inequalities
number theory
prime factorization

everybody know this: if u want to check if a number is a prime, u need to divide it by all the numbers that are smaller or equal to his second root. when can u check until the nth root?

Generally you can't. What does the value of $n$ depend on?

I think what you're thinking of is:

If you want to check if a number, $n$, is prime, you divide it by all the primes up to $\sqrt{n}$

[quote="srulikbd"][hide]maybe I'm wrong  :?  I think that for number that have $n$ factors, u can check until the $n$th root.[/hide][/quote] Well, if $n>2$, it isn't prime. :?
If you're ckecking for prime factors, it may work.
EDIT: Actually it doesn't work. Look at 12. It has 6 factors and $\sqrt[6]{12}<3$.

Um, if you know the number of factors the number has, then you already know if it's prime or not, so a primality test is useless.

that's right. but it helps to find the factors!

There is a way just to look till $\sqrt[3]n$:

For all big $n$ it holds:

$n$ is prime iff:
1. $n$ is not a square
2. there is no nontrivial divisor $p|n$ with $p \leq \sqrt[3]n$
3. $k^{n-1} \equiv 1 \mod n$ for all $k < \sqrt[5]n$

[quote="srulikbd"]right, but if u have a number that have a 100 factors(primes)[/quote]

This is pointless.  If you know its factors you know whether it's prime or not.   :|

u know the number of the primes, not their value...

[quote="perfectnumber628"][quote="ZetaX"]For all big $n$ it holds:[/quote]
how big does $n$ have to be?[/quote]
Don't know, but the proof uses the existence of small primitive roots.

PS: there are also some of complexity about $O(\sqrt[8]n)$.

[quote="srulikbd"]right, but if u have a number that have a 100 factors(primes), and u want to find them, I thins it's quite nice to take the 100nt root and when u find a prime that divide the number, u divide it and then check until the 99nt root. better then the 2th root...[/quote]
but you could just start with the first prime (2) and then test numbers and you'll find one before the 100th root...knowing there is one less than or equal to the 100th root doesn't help you find it, if you were going to start with 2 anyway.

[quote="srulikbd"]it works, $12=2^2\cdot3$ thus we need to check until the 3th root.[/quote]
Well, try it on 24.  ;)

[quote="srulikbd"][quote="t0rajir0u"][quote="srulikbd"]when can u check until the nth root?[/quote]

Never.  If you want to exhaustively confirm that a number is prime, you need to check all primes up to the square root.

Given a factor $f$ of $n$ such that $f \le \sqrt{n}$, it's obvious that $d = \frac{n}{f} \ge \sqrt{n}$.  So we can pair up all of $n$'s factors in this way.

That means in order to check whether $n$ has any factors besides $1$ and $n$, we only need to check the ones $\le \sqrt{n}$, since the ones $\ge \sqrt{n}$ can only exist if they're paired with one $\le \sqrt{n}$![/quote]
right, but if u have a number that have a 100 factors(primes), and u want to find them, I thins it's quite nice to take the 100nt root and when u find a prime that divide the number, u divide it and then check until the 99nt root. better then the 2th root...[/quote]

But then just working with factors $\leq \sqrt{n}$ is exactly the same thing because you're still going to start with 2 and go up. Using the square root is actually simpler because you won't have to find 99 unnecessary roots of the number; you'll only have to find one.

For example, if you want to find the smallest prime factor of a number, and you are given that it has 100 (possibly the same) prime factors, then you don't have to take the 100th root and check all primes less than or equal to that number; you just start with 2 and go up.

Knowing that a number a has n prime factors, you can say that there is a prime factor p such that $p \le \sqrt[n]{a}$. This can be proven pretty easily. Suppose all prime factors $p_1, \cdots, p_n$ are such that $p_i > \sqrt[n]{a}$. Then multiplying all of the inequalities together, we get $p_1 p_2 \cdots p_n > a$ when we know it should be equal, thus there is at least one prime factor $p \le \sqrt[n]{a}$.

This doesn't really help with showing if a number is prime or not, as many people before me have said, if you know the number of prime factors, then you know whether or not it is prime without computation.

[quote="lotrgreengrapes7926"][quote="mathgeniuse^ln(x)"]Are we trying to check whether a number is prime or not?

Because if we know the factors of the number, then what is the point for checking if it is prime or not?  It is not prime, right?[/quote]
I think we're trying to find the prime factorization now.[/quote]

But then of course you'll check starting from smaller numbers first, and you will hit one factor regardless if you know that fact or not.  I guess what he points out is right, but it just comes out completely useless...


Uh...you sure you type this up right darkquantum? The LHS simplifies to 2/cos(x).

Well...

[hide](1+sinx)/cosx + cosx/(1+sinx)

((1+sinx)^2 + (cosx)^2)/(cosx(1+sinx))

(1 + 2sinx + (sinx)^2 + (cosx)^2)/(cosx(1+sinx))

(2 + 2sinx)/(cosx(1+sinx))

2(1 + sinx)/(cosx(1+sinx))

2/cosx

Which does not equal 4.

[/hide]

O shooot.

my bad.

Original question is a "solve" question

Hm ok so maybe it does equal 2/cos(x). Anyway, find the solutions. That isn't very hard from here, is it?

So [hide]x = 60 + 360k, 300 + 360k degrees[/hide]. Wow, this is amazing. This is truly amazing. I, a mere mortal below the land of Olympus, have proved one of the greats, a true master of the art of problem solving, wrong.    I feel so special now.  8-)

hm
think its more like, u got lucky, and Simon made a careless error

Hmm...that is possible...even probable....but my version of things sounds more interesting.  :D

To tell you the truth, it didn't look like 2/cos(x) to Simon, and so he didn't bother to check. He should have. Good catch though Fierytycoon.


For what whole number $ n$ is $ 10^n < 695000 < 10^{n\plus{}1}$?

hello, your inequality is equivalent with
$ \frac{\ln(695000)}{\ln(10)}\minus{}1<n<\frac{\ln(695000)}{\ln(10)}$.
The solution is $ n\equal{}5$.
Sonnhard.

Natural logarithms would be the right way to do this, but we don't have a calc on FTW. ;)


trigonometry
integration
calculus
calculus computations

I'll give a hint as the problem is not so easy. [hide] :idea: Use the theorem of Lagrange for differential calculus.[/hide]

What is n?
It is not true in general, say n = 0, 1, 2 or 3.

[quote="panos_lo"]Prove that: tan(5π/18)>1+π/18[/quote]

I believe that "π" is the $ \pi$ symbol.

$ \tan(\dfrac{5\pi}{18})>1+\dfrac{\pi}{18}$

[hide]Let $ f(x) = \tan(x)$

Note that $ f'(x) = \sec^2(x)$, and $ f''(x) = 2\sec^2(x)\tan(x) > 0$ for all $ x \in (0,\dfrac{\pi}{2})$.

Thus, $ f(x)$ is concave up on $ (0,\dfrac{\pi}{2})$.

So we have:
$ f(\dfrac{\pi}{4}+\dfrac{\pi}{36}) > f(\dfrac{\pi}{4}) + \dfrac{\pi}{36}f'(\dfrac{\pi}{4})$

$ \tan(\dfrac{5\pi}{18}) > \tan(\dfrac{\pi}{4}) + \dfrac{\pi}{36}\sec^2(\dfrac{\pi}{4})$

$ \tan(\dfrac{5\pi}{18}) > 1 + \dfrac{\pi}{18}$ as desired. [/hide]

Essentially same proof as Xantos

$ \tan(\frac {5\pi}{18}) \minus{} 1 \equal{} \int_{0}^{\frac {\pi}{36}}{\sec^{2}(\frac {\pi}{4} \plus{} x)dx} > \int_{0}^{\frac {\pi}{36}}{\sec^{2}(\frac {\pi}{4})dx} \equal{} \int_{0}^{\frac {\pi}{36}}2 dx \equal{} \frac {\pi}{18}$

In fact for $ x \in (0,\frac {\pi}{4})$ we have that

$ \tan(\frac {\pi}{4} \plus{} x) > 1 \plus{} 2x$


geometry
incenter
search
geometry unsolved

You don't use $ P$ anywhere in the problem. Is it correctly stated?

$ I$ $ think$ $ that$ $ the$ $ given$ $ question$ $ is$ $ wrong$. :?: $ Since$ $ in$ $ an$ $ equilateral$ $ triangle$ $ the$ $ angle$ $ between$ $ the$ $ lines$ $ BE$ $ and$ $ AI$ $ is$ $ 120$ $ and$ $ not$ $ 90$. :blush:

[quote="primoz2"]Leta ABC be a triangle. M; K ; E the touching points of the sides AB; BC; CA with the incircle of the triangle ABC. Let P be the intersection of MK and AC
Prove that AI is perpendicular to BE, where I is the incenter of the incircle.

If this problem was already sent, please give me a link to it, because I couldnt find it.[/quote]

He means $ PI$ is perpendicular to $ BE$  :wink: (NOT $ AI$ is perpendicular to $ BE$). If so, this problem can be solved easily by using polar and pole theory.

[quote="primoz2"]Of course its PI. I made a silly mistake while typing the problem, i am still loking for a nice solution.[/quote]

[color=red][b]According to April's hint  , we can say that :[/b][/color]

[b]  a)[/b] $ KM$ is the polar of $ B$.
 [b] b) [/b]Since P belongs to the polar of $ B$ , the polar of $ P$ passes through $ B$.
 [b]c)[/b] Sice $ AE$ is tangent , the polar of $ P$ passes through $ E$ (and of course through $ B$).
  [b]d)[/b] The polar of $ P$ is $ EB$ and since $ I$ is the center of the circle , $ PI$ is perpendicular to the polar $ BE$ of $ P$(This is standard theory of the polar and pole)

   I have not tried for other metric solution. Of course , it could be sufficient to prove that :
  $ {IB}^2 \minus{} {IE}^2 \equal{} {PB}^2 \minus{} {PE}^2$

 But probably this does not work.We must see it most carefully.
[u] Babis[/u]

[quote="primoz2"]Can you give me a link where i can learn somethnig about polars, because i know nothing about them.
I still want to se a nice geometric prove of this problem, because i couldnt solve it that way, i solved it using komplex numbers, but i would like to see a nicer proof.[/quote]

 I think that this :

 http://www.mathlinks.ro/viewtopic.php?search_id=895629111&t=168866

 is a very important place to study nice olympiad geometry. Of course you can search a little in the net(mathforum, or cut the knot  and in  many other places).

 [u]Babis[/u]

[quote="primoz2"]Can you give me a link where i can learn somethnig about polars, because i know nothing about them.
I still want to se a nice geometric prove of this problem, because i couldnt solve it that way, i solved it using komplex numbers, but i would like to see a nicer proof.[/quote]

[color=red][b]Try also this :[/b][/color]

 Let $ Q$ be the projection of $ E$ on $ PI$.Then $ {IE} ^2 \equal{} IQ \cdot IP, {IK}^2 \equal{} IN \cdot IB$ , where $ N$ is the intersection of $ IB,PK$  .Hence :

$ IQ \cdot IP \equal{} IN \cdot IB$ , that is $ PQNB$ is inscribed..Thus $ \angle PQB \equal{} \angle PNB \equal{} 90$   :) .

I hope I have helped.

[u] Babis[/u]


If a,b and c are positive numbers,prove that
$ \frac{a(3a\minus{}b)}{c(a\plus{}b)}\plus{}\frac{b(3b\minus{}c)}{a(b\plus{}c)}\plus{}\frac{c(3c\minus{}a)}{b(c\plus{}a)} \le \frac{a^3\plus{}b^3\plus{}c^3}{abc}$

[quote="rachid"]If a,b and c are positive numbers,prove that
$ \frac {a(3a \minus{} b)}{c(a \plus{} b)} \plus{} \frac {b(3b \minus{} c)}{a(b \plus{} c)} \plus{} \frac {c(3c \minus{} a)}{b(c \plus{} a)} \le \frac {a^3 \plus{} b^3 \plus{} c^3}{abc}$[/quote]
It's equivalent to $ \sum_{cyc}\frac{(a\minus{}b)^2(11a^2\minus{}3ab\plus{}b^2)}{a\plus{}b}\geq0.$ :wink:


I've done badly freshman/sophomore year, including a D (Calc). I plan to get a 4.0 next year, which would bring my grade point avg to 3.5. A 4.0 for the first half of senior year gets me to 3.58. I know I can do this, as I'm smart enough.

Anyways, I've realized all the schools I'd probably dream of are no longer options. What schools don't weight gpa highly? I can write a good essay and get good recs and have lots of ecs. I think the only thing that would be bad would be my sophomore/freshman grades.

What do colleges think of repeating classes? I might take that D again to clear that from my slate.

If I do not retake that D, I have 5 APs next year (excluding French), and 6 APs in 12th. I am planning to self study Macroeconomics, Art History, Human Geography, and Physics B, but this might change if I find myself short on time. Plus, I want to join the tennis team, even if I'll be stuck on JV, so "short on time" might just happen. I also took 2 AP classes in 10th (both As Chem/Environmental), and took an AP test for a class I didn't take (World History). I will probably have the most rigorous courseload of anyone in the school.

Weighted I currently have a 3.44 (I'm doing this in my head so I'm not quite certain on the number), if I take all those APs, it'll be 3.79 weighted by the end of junior year. If I'm correct, I get 3.97 weighted by the middle of senior year. I'll break four by the end of senior rate if I can keep that up. I'm only counting AP classes here. If I count the 5 honors courses I've taken in this calculation, I have a 3.85 weighted. I'd get something like 4.4 weighted by the end of senior year then.

I'm nowhere near the best in my class, but my school doesn't rank. We have maybe 20 students that receive career 4.0's.

First of all, you're speculating way too much.  If you haven't even completed your junior year, there is no reason you should be speculating on what your final GPA will be.  Many colleges look at all 4 years, but they place little emphasis on freshman year grades.  If you can pull a 4.0 in junior year and the first half of senior year, that will impress them, since it shows an upward trend.  Worry about completing next year first.

I agree with JRav. It is more important when there is development and improvement in your results (upward trend as he said). Still the GPA is not the most important criteria when applying to a university. A low GPA in your first two years doesn't mean that you will get rejected by the prestigious universities.

Were there any particular reasons why you had a low GPA? If you had any personal circumstances or something, or had some massive revelation, then you may want to include that in a written statement somewhere in your college applications explaining:

1. What was wrong
2. How you realized it and fixed it
3. What you took away from it

Many colleges will read that and take that into consideration. Of course, this only works if you succeed in making a change in your GPA -- good luck with that! :)

As for retaking it, was it an AP class? If you got a good AP score, I wouldn't worry about retaking the class. If it wasn't an AP class, try taking the AP exam next year to show that you know the material. This reinforces your claim that your early GPA is not a good indicator of your ability to perform in college.


algebra
polynomial
calculus
integration
algebra unsolved

Let $a_1, a_2, a_3, a_4$ be four integer numbers and let $P(x)$ a polynomial with integer coefficients such that

$P(a_1)=P(a_2)=P(a_3)=P(a_4)=1$            $*$

1)Prove that it doesn't exists an integer $n$ such that $P(n)=12$
2)Do a polynomial $P(x)$ satisfing $*$ and an integer $n$ such that $P(n)=1998$ exist?

