Hi all,

First let me tell you that I'm French so my english isn't perfect, I hope you would understand me as well.

I have a little problem to submit to you, this is for a project about making a serious game with an evaluation system. At the end of the game, we have to assign a score to the player.

My problem is how to assign this score, let me tell you the hypotesis:
- The score is between 0 and 100, this score represent a skill, for exemple a score of 20 is egal to a beginner level, 60 to master level, 100 to perfect level.

- This score is based on several evaluation, it means that it has to take in account every evaluation in the game.

- Each evaluation is not based on the same max score. For exemple, the first max is 20, then 60, and finally 100. If the player has got 20 on 20 at the first evaluation, it does only mean that is has reached the beginner level.

- The last evaluation should be more important in the notation than the first one. We can use an exponentiel fonction like this one:
Ts: total score
S: score on the last evaluation
a: 0.66 for exemple

Ts = a * S + ( 1 - a ) * Ts

Now, my problem...
I have to make a total score that tell exactly the level of the player, let me show you some score exemples:
The evaluation in the exemple is in 4 part, the first max is 20, then 30, then 80, and finally 100.
Exemple 1: 20/20 30/30 80/80 100/100 -> total score: 100
Exemple 2: 20/20 25/30 20/80 30/100 -> total socre: might me something like 25
Exemple 3: 10/20 25/30 55/80 60/100 -> total score: something like 50 or 55 would be good...

I have to find some formulas to get the total score from the score of each evaluation.
I hope you see what the difficulties are, and hope you'll find it fun to solve (i have'nt got the solution, i found some good approximation, but somes cases don't work :'(  )

Thank you !

Essentially you are trying to use some weighted average $ Ts\equal{}aS_1/20\plus{}bS_2/30\plus{}cS_3/80\plus{}dS_4/100$ where $ S_i$ is the number of points obtained in the $ i$-th round and $ a\plus{}b\plus{}c\plus{}d\equal{}1$, $ 0<a<b<c<d$. The coefficients may be assigned in many ways and you can play with them to see if you can get a satisfactory result from this simple linear formula. If something is dissatisfactory, you'll have to explain us exactly what it is. It is not really easy to understand what you mean by
[quote]
i have'nt got the solution, i found some good approximation, but somes cases don't work :(
[/quote]


Find all integers $ n$ for which there exists a polynomial $f$ with integer coefficients such that \[ f(i)\equal{}2^i,\] for $ 1\le i\le n$.

[quote="hxy09"]Find all $ n$,so that there exists an integer-coefficient polynomial $ f(x)$ and $ f(i) \equal{} 2^i$ for $ 1\le i\le n$[/quote]

For $ n\le 3$, we have $ f(x)\equal{}x^2\minus{}x\plus{}2$

For $ n\equal{}4$ : $ f(x)\minus{}(x^2\minus{}x\plus{}2)\equal{}(x\minus{}1)(x\minus{}2)(x\minus{}3)g(x)$ with $ g(x)\in\mathbb Z[X]$ and $ 2^4\minus{}14\equal{}6g(4)$ and so $ g(4)\equal{}\frac 13$, impossible.

And, since existence of polynomial for $ n$ implies existence for $ p\le n$, the answer is $ \boxed{n\in\{1,2,3\}}$

Hi my friend,thank you for your nice solution :)
A similar way:if $ n\ge 4$ then $ 4\minus{}1|f(4)\minus{}f(1)$ which implies $ 3|14$ impossible :D


logarithms
floor function
number theory
prime numbers
number theory proposed

Find a constant $ c > 1$ with the property that, for arbitrary positive integers $ n$ and $ k$ such that $ n>c^k$, the number of distinct prime factors of $ \binom{n}{k}$ is at least $ k$.

[i]P. Erdos[/i]

If we use Bertrand's postulate then it's easy to see that c=2 works. There are at least k prime numbers from 1 to $ 2^k\plus{}1$
and since $ 2^k\plus{}1>2k$ all the primes less than $ k$ divide $ n\choose k$.

I don't understand your argument. You say that all primes from 1 to k divide $ \binom{n}{k}$ and that this shows that there must be at least k distinct prime factors ???

Sorry, I've made a stupid mistake. My approach was
wrong, the problems is harder than I've thought at first.

Let's try again:)
Let $ p_1<\dots<p_a$ are all prime numbers not greater than $ k$.
(Here $ a\equal{}\pi(k)$.)
All we have to do is to prove that $ A\equal{}n(n\minus{}1)\dots(n\minus{}k\plus{}1)$ is divisible by
at least $ k$ prime numbers for $ n>c^k$.
Suppose that's not true. There are $ b<k\minus{}a\equal{}k\minus{}\pi(k)$
prime divisor of $ A$ larger than $ k$ and all of them divide
only one of the numbers $ n,n\minus{}1,\dots,n\minus{}k\plus{}1$.
Consider the other $ k\minus{}b>a$ numbers. Choose any
$ p_i,  (i\leq a)$. Notice that $ p_i^t$ divide at most one
of our numbers if $ t\geq\log_{p_i}k$. It follows that there
is at least one number $ X$ between our numbers which has
the form $ X\equal{}p_1^{t_1}\dots p_a^{t_a}$ where $ t_i<\log_{p_i}k$
for each $ i$. For such $ X$ we have that
$ X<k^a\equal{}k^{\pi(k)}$. Now I claim that for suitable $ c$ we have that
$ k^{\pi(k)}<c^k$ and that is the contradiction.

I think this is not enough by itself since we are considering $A/k!$, not just $A$.

But it's easily fixed...

Let $p_1<\cdots<p_\ell<k$ be the prime numbers less than $k$, where $\ell=\pi(k-1)$. If $n-i$ (for some $0\le i\le k-1$) has a prime factor $p\ge k$, then $\gcd(p,n-j)=1$ for all $0\le j\ne i\le k-1$, so for all such $i$ we can get distinct prime factors no less than $k$. Now suppose that $n-i_j=p_1^{\alpha_j(1)}\cdots p_\ell^{\alpha_j(\ell)}$ (for $0\le i_1<\cdots<i_m\le k-1$) only have prime divisors less than $k$.

If $m>\ell$, then as mentioned in the previous post there must be some $j$ for which $p_i^{\alpha_j(i)}<k\forall 1\le i\le \ell$ since $\gcd(n-i_r,n-i_s)<k\forall 1\le r\ne s\le m$ and so
\[c^k-k+1<n-k+1\le n-i_j < k^\ell \le k^{\pi(k)} < k^{\frac{Ck}{\ln(k)}} = e^{Ck}\]for some constant $C$ independent of $k$ (by the PNT), which is false for sufficiently large $c$.

Thus $m\le\ell$. Using the same logic as in the previous paragraph, we can show that the largest prime powers dividing $n-i_1,\ldots,n-i_m$ are pairwise relatively prime. Suppose that $q^r$ is the largest prime power dividing $n-i_j$ so that
\[q^r \ge \sqrt[\ell]{n-i_j} \ge \sqrt[\pi(k-1)]{n-k+1} > n^{\frac{\ln(k)}{C'k}} > c^{\frac{\ln(k)}{C'}} > k\]for sufficiently large $c$ ($C'$ is another constant independent of $k$ again determined by the PNT). Then
\[v_q\left(\binom{n}{k}\right) = \sum_{s\ge0}\left\lfloor\frac{n}{q^s}\right\rfloor-\left\lfloor\frac{k}{q^s}\right\rfloor-\left\lfloor\frac{n-k}{q^s}\right\rfloor \ge \left\lfloor\frac{n}{q^r}\right\rfloor-\left\lfloor\frac{k}{q^r}\right\rfloor-\left\lfloor\frac{n-k}{q^r}\right\rfloor = 1,\](note that $q^r>k$ and $n-k<n-i_j\le n$) where we have used the fact that
\[0\le\lfloor{a+b}\rfloor-\lfloor{a}\rfloor-\lfloor{b}\rfloor\le1\]for all reals $a,b$.

Hence we have at least $(k-m)+m=k$ prime factors, as desired.


n golule manande ham dar yek radif kenare ham gharar darand.mikhahim ruye in goluleha Tedadi golule(be har Tedad va be har chand radif)gharar dahim.albate har golule bayad zire khod 2 golule dashte bashad.in kar be chand tarigh momken ast?


What is the range of the set {112, 124, 136, 148, 160}?


geometry
quadratics
trigonometry
algebra
geometry unsolved

P is a point in the interior of a square ABCD. and PA=5, PB=8, PC=13.
Find the area of ABCD.

Let a = PA, b = PB, c = PC, s = AB the square side and $S = s^{2}$ the square area. Three different solutions for a = 1, b = 2, c = 3 have been posted at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=67307[/url]. Either one can easily be adapted for the general values of a, b, c. The quadratic equation for the square area in the general case comes out as

$S^{2}-(a^{2}+c^{2})\ S-b^{2}(a^{2}+c^{2})+\frac{a^{4}+2b^{4}+c^{4}}{2}= 0$

For a = 5, b = 8, c = 13, the equation becomes $S^{2}-194\ S+6273 = 0$ with the roots S = 153 or 41. The larger root corresponds to the point P inside the square, the smaller one to the point P outside the square.

[color=darkblue]Denote $O\in AC\cap BD$, $AB=BC=\sqrt S$, where $S$ is the area of the square, $PA=a\ ,\ PB=b\ ,\ PC=c$ and $x=m(\widehat{ABP})$, $y=m(\widehat{CBP})$. Then $\cos^{2}x+\cos^{2}y=1\Longleftrightarrow$ $\left(\frac{S+b^{2}-a^{2}}{2b\sqrt S}\right)^{2}+\left(\frac{S+b^{2}-c^{2}}{2b\sqrt S}\right)^{2}=1\Longleftrightarrow$ $\left(S+b^{2}-a^{2}\right)^{2}+\left(S+b^{2}-c^{2}\right)^{2}=4b^{2}S$ $\Longleftrightarrow$ $\boxed{\ 2S^{2}-2\left(a^{2}+c^{2}\right)S+\left(b^{2}-a^{2}\right)^{2}+\left(b^{2}-c^{2}\right)^{2}=0\ }\ .$

[b]Remark 1.[/b] Let $ABCD$ be a square and for any point $P$ denote $PA=a$, $PB=b$, $PC=c\ .$ Then $\left|a^{2}+c^{2}-2b^{2}\right|\le 2ac$, i.e. $|a-c|\le b\sqrt 2\le a+c\ .$
[b]Remark 2.[/b] If $PD=d$, then $2d^{4}-2(a^{2}+c^{2})d^{2}+\left(S-a^{2}\right)^{2}+\left(S-c^{2}\right)^{2}=0$, $b^{2}+d^{2}=a^{2}+c^{2}=2\cdot PO^{2}+S$ and $|a-c|\le \sqrt{2S}\le a+c\ .$[/color]

in this shape $PD^{2}+PB^{2}=PA^{2}+PC^{2}$  Then
$PD^{2}=5^{2}+13^{2}-8^{2}=130$ Thus $PD=\sqrt{130}$
Then we  determine $P'$ and $P''$ points such that $P'$ is on right of $BC$ and $P''$ is on left of $DA$.And also $\triangle PAB\simeq \triangle P'CD$ and

$\triangle PCD\simeq \triangle P''AD$.$P''D=\sqrt{130}$, $P''A=13$, $P'C=5$ and $PB=8$. $\angle PBA=\angle P'BC=\alpha$ therefore $\angle P'BP=90^\circ$ and

$P'P^{2}=P'B^{2}+BP^{2}=64+64=128$.$P'P=8\sqrt{2}$ if we use cosinus theorem in $\triangle P'CP$
$P'C^{2}+P'P^{2}-2\cdot P'C\cdot P'P\cdot Cos \ \widehat{CP'P}=25+128-2.5.8\sqrt{2}.Cos \ \widehat{CP'P}=169$          $Cos \ \widehat{CP'P}=\frac{-\sqrt{2}}{10}$ Hence $Sin \ \widehat{CP'P}=\frac{7\sqrt{2}}{10}$
if we show area ABC with $ABC$;
$A(ABCP)=A(APB)+A(PBC)$ from $\triangle APB\simeq \triangle CP'B$ we can say that $A(ABCP)=A(CPBP')=A(CP'P)+A(BP'P)$
$A(CP'P)=\frac{1}{2}\cdot 5\cdot 8\sqrt{2}\cdot \frac{7\sqrt{2}}{10}=28$ and $A(BP'P)=\frac{8.8}{2}=32$ consequently we can be find $A(ABCP)=32+28=60$
 Similarly,
$A(APCD)=A(PDP''A)=A(PDP'')+A(PP''A)$
$PP''^{2}=130+130=260$ we find that $PP''=2\sqrt{65}$
in $\triangle PP''A$ we use Cosinus Theorem
$25=260+169-2.2\sqrt{65}.13.Cos \ \widehat{PP''A}$  and $Cos \ \widehat{PP''A}=\frac{101}{13\sqrt{65}}$ and therefore $Sin \ \widehat{PP''A}=\frac{28}{13\sqrt{65}}$
now we can calculate $A(PDP''A)$
$A(PP''A)=\frac{1}{2}\cdot 2\sqrt{65}\cdot 13\cdot \frac{28}{13\sqrt{65}}=28$
$A(PDP'')=\frac{\sqrt{130}.\sqrt{130}}{2}=65$

Hence $A(APCD)=A(PDP''A)=65+28=93$

Now, $A(ABCD)=\overbrace{A(ABCP)}^{A(PBP'C)}+\underbrace{A(APCD)}_{A(PDP''A)}=60+93=153$

it became too long!!:) :blush:  :blush: :):):

[img]http://www.yuklerim.com/uploads/6088345b69.gif[/img]


trigonometry
rotation
geometry
geometric transformation

1. $\sqrt {\frac{1 - \cos \theta}{1 + \cos \theta} = }$?

I get $\frac{1 - \cos \theta}{|\sin \theta|}$

But choices were:

A) $\frac{1}{2} \tan \theta$
B) $\cot \frac{\theta}{2}$
C) $\tan \frac{\theta}{2}$
D) $\frac{1}{2} \cot \theta$
E) None of above

2. If a square of side 12 is rotated in its plane about a point on one side 3 units from a vertex, the area of the figure swept out is:

A) 2500
B) $225 \pi$
C) $144 \pi$
D) $81 \pi$
E) None of above

3. A circular segment is cut from a regular pentagon, with one vertex as the center and a side of the pentagon as the radius of the circle. If the area of the segment is 17 then the length of each side of the pentagon is:

A) $\sqrt {17}$
B) $\frac{1}{2} \sqrt {17}$
C) $\frac{5 \pi}{17}$
D) $\sqrt {\frac{85}{\pi}}$
E) None of above

More will be posted.

Right idea with number 1.

[hide]$\tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta} = \frac{\sin \theta}{1+\cos \theta}$  is a pretty useful identity.  Prove it using the $\sin \frac{\theta}{2}$ and $\cos \frac{\theta}{2}$[/hide]

EDIT: under closer inspection, #1 would be easier to initially note the use of half angle identities based on the square roots and the cos + and -'s.  So actually using your method and then using it again but squaring the denominator is how you can prove that identity.

Hint for 2
[hide]Think about it, the region swept out is going to be a circle, and its radius will be the maximum distance of a point in the square at any given moment.[/hide]

Hint for 3
[hide]You've probably already drawn it and called the side s or r or something.  Since the pentagon is regular, you can find the degree measure of the angle of the sector and can thus determine its area in terms of the side length or radius.  Setting this equal to 17 and solving for the variable, you're good to go.[/hide]

For [b]1[/b], the answer is $E$.

Using half angle formula,....

indeed, use half angle identities, but I don't think it works out to E.

