{n-1}} = \frac{1}{n!}$

[quote="pkothari13"][quote="liyi"]
It is known that $(1+x)^\alpha \sim 1+\alpha x$ as $x\to 0$[/quote]

How does one prove that? Binomial theorem?[/quote]
Usually you would refer to the maclaurin expansion. Keeping the remainder term in an Ordo-form will also let you do this with formal reasoning rather than the psuedo-formal "this function approximately behaves like ...".

Is that not an extension of the Bernoulli Inequality?

...and to prove that, here are some ideas:

$f(x)=\left(1+x\right)^{n}-1-nx$

Take some derivatives and find the intervals for which it is greater than 0, and you should have your inequality.

$\lim_{x\rightarrow 0}\left(1+x\right)^{n}=1+nx$

You wanted to prove that?  It seems to be continuous at x=0, so why not just plug it in?


A line, which passes through the incentre $I$ of the triangle $ABC$, meets its sides $AB$ and $BC$ at the points $M$ and $N$ respectively. The triangle $BMN$ is acute. The points $K,L$ are chosen on the side $AC$ such that $\angle ILA=\angle IMB$ and $\angle KC=\angle INB$. Prove that $AM+KL+CN=AC$.
[i]S. Berlov[/i]

Let (I) touch AB, BC, CA at D, E, F. $\triangle IEL \cong \triangle IFM,$ $\triangle IEK \cong IDN$ by ASA, $EL = FM = s-a-AM,$ $EK = DN = s-c-CN.$ $KL = EK+EL = 2s-(c+a)-(AM+CN) = CA-(AM+CN).$

A different and more natural approach to me, but above solution is nice too

Easy to see that $AMIL$ and $CNIK$ are cyclic so because $i$ is  on the angle bisectors of $A$ and $C$ then $IN=IK$ and $IM=IL$. Line from $K$ perpendicular to $AI$ intersects $AB$ at $P$ and lin from $L$ perpendicular to $CI$ intersects $BC$ at $Q$. By simple angle chasing we get that $IPQ$ are collinear. Also $IP=IK=IN$ and $IQ=IL=IM$ (because $IC,IA$ are the perpendicular bisectors of $LQ,KP$). From this we get that $M,N$ are the reflections of $Q,P$ in $AI$ so $MP=NQ$. Now:
$$AM+CN=AM+MP +CN-NQ=AP+CQ=AK+CL$$
which is equivalent to what we wanted.


A wheel has a diameter of x inches and a second wheel has a diameter of y inches. The first wheel covers a distance of d feet in 100 revolutions. How many revolutions does the second wheel make in covering d feet?

A. 100xy
B. 100y - x
C. 100x - y
D. 100y / x
E. 100x / y

This is pretty easyyyyyy.
[hide]The distance d is equal to 100x(pi).  The circumference of the other circle is y(pi), so the number of revolutions is 100x/y, [b][color=darkred][size=200]E[/size][/color][/b]. :lol: [/hide]

Yes...it is quite easy.

[hide]

$ \frac{100\pix}{y\pi}=\frac{100x}{y}$

which gives us and answer: [color=blue][size=200][u][b]E[/b][/u][/size][/color]

[/hide]


I am given two pitch balls, charged with the same amount of charge $q$ and each of the weigh $10 mg$ they are tied with a string of length $30 cm$ and hung from the same point. They repel each other and stay $6 cm$ away.

I have to find the value of $q$ ..
Can anyone help me plz ?

Let $A$ be the point at which the balls are hanged from, call the two charges $B$ and $C$. $\triangle ABC$ is obviously isoscelses and let $\angle BAC$ be $2 \theta$.

We observe that $sin \theta = \frac{1}{10}$ from tis we can see that for the same value of $\theta$, $tan \theta$ will be approximately equal to the value of $sin \theta$. Let $T$ be the tensions in the strings and let $F_{e}$ be the electrostatic force acting on the two charges.

$F_{e}=Tsin \theta$ therefore we have $F_{e}= \frac{kq^{2}}{0.0036}= 0.1T$ (1)

$T=\frac{10 \times 10^{-3}g}{cos \theta}$ since $sin \theta$ and $tan \theta$ are about the same, $T=10 \times 10^{-3}g$. Now substitue $T$ into (1).

