calculus
integration
trigonometry
derivative
calculus computations

Verification of the above integration by parts formula:

Define $u(x) = \int_{0}^{x}{e^{at}}dt$

1) Integration by parts:
$\int_{0}^\pi{\sin }(x)^{n}du+\int_{0}^\pi{u(x)d(\sin }(x)^{n}) = \sin (\pi )^{n}u(\pi )-\sin (0)^{n}u(0) = 0$

2) Stieltjes formula for products
$\int_{0}^\pi{u(x)d(\sin }(x)^{n}) = \int_{0}^\pi{nu(x)\sin }(x)^{n-1}\cos (x)dx = \frac{n}{a}\int_{0}^\pi{\sin }(x)^{n-1}\cos (x)du$

3) $\int_{0}^\pi{\sin }(x)^{n-1}\cos (x)du+\int_{0}^\pi u (x)d(\sin (x)^{n-1}\cos (x)) = \int_{0}^\pi{\sin }(x)^{n-1}\cos (x)du-\int_{0}^\pi{u(x)\sin (x)^{n}+(n-1)\sin (x)^{n-2}\cos (x)^{2})}= 0$

4) $-\int_{0}^\pi{u(x)\sin (x)^{n}+(n-1)\sin (x)^{n-2}\cos (x)^{2})}= \frac{1}{a}\int_{0}^\pi{(n-1)\cos (x)^{2}\sin (x)^{n-2}-\sin (x)^{n}du}$

5) $\frac{1}{a}\int_{0}^\pi{(1-n)(1-\sin (x)^{2})\sin (x)^{n-2}-\sin (x)^{n}du}= \frac{1}{a}\left[{(1-n)\int_{0}^\pi{\sin (x)^{n-2}du+n\int_{0}^\pi{\sin (x)^{n}du}}}\right]$

6) $\int_{0}^\pi{\sin }(x)^{n}du =-\frac{n}{a}\int_{0}^\pi{\sin }(x)^{n-1}\cos (x)du =-\frac{n}{{a^{2}}}\left[{(1-n)\int_{0}^\pi{\sin (x)^{n-2}du+n\int_{0}^\pi{\sin (x)^{n}du}}}\right]$

7) $\left({\frac{{a^{2}+n^{2}}}{{a^{2}}}}\right)\int_{0}^\pi{\sin }(x)^{n}du = \frac{{n(n-1)}}{{a^{2}}}\int_{0}^\pi{\sin (x)^{n-2}du}$

8) $\int_{0}^\pi{\sin }(x)^{n}du = \frac{{n(n-1)}}{{a^{2}+n^{2}}}\int_{0}^\pi{\sin (x)^{n-2}du}$

I am too tired/hungry/lazy to give step by step explanations of how sos440 got his recursive formula but this is the general sequence of steps, if anyone would like to elaborate. Otherwise I am done for the night. All done using Stieltjes integrals, mind you. Therefore "integration by parts" may not mean the usual calculus formula of integrating the derivative of a product.

That's exactly same as my solution. :lol:


Let $ P\in \mathbb{R}[X]$ be a polynomial such that for any rational $ x$, $ \ds P(x)$ is also rational, and for any irrational $ y$, $ P(y)$ is also irrational.

Prove that $ P$ has degree $ 1$.

It can be shown by using idea similar to
http://www.mathlinks.ro/viewtopic.php?p=1310073#1310073.


Given a,b,c>0 and $ [1/(1\plus{}a)]\plus{}[1/(1\plus{}b)]\plus{}[1/(1\plus{}c)] >\equal{} 2$

Prove that abc<=8

This is very similar to a problem I posted in intermediate under "Proof"

$ \frac{1}{1\plus{}a} \plus{} \frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c} \ge 2 \Rightarrow \frac{1}{1\plus{}c} \ge \left(1\minus{}\frac{1}{1\plus{}a} \right) \plus{} \left(1\minus{}\frac{1}{1\plus{}b} \right)$
$ \equal{} \frac{a}{1\plus{}a} \plus{} \frac{b}{1\plus{}b} \ge 2 \sqrt{\frac{ab}{(1\plus{}a)(1\plus{}b)}}$

Likewise
$ \frac{1}{1\plus{}b} \ge  2 \sqrt{\frac{ac}{(1\plus{}a)(1\plus{}c)}}$ and $ \frac{1}{1\plus{}a} \ge  2 \sqrt{\frac{bc}{(1\plus{}b)(1\plus{}c)}}$