[hide]
$\sqrt {\frac{1 - \cos \theta}{1 + \cos \theta}} =$
$\frac{\sqrt{1 - \cos \theta}}{\sqrt{1 + \cos \theta}} =$
$\frac{\sqrt{\frac{1 - \cos \theta}{2}}}{\sqrt{\frac{1 + \cos \theta}{2}}} =$
$\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} =$
$\tan \frac{\theta}{2}$
[/hide]

You made a basic mistake. Notice $\sqrt{A^2}=|A|$ ;)

[hide="number2"]Notice that it's a circle. If the point of rotation is 3 on a side and 3 units away from a vertex, then a line that is the radius of the circle can be drawn that is $\sqrt{(12)^{2}+(12-3)^{2}}$, using the pythagorean theorem.  So the radius of the circel $15$, so the area is $225\pi$.[/hide]

For #3, what is the measure of an angle of a regular pentagon?

[hide]The answer to number 3 is None of the above.  Because each angle is 108 degrees, the area of the circular section is
(3/10)(s^2)pi
where s is the length of one side.  Let this equal to 17 then do a little algebra to get the answer sqrt(170/(3pi))
[/hide]


Two different  numbers are selected simultaneously and at random from the set $ \{1, 2, 3, 4, 5, 6, 7\}$. What is the probability that the positive difference between the two numbers is $ 2$ or greater? Express your answer as a common fraction.

We find that there are $ \binom{7}{2}\equal{}21$ possible ways to pick two numbers. The only time that it does not have a positive difference of $ 2$ or greater is if that difference is $ 1$. There are $ 6$ pairs of this. $ 1\minus{}\frac{6}{21}\equal{}\frac{15}{21}\equal{}\boxed{\frac{5}{7}}$.


I am given $ \lambda_{1} \equal{} \minus{}2$ and $ \lambda_{2} \equal{} \minus{}4$

and $ e_{1} \equal{} e_{2} \equal{}$\[ \left(\begin{array}{c}2 \\1 \\\end{array}\right)\]

Is it possible to find a matrix $ A$ such that $ A \equal{} PDP^{\minus{}1}$ from the given eigenvalues and eigenvectors of $ A$? $ P$ based on the eigenvalues is singular, so what other options do I have?

Thanks.

The same vector cannot be an eigenvector for two different eigenvalues. The statement above makes no sense.

Did you mean to write something else?

Thanks for the reply Kent. I am still curious, if no two different eigenvalue can share the same eigenvector, why is there no theorem detailing the uniqueness of eigenvectors, or at least some statement to that effect? Or is that implied by $ Ax \equal{} \lambda x$ ?

That is implied by $ Ax\equal{}\lambda x.$ It really is that simple.


[quote="shake9991"]how did you get from the:

$b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}$

to
$b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}$[/quote]

Since $b_{n+1}+\frac{1}{b_{n}}=b_{n}+\frac{1}{b_{n-1}}$, it turns out that the value of $b_{n+1}+\frac{1}{b_{n}}$ does not depend on $n$ at all. So we can find this value by plugging in $n=2$.

thanks... just realized that after your post.

wow, i am dumb as far as recursion is concerned... at least I have a day to improve :)

A3=672, so 672/3=224 is easy to guess, but hard to prove.

Not rigorous solution:
Evaluate the first few terms of the sequence: $3, 3, 672, \frac{672^2+2007}{3}.$ Compared to $672^2$, $2007$ is insignificant. So $a_{n+1}a_{n-1}\approx a_n^2$. From this we get $\frac{a_{n+1}}{a_n}\approx \frac{a_n}{a_{n-1}}$. We evaluate a few more terms of the sequence, and get $a_4\approx \frac{672^2}{3}, a_5 \approx \frac{672^3}{9}, a_6 \approx \frac{672^4}{27}$. It's clear that the "common ratio" is $\frac{672}{3} = 224$, so $\frac{a_{2007}}{a_{2006}}\approx 224$. The expression asked for is $\frac{a_{2007}}{a_{2006}} + \frac{a_{2006}}{a_{2007}}$, and the smallest integer equal to or less than this is $\boxed{224}$.


Starting 1:00 AM Friday, October 17 and ending at 1:00 AM Friday, October 24 is THE GAME. Hope you know the rules. You do not have to tell me, but if you want to, you can PM me that you are participating. Otherwise, I count everybody in. If you PM me, do it on Saturday, October 18. Good luck!

Who did you get your Generation 22 from?

Stop trying to post THE GAME. Have you noticed all the onees you posted are locked? That means the mods want you to stop.

Can't somebody like get this guy banned or something?
it's already something like the fourth or fifth time
what's the point of making other people lose the game anyway, you just lost the game too...

while i'm at it, i lost the game, good job

Just revoke his ability to make more topics in this forum...


function
trigonometry
real analysis
real analysis unsolved

Prove that a set of local extremes of an arbitrary real function is finite or countable.

Around each extrema there is an open interval with no more extrema, by definition. So we can choose a disjoint collection of open intervals, each containing one extrema. Since each such interval contains a rational number, it could not be an uncountable collection.

I mean, this is not true if you allow for weak extrema, where the function could be constant.

[quote="Kalle"]Around each extrema there is an open interval with no more extrema, by definition[/quote]
I am not sure. For example $ f(x) \equal{} x^2 \plus{} x \sin(\frac{1}{x})$ has one global extremum at $ x\equal{}0$, but there are local extrema arbitrary close to $ x\equal{}0$.

Yeah, that was a dumb mistake.  :blush: So we need to prove that we can choose an interval around an extrema such that there are only countably many more extremas in it, I guess

or just like you did: you choose one rational interval $ ]p_m, q_m[$ around each local minima $ m$ such that $ m$ is the global minimum of $ f$ over $ ]p_m, q_m[$. Because $ \mathbb{Q}^2$ is countable, that shows that the set of local minima is countable.


The altitude to the hypotenuse of a right triangle cuts it into segments of lengths $p$ and $q$ and $p < q$. If that altitude is $1/4$ the hypotenuse, then $p/q$ will equal $a - b\sqrt{3}$. Compute the ordered pair of positive integers $(a,b)$

-NYSML 1989 Team #5

[hide="Geometric mean"]
Note that the altitude to the hypotenuse of a right triangle is the geometric mean of the two segments the hypotenuse is divided into. So,
$\sqrt{pq}=\frac{p+q}{4} \iff 16pq=(p+q)^2 \iff 14pq=p^2+q^2 \iff \frac{14p}{q}=\frac{p^2}{q^2}+1$.
Setting $\frac{p}{q}=u$ we get $0=u^2-14u+1 \iff u= \frac{14 \pm \sqrt{ 14^2 - 4}}{2} = 7 \pm 4\sqrt 3$. Seeing that $u<1$, $\boxed{\frac{p}{q}=7-4\sqrt{3} \Longrightarrow (a,b)=(7,4)}$[/hide]

[quote="Altheman"][quote="cincodemayo5590"]am i correct?[/quote]

looks good...[/quote]

yay, lol


geometry
circumcircle
radical axis
power of a point
geometry proposed

Given a circle k and a triangle ABC. Let the circle k meet the line BC at $A_1$ and $A_2$, the line $CA$ at $B_1$ and $B_2$, and the line AB at $C_1$ and $C_2$. We construct the three circles with diameters $A_1 A_2$, $B_1 B_2$, $C_1 C_2$. Prove that the radical center of these three circles is the isogonal conjugate of the center of k with respect to triangle ABC.

  Darij

Really cool!

$AB_1\cdot AB_2=AC_1\cdot AC_2$, so $A$ lies on the radical axis of the circles $(B_1B_2),\ (C_1C_2)$. Let the center of $k$ be $K$. The projections $K_b,K_c$ of $K$ on $CA,AB$ are the centers of $(B_1B_2),\ (C_1C_2)$ respectively, so $K_bK_c\perp AK^*$, where $K^*$ is the radical center. This is the necessary and sufficient condition for the cevians $AK,AK^*$ to be isogonal conjugates. We do the same for $BK,BK^*$ and $CK,CK^*$, and we are done.

Oh, now I'm ashamed to post my proof, since it is around four times as long...

In fact, I considered the isogonal conjugate N of the point K with respect to the triangle ABC. Of course, I had to prove that this isogonal conjugate N has equal powers with respect to the circles with diameters $A_1A_2$, $B_1B_2$, $C_1C_2$. Well, I did this using some algebra (length of a median in a triangle) and the fact that the circumcenter of triangle $K_a K_b K_c$ is the midpoint of the segment KN. But now I see this all isn't necessary... thanks!

  Darij


Mandelbrot December 1990 Round 2 #8

If the polynomial $x^8 + 5x^6 + 13x^4 + 20x^2 + 36$ is evenly divisible by the trinomial $x^2 - x + a$, then determine the positive integer $a$.

[hide="Hint...."]$x^2 = x +a$ Then you can go from there.[/hide]

you can just do the long division and set the remainder equal to zero

Al, I can do long division, but that's too long. How does synthetic divison work?

Synthetic division only works for factors in the form of $\ (x-a)$

You can't synthetically divide a non-monic polynomial.

Ok, thanks..... I dozed off that Intermediate Albebra class.

In this problem the top factors after a while of examination.  Take every other term starting with $\ x^8$ and then take the two terms left over and factor to get $\ (x^4 + 4)(x^4 + 5x^2 + 9)$.  The expression on the left factors but it turns out not into a polynomial of the form $\ x^2 - x + a$, so we examine the right expression.  It can be written as $\ (x^2 + 3)^2 -x^2$, which can be factored using difference of two squares as $\ (x^2 +x + 3)(x^2 - x +3)$.  So $\ a=3$.


I did every step here :[url]http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Macintosh[/url] carefully. But it does not work at all. when I type in terminal "asy asyfile" I have this massage:
"-bash: asy: command not found"
what can i do to run asymptote?

If the /usr/local/bin directory is missing from your path, you need to specify the full path to asy:

/usr/local/bin/asy


calculus
derivative
function
binomial coefficients
real analysis
real analysis unsolved

It is known that for all positive integers n $|f^{(n)}(x)|\leq e^{-x}$.Prove that $f(x)=ce^{-x}, |c|\leq 1$.

I'd like to replace $e^{-x}$ with $e^{x}$, which does not change the problem. Let's write $f(x)=e^{x}g(x)$. Since $f^{(n)}(x)=e^{x}\sum_{k=0}^{n}\binom{n}{k}g^{(k)}(x)$, it follows that
\[\left| \sum_{k=0}^{n}\binom{n}{k}g^{(k)}(x)\right|\le 1,\quad \forall x\in\mathbb R,\ \forall n\ge 0\qquad (1)\]
The growth of binomial coefficients makes it easy to believe that all derivatives of $g$ must vanish. However, one cannot prove this by looking at just one point $x=x_{0}$. Indeed, for any sequence of real numbers $(\alpha_{k})$ there is a $C^{\infty}$ function $g$ with $g^{(k)}(x_{0})=\alpha_{k}$. The numbers $\alpha_{k}$ can be chosen so that (1) holds for every $n$.  :|


A solar system contains four planets and a sun as shown. Planet A completes an orbit around the sun in 20 months; planet B completes an orbit in 36 months; planet C completes an orbit in 40 months; and planet D completes an orbit in 48 months. If the four planets lie along a straight line in May of the year 2000, what is the next year that they will again lie along the same line? Assume that the orbits are circular and that a year has exactly 12 months.

We simply take the LCM of the four numbers. Using their prime factorizations, we get that their LCM is $ 2^4(3^2)(5)\equal{}720$. Thus, they will be in a straight line again in $ \frac{720}{12}\equal{}60$ years. That year will be $ \boxed{2060}$.

2030 the person above me forgot one thing -- they dont all have to be on the same sdie of the sun as long as they're in a straight line it works. so 60/2=30

Huh. Strange; they give the answer of 2120


Consider these two geoboard quadrilaterals.
[asy]/* AMC8 2000 #18 Problem */
for ( int x = 0; x <= 4; ++x )
{
    for ( int y = 0; y <= 4; ++y )
    {
        dot((x,y));
    }
}
draw((0,4)--(1,3)--(1,2)--(0,3)--cycle);
draw((2,1)--(4,2)--(3,1)--(3,0)--cycle);
label("I", (.5, 3));
label("II", (2.6, 0.9));[/asy]
Which of the following statements is true? \\ \\

(A) The area of quadrilateral I is more than the area of
quadrilateral II \\ \\
(B) The area of quadrilateral I is less than the area of quadrilateral II. \\ \\
(C) The quadrilaterals have the same area and the same perimeter. \\ \\
(D) The quadrilaterals have the same area, but the perimeter of I is more than the perimeter of II. \\ \\
(E) The quadrilaterals have the same area, but the perimeter of I is less than the perimeter of II. \\ \\
Select the most appropriate letter.

I and II have same area.

perimeter of I : $ 2\sqrt2 \plus{} 2$
perimeter of II : $ 2\sqrt2 \plus{} 1 \plus{} \sqrt {5}$

$ \sqrt5 > 1$

answer : e

EDIT : oops, basic latex mistake  :D  fixed.


induction
combinatorics unsolved
combinatorics

[b]

find the number of the ways to place on chess board NxN K rook so that they did not beat each other. All rooks are considered alike.

for example:
if n=8 and k=8, the answer will be 40320
if n=3 and k=2, the answer will be 18[/b]

Can you find formula

i think that if n=k then answer will be n!

For n=k:

We can set up first rook on $ k^2$ ways, second on $ (k\minus{}1)^2$,...,last on one square. But, as rooks are equal, we divide by $ n!$.
Total: $ (n!)^2: n!\equal{}n!$

The formula is:$ f(n,k)\equal{}\frac{(n!)^2}{k!.[(n\minus{}k)!]^2}$ for $ 1 \leq k \leq n$
Prove it by induction on k.

[quote="No Reason"]by induction on k.[/quote]  Why induct when it's at least as easy to do it directly (and more revealing): you need $ k$ columns and $ k$ rows, and then you've reduced to $ k$ rooks on a $ k\times k$ board.  So the answer is $ \binom{n}{k}^2 k!$, which is what you wrote.  Or, use Bugi's method to get $ \frac{(n_{(k)})^2}{k!}$ where $ n_{(k)} : \equal{} n(n \minus{} 1) \cdots (n \minus{} k \plus{} 1)$, which is also equal to your answer.


Zsigmondy
prime divisors
number theory
prime numbers
Divisibility
IMO Shortlist
power of 2

bn-1 is a Mersenne sequence, meaning that (bm-1, bn-1)=b(m,n)-1. Let d=(m,n). This means that bd-1 and bm-1 have the same prime divisors. Let bd=a, and m/d=M. Then a-1 and aM-1 have the same prime divisors.

Let p|aM-1 be a prime. Then p|a-1, so p|aM-a=a(aM-1-1) so p|aM-1. We continue like this until we get p|a2-1=(a-1)(a+1). We get that a-1 and (a-1)(a+1) have the same prime divisors. It's easy to get from here a+1=2t, so bd+1=2t. If d is even then bd is a perfect square which is 3(mod 4), which is false, so d is odd, so 2t=bd+1 is divisible with b+1, so b+1 is a power of 2.

I'll use a lemma: if $a>1,k>1$ and $a^k-1,a-1$ have the same prime factors, then $k$ is a power of $2$ and so is $a+1$.

Assume the contrary, and let $p$ be an odd prime factor of $k$. Then any prime factor $q$ of $1+a+\ldots+a^{p-1}$ also divides $1+a+\ldots+a^{k-1}|a^k-1$, so it divides $a-1$, meaning that $1+a+\ldots+a^{p-1}\equiv p\pmod q\equiv 0\pmod q$, so $q=p$, so $1+a+\ldots+a^{p-1}=p^t,\ t>1$. Put $a=up+1$ and we'll get a contradiction fast (we find $1+a+\ldots+a^{p-1}\equiv p\pmod{p^2}$). This shows that $k$ is a power of $2$. If we take $r$ to be a prime which divides $a+1$, then it will divide $a^k-1$, and thus $a-1$, so $r=2$, which is what we wanted.

Apply the lemma for $a=b^d$, where $d=(m,n)$, and $k=\frac md$. We find that all prime factors of $b^{kd}-1$ also divide $b^d-1$, where $k$ is a power of $2$. We also know that $b^d+1=2^t,\ t\ge 2$. Since a perfect square cannot be equal to $2^t-1,\ t\ge 2$, it means that $d$ is odd. Take a prime $r|b+1|b^2-1|b^{kd}-1$, so $r|b^d-1\equiv -2\pmod r$, so $r=2$, meaning that $b+1$ is a power of $2$, Q.E.D.

