modular arithmetic
quadratics
inequalities
invariant
integration
geometry
rectangle

Hello, I will be posting problems of days every day at the pre-olympiad level.  If you're interested (hopefully), you'll post some interesting solutions that will enlighten many of us.  The first problem is as follows.

Find all integer solutions $(x,y)$ that satisfy the equation
\[ x^2-y!=2001  \]

Masoud Zargar

[hide="Solution"]Since $m^2\equiv 0,1,2,4\pmod{7}$ for $m\in\mathbb{Z}$ and $2001\equiv -1\pmod{7}$, we have $(x^2-2001)\equiv 1,2,3,5\pmod{7}\Longrightarrow 7\nmid y!$. Hence, $y\leq 6$. Now we can manually check the values $0\leq y\leq 6$ - the only solutions are $(45, 4)$ and $(-45, 4)$.[/hide]

Would someone please confirm?
[hide]
$\\x^2-y!=2001\\ x^2=y!+2001$
For $y\geq 6$, $9|y!$.
$y!+2001\equiv 3\pmod 9$, however $3$ is a quadratic nonresidue (mod 9), meaning that $9\nmid y!=x^2$; the only values that we may consider are $y\in\{0,1,2,3,4,5\}$.  $4$ is the only satisfying solution, resulting in the solutions $(\pm 45, 4)$
[/hide]
Masoud Zargar

You can confirm yourself: $0^2\equiv 0, (\pm 1)^2\equiv 1, (\pm 2)^2\equiv 4, (\pm 3)^2\equiv 2 \pmod{7}$.

True...Thanks.

Since this is solved, I'll just place a new problem, evern though a day has not passed by yet.

[b]2.[/b]  Prove that for any prime $p$, $p^2|\binom{2p}{p}-2$.

Masoud Zargar

[hide]

The last step is a result of the first and last term in the sum being , so .

So , and every term in the sum has a factor of  in it, so the whole sum is divisible by , and we are done.

edit> it didn't settle with me that I didn't clear this up, but  won't ever be divided out of the numerator because all factors in the denominator are less than  and thus are relatively prime to  (trivial by the definition of a prime). So we will always have a factor of  in the numerator of every term.[/hide]

I decided to just make this forum a place for interesting problems.
Here's another one.

[b]3.[/b]Find all positive integers $n$ for which $1+9^n+10^n=5^n+6^n+11^n$.

Masoud

You got that out of the Diophantine equations section of Mathematical Olympiad Challenges. I'm not going to spoil it for anyone though  :)

Yup :) I've learnt to share in kindergarten. :P

Hmm, I'm not good at Diophantine... I know n has to be even, though, and that the RHS eventually increases faster than the LHS, so there can't be more than a few solutions, and know n = 2 works from testing, but can't prove that that's the only one, if it even is.

[hide]First, $n$ odd does not solve the equation, because by sum of odd powers, $1+10^n$ and $5^n+6^n$ is divisible by 11, and of course $11^n$ is divisible by 11. This would mean that $11^n+5^n+6^n-1-10^n=9^n$ would have to be divisible by 11. Contradiction.

$n$ can be bounded above this way: from the equation and since $5^n+6^n$ is (much) greater than 1, $9^n+10^n>11^n\Rightarrow \left( \frac{9}{11}\right)^n+\left( \frac{10}{11} \right)^n>1$. Since the left side is decreasing, once this is false for some $n$, it will be false for greater $n$ as well.

Testing even $n$, find that  $n=6$ makes the inequality false, so the only possible solutions are 2 and 4. Testing these into the equation shows that they both work, so the solutions are $\boxed{2,4}$.[/hide]

[b]4.[/b] Prove that every number that is not congruent to 2 (mod 4) can be written as the difference of squares.

Masoud Zargar

[hide]
All squares are in the form either $0,1 \pmod 4$
$0 \equiv 0 - 0 \pmod 4$
$1 \equiv 1 - 0 \pmod 4$
$3 \equiv 0 - 1 \pmod 4$
[/hide]

Yup.  Here's another one.

Here's an easy one:
[b]5.[/b] On an $1997\times 1997$ chessboard, we have written a $1$ or $-1$ on every square.  Let $R_i$ and $C_i$ be the product the $ith$ row and column, respectively.  Prove that $\sum_{i=1}^{1997}(R_i+C_i)\neq 0$ under any condition.

Masoud Zargar

[hide="5:  Invariants!"]
Assume you have put a $1$ on every square.  Then the sum is equal to $1997^2$.  Next, change one of the $1$s to a $-1$ - you have negated the sum of one column and one row, and the sum becomes $1997^2 - 2(1997)$.  By a similar argument, changing a $-1$ to a $1$ changes the sum by $+2(1997)$ or by $0$ (if the row sum became positive and the column sum became negative).  In any case, $\sum_{i=1}^{1997} (R_i + C_i) \equiv 1 \bmod 2$ is our invariant, which can never equal $0$.  Q.E.D.
[/hide]

I thought I saw thsi problem somewhere on this website, so here it is
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=AIME+1989&t=57748[/url]

I assumed that teh problem was not posted since it wasn't in teh resources.

[quote="boxedexe"]Here's another problem you may enjoy (AIME 1989).

[b]11.[/b]
One of Euler’s conjecture was disproved in the 1980s by three American
Mathematicians when they showed that there is a positive integer n such
that
$n^5 = 133^5 + 110^5 + 84^5 + 27^5$

Find the value of $n$.

Masoud Zargar[/quote]

Ah, I have done this problem before on a test :)

[hide="Solution"]
Observe first that from Fermat's we have $x^5 \equiv x \bmod 10$, so taking $\bmod 5$ we know $n \equiv 4 \bmod 10$, so $n$ ends in $4$.  Observe also that $133 < n < 133 \sqrt[4]{5}$, and we can make a very bad estimation using $133 \sqrt[4]{9} = 133 \sqrt{3} \approx 230$.

Next, notice that $84 + 27 = 111 = 110 + 1$.  We write

$(n^5 + 1) - (84^5 + 27^5) = 133^5 + (110^5 + 1)$

And, taking $\bmod 111$, we conclude $n^5 + 1 \equiv 133^5 \bmod 111 \implies n^5 \equiv 22^5 - 1 \bmod 111$ which implies $3 | n$.

Looking for numbers that satisfy these conditions, we see that we only have to test $144, 174, 204$.  We see that they are distinct $\bmod 8$, so taking that, we have $n^5 \equiv 5^5 + 6^5 + 4^5 + 3^5 \bmod 8 \equiv 6^5 + 4^5 \bmod 8 \equiv 0 \bmod 8$ - hence, our answer is $\boxed{144}$.
[/hide]

[quote="The Zuton Force"]I encountered this very recently on a practice AIME
[hide]
The units digit of $3^5 + 0^5 + 4^5 + 7^5$ is 4. The only units digit that, when raised to the 5th power, gives a units digit of 4, is 4 itself. n > 133, so n can be 134, 144, etc.
Expanding of $(x+1)^5$, it is obvious that 134 is too small. Since the $110^5, 84^5, and 27^5$ are obviously significantly small compared to $133^5$, $154^5$ will be too big, so $n=144$[/hide][/quote]

 :| All remainders mod 5 or 10 repeat are the same as the fifth (or multiple) power, no?

[quote="flanaha14"]Find all sets of four real numbers such that the sum of any one and the product of the other three is 2.[/quote]Tex'd. $a+bcd=2$, $b+acd=2$, $c+abd=2$ and $d+abc=2$.

[hide="Solution"]Subtract the second equation from the first:

$(a-b)(1-cd)=0$

Now it's either $a=b$ or $cd=1$, which breaks down to casework and simplification of the other equations. Since the work is fairly simple yet cumbersome, I'll skip it. If I haven't made any mistakes, the solutions are $(1,1,1,1), (3,-1,-1,-1),(-1,3,-1,-1),(-1,-1,3,-1,),(-1,-1,-1,3).$[/hide]

[quote="Gabriel(fr)"][quote="flanaha14"]Let a and b be positive integers such that ab + 1 divides a^2 + b^2. Show that (a^2 + b^2)/(ab + 1) is a perfect square.[/quote]

[hide]The synthetic division gives:

$a^2+b^2=(ab+1)\frac{a}{b}+(b^2-\frac{a}{b})$
to make this a exact division  we need have:

$b^2-\frac{a}{b}=0 \implies a=b^3$ substituing in the fraction:

$\frac{a^2 + b^2}{ab + 1}=\frac{b^2(b^4+1)}{b^4+1}=b^2$ [/hide]
[/quote]

The proof isn't correct. For example, $a=8, b=30$ satisfy the initial condition, yet $a\neq b^3.$

[quote="Gabriel(fr)"]
[b]PROBLEM 13[/b]

Find $\alpha\ ,\beta\ ,\gamma$ such that the polinomial,  $\alpha x^{\gamma+1}+\beta x^{\gamma}+1$, $x \in \ Q$, is divisible by $(x-1)^2$ and the value of the quotient is $120$ when $x=1$.[/quote]

[hide="calculus is bad..."]
just kidding...i used a lot of it...
we have to satisfy the conditions,
1. there is a double root at $x=1$, so there is a critical point there
$f(x)=ax^{c+1}+bx^c+1$
$f'(x)=a(c+1)x^c+bcx^{c-1}=0$
$ax^{c-1}(a(c+1)x+bc)=0$
$ac+a+bc=0$ (1)

2. when it is at x=1, and we divide out the $x-1)^2$, the value is at $120$, by l'hopital
$\lim_{x\rightarrow 1} \frac{ ax^{c+1}+bx^c+1 }{ x^2-2x+1}$
$=\frac{ a(c+1)x^c+bcx^{c-1} }{2x-2}=\frac{ ac(c+1)x^{c-1}+bc(c-1)x^{c-2}}{2}=120$
(substitute $x=1$ for the limit)
$ac(c+1)+bc(c-1)=240$
$c(-bc)+bc^2-bc=240$
$bc=-240$ (2)
using (1) and (2)
$a(c+1)=240$ (3)

3.
additionally, roots have to be at x=1, so $f(1)=a+b+1=0$ (4)

solve using (2), (3), (4) to get
$c=15$,$b=-16$, $a=15$
so our polynomial is $f(x)=15x^{16}-16x^{15}+1$

i know that every step pretty much used calculus, but this problem directed me that way, also this problem is pretty straight forward with calc...it just takes a while to type out all the statements[/hide]

do you guys want another problem?...here is one that i just thought of...it is fun to make problems  :)

[size=134][b]Problem 14[/b][/size]

Find all ordered pairs of integers $(x,y,z)$ that satisfy the system:

$x^3+8x^2-3x-8=2y$
$y^3+8y^2-3y-8=2z$
$z^3+8z^2-3z-8=2x$

This remained unsolved.

[quote="flanaha14"]Let a and b be positive integers such that ab + 1 divides a^2 + b^2. Show that (a^2 + b^2)/(ab + 1) is a perfect square.[/quote]

I hate to stop you trying, but this is toughest problem ever asked at the IMO. 11 people in the world got it! But keep going!

Link to the proof is [url=http://www.kalva.demon.co.uk/imo/isoln/isoln886.html]here[/url]


calculus
integration
logarithms
trigonometry
calculus computations

I found from a book of mine that $\int_0^\infty \ln (1+x)x^{s+1} dx=\frac{\pi \csc \pi s}{x}$. I'm wondering how to evaluate such an integral.

Oops, $z=x$.

Strange. I thought this is the Mellin transformation of $\ln (1+x)$ and I watched probably a wrong formula. So I have probably change $s+1$ to $s-1$ and think more.


[quote="kamyar.sac"]Let

$ f(n) \equal{} f(1)f(n \minus{} 1) \plus{} g(n \minus{} 1)$
$ g(n) \equal{} f(n \minus{} 1) g(n \minus{} 1)$
$ f(1) \equal{} 2009$
$ g(1) \equal{} 2009$

Find $ f(2009) \plus{} g(2009)$.

Thank you for any help in advance.[/quote]

Are you sure this problem is the correct one ? where is it coming from ?

We have $ f(2009)\plus{}g(2009)$ around $ 2009^{10^{420}}$ !!

I'm pretty sure the statement of the problem is correct. I don't know the source. I found it somewhere on Yahoo Answers.

Maybe it's possible to write the answer in terms of a finite summation or product.


Determine all nonnegative integers $n$ for which $2^n+105$ is a perfect square.


Find the sum of all numbers less than 100 billion that are
1.divisible by 7 , and
2. the reverse of the number is divisible by 7 too.

(e.g. sum of 7,70,77, 161, 168 etc.. )

Exuse me! I should tell you it is not easy & it is so hard :blush:  :blush:
but you can write a program.  :oops:  :oops:

What do you mean by reverse?

You mean to change the order of the digits from left to right
for example:

$123456 \rightarrow 654321$

define your words,preciesly :mad:

hey lupin you are right- its the order of the digits reversed.
123456<->654321

the next question you're going to ask me is -how do you handle reverses of 70, 700,770 etc;
the question/source doesnt say, but i think you can take them as 7, 7 and 77 respectively.


function
calculus
derivative
geometry
geometric transformation
ratio
inequalities

Give example of an interval I and a function f defined on I, derivable on I such that:
a) f is not strictly increasing;
b) the set of points where the derivative is positive is dense in I.

Here is a construction I borrowed from the second edition of  "Selected Problems in Real Analysis" (The first edition was published in AMS 1992, Translations of Mathematical Monographs series, Volume 107, ISBN 0-8218-4559-4; the second edition is not available in English yet. By the way, I highly recommend the book to any college student interested in Analysis. Even the first edition is awesome (in my humble opinion, of course)).

Step 1: there exists a continuous function $h(x)$ that is strictly increasing, maps $R$ onto $R$ and such that the derivative $h'$ (understood as the, possibly infinite, limit of the standard difference ratio) exists at every point, is strictly greater than $1$ everywhere and equals $+\infty$ on a dense set in $R$.
[hide="Click here to see the proof"]
Put $\psi_k(x)=a_k\sqrt[3]{x-q_k}$ and $h(x)=x+\sum_{k=1}^\infty\psi_k(x)$ where $q_k$ is some enumeration of rationals and $a_k>0$ tend to zero so fast that the series converges uniformly on any finite interval $I\subset R$. Then, obviously $h$ is strictly increasing, continuous, and $h$ maps $R$ onto $R$. The trick is to show that the derivative exists at every point.