$k$: Coloumb's Elesctrostatic Constant

$g$: acceleration due to gravity

So $q= 6.260 \times 10^{-8}C$

thanks a lottttt joseph!
your process is completely right, but when i did the calculation the answer i found was $1.9798 \times 10^{-7}C$

thanks again!


http://simfish.googlepages.com/snap000672.jpg

Okay, so I tried to separate the (1-W/b) term by letting it go into the denominator. But then you can't really solve the equation by the traditional dW/W method, since W + constant is raised to a power of n. i could multiply both sides by (1-W/b) to a power of n-1, but then when i integrate, i don't get the desired answer. can anyone help? thanks

Let $ t \equal{} 1 \minus{} \frac {w}{b}$ and $ x \equal{} 1 \minus{} \frac {z}{a}$.

We have:
$ (1 \minus{} \frac {w}{b})^n dw \equal{} \frac {M}{1 \minus{} \frac {z}{a}}dz \implies t^{n}dt \equal{} \frac {Ma}{bx}dx$.

$ \frac {t^{n \plus{} 1}}{n \plus{} 1} \equal{} \frac {Ma}{b}\ln|x| \plus{} C \implies t^{n \plus{} 1} \equal{} (n \plus{} 1)[\frac {Ma}{b}\ln|x| \plus{} C]$

$ \implies ln|t| \equal{} ln[(n \plus{} 1) (\frac {Ma}{b}\ln|x| \plus{} C)]^{\frac {1}{n \plus{} 1}}$

$ \implies t \equal{} [(n \plus{} 1) (\frac {Ma}{b}\ln|x| \plus{} C)]^{\frac {1}{n \plus{} 1}}$

$ \implies 1 \minus{} \frac {w}{b} \equal{} [(n \plus{} 1) (\frac {Ma}{b}\ln|1 \minus{} \frac {z}{a}| \plus{} C)]^{\frac {1}{n \plus{} 1}}$

$ \implies w(z) \equal{} b \minus{} b[(n \plus{} 1)( \frac {Ma}{b}\ln|1 \minus{} \frac {z}{a}| \plus{} C)]^{\frac {1}{n \plus{} 1}}$

With the given condition it seems like $ C \equal{} \frac {1}{n \plus{} 1}$


algebra
polynomial
Rational Root Theorem

Given an integer $n,$ $n>2,$ for how many different rational numbers $x$, does there exist a polynomial of degree $n-1$ with $P(0)=0$ such that $54x^{n}+P(x)=315?$

Do you guys get
[hide="Answer?"]72 by the Rational Root Theorem? (Hope I didnt count incorrectly.

Also, how can you guarantee that all 72 of those roots can be attained?[/hide]

You probably want $P(x)$ to have integer coefficients, for one thing.

Yes.  Without that condition,

[hide="Triviality occurs"] $(x-r)(54x^{n-1}+...+\frac{315}{r}) = 54x^{n}+P(x)-315$ [/hide]


What is the sum of the prime factors of the number represented by $2^{12}+2^{11}+2^{10}+\dots+2^{2}+2^{1}$?

[hide]The prime factorization of the number is $2^{1}\cdot 3^{2}\cdot 5^{1}\cdot 7^{1}\cdot 13^{1}$. Adding that gives us $33$.[/hide]

[hide]Didn't you forget to add a 3? It's $3^{2}.$[/hide]

[quote="bpms"][quote="SplashD"][quote="me@home"]in the prime factorization you don't count a number twice...[/quote]

the question never said distinct prime factors though[/quote]
I think you know what he meant...
[size=59][color=white]then again, maybe not...[/color][/size][/quote]

So who got it right? The question is asking for the sum of the prime factors, and I'm not sure which way that should be interpreted, all or all distinct.

[quote="i_like_pie"][/quote][quote="bpms"][quote="SplashD"][quote="me@home"]in the prime factorization you don't count a number twice...[/quote]

the question never said distinct prime factors though[/quote]
I think you know what he meant...
[size=59][color=white]then again, maybe not...[/color][/size][/quote][quote="i_like_pie"]

So who got it right? The question is asking for the sum of the prime factors, and I'm not sure which way that should be interpreted, all or all distinct.[/quote]
All distinct...

the question is verbatim, so i think it is all, but i may be wrong :maybe:

Well, if know one knows what exactly the question means then let's just forget this...