P.S. The fact that $k$ is a power of $2$ is only useful for showing that $kd$ is even.

It is nice and easy: One can show that if the prime factors of $A^n-1$ and $A^m-1$ coincide, then $n=2, m=1$ and $A+1=2^t$. And also using this one can show that If the prime factors of $A^m-1$ are a subset of the ones from $A^n-1$ then $m|n$ or $m=2$ and $A+1=2^t$ and a list of nice corollaries follows:

The nicest
Corollary:

$A_1^{x_1}-A_2^{x_2} A_2^{x_3}....A_n^{x_n}=1$ with $A_i$ given integers has at most one solution $(x_1,x_2...,x_n)$ with $x_1>1$ and $x_i>0$.

For example $3^x-2^y7^z=1$ has at most one solution $(x,y,z)$

Lets say m>n

Because
\[ (b^m \minus{} 1 , b^n \minus{} 1 ) \equal{} b^{(m,n)} \minus{} 1
\]
then $ b^{(m,n)} \minus{} 1$ and $ b^m \minus{} 1$ must have same prime divisors.
so now put $ (m,n) \equal{} \zeta , m \equal{} k \zeta , a \equal{} b^\zeta$   so now  it is clear that $ a \minus{} 1 , a^k \minus{} 1$  share same prime divisors.
now suppose $ P$ is one of  $ k$ prime divisors.
because $ a^p \minus{} 1 \mid a^k \minus{} 1 , a \minus{} 1 \mid a^p \minus{} 1$ ,   $ a \minus{} 1, a^p \minus{} 1$  also must have same prime divisors.
Now lets put : $ \alpha \equal{} a^{p \minus{} 1} \plus{} a^{p \minus{} 2} \plus{} ... \plus{} 1$
If  $ q$ be a prime divisor of $ \alpha$ , Then $ a \minus{} 1 \equiv 0 (mod q)$   , as a result, $ \alpha \equiv 1 \plus{} 1 \plus{} 1 \plus{} ... \plus{} 1 \equiv p$   so that means $ q \equal{} p$
and the obviously we can say that $ \alpha \equal{} p ^ \phi$
in order to find out what $ \phi$ can be , we write : $ a \equal{} pt \plus{} 1$  ( $ a \minus{} 1 \equiv 0, (mod p )$ )
now if we look closer we easily see that :
$ a^r \equiv (pt \plus{} 1)^r \equiv \sum(j \equal{} 0 ,\rightarrow r) (pt)^j * C(r,j) \equiv C(r,1)pt \plus{} 1 (mod p^2 )$
$ \Rightarrow \alpha \equiv \sum (r \equal{} 0 \rightarrow p \minus{} 1) rpt \plus{} 1 \equiv pt \frac {(p \minus{} 1)p} {2} \plus{} p \equiv p (mod p^2)$

now if $ p$  is prime and odd, $ \phi \equal{} 1 \rightarrow \equal{} a^p \minus{} 1 \plus{} ... \plus{} 1 \equal{} p$  and this contradicts  $ a > 1$  , therefore p should be even, and that is $ p \equal{} 2$
so prime divisors of $ a^2 \minus{} 1 , a \minus{} 1$  are the same.  but $ (a \plus{} 1) \minus{} (a \minus{} 1) \equal{} 2$
and so $ a \plus{} 1 \equal{} 2^ w$ ,,, inother words , $ b^\zeta \plus{} 1$ is a power of $ 2$ ,
and again ofcourse $ \zeta$ cant be even , because if be, then $ b^\zeta \plus{} 1 \equiv 2 (mod 4)$ and because $ b^\zeta \plus{} 1 \equal{} 2 ^ w$ we would conclude that $ b \equal{} 1$  (since b is odd)
therefore $ \zeta$ is odd and $ b \plus{} 1 \mid b^\zeta \plus{} 1$
and we are all set, $ b \plus{} 1$ is a power of $ 2$ .
 :ninja:

[i]A good solution ,Robert_89 , but there are too many mistakes in your solution

  I had tried my best to understand your solution

   In order to take profits from Mathlinks . You should pratise English more . :lol: [/i]

Without loss of generality, suppose $m < n$. If $b+1$ is not a power of 2 and $n \neq 6$, then by [url=http://en.wikipedia.org/wiki/Zsigmondy%27s_theorem]Zsigmondy's theorem[/url] we can find some prime $p$ such that $p | b^n - 1$ but $p \not | b^m - 1$, which is impossible. Furthermore, $2^1 - 1 = 1$, $2^2 - 1 = 3$, $2^3 - 1 = 7$, $2^4 - 1 = 3 \cdot 5$, $2^5 - 1 = 31$, and $2^6 - 1 = 3^2 \cdot 7$, so $2^6 - 1$ does not have the same set of prime factors as any of $2^i - 1$ with $1 \leq i \leq 5$. Hence, $n \neq 6$, so we may conclude that $b+1$ is a power of 2.

Note that Zsigmondy's theorem tells us that $b+1$ must be a power of two [i]and[/i] that $(m,n) = (1,2)$ or $(m,n) = (2,1)$.

[quote=sam-n]Let $ b, m, n$ be positive integers such that $ b > 1$ and $ m \neq n.$ Prove that if $ b^m \minus{} 1$ and $ b^n \minus{} 1$ have the same prime divisors, then $ b \plus{} 1$ is a power of 2.[/quote]

[hide=Solution]WLOG, assume $n>m$. Hence, by Zsigmondy theorem (excluding the exception cases), we have a prime divisor that divides $b^n-1$ but not $b^m-1$, but this contradicts the problem statement. Hence, we take the exception cases

Case 1 : $b=2,n=6$. It's easy to prove that this case fails, since when $1 \leq m \leq 5$, $2^m-1$ does not have all its prime divisors same as $2^6-1=63$

Case 2 : $b+1$ is a power of $2$, since the first case fails this case must hold true, hence the claim. $\blacksquare$[/hide]

Let $d = (m,n)$, with $m = dm_0, n = dn_0$ and $(m_0,n_0) = 1$. We have that $(b^n - 1, b^m - 1) = b^d - 1$, so we get that

$b^{dm_0} - 1 = (b^d - 1)\cdotM$
$b^{dn_0} - 1 = (b^d - 1)\cdotN$
Notice that if there is a new prime $p$ that divides $M$ but does not divide $b^d -1$, then it divides $b^{dm_0} - 1$, which due to the problem implies that this prime divides $b^{dn_0} - 1$. But this is would implie that $p$ divides $(b^{dm_0} - 1, b^_{dn_0} - 1) = b^d - 1$, which is a contradiction.
Hence, every prime that divides $b^d - 1$ must divide $b^{dm_0} - 1$ and vice-versa.

Bruhhh direct zsigmondy;
Let $m>n$ from zsigmondy's theorem there is must be non-common prime divisor. But zsigmondy does not work at these two case
Case 1: $b=2$ and $n=6$ trying for $m=1,2,3,4,5$ we get that no sol.
Case 2: $b+1=2^i$ from this statement we're done.

[quote=sam-n]Let $ b, m, n$ be positive integers such that $ b > 1$ and $ m \neq n.$ Prove that if $ b^m \minus{} 1$ and $ b^n \minus{} 1$ have the same prime divisors, then $ b \plus{} 1$ is a power of 2.[/quote]
WLOG $m>n$:
$gcd(b,1)=1$
By Zsigmondy's Theorem:
$\exists$ prime $p/ p|b^m-1, p\nmid b^n-1$
$(\Rightarrow \Leftarrow)$
But there is an exception:
$b+1=2^t_\blacksquare$

easy one ;)

Without losing generality, suppose that $m<n.$ Since $b^m-1$ and $b^n-1$ share the same prime divisors, there does not exist some prime $p$ such that $p\mid b^n-1$ and $p\nmid b^i-1$ for all $i\in\{1,2,\cdots,n-1\}$, as $m<n$ implies that $m\in\{1,2,\cdots,n-1\}.$ Thus, the triplet $(b,1,n)$ must be an exception to Zsigmondy's Theorem, implying that either $b+1$ is a power of $2$ and $n=2$ or $(b,n)=(2,6).$

If we have the latter case, then $b^n-1=2^6-1=63=3^2\cdot7.$ One can check that there does not exist an element of $\{2^1-1,2^2-1,\cdots,2^5-1\}$ such that its prime divisors are exactly $3$ and $7$. Thus this case fails, so it follows that $b+1=2^k$ for some positive integer $k$ and $n=2$, as desired. $\blacksquare$


factorial
algebra
polynomial
function
combinatorics unsolved
combinatorics

Do any of you know a [b]non-inductive[/b] proof of the following equation using Stirling's number:

$ (a \plus{} b)_n \equal{} \sum_{{j\equal{}0}}^n {n \choose j} (a)_{n\minus{}j}(b)_{j}$

where $ (a \plus{} b)_n$ is the Pochhammer symbol denotes the falling factorial, $ (a \plus{} b)_n \equal{} (a \plus{} b)(a \plus{} b \minus{} 1)(a \plus{} b \minus{} 2)...(a \plus{} b \minus{} n \plus{} 1)\equal{}\sum_{{k\equal{}0}}^n{s(n,k)}x^k$, where $ s(n,k)$ is the stirling number of the first kind.

Thank you very much. :P

fix $ n$ and let $ a,b\geq n$ be integers. the LHS then counts the number of ways to take $ n$ objects from a group of $ a \plus{} b$ objects -- say $ a$ lamps and $ b$ chairs -- and line them up. the RHS counts the same thing, according to how many of the objects are taken to be chairs: if $ j$, $ 0\leq j\leq n$, of the objects are taken to be chairs, there are $ \binom{n}{j}$ ways we can choose the $ j$ places in the line that are to be occupied by chairs, $ (b)_{j}$ ways to arrange chairs in those places, and then $ (a)_{n \minus{} j}$ ways to arrange lamps in the remaining places in the line.

now, fix an integer $ b\geq n$ and consider the polynomials $ P(x,b) \equal{} (x \plus{} b)_n$ and $ Q(x,b) \equal{} \sum_{j \equal{} 0}^n\binom{n}{j}(x)_{n \minus{} j}(b)_j$. by the above argument, $ P(x,b)$ and $ Q(x,b)$ take the same value for infinitely many integers $ x$, and thus, since $ P(x,b)$ and $ Q(x,b)$ only have degree $ n$ as polynomials in $ x$, we must have $ P(x,b) \equal{} Q(x,b)$ for all real $ x$. since $ b$ was chosen arbitrarily, the equality $ P(x,b) \equal{} Q(x,b)$ in fact holds for all $ x$ and all integers $ b\geq n$. now fix $ x$ and view $ P(x,b)$ and $ Q(x,b)$ as polynomials in $ b$. since $ P(x,b) \equal{} Q(x,b)$ for infinitely many integers $ b$, and $ P(x,b)$ and $ Q(x,b)$ only have degree $ n$ as polynomials in $ b$, we must have $ P(x,b) \equal{} Q(x,b)$ for all real $ b$ and the fixed value $ x$. but since $ x$ was chosen arbitrarily, $ P(x,b) \equal{} Q(x,b)$ in fact holds for all real $ x,b$. and this completes the proof.

The exponential generating function of the sequence $ (a)_n$ is
\[ \sum_{i = 0}^\infty \frac{(a)_i}{i!} \: x^i = \frac{1}{0!} + \frac{a}{1!} \: x + \frac{a(a - 1)}{2!} \: x^2 + \frac{a(a - 1)(a - 2)}{3!} \: x^3 + \dots = (1 + x)^a.\]
Similarly, the exponential generating function of the sequence $ (b)_n$ is
\[ \sum_{j = 0}^\infty \frac{(b)_j}{j!} \: x^j = \frac{1}{0!} + \frac{b}{1!} \: x + \frac{b(b - 1)}{2!} \: x^2 + \frac{b(b - 1)(b - 2)}{3!} \: x^3 + \dots = (1 + x)^b.\]
Therefore,
\[ (1 + x)^{a + b} \&= (1 + x)^a (1 + x)^b = \left( \sum_{i = 0}^\infty \frac{(a)_i}{i!} \: x^i \right) \left( \sum_{j = 0}^\infty \frac{(b)_j}{j!} \: x^j \right).\]

But
\[ (1 + x)^{a + b} = \sum_{n = 0}^\infty \frac{(a + b)_n}{n!} \: x^n.\]
Taking the coefficient of $ x^n$ in both expressions, we get
\[ \frac{(a + b)_n}{n!} = \sum_{i = 0}^n \frac{(a)_i}{i!} \cdot \frac{(b)_{n - i}}{(n - i)!},\]
so
\[ (a + b)_n = \sum_{i = 0}^n \frac{n!}{i! (n - i)!} \: (a)_i (b)_{n - i} = \sum_{i = 0}^n \binom{n}{i} (a)_i (b)_{n - i}.\]

Thank you very much, pleurestique and nsato. :P

By the way, anyone know how to prove the relationship $ (x)_n \equal{} \sum_{{k \equal{} 0}}^n{s(n,k)}x^k$, where $ s(n,k)$ is the stirling number of the first kind?

Thank you very much.  :P

a counting argument, using the definition of $ s(n,k)$, should do the trick for this other identity.


[color=purple] Let an acute  triangle ABC in plane prove that :
    $ \frac{h_{a}}{l_{a}} \plus{} \frac {h_{b}}{l_{b}} \plus{} \frac {h_{c}}{l_{c}}\ge 2 \plus{} \frac {2r}{R}$
  Hope all female members of mathlinks a nice day  :D [/color]

[quote="evarist"][color=purple] Let an acute  triangle ABC in plane prove that :
    $ \frac {h_{a}}{l_{a}} \plus{} \frac {h_{b}}{l_{b}} \plus{} \frac {h_{c}}{l_{c}}\ge 2 \plus{} \frac {2r}{R}$
  Hope all female members of mathlinks a nice day  :D [/color][/quote]

The same inequalities:
Prove that in triangle ABC :

$ \frac{h_a}{l_a}\plus{}\frac{h_b}{l_b}\plus{}\frac{h_c}{l_c} \geq 3 \sqrt[3]{\frac{2r}{R}}$

To N. Thảo my love!

[quote="evarist"][color=purple] Let an acute  triangle ABC in plane prove that :
    $ \frac {h_{a}}{l_{a}} \plus{} \frac {h_{b}}{l_{b}} \plus{} \frac {h_{c}}{l_{c}}\ge 2 \plus{} \frac {2r}{R}$
  Hope all female members of mathlinks a nice day  :D [/color][/quote]
 [color=purple] We have the following line for acute triangle:  $ \frac{12}{5}\plus{}\frac{6r}{5R}\ge\frac {h_{a}}{l_{a}} \plus{} \frac {h_{b}}{l_{b}} \plus{} \frac {h_{c}}{l_{c}}\ge 2 \plus{} \frac {2r}{R}$
 :) [/color]


trigonometry
inequalities
Euler
logarithms
LaTeX
pigeonhole principle

Το αγαπημένο πεδίο ασκήσεων σε πολλούς από σας.

 [color=red][b]        Καλή αρχή ! [/b][/color]

Επειδή δε σας άρεσε η συνδυαστική, ομολογουμένως είναι λιγάκι άχαρη, και επειδή έχουμε πολλές γεωμετρίες, να μια εύκολη ανισότητα με μια μικρή παραλλαγή.