It is easy to see that $h'(x)=+\infty$ unless $\sum_k \psi_k'(x)<+\infty$. To show that the derivative of the sum equals the sum of the derivatives at the points where the series converges, use the inequality
\[
0<\frac{\psi_k(y)-\psi_k(x)}{y-x}<100\psi_k'(x)
\]
(an easily verifiable property of the cubic root)
[/hide]
Step 2: there exists a function $g(x)$ that is strictly increasing, maps $R$ onto $R$ and such that the derivative $g'$ (understood as the usual limit of the standard difference ratio) exists at every point, is strictly less than $1$ everywhere and equals $0$ on a dense set in $R$.
[hide="Click here to see the proof"]
Take $g$ to be the inverse function of $h$. The set of points where $g'=0$ is the image under $h$ (which is a homeomorphism of $R$) of the set where $h'=+\infty$, so it is also dense in $R$.
[/hide]
Step 3. Put $G(x)=\delta x-g(x)$ for small $\delta>0$. Then $G$ is not strictly increasing (if $\delta$ is chosen sufficiently small) but $G'>0$ wherever $g'=0$.

You can also use the function $g$ to construct a function differentiable everywhere on $R$ whose derivative changes sign on each interval, but that will require one extra twist.

P.S. See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=244776[/url]


Some quick suggestions:

I think that when you're replying to someone in a PM, I think a list of replies should be below the PM, sort of like in e-mail or when posting in a forum.

I also think that you should be able to search by signature. This might be too chaotic as many people change their sigs, but sometimes, the only memorable part about the user is their signature.

My question is how do you find out which user number you are? I know someone figured it out because they have that they are the 100,000th user or something.

You are user 31250. Searching by signature is not feasible any time soon. PMs might be revamped in the future, but for now we just don't have the time for it.

If you click on your profile, you'll see u=(number) in the URL bar.

Do you mean 10000th user? I'm not sure they're actually the 10000th user, actually...

Yeah, he/she is actually user [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=10153]#10153[/url].

He was the 10000th user at the time he registered. It was because 153 previous users got deleted from the database (so the id is still incremented).


superior algebra
superior algebra solved

Let $G$ be a group. Prove that the following are equivalent:
a) $G\simeq D_{\infty}$
b) There exists a normal infinite cyclic subgoup $H$ of $G$, $H\neq G$, $H=\langle h\rangle$ for some $h\in H$ and $C_G(h^n)=H\ \forall n\in\mathbb{N}^*$.

$a\rightarrow b$: $D_{\infty}=<a,b|a^2=b^2=1>$.
Choose $h=ab$ and $H=\langle h \rangle$.
All the conditions of (b) follow from the fact that the presentation I've chosen for $D_{\infty}$ makes it canonically isomorphic to $\mathbb{Z}*\mathbb{Z}$ and in this group we have the canonical forms for its elements: $1,a,b,abababa...$ or $bababab...$.

$b\rightarrow a$:
Let $g,g'=g^{-1}\in G$ and $k,l\in\mathbb{Z}$ such that $ghg'=h^k$ and $g'hg=h^l$. Then $h=g'ghg'g=g'h^kg=(h^k)^l=h^{kl}$. Because $H$ is inifnite cyclic we get $kl=1$. So $k=l=1$ or $k=l=-1$.

Define $f:G\to U_2=(\pm1,\cdot )\simeq \mathbb{Z}_2$ by $f(g)=\left\{\begin{array}{cc}1&\mbox{, if }ghg'=h\\ -1&\mbox{, if }ghg'=h^{-1}\end{array}\right.$
$f$ is a group morphism and $\ker f=C_G(h)=H$. $f$ is surjective because $H\neq G$.
So $G/H\simeq \mathbb{Z}_2$. Let $x\in G\setminus H$. Then $G=H\sqcup xH$. Because $G/H\simeq \mathbb{Z}_2$, we get $x^2\in H$. But (b) tells us $C_G(x^2)=H$ if $x\neq 1$. On the other hand, obviously $x\in C_G(x^2)$.
So $x^2=1$ otherwise we contradict the disjointness of $H$ and $xH$.
Now $h,x$ generate $G$ and the isomorphism $G\simeq D_{\infty}$ should follow.


At Beaumont High School, there are 20 players on the basketball team.  All 20 players are taking at least one of biology or chemistry.  If 8 players are taking biology and 4 players are taking both sciences, how many players are taking chemistry?

Using PIE:

$ 8\minus{}4\plus{}x\equal{}20 \implies x\equal{}\boxed{16}$.


Ok im not sure where to post this so i posted here because there isnt many posts here. Ok anyway-

A  *WORD* has to meet the following conditions for it to be a *WORD*-

1. It can't start with a vowel
2. It can only have 3,5,or 7 letters in it
3. The middle letter has to be a vowel

For example- dog is a *WORD*

How many *WORD*'s can you make with the letters- G,O,B,R,A,V,E,S

[hide]OKAY, I'M GONNA TYPE IN ALL CAPS:
G,O,B,R,A,V,E,S
IF THE *WORD* HAS 3 LETTERS, THEN THE FIRST LETTER CAN BE CHOOSEN FROM 5LETTERS : G,B,R,V,S, SINCE IT CAN'T START WITH A VOWEL. THE MIDDLE LETTER CAN BE CHOOSEN FROM 3 VOWELS: O,A,E... AND THEN THE LAST LETTER CAN BE CHOOSEN FROM THE 6 REMAIN LETTERS.... SO IT'S 5*3*6 = 90 (I THINK)

IF THE *WORD* HAS 5 LETTERS, IT'S PRETTY MUCH THE SAME THING, SO IT'S GONNA BE 5*7*3*6*5 = 3150

IF THE *WORD* HAS 7 LETTERS, IT'S GONNA BE 5*7*6*3*5*4*3 = 37800

3150 + 37800 + 90 = 41040[/hide]

[hide]
FOR SOME REASON, I THINK I'M TOTALLY WRONG HERE.... [/hide]


calculus
integration
trigonometry
calculus computations

hi,

i  have a question about an integral that i can't solve. i learned calc 1+2 on my own, but i haven't done more than that.

here's the problem:

Int( 1 / sqrt( x^2 - 1 ) ) dx

thanks in advance.

Alternatively, there's the "ordinary" trig substitution instead of the hyperbolic substitution.

Let $x=\sec\theta,\ dx=\sec\theta\tan\theta\,d\theta,\ \sec^2\theta-1=\tan^2\theta.$

Note that both Ramanujian and I are assuming that $|x|>1.$


algebra
polynomial
combinatorics unsolved
combinatorics

Let $d$ be an even integer, and $n$ an integer such that $d\le n<2d$.

Consider the following polynomial in $z$,
\[p(z)=(2d-n)(1+z)^{n}+n(1-z)^{d}(1+z)^{n-d}. \]
Show all coefficients of $p(z)$ are nonnegative.

Equivalently,

\[(2d-n)\binom{n}{m}+n \sum_{i = 0}^{m}(-1)^{i}\binom{d}{i}\binom{n-d}{m-i}\ge 0,\]

for all $0\le m \le n$.

Is there a very specific reason why it's in the Combinatorics Forum. Shouldn't it be in the Algebra Forum?
  Ashwath

Presumably, we are meant to have $d \leq n < 2d$ and to show that the expression has [i]all[/i] coefficients non-negative.  The coefficient of $x^{m}$ in this polynomial is $(2d-n)\binom{n}{m}+n \sum_{i = 0}^{m}(-1)^{i}\binom{d}{i}\binom{n-d}{m-i}$, and so the question is equivalent to showing that each of these expressions is nonnegative.  I also note that the condition that $d$ be even is necessary; the statement is false when $d$ is odd.

kb107,

Showing $p(1)>0$ just shows that the sum of the coefficients is greater than zero. For instance,

$f(x)=-x^{2}+17$
$f(1)=16$[/quote]

The phrase, "has nonnegative coefficients," is ambiguous.  K81o7 read it as, "has some nonnegative coefficients," which is true but not very interesting.

thanks for pointing out the mistakes. have ammended the question as required.

the reason i put it in the combinatorics forum is because the actual problem came in form as JBL described.


inequalities
geometry
geometric transformation
rotation
inequalities unsolved

given $z_{1},\dots,z_{n}\in\mathbb{C}$ prove that there exists a subets B of $\{1,\dots,n\}$ such that
\[\left|\sum_{k\in B}z_{k}\right|\ \ \ge\ \ \frac{1}{\pi}\sum_{k=1}^{n}\left|z_{k}\right|\]

Let $R$ be a random rotation (chosen uniformly among all rotations) of the complex plane. Define the random set $B_{R}$ as follows
\[B_{R}= \{k: \Re(Rx_{k}) > 0\}. \]
Note that for every $z_{k}$, the point $Rz_{k}$ is a random point on the circle of radius $|z_{k}|$. Therefore,
\[\mathrm{Pr}_{R}(k\in B_{R}) = 1/2, \]

\[\mathbb{E}_{R}(\Re(Rz_{k}) | k\in B_{R}) = \frac{2}{\pi}|z_{k}|, \]
So
\[\mathbb{E}_{R}\left(\sum_{k\in B_{R}}\Re(Rz_{k})\right) = \sum_{k=1}^{n}\mathbb{E}_{R}\left(\Re(Rz_{k}) | k\in B_{R}\right) \cdot\mathrm{Pr}_{R}(k\in B_{R}) = \frac{1}{\pi}\sum_{k=1}^{n}|z_{k}|. \]
Now we use that for every complex number $w$, $|w| = |Rw|$ and $|Rw| \geq \Re(Rw)$:
\[\mathbb{E}_{R}\left(\left|\sum_{k\in B_{R}}z_{k}\right|\right) = \mathbb{E}_{R}\left(\left|\sum_{k\in B_{R}}Rz_{k}\right|\right) \geq \mathbb{E}_{R}\left(\sum_{k\in B_{R}}\Re(Rz_{k})\right) = \frac{1}{\pi}\sum_{k=1}^{n}|z_{k}|. \]
Finally, we use that the maximum of a random variable is at least its average. So for some set $B$:
\[\left|\sum_{k\in B}z_{k}\right| \geq \frac{1}{\pi}\sum_{k=1}^{n}|z_{k}|. \]


Let $ \Gamma$ be the circumcircle of a triangle $ ABC$. A circle passing through points $ A$ and $ C$ meets the sides $ BC$ and $ BA$ at $ D$ and $ E$, respectively. The lines $ AD$ and $ CE$ meet $ \Gamma$ again at $ G$ and $ H$, respectively. The tangent lines of $ \Gamma$ at $ A$ and $ C$ meet the line $ DE$ at $ L$ and $ M$, respectively. Prove that the lines $ LH$ and $ MG$ meet at $ \Gamma$.

[url=http://www.mathlinks.ro/viewtopic.php?t=195491]APMO 2008 problem 3[/url]


What do you think about this statement?

$ m$ is a positive integer such that $ 2^m$ is the greatest number not exceeding $ n$.

[hide]For me, it does not make sense and contradictive. For example, take $ n \equal{} 79$, then the greatest number not exceeding $ n$ would be $ 79$. So $ 2^m \equal{} 79$. On the other hand, $ m$ is an integer.
[/hide]
What do you think about this? I just want to convince my friend (or myself).

Yes, the statement has no integer values of $ m$ for some values of $ n$, although that is a rather nugatory point.

If this statement was in a question or problem, it probably means "$ m$ is a positive integer such that $ 2^m$ is the greatest [i]power of 2[/i] not exceeding $ n$.

You can always find such a value for $ m$ for every $ n$, ie $ m\equal{}\lfloor\log_2n\rfloor$...


integration
limit
real analysis
real analysis solved

Here it comes:

Prove that n*int {0 to 1} ln(1+x^n)dx -> Pi^2/12 as n->infinity. I really enjoyed this one.

HELL :D  GR :D BBER,

Is it Critinel problems book ?

A n =n\int_[0;1]ln(1+x n)dx

consider the expansion of ln(1+x n) in power series

A n =\int_[0;1] \sum_{k \geq 1}(-1) k-1nx nk/k dx

= \sum_{k \geq 1}(-1) k-1/k .n/(nk+1)

 \lim A n = 1-1/2 2 +1/3 2 -1/4 2 +....=

dzeta(2) -1/2dzeta(2) = dzeta(2)/2 =  \pi 2/12

Now arrive

Diogene

"Sorry oubi i am not agree with your proof, you did
not justify the  \int  \sum and  \sum  \int "

That's exactly what I did, I wrote ln(1+x) as a Taylor series. Isn't it nice?

P.S.

No, it's not in Cristinel Mortici's book  :D I found it in a magazine.

Yes gr :D bber it is nice
we can also find an asymptotic expansion of A n - \pi /2
when n-->+\infty

haha m :D ubi


geometry
cyclic quadrilateral
geometry proposed

Consider two circles S and S' intersecting at two points A and B. Some line through A meets S, S' again at C, D respectively. Let M be a point on CD. The line through M parallel to BC meets BD at K, and the line through M parallel to BD meets BC at N. The perpendicular to BC at N meets S at the point E on the opposite side of BC to A. The perpendicular to BD at K meets S' at F on the opposite side of BD to A.
Show: AMEF is cyclic quadrilateral with diameter EF.

Is easy to see that ENB~FKB. Therefore we can write EN/NB = BK/KF or EN/MK = NM/KF and finally EN/NM = MK/KF. Futermore <ENM = 90+<BNM = 90+<BKM = <FKM. Then ENM~MFK from which, as NM _|_FK and EN_|_MK, follows that <EMF = 90.

Now <EAF = <EAB + <BAF = <ECB + <BDF. But CEN and FDK are similar rectangular triangles, from which  for instance, <ECB = <DFK and finally <EAF = 90.


logarithms
real analysis
real analysis unsolved

Let two sequences of reals numbers $p_n >1$ and $u_n>0$

$C_n=\sqrt[p_1]{u_1+\sqrt[p_2]{u_2+\sqrt[p_3]{u_3+\dots+\sqrt[p_n]{u_n}}}}$

If $C_n$ converge  then $q_n=\frac{\ln ({u_1}^{p_2...p_n}+{u_2}^{p_3...p_n}+...+{u_{n-1}}^{p_n}+u_n)}{p_1...p_n}$ is bounded above
there exist $K>0$ such that $q_n\leq K$

Proof: Use $\sqrt[k]{a}+\sqrt[k]{b}\geq \sqrt[k]{a+b}$

Question: Does the converse is true ?

The necessary and sufficient condition I got is that $\frac{\ln(1+u_1+\dots+u_n)}{p_1\dots p_n}$ should be bounded (that is exactly what Moubinool once suggested)


For a positive integer $n$  , let
$a_{n}=\left\lfloor{n+\sqrt{n}+\frac{1}{2}}\right\rfloor$

Prove that the sequence $a_{1},\ ...\ ,a_{n}$ is the sequence of the positive integers that are not perfect squares .