1) You would get that $n-a_i | f(n)-f(a_i)=11$, so we have $\{a_1,a_2,a_3,a_4\} = \{n-11 , n-1 , n+1 , n+11\}$, and $P(x)-1$ has the roots $a_i$.
Thus it is $P(n-x) = (x^2-1^2)(x^2-11^2)\cdot p(x)+1$ with $p(x)$ being an integral polynomial again. This gives $P(n)=11^2 \cdot p(n) +1 = 12$, clearly impossible $\mod 121$.

2) Now we need $n-a_i | f(n)-f(a_i)=1997$ (prime), so we have $\{a_1,a_2,a_3,a_4\} = \{n-1997 , n-1 , n+1 , n+1997\}$ and as before $P(n-x) = (x^2-1^2)(x^2-1997^2)\cdot p(x)+1$. The rest is above, too.

$p(x) = (x - a_{1})(x - a_{2})(x - a_{3})(x - a_{4})q(x) + 1$.
$p(n) - 1 = (n - a_{1})(n - a_{2})(n - a_{3})(n - a_{4})q(n) = 11$.
But the factors can only be $\pm 1$ and $\pm 11$ and it's not possible to write $11$ as the product of 4 distinct factors.
The same for $p(1998)$.


Um...

When both quantities are equal to each other or when one of the quantities on each side equals $0.$

That means when $x=y$ or when $(x,y)=(-1,0), (0,-1).$

There are more solutions...
[hide]
$x^2-y^2+x-y=0\Longleftrightarrow (x-y)(x+y+1)=0$

So $x=y$ or $x+y=-1$
[/hide]

4everwise's way is fine, but he just missed a case when one side is a product of two positive numbers and the other side is a product of two negative numbers.

[quote="José"]What are the conditions for this equality: $x(x+1)=y(y+1)$?[/quote]

$(x^2-y^2)+(x-y)=0$

$(x-y)(x+y+1)=0$

[quote="frt"]4everwise's way is fine, but he just missed a case when one side is a product of two positive numbers and the other side is a product of two negative numbers.[/quote]Oops.  :blush:

And what about $x=0,25$ and $y=0,75$? You're missing a case jajaja

Well... $0.25\cdot 1.25\ne 0.75\cdot 1.75$ :huh:

I'm pretty sure that the only conditions are $x=y$ or $x+y=-1$ (as mentioned in my previous post and Altheman's post).

Sorry, I wanted to say $x=0,25$ and $y=0,75$, but in $x(x-1)=y(y-1)$

What are the conditions for this one?

[quote="frt"]Well, using the same approach,
$x^2-y^2-x+y=0\Longleftrightarrow (x+y)(x-y)-(x-y)=0\Longleftrightarrow (x-y)(x+y-1)=0$, so the conditions are $x=y$ or $x+y=1$[/quote]

Cool. Thanks.


quadratics
algebra
polynomial
Rational Root Theorem

Find $x$ and $y$ in this system. (No graphic calculator or software)

$\{\begin{array}{rcl}{x^{2}+y=7}\\{x+y^{2}=11}\end{array}$

[hide="solution?"]First, we solve for x, and then we solve for y.

$x=\sqrt{7-y}=11-y^{2}$

So $y^{2}\leq 11$ and $y \leq 7$. Therefore, $-\sqrt{11}\leq y \leq \sqrt{11}$

$y=\sqrt{11-x}=7-x^{2}$

So $-\sqrt{7}\leq x \leq \sqrt{7}$

$x=\sqrt{7-\sqrt{11-x}}$

$x^{2}=7-\sqrt{11-x}$

$x^{2}-7=-\sqrt{11-x}$

$x^{4}-14x^{2}+x+38=0$

Now using the quartic formula would be just brutal, so we just go back to the original equations and plug in some random numbers, say x=2. Then y=3. Yay.

$x^{4}-14x^{2}+x+38=0$ divided by $x-2$ is $x^{3}+2x^{2}-10x-19$. We check all other integers greater than -3 and less than 3, but none work. So the rest are non-integers. So by the rational root theorem, we try 1/19 / -1/19. 1/19 doesn't work. nerther does -1/19. So the rest of the roots are non-rational. Wah.

So the only roots that I can find are x=2 and y=3.[/hide]

[quote="1=2"]Now using the quartic formula would be just brutal, so we just go back to the original equations and plug in some random numbers, say x=2. Then y=3. Yay.[/quote]

So you can solve it by random numbers?

[quote="Teki-Teki"]
(3,2) is obviously a solution to this.[/quote]

But what about the other answers!?

Well, Mathematica isn't giving pretty answers -- what's most likely the case is that you'll have to use [url=http://en.wikipedia.org/wiki/Cubic_equation]one of the methods here[/url] to solve the cubic that results.

Chessy didn't say that $x$ and $y$ must be integer, they can also be real.
So something has to be done with the cubic equation.
Applying Cardano???

There are solutions. We have shown that we can find them. Just leave it at that...

Here is a slightly different problem:

Determine $(a,b)$, integers such that the system:
$y^{2}=x+a$
$x^{2}=y+b$

has solutions that are solutions to quadratics with integer coefficients.

[I am pretty sure this is solvable...]


algebra
polynomial
superior algebra
superior algebra solved

\[
Having A \in M_n (C), and \exists p \in N ,so, A^p  = O_n .
\]
\[Prove: \[
\det (xA + I_n ) = 1,\forall x \in C
\]

It should go in the "Superior Algebra" section.

Since $A$ is nilpotent, its characteristic polynomial is $x^n$, meaning that $\det(xI_n+A)=x^n,\ \forall x$. Assuming $x\ne 0$, we can divide that relation by $x$ to get $\det(I_n+yA)=1,\ \forall y\ne 0$, where $y=\frac 1x$. $\det(xA+I_n)=1$ holds for $x=0$, of course, so we're done.

The Jordan canonimal form of $A$ is $\mbox{diag}\{J_1, J_2, \dots, J_n\}$. The diagonal elements of all $J_i$ are 0 since $A$ is nilpotent. We can verify the answer directly.


geometry
trapezoid
projective geometry
geometry proposed

Let $ ABC$ be an acute triangle such that $ AB < AC$. $ A_1$ is the foot of the altitude from $ A$ to $ BC$. $ B_1$ and $ C_1$ are defined similary. Denote by $ D \equal{} B_1C_1 \cap BC$. The parallel line from $ D$ to $ AC$ intersects $ AB$ in $ N$. $ M$ is the midpoint of the side $ AC$. Prove that $ M,A_1,N$ lie on the same line.

Nice problem. :)

We have $ \angle NDA_1\equal{}C\equal{}\angle NC_1A_1$ thus $ NA_1C_1D$ is cyclic. Hence $ \angle NA_1C_1\equal{}\pi\minus{}\angle NDC_1\equal{}\pi\minus{}\angle AB_1C_1\equal{}\pi\minus{}B\quad (1)$. Now, since $ A_1,B_1,C_1,M$ lie on the nine point circle, $ \angle A_1MC_1\equal{}\angle A_1B_1C_1\equal{}\pi\minus{}2B$ (well known). Thus since $ ACA_1C_1$ is cyclic with center $ M$ we have $ \angle MA_1C_1\equal{}\angle MC_1A_1\equal{}\frac{\pi\minus{}\angle A_1MC_1}{2}\equal{}B\quad (2)$. Adding $ (1)$ and $ (2)$ we get $ \angle NA_1C_1\plus{}\angle MA_1C_1\equal{}\pi$, which implies $ M,A_1,N$ lie on a straight line.

I did something like Nayel

$ \angle C\equal{}\angle A_1DN\equal{}\angle DC_1N\equal{}\angle DA_1N$. Because of $ A_1M\equal{}MC$ we have $ \angle MA_1C\equal{}C$. Then $ \angle DA_1N\equal{}\angle MA_1C$ so $ M,A_1,N$ are colinear.  :D

Completely different, but with harmonics:

$ D$, $ B$, $ A_1$ and $ C$ are harmonic, so $ \frac{CD}{DB}\equal{}\frac{CA_1}{A_1 B}$, but $ \frac{AN}{NB}\equal{}\frac{CD}{DB}$ by the similarity $ \triangle ABC\sim\triangle NBD$, so $ \frac{AN}{NB}\equal{}\frac{CA_1}{A_1 B}$. Finally, $ \frac{AN}{NB}\frac{A_1 B}{CA_1}\frac{CM}{MA}\equal{}1$ so by Menelaus, $ N$, $ A_1$, and $ M$ are collinear.

great solution tjhance!
when i saw the word menelaus in your solution
i couldnt wait to see the solution
bye  :)

Thanks for the solutions. My solution is a simple one:

$ AA_1$ intersects the parallel from $ D$ to $ AC$ in $ P$.

$ (B,C,D,A_1)$ is an harmonic division $ \Rightarrow$ the pencil $ A(B,C,D,A_1)$ is harmonic $ \Rightarrow$ the division $ (N,\infty,D,P)$ is harmonic $ \Rightarrow$ $ N$ is the midpoint of $ NP$

Since the quadrilateral $ ACPD$ is a trapezoid with $ AC \parallel PD$ $ \Rightarrow$ $ M,A_1,N$ lie on the same line

Dear Mathlinkers,
my synthetic solution:
1. A, C, A1 and C1 are on a Thales circle
2. By a converse of Reim's theorem, A1, C1, D and N are cocyclic (http://perso.orange.fr/jl.ayme)
3. A1, B1, C1 and M are on the nine points circle of ABC
4. By another converse of Reim's theorem, A1, N and m are colinear
Sincerely
Jean-Louis

its true for every three cevian line.
proof is the same as moldovan said.
suppose $ AA_{1},BB_{1},CC_{1}$ are cevian line.
we have $ N(DMAC)\equal{}\minus{}1$ so if $ NM$ intersect $ BC$ in $ K$ we hav $ (DKBC)\equal{}\minus{}1$ but $ (DA_{1}BC)\equal{}\minus{}1$ means
$ A_{1} \equiv K$


linear algebra
matrix
linear algebra unsolved

When I see this problem hydrogen bomb was going to detonate in my brain :D
Show that for every even $n$ the matrice $A_{n\times n}$, such that $A_{i,j}=\frac{1}{C_{2i+j-1}^{i}}$, is invertible.

[i]Edited by Myth[/i]

Also, please learn to follow the rules of the forums. The source for this problem [u]is not[/u] well documented. Magazine? What magazine? That magazine deserves the proper documentation as well as the person who solves the problem. Below you will find the link on how to properly put the source in this forum by our great Valentin Vornicu. Thank you.

http://www.mathlinks.ro/viewtopic.php?t=3518

lich...

excuse me plz :(  the source is sharif university of technology magazine

OK, I move it to Superiour Algebra section.

It seems I met this problem some day, and maybe on this forum.

Harazi said: "Please, from now on post linear algebra problems in the superiour algebra section. Thanks."
    This seems to me very unclear; in a second time do you shut the matrix-linear algebra section?

It was in Olympiads - Algebra Unsolved problems.

This problem predates the (imo poorly named) 'Matrix-Linear Algebra' section


calculus
derivative
function
calculus computations

[quote="!!!"]How can I get from the problem to the answer?
[img]http://img301.imageshack.us/img301/2637/19634307qi7.jpg[/img][/quote]
The product rule works due to a great little technique:

$ \dfrac{d}{dx}(fg) \equal{} lim_{h > > 0}\dfrac{f(x \plus{} h)g(x \plus{} h) \minus{} f(x)g(x)}{h}$
Now we subtract and add f(x+h)g(x) in the numerator:
$ \equal{} lim_{h > > 0}\dfrac{f(x \plus{} h)g(x \plus{} h) \minus{} f(x \plus{} h)g(x) \plus{} f(x \plus{} h)g(x) \minus{} f(x)g(x)}{h}$
And undistribute:
$ \equal{} lim_{h > > 0}[f(x \plus{} h)\dfrac{g(x \plus{} h) \minus{} g(x)}{h} \plus{} g(x)\dfrac{f(x \plus{} h) \minus{} f(x)}{h}]$
And use limit rules to separate:
$ \equal{} lim_{h > > 0}f(x \plus{} h)*lim_{h > > 0}\dfrac{g(x \plus{} h) \minus{} g(x)}{h} \plus{} g(x)lim_{h > > 0}\dfrac{f(x \plus{} h) \minus{} f(x)}{h}$
Now we see that each of these two limits are the definitions of the derivatives of their respective functions, and we can claim that:
$ \dfrac{d}{dx}(fg) \equal{} u\dfrac{dv}{dx} \plus{} v\dfrac{du}{dx}$

The chain rule (which I don't care to prove):
$ \dfrac{d}{dx}f(g(x)) \equal{} (\dfrac{df}{dx}$ evaluated at $ g(x))*(\dfrac{dg}{dx}$ evaluted at $ x)$

The power rule for ANY exponent n and coefficient a (memorize it!):
$ \dfrac{d}{dx}af(x)^n \equal{} anf(x)^{n \minus{} 1}\dfrac{df}{dx}$

How does this apply to our problem?  Well, we see that we have the derivative of a product, so we can use the product rule:
$ f \equal{} x^2 \plus{} 1$
$ \dfrac{df}{dx} \equal{} 2x$
$ g \equal{} (x^2 \plus{} 2)^{1/3} \equal{} h(x)^{1/3}$, where $ h(x) \equal{} x^2 \plus{} 2$
$ \dfrac{dg}{dx} \equal{} \dfrac{1}{3}h(x)^{ \minus{} 2/3}\dfrac{dh}{dx} \equal{} \dfrac{1}{3}(x^2 \plus{} 2)^{ \minus{} 2/3}(2x)$
Now you can just plug in $ f,\dfrac{df}{dx},g,$ and $ \dfrac{dg}{dx}$ into the chain rule equation and you'll see the first part of $ y^{'}$ there - just undistribute and see the final part!


i think most r (since i'm the average at 19...i scored 115th). that makes sense, but there r only like 10 ppl that voted...

-rorri :?

[quote="solafidefarms"]Just wondering. Are AoPSers smarter than the average?[/quote]

We can hardly say that AoPSers are smarter than the average at nats. After all, some (no, a lot) of us didn't even make nationals. They aren't smarter than the average at nats.

[quote="nat mc"]Mr Fish sent me the test to take and I got 33 (19/7) but obviously I wasn't as nervous as a competitor[/quote]

Can you send me the test please?  :) I want to see if I would get last place or not :D

It was n't by email, but you coulsd PM your adress and I could make a copy to send to you

The Mathcounts site only posts the to 25% of scores, or the top 57 kids.  All other kids can find out about their own scores from the state coach who received their rankings from the MC folks.

My sprint wasn't too bad, but I bombed the target.

Same. I bet my bomb was better than yours, 45.

[quote="uberl33tmage"]i got an 18/6, barely missed CD. Oh well, i have next year[/quote]

18/6 didn't barely miss countdown, that's what I got and that score created a nine-way tie from 57th to 65th. Unfortunately, I got 58th :(

I think he means 18/12, 6 target questions right

I think 18/6 is the right way to say it. In which case he lost CD by a tiebraker.