Δείξτε ότι:

$ cbrt(tanx\plus{}6*coty)\plus{}cbrt(tany\plus{}6*cotz)\plus{}cbrt(tanz\plus{}6*cotx)\leq tanx*tany*tanz$

για χ, ψ, ζ γωνίες τριγώνου

ΥΓ: cbrt=3η ρίζα  :wink:

Καταρχήν για να γράψεις τρίτη ρίζα του α ας πούμε κάνε το εξής γράψε \sqrt[3]{a} και το τρίγωνο πρέπει να είναι οξυγώνιο
Τώρα θέτουμε $ \cot x \equal{} a$ , $ \cot y \equal{} b$ , $ \cot z \equal{} c$ και έχουμε νδο

[color=red][b]ΑΣΚΗΣΗ 1[/b][/color]

α)  Αν  $ a,b,c > 0$  και $ ab\plus{}bc\plus{}ca \equal{} 1$ , να αποδείξετε ότι :
$ \sqrt [3]{\frac{1}{a}\plus{}6b}\plus{}\sqrt [3]{\frac{1}{b}\plus{}6c}\plus{}\sqrt [3]{\frac{1}{c}\plus{}6a}\leq\frac{1}{abc}$
β)  Να αποδείξετε ότι :

$ \sqrt[3]{tanx\plus{}6*coty}\plus{}\sqrt[3]{tany\plus{}6*cotz}\plus{}\sqrt[3]{tanz\plus{}6*cotx }\leq{tanx*tany*tanz}$

  όπου $ x , y , z$  είναι γωνίες  οξυγωνίου τριγώνου .

[b] ΛΥΣΗ[/b]

Από Holder έχουμε ότι :

$ \sqrt [3]{\frac{1}{a}\plus{}6b}\plus{}\sqrt [3]{\frac{1}{b}\plus{}6c}\plus{}\sqrt [3]{\frac{1}{c}\plus{}6a}\leq\frac{3}{\sqrt [3]{3}}\sqrt [3]{\frac{ab\plus{}bc\plus{}ca}{abc}\plus{}6(a\plus{}b\plus{}c)}$

Άρα μένει να αποδείξουμε ότι $ \frac{3}{\sqrt [3]{3}}\sqrt [3]{\frac{ab\plus{}bc\plus{}ca}{abc}\plus{}6(a\plus{}b\plus{}c)}\leq\frac{1}{abc}$
ή ισοδύναμα
$ 9a^{2}b^{2}c^{2}\plus{}54(a\plus{}b\plus{}c)(abc)^{3}\leq 1$ (1)

Όμως
 $ (ab\plus{}bc\plus{}ca)^{2}\geq 3abc(a\plus{}b\plus{}c)$

Άρα $ abc(a\plus{}b\plus{}c)\leq\frac{1}{3}$ (2) και από ΑΜ-ΓΜ $ a^{2}b^{2}c^{2}\leq\frac{1}{27}$ (3)

Επομένως  , από  (2) , (3)  παίρνουμε

 $ (abc)^{3}(a\plus{}b\plus{}c)\leq\frac{1}{81}$ .Άρα

 $ 54(abc)^{3}(a\plus{}b\plus{}c)\leq\frac{54}{81}\equal{}\frac{2}{3}$ (4)

kai $ 9a^{2}b^{2}\plus{}c^{2}\leq\frac{1}{3}$ (5)

(4)+(5)=> (1) , το ζητούμενο

 β) Είναι εφαρμογή του α) αν θέσουμε  $ \cot x \equal{} a$ , $ \cot y \equal{} b$ , $ \cot z \equal{} c$

[i]Γιατί βρε χάλασες την άσκηση  :P[/i]

Ευχαριστώ για το ρίζα 3. :)  Ε, εντάξει απλά γιατί στην αρχή έτσι προσπάθησα να τη λύσω (με τριγωνομετρία) και δε μου βγήκε και μετά την έλυσα με ανισότητα δυνάμεων. Ουσιαστικά ανάγεται κατ'ευθείαν... :P

Παραθέτω μια ωραία εύκολη άσκηση αλλά λίγο παράξενη .
[color=red]ΑΣΚΗΣΗ 2[/color]
Αν $ a,b,c,d > 0$ πραγματικοί αριθμοί και $ d \equal{}\min{{a,b,c,d}}$ να δειχθεί ότι
$ a^{4}\plus{}b^{4}\plus{}c^{4}\plus{}d^{4}\minus{}4abcd\geq 4d((a\minus{}d)^{3}\plus{}(b\minus{}d)^{3}\plus{}(c\minus{}d)^{3}\minus{}3(a\minus{}d)(b\minus{}d)(c\minus{}d))$

Έστω $ a\minus{}d\equal{}x, b\minus{}d\equal{}y, c\minus{}d\equal{}z$ με $ x,y,z\geq 0$. Τότε η προς απόδειξη ανισότητα ισοδυναμεί με την
$ \sum x^{4}\plus{}2d^{2}\sum x^{2}\plus{}2d^{2}\sum (x\minus{}y)^{2}\plus{}8xyzd\geq 0$ που προφανώς ισχύει. H ισότητα ισχύει όταν $ x\equal{}y\equal{}z\equal{}0$.

Αυτό είναι, πολύ ωραία ... :)   . Να μια ακόμα καλή

[color=red]ΑΣΚΗΣΗ 3[/color]
Έστω $ a,b,c$ μη αρμητικοί αριθμοί ώστε δύο από αυτούς να μην είναι συγχρόνως 0 .
Να δειχθεί ότι
$ \frac{1}{b^{2}\minus{}bc\plus{}c^{2}}\plus{}\frac{1}{c^{2}\minus{}ca\plus{}a^{2}}\plus{}\frac{1}{a^{2}\minus{}ab\plus{}b^{2}}\geq\frac{3}{ab\plus{}bc\plus{}ca}$

Ωραία άσκηση, silouan :) .
Χωρίς βλάβη της γενικότητας υποθέτουμε ότι $ a\geq b\geq c$. Έστω $ f(a, b, c)\equal{}\sum\frac{1}{a^{2}\minus{}ab\plus{}b^{2}}\minus{}\frac{3}{ab\plus{}bc\plus{}ca}$. Εύκολα μπορούμε να δείξουμε ότι $ f(a, b, c)\geq f(a\plus{}c, b, 0)$. Αρκεί επομένως $ f(a\plus{}c, b, 0)\geq 0$. Όμως αυτή είναι ισοδύναμη με την $ \frac{(a\plus{}c\minus{}b)^{4}}{(a\plus{}c)^{2}b^{2}[(a\plus{}c)^{2}\minus{}(a\plus{}c)b\plus{}b^{2}]}\geq 0$, που ισχύει.

Τελικά άλλος έχει το όνομα και άλλος τη χάρη  :wink:  . Πολύ καλά Michael ....
Ακόμη μία
[color=red]ΑΣΚΗΣΗ 4[/color]
Αν $ a,b,c,d>0$ πραγματικοί αριθμοί ώστε $ abcd\equal{}1$ να δείξετε ότι
$ (1\plus{}a^{2})(1\plus{}b^{2})(1\plus{}c^{2})(1\plus{}d^{2})\geq (a\plus{}b\plus{}c\plus{}d)^{2}$

Κάνοντας τις πράξεις παίρνουμε ότι η αποδεικτέα είναι ισοδύναμη με την $ 2\plus{}\sum a^{2}b^{2}\plus{}\sum a^{2}b^{2}c^{2}\geq 2\sum ab$, δηλαδή με την $ \sum (ab\minus{}1)^{2}\plus{}(\sum a^{2}b^{2}c^{2}\minus{}4)\geq 0$, που ισχύει αφού από $ AM\minus{}GM$ έχουμε $ \sum a^{2}b^{2}c^{2}\geq 4\sqrt[4]{a^{6}b^{6}c^{6}d^{6}}\equal{}4$. Η ισότητα ισχύει για $ a\equal{}b\equal{}c\equal{}d\equal{}1$.

Α δεν θα τα πάμε καλά .... :)
An $ a,b,c>0$ με $ abc\equal{}1$ vdo
$ \frac{1}{(1\plus{}a)^{2}}\plus{}\frac{1}{(1\plus{}b)^{2}}\plus{}\frac{1}{(1\plus{}c)^{2}}\plus{}\frac{2}{(1\plus{}a)(1\plus{}b)(1\plus{}c)}\geq 1$

Αν λυθεί και αυτή θα βάλλω καμιά πυρινική μετά  :P

Πολύ καλή άσκηση...  :) . Βρήκα μια περίεργη σχετικά λύση.

Θέτουμε $ a \equal{}\frac{1\minus{}x}{1\plus{}x}, b \equal{}\frac{1\minus{}y}{1\plus{}y}, c \equal{}\frac{1\minus{}z}{1\plus{}z}$ με $ \minus{}1 < x, y, z < 1$. Τότε η συνθήκη γράφεται $ x\plus{}y\plus{}z\plus{}xyz\equal{}0$ και η προς απόδειξη $ \sum x^{2}\plus{}2\sum x\plus{}\sum xy\geq 0$. Tώρα $ 2$ τουλάχιστον από τους $ x, y, z$ θα έχουν το ίδιο πρόσημο. Έστω ότι αυτοί είναι οι $ x, y$. Tότε $ xy\geq 0$. Όμως με βάση τη συνθήκη
\[ LHS\equal{}\sum x^{2}\plus{}\sum xy\minus{}2xyz\equal{}\sum x^{2}\plus{}xy\plus{}z(x\plus{}y)\minus{}2xyz\equal{}\sum x^{2}\plus{}xy\minus{}z^{2}(1\plus{}xy)\minus{}2xyz\equal{}x^{2}\plus{}y^{2}\plus{}xy\minus{}xyz^{2}\minus{}2xyz\geq 3xy\minus{}xyz^{2}\minus{}2xyz\equal{}xy(3\minus{}2z\minus{}z^{2})\equal{}xy(1\minus{}z)(z\plus{}3)\geq 0\]

Καλά φυσικά δε συγκρίνεται με την προηγούμενη λύση που είναι πανέξυπνη αλλά αν απλά κάνουμε απαλοιφές και τις πράξεις (lol) :lol: καταλήγουμε στο ΑΜ-GMκό $ a^{2}\plus{}b^{2}\plus{}c^{2}\geq 3$.

Θέλω να πω ότι οι άλλες δεν έβγαιναν έτσι brutally, ειδικά η πρώτη με το $ dmin$ μου φάνηκε ότι ήθελε πολύ καλή γνώση ανισοτήτων...

Αλήθεια Σιλουανέ πού βρήκες (σε ποιο βιβλίο) αυτή με το $ dmin$? Ήταν από τις καλύτερες που έχω δει τελευταία.

ΑΣΚΗΣΗ 6
(εύκολη για σας)

a, b, c > 0
Να δειχθεί ότι:

$ sqrt{(b\plus{}c)}/a\plus{}sqrt{(c\plus{}a)}/b\plus{}sqrt{(a\plus{}b)}/c\geq 4*(a\plus{}b\plus{}c)/sqrt{((a\plus{}b)*(b\plus{}c)*(c\plus{}a))}$

Σόρρυ αλλά χωρίς το Λάτεξ έτσι είναι το καλύτερο που μπορώ να γράψω.

Βάζω μια σχετικά μακροσκελή λύση με αρκετούς μετασχηματισμούς.

Έστω $ a\plus{}b \equal{} x^{2}, b\plus{}c \equal{} y^{2}, c\plus{}a \equal{} z^{2}$. Τότε oι αριθμοί $ x, y, z$ αποτελούν πλευρές τριγώνου (οξυγωνίου) αφού το ίδιο ισχύει για τα τετράγωνά τους και η προς απόδειξη γράφεται $ \sum\frac{x}{y^{2}\plus{}z^{2}\minus{}x^{2}}\geq\frac{\sum x^{2}}{xyz}$. Με βάση τώρα τους νόμους ημιτόνων και συνιμητόνων $ x \equal{} 2R\sin X,\cos X \equal{}\frac{y^{2}\plus{}z^{2}\minus{}x^{2}}{2yz}$ η αποδεικτέα παίρνει τελικά τη μορφή $ \sum (\frac{1}{\cos X}\plus{}\frac{1}{2}\minus{}\cos X\plus{}(\cos X)^{2})\geq\frac{15}{2}$, δηλαδή $ \frac{3}{4}\sum\frac{1}{\cos X}\plus{}\sum (2(\cos X)^{2}\plus{}\frac{1}{4\cos X}\plus{}\frac{1}{2}\minus{}\cos X)\geq\frac{15}{2}$. Έχουμε όμως $ \sum\frac{1}{\cos X}\geq\frac{9}{\sum\cos X}\geq 6$, αφού για τις γωνίες τριγώνου ισχύει ότι $ \sum\cos X\leq\frac{3}{2}$, και $ 2(\cos X)^{2}\plus{}\frac{1}{4\cos X}\plus{}\frac{1}{2}\minus{}\cos X\geq\cos X\plus{}\frac{1}{4\cos X}\geq 1$, από τις οποίες προκύπτει το ζητούμενο.

[quote="savvas_michael"]
Έστω $ a\plus{}b \equal{} x^{2}, b\plus{}c \equal{} y^{2}, c\plus{}a \equal{} z^{2}$. Τότε oι αριθμοί $ x, y, z$ αποτελούν πλευρές τριγώνου (οξυγωνίου) αφού το ίδιο ισχύει για τα τετράγωνά τους και η προς απόδειξη γράφεται $ \sum\frac{x}{y^{2}\plus{}z^{2}\minus{}x^{2}}\geq\frac{\sum x^{2}}{xyz}$[/quote]

Έχω βρει μια άλλη λύση μόνο Cauchy kai AM-GM αλλά προς το παρόν θα βάλλω μια διαφορετική συνέχεια στην προσέγγιση του Μιχάλη.
Here it goes
Από Cauchy Swartz έχουμε $ \sum\frac{x}{y^{2}\plus{}z^{2}\minus{}x^{2}}\geq\frac{(x\plus{}y\plus{}z)^{2}}{\sum x(y^{2}\plus{}z^{2})\minus{}\sum x^{3}}$
Επομένως αρκεί νδο

$ \frac{(x\plus{}y\plus{}z)^{2}}{\sum x(y^{2}\plus{}z^{2})\minus{}\sum x^{3}}\geq\frac{\sum x^{2}}{xyz}$

ή ισοδύναμα αρκεί νδο $ x^{5}\plus{}y^{5}\plus{}z^{5}\plus{}xyz(x^{2}\plus{}y^{2}\plus{}z^{2})\geq xy(x^{3}\plus{}y^{3})\plus{}yz(y^{3}\plus{}z^{3})\plus{}zx(z^{3}\plus{}x^{3})$
που ισχύει αφού είναι η ανισότητα Schur  :)

Το μόνο που θέλει είναι να δούμε ότι $ 4tRr\equal{}abc$ και μετά με SOS

$ \sum_{cyc}\frac {\left((a \minus{} b)^2 \plus{} c^2\right)^3}{c^3} \geq 8t(t^2 \minus{} 12Rr)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {(a \minus{} b)^6}{c^3} \plus{} \frac {3(a \minus{} b)^4}{c}\right)\geq0.$

qed  :D

Α το φαντάστηκα ότι θα βγαίνει με ΣΟΣ γιατί στηρίχτηκα στο λήμμα mildorf για κ=3. :wink:  Μπράβο πάντως... :)

[quote="Protonios"]Α το φαντάστηκα ότι θα βγαίνει με ΣΟΣ γιατί στηρίχτηκα στο λήμμα mildorf για κ=3. :wink:  Μπράβο πάντως[/quote]

Κώστα θα μπορούσες να μας πεις τη λύση σου?

Οταν λες [i]λήμμα mildorf[/i] εννοείς το:

$ \frac{(1\plus{}k)(a\minus{}b)^2\plus{}8ab}{4(a\plus{}b)}\geq\sqrt[k]{\frac{a^k\plus{}b^k}{2}}$,  $ k\geq\minus{}1$    :?:

γιατί μόνο αυτό έχω υπόψη αλλά δεν βλέπω να κολλάει πουθενά...(ίσως είναι και κάτι σχετικό με τη Schur αλλά δεν είμαι σίγουρος...)

Mπράβο για την άσκηση!