Let $m$ be a positive integer, so that $m$ isn`t in ${\left\lfloor{k+\sqrt{k}+\frac{1}{2}\right\rfloor}}$ ($\star$)

Because $f(n)=n+\sqrt{n}+\frac{1}{2}$ is increasing, there is a number $n$ for which we have:

$n+\sqrt{n}+\frac{1}{2}<m\leq n+1+\sqrt{n+1}+\frac{1}{2}$

${\Rightarrow m\leq \left\lfloor{n+1+\sqrt{n+1}+\frac{1}{2}}\right\rfloor}$ but from ($\star$) $m$ can`t be ${\left\lfloor{n+1+\sqrt{n+1}+\frac{1}{2}}\right\rfloor}$

${\Rightarrow m+1\leq \left\lfloor{n+1+\sqrt{n+1}+\frac{1}{2}}\right\rfloor}\leq n+1+\sqrt{n+1}+\frac{1}{2}$

Therefore, $\sqrt{n}+\frac{1}{2}<m-n<\sqrt{n+1}+\frac{1}{2}$.
Let $x=m-n$, so : $n<(x-\frac{1}{2})^{2}\leq n+1$ and $-\frac{1}{4}<x^{2}-x-n\leq \frac{3}{4}$

But $x^{2}-x-n$ is an integer, so we have that : $x^{2}-x-n=0$ and
$n=x^{2}-x$ , $m=n+x=x^{2}$.


abstract algebra
group theory
superior algebra
superior algebra unsolved

I know how to prove that the only abelian group is $ Z /10Z$, but I am stuck trying to prove that the only nonabelian group of order 10 is the dihedral group $ D_5\equal{}\{x^i y^j : x^2\equal{}1, y^5\equal{}1, xy\equal{}y^{\minus{}1}x\}$.

Let $ G$ be a nonabelian group. By Cauchy's Theorem there are two elements $ x,y \in G$ such that $ x^2\equal{}1$ and $ y^5\equal{}1$. What I need to prove is that the order of $ xy$ is two, so $ xy\equal{}y^{\minus{}1}x.$

The teacher gave us the hint that we should use the formula  $ |Orbit(xy)| \equal{} [G: C_G(xy)]$ , where $ C_G(xy)$ is the centralizer of $ xy$ and $ Orbit(xy)\equal{}\{a \in G: axya^{\minus{}1}\equal{}xy\}$.   With this in mind I proceeded as this:

By using Lagrange Theorem and the fact that if $ G/Z(G)$ is cyclic implies $ G$ is abelian ($ Z(G)$ is the center of $ G$), we can obtain that $ Z(G)\equal{}\{1\}.$

It is easy to see that $ xy \neq 1$, so $ |C_G(xy)|>1$. The order of $ C_G(xy)$ cannot be $ 10$ since in that case $ G\equal{}C_G(xy)$, which means that $ xy$ is in $ Z(G)\equal{}\{1\}.$   Hence, $ |C_G(xy)|\equal{}5.$  Thus, $ |Orbit(xy)| \equal{} [G: C_G(xy)]\equal{}2.$ Then, I do not know how to continue.

Am I proceeding correctly?

Morova

[quote="morova"]$ Orbit(xy) \equal{} \{a \in G: axya^{ \minus{} 1} \equal{} xy\}$[/quote]
I think you mean $ C_G(xy)\equal{} \{a \in G: axya^{ \minus{} 1} \equal{} xy\}$.
The orbit is the conjugacy class $ \{axya^{ \minus{} 1}: a \in G\}$.
[quote="morova"]
Thus, $ |Orbit(xy)| \equal{} [G: C_G(xy)] \equal{} 2.$ Then, I do not know how to continue.[/quote]
Then $ C_G(xy)\unlhd G$ and by Sylow it is the only subgroup of order $ 5$.
So $ C_G(xy)\equal{}\langle y\rangle\equal{}\langle xy\rangle$ contains $ (xy)y^{\minus{}1}\equal{}x$. Contradiction!

But I think this is too much work.
$ |G: \langle y\rangle|\equal{}2$ gives $ \langle y\rangle\unlhd G$ and $ xyx^{\minus{}1}\in\langle y\rangle$.
Then conjugation with $ x$ gives an automorphism of $ \langle y\rangle$ of order $ \le 2$.
But $ \langle y\rangle\cong \mathbb{Z}/5\mathbb{Z}$ has automorphism group $ (\mathbb{Z}/5\mathbb{Z})^\times$ cyclic of order $ 4$
and the only automorphisms of order $ \le 2$ are $ y\mapsto y^{\pm 1}$.
So $ G$ is epimorphic image of $ \langle x,y: x^2\equal{}y^5\equal{}1,xyx^{\minus{}1}\equal{}y^{\pm 1}\rangle$,
which is $ \mathbb{Z}/10\mathbb{Z}$ or $ D_{10}$ both of order $ 10$.


[quote="tennisstar"][b]Problem #1:[/b]
When I was in the lobby, a message popped up randomly that said that I had specified a room that didn't exist.
[b]
Problem #2:[/b]
This was the question: In any month with a ___ Friday the Thirteenth ___ what day of the week is the first day of the month?

In the ___ spaces, there was a "a" with some symbol on top of it.

Thanks![/quote]

Å. Was that the symbol? or was it a with : ¨ on top

There was also one, I reported an error, but it said
$ \text{How many bottles are in the flat}\bar{a}\text{s?}$

When you see an error with a problem, please click the review button either at the end of the game or in the lobby after the game, browse for the problem, and click "Report Error."  This allows us to easily know which problem you are reporting.


integration
search
calculus
calculus computations

Does the sum of this series have limit.

$ \sum_{n \equal{} 1}^{\infty}\frac {1}{n}$ :blush:

$ \sum_{i\equal{}1}^n\frac{1}{i}\equal{}\plus{}\infty$
Posted before.

No it would be a good exercise for you to show that the series (the harmonic series) diverges.

I tried to do in this way $ \int _{0}^{\infty}\frac{1}{x}dx\equal{}lnx|^{\infty}_{0}\equal{}\infty$.

Note that $ \sum_{n \equal{} 1}^{\infty}\frac{1}{n} \equal{} \int_{0}^{1} \frac{1}{1\minus{}x}dx$


modular arithmetic
algebra
polynomial
group theory
number theory
relatively prime
abstract algebra

What value of k gives the most distinct kth power residues mod 29?
In particular, I'm looking for a k such that for n=0-28 $ n^{k}$ is unique.

You actually don't need a primitive root to show that if $ k$ is relatively prime to 28, then the $ k$th powers are a rearrangement of the original elements mod 29.

To show this, we only need to show that raising to the $ k$th power is 1-1. If $ a^k\equiv b^k$, we must show $ a\equiv b$. This is certainly true if $ a\equiv 0$ or $ b\equiv 0$. Otherwise, $ a^{28}\equiv b^{28}\equiv 1$. Now choose $ m$ such that $ km\equiv 1\pmod{28}$ and raise both sides of $ a^k\equiv b^k$ to the $ m$th power to get $ a\equiv b$.

To prove the converse, since $ (\minus{}1)^2\equiv 1$, $ k$ must be odd. To show $ k$ not a multiple of 7, you would need to find $ \omega \not\equiv 1$ such that $ \omega ^7\equiv 1$. This can certainly be done with primitive roots, but in this case it's easy to find an $ \omega$ (namely, 16).

[quote="t0rajir0u"]$ k \equal{} 1$?  :P

[hide="Discussion"] $ \mathbb{Z} / 29 \mathbb{Z}$ has a primitive root (because $ 29$ is prime); its multiplicative structure is therefore that of a cyclic group, so for all $ k$ such that $ \gcd(k, 28) \equal{} 1$ there will be precisely $ 28$ distinct (non-zero) $ k^{th}$ powers. [/hide][/quote]
I understand what primitive roots and cyclic groups are, but how does that imply that for k such that $ (k,28) \equal{} 1$, $ x^{k} \equal{} y^{k} \Rightarrow x \equal{} y$?

[quote="Altheman"]
Do we need to say that it is a cyclic group? I think that the existence of a primitive root is sufficient for this problem.

How do we prove either of these propositions?[/quote]
All moduli of the form $ 2 \cdot p^{k}$ have primitive roots (I think).
[quote="scorpius119"]You actually don't need a primitive root to show that if $ k$ is relatively prime to 28, then the $ k$th powers are a rearrangement of the original elements mod 29.

To show this, we only need to show that raising to the $ k$th power is 1-1. If $ a^k\equiv b^k$, we must show $ a\equiv b$. This is certainly true if $ a\equiv 0$ or $ b\equiv 0$. Otherwise, $ a^{28}\equiv b^{28}\equiv 1$. Now choose $ m$ such that $ km\equiv 1\pmod{28}$ and raise both sides of $ a^k\equiv b^k$ to the $ m$th power to get $ a\equiv b$.

To prove the converse, since $ ( \minus{} 1)^2\equiv 1$, $ k$ must be odd. To show $ k$ not a multiple of 7, you would need to find $ \omega \not\equiv 1$ such that $ \omega ^7\equiv 1$. This can certainly be done with primitive roots, but in this case it's easy to find an $ \omega$ (namely, 16).[/quote]
But aren't we assuming that raising to the mth power is 1-1?

Sorry if my questions have obvious answer, but I'm new to a lot of this stuff  :oops: .

We actually don't use the fact that raising to the $ m$th power is 1-1, just that it's well-defined:
\[ a^k\equiv b^k\Rightarrow (a^k)^m\equiv (b^k)^m\Rightarrow a\equiv b\]

[quote="diophantient"]I understand what primitive roots and cyclic groups are, but how does that imply that for k such that $ (k,28) \equal{} 1$, $ x^{k} \equal{} y^{k} \Rightarrow x \equal{} y$?[/quote]

Multiplication $ \bmod 29$ is isomorphic to addition $ \bmod 28$ (the explicit isomorphism is given by $ \phi(p^k) \equal{} k$ for $ p$ a primitive root).  This is the content of scorpius' solution.  In particular, raising to the power of $ k \bmod 29$ is isomorphic to multiplication by $ k \bmod 28$, which is clearly one-to-one since the inverse of $ k$ exists.

[quote="diophantient"][quote="Altheman"]
Do we need to say that it is a cyclic group? I think that the existence of a primitive root is sufficient for this problem.

How do we prove either of these propositions?[/quote]
All moduli of the form $ 2 \cdot p^{k}$ have primitive roots (I think).[/quote]

Only for $ p > 2$.  The set of $ m$ such that $ \mathbb{Z}/m\mathbb{Z}$ has a primitive root is precisely $ \{ 1, 2, 4, p^k, 2p^k \}$ over all odd primes $ p$.

This general theorem is not easy to prove.  It is, however, fairly easy to prove that a primitive root exists $ \bmod p$.  In particular, we need only to prove Lagrange's Theorem (a polynomial of degree $ n$ has at most $ n$ roots), and then consider the roots of the polynomials

$ x^{\frac {p \minus{} 1}{d} } \minus{} 1 \equiv 0 \bmod p$

over all divisors $ d$ of $ p \minus{} 1$.

[quote="diophantient"][quote="scorpius119"]We actually don't use the fact that raising to the $ m$th power is 1-1, just that it's well-defined:
\[ a^k\equiv b^k\Rightarrow (a^k)^m\equiv (b^k)^m\Rightarrow a\equiv b
\]
[/quote]
But can we go backwards from $ a^{km} \equiv b^{km}$ to $ a^{k} \equiv b^{k}$?[/quote]

We don't need to, although we can because $ km \equal{} 1$.  Then we just raise to the power of $ k$ again.

[quote="t0rajir0u"]
[quote="diophantient"][quote="scorpius119"]We actually don't use the fact that raising to the $ m$th power is 1-1, just that it's well-defined:
\[ a^k\equiv b^k\Rightarrow (a^k)^m\equiv (b^k)^m\Rightarrow a\equiv b
\]
[/quote]
But can we go backwards from $ a^{km} \equiv b^{km}$ to $ a^{k} \equiv b^{k}$?[/quote]

We don't need to, although we can because $ km \equal{} 1$.  Then we just raise to the power of $ k$ again.[/quote]
Okay, I understand now :)
2 more questions:
1) So we can generalize this holding for all k $ (k,p\minus{}1)\equal{}1$ in mod p?
2) How does the solution with primitive roots work?


(1) Is there is any short cut method to find the bond order of any compound like CO.
N2,O2

Finding the number of electrons in bonding and anti-bonding orbitals is normally the safest method.


Ehsan2004! It is not a first time I ask you to name topics properly! ;)

Rrr... [b]M[/b]yth.

http://www.mathlinks.ro/viewtopic.php?p=140098

[quote="ehsan2004"]Find all [b](p,x,y,z)[/b] such that: [b]p[/b] is a prime number and [b]x,y,z[/b] is natural number and [b]x p+y p=p z.   [/b][/quote]
If $p\mid x$ or $p\mid y$ then $p\mid x,y$. So we may assume that $(x,y)=1$.

Case 1. $p=2$. Then $x=y=z=1$.

Case 2. $p>2$. Then $x+y=p^l$, thus $x^p+y^p=p^{2l+1}M+p^{l+1}x=p^z$. Since $p\nmid x$ we conclude $z=l+1$. But $p^{l+1}=x^p+y^p\geq 2\left(\frac{x+y}{2}\right)^p=\frac{p^{pl}}{2^{p-1}}$ -- impossible for $l\geq 2$. And for $l=1$ we obtain $2^{p}+1\leq x^p+(p-x)^p=p^2$, thus $p=3$.

I thought an interger answer was incorrect but you have same exact answer I had come up with. Your proof is much more rigorous though.


I've seen things concerning what is allowed and what is not i.e. rulers, protractors, ...

But I never found out if colored writing implements (colored markers, highlighters, and the like) are acceptable to use on AIME for things like highlighting key words in a problem (BESIDE for #2 on AIME I 2007) or outlining specific figures in a diagram.

Can someone (AMCDirector maybe?) give me a ruling on this?

I have no idea why colored writing implements would not be allowed.  It would be absurd, IMHO, to allow protractors and such, but not colored pencils.  Protractors give you much more of an advantage.

hmm it seems like colored writing utensils would be helpful. Did they say we can use pens? If so, why don't we just use colored pens?

Or, you could wait until the AMC director makes a reply :idea: .

Colored utensils are [i]really[/i] helpful on certain geometry problems with too many lines. If allowed I'd definately bring a set to the USAMO.

I don't see what the big deal is, honestly. I think I've seen people use colored pencils, even on the AIME.

If you missed beside in #2 when you didn't have a highlighter, what makes you think you wouldn't miss it if you did have one?