[quote="Assassin of Death"]Wow, if that poll is accurate it would suck for Utah. Our best guy got a 16. Then in succession, 15, 14, 13. (I think at least :blush: ) Are the listing of all 228 participants on the MC site or not?  With score and state? I'm not sure how to navigate that site. Its wierd.[/quote]
16:  Non-AoPSer
15:  Me
13:  AoSD
11:  Not available....

Not really. If anyone from Indiana had scored just 1 more point, then they would have won!

However, if anybody from Texas had scored just one point, they still would have won.

if one person from texas scored 1 then indiana would have needed to score 2 unless u got one team :(  then texas would have died

Texas is really good, as noneoftheabove would know  :D . Actually I would say the problem that won the competition for Texas was not the one team problem but they got and Indiana didn't but it was the one sprint question that Zhang was the only person in the nation to get. And you AoPSers should sopport Texas, almost all of their team members are members of AoPS!

So Mark got #30? But I don't think it matters for the team score if you solve the more difficult individual problems, as long as you get the same score. I heard that he got a 25/5, which is a rather low target compared to his sprint.

Any other day Indiana could've won, and won big. And I mean really big. It was just an unlucky day. BTW, 1,2,3,5 in state in IN are AoPSers too

[quote="sirorange"]Actually, any other day, Massachusetts could have probably gotten 2nd with a very slight chance of first. But honestly, it's not how you do in practice, it's how you do in the actual competition, and you can never change that.[/quote]

You guys are forgetting about CA... we had a horrible day... (at least i did.... I mean, a 3 on target?!?!?!? come on...)

[quote="superman"]and 24

thats me

Pardha is here?[/quote]
bigHC

[quote="sirorange"]Actually, any other day, Massachusetts could have probably gotten 2nd with a very slight chance of first. But honestly, it's not how you do in practice, it's how you do in the actual competition, and you can never change that.[/quote]

Why?

Anway, any day CT could win. After all, we are full of surprises. That's why they didn't put any of us in the CountDown or the team in the top 3; it wouldn't be surprising  ;)

Lol, I t hink everyone knows hat your full of surprises

any other day, i couldve gotten first, but i didnt even make it to nats


geometry
perimeter
function
number theory
prime numbers
national olympiad

\[\text{Day 1}\]
1. http://www.mathlinks.ro/Forum/viewtopic.php?p=798074#798074
Squares $ABB_{1}A_{2}$ and $BCC_{1}B_{2}$ are externally drawn on the hypotenuse $AB$ and on the leg $BC$ of a right triangle $ABC$ . Show that the lines $CA_{2}$ and $AB_{2}$ meet on the perimeter of a square with the vertices on the perimeter of triangle $ABC .$

2. http://www.mathlinks.ro/Forum/viewtopic.php?p=798081#798081
Let $f$ be an increasing mapping from the family of subsets of a given ﬁnite set $H$ into itself, i.e. such that for every $X \subset Y\subset H$  we have $f (X )\subset f (Y )\subset H .$  Prove that there exists a subset $H_{0}$ of $H$ such that $f (H_{0}) = H_{0}.$

3. http://www.mathlinks.ro/Forum/viewtopic.php?p=798082#798082
Find all sequences $x_{1},x_{2},...,x_{n}$ of distinct positive integers such that
 $\frac{1}{2}=\sum_{i=1}^{n}\frac{1}{x_{i}^{2}}$.
\[\text{Day 2}\]
4. http://www.mathlinks.ro/Forum/viewtopic.php?p=798090#798090
Does there exist a sequence of $2005$ consecutive positive integers that
contains exactly $25$ prime numbers?

5. http://www.mathlinks.ro/Forum/viewtopic.php?p=798097#798097
 Let $F_{n}$ be the $n-$ th Fibonacci number (where $F_{1}= F_{2}= 1$). Consider the functions
$f_{n}(x)=\parallel . . . \parallel |x|-F_{n}|-F_{n-1}|-...-F_{2}|-F_{1}|, g_{n}(x)=| . . . \parallel x-1|-1|-...-1|$ ($F_{1}+...+F_{n}$ one’s).
Show that $f_{n}(x) = g_{n}(x)$ for every real number $x.$

6. http://www.mathlinks.ro/Forum/viewtopic.php?p=799074#799074
There are seven rods erected at the vertices of a regular heptagonal area. The top of each rod is connected to the top of its second neighbor by a straight piece of wire so that, looking from above, one sees each wire crossing exactly two others. Is it possible to set the respective heights of the rods in such a way that no four tops of the rods are coplanar and each wire passes one of the crossings from above and the other one from below?

Funny to see this, 'cause today was the ending day of this year's Gillis-Turán Hungary-Israel Math Contest. :P

Another thing is that this year's competition was the XVIIth so i should think that last year's was the XVIth (because there was one last year)

[quote="Megus"]kdano, could you post up questions?  :)[/quote]

Yes. I'll try tomorrow :P


calculus
integration
calculus computations

Let $ n,\ m$ be positive integers and $ \alpha ,\ \beta$ be real numbers.
Prove the following equations.

(1) $ \int_{\alpha}^{\beta} (x \minus{} \alpha)(x \minus{} \beta)\ dx \equal{} \minus{} \frac 16 (\beta \minus{} \alpha)^3$

(2) $ \int_{\alpha}^{\beta} (x \minus{} \alpha)^n(x \minus{} \beta)\ dx \equal{} \minus{} \frac {n!}{(n \plus{} 2)!}(\beta \minus{} \alpha)^{n \plus{} 2}$

(3) $ \int_{\alpha}^{\beta} (x \minus{} \alpha)^n(x \minus{} \beta)^mdx \equal{} ( \minus{} 1)^{m}\frac {n!m!}{(n \plus{} m \plus{} 1)!}(\beta \minus{} \alpha)^{n \plus{} m \plus{} 1}$

(1) intergration  by part

(2) recurrence for n

(3) we fixe  n  , and we do  a reccurence for m

$ \int_{\alpha}^{\beta}(x\minus{}\alpha)^{n}(x\minus{}\beta)^{m}dx  \equal{} |t\equal{}\frac{x\minus{}\alpha}{\beta\minus{}\alpha}| \equal{} (\minus{}1)^m (\beta \minus{}\alpha)^{m\plus{}n\plus{}1}\int_0^1 t^n(1\minus{}t)^m dt \equal{} (\minus{}1)^m (\beta \minus{}\alpha)^{m\plus{}n\plus{}1} B(n\plus{}1,m\plus{}1) \equal{} (\minus{}1)^m (\beta \minus{}\alpha)^{m\plus{}n\plus{}1} \frac{n!m!}{(n\plus{}m\plus{}1)!}$

[quote="kunny"]Let $ n,\ m$ be positive integers and let $ \Re \alpha > 0,\Re \beta >0$.
Prove the following equations.

(1) $ \int_{\alpha}^{\beta} (x \minus{} \alpha)(x \minus{} \beta)\ dx \equal{} \minus{} \frac 16 (\beta \minus{} \alpha)^3$

(2) $ \int_{\alpha}^{\beta} (x \minus{} \alpha)^n(x \minus{} \beta)\ dx \equal{} \minus{} \frac {n!}{(n \plus{} 2)!}(\beta \minus{} \alpha)^{n \plus{} 2}$

(3) $ \int_{\alpha}^{\beta} (x \minus{} \alpha)^n(x \minus{} \beta)^m\ dx \equal{} ( \minus{} 1)^{m}\frac {n!m!}{(n \plus{} m \plus{} 1)!}(\beta \minus{} \alpha)^{n \plus{} m \plus{} 1}$[/quote]

Put (\beta-\alpha) t = x-\alpha in (1) to get

\[ (1) \int_{\alpha}^{\beta} (x \minus{} \alpha)(x \minus{} \beta)\ dx \equal{} (\beta\minus{}\alpha)^3\int_{0}^{1}t(t\minus{}1)\, dt\]
\[ \equal{} \minus{}\frac{1}{6} (\beta \minus{} \alpha)^3\]

(2) and (3) follow naturally with the same substitution.


integration
limit
real analysis
real analysis unsolved

Are $ \{X(i)\}_{i}$ a partition of $ [0,1]$ or what?

Dear dphexen, the title says recurrent relation  :lol:

Kunny, i really dont understand your "solution"

I made a simple one, some small assumptions for $ X$ I hide because is too obviously.

[hide]$ f(t)\equal{}t^{3/2}$ , $ \int\sqrt{f'(t)^{2}\plus{}1}\, dt\equal{}\left(\frac{2}{3}\right)^{3}\left(\frac{9 t}{4}\plus{}1\right)^{3/2}$

$ \int_{0}^{1}\sqrt{f'(t)^{2}\plus{}1}\, dt\equal{}\frac{1}{27}\left(\minus{}8\plus{}13\sqrt{13}\right)\equal{}c_{1}$

$ X(n)\geq 0\Rightarrow Y(n)\geq 0$ , $ X(n)\equal{}\frac{1}{9}\left(Y(n)^{2/3}\minus{}4\right)$

$ \int_{X(n)}^{X(n\plus{}1)}\sqrt{f'(t)^{2}\plus{}1}\, dt\equal{}\frac{1}{27}(Y(n\plus{}1)\minus{}Y(n))\equal{}c_{1}$

The relation is linear, so, the solution is $ Y(n)\equal{}n\alpha\plus{}\beta$

$ \frac{X(n\plus{}1)}{X(n)}\equal{}\frac{((n\plus{}1)\alpha\plus{}\beta )^{2/3}\minus{}4}{(n\alpha\plus{}\beta )^{2/3}\minus{}4}$

And the limit above for $ n\to\infty$ is alwats $ 1$[/hide]


inequalities
geometry
power of a point
inequalities open

Let $ O_1,O_2,O_3,O_4$ be the centers of four nonoverlapping coplanar circles of unit radius. Find the optimal constant $ k$ such that if $ P$ is any point in the plane of the circles then $ \sum_{i\equal{}1}^4(PO_i)^2\geq k$.

Ofcourse, power of a point with respect to the 4 circles is given by
$ \sum |PO_i^2 \minus{} R^2| \ge  |\sum PO_i^2 \minus{} 4R^2 | \equal{} |\sum PO_i^2 \minus{}4| \ge |k\minus{}4| \ge 0$ which gives $ k$.


Define the operation $ \star$ by $ a\star b \equal{} (a \plus{} b)b$. What is $ (3\star 5) \minus{} (5\star 3)$?

$ \textbf{(A)}\ \minus{}16\qquad
\textbf{(B)}\ \minus{}8\qquad
\textbf{(C)}\ 0\qquad
\textbf{(D)}\ 8\qquad
\textbf{(E)}\ 16$

[hide="Observations"]
It's not hard to show that $ (a\star b)\cdot\frac{a}{b}\equal{}(b\star a)$

So $ (3\star 5)\minus{}(3\star 5)\cdot\frac{3}{5}\equal{}\minus{}\frac{2}{3}(3\star 5)\equal{}\frac{2}{5}\cdot 40\equal{}\boxed{\textbf{(E)}\ 16}$.

It's also not hard to show that $ (a\star b)\pm(b\star a)\equal{}(a\plus{}b)(b\pm a)$, from which you get $ (3\star 5)\minus{}(5\star 3)\equal{}(3\plus{}5)(5\minus{}3)\equal{}\boxed{\textbf{(E)}\ 16}$.
[/hide]


calculus
integration
limit
calculus computations

Let $ f(x) \equal{} e^{(p \plus{} 1)x} \minus{} e^x$ for real number $ p > 0$. Answer the following questions.

(1) Find the value of $ x \equal{} s_p$ for which $ f(x)$ is minimal and draw the graph of $ y \equal{} f(x)$.

(2) Let $ g(t) \equal{} \int_t^{t \plus{} 1} f(x)e^{t \minus{} x}\ dx$. Find the value of $ t \equal{} t_p$ for which $ g(t)$ is minimal.

(3) Use the fact $ 1 \plus{} \frac {p}{2}\leq \frac {e^p \minus{} 1}{p}\leq 1 \plus{} \frac {p}{2} \plus{} p^2\ (0 < p\leq 1)$ to find the limit $ \lim_{p\rightarrow \plus{}0} (t_p \minus{} s_p)$.

hello
[hide="1"]
after differentiating $ f(x)$ we will get
$ f'(x)\equal{}e^{x}((p\plus{}1)e^{px}\minus{}1)\equal{}0$
$ s_{p}\equal{}\minus{}\frac{ln(p\plus{}1)}{p}$
[hide="Drawing the graph"]
$ \lim_{x\to\minus{}\infty}f(x)\equal{}0$
$ \lim_{x\to\infty}f(x)\equal{}\infty$
$ f(0)\equal{}0$
it has one minima at $ x\equal{}s_{p}$
so its graph would like this
at $ \minus{}\infty f(x)$ tends to zero.Then decreasing with increasing x -value in $ III$rd quad it will come to its minimal value at $ x\equal{}s_{p}$.From there it will start increasing passing through origin tends to go to $ \infty$ with x tends to $ \infty$
[/hide]
[/hide]
thank u

[hide="2"]$ g(t)\equal{}e^t\int_t^{t\plus{}1} (e^{px}\minus{}1)dx\equal{}\frac{e^t}{p}\left[e^{pt}(e^p\minus{}1)\minus{}p\right]$

$ g'(t)\equal{}\frac{p\plus{}1}{p}[e^{pt}(e^p\minus{}1)\minus{}1]e^t$

$ t_p\equal{}\minus{}\frac{\ln(e^p\minus{}1)}{p}$[/hide]

[quote=" kunny"]

(1) $ f'(x) \equal{} (p \plus{} 1)e^x\left(e^{px} \minus{} \frac {1}{p \plus{} 1}\right)$ より, $ s_p \equal{} \boxed{ \minus{} \frac {\log (1 \plus{} p)}{p}}$

(2) $ g(t) \equal{} \int_t^{t \plus{} 1} e^t(e^{px} \minus{} 1)\ dx \equal{} \frac {1}{p}\{(e^p \minus{} 1)e^{pt} \minus{} p\}$ より, $ p > 0\Longleftrightarrow e^p > 1$ であるから,

$ g'(t) \equal{} \frac {1 \plus{} p}{p}e^{pt}(e^p \minus{} 1)\left\{e^{pt} \minus{} \frac {p}{(1 \plus{} p)(e^p \minus{} 1)}\right\}$ より,

$ t_p \equal{} \boxed{\frac {1}{p}\log \frac {p}{(1 \plus{} p)(e^p \minus{} 1)}}$

(3) $ t_p \minus{} s_p \equal{} \minus{} \frac {1}{p}\log \frac {e^p \minus{} 1}{p}\ (p > 0)$

すると, 与えられた不等式から, $ \log \left(1 \plus{} \frac {p}{2}\right)\leq \log \frac {e^p \minus{} 1}{p}\leq \log \left(1 \plus{} \frac {p}{2} \plus{} p^2\right)$

$ \therefore \minus{} \frac {1}{p}\log \left(1 \plus{} \frac {p}{2} \plus{} p^2\right)\leq \minus{} \frac {1}{p}\log \frac {e^p \minus{} 1}{p}\leq \minus{} \frac {1}{p}\log \left(1 \plus{} \frac {p}{2}\right)$

$ \boxed{\lim_{h\rightarrow 0} (1 \plus{} h)^{\frac {1}{h}} \equal{} e}$ より, $ \frac {1}{p}\log \left(1 \plus{} \frac {p}{2}\right) \equal{} \frac {1}{2}\log \left\{1 \plus{} \frac {p}{2}\right)^{\frac {2}{p}}\longrightarrow \frac {1}{2}\ (p\rightarrow \plus{} 0)$,

$ \frac {1}{p}\log \left(1 \plus{} \frac {p}{2} \plus{} p^2\right) \equal{} \left(\frac {1}{2} \plus{} p\right)\log \left\{\left(1 \plus{} \frac {p}{2} \plus{} p^2\right)^{\frac {1}{\frac {p}{2} \plus{} p^2}}\right\}$

$ \longrightarrow \frac {1}{2}\log e \equal{} \frac {1}{2}\ (p\rightarrow \plus{} 0)$.