[quote="anirudh"]ok. It means all numbers, [i]not [/i]all distinct, [i]but [/i][b]all numbers[/b][/quote]

how do you know?

Anirudh said [i]all[/i] numbers, not just distinct. [hide]$2+3+3+5+7+13=33$[/hide]


MATHCOUNTS email geometry perimeter incenter percent AMC

Specify the grade youre in and send them via pm or email me at ddrmathskittles@hotmail.com

1) A floor, 9 feet by 12 feet, is to be tiled with 4 inch by 6 inch tiles.  How many tiles are needed to cover the floor?

2) To swim a mile in a certain rectangular swimming pool, one must either swim the long length 80 times, or negotiate the perimeter of the pool 22 times.  What is the number of square yards in the area of the swimming pool?

3) What is the degree measure of the acute angle formed by the hands of a clock at 3:20?

4) The graphs of the equations x+3y=6 and kx+2y=12 are perpendicular.  What is the value of k?

5) If triangle PQR has sides 40, 60, and 80, then the shortest altitude is K times the longest altitude.  Find the value of K.

6) Given that I is the incenter of triangle ABC, AB=AC=5, and BC=8, find the distance AI

7) Express the absolute value of the difference of .3636363636 (repeating) and .36 as a common fraction.

8) Find x^4+1/x^4 if x-1/x=5

9) If 9n^2-30n+c is a perfect square for all integers n, what is the value of c?

10) In how many ways can a debt of $69 be paid exactly using only $5 bill and $2 bills.

11) If p is 50% of q and r is 40% of q, what percent of r is p?

12) Twenty-five women did 1/5 of a job in 8 days.  Then, because of an emergency, it became necessary to complete the job in the next 20 days.  How many additional women needed to be added to the crew of 25 to accomplish this?

13) Two dogs, each traveling 10 feet/second, run toward each other from 500 feet apart.  As they run, a flea flies from the nose of one dog to the nose of the other at 25 feet/second.  The flea flies between the dogs in this manner until it is crushed when the dogs collide.  How far did the flea fly?

14) The medians of a right triangle which are drawn from the vertices of the acute angles are 5 and  :sqrt: 40.  What is the length of the hypotenuse in simplest radical form?

15) How many factors of 2^95 are there which are greater than 1,000,000?

16) How many odd positive integers are factors of 480?

17) Find the greatest n for which 12^n evenly divides 20!

18) If the sum of all the angles except one of a convex polygon is 2190 degrees, then how many sides does the polygon have?

The following are somewhat challenging for mathcounts, but still fun nonetheless. ill award 2 points for these  problems out of the kindness of my heart. and 0 points for a wrong answer. put your solutions to these questions too. no partial credit

19) What is the length of the common external tangent segment of two externally tangent circles whose radii are 8 and 11 in simplest radical form?

20) For how many k is the least common multiple of 6^6, 8^8, and k equal to 12^12?

21) The complex numbers w and z form the equations of w^2+z^2=7 and w^3+z^3=10.  What is the largest possible real value of w+z?

22) Find the greatest integer x for which 3^20 > 32^x.

if you sent in your answers prior to 11:04 pm central time, im sorry. and please send it again because my pm box was all full. sorry again.

Well I have no idea what an inch, foot, or yard is, so unless I can be bothered searching, I won't enter..

You are a bit too old for a junior high school contest, don't you think?

foot=12 inches, thats all you really need for the contest thing. Triple M can enter if he wants, at least its someone compared to the current 0 submissions.

I'm in junior high...9th grade's junior high in some places, is it not?

anybody can do this contest. thats why if you click on the scoreboard there is two diff ones.

Yeah theres two scoreboards. And I just like entering contests. But I probably won't.

[hide]

partially cos NOBODY ANSWERED MY QUESTION! well syntax tried. aaaaaaaah number 2 you need to know how many yards in a mile lol

[/hide]

1760 yards=1 mile

Don't they use Imperial/Yard-Pound in New Zealand? I recall Australia was a British colony so I thought they used yard so New Zealand people used yards too...