Hence
$ \frac{1}{1\plus{}a} \ \frac{1}{1\plus{}b} \ \frac{1}{1\plus{}c} \ge \frac{8abc}{(1\plus{}a)(1\plus{}b)(1\plus{}c)}$

so that $ abc \le \frac{1}{8}$

The following stronger inequality is true too.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ \frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c}\geq2.$ Prove that
\[ (ab\plus{}ac\plus{}bc)^3abc\leq\frac{27}{512}\]


[quote="perfect628"]a man is 2/5 the way through a tunnel, when he hears a train approaching. No matter which end of the tunnel he runs toward, he will reach that side at the exact same time as the train. The train is going at 60 MPH. What is the man's speed?[/quote]
Poor guy, doomed to death no matter where he runs...
[hide]Since we don't have more information it is safe to assume that the length of the tunnel doesn't matter. Lets say that it is 60 miles long. So the guy goes 36 miles in the same time the train goes 60 miles, which is one hour. So it takes him 1 hour to go 36 miles. So our answer is 36 mph?[/hide]
I might be wrong...

Oh no, no, no, no.... :o
[hide]
X------|--M---|

The train starts out of the tunnel at point x.  The guy goes to the front of the tunnel, 2/5 of the total length.  The train gets there the exact same time as he does.

Lets say he runs the other way.  He runs 3/5 of the tunnel, 1/5 more than before.  In that same time, the train goes to the front of the tunnel [i]plus[/i] the distance to the other end.
At the time when the man has made it 4/5 of the way through the tunnel, we know from before, when he ran 2/5 of the tunnel, the train would be at the front.  This means that in the time the man travels his last fifth of the tunnel, the train will travel the whole tunnel.

Thus the train has gone 5 times the distance of the man in the same amount of time. ie: the train in going 5 times faster than the man.  Since the train had a speed of 60mph, 1/5 of that is [b]12mph[/b] for the man.  :wink: [/hide]

Correct!That's the good way to do it, too.

Hadn't someone posted this in previous Questions?
[hide]The train travels at 60mph. We don't really have to define the distance of the tunnel. It is known as one. A key thing is, we don't know the direction of the train so we have to superimpose it both ways
[b]If the train travels from the left and the man runs towards the left as well  [/b], then it would take 3/5. The train has to travel 1. So thus, it would take 1/60 hours for the train and the man to reach the other side.
[b]S=VT[/b] :arrow: Thus the speed the man takes would be 3/5*60=36.
I think this is correct but the other speed would be 24. (other direction)
 :wink: [/hide]


inequalities
geometry
trigonometry
inradius
circumcircle
inequalities unsolved

Suppose that the incircle of $\triangle ABC$ is tangent to the sides $BC, CA, AB,$ at $D, E, F$, respectively. Prove that $EF^2+FD^2+DE^2\le \frac{s^2}{3}$, where $s$ is the semiperimeter of $\triangle ABC$
[hide]Now this must be true :)[/hide]

Let $AE=AF=x$, $BD=BF=y$, $CD=CE=z$. Then, by Cosine Rule,
$EF^2=x^2+x^2-2(x)(x)\cos A$
$=2x^2(1-\cos A)$
$=2x^2(1-\frac{(x+y)^2+(z+x)^2-(y+z)^2}{2(x+y)(x+z)})$
$=\frac{4x^2yz}{(x+y)(x+z)}$
Now we need to prove
$\sum_{cyc}\frac{12x^2yz}{(x+y)(x+z)}\leq (x+y+z)^2$
$\Leftrightarrow \sum_{cyc}12x^2yz(y+z)\leq (x+y)(y+z)(z+x)(x+y+z)^2$
$\Leftrightarrow \sum_{sym}x^4y+3\sum_{sym}x^3y^2+3\sum_{sym}x^3yz\geq 7\sum_{sym}x^2y^2z$
which is true by Muirhead Inequality.