$ \therefore \lim_{p\rightarrow \plus{} 0} (t_p \minus{} s_p) \equal{} \boxed{ \minus{} \frac {1}{2}}$

[/quote]


A cyclic quadrilateral is that that is inscribed in a circumference. Its main characteristic is that each pair of opposite angles are supplementary.

And two other cool, important, useful things:
1. Ptolemy's Theorem: In a cyclic quadrilateral, the product of one pair of opposite sides plus the product of the other two opposite sides is equal to the product of the diagonals. If we label the sides a, b, c, d, (in order) and the diagonals e and f respectively, we get $AC+BD = EF$.

2. Brahmagutpa's Area Formula:
This can be generalized to non-cyclics, but there's a nice formula for cyclics.
The area of a cylic quad with sides A, B, C, and D (and semiperimeter $S = \frac{A+B+C+D}{2}$) is equal to
$\sqrt{(S-A)(S-B)(S-C)(S-D)}$


Does anyone know how to hide the "Table of Contents" box for one particular page in a MediaWiki?

Write an extension that, if selected will implement the following code for css:

[code]#toc {display: none;}[/code]


AMC USA(J)MO USAMO algebra polynomial function

Let a,b,c be 3 distinct integers, and let P be a polynomial with integer coefficients. Show that in this case, the condition P(a)=b,P(b)=c,P(c)=a cannot be satisfied simultaneously.

[hide="My Solution"]Lets assume that the three conditions are satisfied at the same time, now because $ P$ is a polynomial with integer coefficients we have:

$ a-b|P(a)-P(b)=b-c$
$ b-c|P(b)-P(c)=c-a$
$ c-a|P(c)-P(a)=c-a$

from these three we can get that $ a=b=c$(why?) but that is a contradiction because $ a,b,c$ must be distinct! :)[/hide]

[b]Follow-up:[/b]  For what real functions $ F(x)$ does the result continue to hold over reals?

[quote="mdk"]yep, that's basically how I did it. Some of those archaic problems are relatively easy.[/quote]

It is only easy because we are familiar with the theorem.


I need to find a site that with free sub-domain names that [b]lets you use html[/b] (and preferably php, as well).  Anyone know of any?  All the sites I've seen just give you formats and a bunch of other worthless stuff and don't let you do anything.  :(

Well you could (dont recommend it) download apache webserver (google it) follow its instructions configure php and such. After that you can use the mod security, and make it live (note that this may expose your computer to security risks).

If you don't want to go through the hassle of this, then I would go with free or minimum pay web hosting. I recommend Microsoft Office Live's website hosting. Its really meant for businesses and such but they DO give you 500mb for free so long as you have a live or hotmail account. It doesnt support php and it sadly forces you to their limited designs, but it [b]does [/b] allow you to input your own html modules.


integration
real analysis
real analysis unsolved

$ {\rm{Evaluate:  }}\sum\limits_{n \equal{} 0}^\infty  {\frac{{\left( {2n} \right)!}}{{\left( {2^n n!} \right)^2 \left( {2n \plus{} 1} \right)^2 }}}$


geometry probability AMC AIME USA(J)MO USAMO MATHCOUNTS

Here's my study schedule:

Intro to C&P
Intro to Geometry
AoPS volume 2
ACoPS
(possibly)intermediate probability
(possibly)problem solving strategies

Is this a good study schedule for someone who hopes to score high on AIME and make USAMO?

This is the Mathcounts/ MOEMS Forum.  Post this in the AMC forum

EDIT: Lol I just read that you needed help from successful competitors. Im definitely not a successful competitor at all.

Heh what are the chances you'll actually do all of that? Believe me, if you do, you should be scoring relatively highly on USAMO.

I almost made USAMO this year despite only going through Volume 1 thoroughly, and through Volume 2 and Intro to CP not so thoroughly . I had done random chapters in Intro to Geometry. The thing is, I had done lots of problems, which is definitely the most helpful.

You don't really need the Intermediate Books to make USAMO. If you do go through them, you should be able to solve almost all the questions on the AIME that are algebra/combinatorics. I personally think the material in Volume 2 gives you more than enough to make USAMO, [b]if you know it very well[/b]. There's a difference between going through everything in the book and being able to use everything in the book.

The Intro books are great to get a good foundation, which is extremely necessary (infact, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1558908#1558908]this[/url]; most USAMO problems rely on only the basics and clever ideas). I think Volume 2 is good too. Intermediate CP might be good if you're not very good at the combo questions, but I don't think it's necessary. ACoPS is more for people new to olympiads,but I suppose some parts would be useful for AIME, so go through it if you want. On the other hand, Engel (PSS) is for olympiad preparation. I don't think it'll really help on the AIME. I personally think that Engel, and to a point ACoPS, are more of problem books, since you'll get most of the experience from doing the problems than actually reading the information in the sections.

So, if you're aiming just to make USAMO, I'd probably work through the Intro and Volume 2 and make sure you understand everything in there. If you have enough time for another book, I'd probably do ACoPS, unless you're really bad at combo. If you do manage to get through everything on that list [b]thoroughly[/b], you'd be pretty beast. However, do note that solving everything in Engel is pretty much impossible  :D .

The best advice is just to do lots of problems. Also, I'd rather go through only 2-3 of those books and understand all the material completely than work through 5 and not understand everything.

XPmath... You are amazing... You made USAMO with just that!? Don't worry, I'm already half way done with my schedule, and I have already read AoPS volume 1 twice!!!

oh, yeah... It doesn't matter if you are a successful competitor or not... I just wanted some[b] good[/b] advice...

I plan to score high on the AIME this year!!!! :yup:  :thumbup:  :jump:

Sorry for stealing your thread, but do you guys think just doing Inter. Algebra and AoPS Volume will qualify me for USAMO?

It should probably qualify you for USAMO, but I would suggest interm. prob over interm. alg... Do you mean AoPS volume 2? Well... Yeah it should probably... But as xpmath said, doing problems is also very helpful!

gl pytheagle... Your ambitions are similar to mine... :D (that makes you my competitor)

Er...I didn't quite make USAMO but I will blame Mathcounts for that  :D  except I won't.

My point is that those books are more than enough to qualify, but it'll be better if you do problems as well.

You could be like a teacher y'know? You're pretty amazing xpmath.... :lol:


[quote="hoangvn.fix"]$ 2sin3x \minus{} 3cos3x \plus{} 10cos2x \plus{} 2sin2x \plus{} 2sinx \plus{} 3cosx \minus{} 2 \equal{} 0$[/quote]

$ 1.08264952471866...\plus{}2k\pi$
$ 2.55359005004222...\plus{}2k\pi$
$ 3.62973945566603...\plus{}2k\pi$
$ 5.69518270363202...\plus{}2k\pi$

[i]Please, could you indicate us in what contest you found these crazy problems ... ?[/i]

this is a problem from my teacher
$ \Leftrightarrow (2cosx\plus{}3sinx)(sinx\minus{}cosx\minus{}sinxcosx)\equal{}0$

hello, you can rewrite your equation in the following form
$ 2(2\cos(x)\plus{}3\sin(x))(2\sqrt{2}\sin(\frac{\pi}{4}\minus{}x)\plus{}\sin(2x))\equal{}0$
Sonnhard.


I didn't print the problems yet!

Thank you
Billy

Do you not see the Archives down at the bottom of the class forums list?

In the online school section? No.  :(  :?

Billy

Thank you! :lol:  :lol:  :lol:  :lol:  :lol:  :lol:


search
Support
inequalities unsolved
inequalities

If $ x$, $ y$ and $ z$ are three real numbers, all different from 1, such that $ xyz \equal{} 1$, then prove that
$ \frac {x^{2}}{\left(x \minus{} 1\right)^{2}} \plus{} \frac {y^{2}}{\left(y \minus{} 1\right)^{2}} \plus{} \frac {z^{2}}{\left(z \minus{} 1\right)^{2}} \geq 1.$
I'm sorry if it has been posted. :oops:

I'm so sorry because of posting such a stupid spam.
Can mod delete this topic, please?

You don't need to use the search engine if you know the source.On this web page,you can see "Resources" on the top.Click it and you can find the problem easily.

[quote="Unidranix"]:P
Oh, I see. Thanks for your support[/quote]

It's very nice..But It's old problem  :P


Find all $ f: R→R$ that satisfies that condition.

$ f(x+y)+f(x-y)=2f(x)f(y)$

[quote="pco"][quote="Dimitris X"]Thats the well known d'alembert functional equation :wink: ....[/quote]

which is generally proposed with the supplementary constraint (likely forgotten here) of continuity[/quote]

exactly.... :)


function
LaTeX
calculus
calculus computations

Cauchy's Theorem:
[f(b)-f(a)]/[g(b)-g(a)] = df(x*)/dg(x*)
for some a<x*<b

Prove Cauchy's Theorem by applying Rolle's Theorem to the function
[f(b)-f(a)]g(x) - [g(b)-g(a)]f(x)

Note: In proving this result, all the work lies in thinking up the appropriate function to which the Mean Value Thm is applied.

Cauchy's Theorem:
$ \exists x^*, a < x^* < b : \frac{f(b) \minus{} f(a)}{g(b) \minus{} g(a)} \equal{} \frac{df}{dg}(x^*)$

Prove Cauchy's Theorem by applying Rolle's Theorem to the function:
$ \left[ f(b) \minus{} f(a) \right] g(x) \minus{} \left[g(b) \minus{} g(a)\right] f(x)$.

Right, I made an error while formatting it into LaTeX: $ \frac {df}{dg}(x^*)$ should be $ \frac {f^{'}(x^{*})}{g^{'}(x^{*})}$

So let's see. I will define a slightly different function as yours seems to not do what we want.

Note that we assume $ f$ and $ g$ to be differentiable on $ (a,b)$, and continous on $ [a,b]$, where $ a$ and $ b$ are distinct.

Define $ h(x) : \equal{} \left[ f(b) \minus{} f(a) \right] \left [g(x) \minus{} g(a) \right ] \minus{} \left[g(b) \minus{} g(a)\right] \left[ f(x) \minus{} f(a) \right ]$.

Then $ h(x)$ is continuous on $ [a,b]$ and differentiable on $ (a,b)$ by linearity (and that the constant function is too).

Note that:
$ h(b) \equal{} \left[ f(b) \minus{} f(a) \right] \left [g(b) \minus{} g(a) \right ] \minus{} \left[g(b) \minus{} g(a)\right] \left[ f(b) \minus{} f(a) \right ] \equal{} 0,$

$ h(a) \equal{} \left[ f(b) \minus{} f(a) \right] \left [g(a) \minus{} g(a) \right ] \minus{} \left[g(b) \minus{} g(a)\right] \left[ f(a) \minus{} f(a) \right ] \equal{} 0$.

So $ h(b) \equal{} h(a)$ (in retrospect, yours would do fine aswell).

Rolle now gives that:

$ \exists \, c \, \in \, (a,b) : h^{'}(c) \equal{} 0.$

Furthermore, note that: $ h^{'}(x) \equal{} \left[ f(b) \minus{} f(a) \right] g^{'}(x) \minus{} \left[g(b) \minus{} g(a)\right] f^{'}(x)$.

As such we get for that $ c$:

$ \left[ f(b) \minus{} f(a) \right] g^{'}(c) \minus{} \left[g(b) \minus{} g(a)\right] f^{'}(c) \equal{} 0$

$ \left[ f(b) \minus{} f(a) \right] g^{'}(c) \equal{} \left[g(b) \minus{} g(a)\right] f^{'}(c)$

Assuming $ g(b) \neq g(a)$, $ g^{'}(c) \neq 0$, we get:

$ \frac {f(b) \minus{} f(a)}{g(b) \minus{} g(a)} \equal{} \frac {f^{'}(c)}{g^{'}(c)}$

And we are done.


If you express $ \frac{1}{13}$ in binary, it will look like below:

$ 0.\overline{a_1 a_2 a_3\cdots a_r}$.

Find the smallest value of $ r$.

(i.e. Find the length of the smallest repeating block when you express $ \frac{1}{13}$ in binary.)

Write $ 0.\overline{a_1 a_2 ... a_r}$ as $ \frac{\overline{a_1 a_2 ... a_r} }{2^r \minus{} 1}$.  What does this tell you about $ 2^r \minus{} 1$?

[hide]
The largest value of $ 1/2^n$ can be subtracted from $ 1/13$ is $ n \equal{} 4$. Subtracting, we obtain $ \frac {3}{208}$. The next largest $ 1/2^n$ that can be subtracted is $ 1/2^8$. Noticing the pattern, we find that $ \frac {1}{13} \equal{} 0.\overline{0001}_2$. Therefore, $ r \equal{} \boxed{4}$.
[/hide]

EDIT: Think about how you find a repeating decimal in base ten. You divide by $ 10^n \minus{} 1$ for $ n$ repeating digits. The same thing works for other bases. If you want to prove it, let $ x$ be the repeating decimal and apply the same process you would when you would find the fraction that represents a repeating decimal in base $ 10$.

[quote="sunehra"][hide]
The largest value of $ 1/2^n$ can be subtracted from $ 1/13$ is $ n \equal{} 4$. Subtracting, we obtain $ \frac {3}{208}$. The next largest $ 1/2^n$ that can be subtracted is $ 1/2^8$. Noticing the pattern, we find that $ \frac {1}{13} \equal{} 0.\overline{0001}_2$. Therefore, $ r \equal{} \boxed{4}$.
[/hide][/quote]
Sorry, that answer is wrong.

I have the correct answer, but I do not know how to get it...

[quote="FantasyLover"][quote="t0rajir0u"]Write $ 0.\overline{a_1 a_2 ... a_r}$ as $ \frac {\overline{a_1 a_2 ... a_r} }{2^r \minus{} 1}$.  What does this tell you about $ 2^r \minus{} 1$?[/quote]
Er, can you explain how you got that?[/quote]
Do you know how to express repeating decimals as fractions in base $ 10$?  It's the same thing.  If you don't, $ 0.\overline{a_1 a_2 ... a_r}$ is just short for

$ \frac{\overline{a_1 a_2 ... a_r}}{2^r} \plus{} \frac{\overline{a_1 a_2 ... a_r}}{2^{2r}} \plus{} ...$

which is a geometric series.