Meh technically I'm in 8th but I'm moving up mid-January...?

12 inches = 1 foot
3 feet = 1 yard
1760 yards = 1 mile
5280 ft = 1 mile
63360 inches = 1 mile
36 inches = 1 yard

there...

Well if you're gonna name'em all then you need furlong in there somewhere...(furlong is 220 yards or one-eighth of a mile)

Neal, according to the proportions of time for each problem, they should get at least approx. 36 min... but u can't enforce 25 anyways... maybe that's why they're not getting too high of scores

well it was supposed to be 25 minutes when i had originally planned for 15 question, but then i added some harder questions and stuffs but forgot to change the time. i doubt they went by the time anyways

hehe, technically they should have way more than 36 minutes

hm. now that you say that, i'm going to wait until you get online and then try to do them in 15.

he meant 36 for mathcounts level lpeople, not you mystic

u get 4/3 for the 18 questions...
that's 24 minutes...
you get 3 for the 4 questions, that's 12, so 36

2 of the last 4 questions were like the first questions of aimes. so technically 24 minutes for those 2 at least

20 and 21 were the easy aime ones. thats why i saw them fit as bonus questions


topology
geometry
real analysis
combinatorics unsolved
combinatorics

Let us say that a triad contains a point $ A\in\mathbb{R}^2$ and three different paths starting from $ A$. Prove that $ \mathbb{R}^2$ is not an union of disjoint triads.

[quote="puuhikki"]Let us say that a triad contains a point $ A\in\mathbb{R}^2$ and three different paths starting from $ A$. Prove that $ \mathbb{R}^2$ is not an union of distinct triads.[/quote]

What exactly do you name a "path starting from $ A$" ?

[quote="puuhikki"]I mean that it is the graph of a continuous mapping $ f: [0,1]\to \mathbb{R}^2$ s.t. $ f(0) \equal{} A$.[/quote]
OK, thanks, it was not clear that it was $ [0,1]$ and not $ [0,1)$, for example.

"Distinct" should be "disjoint," I assume.

This works if we define a curve to be a subset of $ \mathbb{R}^2$ to be homeomorphic to $ \{(x,0): x \in [0,1]\}$.
Let the set $ S$ be such that it contains the circles satisfying a certain circles:

a) The center of the circle belongs to $ \mathbb{Q}^2$.
b) The radius is rational.

If one chooses from the circle of $ S$ three different rational arcs (the angles bounded by arcs are rational multiples of $ \pi$), one can form the set $ R$ of the members of $ S$ having a rational arc triplet. Then $ R$ is countable. Not that the number of triads have to be uncountable, because the area of any triad has a Lebesgue measure $ 0$ as well as their countable union has a Lebesgue measure $ 0$. For every triad $ A$, choose an arc triplet of $ R$ such that the three curves of the triad meets the corresponding circle of $ S$ first time at the same point where they meet the disjoint arcs, and the middle point of the triad lies in the interior of the disc bounded by $ S$. Some arc triplet has chosen at least for two triads, because otherwise the number of triads is countable. This is a contradiction sinve those two triads are disjoint.

[quote="puuhikki"] because the area of any triad has a Lebesgue measure $ 0$ as well as their countable union has a Lebesgue measure $ 0$. [/quote] Ah, but this step is false! One can draw a Jordan curve on the plane that has positive Lebesgue measure ;)


Just out of curiousity, what languages do all of you speak? I tried to include as many as I could think of. Latex isnt included  :lol: . I think i spelled bilingual and trilingual etc correctly. I speak English and hebrew, the former more fluently than the latter. I also speak little bits of Japanese and Spanish.. but I wouldnt count that as 'speaking the language'.

I used to be billingual, in English and Marathi, an Indian language.  But then I forgot Marathi.

English and Russian. How could you forget the latter? INFIDEL ALERT!!!! :P

i used to be one and half-lingual. but i forgot all the french i learned so now i'm back to being monolingual.

Um english, mandarine chinese (only speak, I can read/write like 25 words), and I am taking Spanish III right now in school.  Should I count Spanish?