Can somebody prove Muirhead in details PLEASE

[quote="Beat"]Suppose that the incircle of $\triangle ABC$ is tangent to the sides $BC, CA, AB,$ at $D, E, F$, respectively. Prove that $EF^2+FD^2+DE^2\le \frac{s^2}{3}$, where $s$ is the semiperimeter of $\triangle ABC$[/quote]

Let r be the inradius of triangle ABC. Then, since the incircle of triangle ABC is the circumcircle of triangle DEF, the inradius r of triangle ABC must be the circumradius of triangle DEF, so that, applying the inequality 3 from http://www.mathlinks.ro/Forum/viewtopic.php?t=17684 to the triangle DEF, we get $EF^2+FD^2+DE^2\leq 9r^2$. On the other hand, the inequality $3\sqrt3r\leq s$ (well-known; for a proof, see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=191721#p191721]http://www.mathlinks.ro/Forum/viewtopic.php?t=19316 post #4[/url], somewhere during the proof of Theorem 3) yields $9r^2=\frac{\left(3\sqrt3r\right)^2}{3}\leq\frac{s^2}{3}$. Thus we are done.

  Darij

Here is my solution:
 Let I be the center of the incircle of ABC, and r the lenght of this circle's radio. Because D, E and F are tangency points, we get that $AF=AE=x$, $BF=BD=y$ and $EC=CD=z$.  Also, $FI=EI=DI=r$.

 As $ID\bot BC$, $IE\bot AC$ and $IF\bot AB$, we know that $\angle FID=180-B$, $\angle FIE=180-A$ and $\angle EID=180-C$.

 Then, by cosine law, we get:
 $FD^2=r^2+r^2-2r^2(180-cosB)\Rightarrow FD^2=2r^2+2r^2cosB$. Also, $FE^2=2r^2+2r^2cosA$ and $ED^2=2r^2+2r^2cosC$.

 So, to prove our inequality, we must prove that

$6r^2+2r^2cosA+2r^2cosC+2r^2cosC\leq s^2/3$
$\Leftrightarrow18r^2+6r^2(cosA+cosB+cosC)\leq s^2$ (1)

 Now, notice that $(sr)^2 = s(s-a)(s-b)(s-c)$, then $r^2=(s-a)(s-b)(s-c)/s$ But $s=x+y+z$, so we get that $r^2=xyz/s$. By replacing this result in (1), we get

 $18xyz+6xyz(cosA+cosB+cosC)\leq s^3=(x+y+z)^3$
 $\Leftrightarrow12xyz+6xyz(cosA+cosB+cosC)\leq x^3+y^3+z^3+3x^2y+3xy^2+3x^2z+3xz^2+3y^2z+3yz^2$. (2)

By Schur, we have that for the angles A, B, and C of ABC, $cosA+cosB+cosC\leq 3/2\Rightarrow6xyz(cosA+cosB+cosC)\leq9xyz$  (3)

 Also, by AM-GM, we have

 $x^3+y^3+z^3+3x^2y+3x^2z+3y^2z \geq 12xyz$ (4) and
 $3xy^2+3xz^2+3yz^2 \geq9xy \geq 6xyz(cosA+cosB+cosC)$ (5)

 From (4) and (5) it follows that (2) is true and we are done.
I hope it's ok :)


E = mc^2 = hf
replacing f with c/w (w = lambda = wavelength...by the way, how do use the latex font or whatever?), we get mc^2 = hc/w
or mc = h/w
or w = h / mc
replacing c with v, we have the de broglie equation/hypothesis, w = h / mv
so, is there a conceptual flaw with this derivation?
if so, im guessing it lies in equation mc^2 with hf or replacing c with v

thanks,
new aops member

Here's your derivation:

http://en.wikipedia.org/wiki/De_Broglie_hypothesis

Here's your [url=http://www.artofproblemsolving.com/Wiki/index.php/Latex]LaTex Help[/url].

\begin{align*} E = mc^{2} = hf & \Leftrightarrow \\
mc^{2} = \frac {hc}{\lambda} & \Leftrightarrow \\
mc = \frac {h}{\lambda} & \Leftrightarrow \\
\lambda = \frac {h}{mc} \end{align*}

[hide="Code"][code]\begin{align*} E = mc^{2} = hf & \Leftrightarrow \\
mc^{2} = \frac {hc}{\lambda} & \Leftrightarrow \\
mc = \frac {h}{\lambda} & \Leftrightarrow \\
\lambda = \frac {h}{mc} \end{align*}[/code][/hide]


[url=http://img515.imageshack.us/my.php?image=limithw7.png][img=http://img515.imageshack.us/img515/9440/limithw7.th.png][/url]

What are the limits??