Ναι για k=3 παίρνει τη μορφή $ (x^2\plus{}y^2/x\plus{}y)^3>\equal{}(x^3\plus{}y^3)/2$
Εφαρμόζεις κυκλικά για τις 3 μεταβλητές και θέτεις $ x\plus{}y\equal{}a$ κλπ. :wink:  Γι' αυτό βγαίνουν πλευρές τριγώνου. :)  Σόρρυ για την καθυστέρηση και για την περιληπτικότητα. :(

Υπολογίστε το ακέραιο μέρος του:

$ A \equal{} \frac {1}{\sqrt {2}} \plus{} \frac {1}{\sqrt {3}} \plus{} \ldots \plus{} \frac{1}{\sqrt{10^4}}$

Πολύ καλή άσκηση Στέλιο,

Ισχύει $ \sqrt{n\plus{}1}\minus{}\sqrt{n}\equal{}\frac{1}{\sqrt{n\plus{}1}\plus{}\sqrt{n}}$ (απόδειξη εύκολη με συζυγείς).

Όμως $ \frac{1}{2\sqrt{n\plus{}1}} < \frac{1}{\sqrt{n\plus{}1}\plus{}\sqrt{n}} < \frac{1}{2\sqrt{n}}$, άρα

$ \frac{1}{2\sqrt{n\plus{}1}} < \sqrt{n\plus{}1}\minus{}\sqrt{n} < \frac{1}{2\sqrt{n}}$

Από την $ \frac{1}{2\sqrt{n\plus{}1}} < \sqrt{n\plus{}1}\minus{}\sqrt{n}$, προσθέτωντας για $ n\equal{}1$ έως $ n\equal{}10^4\minus{}1$ παίρνουμε

$ \frac{1}{2}A < 99$ κι έτσι $ \boxed{ A < 198, \ \ (1)}$.

Από την $ \sqrt{n\plus{}1}\minus{}\sqrt{n} < \frac{1}{2\sqrt{n}}$, προσθέτωντας για $ n\equal{}2$ έως $ n\equal{}10^4$ παίρνουμε

$ \frac{1}{2}A > \sqrt{10^4\plus{}1} \minus{}\sqrt{2} >$ κι έτσι $ A > 2\left(\sqrt{10^4\plus{}1} \minus{}\sqrt{2}\right) > 197$ άρα $ \boxed{ A > 197, \ \ (2)}$.

Τελικά λοιπόν $ 197 < A < 198$ δηλαδή το ακέραιο μέρος του $ A$ είναι το $ 197$.

Αλέξανδρος

Μια και είμαστε πριν από τον Ευκλείδη σας αφήνω μία ενδιαφέρουσα άσκηση για εξάσκηση:

Εαν $ x_1,x_2,x_3 \in [a,b]$ όπου $ 0<a<b$ να αποδείξετε ότι

$ (x_1\plus{}x_2\plus{}x_3)\left(\frac{1}{x_1}\plus{}\frac{1}{x_2}\plus{}\frac{1}{x_3}\right) \leq \frac{9(a\plus{}b)^2}{4ab}$

Αλέξανδρος

Η ανισότητα είναι ειδική περίπτωση της Katrovici -Sweitcher .

Η συνάρτηση του πρώτου μέλους είναι κυρτή ως προς κάθε μεταβλητή της επομένως παρουσιάζει ακρότατο στα άκρα του διαστήματος που ορίζεται .Έστω λοιπόν ότι $ k$ μεταβλητές είναι ίσες με $ a$ και τότε $ 3 \minus{} k$ μεταβλητές είναι ίσες με $ b$

Τότε αρκεί νδο $ (ka \plus{} (3 \minus{} k)b)(\frac {k}{a} \plus{} \frac {3 \minus{} k}{b})\leq\frac {9(a \plus{} b)^2}{4ab}$

ή ισοδύναμα $ (a \minus{} b)^2(2k \minus{} 3)^2\geq 0$ που πράγμτι ισχύει άρα ιανισότητα αποδείχθηκε  :D

Πολύ σωστά Σιλουανέ! Είσαι άψογος! (Φυσικά με την συγκεκριμένη ανισότητα, η παραπάνω γενικεύεται και για $ n$ μεταβλητές).

Θα ήθελα όμως μία λύση απλή η οποία να μην βάζει σκληρά μέσα της ανάλυσης! Έχω μία τέτοια υπόψιν και γι'αυτό την κατέθεσα σαν άσκηση προετοιμασίας για Ευκλείδη! Μην αγχώσουμε και αυτούς που δεν γνωρίζουν αυτές τις ανισότητες την τελευταία στιγμή!! Γι'αυτό προσπαθήστε να βρείτε μία απλούστερη εξήγηση για την παραπάνω ανισότητα για την δική σας εξάσκηση! Νομίζω ότι οι ασκήσεις αυτού του είδους είναι πολύ πρωτότυπες για διαγωνισμούς και παράλληλα εξετάζουν την κριτική σκέψη του διαγωνιζόμενου χωρίς να χρειάζεται η απομνημόνευση πολλών γνωστών ανισοτήτων. Αυτό δεν είναι και το νόημα των διαγωνισμών?

Φιλικά,
Αλέξανδρος

Παραθέτω μια διαφορετική λύση στο παραπάνω πρόβλημα, νομίζω ενδιαφέρουσα καθώς η λύση του είναι κατανοητή και από κάποιο καλό μαθητή της Α' Λυκείου που δεν γνωρίζει ανισότητες.

Για κάθε αριθμό $ x\in [a,b]$, το τριώνυμο με ρίζες τα $ a,b$, δηλαδή το $ x^2-(a+b)x+ab$ είναι $ \leq0$. Άρα για $ x>0$ παίρνουμε

$ x+ab\frac{1}{x}\leq a+b$, συνεπώς αφού $ x_1,x_2,x_3\in [a,b]$ άρα

$ x_1+ab\frac{1}{x_1}\leq a+b$
$ x_2+ab\frac{1}{x_2}\leq a+b$
$ x_3+ab\frac{1}{x_3}\leq a+b$

οπότε λόγω της γνωστής ανισότητας $ 2\sqrt{ab}\leq a+b$ ισχύει

$ \begin{eqnarray}
2\sqrt{ab(x_1+x_2+x_3)\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\right)} &\leq& (x_1+x_2+x_3) + ab\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\right) \nonumber \\ &\leq& 3(a+b)$

Με ύψωση στο τετράγωνο της παραπάνω παίρνουμε το ζητούμενο!

Ελπίζω να σας άρεσε! Καλή επιτυχία και στον Ευκλείδη!!

Αλέξανδρος

Να μια ανισότητα που κατασκεύασα σήμερα επιπέδου πιστεύω μάλλον Αρχιμήδη-Μεσογειακού.

Έστω οξυγώνιο τρίγωνο στο επίπεδο $ ABC$ με πλευρές $ a,b,c$ και αντίστοιχες διαμέσους $ m1,m2,m3$. Να βρεθεί η ελάχιστη σταθερά $ k$ ώστε να ισχύει:

$ [a(2m1 \minus{} a)(2m1 \plus{} a)](^1/4) \plus{} [b(2m2 \minus{} b)(2m2 \plus{} b)](^1/4) \plus{} [c(2m3 \minus{} c)(2m3 \plus{} c)](^1/4) < \equal{} k(2abc)(^1/4)$

Αυτό το 1/4 εννοεί 4η ρίζα. Συγγνώμη για τη μορφή.

[quote="Protonios"]Να μια ανισότητα που κατασκεύασα σήμερα επιπέδου πιστεύω μάλλον Αρχιμήδη-Μεσογειακού.

Έστω οξυγώνιο τρίγωνο στο επίπεδο $ ABC$ με πλευρές $ a,b,c$ και αντίστοιχες διαμέσους $ m1,m2,m3$. Να βρεθεί η ελάχιστη σταθερά $ k$ ώστε να ισχύει:

$ \sqrt[4]{a(2m_1 \minus{} a)(2m_1 \plus{} a)} \plus{}\sqrt[4]{b(2m_2 \minus{} b)(2m_2 \plus{} b)} \plus{} \sqrt[4]{c(2m_3 \minus{} c)(2m_3 \plus{} c)}\leq \sqrt[4]{k(2abc)}$

[/quote]

[color=red][size=150]ΚΑΛΗ ΕΠΙΤΥΧΙΑ ΣΕ ΟΛΟΥΣ ΓΙΑ ΤΟΝ ΑΡΧΙΜΗΔΗ!!![/size] [/color]:)

[quote="r_boris"][quote="bilot4"][quote="Protonios"]Να μια ανισότητα που κατασκεύασα σήμερα επιπέδου πιστεύω μάλλον Αρχιμήδη-Μεσογειακού.

Έστω οξυγώνιο τρίγωνο στο επίπεδο $ ABC$ με πλευρές $ a,b,c$ και αντίστοιχες διαμέσους $ m1,m2,m3$. Να βρεθεί η ελάχιστη σταθερά $ k$ ώστε να ισχύει:

$ \sqrt [4]{a(2m_1 \minus{} a)(2m_1 \plus{} a)} \plus{} \sqrt [4]{b(2m_2 \minus{} b)(2m_2 \plus{} b)} \plus{} \sqrt [4]{c(2m_3 \minus{} c)(2m_3 \plus{} c)}\leq \sqrt [4]{k(2abc)}$

[/quote]

[color=red][size=150]ΚΑΛΗ ΕΠΙΤΥΧΙΑ ΣΕ ΟΛΟΥΣ ΓΙΑ ΤΟΝ ΑΡΧΙΜΗΔΗ!!![/size] [/color]:)[/quote]
Μηπως καταληγεις να ζητας το μέγιστο του$ \sqrt {cos(A/2)}$+ κυκλικά και μετά Jensen για την κοίλη ρίζα , με γνωστό το max του $ \Sigma cos(A/2)$?ή κάνω λάθος? γιατί βιάζομαι[/quote]
typos όπου Α/2 βάλε Α και κυκλικά

Κύριε Μπόρη αυτό ακριβώς εννοώ. Συγγνώμη που δεν απάντησα νωρίτερα. :)


superior algebra
superior algebra solved

Let $A$ be a commutative ring.Show that $A$ has idempotent elements other than $0$ and $1$  if  $A$ is isomorphic with a ring of the type $B\times C$ where $B$ and $C$ are $2$ rings(not reduced to $0$).

I have a question:

By $BxC$ do you mean $B\times C$, i.e. the direct product of the two rings?

just let $e_1^2=e_1$ be an idempotent, let $e_2=1-e_1$. it is easy to verify that $e_1e_2=0$, $e_2^2=e_2$. just let $B=e_1A$, $C=e_2A$. everything else are simple computations(multiplying with $e_1$ vanishes all multiples of $e_2$ and the other way round, but $(e_1+e_2)x=x$).


geometry
geometric transformation
reflection
analytic geometry

A general question about a reflection:

If you have a point $(p,q)$ and want to reflect it across the line $y=ax+b$
how do you find the coordinates of the new point?

Thanks

draw a perpendicular line to the given line that goes through that point
find the intersection
think if you have point P and point P' (the reflection of the point P), then the midpoint of those 2 points will be the intersection you found earlier


Show that the distance between the incenter and circumcenter of a triangle is $\sqrt{R(R-2r)}$, where $R$ and $r$ are the circumradius and inradius, respectively.

[color=blue]Euler's Formula: The distance between the incenter and circumcentre of any triangle is $d=\sqrt{R(R-2r)}$.[/color]

Proof: Let $D$ be the intersection of $AI$ and the circumcircle, $E$ be the tangent point of the incircle and the side $AB$ and $F$ be the intersection of the line $DO$ and the circumcircle.

$\triangle{AIE} \sim \triangle{FDB}$ gives: $\frac{AI}{IE}=\frac{FD}{DB}$, since $BD=ID$, $IE=r$ and $DF=2R$, so $AI \cdot ID = 2Rr$. Use the Power of the point theorem on point $I$, we get that:
\[ (R+d)(R-d)=2Rr \]
\[ d=\sqrt{R(R-2r)} \]

Remark: If $AB=AC$, then we get the problem 6 of IMO 1962.

http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=1962


function
induction
algebra unsolved
algebra

Find all functions $f: N \rightarrow N$ such that :

$f(f(n)) + f(n) \in \{2n+2001, 2n+2002\}$

for all $n \in N$.

We define the sequence of positive integers $\left( a_{n}\right) _{n\geq0}$
as follows: $a_{0}$ is an arbitrary positive integer and $a_{n+1}=f\left( a_{n}\right) ,$ for all $n.$ Let
\[ c_{n}=a_{n}-a_{n-1}-667,  \]
for all $n\geq1.$ Then
\[ c_{n+1}+2c_{n}=a_{n+1}+a_{n}-2a_{n-1}-2001,  \]
therefore for all $n\geq1,$ $c_{n+1}+2c_{n}$ is an integer, satisfying
\[ 0\leq c_{n+1}+2c_{n}\leq1.  \]

Suppose that $c_{1}>0;$ then $c_{1}\geq1$ and $c_{2}\leq-2c_{1}+1\leq-1.$
Moreover, $c_{3}\geq-2c_{2}\geq2.$ We claim that $c_{2k+1}\geq2^{k}$ and we prove it by induction$.$ If $c_{2k+1}\geq2^{k}$ holds, then $c_{2k+2} \leq-2c_{k+1}+1\leq-2^{k+1}+1$ and $c_{2k+3}\geq-2c_{k+2}\geq2^{k+2} -2\geq2^{k+1},$ and the claim is proved.

Observe that
\[ a_{2k+2}-a_{2k}-1334=c_{2k+2}+c_{2k+1}\leq-2^{k}+1,  \]
therefore, for $k\geq11,$ we have
\[ a_{2k+2}<a_{2k}.  \]
This is a contradiction, since all $a_{k}$'s must be positive integers.

If $c_{1}<0,$ then $c_{2}\geq-2c_{1}>0$ and we repeat our argument until we obtain
\[ a_{2k+3}<a_{2k+1},  \]
for all $k\geq11,$ another contradiction.

We conclude that $c_{1}=0,$ that is, $a_{1}=a_{0}+667.$ It follows that
\[ f\left( n\right) =n+667,  \]
for all $n.$

[quote=peeta]idk if it helps any but the only solution for
\[ f(f(n))+f(n)=2n+2001 \]
is \[ f(n) = n + 667 \]

this is, since, if $f(1)=a$ and $a_n=f^n(1) \implies a_0=1\text{ and }a_1 =a$ as well as $a_{n+2}= 2a_n+2001-a_{n+1}$ then
\[ a_n = \left(\frac{668-a}{3}\right)(-2)^n + 667n + \frac{a-665}{3} \]
which is only positiv for all $n$ if $a=668$.

similary for all other $n=2,\dots$[/quote]

How did you get the recursive relation?


search
number theory
Diophantine equation
number theory open

This is a very beautiful, yet not easy, problem.

Solve the diophantine equation:

$ (1^{2}\plus{}1)(2^{2}\plus{}1)\cdots(n^{2}\plus{}1)\equal{}k^{2}$.

This http://www.mathlinks.ro/viewtopic.php?search_id=861007680&t=126781 shows that the problem has only a finite number of solutions and if I remember correctly it can give an effective upper bound for the variables. After that, it's just computations! I don't like the approach, but nature isn't always perfect.  :D

It has been proved recently that the equation has only one solution, namely, $ (n,k)\equal{}(3,10)$  by someone from Spain.

I posted the problem on the net, not knowing that it has been discussed on the forum before, hoping that some of you can provide an *elementary* solution to this nice problem.

Thanks for the link.


How many different integers can be expressed as the sum of three different numbers in the set $ \{1,2,3,4,5,6,7\}$?

To find this, you must find the lowest sum and the highest sum.

lowest: 1+2+3 or 6.
highest: 5+6+7 or 18.

then subtract the two and add one, for a total of [b]13[/b]


Given are a triangle $ABC$ for which $CA=CB=2\cdot AB$ and
two points: $E\in (AB)$, $EA=m\cdot EB$; $D\in BC$, $DC=n\cdot DB$.
Ascertain the natural numbers $m$, $n$ so that $AD\perp CE$.