Source: CSU math day competition 2006

Three unit circles are mutually tangent and enclose a triangular like region R. Find the area of R.

[hide="answer"]
$\sqrt{3}-\frac{\pi}{2}$

[/hide]

[hide="Hint"] Consider the triangle formed by the centers of the circles. [/hide]

Another way to do it is to draw a large square around the circles and then break it into smaller squares such that you can find the area of a smaller square and subtract it from the area of a part of the circle if you know what I mean.


geometry
calculus
analytic geometry
geometric transformation
rotation
trigonometry

can anyone find a relatively simple method (no calculus) to find the coordinates of a point Q (x,y) which is the rotation of point P (a,b) through an angle of $l$ about the origin? (x,y) in terms of a,b,$l$

Let $a=r\cos \theta$, $b=r\sin \theta$ (where $r=\sqrt{a^2+b^2}$). Then:
\[ x=r\cos (\theta+l)=r\cos \theta \cos l-r\sin \theta \sin l=a\cos l-b\sin l \]
\[ y=r\sin (\theta+l)=r\sin \theta \cos l+r\cos \theta \sin l=b\cos l+a\sin l \]


modular arithmetic
algebra unsolved
algebra

this is more number theory than algebra problem!

[quote="Facci"]For every prime $p>5$
and $a,b \in \{1,2,3,4...,p-2\}$
$c,d \in \{1,2,3,4...,p-1\}a \neq b$ prove:
if
$ac \equiv a+1 \pmod p$
$bd \equiv b+1 \pmod p$
then $c \neq d$[/quote]
It is obviosly $ab(c-d)=b-a(mod p)$. If $c=d(mod p)$, then contradition $a=b(mod p)$. It is true for any p (p can be not prime).


Let a be a positive real number, n a positive integer, and define the power tower a|n recursively with a|1=a, a|(i+1)=a^(a|i) for i=1, 2...For example, 4|3=4^(4^4)=4^256, a number which has 155 digits. For each positive integer k, let x_k denote the unique positive real number solution of the equation x|k=10|(k+1). Which is larger: x_42 or x_43?

[color=cyan]x_k is a decreacing function of k.
Proof:
Consider x_k.  This is the number which satisfies
x_k|k = 10|(k+1).  Now, clearly x_k is more than 10.  Therefore,
x_k|(k+1) = (x_k)^(10|(k+1)) > 10|(k+2) = x_(k+1)|(k+1).

Thus, x_k > x_(k+1).

Finis.[/color]


floor function
ceiling function
quadratics
probability
expected value
combinatorics unsolved
combinatorics

Let $X$ be a set with $n$ elements, and let $A_{1}$, $A_{2}$, ..., $A_{m}$ be subsets of $X$ such that:

1) $|A_{i}|=3$ for every $i\in\left\{1,2,...,m\right\}$;
2) $|A_{i}\cap A_{j}|\leq 1$ for all $i,j\in\left\{1,2,...,m\right\}$ such that $i \neq j$.

Prove that there exists a subset $A$ of $X$ such that $A$ has at least $\left[\sqrt{2n}\right]$ elements, and for every $i\in\left\{1,2,...,m\right\}$, the set $A$ does not contain $A_{i}$.

[i]Alternative formulation.[/i] Let $X$ be a finite set with $n$ elements and $A_{1},A_{2},\ldots, A_{m}$ be three-elements subsets of $X$, such that $|A_{i}\cap A_{j}|\leq 1$, for every $i\neq j$. Prove that there exists $A\subseteq X$ with $|A|\geq \lfloor \sqrt{2n}\rfloor$, such that none of $A_{i}$'s is a subset of $A$.

We will show a slightly stronger bound:

[color=blue][b]Theorem 1.[/b] Let X be a set with n elements, and let $A_1$, $A_2$, ..., $A_m$ be subsets of X such that $\left|A_i\right|=3$ for every $i\in\left\{1;\;2;\;...;\;m\right\}$ and $\left|A_i\cap A_j\right|\leq 1$ for any two different i and j.

A subset of X is called [i]good[/i] if it does not contain any of the sets $A_1$, $A_2$, ..., $A_m$.

Then, there exists a good subset of X whose number of elements is $\geq\left\lceil\sqrt{2n}-\frac12\right\rceil$.[/color]

This theorem implies the problem, since we always have $\left\lceil\sqrt{2n}-\frac12\right\rceil\geq\left\lfloor\sqrt{2n}\right\rfloor$ (in fact, $\left\lceil\sqrt{2n}-\frac12\right\rceil=\left\lfloor\sqrt{2n}\right\rfloor$ if $0\leq\sqrt{2n}-\left\lfloor{\sqrt{2n}}\right\rfloor<\frac12$ and $\left\lceil\sqrt{2n}-\frac12\right\rceil=\left\lfloor\sqrt{2n}\right\rfloor+1$ if $\frac12\leq\sqrt{2n}-\left\lfloor{\sqrt{2n}}\right\rfloor<1$).

[i]Proof of Theorem 1.[/i] We will show that, if we have a good subset T of X with u elements and $u<\sqrt{2n}-\frac12$, then we can find an element r of X \ T such that $T\cup\left\{r\right\}$ is also a good subset of X. Once this will be proved, it will be possible to construct stepwise a good subset of X, starting with the empty set (which is trivially good) and adding a new element in each step, until we reach a good subset of X whose number of elements will be $\geq\sqrt{2n}-\frac12$. Since the number of elements of a set is always integer, it will thus follow that the number of elements of this good subset is $\geq\left\lceil\sqrt{2n}-\frac12\right\rceil$. But this will obviously prove Theorem 1.

So it remains to show that, if we have a good subset T of X with u elements and $u<\sqrt{2n}-\frac12$, then we can find an element r of X \ T such that $T\cup\left\{r\right\}$ is also a good subset of X. Well, assume the contrary, i. e. assume that, for any element r of X \ T, the set $T\cup\left\{r\right\}$ is not a good subset of X. Then, it must contain one of the sets $A_1$, $A_2$, ..., $A_m$. Let $i_r$ be such that the set $T\cup\left\{r\right\}$ contains the set $A_{i_r}$. So all three elements of the set $A_{i_r}$ are in the set $T\cup\left\{r\right\}$. Obviously, one of these three elements must be r (else, the set T would already contain the set $A_{i_r}$, but this is impossible, since the set T is good). The other two elements must be from the set T.

Hence, for any element r of X \ T, we can find a pair of two different elements of the set T such that r, together with these two elements, forms one of the sets $A_1$, $A_2$, ..., $A_m$ (namely, the set $A_{i_r}$). Moreover, for two different elements r and s of X \ T, we get different pairs of elements of T (in fact, if we would get the same pair of elements for r and s, then the sets $A_{i_r}$ and $A_{i_s}$ would satisfy $\left|A_{i_r}\cap A_{i_s}\right|\geq 2$, but this contradicts $\left|A_{i_r}\cap A_{i_s}\right|\leq 1$, which must be fulfilled since $i_r\neq i_s$, as the sets $A_{i_r}$ and $A_{i_s}$ are different because of different r and s). Thus, we have an injective transformation from the set X \ T to the set $\binom{T}{2}$ of pairs of two different elements of T. Hence, $\left|X\backslash T\right|\leq\left|\binom{T}{2}\right|$. Since $\left|X\backslash T\right|=\left|X\right|-\left|T\right|=n-u$ and $\left|\binom{T}{2}\right|=\binom{\left|T\right|}{2}=\binom{u}{2}=\frac{u\left(u-1\right)}{2}$, this becomes $n-u\leq\frac{u\left(u-1\right)}{2}$, what transforms into $u^2+u-2n\geq 0$. But since $u<\sqrt{2n}-\frac12$, we have $u^2+u-2n<\left(\sqrt{2n}-\frac12\right)^2+\left(\sqrt{2n}-\frac12\right)-2n=-\frac14<0$, so we get a contradiction. Thus, our assumption was wrong, and we can find an element r of X \ T such that $T\cup\left\{r\right\}$ is also a good subset of X. This, as explained above, yields the proof of Theorem 1, and thus the problem is solved.

  darij

Nice solution darij! I swear I have seen this problem somewhere before on the forum, but maybe it wasn't solved on there, can't find the link... :?

let us try to construct such a set. for a pair of points in $A$ we consider as a 'bad' point the third(if there is one such) point that determines a block.(note that the third point if exists,for a pair, is uniquely determined). Thus if we have chosen $k$ points, then we have atmost $\frac{k(k-1)}{2}$ bad points. if however we have a maximal such set, then this set does not admit an extension to $k+1$ points which means, that apart from the $k$ points in the set $A$, all the points of the set $X\setminus A$ are 'bad' points.
this gives a quadratic in $k$ and simplification of that gives the required bound.

looks like my solution is the same as darij's, and his has all the details that i have skipped.

I did exactly the same as sesahdri and darij, but my first idea was to consider a graph that has n vertices and m triangles such that every edge is contained in exactly one triangle.  :D maybe i am to obbssesed with graphs, maybe there is something to it. LOL..
i first heard this problem like 6 months ago because it is from Italy TST 1999, then I heard it was part also of a Iranian TST, now Valentin posts it as Romanian TST's 1999!
3 TST's in one year! is this problem from 1998 IMO Shorlist?

I just wanna ask if there are any boundaries for $m$, because if $m>\binom{n}{3}$ then there will be too many triangles so I wonder what's the maximal value of $m$ for which this problem holds :)

I just wanna ask if there are any boundaries for $m$, because if $m>\binom{n}{3}$ then there will be too many triangles so I wonder what's the maximal value of m for which this problem holds ?

[quote="Loser"]I just wanna ask if there are any boundaries for $ m$, because if $ m > \binom{n}{3}$ then there will be too many triangles so I wonder what's the maximal value of $ m$ for which this problem holds :)[/quote]

This is being / has been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=70146 (don't know if it was really answered there).

  darij

I have a solution for $\frac{2}{3}$ times te desired quantity.
Let's first count the number of triples $(a,b,c)$ where $a$ and $b$ are elements of $X$ and $c$ is one of the $A_i$ such that $a$ and $b$ belong to $c$.  The triples $(a,b,c)$ and $(b,a,c)$ are equivalent. The number of such triples is evidently $3m$. On the other hand we can get a lower bound on these triples. We know that each of the $\binom{n}{2}$ pairs can be in at most one of the $A_i$, therefore the number of triples is at most $\binom{n}{2}$. Hence $3m \leq \binom{n}{2} \leq \frac{n^2}{2}$.
Now we can translate the problem into hypergraph theory. The set $X$ will be our set of vertices. The $A_i$ will be our edges of our three-uniform hypergraph.
Of course if the number of edges is $< \frac{n}{3}$ then we can take our set $A$ to be the set of all $n$ vertices minus one from each edge, which would yield at least $\frac{2n}{3}$ vertices in our independent set whoch is for $n$ at least $2$ clearly more than $\frac{2\sqrt{2n}}{3}$. So we can assume $m \geq \frac{n}{3}$. Now we apply the Lemma and we are done.
[b]Lemma[/b]
Let $H$ be a three-uniform hypergraph with $m\geq \frac{n}{3}$ edges and $n$ vertices. Then it contains an independent set of size at least \[\frac{2n\sqrt{n}}{3\sqrt{3}\sqrt{m}}\] edges.
[b]Proof[/b]
[hide]the proof will be added soon, I can give you a hint now. Choose a subset of vertices by choosing each vertex with probability p. Then remove one vertex per every edge that you find in the chosen subset. What is the expected number of vertices in the  new (altered) subset? [/hide]

Our set $A$ will be this independent set which has size at least \[\frac{2n\sqrt{n}}{3\sqrt{3}\sqrt{m}} \geq \frac{2n\sqrt{n}}{3\sqrt{3}\sqrt{\frac{n^2}{6}}}=\frac{2\sqrt{2n}}{3}\]

Note that if anyone improves the statement of this lemma the improvement would maybe enable my solution to reach the desired bound. The improvement of this lemma is discussed here [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=599714&p=3560533#p3560533[/url]


function
number theory proposed
number theory

a) Find all multiplicative functions $f: \mathbb Z_{p}^{*}\longrightarrow\mathbb Z_{p}^{*}$ (i.e. that $\forall x,y\in\mathbb Z_{p}^{*}$, $f(xy)=f(x)f(y)$.)
b) How many bijective multiplicative does exist on $\mathbb Z_{p}^{*}$
c) Let $A$ be set of all multiplicative functions on $\mathbb Z_{p}^{*}$, and $VB$ be set of all bijective multiplicative functions on $\mathbb Z_{p}^{*}$. For each $x\in \mathbb Z_{p}^{*}$, calculate the following sums :\[\sum_{f\in A}f(x),\ \ \sum_{f\in B}f(x)\]

$\mathbb{Z}_{p}$ is multiplication group of all unit of field $\mathbb{Z}_{p}$, so it is cyclic. If $a$ is lies $\mathbb{Z}_{p}$ such that $\mathbb{Z}_{p}=<a>$ then we only need dertemine value of $f(a)$. That is all!

So, working as N.T.Tuan did, then answers should be:
a) p-1; just choose the image of a generator
b) $\phi(p-1)$: the image of a generator has to be a generator!

$\sum_{f \in A}f(x) = 0$. It follows from the fact that:
$S = 1^{k}+2^{k}+\ldots+(p-1)^{k}= (1g)^{k}+(2g)^{k}+\ldots+((p-1)g)^{k}= g^{k}S$, since $g^{k}-1 \not \equiv 0$, $S \equiv 0$.

I'll let you prove that $\sum_{f \in B}f(x) = \mu (\mathrm{ord}x)$, this is very nice  :)


algebra
polynomial
quadratics
quadratic formula

Let $x_{1},x_{2},x_{3},x_{4}$ the roots of the following equation $(a^{2}-4)x^{4}-x^{3}(3a^{2}+4)+x^{2}(a^{2}+12)+x(3a^{2}+20)-2(a^{2}-4)=0$ where $a\in\mathbb{C}$ \ $\mathbb{R}$ and $|a|=2$. Knowing that $x_{1}$ and $x_{2}$ are the roots independent form $a$ of the equation, compute $T=Im(x_{3})\cdot Im(x_{4})$.

[hide="Hint"]
It's not so obviously but we see that $x_{1}=-1$ and $x_{2}=2$. So I've divided the equation by $(x-2)(x+1)$ and I got a 2nd degree equation $(a^2-4)x^2-x(2a^2+8)+(a^2-4)=0$ with the roots $x_{3}=\frac{5a^2-4}{a^2-4}$ and $x_{4}=\frac{4-3a^2}{a^2-4}$.