[quote="JRav"]The limits as $ x$ approaches which value?[/quote]

it doesn't state it that's the problem :S ....... but i assume it's positive and negative infinity

[quote="JRav"]Well you can't really assume.  It appears to have different limits depending on what $ x$ approaches.[/quote]

hmmm.... yah i'm pretty new to calculus so i have really no idea what's going on...... but would u happen to know how to figure it out....
cause i'm pretty sure there is more than one limit caus the question is "what LIMITS did you find? use limit notation"

Sorry to do this to you but I can't give you a definitive answer.  For some of them it approaches -1, for others it approaches 0, and for some it varies based on what $ x$ approaches.

what if it was positive and negative infinite then??


Prove that if $ \alpha_i\in\mathbb{R}$ such that $ \alpha_i > \alpha_{i \plus{} 1} > 0,\, i \equal{} 0,...,n \minus{} 1,$ and $ z\in\mathbb{C}$ satisfies $ \alpha_nz^n \plus{} \alpha_{n \minus{} 1}z^{n \minus{} 1} \plus{} ... \plus{} \alpha_1 z \plus{} \alpha_0 \equal{} 0$, then $ |z| > 1.$

At least this is correct for $ n\equal{}1.$ Others?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Hello,

let $G$ be a group and $N$ a normal subgroup, such that :

-all nontrivial elements of $N$ are conjugated (by which I mean that ALL inner automorphisms defined by elements of $G$ together work transitive on $N\backslash\{e\}$
- all cosets of $N$ apart from $N$ itself is are exactly one orbit under conjugation

Prove that $N$ is elementary abelian.  What worries me is that I don't know whether or not $G$ is finite.  If it is assumed, it should be no problem.  But what if it is not? :maybe:

[quote="Peter VDD"]Is it necessary that $G$ is finite, is it not enough that $N$ is finite for the reasoning you mentioned?[/quote]
Yeah, I think you are correct. The problem is only difficult is $N$ is not finite as well.

You really need $N$ to be finite (or atleast some other conditions). The main problem is the following: There is an infinite group $G$ which has exactly two conjugacy classes namely $\{e\}$ and $G-\{e\}.$ You can then consider the group $G\times G$ with $N=G\times \{e\}.$

The construction of such a group $G$ can be carried out by using HNN extensions (Higman, Neumann, Neumann).


For positive reals $x,y,z$ with $xyz=1$ prove

\[ \frac{x^4}{y^4+y+z}+\frac{y^4}{z^4+z+x}+\frac{z^4}{x^4+x+y} \geq 1 \]

Using CS inequality we get \[ \left(\frac{x^4}{y^4+y+z}+\frac{y^4}{z^4+z+x}+\frac{z^4}{x^4+x+y}\right)(x^4+y^4+z^4+2(x+y+z))\geq (x^2+y^2+z^2)^2 \] so we only need to prove $(x^2+y^2+z^2)^2\geq x^4+y^4+z^4+2(x+y+z)xyz$, and this true by Muirhead inequality ($[2,2,0]\geq [2,1,1]$).


limit
real analysis
real analysis unsolved

Let $ (a_n)_{n \in \mathbb{N}^*}$ be a sequence of real positive numbers such that $ a_n>a_0,n\in \mathbb{N}$. Prove that $ \displaystyle\lim_{n\to\infty}\displaystyle\sum_{k\equal{}0}^{n}(\frac{a_k}{a_{n\minus{}k}})^k\equal{}\infty$.

arithmetic mean is greater harmonic mean

$ u(n)\equal{}\sum_{k\equal{}0}^{n}(\frac{a(k)}{a(n\minus{}k)})^k$

$ \frac{u(n)}{n}\geq \frac{n}{u(n)}$

$ u(n) \geq n$


AMC AIME USA(J)MO USAMO HMMT geometry MATHCOUNTS

How was it? From what I've heard...

Me - 10
Arnav - 10 (11?)
Jeremy - 8
Yasha - AIME didn't come in the mail... lol

That's all I know. Answers are on the AMC site and the AMC forum. Whee, now I need to actually convince myself to start practicing for the USAMO again.

I'm thinking a floor of 6, because of the 150/6 that apparently exists, and probably around 200 for the index... a little over that. I bet AIME IIers will have an enormous advantage.