Let $ a,b,c > 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \equal{} 4$
Prove that    $ b \sqrt {\frac {a^2 \plus{} 1}{a^2 \plus{} b^2}} \plus{} c \sqrt {\frac {b^2 \plus{} 1}{b^2 \plus{} c^2}} \plus{} a \sqrt {\frac {c^2 \plus{} 1}{c^2 \plus{} a^2}} \leq 3$

[quote="RFZ"]My dear friends! Can anyone solved his hard inequality???[/quote]
we have stronger ineq
 $ \sum_{cyc}{\frac {b^2(a^2 \plus{} 1)}{a^2 \plus{} b^2}} \leq 3$


[hide]Trick question or something??? I got 5/12[/hide]

[quote="MCrawford"]Find the fraction which, when added to $\frac{1}{3}$, equals $\frac{3}{4}$.[/quote]
[hide="solution"]

1/3 + x = 3/4
4/12 + x = 9/12
x = [b]5/12[/b]
[/hide]

well!
u can do it the other way around which
3/4-1/3= 5/12
but i like algebra more and is better!(4 me)

[hide]find the lcd (least common denominator) which is 12.  9/12-4/12=5/12.[/hide]


calculus
integration
logarithms
real analysis
real analysis unsolved

first use the substitution $ y^{2}\equal{} 1\minus{}x$ to get

$ \textbf I\; \equal{}\;8\;\int_{\frac{1}{\sqrt 2}}^{1}\;\frac{\ln^{2}y}{y\cdot(1\minus{}y^{2})}\;dy$

after binomial expansion of the denominator, separate out
$ k \equal{} 0$ and $ 1\leq k <\infty$,
$ \textbf I\; \equal{}\;8\;\left[\int_{\frac{1}{\sqrt 2}}^{1}\;\frac{\ln^{2}y}{y}\;dy\;\plus{}\;\sum_{k \equal{} 1}^{\infty}\;\left(\int_{\frac{1}{\sqrt 2}}^{1}\;y^{2k\minus{}1}\cdot\ln^{2}y\;dy\right)\right]$

then use repeated integration by parts to compute the 2nd integral
$ \int_{\frac{1}{\sqrt 2}}^{1}\;y^{2k\minus{}1}\cdot\ln^{2}y\;dy$

$ \; \equal{}\;\ln^{2}y\cdot\frac{y^{2k}}{2k}\bigg|_{\frac{1}{\sqrt 2}}^{1}\minus{}2\ln y\cdot\frac{y^{2k}}{(2k)^{2}}\bigg|_{\frac{1}{\sqrt 2}}^{1}\plus{}2\cdot\frac{y^{2k}}{(2k)^{3}}\bigg|_{\frac{1}{\sqrt 2}}^{1}$

then you have
$ \textbf I\; \equal{}\;\left[\frac{8}{3}\ln^{3}y\bigg|_{\frac{1}{\sqrt 2}}^{1}\minus{}\ln^{2}2\sum_{k \equal{} 1}^{\infty}\;\frac{1}{k\cdot 2^{k}}\minus{}2\ln 2\sum_{k \equal{} 1}^{\infty}\;\frac{1}{k^{2}\cdot 2^{k}}\plus{}2\zeta{(3)}\minus{}2\sum_{k \equal{} 1}^{\infty}\;\frac{1}{k^{3}\cdot 2^{k}}\right]$

there is a thread where others and myself have posted [i]closed [/i]forms
for all of the 3 sums, i.e.

$ \sum_{k \equal{} 1}^{\infty}\;\frac{1}{k\cdot 2^{k}},\;\;\;\quad\sum_{k \equal{} 1}^{\infty}\;\frac{1}{k^{2}\cdot 2^{k}},\;\;\;\quad\text{and}\;\;\;\quad\sum_{k \equal{} 1}^{\infty}\;\frac{1}{k^{3}\cdot 2^{k}}$

sorry, i couldn't locate it... anyhow, if you plug them in and simplify,
the result should follow.....

let me know if that helped... :|

How can one compute the series $ S \equal{}\sum_{k\equal{}1}^{\plus{}\infty}\frac{1}{k^{3}2^{k}}$?

Or can someone provide the link to the post where it is solved?

See [url=http://www.mathlinks.ro/viewtopic.php?t=159703]here[/url] (mainly didilica's post).

It's not exactly spelled out there, so here's how you do it:
Let $ f(x) \equal{}\sum\frac{1}{k^{3}x^{k}}$. Next, differentiate, multiply by $ x$, differentiate etc. You'll get the answer in terms of a double integral (for $ k^{2}$ you have only a simple one). Now switch the order of integration to reach didilica's computation.


Math club membership grew from 75 to 99 members at West Junior High. The increase represents a growth of $ 25\%$. for girls and $ 40\%$ for boys. How many girls were in the math club last year?

Set up two equations.  Let $ B$ represent the initial number of boys, and $ G$ the initial number of girls.  Then
$ B\plus{}G\equal{}75$
$ 1.4B\plus{}1.25G\equal{}99$
Solving,
$ B\equal{}75\minus{}G\Rightarrow 105\minus{}1.4G\plus{}1.25G\equal{}99\Rightarrow 6\equal{}.15G\Rightarrow G\equal{}40$.


arithmetic sequence
Diophantine equation
number theory unsolved
number theory

Prove that there exist infinitely many arithmetic progressions of positive integers $a, b, c$ such that
each of $2005^2+ab; 2005^2+bc; 2005^2+ca$ is perfect squares

Here is my proof:
we will show that:there exist a arithmetic progression ,namely $x,y,z$ such that $1+xy,1+yz,1+zx$ is all perfect squares,then with $a=2005x,b=2005y,c=2005z$ be a a.p. satisfying the require condition...
  Indeed,Take $(m,n)$ is one root of the Pell-Equation:$m^2-3n^2=1$.
Take:$x=2n-m,y=2n,z=2n+m$(they are all positive for $m,n$ large sufficient) will work(check it!!)
  Because this Pell-Equation has infinitely many roots,so there are also infinitely many such a.p.So we are done!!!
  Happy New Year,to Hong Quy and all of Mathlinkers!!!
:D  :D :D


If i take AMC 12a and 10b, and I score higher on 10b, which score will be used for USAMO index?

The higher score will be used to calculate your USAMO index score.

Frequently Asked Question AIME #2
at
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions


Let $ M$ and $ N$ denote the greatest and least positive two-digit
multiples of 13. What is the value of $ M \plus{} N$?

The greatest two-digit multiple of $ 13$ is $ 7(13)\equal{}91$. The least is simply $ 13$ itself. Thus, the sum of the two is $ 91\plus{}13\equal{}7(13)\plus{}13\equal{}8(13)\equal{}\boxed{104}$.

can't the least multiple of 13 be negative?

[quote="GameBot"]Let $ M$ and $ N$ denote the greatest and least [b][color=#FF0000]positive two-digit
multiples[/color][/b] of 13. What is the value of $ M \plus{} N$?[/quote]


induction
inequalities
inequalities unsolved

Let $a_1,a_2,\ldots,a_k$ be  real  numbers  and $a_1+a_2+\ldots+a_k=0$.  Prove that
\[ \max_{1\leq i \leq k} a_i^2 \leq \frac{k}{3} \left( (a_1-a_2)^2+(a_2-a_3)^2+\cdots +(a_{k-1}-a_k)^2\right).  \]

if I am not wrong, we can proceed by induction.
For two variable case it's trivial.
Suppose it's true for k variables.
We define an operation on $a_{k+1}$ as a replacement of $a_1,...,a_k$ by $a_1 + \frac{a_{k+1}}{k}$ and so forth.
Since $\sum a_1 + \frac{a_{k+1}}{k} = 0$, by induction hypothesis, $\frac{k}{3} ((a_1 - a_2)^2 + ... + (a_{k-1} - a_k)^2 \ge \max \left(a_1 + \frac{a_{k+1}}{k}\right)^2$
Similarly we have the analogous inequalities by operations on $a_1,...,a_k$, sum them up and if we let $S = (a_1 - a_2)^2 + ... + (a_k - a_{k+1})^2$, we consider a variable $a_1$ (and similarly, for $a_2,...,a_{k+1}$,
$\frac{k(k-1)}{3} S \ge \left(a_1 + \frac{a_2}{k}\right)^2 + \left(a_1 + \frac{a_3}{k}\right)^2 + ... + \left(a_1 + \frac{a_{k+1}}{k}\right)^2$
By Cauchy-Schwarz,
$\ge \frac{1}{k} \left(k a_1 + \frac{a_2 + ... + a_{k+1}}{k} \right)^2 = \frac{(k^2-1)^2}{k^3}a_1^2$
Thus we have $S \ge \frac{3(k-1)}{k^4}(k+1)^2a_1^2$
Now, $\frac{k+1}{3}S \ge \frac{(k+1)^3(k-1)}{k^4} a_1^2$, so it remains to show $(k+1)^3(k-1) \ge k^4$, which is equivalent to $2k^3 \ge 2k + 1$, which is true for $k \ge 2$.

As $\frac{k+1}{3} S \ge a_i^2$ for all $i = 1,...,{k+1}$, the induction is completed. QED

the original problem is:
$\max_{1 \le i \le k} (a_i^2)\leq [(a_1-a_2)^2+(a_2-a_3)^2+...+(a_{k-1}-a_k)^2]\frac{k}{3}$
same condition.

to soarer :you surely gived a wonderful solution.
Have you ever had a chance to view the official answer which is wonderful as well.
I consider the official answer as a little weird to me for it's quite beyond me.
Would you try to have a glance at it and give your comments?
I'd also like to see how you managed to tackle the problem via your approach,I mean,how you get the idea of induction and so forth?

[quote="mysteryfigure"]Do you have a chance to view the High School Mathematics (a Chinese magazine)? If so you you  will find the very answer.[/quote]

No...  :(

[quote="Summerburn"][quote="Soarer"]
$\frac{k(k-1)}{3}S \ge \left(a_{1}+\frac{a_{2}}{k}\right)^{2}+\left(a_{1}+\frac{a_{3}}{k}\right)^{2}+...+\left(a_{1}+\frac{a_{k+1}}{k}\right)^{2}$
[/quote]

I think this is mistaken. There's no way to get that sum.[/quote]
?what do you mean? :huh:

As [b]Soarer[/b] did,we will prove that $ a_i^2\leq\frac{n}{3}((a_1\minus{}a_2)^2\plus{}(a_2\minus{}a_3)^2\plus{}\dots\plus{}(a_{n\minus{}1}\minus{}a_n)^2)$,for all $ 1\leq i\leq n$.
Define $ d_i\equal{}a_i\minus{}a_{i\plus{}1}$,for all $ 1\leq i\leq n$ and $ a_{n\plus{}1}\equal{}a_1$.Obviously $ d_1\plus{}d_2\plus{}\dots\plus{}d_n\equal{}0$.
Also it is easy to understand that $ d_1\plus{}2d_2\plus{}3d_3\plus{}\dots\plus{}nd_n\equal{}a_1\plus{}a_2\plus{}\dots\plus{}a_n\minus{}na_{1}$,hence
$ a_1\equal{}\minus{}\frac{d_1\plus{}2d_2\plus{}\dots\plus{}nd_n}{n}$,but as we know $ d_n\equal{}\minus{}(d_1\plus{}d_2\plus{}\dots\plus{}d_{n\minus{}1})$,so finally
$ \boxed{a_1\equal{}\frac{d_{n\minus{}1}\plus{}2d_{n\minus{}2}\plus{}\dots\plus{}(n\minus{}1)d_1}{n}}$.
Now let's find similar formula for every $ a_k$, where $ 2\leq k\leq n$:
$ a_2\equal{}a_1\minus{}d_1$,$ a_3\equal{}a_2\minus{}d_2\equal{}a_1\minus{}d_1\minus{}d_2,\dots,\Rightarrow$,$ a_k\equal{}a_1\minus{}d_1\minus{}d_2\dots\minus{}d_{k\minus{}1}$,combining it with boxed formula,we finally obtain:
$ \boxed{a_k\equal{}\frac{\minus{}d_1\minus{}2d_2\minus{}\dots\minus{}(k\minus{}1)d_{k\minus{}1}\plus{}(n\minus{}k)d_k\plus{}(n\minus{}k\minus{}1)d_{k\plus{}1}\plus{}\dots\plus{}d_{n\minus{}1}}{n}}$.
We need to prove that
$ (a_k)^2\equal{}\frac{(\minus{}d_1\minus{}2d_2\minus{}\dots\minus{}(k\minus{}1)d_{k\minus{}1}\plus{}(n\minus{}k)d_k\plus{}(n\minus{}k\minus{}1)d_{k\plus{}1}\plus{}\dots\plus{}d_{n\minus{}1})^2}{n^2}\leq\frac{n}{3}(d_1^2\plus{}d_2^2\plus{}\dots\plus{}d_{n\minus{}1}^2)$
Using CBS:
$ \frac{(\minus{}d_1\minus{}2d_2\minus{}\dots\minus{}(k\minus{}1)d_{k\minus{}1}\plus{}(n\minus{}k)d_k\plus{}(n\minus{}k\minus{}1)d_{k\plus{}1}\plus{}\dots\plus{}d_{n\minus{}1})^2}{n^2}\leq (d_1^2\plus{}d_2^2\plus{}\dots\plus{}d_{n\minus{}1}^2)(\frac{1^2\plus{}2^2\plus{}\dots\plus{}(k\minus{}1)^2\plus{}1^2\plus{}2^2\plus{}\dots\plus{}(n\minus{}k)^2}{n^2})$,so it is enough to prove that
$ \frac{1^2\plus{}2^2\plus{}\dots\plus{}(k\minus{}1)^2\plus{}1^2\plus{}2^2\plus{}\dots\plus{}(n\minus{}k)^2}{n^2}\leq\frac{n}{3}$,or
$ 1^2\plus{}2^2\plus{}\dots\plus{}(k\minus{}1)^2\plus{}1^2\plus{}2^2\plus{}\dots\plus{}(n\minus{}k)^2\leq\frac{n^3}{3}$,but it is easy to understand that:
$ 1^2\plus{}2^2\plus{}\dots\plus{}(k\minus{}1)^2\plus{}1^2\plus{}2^2\plus{}\dots\plus{}(n\minus{}k)^2\leq 1^2\plus{}2^2\plus{}\dots\plus{}(n\minus{}1)^2\equal{}\frac{n(n\minus{}1)(2n\minus{}1)}{6}\leq\frac{n^3}{3}$.
So we are done.

Check my solution, I used convex functions:
$\[ \max_{1\leq i \leq k} a_i^2 \leq \frac{k}{3} \left( (a_1-a_2)^2+(a_2-a_3)^2+\cdots +(a_{k-1}-a_k)^2\right). \]$
This can be written as a quadratic function depending on $a_i$ for every $1\leq i \leq k$. The first coefficient is positive and as quadratic function s convex, it attains it's minimum at the border meanings of $a_i$. Actually, these borders are $\min_{a_i}$ and $\max_{a_i}$. Denote them as $b$ and $a$ respectively. So, $k-1$ of $a_i$ must equal $a$ or $b$. Let $p$ of the numbers be $a$ and $q$ be $b$. So, the last one equals $-pa-qb$ in order to sum be $0$. And we must prove that $\max_{a^2,c^2} \leq \frac{k}{3}((a-ap+cq)^2+(ap-cq+c)^2)$, where $-b=c$(Note that, $c$ is positive)
We can assume that $b^2 \leq a^2$.
$c \leq -ap+cq \leq a$
$\frac{q-1}{p} \geq \frac{a}{c} \geq \frac{q}{p+1}$
Now, let's prove that $(a-ap+cq)^2\frac{p+q}{3} \geq a^2$
$(1-p+\frac{c}{a}q)^2 \frac{p+q}{3}\geq (1-p+\frac{pq}{q-1})^2\frac{p+q}{3}=(\frac{p+q-1}{q-1})^2\frac{p+q}{3}\geq 1$
OR:  $(p+q-1)^2(p+q)\geq 3(q-1)^2$
Cases when $q=1$ or $0$ are easy.
Q.E.D.