[quote="Derek"]Cantonese, Mandarin, English, Japanese.[/quote]
Wow, quadrilingual! :o

My mother language is Japanese, but I can speak English to some extent because I've been in the U.S. for about 2 years. :roll:

I'm about 1.03-lingual total. Fluent in English, but I barely know any Spanish, Latin, Italian, and Russian.

English, some Korean, and a tiny bit of Chinese.


For$a,b,c,d,e,f$ positive reals prove Shapiro's ineq

$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}{e+f}+\frac{e}{f+a}+\frac{f}{a+b} \geq 3$

[quote="Peter VDD"]I think even the generalised case is quite well-known: (0<i<k<n)

$\sum_{cyc} \frac{a_1+...+a_i}{a_{i+1}+...+a_k} \geq \frac{n\cdot i}{k}$[/quote]

You probably meant

$\sum_{cyc} \frac{a_1+...+a_i}{a_{i+1}+...+a_{i+k}} \geq \frac{n\cdot i}{k}$.

This is well-known indeed, but it is wrong (in general)... The special case Manlio posted is correct.

  Darij

[quote="darij grinberg"]You probably meant

$\sum_{cyc} \frac{a_1+...+a_i}{a_{i+1}+...+a_{i+k}} \geq \frac{n\cdot i}{k}$.

This is well-known indeed, but it is wrong (in general)... The special case Manlio posted is correct.[/quote]

That one is false? :?

Dam n, it's on one of our IMO training sheets :D

Anyway it's quite simple this one, Jensen for 1/x does the job...

[quote="Peter VDD"]Dam n, it's on one of our IMO training sheets :D[/quote]

Oh, well, it's in our training script as well, but with the remark that it's generally wrong. Actually, the special case when i = 1 and k = 2, namely the inequality

$\sum_{cyc} \frac{a_1}{a_2+a_3} \geq \frac{n}{2}$,

is the famous "Shapiro inequality". It is generally true for some "little" values of n (namely, for n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 17, 19, 21, 23, including the value n = 6 which is equivalent to Manlio's problem), while for all other values of n, counterexamples can be found. See http://mathworld.wolfram.com/ShapirosCyclicSumConstant.html .

[quote="Peter VDD"]Anyway it's quite simple this one, Jensen for 1/x does the job...[/quote]

Actually, Jensen for 1/x is the same as HM-AM or Cauchy-Schwarz...  ;)

  Darij


calculus
integration
trigonometry
real analysis
limit
real analysis unsolved

$ \textbf{(i)}$ Calculate $ \int_{\minus{}\infty}^{\infty}{e^{\minus{}x^2}\cos (2 x^2)\; dx}$.
$ \textbf{(ii)}$ Show that when $ k\to \infty$ then $ \int_{\minus{}\infty}^{\infty}{e^{\minus{}x^2}\cos (k x^2)\; dx}\equal{}O(k^{\alpha})$ and find $ \alpha\in\mathbb{R}$.

Taking on part (ii):

$ f(k) \equal{} \int_{ \minus{} \infty}^{\infty}e^{ \minus{} x^2}\cos(kx^2)\,dx \equal{} \int_0^{\infty}\frac {e^{ \minus{} u}}{\sqrt {u}}\,\cos(ku)\,du.$

That makes it essentially the Fourier transform of $ \frac {e^{ \minus{} u}}{\sqrt {u}},$ which is an $ L^1$ function. Hence by the Riemann-Lebesgue Lemma, $ f(k)\to0$ as $ k\to\infty.$

But we can do more. Make the further substitution $ t \equal{} ku:$

$ f(k) \equal{} \frac {1}{\sqrt {k}}\int_0^{\infty}\frac {e^{ \minus{} t/k}}{\sqrt {t}}\,\cos t\,dt.$

Let $ C \equal{} \int_0^{\infty}\frac {\cos t}{\sqrt {t}}\,dt,$ noting that this integral does converge (conditionally, as an improper integral).