I believe they use a different index for the AIME I and AIME II.  Yeah... my score was 4 points lower than last year's, and after HMMT I thought I had improved and whatnot...

Anyways, how many of you folks think you might have made USAMO?  :lol:

Predicted floor: 6
Predicted index: 195-205

w00t, I'm so excited!  I got 9!!!   :jump:  :jump:  :starwars:

I didn't even bother trying any of 12-15 and I only missed one of the ones I answered (4).

I had no clue how to do 9, and most of the others I thought were easy as in straight forward in the casework or whatever.

Dude yeah NCAA is much more math-related than 2007 AIME I.

[quote="MysticTerminator"]arrgh. why has this forum been perverted to reporting our bad AIME scores

it used to be a bastion of sparkling debate about NCAA[/quote]

Which questions did you get correct/miss?

HA I'M TAKING THE AIME II

how are scores so bad!?

Perhaps Arnav DOES deserve to be banned.

@nottrig: ...because it was hard? Or rather, not hard, but messy. Look at the problems; they're disgusting!

I thought a lot of the problems were quite nice.  The computations weren't very difficult, with exceptions to on 11, 12 and 15.

I thought that 4, 8, and 11 were very nice problems (especially 8), though 11 could be solved by pattern guessing.

The problems weren't that much more difficult, it's just that the problems had more places to mess up in.

In general, I thought this was a good test, though they might have put a geometry question in the first part & swapped one of the later geometry ones for something else.

yeah 12 is disgusting :wacko:

 :D

apparently everyone's like "OH NO s ! I MADE STUPID MISTAKE  s !" (homage to Arnav)

[quote="b-flat"]I thought that 4, 8, and 11 were very nice problems (especially 8), though 11 could be solved by pattern guessing.[/quote]

#4 was nothing but a trick question. People got 14s missing #4. #8 was pretty easy without much calculation, and #11 I've seen a million times; besides, it was again hefty on calculation. The only problem I liked was #9. If you haven't solved it (b-flat), you should really try to. Also, the early problems took too much work (I'm not saying they're too hard).

I really liked number 4 on the 2007 AIME I and appreciated DPatrick's foolproof solution on the AIME math jam.  It exposed some muddled thinking on my part (and probably some others) about an old mathcounts problem that was similar.  I suspect that was the motivation behind the AIME problem.

1999 State Sprint #10.

A solar system contains four planets and a sun as shown (A,B,C, and D are shown increasingly distant from the sun but not in a straight line.)  Planet A completes an orbit around the sun in 20 months; planet B completes an orbit in 36 months; planet C completes an orbit in 40 months; and planet D completes an orbit in 48 months.  If the four planets lie along a straight line in May of the year 2000, what is the next year that they will again lie along the same line?  Assume that the orbits are circular and that a year has exactly 12 months.

The published solution in the "All Time Greatest MathCounts Problems" appears below:

"Each planet will again lie along the same line when it completes a full orbit.  However, note that it will also lie along the same line when it makes a half-orbit, that is, when it is on the other side of the sun.  Consequently the number of months that pass until all planets lie on the same line again can be found by computing the least common multiple of the half-orbits, which are 10, 18, 20, and 24 months. (computation shown)....360 months or 30 years."

Comment:  This solution assumes that the sun is on the same line as the planets in May 2000.  It seems possible that the planets could be on a line that doesn't include the sun in May of 2000.  If the sun is not in line with the planets in May 2000, does this problem necessarily have a unique solution?  If not, what are all other possible solutions?

Checking the MathCounts "errata," no errors in this problem have been reported.

Compare the above with the 2007 AIME I Problem 4.

Three planets revolve about a star in coplanar circular orbits with the star at the center.  All planets revolve in the same direction, each at a constant speed and the periods of their orbits are 60, 84, and 140 years.  The positions of the star and all three planets are currently collinear.  They will next be collinear after n years.  Find n.

In the AIME problem, the sun must be collinear with the planets.     However, it asks when the next time the planets and sun will be collinear, not necessarily when they will again be on the original line.


function induction calculus derivative inequalities logarithms

[quote="white_horse_king88"]For $n = 2, 3, 4\ldots$, prove $n! < \left(\frac{n+1}{2}\right)^n$.[/quote]

Take the natural log of both sides.  :)

You need to prove $\left( \frac{n+1}{n} \right)^n$ is an increasing function if you want to do it your way.