Now we have to calculate $Im\left(\frac{5a^2-4}{a^2-4}\right)\cdot Im\left(\frac{4-3a^2}{a^2-4}\right)$.
[/hide]

[hide="partial solution"]By trial and error, -1 and 2 are roots regardless of $a$. (The possibilties are similar to Rational Root Theorem.)
On dividing by $(x+1)(x-2)$, the new polynomial, $(x-x_3)(x-x_4)=(a^2-4)x^2-(2a^2+8)x+(a^2-4)=0$.

From quadratic formula, $x=\frac{a^2+4\pm \sqrt{(a^2+4)^2-(a^2-4)^2}}{a^2-4}=\frac{a^2\pm 4a+4}{a^2-4}$
$\Rightarrow x_3=\frac{a+2}{a-2},x_4=\frac{a-2}{a+2}$.

Someone else will have to finish this...[/hide]

Oupss :blush: I calculate wrong $x_{3}$ and $x_{4}$. now it's easy! Stupid mistake :P!


function
inequalities
algebra proposed
algebra

[hide="Uh"]$ x \equal{} 1/y$, then $ y \equal{} 1$ gives $ xf(1) \leq f(x) \leq xf(1)$ for all nonzero $ x$. Put in $ y \equal{} 0$ to see that $ f(0) \equal{} 0$. Done? $ f(x) \equal{} cx$[/hide]

Dear [b] MellowMelon [/b]

In case $ y<0$, putting $ x \leftarrow \frac1y$ yields nothing other than $ y f(1) \geq f(y)$. What you proved is that $ f(x) \equal{} cx$ for all $ x \geq 0$.

[quote="moldovan"]Determine all functions $ f: \mathbb{R} \to \mathbb{R}$ satisfying the condition $ f(xy) \le xf(y)$ for all real numbers $ x$ and $ y$.[/quote]
For all x,y is different from 0 we have :
$ \frac{f(xy)}{xy}\leq \frac{f(y)}{y}$
For each pairs of $ (u,v)$ ,each of them is different from 0  we can find $ x,y$ such that
$ u\equal{}x,v\equal{}xy$ (indeed $ u\equal{}x,y\equal{}\frac{v}{u}$)
Therefore we have $ \frac{f(u)}{u}\geq\frac{f(v)}{v}\geq \frac{f(u)}{u}$ ,it means that $ f(u)\equal{}au$ forall u is different from 0.
Now take $ x\equal{}0$ we have $ f(0)\leq 0$
Take $ y\equal{}0$ and $ x>1$ then we have $ f(0)\geq 0$
So $ f(0)\equal{}0$
From above result we have $ f(x)\equal{}cx,\forall x\in R$

Hi There,

  Unfortunately, TT sphn's solution is incorrect. His claim that
$ \frac{f(xy)}{xy} \leq \frac{f(y)}{y}$ is incorrect. This is because $ f(xy) \leq xf(y)$. But dividing the inequality by $ xy$ need not preserve the inequality as $ xy < 0$ is possible.

The correct solution is as follows:
$ f(x) \equal{} xf(1) \;\;\forall x \geq 0$
$ f(x) \equal{} \minus{}xf(\minus{}1) \;\;\forall x < 0$
$ f(1) \leq \minus{}f(\minus{}1)$.
One can verify that this function satisfies the given functional equation. One may have to verify four cases
$ (i) x \geq 0, y\geq 0$
$ (ii) x \geq 0, y < 0$
$ (iii) x < 0, y < 0$
$ (iv) x < 0, y\geq 0$
In all four cases, the above function satisfies the given functional equation.

Setting y=1 we get that $f(x) \leq x f(1)$. So $f(xy) \leq xf(y) \leq xyf(1) $.Let $y=\frac{1}{x}$. We have that $f(1) \leq xf(\frac{1}{x}) \leq f(1) $, so $f(x)=xf(1)$, for all x.


inequalities
geometry
trigonometry
area of a triangle
Heron\u0027s formula

this is from australian maths olympiad 2004

prove that: $(n+1)^n \geq  2^n(n!)$

Rewrite it as $n! \leq \left( \frac{n+1}{2} \right)^n$, then it becomes:
http://www.mathlinks.ro/Forum/viewtopic.php?p=336039#p336039

[hide="Proof"]From AM-GM inequality we have:

\begin{eqnarray*}{{1+2+\dots+n}\over n} &\geq& \sqrt[n]{1\cdot2\cdot\dots\cdot n}\\{{{n(n+1)}\over 2}\over n} &\geq& \sqrt[n]{n!}\\{{n+1}\over 2} &\geq& \sqrt[n]{n!}\\n+1 &\geq& 2\sqrt[n]{n!}\\(n+1)^n &\geq& 2^nn!\end{eqnarray*}[/hide]

while we are still on the topic .. here are some more questions

1) Use AM-GM to prove that $a^2 + b^2 + c^2 \geq ab + bc + ac$

2) For a real number $a$ and positive real number $x$, prove that $x(a-x) \leq \frac{a^2}{4}$

3) For all positive real numbers$a$, prove that $a + \frac{1}{a} \geq 2$

4) For all positive real numbers a, b, c, d such that $a+b+c+d=1$, prove that $\sqrt{4a+1} + \sqrt{4b+1} + \sqrt{4c+1} + \sqrt{4d+1} < 6$

1. Also, without AM-GM: Expand $\frac{1}{2}\left((a-b)^2 + (b-c)^2 + (c-a)^2\right) \geq 0$.

2. Clearly, if $(a-x) < 0$, then the inequality holds, so let's assume that $(a-x) \geq 0$. As $x$ is positive, we can use AM-GM on $\{x, a-x\}$: \[ \frac{a}{2}=\frac{x+(a-x)}{2}\geq \sqrt{x(a-x)}, \]which gives us the result immediately.

3. Use AM-GM to show that \[ a+\frac{1}{a}\geq 2\cdot\sqrt{a\cdot\frac{1}{a}} = 2 \]

4. According to Cauchy-Schwarz-Bunjakowsky \[ \sqrt{4a+1} + \sqrt{4b+1} + \sqrt{4c+1} + \sqrt{4d+1} \leq \sqrt{\left(1^2+1^2+1^2+1^2\right)\cdot\left(4a+1+4b+1+4c+1+4d+1\right)} = \sqrt{4\cdot(4+4)} = \sqrt{32} < 6 \]

2. $({a\over 2}-x)^2\geq 0\implies {{a^2}\over 4}\geq ax-x^2$

3. $(\sqrt{a}-{1\over\sqrt{a}})^2\geq 0\implies a+{1\over a}\geq 2$

4. From AM-QM it follows

\begin{eqnarray*}\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}+\sqrt{4d+1}\leq 4\sqrt{{4a+1+4b+1+4c+1+4d+1}\over 4}}=4\sqrt{2}<6\end{eqnarray*}

1)

${RTP: } a^2 + b^2 + c^2 \geq ab + bc + ac$

from AM-GM

$\frac{a^2 + b^2}{2} \geq ab \implies a^2 + b^2 \geq 2ab$

Similarly $b^2 + c^2 \geq 2bc$ and $c^2 + a^2 \geq 2ac$

Adding the three inequalities together:

$2a^2 + 2b^2 + 2c^2 \geq 2ab + 2ac + 2bc \implies a^2+b^2+c^2 \geq ab+bc+ac$

2)

${RTP: } x(a-x) \leq\frac{a^2}{4}$

Rearranging gives: $4ax \leq a^2+4x^2$

From AM-GM inequality

$\frac{a^2 + 4x^2}{2} \geq \sqrt{a^2\cdot4x^2}$

$\implies a^2 + 4x^2 \geq 4ax$

3)

${RTP: }a + \frac{1}{a} \geq 2$

Using AM-GM inequality we obtain: $\frac{a + \frac{1}{a}}{2}\geq\sqrt{a\cdot\frac{1}{a}}$

$\implies a + \frac{1}{a}\geq 2$

4) ${RTP: }\sqrt{4a+1} + \sqrt{4b+1} + \sqrt{4c+1} + \sqrt{4d+1} < 6$

$\sqrt{4a+1} = \sqrt{2(2a + \frac{1}{2})}$

By AM-GM inequality,

$\frac{2+(2a+\frac{1}{2})}{2}\geq\sqrt{2(2a + \frac{1}{2})}$

$\therefore \sqrt{4a+1}\leq\frac{2+(2a+\frac{1}{2})}{2}$

Similarly, $\sqrt{4b+1}\leq\frac{2+(2b+\frac{1}{2})}{2}$

$\sqrt{4c+1}\leq\frac{2+(2c+\frac{1}{2})}{2}$

$\sqrt{4d+1}\leq\frac{2+(2d+\frac{1}{2})}{2}$

Now adding the four inequalities together,

$\sqrt{4a+1} + \sqrt{4b+1} + \sqrt{4c+1} + \sqrt{4d+1}\leq \frac{10+2a+2b+2c+2d}{2}$

But $a+b+c+d = 1 \implies 2a+2b+2c+2d = 2$

$\implies \sqrt{4a+1} + \sqrt{4b+1} + \sqrt{4c+1} + \sqrt{4d+1} < 6$

Here are two geometric inequalities...im new at inequalities and need a bit of help.

1) Let a, b, c be the sides of a triangle; prove;

$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} < 2$

2) Let x, y, zbe sides of a triangle, prove.$\frac{1}{x+y-z} + \frac{1}{y+z-x} + \frac{1}{z+x-y} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$

[quote="sen"]does it matter whether a, b, c are sides of a triangle? that means the inequality holds for when a, b, c are positive.[/quote]

It has to be side of triangle . Or else when $a=b=1$ and pick a very large $c$ say $c=100$ then the inequality doesnt hold .

Problem 2
Let $a=y+z-x$, $b=z+x-y$, $c=x+y-z$. Since $x,y,z$ are the sides of a triangle, $a,b,c$ are positive real numbers.
The original inequality is equivalent to
$\frac1a+\frac1b+\frac1c\geq\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}$
By well-known inequaltiy (e.g. AM-HM or Cauchy-Schwarz or any other else) we have $\frac1a+\frac1b\geq\frac{4}{a+b}$. Adding analogous inequalties gives the result.

um, $a_k$ are positives, right? (There are counterexamples for negatives, such as -1/2, -1/2, 4.)

[hide]Let $\sigma_j$ be the sum of all products of $j$ of the $a_k$, so that $\sigma_1=\sum_k a_k$, $\sigma_2=\sum_{k<l} a_ka_l$, ..., $\sigma_n=\prod_k a_k$.

$\prod_k (1+a_k)=1+\sum_j \sigma_j$. Now each of the $a_k$ appears an equal number of times for each $\sigma_j$, so the geometric mean after expanding $\sigma_j$ is $\prod_k a_k^\delta$ (where exact value of $\delta$ is irrelevant), which equals 1.

The arithmetic mean of $\sigma_j$ is $\frac{\sigma_j}{\binom{n}{j}}$ because there are $\binom{n}{j}$ terms.

By AM-GM, $\frac{\sigma_j}{\binom{n}{j}}\geq 1\Rightarrow \sigma_j\geq \binom{n}{j}$.

Combining, $1+\sum_j \sigma_j\geq 1+\sum_{j=1}^n \binom{n}{j}=\sum_{j=0}^n \binom{n}{j}=(1+1)^n=2^n$.

And so it follows that $\prod_k (1+a_k)\geq 2^n$.[/hide]

[quote="sen"]Problem:

Prove that $(1+a_1)(1+a_2)...(1+a_n)\geq2^n$ if $a_1\cdot a_2\cdot ... \cdot a_n=1$[/quote]

[hide]
Just take $\frac{1+a_i}{2} \ge \sqrt{a_i} \Rightarrow 1+a_i \ge 2\sqrt{a_i}$.

Then $\prod (1+a_i) \ge 2^n\sqrt{\prod a_i} = 2^n$.[/hide]

this is similar to one problem I remember

for any positive $a_k$ such that $\prod_{k=1}^na_k=1$ , find the minimum of

$(2+a_1)(2+a_2)\cdots (2+a_n)$

Yep paladin8 thats the way to go.. very quick indeed.

Another problem:

prove $a^2+b^2+c^2 \geq (a-b)^2 + (b-c)^2 + (c-a)^2 +4\sqrt{3}A$

where a, b, c are sides of a triangle and area A.

[quote="sen"]Yep paladin8 thats the way to go.. very quick indeed.

Another problem:

prove $a^2+b^2+c^2 \geq (a-b)^2 + (b-c)^2 + (c-a)^2 +4A\sqrt{3}$

where a, b, c are sides of a triangle and area A.[/quote]

[hide]
Don't know if this is the best solution, but set $a = x+y, b = y+z, c = z+x$ and the inequality becomes

$(x+y)^2+(y+z)^2+(z+x)^2 \ge (x-z)^2+(y-x)^2+(z-y)^2+4A\sqrt{3}$
$4(xy+yz+zx) \ge 4A\sqrt{3}$
$xy+yz+zx \ge A\sqrt{3}$

By Heron's we have $A = \sqrt{xyz(x+y+z)}$, so it remains to show that

$xy+yz+zx \ge \sqrt{3xyz(x+y+z)}$.

Let $p = xy$, $q = yz$, $r = zx$ to get

$p+q+r \ge \sqrt{3(pq+qr+rp)}$
$(p+q+r)^2 \ge 3(pq+qr+rp)$
$p^2+q^2+r^2 \ge pq+qr+rp$,

which is true by AM-GM, Trivial, Rearrangement, etc.[/hide]

Hmmm I should try using [hide="elephant tusks"]
So use some Law and Cosine and stuff to be like "let's try to prove:"

$ab + bc + ca \ge ab \cos C + bc \cos A + ac \cos B + 2 \sqrt{3} A$

Now, now... I'm sensing a conspiracy here let's do something like

$ab + bc + ca \ge \sum ab \left( \cos C + \frac{1}{\sqrt{3}} \sin C \right)$

$ab + bc + ca \ge \frac{2}{\sqrt{3}} \sum ab ( \sin 60 \cos C + \cos 60 \sin C )$

$ab + bc + ca \ge \frac{2}{\sqrt{3}} \sum ab \sin (60 + C)$

Errr... darn now I think you have to use Jensen, which is like orznorz too bad. But it looks really cool ;). Oh wait one last try geometrically... construct those equilateral triangles on the sides... wait darn that goes back to the original equation.
[/hide] to solve this..

It's not necessary to put a=x+y etc...
By AM-GM on Heron's formula you get $A\leq \frac{(a+b+c)^2}{12\sqrt{3}}$ and finally it remains to prove $ab+bc+ac\leq a^2+b^2+c^2$.