There is an integral analogue of Abel's theorem for series, proved in much the same way. By that theorem,

$ \lim_{k\to\infty}\int_0^{\infty}\frac {e^{ \minus{} t/k}}{\sqrt {t}}\,\cos t\,dt \equal{} \int_0^{\infty}\frac {\cos t}{\sqrt {t}}\,dt \equal{} C.$

Hence, $ f(k)$ is asymptotic to $ \frac {C}{\sqrt {k}}$ as $ k\to\infty.$

I'm putting the theorem I quoted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=216629]here[/url].

$ \int_{\minus{}\infty}^{\infty} e^{\minus{}x^2} \cos(k x^2) \, dx \equal{} \frac{1}{\sqrt{\pi}} \int_{\minus{}\infty}^{\infty} \frac{1\plus{}t^2}{k^2 \plus{} (1\plus{}t^2)^2} \, dt$ $ \equal{} \sqrt{\frac{\pi}{2}} \frac{\sqrt{\sqrt{k^2\plus{}1}\plus{}1}}{\sqrt{k^2\plus{}1}} \sim \sqrt{\frac{\pi}{2 k}}$ as $ k \to \infty$.

[quote="sos440"]$ \int_{ \minus{} \infty}^{\infty} e^{ \minus{} x^2} \cos(k x^2) \, dx \equal{} \frac {1}{\sqrt {\pi}} \int_{ \minus{} \infty}^{\infty} \frac {1 \plus{} t^2}{k^2 \plus{} (1 \plus{} t^2)^2} \, dt$[/quote]

Where did this step come from?


A child throws a ball up a hill that makes an angle of 45 degrees with the
horizontal. The ball lands 100 feet up the hill. Its trajectory is a
parabola with the equation y = -x^2 + ax for some number a. Find a.

The problem is lacking information.
How far is the hill?
Is $100$ft the highest height the ball reaches?

I took Advanced Physics this year, -I GOT THIS. (Not that I used any haha).

 :play_ball:
[hide="On the lock"]
[i]IMPORTANT:[/i]

1) The hill makes an angle of $45^{\circ}$ with the horizontal. A 45-45-90 Triangle.
2) The ball lands 100 feet up the hill. Hypotenuse $c = 100$.
3) OBJECTIVE FUNCTION :$f(x) =-x^{2}+bx$
[hide="-FIND INTERCEPT-"]
In a 45-45-90 Triangle legs $a$ and $b$ are identical. Thus,
$100 = \sqrt{2a^{2}}$
$100^{2}= 2a^{2}$
$\frac{100^{2}}{2}= a^{2}$
$\sqrt{\frac{100^{2}}{2}}= a = \frac{100}{\sqrt{2}}= 50\sqrt{2}$

POINT (X, Y) : ($50\sqrt{2}$, $50\sqrt{2}$)
[/hide]
[hide="-PLUG N CHUG-"]
In this case, X and Y have the same values so;
$y = x =-x^{2}+bx$
$\frac{x+x^{2}}{x}= \frac{x(x+1)}{x}= x+1 = b$

$b = x+1 = 50\sqrt{2}+1$[/hide]
Graph $y = x$, and $y =-x^{2}+(50\sqrt{2}+1)x$
Do they intersect at ($50\sqrt{2}$, $50\sqrt{2}$)?

mmmmmhm.   :D [/hide]

EDIT: Typo like always...


Show that if a,b,c are the side of a triangle and A,B,C are its angles
then

(a+b-2c)/sin(C/2) + (b+c-2a)/sin(B/2) + (a+c-2b)/sin(C/2) >=0

The inequality you have to prove is

(a+b-2c)/sin(C/2) + (b+c-2a)/sin(A/2) + (a+c-2b)/sin(B/2) >=0

Another solution of this problem

The expression can be rewrite as

(a-b)(1/sin(B/2)-1/sin(A/2)) + (a-c)(1/sin(C/2) - 1/sin(A/2)) +

(b-c)(1/sin(C/2) - 1/sin(B/2))

Each term is nonnegative.

Consider a triangle ABC with sides a=BC,c=AB,b=AC.
Then b>=a  if and only if angle(B) >= angle(A), and since
angle(A),angle(B) < 180, if and only if sin(B/2) >= sin(A/2)
It follows that (a-b)(1/sin(B/2) - 1/sin(A/2)) >= 0
This is true for any two sides and corresponding vertices, it holds
for the other terms.