[quote="sunehra"][hide]
The largest value of $ 1/2^n$ can be subtracted from $ 1/13$ is $ n \equal{} 4$. Subtracting, we obtain $ \frac {3}{208}$. The next largest $ 1/2^n$ that can be subtracted is $ 1/2^8$. Noticing the pattern, we find that $ \frac {1}{13} \equal{} 0.\overline{0001}_2$. Therefore, $ r \equal{} \boxed{4}$.
[/hide][/quote]
Your second step is incorrect; it's $ \frac{1}{2^7}$.

Oh, I understand now.

As for the problem, I think $ 2^r\minus{}1\equiv 0 \mod 13$, is it so?

Yes, but more than that; $ r$ is the smallest positive integer for which this is true, sometimes known as the [url=http://en.wikipedia.org/wiki/Order_(number_theory)]order[/url] of $ 2 \bmod 13$.  Are you familiar with [url=http://en.wikipedia.org/wiki/Fermat%27s_little_theorem]Fermat's little theorem[/url]?  It tells you another property of $ r$.

I got it;

$ 2^6\equiv \minus{} 1 \mod 13 \implies 2^{12}\equiv 1\mod 13 \implies r\equal{}12$, which is the correct answer.

Thanks, t0rajir0u!

Yet another "by hand" approach:

Find the smallest power of $ 2$ such that the product of $ {1\over 13}$ and that number exceeds $ 1$. Obviously, it would be $ 16\equal{}2^4$. Now take the fractional part of the product and repeat the process:

$ {1\over 13}\cdot 2^4\equal{}{16\over 13}$, frac. part is $ {3\over 13}$

$ {3\over 13}\cdot 2^3\equal{}{24\over 13}$, frac. part is $ {11\over 13}$

$ {11\over 13}\cdot 2^1\equal{}{22\over 13}$, frac. part is $ {9\over 13}$

$ {9\over 13}\cdot 2^1\equal{}{18\over 13}$, frac. part is $ {5\over 13}$

$ {5\over 13}\cdot 2^2\equal{}{20\over 13}$, frac. part is $ {7\over 13}$

$ {7\over 13}\cdot 2^1\equal{}{14\over 13}$, frac. part is $ {1\over 13}$

Hence we're back at the beginning, so now we simply add up the exponents of $ 2$ obtained in the process:

$ 4\plus{}3\plus{}1\plus{}1\plus{}2\plus{}1\equal{}12$, which is of course the same result gotten by the previous methods.


function
algebra
functional equation
algebra proposed

Find all $f(x) : \mathbb{R} \to \mathbb{R}$ that satisfy

$f \left( \frac{x-3}{x+1} \right) + f \left( \frac{3+x}{1-x} \right) = x$

I am not confident about my solution to this, so I would like a second one :D

[hide="Another opinion"]First, it's obvious that $x\neq\pm 1$. We'll substitute $t={{x-3}\over{x+1}}$, and then $t={{3+x}\over{1-x}}$. Along with the initial equation we'll get:

\begin{eqnarray}f(t)+f\left( {{t-3}\over{t+1}}\right) &=& {{3+t}\over{1-t}} \\ f(t)+f\left( {{3+t}\over{1-t}}\right) &=& {{t-3}\over{t+1}} \\f\left( {{t-3}\over{t+1}}\right) +f\left( {{3+t}\over{1-t}}\right) &=& t\end{eqnarray}

Now calculate ${{(1)+(2)-(3)}\over 2}$ to obtain $f(t).$ If I'm not mistaken, the result is $f(t)={{t(t^2+7)}\over{2(1-t^2)}}$[/hide]

[quote="t0rajir0u"]Yes, that's how I solved it too... and... it doesn't work in the original functional equation :(  This should imply no solution, I know, but the problem seems too neat not to have a solution...[/quote]Farenhajt's solution (or your solution) is correct. I substitute in $f$ and it works in original equation... did you check thoroughly the algebra when simplifying the expression while plugging in $f(x)=\frac{x(x^2+7)}{2(1-x^2)}$ ? Because at first, I substitute function and did not get original equation. I went back to check on my work and found an error (a minus sign) and got it correct. I checked by using a numerical example $x=2$ and found $f(2)=-\frac{11}3$, $f\left(-\frac13\right)=-\frac43$, and $f(-5)=\frac{10}3$ and got $f\left(-\frac13\right)+f(-5)=2$. I strongly believe that's the answer unless someone PROVES it wrong  :D

Yes, 10000th user is correct - I've checked again (without haste :)) and it works.


A warehouse contains 175 boots of size 8, 175 boots of size 9 and 200 boots of size 10. Of these 550 boots, 250 are for the left foot and 300 are for the right foot. Let $n$ denote the total number of usable pairs of boots in the warehouse. (A usable pair consists of a right boot and a left boot of the same size.)

(a) Can we have $n = 50$?
(b) Can we have $n = 51$?

I dont really understand the question is it something like ... this ?

[b]If[/b]

Size 8  :arrow:  Left  $50$ ; Right $125$

Size 9  :arrow: Left  $0$ ; Right $175$

Size 10 :arrow: Left $200$ ; Right $0$

Hence total usable pair of boots =$50$

So the question want us to find EXACTLY $50$ and $51$ pairs of boots ?

I think I got the question , so now i will prove part (b)

[hide]
We shall prove that $n=51$ wont work .

Assume it exists for such $n=51$

First letting the left-right side of each size of boots (size 8 , size 9 ,size 10) be $L_1,R_1$ ; $L_2,R_2$ ; $L_3,R_3$ respectively .

For any two $L_i$ , we denote their sum as $L$ .If two of them sum up to $L$ then the third one should be $250-L$ . Since this value cannot exceed $200$(which is the max number of each size of boot) , we will get $L\geq 50$ $[*]$.

Similar argument yields for $R$ where $R$ is the sum of any two $R_i$ . We will get $R\geq 100$ $[**]$

Since we are to choose pair of boots with same size , it means we must choose $min\{L_i,R_i\}$ for each size .We also know that $min{\L_i,R_i\}\leq \frac{175}{2}<88}$ .

If any one of $L_i$ or $R_i$ with same size is greater or equal than $52$ , then the other one with same size must be less or equal than $175-52=123$ . But these will give us $min\{L_I,R_i\}\geq 52$ which contradict with our assumtion. Hence $L_i,R_i$ must range in $[0,51]U[124,200]$ $[\#]$

Now we shall divide cases
(i)If all $min\{L_i,R_i\}=L_i$ , then all $L_i\leq 51$ (from $[\#]$) . We want to find $\sum L_i=51$ . But from $[*]$ , we obtain $\sum L_i\geq \frac{150}{2}=76.5$ . Hence there is no way to get $51$ in this case .

(ii)If all $min\{L_i,R_i\}=R_i$ , then all $R_i\leq 51$ . We wish to find $\sum R_i=51$ but $[**]$ gives us $\sum R_i\geq \frac{300}{2}=150$ . So again there is no solution here .

(iii)For one $L_i$ and two $R_j$ , since we have $[**]$ , so there will be at least $100$ pairs here which also no solution.

(iv)For two $L_i$ and one $R_j$ , we have $[*]$ , this implies $R_j=0$ or $1$ .
 :arrow: When $R_j=1$ , the two $L_i$ must be $50$ . But $R_j=1$ gives us the number of left boot of same size with $R_j$ is $174$ or $199$ pairs . So we will obtain the remaining left boots to be either $76$ or $51$ . But both of this doesnt gives us $50$ .
 :arrow: When $R_j=0$ , the two $L_i$ must be $51$ . The number of remaining left boot is $75$ or $50$ . Again we doesnt obtain $51$ here . So no solution.

Combining all the cases above , there are no way we can choose $51$ pairs of boots .

:)
[/hide]


analytic geometry
graphing lines
slope
algebra
linear equation

[quote="symmetry"]Hi, I need help finding the x and y intercepts for problems like these..

1.Passes through points (-1,5) and (2,-1)

2.Slope -4 and x-intercept 7.

3. slope 2/3 and passes through point (5,7)[/quote]
Do you know what a point/slope equation is?

If you don't know about the point slope form of a linear equation, you can always find the equation that the two points work for and then find the intercepts.

For number question 1, the slope = -2.

I then used the point-slope formula:

y - y1 = m(x - x1) y - 5 = -2(x -(-1) ) y-5 = -2x -2 y = -2x + 3 3 is the y-intercept.

Set y to 0 and solve for x. 0 = -2x + 3 -3 = -2x -3/-2 = x 3/2 = x = intercept.

In Summary:
y-intercept is 3 in the form (o,3)
x-intercept is 3/2 in the form (3/2,0)
The slope is -2.

*******************************

For question 3,

y - y1 = m(x - x1) y - 7 = 2/3(x - 5) y - 7 = 2/3(x)-10/3 y = 2/3(x) + 11/3 11/3 is the y-intercept.

Set y to 0 and solve for x. 0 = 2/3(x) + 11/3. -11/3 = 2/3(x) -11/2 divided by 2/3 = x -11/2 = x = x-intercept.

In Summary: y-intercept is 11/3 in the form (0,11/3)

x-intercept is -11/2 in the form (-11/2,0)

Slope is 2/3.

***************************

For question 2, the equation of a line is in the form y = mx + b where m is the slope and b is the y-intercept.

You was given m as -4 and given the x-intercept 7. I Substituted 7 for x and 0 for y in the general equation of a line: (y = mx + b) and solved for b.

0 = -4x + b 0 = -4(7) + b 0 = -28 + b 28 = b = y-intercept In

Summary:

x-intercept is 7 in the form (7,0)
y-intercept is 28 in the form (0,28)
Slope is -4.

My question: Is any of this correct?
If not, where did I mess up?

I responded to your post in the intermediate section, but... these are all correct!!  :D

Good Job!

[hide]Am I a good moderator or what?  :D  :D [/hide]


For $c\geq 1,$ we have $\frac{1}{\sqrt{c+1}+\sqrt{c}}<\frac{1}{\sqrt{c}+\sqrt{c-1}}\Longleftrightarrow \sqrt{c+1}-\sqrt{c}<\sqrt{c}-\sqrt{c-1}.$

also note that $\displaystyle\lim_{n\to\infty}\sqrt{n+1}-\sqrt{n}=0$
so as $n$ increasses it gets cloer to 0.
or
let $f(n)=\sqrt{n+1}-\sqrt{n}$ and show that
$f'(n)$ is negative for all $n$ so $f(n)$ is a decreasing function.

This reminds me of a limit evaluating technique. :)

[hide="Solution"]Multiplying and dividing $\sqrt{c+1}-\sqrt c$ by its conjugate, \[ \sqrt{c+1}-\sqrt c=\frac{(\sqrt{c+1})^2-(\sqrt c)^2}{\sqrt{c+1}+\sqrt{c}}=\frac1{\sqrt{c+1}+\sqrt{c}}.  \]Similarly, $\sqrt c-\sqrt{c-1}=\frac{1}{\sqrt c-\sqrt{c-1}}.$ We know that  $\frac1{\sqrt{c+1}+\sqrt{c}}<\frac{1}{\sqrt c-\sqrt{c-1}}$ for all positive $c$, so $\sqrt{c+1}-\sqrt c <\sqrt c-\sqrt{c-1}.$[/hide]

[quote="chess64"]Determine which of the two numbers $\sqrt{c+1}-\sqrt{c}$, $\sqrt{c}-\sqrt{c-1}$ is greater for any $c\ge 1$.[/quote]

[hide]
\[ \sqrt{c+1} - \sqrt{c} \text{ ? } \sqrt{c} - \sqrt{c - 1} \]
\[ \sqrt{c+1} + \sqrt{c-1} \text{ ? } 2\sqrt{c} \]
\[ 2c + 2\sqrt{c^2 - 1} \text{ ? } 4c \]
\[ \sqrt{c^2 - 1} \text { ? } c \]
\[ c^2 - 1 < c \] so ? is $<$ [/hide]

[quote="buinamnghia"][quote="maokid7"]also note that $\lim_{n\to\infty}\sqrt{n+1}-\sqrt{n}=0$
so as $n$ increasses it gets cloer to 0.
or
let $f(n)=\sqrt{n+1}-\sqrt{n}$ and show that
$f'(n)$ is negative for all $n$ so $f(n)$ is a decreasing function.[/quote]
Oh ! what did you do?
 I think you did a bad solution.[/quote]

the first limit does not really prove anything...only that it is bounded but i don't see why that is important, but the second part, the calculus way works fine, once you have that it is a decreasing function, it is immediately true that

$f(n)<f(n-1)$

which was what we wanted ($W^{5}$ for acops readers  :P )

Here is a very simple solution since it does not require a proof or anything.
[hide]
Let $c=1$.
You have the two numbers $\sqrt{2}-\sqrt{1}$ and $\sqrt{1}-\sqrt{0}$. The first number becomes $\sqrt{2}-1$ amd the second $1$. Now, for $\sqrt{2}-1$ to be greater than 1, $\sqrt{2}$ must be greater than 2. That means $\sqrt{2}$ must be greater than $\sqrt{4}$ since 2 is the square root of 4, because it is not, the answer is clearly $\sqrt{c}-\sqrt{c-1}$.[/hide]

[quote="sen"]im just wondering, is that sorta proof allowed in competitions?[/quote]

I didn't think a proof was required given the wording of the original problem.

[quote="lotrgreengrapes7926"]It was a Canadian National Olympiad...[/quote]

I didn't really pay attention to that and all it said was "Canada" That could mean a lot of things. Anyway, I hope my way would at least score me 1 point assuming they have the same scoring rules as USAMO.

[quote="Altheman"]$\frac{\sqrt{c+1}+\sqrt{c-1}}{2}<\sqrt{c}$ by jensen, so $\sqrt{c+1}-\sqrt{c}<\sqrt{c}-\sqrt{c-1}$[/quote]

Hmm... could you elaborate more upon how Jensen's was used, please?

Anyhow, here's my solution:

[hide]Let's assume that $\sqrt{c+1}-\sqrt{c}> \sqrt{c}-\sqrt{c-1}$.

$\sqrt{c+1}-\sqrt{c}> \sqrt{c}-\sqrt{c-1}$
$\sqrt{c+1}-\sqrt{c-1}> 2\sqrt{c}$
$2c+2\sqrt{c^{2}-1}>4c$
$\sqrt{c^{2}-1}>c$
$c^{2}-1>c^{2}$
$0>1$

Quite obviously this is a contradiction. However, if the inequality sign is reversed, the inequality is satisfied. Thus, we have that $\sqrt{c}-\sqrt{c-1}> \sqrt{c+1}-\sqrt{c}$.[/hide]

[quote="Karth"][quote="Altheman"]$\frac{\sqrt{c+1}+\sqrt{c-1}}{2}<\sqrt{c}$ by jensen, so $\sqrt{c+1}-\sqrt{c}<\sqrt{c}-\sqrt{c-1}$[/quote]

Hmm... could you elaborate more upon how Jensen's was used, please?
[/quote]

Consider $f(x) = \sqrt{x}$, so $f''(x) = \frac{-1}{4\sqrt[3]{x^{2}}}$, so it's concave on $\mathbb{R}^{+}$. Therefore, Jensen says that \[\begin{aligned}& \frac{f(c+1)+f(c-1)}{2}\leq f\left(\frac{(c+1)+(c-1)}{2}\right)\ \\ & \Longleftrightarrow\ \sqrt{c+1}+\sqrt{c-1}\leq 2\sqrt{c}\ \\ & \Longleftrightarrow\ \sqrt{c+1}-\sqrt{c}\leq \sqrt{c}-\sqrt{c-1}.\end{aligned}\] But equality occurs iff $c+1=c-1$, i.e. $2=0$. Therefore, the inequality is strict, so we're done.