[quote=Summerburn]How that line comes? I can't see the left side being $\frac{k(k-1)}{3}S$[/quote]

You're right.That is a mistake in this solution.


inequalities
rearrangement inequality
inequalities proposed

a,b,c > 0 then prove that:

$ \boxed {a^4c \plus{} b^4 a \plus{} c^4b \plus{} a^2 b^3 \plus{} b^2 c^3 \plus{} c^2 a^3 \ge 2(a^3 bc \plus{} c^3 ca \plus{}c^3 ab)}$

with rearrangement inequality :

$ a^4c \plus{} b^4a \plus{} c^4b \equal{} a^3ac \plus{} b^3ba \plus{} c^3cb\geq\,a^3bc \plus{} b^3ca \plus{} c^3ab$
and:
$ a^3c^2 \plus{} b^3a^2 \plus{} c^3b^2\equal{}a^3cc \plus{} b^3aa \plus{} c^3bb\geq\,a^3bc \plus{} b^3ca \plus{} c^3ab$
Done  :D

your first inequality is true

but the second is not true by rearangement.

[quote="FOURRIER"]a,b,c > 0 then prove that:

$ \boxed {a^4c + b^4 a + c^4b + a^2 b^3 + b^2 c^3 + c^2 a^3 \ge 2(a^3 bc + b^3 ca + c^3 ab)}$[/quote]
  by cauchy's  ineq$ a^4c + c^2a^3 + a^2b^3 \geq 3a^3bc.$
  similar   we have :$ b^4a + a^2b^3 + c^3b^2 \geq 3b^3ac$
 and $ c^4b + b^2c^3 + a^3c^2  \geq  3c^3ab.$
final  we receive ineq : $ a^4c + b^4a + c^4b + 2(a^2b^3 + b^2c^3 + c^2a^3) \geq 3abc(a^2 + b^2 + c^2)$
 suppose: $ %Error. "suma" is a bad command.
^2b^3 \leq abc.\sum a^2$
   still if $ %Error. "suma" is a bad command.
^2b^3 \geq abc.\sum a^2$   then mehdi cherif solved


Let's call a subset $B$ satisfying the conditions of the problem [i]proper[/i]. We are going to prove that the number of proper subsets is $2^{n+1}$. Considering the case $n=2$ it is easy to see that the proper subsets are $\{1\},\ \{3\},\ \{1, 2\},\ \{1, 4\},\ \{1, 2, 4\},\ \{3, 2\},\ \{3, 4\},\ \{3, 2, 4\}$ so there are $8$ proper subsets. Now assume we have proven the result for all naturals up to $n$ and we wish to prove it for $n+1$. Take a proper subset $B_n$ for $n$ and note that if $a\in [1; 2^n-1]$ and $a\in B_n$ then $2^{n+1}-a\notin B_{n+1}$ ($B_{n+1}$ is a proper subset for $n+1$). And if $a\notin B_n$ then $2^{n+1}-a\in B_{n+1}$. And we may choose freely whether $2^{n+1}\in B_{n+1}$ or not.  In such a way we get that every subset $B_n$ corresponds to exactly two subsets $B_{n+1}$ (which clearly satisfy the conditions). And every $B_{n+1}$ corresponds to some $B_n$. The result follows.

Later posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1365040#p1365040

[hide=Graph-Theoretic Reformulation/Solution]
Consider a graph $G_n$ on $2^n$ vertices labelled from $1$ to $2^n$, where $i$ and $j$ are joined if their sum is a power of $2$.

-> We can show that $G_n$ has no cycles.
-> We can show that the vertices with the same highest power of $2$ form the connected components of $G_n$.
-> Thus there are $n+1$ disjoint trees that make up $G_n$, and the sizes of the trees are $1,1,2,\cdots,2^{n-1}$.
-> We want to choose exactly one of the endpoints for every edge for our set $B$. It is easy to see that for each tree, we have only $2$ choices.

Thus our answer is $2^{n+1}$.
[/hide]

[quote=bilarev]Consider the set $A=\{1,2,3\ldots ,2^n\}, n\ge 2$. Find the number of subsets $B$ of $A$ such that for any two elements of $A$ whose sum is a power of $2$ exactly one of them is in $B$.

[i]Aleksandar Ivanov[/i][/quote]

4(2^n-2)


I'm trying to make a decent prediction on the winners at nationals. Everyone state the following:

Name

State

State Score

I guess I'll start...
Carl Lian
Massachusetts
41 at state

Thanks, let's have less spam here, you can make more comments in the "Getting to know eachother" thread, restrain from small 1-4 line posts, and just include them as comments in a bigger one, it gets hard to compile this stuff when there's so much stuff in here.

Name : $\text{Allen Yuan}$

State : $\text{Michigan}$

State Score : Arg, 27/6, $\boxed{39}$.

Goal : National Countdown, but I can feel I'm getting 13th this year, noting that it has been a 7th grader each year.

oh yeah if you know the rest of your team's scores post those too

Elizabeth Synge-38
George Sivlis-38
Sam Trabucco-39

Neil Gurram
Michigan
Score at States 43 (29/14)

Kansas
-Jason Hoffman- 32
-David Wang- 34+
-David Ke- 34
-Stephen Rong- 32 18/14 (me)

Does it really matter who wins individually? I for one am much more interested in the team.

[quote="math92"][i'm not going to post my name for security reasons]

score: sadly, its a 28[/quote]

don't worry, i got the same thing!  :lol:
i'm ben cassidy from duncan, oklahoma.

[quote="math92"][i'm not going to post my name for security reasons]
[/quote]
Me neither. :ninja:
State: Maine (that narrows it down to 4)
and if I'm confident enough to have a goal of top 57, I probably went last year. (hint, hint) Not that anybody would want to know who I am.
State Score: 36.
I don't know what happened that day. My brain just died. :stretcher:

Name: Andrew Ardito
State: NY
Score: 43
Other scores: don't know, 29, don't know, don't know
The 29 is me, and I didn't look at the other scores. The first don't know got 6th in NY states while Andrew got 1st.
This is my first time at states and I'm in 6th grade so it's obvious that I should get into the 20s.
I took the test again so I got a score of 30.

NAME- Santhosh Narayan
STATE- Indiana
Score-(19 sprint, 16 target) 35
Rank-10

[quote="fireemblem13"]*sigh*
35, missed a target i shouldn't have, didn't do very well overall, but why do i have to live in Texas?[/quote]

Ha! I missed two targets I shouldn't have, I am one up!

score: 38

[quote="pbauer"]Philip B., FL--i got 40 at nationals, but i am averaging 41 on nat tests.  Then again I got 37 at chapter.  :wallbash:[/quote]

Wait, but the highest score at Nationals last year was a 39, so you couldn't have gotten a 40 on nationals.

What did you get at states though?

He meant a 40 at states. Glad you finally got an AOPS account :D.
Preet Patel
4th written
35 (23/12)
I crashed that day and didn't even expect countdown...

[anonymous]

North Carolina

36 :(

22 Sprint, 7 Target

I would have made it to Nationals with that last Target...
I did 10^3 instead of 10 C 3!!!!!!

:mad: :mad:

I'm Sean S.  I can't remember my state score; somewhere in the high 30s.  I'm from SC, and I placed 3rd.

It's a good thing for you that I wasn't in your state, I would've gotten first place written  ;) , but would've probably gotten 2nd place countdown... :(

Lol... you're... Neil?

ohhhh neil

>_>

good luck... i guess

[Ma'am, we do not talk about any deity's *cough* practices. See biffanddoc's blog's shoutbox. Thank you.]

Good Luck at Nationals to everyone.

I hope to see you all there, especially people I saw face-to-face already.

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

Do you know what Eric McCabe got? (Since you are from NC).[/quote]

Sorry, I don't. :(

[quote="Iron Fist"]Daesun Yim... the infamous Daesun Yim for those of you with long memory spans...

27/8, 1st written and countdown. Go me, I guess.[/quote]

What happened to Treething?

Then he came from underground, and know he is going to start whacking people with his trophies, and is going to start eating people as well.  He is a zombie, ahhh.

At states:

Name: Matthew Babbitt

State: New York

Score: 29

Place: 25th

Grade: 6th

My team mates places: 1st, 6th, 48th, don't know

Team place: 3rd

At chapters:

Score: 36

Place: 19th

My team mates places: 1st, 2nd, 3rd, don't know for the rest.

Team place: 1st

P.S. I activate the cut-off spell, and Treething goes back into his coffin, coughin'.

Hey! I should make a game with spells.

Eric Wang

7th Grade

Got 30 at State  :(

1st place team at State  :D


h(x)=CBRT(x)-2
What is the algebraic expression for the inverse function?
i know you switch the x and y values and then solve for y so would the inverse function be (x+2)^3=y

yes, that's correct. Note that the definition of an inverse function is that f(x) and g(x) are inverses if and only if f(g(x))=x.


analytic geometry
combinatorics unsolved
combinatorics

I have an ( n x floor( (n+1) / 2 ) ) lattice. I must connect points (0,0) and (n,0) by using directed connections.

I may only move from left to right, horizontally, North-East or South-East.

I want to work out how many ways there are of connecting said points for any n. I have found an answer on the Internet but I would much rather derive it. I am having difficulty managing this though.

If any one could help me along I would very much appreciate it.

H

This is [url=http://www.artofproblemsolving.com/Forum/index.php?f=5]combinatorics[/url], not calculus. A moderator will probably move it soon.

Your statement is somewhat ambiguous- what are the legal moves in coordinates, and where are the outer edges of your lattice?

I apologise for putting it in the wrong section  :(

As I said, the lattice has n columns and floor( (n+1) / 2 ) ) rows.

Legal moves are

(i, j) -> (i+1, j+1)
(i, j) -> (i-1, j-1)
(i, j) -> (i+1, j)

EDIT : Sorry, i mean it ends at (n,0), not (0,n) :(

Ok I've found that it's a Motzkin Path.. I can't seem to find a proof of the closed formula for this though :(

The other ambiguity I was talking about was where the bottom of the allowable points is- from context, that appears to be $ y\equal{}0$.

Count by the number of "up" edges- if there are $ k$ of them, there are $ k$ "down" edges and $ n\minus{}2k$ flat edges. The up and down edges are a Dyck path, well known to be counted by the Catalan numbers $ \frac1{k\plus{}1}\binom{2k}{k}$, and the flat edges are interspersed in any order- there are $ \binom{n}{2k}$ ways to do this. That gives an answer $ \sum_k \frac1{k\plus{}1}\binom{2k}{k}\binom{n}{2k}$. From here, there are a lot of tricks worth trying.

Ahh Thanks.

I didn't know what a Dyck path was or that the Catalan numbers represented it, but it is valuable to me as it gives me another idea in future for solving problems

Thanks again :)

H

How can I understand the nested recurrence relation shown here: [url]http://mathworld.wolfram.com/MotzkinNumber.html[/url] as equation 6?


function
integration
Support
limit
real analysis
real analysis unsolved

Let $ f\in$ $ L^1$$ (\mathbb{R})$ . For any$ \epsilon>0$ , Show that there exist a simple function $ \phi$ with $ \int_\mathbb{R}|f\minus{}\phi|<\epsilon$

How can we begin? thanks

Start by assuming $ f$ is nonnegative.

Now: there's a well-known construction that produces an increasing sequence of nonnegative simple functions $ \phi_k$ that converge to $ f$ pointwise.

Do you know that construction?

Can you then see how $ \phi_k\to f$ in $ L^1?$ (Hint: MCT.)

Can you see then how to deal with the case of $ f$ not necessarily positive?

------

Followup exercises:

1. Show that there is a step function $ s$ such that ${ \int_{\mathbb{R}}|f-s|<\epsilon}.$ (A step function is a simple function for which the inverse image of any particular value is a finite union of intervals.)

2. Show that there is a continuous, compactly supported function $ g$ such that $ \int_{\mathbb{R}}|f-g|<\epsilon.$

There are many ways to show this.

For example, let $ \mathcal{C}$ be the set of $ L^1$ functions on $ \mathbb{R}$ and show that $ \mathcal{C}$ actually coincides $ L^1(\mathbb{R})$. This can be achieved by several steps. First, you can show that any integrable bounded function with compact support is contained in $ \mathcal{C}$. There are some well-known construction methods for this step. Then show that any $ L^1$ function $ f$ can be approximated by such functions in the sense that there exists a sequence $ \{ f_k \}$ of integrable bounded functions with compact support such that $ \textstyle \lim_{k\to\infty} \int_{\mathbb{R}} |f \minus{} f_k| \equal{} 0$. These two steps are sufficient to prove the statement.

[quote="sos440"]For example, let $ \mathcal{C}$ be the set of $ L^1$ functions on $ \mathbb{R}$ and show that $ \mathcal{C}$ actually coincides $ L^1(\mathbb{R})$.[/quote]

What do you mean? :blush:

Is it the construction

$ \phi_n \equal{} \sum^{n2^n \minus{} 1}_{k \equal{} 0}a_i\chi_{f^{ \minus{} 1}[\frac {k}{2^n},\frac {k \plus{} 1}{2^n})} \plus{} a_{n2^n}\chi_{f^{ \minus{} 1}[n,\infty)}$

where $ \chi_A$ is the characteristic function of $ A$

From MCT we know that $ \int f \equal{} \lim\int \phi_n$
So $ \lim\int(f \minus{} \phi_n) \equal{} 0$ , and we can claim that for any $ \epsilon > 0$ there exist $ N$ such that $ |\int(f \minus{} \phi_N)| < \epsilon$ ?

Am I at the right track? I cant see how to make $ \int|(f \minus{} \phi_N)| < \epsilon$ continue from my steps?

If I can solve this step, then for the case where $ f$ not neccesary positive we just write $ f\equal{}f^\plus{}\minus{}f^\minus{}$ and we constructs 2 simple funtions which converges to $ f^\plus{}$ and $ f^\minus{}$ respectively.

[quote="shyong"]Am I at the right track? I cant see how to make $ \int|(f \minus{} \phi_N)| < \epsilon$ continue from my steps?[/quote]
Since $ \phi_N$ is an increasing sequence, we have that $ |f\minus{}\phi_N|\equal{}f\minus{}\phi_N.$


function
combinatorics unsolved
combinatorics

Let A be the set of permutions p of (1,2,..,n) s.t. p^2 is identity permutation. Let B be the set of permutations p of the same set s.t. for k=1,2,...,n-2, its not true that p(k)>p(k+1) and p(k)>p(k+2). Prove that size of A=size of B.

Bomb

I guess p(k) is the image of k under the permutation.

I misunderstood: [b]p^2[/b] is [b]identity[/b] permutation???

I guess he means that permutating twice gives the orginal "sequence" (p(p(k)) = k for all k)?

We have the result that :A(n)=A(n-1)+(n-1)(n-2)*A(n-3) with A1=1,A2=2.

anybody has already known the way to cout B

Hmm... I think I have a bijection!

Consider the set $A$ of involutions (that is, the set of permutations $p$ such that $p^2$ is the identity). A cycle of a permutation is a sequence $(a_1,a_2,\ldots,a_k)$ such that $a_{i+1} = p(a_i)$ for $i = 1, 2, \ldots, k-1$ and $a_1 = p(a_k)$. Since $p^2$ is the identity, all cycles of $p$ have length $1$ (fixed points) or $2$.

Now consider the representation of permutations as product of its cycles, that is, instead of writing, say, $(1,3,2,4)$ as a permutation of $(1,2,3,4)$ we write $(1)(2\ 3)(4)$. Note that $(2\ 3)$ is a cycle and that the order of the cycles doesn't matter, that is, $(1)(2\ 3)(4) = (4)(1)(2\ 3)$. Thus write each cycle of length $2$ in descending order and write the cycles in ascending order of the first element of the cycle. For example, write $(1\ 5)(4)(2\ 3)$ as $(3\ 2)(4)(5\ 1)$.

Now forget the brackets and write the numbers in order of appearance: $(3,2,4,5,1)$. We obtain a permutation from $B$: note that $p(k+1) < p(k)$ if and only if $(p(k)\ p(k+1))$ is a cycle. The next number, $p(k+2)$, is bigger than $p(k)$, because we are writing cycles in ascending order of the first element of the cycle. Thus the function "forget the brackets" is well defined.