Consider we have a sequence like the below and we have to evaluate the sum

$5+7+13+31+85.......$

Now we subtract consecutive terms...we get

$2 , 6, 18 ,54....$

Now we could subtract them till we hit a repeating number but that would be too big.In the above equation,we can see that its a $G.P$.So how do we write the $nth$ term for such an occurence?What if we subtracted twice,thrice....to get a G.P(what would be the nth sequence then)?

Here's one way to find a formula for the sum of the first $n$ terms:

[hide]We can define the sequence recursively by $t_{1}=5$, $t_{n}=t_{n-1}+2\cdot 3^{n-2}$.  Then we can use the recursive definition repeatedly to find a explicit definition.

$t_{n}=t_{n-1}+2\cdot 3^{n-2}$
$t_{n}=t_{n-2}+2\cdot 3^{n-3}+2\cdot 3^{n-2}$

Eventually we get

$t_{n}=t_{1}+(2\cdot 3^{0}+2\cdot 3^{1}+\cdots +2\cdot 3^{n-2})$
$t_{n}=5+2(1+3+9\cdots +3^{n-2})$
$t_{n}=5+2\cdot\frac{3^{n-1}-1}{3-1}$
$\boxed{t_{n}=3^{n-1}+4}$

Now we want to find the sum.  Let $S_{n}$ be the sum of the first $n$ terms.

$S_{n}=\sum_{k=1}^{n}(3^{n-1}+4)=4n+\sum_{k=1}^{n}(3^{n-1})$
$S_{n}=4n+(1+3+9+\cdots +3^{n-1})$
$S_{n}=4n+\frac{3^{n}-1}{3-1}$
$\boxed{S_{n}=4n+\frac{3^{n}-1}{2}}$[/hide]


cot 16 cot 44 +cot44 cot 76 -cot 76 cot 16=                         (ANGLES ARE IN DEGREE)

2.    cosec(x-a), cosec x,  cosec(x+a) are in   AP   and  " a"     belongs to  (-pi /2 to pi/2)  then x belongs to --------

3.SEC(X-A), SEC(X), SEC(X+A) ARE IN AP  THEN COS X  SEC(A/2) =

4. SIN X+COSX=K THEN VALUE OF X FOR WHICH K IS INTGER

Latexified:

$ 1. \cot{16^{\circ}}\cot{44^{\circ}}+\cot{44^{\circ}}\cot{76^{\circ}}-\cot{76^{\circ}}\cot{16^{\circ}}=x$
Solve for $ x$.

$ 2.$ If $ (\csc{(x-a)},\csc{x},\csc{(x+a)})\in\text{AP}$ and $ a\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, then $ x\in{ ?}$ (What's AP?)

$ 3.$ If $ (\sec{(x-a)},\sec{x},\sec{(x+a)})\in\text{AP}$ then $ \cos{x}\sec{\left(\frac{a}{2}\right)}={ ?}$

$ 4. \sin{x}+\cos{x}=k$ Solve for $ x$ such that $ k\in\mathbb{Z}$


Notice that the number of digits in any integer $ n$ is $ 1 \plus{} \lfloor\log n\rfloor$.

Thus, a number $ \log x$ that satisfies $ \lfloor\log(\log x)\rfloor \equal{} 3$ can have at most $ 4$ digits, which means that $ \log x < 10^4$, so

$ 1 \plus{} \lfloor\log x\rfloor\leq(10^4 \minus{} 1) \plus{} 1 \equal{} 10^4 \equal{} \boxed{10000}$.

[hide="solution"] So $ \log(\log x)<4$ so $ \log x<10^4$ and x is less than $ 10^{10000}$. Since $ 10^{10000}$ has 10001 digits, the answer is 10000.[/hide]


For a V atom in its ground electronic state:
How many electrons are there in the 4s orbital? ____ (

I just want to know what they mean by ground electronic state? How do i calculate this? what do i need to know in order to calculate this/

It just means the lowest energy state the atom can have so none of the electrons are "jumping" levels.


$\displaystyle 2n$ students $\displaystyle (n \geq 5)$ participated at table tennis contest, which took $\displaystyle 4$ days. In every day, every student played a match. (It is possible that the same pair meets twice or more times, in different days) Prove that it is possible that the contest ends like this:

- there is only one winner;

- there are $\displaystyle 3$ students on the second place;

- no student lost all $\displaystyle 4$ matches.

How many students won only a single match and how many won exactly $\displaystyle 2$ matches? (In the above conditions)

This problem was also on Croatian MEMO Selection Test 2007. My solution is as follows:

There is obviously 4n victories altogether. Let p be the number of winners' victories. Let d be the number of victories of one of second placed players. Let t be the number of victories of one of the third placed players and let c be the number of victories of one of the fourth placed players (if there are any). Obviously, c<t<d<p and p$\ge$3, d$\ge$2. There are four cases.

1. p=3, d=2, t=1 --- 1*p+3*d+(2n-4)*t=4n, and this leads to 2n=5. Contradiction.
2. p=4, d=2, t=1 --- 1*p+3*d+(2n-4)*t=4n, which leads to n=3. Problem states that n>4. Contradiction.
3. p=4, d=3, t=1 --- 1*p+3*d+(2n-4)*t=4n, 2n=9. Contradiction.
4. p=4, d=3, t=2, c=1 --- let k be the number of third-placed people. Then, 1*p+3*d+k*t+(2n-k-4)*c=4n, therefore k=2n-9.

There are 2n-9 players with 2 victories and 5 players with 1 victory.

We only have to show that this is actually achievable.
Let 1 be the number of first placed player
Let 2,3,4 be the numbers of second placed players
Let 5..2n-5 be the numbers of third placed players
Let 2n-4..2n be the numbers of fourth placed players

Winner - loser - day
1 - 2n-4 - 1
1 - 2n-3 - 2
1 - 2n-2 - 3
1 - 2n-1 - 4
2 - 2n - 1
2 - 2n-4 - 2
2 - 2n-3 - 3
3 - 2n-2 - 4
3 - 2n-1 - 1
3 - 2n - 2
4 - 2n-4 - 3
4 - 2n-3 - 4
4 - 2n-2 - 1
5 - 2n-1 - 2
5 - 2n - 3
2n - 2 - 4
2n-1 - 3 - 3
2n-2 - 4 - 2
2n-3 - 5 - 1
2n-4 - 5 - 4

Now we have an even number of players (6..2n-5). Each of them with an even number can play, say, with the player with the following number and win on days 1 and 2 and lose on days 3 and 4.

That's, all, folks...  :lol:


calculus
trigonometry
search
real analysis
real analysis unsolved

You might want to consult [url=http://en.wikipedia.org/wiki/Pythagorean_trigonometric_identity]this[/url].

[quote="giannis"]Proove using calculus $ sin^{2}x+cos^{2}x=1$ :lol:[/quote]

$ \sin x = x-\frac{x^{3}}{3!}+...$ and $ \cos x = 1-\frac{x^{2}}{2!}+...$

Show $ (\sin x)' = \cos x$ and $ (\cos x)' =-\sin x$.

Show that $ (\sin^{2}x+\cos^{2}x)'=0$

So $ \sin^{2}x+\cos^{2}x = C$ and show $ C=1$.

Another related topic: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142926


Find the natural number $ a$ such that $ \displaystyle\sum_{k \equal{} 1}^{\ n }f(a \plus{} k) \equal{} 16(2^{n} \minus{} 1)$,where the function $ f$ satisfies $ f(x \plus{} y) \equal{} f(x)f(y)$ and $ x,y \in \mathbb{N}$ and $ f(1) \equal{} 2$

[quote="snowingforest"]$ f(x) \equal{} f(1 \plus{} 1 \plus{} 1 \plus{} ... \plus{} 1) \equal{} f(1)*f(1)*f(1)*...f(1) \equal{} 2^x$

so we just need to find a that satisfied $ 2^a(2^n \minus{} 1) \equal{} 16*(2^n \minus{} 1)$

so the a is 4[/quote]
I think snowingforest is wrong in his  answer and I dont understand his method ,maybe he could elaborate it .The answer is [hide="this"]$ 3$[/hide]

Well, you made a little mistake: the answer is not $ 4$ but $ 3$: the case is that the final equation is in form $ 2^{a \plus{} 1}(2^n \minus{} 1) \equal{} 16(2^n \minus{} 1)$.

And just an offtop: it is a well known fact (and can be proved using Cauchy equation) that the only function which satisfies $ f(x \plus{} y) \equal{} f(x)f(y)$ is $ f(x) \equal{} b^x$, where $ b \equal{} const$. So in this case $ b \equal{} 2$.

Edit: [b]srinath.r[/b], i've been writing this post in the same time as you.   :roll:

Given that $ f$ is continuous, of course. But over the integers, it's just a simple matter of induction.

Well, if there are some restrictions in the statement, they cannot prohibit using $ a^x$ function, because it is 'universal' one. We just apply some of these restrictions to this function, in this case, at instance, that it is defined only for natural numbers.


I'm going to be a senior next year, and I go to a charter school, so the middle school is right across the street from the high school. I'm starting MAO at the high school for their competitions as well as AMC's, and I figured I would start mathcounts for the middle school students. How would I go about doing this, and what resources would I need?

I wouldn't know, but my guess is problems and money and more problems, among other things.

Well, aside from problems to do, food is pretty important. The only reason I went to my math club originally was the promise of free donuts for top scorers. Then, by the time the donuts were no longer offered, a love of math had been instilled in me (although I still liked food!)

You need something non-math related to lure students, like food. My math club has food and we play a card game (like 24 or Mao) for half an hour before a lesson or worksheet. That was what made me start going and then I started liking the math part. We only managed to get about 7 or so people out of 500 students, so we aren't experts.

You also need to advertise it well.


search
number theory proposed
number theory

Lets consider the sequence of digits $a_1,a_2,...$ for which $a_i=i$ for $i>0$. Find the least positive integer $k$ such that  from among the first $k$ digits of this sequence the number of every two nonzero digit is diffetent.

[quote="epitomy01"]could you rephrase the problem? don't quite understand it.
anyway, what happened to day 2 problem 3 bulgaria 06? could you please post it, i like bulgaria problems. sorry if it's already posted and my search was faulty.[/quote]
Here http://www.mathlinks.ro/Forum/viewtopic.php?t=88585

Nobody solved this problem :( I know that it is hard to write it but any ideas...


inequalities
inequalities solved
3-variable inequality
cauchy schwarz

If $x$, $y$, $z$ are positive reals, prove that

$\frac{\left(y+z\right)^{2}}{y+z+2x}+\frac{\left(z+x\right)^{2}}{z+x+2y}+\frac{\left(x+y\right)^{2}}{x+y+2z}\geq\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^{2}}{3}$.

Really straightforward: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15125]The Cauchy-Schwarz inequality in Engel form[/url] yields

$\frac{\left(y+z\right)^2}{y+z+2x}+\frac{\left(z+x\right)^2}{z+x+2y}+\frac{\left(x+y\right)^2}{x+y+2z}\geq\frac{\left(\left(y+z\right)+\left(z+x\right)+\left(x+y\right)\right)^2}{\left(y+z+2x\right)+\left(z+x+2y\right)+\left(x+y+2z\right)}$
$=\frac{4\left(x+y+z\right)^2}{4\left(x+y+z\right)}=x+y+z$.

Now, by the AM-QM inequality for the numbers $\sqrt{x}$, $\sqrt{y}$ and $\sqrt{z}$, we have

$\sqrt{\frac{\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+\left(\sqrt{z}\right)^2}{3}}\geq\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}$
$\Longleftrightarrow\ \ \ \ \ \sqrt{\frac{x+y+z}{3}}\geq\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}$
$\Longleftrightarrow\ \ \ \ \ \frac{x+y+z}{3}\geq\left(\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}\right)^2$
$\Longleftrightarrow\ \ \ \ \ x+y+z\geq 3\left(\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}\right)^2=\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}$.

Thus, we get

$\frac{\left(y+z\right)^2}{y+z+2x}+\frac{\left(z+x\right)^2}{z+x+2y}+\frac{\left(x+y\right)^2}{x+y+2z}\geq x+y+z\geq\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}$,

and it's done.

  Darij

Yes. this was my solution too. With a little difference. After the Causcy in Engel's form

I put $x=a^2$ , $y=b^2$ , $z=c^2$ so it suffices to prove that

$3(a^2+b^2+c^2)\geq (a+b+c)^2$ which is trivial


Gauss
Euler
ARML
articles
irrational number

WHO'S BETTER??

I think Euler is much more manly :D

To the extent that this is in part a running inside joke, it probably should stay within the Southern California ARML forum.

I don't want to get into ranking them - they're from different generations, subject to different prior knowledge and different standards of what constituted mathematics. There's such a sharp contrast between their attitudes toward their published work - the polished perfectionism of Gauss, with his model of "Few, but ripe," versus the wild and happy profusion of Euler, whose previously unpublished works were still filling journals 50 years after his death.

To me, an "Eulerian" approach to an analysis problem means to compute first and worry about the justification later - and I'm happy with that. I've seen good students get so hamstung by theory and justification that they're afraid to move - I don't want to be afraid. In Fourier analysis, I like distribution theory - that's a subject not even developed until the 20th century, so it doesn't belong to either Euler or Gauss, but the freedom to calculate that it entails seems to be something Euler would have appreciated. So without attempting any sort of ranking, I'll say that Euler appeals to me more.

I once read an AMM article that lightheartedly compared ages in the history of mathematics to ages in art and literature. One of the author's findings: the visual and literary arts tended to experience a certain age well before mathematics did - and that music often also trailed in time. For instance, the artistic Renaissance was in full swing by the middle of the 15th century, while the first real Renaissance mathematicians were probably Tartaglia, Cardano, and Ferrari. The Athens of Pericles and the Parthenon, of [i]Oedipus Rex[/i] and the comedies of Aristophanes - that was a century or two before Euclid and Archimedes and Apollonius. Part of my response was to start developing my own lists in which I matched a mathematician with someone from the arts - often from music. In part, that's becuase I know music; in part it's because music matches the timing of mathematics a little better.

Guass is maybe Beethoven - self-consciously aware of his revolutionary importance, dedicated to rewriting and perfection (although Beethoven did churn out a lot of "potboilers" to pay the bills.) But Euler is Haydn. Not trying to be a revolutionary, but so vast and prolific that previous forms couldn't hope to contain him. Mobile across Europe through both noble and royal patronage and his own talent. Generally tranquil domestic life. Hollywood made [i]Amadeus[/i] about Mozart and [i]Immortal Beloved[/i] about Beethoven, but I doubt that anyone has a similar project devoted to Haydn - where's the dramatic conflict there? (If you want me to match someone to Mozart, I might go with Abel - utter genius, gone too soon.)