[hide="Hint"]Take reciprocals of both of the terms.

Then it should be pretty obvious which one is larger and then hence the reciprocal of the larger reciprocal is smaller than the reciprocal of the smaller reciprocal.[/hide]

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.CMO1969Problem2

Vo Duc Dien

[quote="chess64"]Determine which of the two numbers $\sqrt{c+1}-\sqrt{c}$, $\sqrt{c}-\sqrt{c-1}$ is greater for any $c\ge 1$.[/quote]

$(\sqrt{c+1}+\sqrt{c-1})^2+(\sqrt{c+1}-\sqrt{c-1})^2=4c\ (c\geq 1)$

$\therefore (\sqrt{c+1}+\sqrt{c-1})^2<4c$.

$\therefore \sqrt{c+1}+\sqrt{c-1}<2\sqrt{c}$

$\Longleftrightarrow \sqrt{c+1}-\sqrt{c}<\sqrt{c}-\sqrt{c-1}$

revive, but my approach : consider of graph y = $sqrt(x)$

$\sqrt{c+1}-\sqrt{c}*\sqrt{c}-\sqrt{c-1}$
$\sqrt{c+1}+\sqrt{c-1}*2\sqrt{c}$
Squaring we have $<$ instead of ${*}$
$2c-2\sqrt{c^2-1}<2c$

We claim that $\sqrt{c}-\sqrt{c-1} \geq \sqrt{c+1}-\sqrt{c}$ for $c \geq 1$. Squaring both sides gives $2c-1-2\sqrt{c-1}\sqrt{c} \geq 2c+1-2\sqrt{c+1}\sqrt{c}$, or $-2\sqrt{c-1}\sqrt{c} \geq 2-2\sqrt{c+1}\sqrt{c}$. This is equivalent to $-\sqrt{c-1}\sqrt{c} \geq 1-\sqrt{c+1}\sqrt{c}$, or $\sqrt{c}\left(\sqrt{c+1}-\sqrt{c-1}
\right) \geq 1$. Squaring both sides gives $c(2c-2\sqrt{c-1}\sqrt{c+1}) \geq 1$, or $2c^2-2c\sqrt{c-1}\sqrt{c+1} \geq 1$. This is equivalent to $(1-2c^2)^2 \geq 4c^2(c-1)(c+1)$, which is always true if $c \geq 1$, as desired. All steps are reversible, so we're done.

We claim that the answer is $\sqrt{c}-\sqrt{c-1}$.
An alternate way to do this would be to claim that $\sqrt{c+1}+\sqrt{c-1}<2\sqrt{c}$, which would follow by squaring both sides to get $c+1+c-1+2\sqrt{c^2-1}<4c$, <=> $2\sqrt{c^2-1}<2c$, which is clearly true since $\sqrt{c^2-1}<\sqrt{c^2}=c$.

[hide=solution]
We claim that $\sqrt{c+1} - \sqrt{c} < \sqrt{c} - \sqrt{c-1}$ for all $c \geq 1.$
Clearly, we have $$c^2 - c + \frac{1}{4} > c^2 - c.$$ Since $c \geq 1,$ we have $c^2 - c \geq 0,$ so both sides are positive. Thus, we can take the square root of both sides to get $$c - \frac{1}{2} > \sqrt{c(c-1)}.$$ Multiplying both sides by $4$ gives $$4c-2 > 4\sqrt{c(c-1)}.$$ We can rearrange this to $$c + 1 < 4c + 4\sqrt{c(c-1)} + c - 1.$$ Rewriting this yields $$(\sqrt{c+1})^2 < (2\sqrt{c} - \sqrt{c-1})^2.$$ Clearly, both $\sqrt{c+1}$ and $2\sqrt{c} - \sqrt{c-1}$ are nonnegative for $c \geq 1,$ so we can take the square root of both sides to get $$\sqrt{c+1} < 2\sqrt{c} - \sqrt{c-1}.$$ Rearranging gives us $$\sqrt{c+1} - c < \sqrt{c} - \sqrt{c-1}.$$ $\square$
[/hide]

MVT's  $f_c=\sqrt{c+1}-\sqrt c=1/(2\sqrt a), \; c<a<c+1\Rightarrow~f$ decreases


Daca intr-un plan avem un cerc care il proiectam pe alt plan, atunci
     se obtine o elipsa.

Este de asemenea si o chestie super-cunoscuta.

Sa presupunem ca cercul este in planul $\mathcal P_{1}$, iar planul pe care se proiecteaza este $\mathcal P_{2}$. Daca mutam $\mathcal P_{2}$ (adica consideram un plan $\|$ cu el) astel incat $\mathcal P_{2}$ taie cercul intr-un diametru, atunci figura proiectata pe acest ultim plan e congruenta cu cea proiectata pe $\mathcal P_{2}$, deci in continuare o sa presupunem ca $\mathcal P_{2}$ chiar indeplineste conditia asta.

Fie $AB$ diametrul de care vorbeam si $CD \in \mathcal P_{1}$ un diametru perpendicular pe $AB$ (in $O$, centrul cercului). Consideram si dreapta $d \in \mathcal P_{2}$ perpendiculara pe $AB$ in $O$. Fie $\alpha = \angle ( \mathcal P_{1}, \mathcal P_{2})$.
Daca $P$ este un punct de coordonate $(a,b)$ ($a$ = distanta de la $P$ la $CD$; $b$ = distanta de la $P$ la $AB$), atunci $P^\prime$ are coordonate $( \frac{a}{\cos \alpha}, b )$
$( \frac{a}{\cos \alpha}$ = distanta de la $P^\prime$ la $CD$; $b$ distanta de la $P^\prime$ la $ d )$.

Din moment ce ecuatia cercului este $x^{2}+y^{2}= 1$, rezulta ca ecuatia proiectiei sale pe $\mathcal P_{2}$ are ecuatia $x^{2}\cos^{2}\alpha+y^{2}= 1$, care se stie ca e o elipsa.

Sper ca se intelege ce-am vrut sa zic...


I decided to use my username because it is part of my email address (stupididiot1337@gmail.com). Why is it inappropriate? I can't seem to log onto FTW.

I've seen many MORE offensive usernames than that, such as Pen_Is_Land...

It's probably "idiot" that offends the admins.

The consideration of "idiot" as offensive should really be removed. "Idoit" (and many other similar ones) can easily replace it while causing just as much offense as "idiot" could possibly do. Plus, it's not that bad in the first place. :wink:

Well my parents used to ground me for saying "idiot", but that was in first grade...

Now we all use idiot or something else every day :wink:
Inappropriate forum, inappropriate topic.
Vallon,Isabella,3rd person- lock this now.


the proof given in AoPS was extremely confusing, please help

it is like if you do the intersection, use the discriminant to make 1 solution for the y value...

[quote="srush13"]This problem is solvable with calculus, but how can you find it without calculus:

A circle rests in the interior of the parabola with equation  y=x^2  so that it is tangent to the parabola at two points. How much higher is the center of the circle than the points of tangency?[/quote]
Let the circle and parabola are touching each other at A and B.  By symmetry, we can let the equation to the circle as $x^2+y^2+2fy+c=0$ centered at $C(0, -f)$.
For the tangency, the equation resulted by substituting y= x2 in the above equation must have two pairs of equal roots.

Hence the equation $y+y^2+2fy+c=0$ must have equal roots, which can happen only when $(2f+1)2=4c$ and the root is $y=-(2f+1)$….(I)
We know, the ordinates of A and B are both $-(2f+1)$ and are positive.
Therefore, it is necessary that $f<-1/2$.

And the difference between the heights of center and ordinate of the point of contact is $(2f+1)-f=f+1<1/2$
Hence the maximum number by which the center is higher than the point of tangency is $1/2$.


y= $ \sqrt {x \plus{} 1}$
$ x \ge \minus{} 1$
$ y \ge 0$

Whats the domain and range in Interval Notation?

the $ x$ less equal to.... is not nesecary because otherwise it would be undefined. and really what you did there was giving the answer away.

Domain:$ [\minus{}1,\infty)$
Range:$ [0,\infty)$


Find two square roots of x+iy.

I got one to be:
the real part: the square root of x plus the square root of x^2 + y^2, all over the square root of 2.
the imaginary part: the square root of -x plus the square root of x^2 + y^2, all over the square root of 2.

How do I get the other square root?

The real and imaginary parts are both meant to be real numbers, but of the two numbers you've written down, at most one can be real (unless $ x \equal{} 0$) -- do you see why?

By the way, what's $ ( \minus{} 2)^2$?

While one could do this problem they way you're approaching it, probably it makes more sense to do everything in polar.

Finally, you've been around here long enough that you really should be learning some [[LaTeX]] -- see that link and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690


Μπορεί κάποιος να παραθέση μια απόδειξη πως τα Συμμετρικά του Ορθόκεντρου ως προς τις πλευρές τριγώνου είναι σημεία του περιγγεγραμμένου κύκλου.

Ευχαριστώ πολύ :oops:

Αυτό που ζητάς φίλε είναι το θεώρημα Carnot λοιπόν θα σου το αποδείξω για ένα συμμετρικό ομοίως για όλα τα άλλα
Έστω τρίγωνο ΑΒΓ εγγεγραμμένο σε κύκλο Κ κι τα ύψη ΑΔ ΒΕ ΓΖ τέμνονται στο Η κι έστω Η' το συμμετρικό του Η ώς προς την ΒΓ νδο το Η' ανήκει στον περιγγεγραμμένο κύκλο του ΑΒΓ
Λύση
Αρκεί το ΑΒΗ'Γ να είναι εγγράψιμο  ή
ΒΑΓ+ΒΉΓ=180  όμως ΒΉΓ=ΒΗΓ (Λόγω συμμετρίας)
[hide]γιάννη αν θες περαιτέρω εξήγηση για αυτό πές το [/hide]
αλλά στο τρίγωνο ΒΉΓ βλέπουμε εύκολα υπολογιστικά ότι
ΒΉΓ-180-Α  άρα
κι αφού ΒΑΓ=Α
ΒΑΓ+ΒΉΓ= 180-Α+Α=180
Άρα το ΑΒΉΓ εγγράψιμο άρα ισχύει το ζητούμενο

Ευχαριστώ πολύ , αν και απλή η απόδειξη δεν μπορούσα να βρω από που να ξεκινήσω. :blush:

Ολοκληρωμένη την απόδειξη του Carnot την έχει στο http://www.telemath.gr στο section της γεωμετρίας μαζί με το σχήμα..

PS: Τώρα μόλις είδα ότι είναι down το site, αλλά λογικά σε μερικές ωρίτσες θα είναι ok.

[quote="giannis"]Μπορεί κάποιος να παραθέση μια απόδειξη πως τα Συμμετρικά του Ορθόκεντρου ως προς τις πλευρές τριγώνου [color=red]ΚΑΙ ΩΣ ΠΡΟΣ ΤΑ ΜΕΣΑ ΤΩΝ ΠΛΕΥΡΩΝ[/color]  είναι σημεία του περιγγεγραμμένου κύκλου.

Ευχαριστώ πολύ :oops:[/quote]

 Συμπλήρωσα κάτι , διότι όλο μαζί είναι σπουδαία πρόταση.Με την πρόταση αυτή προκύπτει εύκολα και ο κύκλος του Euler.Αν έχει χρόνο ο Αμβρόσιος ας συμπληρώσει την απόδειξη.

     Μπάμπης

Πολύ σωστά έτσι  είναι, μετά από μια επανάληψη στα εγγεγραμμένα και εγράψιμα τετράπλευρα , παραθέτω μια απόδειξη για το δεύτερο :

Έστω τρίγωνο ΑΒΓ . Να αποδέιξετε ότι τα συμμετρικά του ορθόκεντρου Η ως προς τα μέσα του τριγώνου είναι σημεία του περιγεγγραμμένου κύκλου του τριγώνου.

[color=red]Απόδειξη[/color]

Έστω τρίγωνο ΑΒΓ εγγεγραμμένο σε κύκλο και ΑΔ, ΒΕ , ΓΖ τα ύψη του και Η το ορθόκεντρο του.Αν Η' είναι το συμμετρικό του Η ως προς το μέσο Μ της ΒΓ , τότε αρκέι να δείξουμε ότι το τετράπλευρο ΑΒΓΗ είναι εγγράψιμο

Είναι : ΒΜ=ΜΓ και ΜΗ=ΜΗ' (αφού τα Η Η' είναι συμμετρικά)
Άρα το τετράπλευρο ΗΒΓΗ' είναι παραλληλόγραμμό αφού ΒΜ=ΜΓ και ΜΗ=ΜΗ' (οι διαγώνιοι του διχοτομούνται). Άρα γων.ΒΗ'Γ=γων.ΒΗΓ (1)
ως απέναντι πλευρές παραλληλόγραμμού.Το τετράπλευρο ΑΖΕΗ είναι εγγράψιμο αφού Ε+Ζ=90+90=180 , άρα Α+ΖΗΕ=180 δηλ. Α+ΒΗΓ=180 (2)
Η (1) λόγω (2) γίνεται Α+ΒΗΓ=180 αν και μόνο αν το ΑΓΗ'Β είναι εγγράψιμο , άρα τα Α Β Γ Η' είναι ομοκυκλικά , άρα τα συμμετρικά του ορθόκεντρου Η ως προς τα μέσα του τριγώνου είναι σημεία του περιγεγγραμμένου κύκλου του τριγώνου.
Ομοίως και για τα άλλα μέσα. :P

[img]http://img129.imageshack.us/img129/1964/2170rn4.png[/img]

P.S Πως με την πρόταση αυτή προκύπτει και ο κύκλος του Euler?

Με προλάβατε από το να συμπληρώσω την απόδειξη αλλά χαλάλι σας γιατί είπατε όλα όσα θα ήθελα κι να συμπλήρωσω για τον κύκλο του Euler
χαιρετώ ξανά :D  :wink:

[quote="stergiu"][i]Σου είχα φτιάξει τη λύση αναλυτικά ,αλλά το antivirus ενεργοποιήθηκε και όλα έσβησαν. [/i]

 Λοιπόν   με βάση το σχήμα σου , πάρε $O, I , K$ το περίκεντρο , το μέσο του $OH$  και το μέσο του  $HB$ αντίστοιχα.
Από το τραπέζιο $HDMO$ παίρνεις $ID = IM$.
 Από το τρίγωνο $HOH'$ παίρνεις¨:
     $IM = \frac{1}{2}OH' =\frac{1}{2}R$.