It's not hard to see that this function is also invertible: consider a permutation from $B$; if $p(1) < p(2)$, $p(1)$ is a fixed point; else, $(p(1)\ p(2))$ is a cycle; assign "fixed point" or "cycle" to the leftmost unassigned number of the permutation, according to this criterion; since we cannot have $p(k) > p(k+1) > p(k+2)$ every cycle has length at most $2$ and the cycles are in ascending order of the first element of the cycle because if $(p(k)\ p(k+1))$ is a cycle then $p(k+1) < p(k)$ and thus $p(k+2) > p(k)$, that is, the next cycle will begin with a bigger number.

So this function is a bijection and the number of elements of $A$ and $B$ are equal.


[b]Question courtesy Chronoz, my friend, whose keyboard is dead[/b] :D

Two waves which are different only in amplitude are propagated in opposite directions, produce standing waves, node not being a position of zero amplitude. Genereally in standing waves, energy is not transported in a medium, but my book says that in this case as amplitudes are different, energy is transpoted across the medium. Plz explain  :?:

if energy was 2 b transported,y wud they b called as "Standing waves"  :idea:

Energy can be transmitted as in this case nobody is at rest. No nodes na?


abstract algebra
calculus
integration
LaTeX
inequalities
induction
quadratics

well lets have a marathon for olympiad with the exams cmg up so every time a person answers a question he is expected to post a neq question and please number the question.
So here goes the first question-
$ 1.$ Let $ x,y,z$ be positive integers that satisfy $ \frac{1}{x^2} \plus{} \frac{1}{y^2} \equal{} \frac{1}{z^2}$ then $ 20 | xy$

[quote]Let  x,y,z be positive integers that satisfy  1/x^2 + 1/y^2 = 1/z^2 then
20|xy
[/quote]

In this what is the questioN? Are we required to prove that 20|xy

oh i am so sorry yes the question is to prove $ 20|xy$

I am not too sure but I think you can make the problem stronger by telling to prove $ 60 \mid xy$

We have $ z^{2}(x^{2}\plus{}y^{2})\equal{}x^{2}y^{2}$,
So $ z|xy$
First suppose $ x^{2}y^{2}\equiv 1,\minus{}1(mod 5)$
then we consider four cases $ x^{2}\equiv 1(mod 5)$,$ y^{2} \equiv 1$ etc.... And we get contadiction in each case so
so $ 5|xy$,and yes similarly one can also show that $ 3|xy$
now It is easy to see that $ x,y$ both can't be odd.Suppose some of them is odd say $ y$ and the other is a even which is not divisible by $ 4$.So $ x^{2}\equiv 4(mod 8)$,$ y^{2}\equiv 1 (mod 8)$ and this will also lead to a contradiction.

Here is my problem-$ P$ and $ Q$ are centers of two circles
$ C_{1}$ and $ C_{2}$ two tangents are drawn from the point $ P$ to circle $ C_{2}$ which intersects $ C_{1}$ at $ A$ and $ B$.Similarly two tangents are drawn from $ Q$ to circle $ C_{1}$ which intersect the circle $ C_{2}$ at $ M$ and $ N$.
Show that $ AB\equal{}MN$.

some1 solve dis
if 2n+1 and 3n+1 r both perfect squares,prove 40|n

$ 2n\plus{}1$ is perfect square.since it is a odd square it must be $ 1(mod 8)$ so $ 4|n$. now $ 3n\plus{}1$ is perfect square and it is also odd so $ 8|n$.now let $ n\equal{}8k$.The we have $ 2n\plus{}1\equiv k\plus{}1(mod 5)$ ,$ 3n\plus{}1\equiv 1\minus{}k(mod 5)$.Now any square is either 1 or -1 or 0 module 5.So $ 5|k$.
So $ 40|n$

Another similar problem
If $ 4n \plus{} 1$ and $ 5n \plus{} 1$  are perfect squares then prove that $ n$ is divisible by 56

@apatimgtr Good one,I went upto 4|n,cudn't do the rest :(

@ scary post dat n int.s question..i bet its quite touf....

let me revive this
Find the prime $ p$ if $ p^{2} \plus{} 11$ has exactly 6 positive integral divisors .

Was asked in the russian math olympiad

Let $ p1$ and $ p2$ be primes.
$ p^{2} \plus{} 11 \equal{} p1 .p2^{2}$
p=2 does not work, so either p1 or p2 is 2(odd+odd).
if,$ p^{2} \plus{} 11 \equal{} 2 .p2^{2}$
putting p=3 works and since a square is either 0 or 1 mod 3, 3 is only answer.
If , $ p^{2} \plus{} 11 \equal{} 4p1$ and $ p^{2}$=1 mod 3 ,p1=3.
This gives p=1(not possible).
So 3 is the answer.
Divisors of 20 are :-
1,2,4,5,10,20

for what smallest integer n is $ 3^n \minus{} 1$ divisible by $ 2^(1995)$

I dunno latex
probably phi(2^1995) which I guess is (2^1995 - 2)/2,if I have not over looked any thing and have not made silly mistakes.
Ones having Challenge and thrill of precollege mathematics,plz try sum no 9 of pg 553,I am unable to post the sum here.

Scary,why you changed your name?Is this you real name or do you mean a bangla word which means rotten. :lol:
Anyway I saw the sum in challenge and thrill.It is the very powerful jensen inequality.You can easiliy prove it by a tricky induction.It is true for 2 variables(given).Suppose it to be true for 1,2,.....,n-1 variables.The by using the induction hypothesis  two times prove it for n variables.Then the induction will be complete.

Its nice to see that you guys are really preparing for the olympiad.

Here is one problem .....Hope you like solving it.Try it on your own.

For what values of $ n > 1$ does there exist a number $ m$ that can be written in the form $ a_1 \plus{}... \plus{} a_n$ $ ($ with $ a_1\in \{1\},a_2 \in \{1,2\},.. .,a_n \in \{1,.. .,n\}$ $ )$ in $ (n \minus{} 1)!$ or more ways?

@apratimgtr
several reasons
1.Classmates came to know my username and were often reading my posts and making fun of me at skool :( I'll soon change the avatar too :P
2.Din feel like typing dat entire scary math evry time I logged in
3.My ex-classmate and now a new member at mathlinks has chosen a similar username,so I decided to change mine to avoid confusion. :D
And dat is really my nickname  :blush:  :blush: I dunno abt what u guys feel,but I really dont feel it to be dat bad,there r worse ones like potol/jhepa/tepa/bhola..........:rotfl:  :rotfl:
And plz post the entire sol,I don't think this 9th prob is rmo level,is it?????????? :huh: And is my answer to aadil's ques ok or needs correction?

can anyone plz proove this using non-calculus methods(just for the sake of it...althou calculus is allowed in RMO)
a1,a2,a3...a7 is an AP(with 1,2..7 being subscripts) with a7=8
find the min value of a1*a2*a7
...n btw pocha aka scary y dont y post that n integers set having subsets walla question???i guess u n i dunno d answer :blush:  :blush:

stop calling me scary man!I've lost the purpose of changing my  username :rotfl:

sorry man..but i dunt like pocha!!!its so like mower in hindi!!!!...i cant use dat for a person...i'll call u BB :lol:  :rotfl:

@pocha
see this link for solution to that question,http://en.wikipedia.org/wiki/Jensen_inequality

Actually i never meant that your nickname is bad ,it is very nice.But i do not think that u will be able to get rid of your classmate in this way.Unless he is really really dumb.

Isnt $ \phi(2^{1995}) \equal{} 2^{1994}$ ?
How do you show that this is the minimal solution?
$ 3^{2^{1994}}$ is huge.
@sahilsharma94
Are there restrictions on a1,a2 etc? (Them being positive,or integers or real or something?)

The first sum in this can be done without casework:
$ (xz)^{2}\plus{}(zy)^{2}\equal{}(xy)^{2}$
We know that(it can be easily proved) that 60|abc if a,b,c form a pythagorean triplet.
so, $ 60|x^{2}y^{2}z^{2}$
From the first eqn, we see that anything which divides z divides xy also(take $ z^{2}$ common).
so, $ 60|xy$

there are no restrictions..beautiful question...isnt it???obviously a1,a2...a7 are real no.s(but there are no all +ve..all -ve type restrictions)

@graffiti
Plz forgive my lil silly mistake,I tried to edit the post,cudn't do it any more.
@apratimgtr
Well,they only visit general discussion :P
And btw,if they continue to tease me,I'll have to quitely submit :( No other option is there,I've almost stopped visiting general dicussion,so that no one over there again starts exclaiming at me -"Oh!it is u ------",as long as this thing remains confined to math olympiad section,it is b8r.And I dont think they'll have patience enaf to read thru all the posts to discover who pocha really is,I know them pretty well.And I thought it wud be b8r to change my location,some of the members mistake me to be an iit student,and fail to realise dat I'm actually a resident of the campus. :(

i dont mean to offend or something but u guys have to stop this kind of spamming! this is the olympiad section we dont really want to know much abt ur nicknames or anything and u are free to discuss all this in the genral discussion or PM the users. i am not referring to only pocha! this marathon was started not for this.
I hope someone posts solution for the problem posted by Pocha.
[u]Problem5[/u]
Prove the Nesbitts Inequality.

this is pretty easy posted only to revive the topic and please dont provide age-old solutions try something new!

[hide="Solution to 5"]$ R \equal{} \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b}$

$ S \equal{} \frac {b}{b \plus{} c} \plus{} \frac {c}{a \plus{} c} \plus{} \frac {a}{a \plus{} b}$

$ V \equal{} \frac {c}{b \plus{} c} \plus{} \frac {a}{a \plus{} c} \plus{} \frac {b}{a \plus{} b}$

$ S \plus{} V \equal{} 3$
$ R \plus{} S\ge3$ By A.M-G.M
and $ R \plus{} V\ge 3$
Adding both we get $ 2R\ge3$[/hide]

fine raghav, wat abt -
[u]Problem6[/u]
Prove AM-GM for n-1 positive reals if it is known to be ture for n reals

[hide="Sahil Sharma Problem"]
We have to find the minimum value of $ a_{1}a_{2}a_{7}$.
$ a_{1}\equal{}a$,$ a_{2}\equal{}a\plus{}d$,$ a_{7}\equal{}8$ but $ a\plus{}6d\equal{}8$ From here Put the value of a and get a quadratic equation whose minimum value is $ \frac{\minus{}D}{4a}$

[/hide]

and why is the Min value of d given by the -D/2a ?? i mean how can we prove it???
n btw another unsolved question...
f(x)= ax^8 + bx^7 + cx^6 + dx^5 ex^4 + fx^3 + gx^2 + hx + i
value of F(x) for diff values of x are given below:
x  f(x)
1  1
2  1/2
3  1/3
4  1/4
5  1/5
6  1/6
7  1/7
8  1/8
9  1/9

find the value of f(x) for x=10 (and it is not equal to 1/10 )

If u don't want to use that formula then u can also go with the method of completing square.

[hide="Hint for your next question"]

Let $ f(x)\equal{}\frac{1}{x}\plus{}\frac{1}{x^{2}}(x\minus{}1)(x\minus{}2)(x\minus{}3).....(x\minus{}9)$
[/hide]

its not about not using the formula...i'm just asing that WHY is -D/a the Min value of this Quad

[b]@ Maths Mechanic:[/b] $ f(x)$ is a polynomial.

[hide="Solution"]
Let $ g(x) \equal{} x \cdot f(x) \minus{} 1$. Then $ 1,2, \cdots 9$ are the only roots of $ g(x)$.

So, let $ g(x) \equal{} k(x\minus{}1)(x\minus{}2).....(x\minus{}9)$. $ \implies g(10) \equal{} k \cdot 9!$

Also, $ g(0) \equal{} k (\minus{}1)(\minus{}2).....(\minus{}9) \equal{} \minus{} k \cdot 9!$. But, from the original definition $ g(0) \equal{} \minus{}1$.

$ \implies k \equal{} \frac 1 {9!}$ $ \implies g(10) \equal{} 1$ $ \implies 10 \cdot f(10) \minus{} 1 \equal{}1$

$ \implies \boxed{f(10) \equal{} \frac 1 5}$[/hide]

i'm sorry to interupt...but it has not been given that [quote]But, from the original definition f(0)=-1 [/quote]
so u cant use it... :(

[quote="keyree10"]But, from the original definition [b]g(0)[/b] = -1. [/quote]

I wrote $ g(0) \equal{} \minus{}1$ not $ f(0) \equal{} \minus{}1$. Notice that $ g(x) \equal{} x \cdot f(x) \minus{}1 \implies g(0) \equal{} 0 \cdot f(0) \minus{}1 \equal{} \minus{}1$

Is my solution very wrong???????
Some1 plz tell :(

$ 1/a \plus{} 1/b \plus{} 1/c$=$ \frac {1}{a \plus{} b \plus{} c}$
or,$ 1/c \plus{} 1/b \equal{} \frac {1}{a \plus{} b \plus{} c} \minus{} 1/a$
or,$ 1/bc \equal{} \minus{} \frac {1}{(a \plus{} b \plus{} c)a}$
or,(after factorisation) ,c=-a,b=-a.
Then we are done.

[size=150]क्या मैं सही हूँ ?[/size]

[quote="harekrishna"]@nobelium, well, your idea was right. Please write your solutions neatly. For example, i didn't understand why you wrote 1/bc = whatever.
Maybe you should try reading solutions of other mathlinkers to get an idea of formal proofs.[/quote]

Thanks Harekrishna.I just skipped a bit intentionally.I am not preparing for RMO.I just did that using the basics I have learnt.I will start soon,though.

[quote="harekrishna"]Prove that if one term of an infinite positive integer arithmetic progression is a perfect square, then there are infinite perfect squares in the sequence.[/quote]

Assume without WLOG that the $ (l\plus{}1)$st term of the sequence is a perfect square.

Let the first term of the sequence be $ a$, and common difference be $ d$.

$ a \plus{} ld \equal{} m^2$ (1)

Now take all squares in the residue class congruent to $ m^2$ mod d.

Such squares take the form $ (sd \plus{} m)^2$ where $ s$ is any natural number

So if we prove that $ a \plus{} jd \equal{} (sd \plus{} m)^2$ (2) is true when $ s$ is any specific natural number and $ j$ is an integer, we are done.

But subtracting (2) - (1) $ (j\minus{}l)d \equal{} s^2d^2 \plus{} 2sdm$

But $ d$ divides the right hand side, and $ l$ is already and integer, this implies that $ j$ is an integer.

By letting $ s$ range over all natural numbers, we have essentially constructed an infinite square sequence in the arithmetic progression.

P.S. I will contribute a question to the marathon very soon. I don't have any question in mind right now. Sorry.


Let an acute triangle ABC. D,E,F are the foots of the altitude from A,B,C respectively. Choose a point M be inside triangle DEF. Let $ (A_1,A_2), (B_1,B_2), (C_1,C_2)$ be the orthogonal projections of M respectively on the lines $ (AD,BC), (BE,CA), (CF,AB)$. Prove that $ A_1A_2,B_1B_2,C_1C_2$ intersect in a single point.

See at the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=126138]Perpendicular feet[/url].

Kostas Vittas.

Thank [b]vittasko[/b]. Your link and last posted was very nice.

-----------
VanQuy


[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=274251]Puerto Rico PRMO Team Selection Test 2009[/url]

On an arbitrary triangle $ ABC$ let $ E$ be a point on the height from $ A$. Prove that $ (AC)^2 \minus{} (CE)^2 \equal{} (AB)^2 \minus{} (EB)^2$.

[hide="hint"]What does the right hand side simplify to via pythag? The left hand side?[/hide]

as,
AM^2=AM^2
we write, AM^2+MC^2-MC^2=AM^2+BM^2-BM^2
AC^2-MC^2=AB^2-BM^2
subtracting EM^2 from both sides
AC^2-(MC^2+EM^2)=AB^2-(BM^2+EM^2)
AC^2-EC^2=AB^2-EB^2


prove that the number of factors $p$ in the prime factorization of$a!$ where$a=(a_1...a_m)_p$is$\dfrac{a-(a_1+...a_m)}{p-1}$

This is Legendre's formula.
Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=legendre&t=19834

Pierre.