[quote="Xevarion"]1) This DEFINITELY belongs in GFF.

2) Why does one of them have to be better than the other? I don't think there is a total order on mathematicians. (Well, one that corresponds to quality, that is. Certainly you could order them according to when they were born or something.)

3) If coerced, I'd probably go for Euler.[/quote]

I lol'd when I saw the Definitely and your avatar, I dunno why.

I'm sorry I'm being off topic but this is heading toward G&FF anyway.

but despite that we can classify their intelligence degree according to their works and i think that euler is more geniuis and more creatif but his only drawback is that he is not rigourous he don't give perfect proof

However,euler is not the best at all but i prefer riemann

Euler! He "invented" another irrational number that I can waste time and memorize!

2.71828185...well that's all I've gotten to atm  :D .

[quote="7h3.D3m0n.117"]2.71828185...well that's all I've gotten to atm  :D .[/quote]
You're either wasting your time, or not spending enough time on it, I'm not sure which.  It's 2.7182818[b]28[/b]45...

um woo go euler
according to dr. z gauss was a bit soft in the head
didnt know how to tie his shoes
or fill a bucket with water

Guys, Galois would totally KICK BOTH THEIR ASSES!

I'm not sure I can find a good match on the art/music side for Galois - someone who died so very young after producing a single great work. (Of course, we only really know about Galois's mathematics because Dirichlet took the trouble to digest, decypher, and explain his work.)

Someone mentioned Riemann. Riemann is the great Romantic mathematician - someone like Liszt or Schumann, only of even greater stature.


[quote="iura"]Excuse me if it's wrong or has been posted before:

If $x_{1},x_{2},\ldots,x_{n}>0$ and $\prod_{i=1}^{n}x_{i}=1$ and $p,q>0$ then

$\sum_{i=1}^{n}\sqrt{px_{i}^{2}+q}\leq \sqrt{p+q}\sum_{i=1}^{n}x_{i}$[/quote]
I think if one post a problem in the Section of Inequalities Proposed & Own Problems,he or she should make sure the problem is true and have the solution .

An old topic before, in Unsolved section said that this inequality is not true for any $p,q>0$.
But the following is true:

[b]Let $a,b,c>0$ whose product $abc=1$, then: \[\sqrt{3a^{2}+4}+\sqrt{3b^{2}+4}+\sqrt{3c^{2}+4}\le \sqrt{7}(a+b+c).\] [/b]

I also thought it msut not be true all the time. Then where is the msitake in the proof?

We try to break the inequality into $\sqrt{mx_{1}^{2}+n}\leq ax+\frac{\sqrt{m+n}-a}{n-1}(x_{2}+x_{3}+\ldots+x_{n})$. As the LHS
depends only on $_{1}x$, we apply AM-GM to the RHS to get the
inequality $ax_{1}+\frac{\sqrt{p+q}-a}{n-1}(x_{2}+x_{3}+\ldots+x_{n})\geq ax+\frac{\sqrt{p+q}-a}{x^{\frac{1}{n-1}}}$ (we let $x=x_{1}$ for
simplicity). Now square to get $a^{2}x^{2}+2a(\sqrt{p+q}-a)x^{1-\frac{1}{n-1}}+(p+q-2\sqrt{p+q}a+a^{2})x^{-\frac{2}{n-1}}\geq px^{2}+q$,
or $(a^{2}-p)x^{2}+2a(\sqrt{p+q}-a)x^{1-\frac{1}{n-1}}+(p+q-2\sqrt{p+q}a+a^{2})x^{-\frac{2}{n-1}}\geq q$. We
apply now weighted Am-GM to get
$(a^{2}-p)x^{2}+2a(\sqrt{p+q}-a)x^{1-\frac{1}{n-1}}+(p+q-2\sqrt{p+q}a+a^{2})x^{-\frac{2}{n-1}}\geq ((a^{2}-p)+2a(\sqrt{p+q}-a)+(p+q-2\sqrt{p+q}a+a^{2}))x^{\frac{2(a^{2}-p)+(1-\frac{1}{n-1})2a(\sqrt{p+q}-a)+(-\frac{2}{n-1})(p+q-2\sqrt{p+q}a+a^{2})}{(a^{2}-p)+2a(\sqrt{p+q}-a)+(p+q-2\sqrt{p+q}a+a^{2})}}=qx^{\frac{\frac{2n}{n-1}\sqrt{p+q}a-2(p\frac{n}{n-1}+\frac q{n-1})}{q}}$.
So now for the desired inequality to take place, we must take
$a=\frac{\frac{n}{n-1}p+\frac{1}{n-1}q}{n\sqrt{p+q}}=\sqrt{p+q}\frac{np+q}{n(p+q)}$. We can
see that $0<a<\sqrt{p+q}$ so for this value the desired breaking
holds and the conclusion follows.

I think that the solution is correct.

Also another approach maybe to observe that
$\sqrt{p+q}x-\sqrt{px^{2}+q}$ has a single inflection point and so we may put n-1 equal variables. this reduces to a one variable inequality.
Of course in this case one should carry out some calculations...

I think the proof above is quite nice and needs only some little corrections of some typos.

Lovasz do you have a counter example to this inequality ?

should we have:$a^{2}>p$?
But I think it is not always true.
 :?:

Indeed, the condition $a^{2}\geq p$ implies
$p \leq \frac{1}{n(n-2)}$ for $q=1$,
which is a weak condition.
See
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=vasc&t=28732


why?  :huh:

In the Iranian forum you do write in Iranian, I suppose, so why shouldn't we write in Italian? Anyway if you have something to ask, all of us can speak English, although we are not so active here  :P


Let a triangle ABC. On side BC, let two points I,K such that BI=CK. The circle (ABI) meets AC at N, circle (ACK) meets AB at M. Prove the midpoints of AB,AC,BN,CM form a cyclic quadrilateral.

My own solution to this problem.
[url]http://www.mathlinks.ro/viewtopic.php?t=218457[/url]


Ok i have a figure like so:
[asy]import graph;
draw((0,0)--(5,5)--(10,0)--cycle);
label("A",(0,0),W);
label("B",(5,5),N);
label("C",(10,0),E);[/asy]
I want to label $ \overline{AB}$ As 5 inches long.
How would i accomplish this

Thanks

Alternatively, you can draw all the lines, and then:

[code]
label("5 in",(2.5,2.5),NW);
[/code]

Although, I do believe my code is optimized. See ph. ivaldi's post here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=881188386&t=252047[/url]


inequalities
function
induction
analytic geometry
graphing lines
slope
geometry unsolved

Well we all know the corner point theorem for solving a Linear programming problem.
For all who dont know it, it states that:
Let R be the feasible region (convex polygon) for a linear programming problem and
Let Z = ax + by be the objective function. When Z has an optimal value (max or min), where the variables x and y are subject to constraints described by linear inequalities, this optimal value must occur at a corner point (vertex) of the feasible region.

How do we prove this theorem ?  :)

It's enough that the objective function be convex in both variables; the one-dimensional analogue is that a convex function on an interval $ [a, b]$ takes on its maximal value at an endpoint.  You can prove by induction on the dimension.

Sorry for the trouble but can you please provide the proof? :)

Thanks

Linear equations are continuous and differentiable...
So in closed region extrema has to be attained...

Use Kuhn-Tucker's algorithm....
If extrema occurs within feasible region, gradient = 0
But for linear equations gradient != 0(since a!= 0 or b!=0) so extrema only on boundary...

Boundary = open line segments + corner points...

On line segment if extrema occurs, gradient is perpendicular to line...
Objective function is ax+by
gradient = a(i)+b(j);
slope (b/a)
Case 1:
slope of constraint function is not (-a/b)...
Then gradient has component along line segment...
So we are not at extrema point

Case 2: slope of constraint function is (-a/b)
So equation at boundary of constraint function:
ax+by=c;
the objective function on this line has same value at each point...
So also at the corner point...

So we conclude that the extrema are attained at corner points..


[quote="cezar lupu"]Let $a,b,c>0$.Prove that:

                      $\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+(a+b+c)^2\geq4\sqrt{3abc(a+b+c)}$ ;)[/quote]

$\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+(a+b+c)^2 = \sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+3\frac{(a+b+c)^2}{3} \ge 4\sqrt[4]{\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c}) \cdot \left(\frac{(a+b+c)^2}{3}\right)^3} = 4\sqrt{a+b+c}\sqrt[4]{\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c}) \cdot \frac{(a+b+c)^4}{3^3}} \ge 4\sqrt{a+b+c}\sqrt[4]{\sqrt{abc} \cdot (3\sqrt[6]{abc}) \cdot \frac{(3\sqrt[3]{abc})^4}{3^3}} = 4\sqrt{3abc(a+b+c)}$

Yes,but it can be also proved by ;)  Schur.

[quote=Cezar Lupu]Let $a,b,c>0$.Prove that:
$$\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+(a+b+c)^2\geq4\sqrt{3abc(a+b+c)}$$[/quote]
[b](2004 Chinese National Team Training Test  Problems)[/b]
[b]Proof:[/b]
$$\sqrt{abc}(\sqrt a+\sqrt b+\sqrt c)\geq \sqrt{abc}\cdot 3\sqrt[3]{\sqrt a\cdot \sqrt b\cdot\sqrt c}=3\sqrt[3]{a^2b^2c^2},$$
$$(a+b+c)^2\geq 3\sqrt[3]{a^2b^2c^2}+\frac{1}{3}(a+b+c)^2\geq 3\sqrt[3]{a^2b^2c^2}+2\sqrt[3]{abc}(a+b+c),$$
$$\sqrt{abc}(\sqrt a+\sqrt b+\sqrt c)+(a+b+c)^2\geq 2(3\sqrt[3]{a^2b^2c^2}+\sqrt[3]{abc}(a+b+c))\geq 4\sqrt{3abc(a+b+c)}.$$
[url=https://artofproblemsolving.com/community/c6h19813p132008]h[/url] [url=https://artofproblemsolving.com/community/c6h197908p1088429]h[/url] [url=https://artofproblemsolving.com/community/c6h452217p2542716]h[/url]
Let$a,b,c $ be non-negative numbers. [url=https://artofproblemsolving.com/community/c6h295331p1598845]Prove that[/url]
$$ 15\sqrt {abc}(\sqrt {a} + \sqrt {b} + \sqrt {c}) + (a + b + c)^2\ge 18\sqrt {3abc(a + b + c)}$$
$$17\sqrt {abc}(\sqrt {a} + \sqrt {b} + \sqrt {c}) + (a + b + c)^2\ge 20\sqrt {3abc(a + b + c)}$$

[quote=Cezar Lupu]Let $a,b,c>0$.Prove that:
$$\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+(a+b+c)^2\geq4\sqrt{3abc(a+b+c)}$$[/quote]
[b]Strengthening:[/b]
Let $a,b,c>0$. [url=https://artofproblemsolving.com/community/c6h1938042p13340641]Prove that:[/url]
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\geq 2.$$
[hide=P]Let  $a\leq{b}\leq{c} . $ Then
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(a+c)(b+c)} \geq{\frac{2a^2+b^2+c^2}{(a+b)(a+c)}+\frac{8abc}{(a+b)(a+c)(b+c)} =\frac{(b+c-2a)(b-c)^2}{(a+b)(b+c)(c+a)}+2}\geq{2}$$
https://artofproblemsolving.com/community/c6h77613p451016:
 $$\frac{ab+bc+ca}{a^2+b^2+c^2}+\frac{\sqrt{2}\cdot (a+b)(b+c)(c+a)}{2abc} \ge 1+4\sqrt{2}.$$
[/hide]
$$\iff$$
Let $A,B,C$ are angles of a triangle. [url=https://artofproblemsolving.com/community/c6h491234p2755113]Prove that[/url]
$$\tan^2\frac{A}2+\tan^2\frac{B}2+\tan^2\frac{C}2 \ge 2-8\sin \frac{A}2\sin \frac{B}2 \sin \frac{C}{2} $$
[url=https://artofproblemsolving.com/community/c6h80859p463010]h[/url] [url=https://artofproblemsolving.com/community/c6h554833p3223553]h[/url] [url=https://artofproblemsolving.com/community/c6h603226p3583033]h[/url]


conics
ellipse
parabola
hyperbola
quadratics
algebra
quadratic formula

Find the equation of the path of a point that moves so that the difference of its distances from (2,-1) and (-1,0) is equal to 2.  Identify the curve.

Thanks to any posts.

oh sorry about that.

thanks for putting it in the right forum

that's one of the properties of conic sections
actually it is only true for ellipses, although it is sort of true for parabolas and hyperbolas if you are allowed to immagine the focus going of to infinity and coming back around the other way.

so the answer is an ellispe with foci at -2,-1 and -1,0

the major axis is length 2 and the minor axis is length $ \frac{\sqrt{3}}{2}$
after that you just plug in some numbers and solve for the actual equation

Wait, isn't the [i]sum[/i] of dist. to the foci constant for an ellipse?  I thought the constant diff. was for hyperbolas...

The textbook says it is not an ellipse, so I don't think your answer is right.

Thanks for posting though

The constant difference is, in fact, for hyperbolas.

[hide="Equation"]$ \sqrt{(x\minus{}2)^{2}\plus{}(y\plus{}1)^{2}}\minus{}\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}\equal{} 2$
$ \implies\sqrt{(x\minus{}2)^{2}\plus{}(y\plus{}1)^{2}}\plus{}\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}\equal{} 2\plus{}2\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}$

Multiplying the respective LHS's and RHS's above, we get a difference of squares:
$ (\sqrt{(x\minus{}2)^{2}\plus{}(y\plus{}1)^{2}})^{2}\minus{}(\sqrt{(x\plus{}1)^{2}\plus{}y^{2}})^{2}\equal{} 4\plus{}4\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}$
$ (x\minus{}2)^{2}\plus{}(y\plus{}1)^{2}\minus{}(x\plus{}1)^{2}\minus{}y^{2}\equal{} 4\plus{}4\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}$

Expanding and collecting on the left, we get:
$ 4\minus{}6x\plus{}2y \equal{} 4\plus{}4\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}$
$ y\minus{}3x \equal{} 2\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}$

Squaring both sides:
$ 9x^{2}\minus{}6xy\plus{}y^{2}\equal{} 4 x^{2}\plus{}8 x\plus{}4\plus{}4y^{2}$
$ \iff\boxed{5x^{2}\minus{}8x\minus{}6xy\minus{}3y^{2}\equal{} 4}$
[/hide]

[hide="Demonstraton that this is a hyperbola"]
Using the quadratic formula, we get:
$ \iff y \equal{}\minus{}x\pm\frac{2}{\sqrt{3}}\sqrt{2x^{2}\minus{}2x\minus{}1}$
$ \iff y \equal{}\minus{}x\pm\frac{\sqrt{6}}{3}\sqrt{\left( x\minus{}\frac{1}{2}\right)^{2}\minus{}\frac{3}{4}}$

As $ x\to\infty$, we get that:
$ y\approx\minus{}x\pm\frac{\sqrt{6}}{3}\left( x\minus{}\frac{1}{2}\right)\implies\boxed{ y\approx\left(\minus{}1\pm\frac{\sqrt{6}}{3}\right) x\mp\frac{\sqrt{6}}{6}}$

These are our two asymptotes.