 Από το τρίγωνο $HOB$ παίρνεις $IK = \frac{1}{2}OB =\frac{1}{2}R$.
 .Άρα τα σημεία $K , D , M$ βρίσκονται στον κύκλο με κέντρο το $I$ που έχει μάλιστα ακτίνα ίση με $\frac{1}{2}R$ , κάτι που έιναι επίσης χρήσιμο.  Ο κύκλος αυτός θα περάσει ακόμα και από τις άλλες αντίστοιχες τριάδες σημείων.
[u] Μπάμπης[/u][/quote]

Ευχαριστώ πολύ, αλλά για να δείξουμε πως ΙΔ=ΙΜ θα χρειστεί να φέρουμε τη διάμεσο του τραπεζίου , ώστε το ΙΔΜ να είναι ισοσκελές ?

[img]http://img403.imageshack.us/img403/2328/2170rn4fh8.png[/img]

[quote="giannis"][quote="stergiu"][i]Σου είχα φτιάξει τη λύση αναλυτικά ,αλλά το antivirus ενεργοποιήθηκε και όλα έσβησαν. [/i]

 Λοιπόν   με βάση το σχήμα σου , πάρε $O, I , K$ το περίκεντρο , το μέσο του $OH$  και το μέσο του  $HB$ αντίστοιχα.
Από το τραπέζιο $HDMO$ παίρνεις $ID = IM$.
 Από το τρίγωνο $HOH'$ παίρνεις¨:
     $IM = \frac{1}{2}OH' =\frac{1}{2}R$.

 Από το τρίγωνο $HOB$ παίρνεις $IK = \frac{1}{2}OB =\frac{1}{2}R$.
 .Άρα τα σημεία $K , D , M$ βρίσκονται στον κύκλο με κέντρο το $I$ που έχει μάλιστα ακτίνα ίση με $\frac{1}{2}R$ , κάτι που έιναι επίσης χρήσιμο.  Ο κύκλος αυτός θα περάσει ακόμα και από τις άλλες αντίστοιχες τριάδες σημείων.
[u] Μπάμπης[/u][/quote]

Ευχαριστώ πολύ, αλλά για να δείξουμε πως ΙΔ=ΙΜ θα χρειστεί να φέρουμε τη διάμεσο του τραπεζίου , ώστε το ΙΔΜ να είναι ισοσκελές ?

[img]http://img84.imageshack.us/img84/7536/ytud3.jpg[/img][/quote]

 [color=red]ΑΚΡΙΒΩΣ ![/color]
 [color=blue]Δεν το έγραψα αναλυτικά για σου αφήσω τη χαρά να το βρεις μόνος σου , κάτι που το έκανες ! Η διάμεσος αυτή είναι , όπως παρατήρησες , μεσοκάθετος του $DM$ .  Βλέπεις πόσα πράγματα προκύπτουν συγχρόνως για τον κύκλο του Euler με αυτή την απόδειξη ;Σου επισημαίνω ακόμα ότι αξίζει να θυμάσαι πως  [b]το βαρύκεντρο , το ορθόκεντρο , το περίκεντρο και το κέντρο του κύκλου του Euler βρίσκονται στην ίδια ευθεία , την ευθεία Euler του τριγώνου.[/b] Πρόκειται για στοιχειώδη και  πολύ  βασική πρόταση στα τρίγωνα. Α!!! Ξέχασα να σου γράψω ότι στο σχήμα σου ισχύει και η εξής πολύ χρήσιμη  πρόταση  : $AH = 2 OM$ .Προσπάθησε να την αποδείξεις. Αν δυσκολευτείς  , μη διαστάσεις να το ζητήσεις.Θα στο αποδείξουν τα άλλα παιδιά ή και εγώ αν δε με ...προλάβει ο Αμβρόσιος ή κανένας άλλος  :) .Αν δεν έχεις  όμως υπομονή , μόνο τότε δες  την

 [hide] Η γωνία $ABH'$ είναι ορθή ( $\angle H' BC = \angle BCZ$ ...), οπότε η $AH'$ είναι διάμετρος . Αρα τα $A , O , H'$ είναι συνευθειακά και το ζητούμενο είναι πια απλό(δες το $\triangle AHH'$ ) ! [/hide]
 [/color]
                        ΜΠΑΜΠΗΣ

[color=blue]Tο βαρύκεντρο , το ορθόκεντρο , το περίκεντρο και το κέντρο του κύκλου του Euler βρίσκονται στην ίδια ευθεία , την ευθεία Euler του τριγώνου και HG=2GO.[/color]

Κατάφερα να βρω κι άλλη απόδειξη γι αυτό πως 2ΟΜ=ΑΗ :

[color=red]Απόδειξη[/color]

Φέρνω το αντιδιαμετρικό του Γ και ας είναι Γ' :
Γ'ΒΓ=Γ'ΑΓ=90 ως εγγεγραμμένες που βαίνουν σε ημικύκλιο

Γ'ΒΓ κάθετη στην ΒΓ και ΑΔ κάθετη στην ΒΓ άρα η ΑΔ είναι παράλληλη με την ΒΓ' . Ομοίως Γ'Α παράλληλη με την ΒΕ.Άρα το τετράπλευρο ΑΗΒΓ' είναι παραλληλόγραμμο , άρα ΑΗ=ΒΓ'(1).

Ο και Μ μέσα των ΓΓ' άρα 2ΟΜ=ΒΓ'  (2)

==>(1) , (2) 2ΟΜ=ΑΗ

[img]http://img130.imageshack.us/img130/470/2170rn4it6.png[/img]
Σωστά?

[quote="giannis"][color=blue]Tο βαρύκεντρο , το ορθόκεντρο , το περίκεντρο και το κέντρο του κύκλου του Euler βρίσκονται στην ίδια ευθεία , την ευθεία Euler του τριγώνου και HG=2GO.[/color][/quote]

[color=red]

 Αυτό είναι μια σπουδαία  πρόταση που πρέπει να γνωρίζει κάθε άνθρωπος που αγαπάει τη γεωμετρία !!! Δυστυχώς θα συναντήσεις και μαθηματικούς ακόμα που την αγνοούν και το χειρότερο : που δε θέλουν να τη μάθουν !!! [/color]

[u]    Μπάμπης[/u]

Επίσης ο Euler έχει και μια απόδειξη για την σχέση που συνδέει την απόσταση των κέντρων του περιγεγγεγραμμένο κύκλου και του εγγεγραμμένου του τριγώνου, μπορεί να την παραθέση κανείς όσο πιο απλά γίνεται , ευχαριστώ :P

Εννοείς φαντάζομαι τη σπουδαία σχέση:

  $OI = \sqrt{R\cdot (R-2r)}$

 Αυτή μάλλον πρέπει να τη μελετήσεις σε κάποιο βιβλίο , διότι θέλει σχήμα.Νομίζω ότι υπάρχουν δυο καλές αποδείξεις , τελείως όμως απλή δε νομίζω να υπάρχει καμία.  Αν κάποιος γνωρίζει κάτι πιο απλό , ας μας το γνωρίσει ! Εγώ έχω την απόδειξη σε βιβλία των  Ντάνη , Κανέλλου , Σκιαδά , Πάλλα κλπ. Ολες μοιάζουν.

 Μπάμπης

Ορίστε ένα σχήμα

[img]http://img358.imageshack.us/img358/3666/eje9.png[/img]

[color=blue]Εκφώνηση[/color]

Να αποδειχτεί πως η απόσταστη του περιγγεγραμμένου κύκλου τριγώνου από τον εγγεγραμμένο δίνεται από τον τύπο [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/0162816795f71cae154ed92f80f0a805.gif[/img]
όπου R η ακτίνα του περιγγεγραμμένου κύκλου και  r
η ακτίνα του εγγεγραμμένου.

[color=red]Απόδειξη[/color]

Έστω τρίγωνο ΑΒΓ με περίκεντρο Ο και έκκεντρο Ι και ο περιγγεγραμμένος και ο εγγεγραμμένος κύκλος του τριγώνου.Τότε :

Φέρνουμε την διχοτόμο ΑΕ και το αντιδιαμετρικό του Ε , έστω Η
και η ακτίνα ΙΖ=r του εγγεγραμμένου κύκλου.
γων ΒΑΕ = γων ΕΑΓ (ΑΕ διχοτόμος)
άρα τα τόξα ΒΕ και ΕΓ (τα μικρά) είναι ίσα άρα και οι χορδές τους είναι ίσες άρα ΒΕ=ΕΓ ,
και οπότε γων ΒΑΕ =γων ΕΗΓ (1)

γων ΕΒΓ = γων ΕΗΓ  (2)(εγγεγραμμένες που βαίνουν στο ίδιο τόξο)

άρα απο (1) και (2)   γων ΕΒΓ = γων ΕΗΓ =γων ΒΑΕ

γων ΒΙΕ =Α+Β/2 (σαν εξωτερική)

γων ΙΒΕ = Α+Β/2 ( με λίγες  πράξεις)

άρα το ΒΙΕ είναι ισοσκελές άρα ΒΕ = ΙΕ=ΕΓ

Τα τρίγωνα ΗΓΕ και ΑΖΙ  είναι όμοια άρα

ΙΑ * ΙΕ = ΙΖ * ΗΕ =2Rr (3)

αν ΟΙ = d τότε απο την δύναμη σημείου ως προς κύκλο και με βάση το (3) προκύπτει

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/0162816795f71cae154ed92f80f0a805.gif[/img]


Does anybody have tips on memorizing all of the Fridrich F2l algs?  There are way more than any other method I have used and it seems like there would be a lot of time wasted in between algs.  Thanx

You could try solving F2L intuitively.If it's not under construction now, and page about pairing up edges and corners on Dan Harris's site(http://www.cubestation.co.uk)should help.

I have actually figured out an F2L method for myself, but I think that Fridrich would be much faster  :wink:

Hi,

The F2L is the hardest part to master of the Fridrich solution, by a long way. I learnt the algorithms brute force way, and in the beginning as you say, a lot of time was wasted searching for pieces. But with lots of practice, and experience, you develop a feel for the algorithms, you know whats going to come up, you know how to deal with all cases instinctively. It's then easy to look back over the algorithms and see how they are intuitive.

But mainly, don't give up!

Dan :)


[quote="can_hang2007"][quote="Hatucdao"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that
\[ \frac {1}{\sqrt [3]{8b^2 \plus{} 11bc \plus{} 8c^2}} \plus{} \frac {1}{\sqrt [3]{8c^2 \plus{} 11ca \plus{} 8a^2}} \plus{} \frac {1}{\sqrt [3]{8a^2 \plus{} 11ab \plus{} 8b^2}} \ge \frac {4}{3}.
\]
I got a very very ugly solution for it :( So I wait for another nicer solution.
:)[/quote]
Thank you a lot, can_hang! I just want to come back for relaxing but it indeed takes a lot of time. Anyway, your problem is so nice and it is not easy to solve it. Here is the solution.[/quote]
Very nice your solution, Hatucdao. It is of course nicer than mine and that's what I am waiting for. :)  :lol:[/quote]
Since my inequality is solved now. It is no secret to present my proof. See it at: http://canhang2007.wordpress.com/2008/12/14/cube-inequality-vo-quoc-ba-can/

P/s: I have edited some errors in your typing. :)


topology
function
advanced fields
advanced fields unsolved

let $ f : X \to Y$ be a proper, continuous map between locally-compact hausdorff spaces. why is the pull-back $ \phi : C_{0}(Y) \to C_{0}(X)$ of $ f$ non-degenerate, i.e. that the span of $ \phi(C_{0}(Y)) C_{0}(X)$ is dense in $ C_{0}(X)$?

According to a hint, this should be easy considering one-point-compactifications. indeed, $ f$ proper means that we have a continuous extension $ f : \hat{X}\to \hat{Y}$, and this induces a hom. $ C(\hat{Y}) \to C(\hat{X})$ between the unitalizations. but it seems to contain no new information about $ \phi$.

ah, I got it:

$ f$ proper just implies that $ \phi$ is well-defined.

now let $ I$ be the closed span of $ \phi(C_{0}(Y)) C_{0}(X)$. then $ I$ is an ideal of $ C_{0}(X)$, so there's a closed subset $ A \subseteq X$ s.t. $ I$ consists of those functions in $ C_{0}(Y)$ that vanish on $ A$. it suffices to show that $ A$ is empty.

pick $ x \in A$, i.e. $ f(x)=0$ for all $ f \in I$. since $ C_{0}(X)$ has an approximate identity, $ \phi(C_{0}(Y))$ lies in $ I$. in particular $ g(f(x))=0$ for all $ g \in C_{0}(Y)$, which is a contradiction to the fact that $ C_{0}(Y)$ separates points.


function
inequalities unsolved
inequalities

[quote="friendlist"]Find the minimum value of the function:
$ f(x) \equal{} (32x^5 \minus{} 40x^3 \plus{} 10x \minus{} 1)^{2006} \plus{} (16x^3 \minus{} 12x \plus{} \sqrt {5} \minus{} 1)^{2008}$[/quote]
$ \min f\equal{}f\left(\sin6^{\circ}\right)\equal{}0.$

Ohh, it's very hard and nice. But i haven't any solution!
arqady: Please post full-solution for it!

Please lock this topic. It's running problem in MYM.

[hide="To friendlist and hotboy"]Tạm dừng việc post bài trên THTT nhé[/hide]


Given convex hexagon $ ABCDEF,M,N,P$ is midpoint of $ AD,BE,CF$ resp. Prove that $ M,N,P$ are colinear $ \Leftrightarrow S_{ABCDEF} \equal{} S_{ACE} \plus{} S_{BDF}$
[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=criterion&t=17902]similar  problem[/url]

Again, I seek for a cleaner proof to it, so I revive it.


i have a doubt................
How will you bring abt the cyclopropanation of alkenes???

do u mean a syclopropane kind molecule in the molecule....
then Simmon-Smith should do

There are several other methods, of course: diazoketones, phosphorus ylides, esthers and Grignard reagents.


Prove that the shortest altitude of a triangle connects the vertex to the longest side.

This is actually very simple if you think about it. If the aread of any given triangle is A, this number is constant for that triangle, so if the area equals the base times the height divided by two, which means the side length you use for the base times the altitude drawn to that base, you have A = (1/2)b*h, so to minimize h, the altitude, you must have the biggest b possible, which means the longest side. So the shortest altitude is drawn to the longest side.


show that if $ E_1$ and $ E_2$ are measurable, then $ m/(E_1\cup E_2) + m(E_1 \cap E_2)=mE_1 + mE_2$.