Hyperbolas are the only type of conic section that have asymptotes.
[/hide]

[hide="Notice on the problem"]
The equation derived is non-continuous; that is, it has two distinct parts. Because the original statement is ambiguous as to which branch the point starts on, we can't really tell which branch it's path lies on. But it is important to realize that the point will not be able to reach all parts of the graph.

For precision, you will need to state that the point will always remain on the same side of the perpendicular bisector of the two foci. The perpendicular bisector of (2,-1) and (-1,0) is y=3x-2. So, either $ y > 3x\minus{}2$ for all points on the path, or $ y < 3x\minus{}2$ for all points on the path, depending on which branch you want to stay on. [/hide]

[quote="A M Y"]The constant difference is, in fact, for hyperbolas.

[hide="Equation"]$ \sqrt{(x\minus{}2)^{2}\plus{}(y\plus{}1)^{2}}\minus{}\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}\equal{} 2$
$ \implies\sqrt{(x\minus{}2)^{2}\plus{}(y\plus{}1)^{2}}\plus{}\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}\equal{} 2\plus{}2\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}$

Multiplying the respective LHS's and RHS's above, we get a difference of squares:
$ (\sqrt{(x\minus{}2)^{2}\plus{}(y\plus{}1)^{2}})^{2}\minus{}(\sqrt{(x\plus{}1)^{2}\plus{}y^{2}})^{2}\equal{} 4\plus{}4\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}$
$ (x\minus{}2)^{2}\plus{}(y\plus{}1)^{2}\minus{}(x\plus{}1)^{2}\minus{}y^{2}\equal{} 4\plus{}4\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}$

Expanding and collecting on the left, we get:
$ 4\minus{}6x\plus{}2y \equal{} 4\plus{}4\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}$
$ y\minus{}3x \equal{} 2\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}$

Squaring both sides:
$ 9x^{2}\minus{}6xy\plus{}y^{2}\equal{} 4 x^{2}\plus{}8 x\plus{}4\plus{}4y^{2}$
$ \iff\boxed{5x^{2}\minus{}8x\minus{}6xy\minus{}3y^{2}\equal{} 4}$
[/hide]
[/quote]

How did you get from step 2 to step 3?  Like what do you mean by "multiply LHS and RHS"?

By multiplying respective sides, she means that:
If $ A \equal{} B$ and $ C \equal{} D$, then $ AC \equal{} BD$.

In this case, we have:
$ A \equal{}\sqrt{(x\minus{}2)^{2}\plus{}(y\plus{}1)^{2}}\minus{}\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}$
$ B \equal{} 2$
$ C \equal{}\sqrt{(x\minus{}2)^{2}\plus{}(y\plus{}1)^{2}}\plus{}\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}$
$ D \equal{} 2\plus{}2\sqrt{(x\plus{}1)^{2}\plus{}y^{2}}$

The first equation is $ A \equal{} B$, and the second equation (which is really just the first equation rewritten) says $ C \equal{} D$

Then, she multiplied and got $ AC \equal{} BD$. Since $ AC$ is a difference of squares, it simplifies very neatly.

(Does that help?)


geometry
geometric transformation
homothety
perpendicular bisector
exterior angle

Circles $O_{1},O_{2}$ are internally tangent at $X$, with $O_{1}$ smaller. The tangent to $O_{1}$ at point $R$ intersects $O_{2}$ at $A,B$. Show that $XR$ bisects angle $AXB$.

[quote="rnwang2"]Does this not follow directly from the SAS congruence of $\triangle{XRB}$ and $\triangle{XRA}$?[/quote]

Only if $R$ happens to be opposite $X$ on $O_{1}$. This doesn't have to be the case.

[hide]
Let $P$ be the intersection of $XA$ with $O_{1}$ other than $X$ and $S$ be the intersection of $XR$ with $O_{2}$ other than $X$. We then have:

-$\angle BXS=\angle BAS$ since both intercept arc $BS$ on $O_{2}$

-$\angle BAS=\angle PRA$ because $RP\parallel SA$ due to homothety of the two circles

-$\angle PRA=\angle RXP$ because the arc cut off by the first angle is the same as the arc intercepted by the second

So we obtain $\angle BXS=\angle RXP$, or $\angle BXR=\angle RXA$.
[/hide]

See the attached diagram. Let $M=(O_{1})\cap XA, N=(O_{1})\cap XB$.

Homothety $\mathcal{H}[X,{XO_{2}\over XO_{1}}]$ transforms one circle into the other, hence

\begin{eqnarray*}\mathcal{H}[X,{XO_{2}\over XO_{1}}](M)=A\land \mathcal{H}[X,{XO_{2}\over XO_{1}}](N)=B\implies MN\parallel AB\implies MN\perp O_{1}R\end{eqnarray*}.

Therefore $O_{1}R$ is the perpendicular bisector of $MN$, which gives $RM=RN\implies \angle RXM=\angle RXN$. QED

Farenhajt vey nice and cute solution .

@Pontios : Η άσκηση αυτή (σίγουρα το ξέρετε είναι στο βιβλίο γεωμετρίας της Α λύκειου του σχολείου ! :) )

Όχι Σιλουανέ, δεν το ήξερα. Ποια άσκηση είναι;   :huh:

A solution similar to scorpius's,

[hide]
WLOG, $A$ is closer to $X$ than $B$.  Extend $XR$ to meet $O_{2}$ at $T$.  Let $XA$ and $XB$ intersect the tangent to $O_{2}$ at $T$ at $C$ and $D$, respectively.  By homothety, $AB\parallel CD \implies m\angle{ABX}=m\angle{CDX}\implies m\widehat{XA}=m\widehat{XT}-m\widehat{TB}$
$\implies m\widehat{TB}=m\widehat{TA}\implies m\angle{AXT}=m\angle{BXT}$ as desired.[/hide]

@Pontios :Κεφάλαιο 6 σελ .130 (Σύνθετα θέματα 2)  :wink:

Here is a more elementary solution.

i) Suppose that $ AT=BT$, the theorem is trivial by HL congruency.

ii) WLOG $ AT<BT$. Let the common tangent at $ T$ intersect $ AB$ at $ Y$. Let $ \angle YXA=a$ and $ \angle AXR=b$. Due to tangents, $ XY=YR$, so $ \angle YRX=a+b$. Due to tangent secant angle theorem, $ \angle XBY=a$. By the exterior angle theorem, $ a+b=\angle RXB+a$. It follows that $ \angle AXR=\angle BXR=b$, as desired.

[quote="Virgil Nicula"]See also http://www.mathlinks.ro/Forum/viewtopic.php?t=155803[/quote]
Now known as http://www.mathlinks.ro/Forum/viewtopic.php?t=155800 (you accidentally posted this topic three times).

  Darij

and also the [i]externaly version[/i] is the following which is true:
[b]circles $ w_{1},w_{2}$ are externaly tangent at $ D$ and $ A$ is a point on $ w_{1}$,the tangent to $ w_{1}$ at point $ A$ intersects $ w_{2}$ at $ E,B$.prove that $ AD$ bisects the arc $ EB$ of circle $ w_{2}$...[/b]


A friend of mine sent me this problem and, for some reason, I can't quite seem to figure out where to begin.  I'm sure it's much more simpler than I'm making it out to be; hopefully discussing this with others will lead me to the insight I need.  Any help would be greatly appreciated.

[img]http://img.photobucket.com/albums/v144/Hate5Six/problem.jpg[/img]

The black circle is tangent to the edge of the outer circle.  Given the above configuration, what is the area, in square mm, of the 'white' region (area of outer circle - area of inner circle), provided that the origin is the center of the outer circle.

[quote="shobber"]The radius of the outer circle is 25.

I think that this problem should be moved to the Intermediat Section.[/quote]

Yes, the outer radius is 25, the inner is 20.5, and the area in question is 643.241 mm^2.  Sorry for posting this in the wrong section; I'm not entirely accustomed to the board's layout yet.


If $ a,b\in \mathbb {R}$

solve

$ 2ab \minus{} 2(a^{2}\plus{}b^{2}) \equal{} \frac{5}{8}$

It is equivalent with $ (a\minus{}b)^2\plus{}a^2\plus{}b^2\plus{}\frac{5}{8}\equal{}0$ , which is impossible .

I am sorry.

The equation should read

$ 4a \plus{} b \minus{}2(a^{2} \plus{} b^{2}) \equal{} \frac {5}{4}$

Easy to notice inequality  :)
$ 2a^2\minus{}4a\plus{}2b^2\minus{}b\equal{}\frac{\minus{}5}4$
$ (\sqrt2a\minus{}\sqrt2)^2\plus{}(\sqrt2b\minus{}\frac{1}{2\sqrt2})\equal{}\frac{7}{8}$

This has infinitely many solutions

Amazing.
[hide="I got this answer"]
$ a \equal{} \frac{1}{2}$

$ b \equal{} \frac{1}{4}$
[/hide]

Wait, what?

$ 4a\plus{}b\minus{}2(a^{2}\plus{}b^{2}) \equal{}\frac{5}{4}$

$ 4a\plus{}b\minus{}2a^2\minus{}2b^2 \equal{}\frac{5}{4}$

$ \minus{}2a^2 \plus{} 4a \minus{} \frac{9}{8} \equal{} 2b^2 \minus{} b \plus{} \frac{1}{8}$

$ \minus{}2(a\minus{}1)^2 \plus{} \frac{7}{8} \equal{} 2\left(b\minus{}\frac{1}{4}\right)^2$

Oh, I see how you got there. Now it's obvious that there are infinitely many solutions (Crudely, this is a consequence from the fact that b can now be written as a function of a), but where's the inequality?

$ - 2(a - 1)^{2} + \frac {7}{8} = 2\left(b - \frac {1}{4}\right)^{2} \geq 0 \Longleftrightarrow \\
\boxed{\frac {4 - \sqrt {7}}{4} \leq a \leq \frac {4 + \sqrt {7}}{4}}$,

and

$ 2\left(b - \frac {1}{4}\right)^{2} \leq \frac{7}{8} \Longleftrightarrow
\boxed {\frac{1 - \sqrt{7}}{4} \leq b \leq \frac{1 + \sqrt{7}}{4}}$

Are there further restrictions on $ a, b$?


calculus
integration
trigonometry
logarithms
calculus computations

wassup ppl ?  :D

[u][b]without[/b][/u] the aid of complex variables, show, for $ y>0$,

$ \int_0^\infty\;\dfrac{\tan^{\minus{}1}\left(\dfrac{x}{y}\right)}{e^{2\pi\,x}\minus{}1}\;\textbf dx\;\equal{}\;\boxed{\dfrac{1}{2}\left[y\plus{}\ln\left(\Gamma(y)\right)\minus{}\left(y\minus{}\dfrac{1}{2}\right)\cdot\ln\,y\minus{}\dfrac{1}{2}\ln\,(2\pi)\right]}$

I moved the post about 7th graders to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1068985#1068985]India forum thread[/url]. Please, stick to mathematical discussions in College Playground (occasional puns, jokes, etc. are OK, of course, but this particular one is not so interesting any more :))


vector
advanced fields
advanced fields unsolved

Let X be a Banach space. Prove that every closed, convex and bounded subset of X satisfies the almost fixed point property.

do you mean that for every map $f: C \to C$, $\inf_{x \in C}\|f(x)-x\|=0$ ?

For every [img]http://www.hapetek.co.il/forums/latexrender/pictures/8ec926d88d20457fb428652391f36e1f.gif[/img] which satisfies [img]http://www.hapetek.co.il/forums/latexrender/pictures/0ab52544fdccbb1b259b0bae76c07710.gif[/img] , we have that [img]http://www.hapetek.co.il/forums/latexrender/pictures/3ff75448396d8f620f6b96fc424bedcf.gif[/img].

That's the definition of an almost fixed point.

are you sure there is no reflexivity hypothesis ?

Nope. The theorem is mentioned briefly [url=http://links.jstor.org/sici?sici=0002-9939%28198305%2988%3A1%3C44%3ATAFPPF%3E2.0.CO%3B2-F&size=LARGE&origin=JSTOR-enlargePage]in this article[/url] but its proof is omitted.

Anyone? I believe that Banach's fixed point theorem is needed somehow (seeing that T is "almost" a contraction mapping).

1) If $0 \in C$, then for every $\epsilon \in (0,1)$ consider the map $g_\epsilon = (1-\epsilon)f$. Since $C$ is convex, $g_\epsilon \colon C \to C$ and is a contraction, hence by Banach's Fixed Point Theorem $g_\epsilon$ has a fixed point $x_\epsilon \in C$. It remains to note that since $C$ is bounded, $f(x_\epsilon)-x_\epsilon = \epsilon f(x_\epsilon) = o(1)$ as $\epsilon \to 0$.

2) If $0 \notin C$, then take arbitrary $x_{0}\in C$ and set $C_{0}= C-x_{0}$. One sees that $C_{0}$ satisfies the same hypotheses as $C$. Consider a map $g(x) = f(x_{0}+x)-x_{0}$, $g \colon C_{0}\to C_{0}$, then by 1) there exists $x_\epsilon \in C_{0}$ with $g(x_\epsilon)-x_\epsilon = \epsilon g(x_\epsilon)$, so it remains to set $x_\epsilon' = x_{0}+x_\epsilon \in C$ to have $f(x_\epsilon')-x_\epsilon' = \epsilon f(x_\epsilon') = o(1)$ as $\epsilon\to0$.

Why does $0\in C_{0}$ ? (if there were no zero vector in $C$)

Recall that $C_{0}: = C-x_{0}$ consists of all elements of the form $x-x_{0}$ as $x$ ranges over $C$. But we picked $x_{0}\in C$, so it remains to take $x=x_{0}$ to obtain that $0=x_{0}-x_{0}\in C_{0}$.


geometry
parallelogram
The Geometry of Numbers

Prove that if a lattice parallellogram contains an odd number of lattice points, then its centroid.

[quote="archimedes1"]...then its centroid is... (a lattice point?)[/quote]
If this is what Peter meant, the problem is obviously wrong. Please review the statement (I don't have the book).

Very awkward indeed. Hojoo, do you have the book?