If $ E_1\cap E_2\equal{}\emptyset$. Then $ m(E_1\cap E_2)\equal{}0$ and $ m(E_1\cup E_2)\equal{}m(E_1)\plus{}m(E_2)$.
So suppose $ E_1\cap E_2\neq\emptyset$. Then $ E_1/(E_1\cap E_2)$, $ E_2/(E_1\cap E_2)$ and $ E_1\cap E_2$ is a disjoint partition of $ E_1\cup E_2$. This gives
$ m(E_1\cup E_2)\equal{}m(E_1/(E_1\cap E_2))\plus{}m(E_2/(E_1\cap E_2))\plus{}m(E_1\cap E_2)
\equal{}m(E_1)\minus{}m(E_1\cap E_2)\plus{}m(E_2)\minus{}m(E_1\cap E_2)\plus{}m(E_1\cap E_2)\equal{}m(E_1)\plus{}m(E_2)\minus{}m(E_1\cap E_2)$.


[quote="FOURRIER"]Can some one prove this lema:

If $ Un$ is an increasing and convergent sequence and has a limit $ l$

Then $ Un\leq l$[/quote]
When you say $ Un\leq l$, I assume you mean $ U_k\leq l$ for all $ k$.

If $ U_k\xrightarrow{k\to\infty}l$, this means that for any $ \epsilon>0$ there exists some $ K$ such that
\[ |U_k\minus{}l|<\epsilon\quad\mbox{ whenever }k>K\]
Moreover, we know that $ U_n$ is increasing, which means that $ U_k\geq U_j$ for all $ k\geq j$. Assume that there is some $ U_{k_0}>l$, and seek a contradiction. Well, that means that all $ U_k>l\implies U_k\minus{}l>0$ for $ k>k_0$. In particular, for $ k>\max{(K,k_0)}$, we have both $ U_k>l$ and $ |U_k\minus{}l|<\epsilon$ for all $ \epsilon>0$. But this is a contradiction - choose $ \epsilon\equal{}\frac{U_k\minus{}l}{2}$ and you cannot possibly satisfy both conditions.

Therefore our assumption that $ U_{k_0}>l$ must have been false, and so we are done.


limit
LaTeX
function
algebra unsolved
algebra

$k$ is a fixed positive integer. Let $a_n$ be the number of maps $f$ from the subsets of $\{1, 2, ... , n\}$ to $\{1, 2, ... , k\}$ such that for all subsets $A, B$ of $\{1, 2, ... , n\}$ we have $f(A \cap B) = \min (f(A), f(B))$. Find $\lim_{n \to \infty} \sqrt[n]{a_n}$.

I did it .I remember that $a_{n}$<= n so lim = 1 !
 :oops: (sorry for my latex)

[quote="Megus"]If you have a solution, post it - just a note about possesing one is useless[/quote]

This is a free Internet forum, and there is nothing wrong in postings that do not contain a solution.

Anyway, it is a beautiful problem and after producing two different solutions I still feel that there is much more that can be said about it.

Solution 1.

$f$ is determined by its values on sets of size $(n-1)$ and $n$.   The only constraint on the possible values of $f$ on these sets, is that the set of size $n$ has the largest value of $f$.  It is easy to see that the number of possible values of $f$ on these sets is $k^n + (k-1)^n + \dots + 1^n$  (assign the value of $f$ on the largest set and then independently choose values at most this large, for the other sets).    The formula for $f$ extends to intersections of any number of subsets, and there is a unique way to write any subset of size $\leq (n-1)$ as an intersection of sets of size $(n-1)$. Using the intersection formula minimum gives a unique and consistent extension of $f$ to all sets; the number of maps is equal to the number of assignments of $f$ to the sets of size $\geq (n-1)$.   The limit is therefore $k$.

Solution 2.  (easier description of $f$, but harder to enumerate)
Any $f$ has the following form:
There is a chain of sets $\emptyset = X_0 \subset X_1 \subset \dots X_p$,
where $p < k$,  $f(X_i) < f(X_{i+1})$ for all $i$, and for any $A$,  $f(A) = f(X_t)$ where $X_t$ is the largest of the $X$ sets contained in $A$.
Clearly any $f$ of this type satisfies the conditions.
Given an $f$ satisfying the conditions, one can construct a chain of $X_i$ by letting $X_0 = \emptyset$, and let $X_{i+1}$ be any set of smallest possible size (it will be unique, but this is not immediately necessary for the construction) with $f(X_{i+1}) > f(X_i)$, if such a set exists (otherwise terminate the chain).  One can show that the function associated to this chain equals $f$.

It is interesting to try to extract the enumeration directly from this description, and to show that it is equivalent to Solution 1.


Euler
geometry
circumcircle
parallelogram
geometric transformation
reflection
symmetry

Prove that the Euler circle of $ \Delta ABC,AHO,BHO,CHO$ concurrent at the Jerabek point of $ \Delta ABC$,where $ H$ and $ O$ is the orthocenter and the circumcenter of $ \Delta ABC$,respectively.
(the Jerabek point of triangle ABC is the antipole of $ X$ wrt the Euler circle $ (E)$ of triangle ABC,where $ X$ is the Anti-Steiner point of the Euler line wrt $ (E)$)

PS:I've edited it.Thanks nsato ^^

Denote $ A',B',C'$ the midpoints of $ BC, CA, AB, X$ the Anti-Steiner of triangle $ ABC$ wrt $ OH, E$ the center of Nine-point circle of triangle $ ABC, J, Q$ the midpoints of $ AO, AH$.
Since $ C'QB'O$ is a parallelogram we get $ S\equal{}QO\cap B'C'$ is the midpoint of $ B'C'$.
$ \Rightarrow S,E,J$ are collinear.
But $ S$ is the midpoint of $ EJ$ then $ (EJQ)$ lies on $ B'C'$.
Denote $ P$ the projection of $ A$ onto $ OH, H_a$ the projection of $ A$ onto $ BC, X'$ the reflection of $ P$ across the line $ B'C', A''$ the reflection of $ O$ across the line $ B'C'$.
We have $ H_a,E,A''$ are collinear.
$ \Rightarrow 90^o\equal{}\angle APO\equal{}\angle H_aX'A''$
$ \Rightarrow X'$ lies on $ (E)$ or $ X'\equiv X$.
Therefore $ X$ is the reflection of $ P$ across the line $ B'C'$. But $ P\in (EJQ)$ so $ X \in (EJQ)$.
Similarly we are done.

9-point circle $ (N_1)$ of $ \triangle AHO$ goes through the 9-point center $ N$ of $ \triangle ABC,$ i.e. the midpoint $ F$ of $ AH$ and the orthogonal projection $ A'$ of $ A$ on $ OH.$ If $ M,L$ are the midpoints of $ AC,AB$ and $ D,E$ the projections of $ B,C$ on $ AC,AB,$ then $ J\equiv ML \cap ED$ is the A-vertex of the side triangle determined by the pedal triangles of $O,H.$ Since they are isogonal conjugates WRT $ \triangle ABC,$ then $A,J,A'$ are collinear. Let  $ P$ be the second intersection of $ (N_1)$ and $ (N)$ and $ Q_1 \equiv LD \cap EM.$ Note that $ AJ$ is the polar of $ Q_1$ WRT $ (N),$ hence $ Q_1A' \cdot Q_1N \equal{} Q_1D \cdot Q_1L$ $ \Longrightarrow$ $ Q_1$ lies on the radical axis of $ (N_1),(N)$ $ \Longrightarrow$ $ Q_1 \in FP.$ From the harmonic pencil $ A'(F,A,P,Q_1)$ and $ AA'\perp OH,$ we have that $ A'A$ bisects $ \angle FA'P$ $\Longrightarrow$ $ \angle PA'A \equal{} \angle FA'A \equal{} \angle FAA'$ $ \Longrightarrow$ $ PA' \parallel AH.$ From cyclic quadrilateral $AA'OS,$ where $ S$ is the second intersection of $ AH$ with $ (N_1),$ we get that $ \angle ASO$ is right $ \Longrightarrow$ $ OS \parallel ML.$ On the other hand, since $ F$ is the orthocenter of $ \triangle ALM$ and $ O$ is the antipode of $ A$ in the circumcircle $ \odot(ALM),$ it follows that $ ML$ bisects $ FO$ $ \Longrightarrow$ $ ML$ is the symmetry axis of the isosceles trapezoid $ FPA'S.$ Hence, $ (N_1),( N)$ and the reflection of $ OH$ onto $ ML$ concur. By similar reasoning the result follows.


If f(0)=0 και f'(0)=2 and
(χ+1)f " (χ)+4xf ' (x)+2f(x)=0
Could you find the function f(χ)?

We can solve using series:
$ f(x) \equal{} 2x\minus{}\frac{2}{3}x^3\minus{}\frac{1}{3}x^4\plus{}\frac{4}{15}x^5\plus{}O(x^6)$

Dear god...

Mathematica spits out the following:
$ f(x) \equal{} \frac{512}{945} e^{\minus{}4 (1\plus{}x)} \sqrt{\pi } (1\plus{}x)^5 * \mathbf{G(x)}$

Where

$ \mathbf{G(x)} \equal{}  e^2 (\minus{}109 \text{BesselI}[0, 2] \plus{} 152 \text{BesselI}[1, 2]) \text{HypergeometricU}[1/2, 6, 4 \plus{} 4 x] \plus{} 128 \text{HypergeometricU}[1/2, 6, 4] \text{LaguerreL}[\minus{}(1/2), 5, 4 \plus{} 4 x]$


abstract algebra
superior algebra
superior algebra unsolved

Let $ A$ be a ring.  The following is a commutative diagram of $ A \minus{}$module homomorphisms, and the rows are exact.

[img]http://i719.photobucket.com/albums/ww191/xianghu21/2-1.png[/img]

Suppose that $ f_1$ is surjective and that $ f_2$ is an isomorphism.  Prove that $ f_3$ is an isomorphism.

Attempt
Onto- $ f_3 \equal{} t \circ f_2$ and both $ f_2$ and $ t$ are both onto so $ t \circ f_2$ is onto.
1-1- This one doesn't seem as trivial.  I am a bit confused on showing this.

Thanks in advance for any hints.

The onto part isn't right: we have $ f_3 \circ h \equal{} t \circ f_2$ not $ f_3 \equal{} t\circ f_2$. But you're right that it's easier than 1-1.

Hint for 1-1: we wish to show $ Ker(f_3) \equal{} 0$. Suppose $ f_3(c) \equal{} 0$ for $ c \in C$. Pick $ b \in B$ such that $ h(b) \equal{} c$. Then $ t f_2(b) \equal{} f_3 h(b) \equal{} f_3(c) \equal{} 0$ so $ f_2(b) \in Ker(t) \equal{} Im(s)$. Hence we can pick $ a' \in A'$ such that $ s(a') \equal{} f_2(b)$. Now do something with $ f_1$ ...


To proof,that there are [b]no[/b] natural numbers [b]a[/b] and [b]b[/b] ,such that the number:
a) $ (36a\plus{}b)(36b\plus{}a)\equal{}2^n$
b) $ (46a\plus{}b)(46b\plus{}a)\equal{}2^n$

$ 36a\plus{}b\equal{}2^x$    $ 36b\plus{}a\equal{}2^y$
$ b\equal{}2^x\minus{}36a$   $ 2|b 2|a$ $ b\equal{}2b1  a\equal{}2a1$
$ a\equal{}2^z.q  b\equal{}2^z.p$  $ (p,2)\equal{}(q,2)\equal{}1$
$ 2|q 2|p$ contradiction!  a=0,b=0 we get any solution

Let $ a \equal{} 2^{a_1}k$, $ b \equal{} 2^{b_1}l$, $ k, l$ - odd. Without loss of generality, we can assume that $ a_1\geq b_1$.

So, $ (9*2^{2 \plus{} a_1}k \plus{} 2^{b_1}l)*(9*2^{2 \plus{} b_1}l \plus{} 2^{a_1}k) \equal{} 2^n$.
$ (9*2^{2 \plus{} a_1}k \plus{} 2^{b_1}l)$ is a power of 2.
$ (9*2^{2 \plus{} a_1 \minus{} b_1}k \plus{} l)$ is a power of 2, too.
But $ (9*2^{2 \plus{} a_1 \minus{} b_1}k \plus{} l)$ is odd ($ a_1\geq b_1$, $ l$ is odd), and not equal to 1. Contradiction.

The second is the same as first.


geometry
geometric transformation
rotation
linear algebra
linear algebra unsolved

I got this question from the book "Linear Algebra and Its Applications". Hope anybody can help :(.

Any two nonzero $ 1 \times 1$ have the same four fundamental subspaces.  For a slightly less trivial counterexample, any two $ 2 \times 2$ rotation matrices have trivial nullspace, trivial transpose nullspace, and row- and column-space $ \mathbb{R}^2$.

Edit:  $ n \times n$ invertible matrices all have trivial nullspace, trivial transpose nullspace, and row- and column-space $ \mathbb{R}^n$ by the rank-nullity theorem.


AMC USA(J)MO USAMO Putnam USAMTS AIME calculus

two questions

1) does proof by intimidation seem to work? this a part joke-part serious question. if I have a ton of time left over should i write some random 'proof' in the hope to receive partial credit?

2)How rigorous are solutions (that receive full credit)? are they comparable to the rigor of USAMTS solutions?(although USAMTS is easier)?

3) How does one prepare for the Putnam? although I'm too young for it I might take it unofficially...

Also, i just got the book "the art and craft of problem solving", by Paul Zeitz. What chapters, if any, are suggested reading for USAMO/AIME?

For #1, i would say no.  the graders are quite experienced, although thanks for giving me that idea.  i'll probably use it at some local competitions :P

[quote="mathking123"]1) does proof by intimidation seem to work?[/quote]
With respect to the Putnam, let me put it this way: The Putnam organization each year releases a question-by-question scoring breakdown for the top 200 students. This breakdown distinguishes between zeros and no answer sheet being submitted. Last year, the top 206 competitors collectively had 494 grades of zero. You've got to be pretty amazing to be at this level, and even these elite competitors averaged nearly two and a half zero-credit questions on the 12-question exam.

That's the long answer to your question. The short answer: no, it doesn't work.

ok thanks everybody! any more opinions/facts would be appreciated!

[quote="worthawholebean"][quote="mathking123"]Also, i just got the book "the art and craft of problem solving", by Paul Zeitz. What chapters, if any, are suggested reading for USAMO/AIME?[/quote]This is probably my favorite math book ever. Seriously, the book as a whole forms a really coherent unit and you'd be really missing out to skip anything.[/quote]

It is a great book, but for AIME and USAMO, i don't think the chapter on calculus is very helpful, but that is just me, i might be wrong...

[quote="cognos599"]It is a great book, but for AIME and USAMO, i don't think the chapter on calculus is very helpful, but that is just me, i might be wrong...[/quote]The calculus chapter is really cool, though, so I think you should read it even if it's not helpful.

[quote="AIME15"]There's no calculus on AMC, AIME, usamo i think.[/quote]

You are correct.

[quote="AIME15"]There's no calculus on AMC, AIME, usamo i think.[/quote]

Not entirely.

I have seen several USAMO problems where you can either use derivatives to prove that functions are increasing or convex or use Lagrange Multipliers to bash an inequality.

[quote="AIME15"]Of course, you CAN use calculus but there are no problems that REQUIRE calculus.[/quote]

This is correct on AMC-AIME-USAMO. However, there are problems that require calculus on Putnam and HMMT.


[color=cyan]When people start sending in problems, of course.[/color]