Edit:  No, no, you can just prove $(n+1)^n \ge 2n^n$.  You might be able to do an induction on this.

[quote="white_horse_king88"]For $n = 2, 3, 4\ldots$, prove $n! < \left(\frac{n+1}{2}\right)^n$.

I'm trying an inductive proof for this, and applying the inductive hypothesis after showing the base case, I have:

$(n+1)n! < \left(\frac{n+1}{2}\right)^n(n+1) = \frac{(n+1)^{n+1}}{2^n}$

So it suffices to prove $\frac{(n+2)^{n+1}}{(n+1)^{n+1}} \ge 2$, which if we make more strict, $\left(\frac{n+1}{n}\right)^n \ge 2$.

For this, The base case for positive integers checks, but can I just say that this value approaches $e$? Will this work in a proof?[/quote]

Well, it's enough to show that $f(n)=\left(1+\frac 1n\right)^n$ is increasing in $n$ (a simple derivative check establishes this) and $f(1)=2$.  I believe there is a way with AM-GM, but I've forgotten it.  The key point is that it approaches $e$, but it also monotonically increases to $e$.

How about $\left(1+\frac{1}{n}\right)^n = 1+n \cdot \frac{1}{n} + \cdots \ge 2$ by Binomial Theorem?

Well, Bernoulli's inequality, but point taken.

[b]Problem:[/b]
$n! < \left(\frac{n+1}{2}\right)^n$
[hide="Solution"]By AM-GM we have:
$\frac{1+2+\cdots + n}{n}>\sqrt[n]{1\cdot2\cdot3\cdots \cdot n}$
$\frac{\frac{n(n+1)}{2}}{n}>\sqrt[n]{n!}$
$\left(\frac{n+1}{2}\right)^n> n!$
 :)
[/hide]

[quote="white_horse_king88"]...For $n = 2, 3, 4\ldots$, prove $n! < \left(\frac{n+1}{2}\right)^n$....
[/quote]
[A direct proof] The function $f : (0,\infty)\to {\mathbb R} \; ,\; f(x)=\ln{x} ,$ is (strictly) concave  on its domain. Therefore
\[ (*)\; \; \; \; \; f\left(\sum\limits_{k=1}^{n}w_k x_k \right)> \sum\limits_{k=1}^{n}w_k\cdot \ln{x_k}  \]
for any system $\{x_1,x_2,...,x_n\} \subset (0,\infty)^n$ and any positive weights $\{w_1,w_2,\cdots, w_n\}$  with $\sum\limits_{k=1}^{n}w_k=1\; .$ For $k\in \{1,2,...,n\}$ consider $w_k=\frac{1}{n} \; ,\; x_k= k\; .$ Then from $(*)$
\[ (*')\; \; \; \; \; \ln\left(\frac{1}{n}\sum\limits_{k=1}^{n}k\right) > \frac{1}{n}\sum\limits_{k=1}^{n}\ln{k} \; ,  \]
or $n\cdot \ln\left(\frac{n+1}{2}\right) > \ln{n!},$ that is  $\left(\frac{n+1}{2}\right)^n > n!\;$ as desired.

Dondigo, your solution is beautiful. How did you come up with it?

In hindsight, Dondigo could simply have made the simple transformation $\frac{n+1}{2} > \sqrt[n]{n!}$ and then noticed "Hey, that's a geometric mean on the RHS... " :P

Well, I don't want to play genius, but It was quite simple to notice, because most of the inequalities can be easily solved using AM-GM. So the first thing that I do while solving inequality problems is trying to use AM-GM. This time it worked :)

Write LS as 1*2*3*...*n.
Pair up i with (n + 1 - i),i. e. terms ((n + 1) / 2 - k) and
((n + 1) / 2 + k) get paired up.Then the product of the terms in each pair is:
(n + 1) ^ 2  /  4 - k ^ 2 < (n + 1) ^ 2 / 4
=((n + 1) / 2) ^ 2
Multiplying out this for every k in range 0 to ((n+1)/2 - 1) gets the desired result. :gleam:

This problem has been posted so many times, and solved in so many times.  By the way Dodingo, I'm not saying you copied, so nice solution :)

Masoud Zargar


geometry
circumcircle
geometric transformation
reflection
parallelogram
incenter
trigonometry

Let $AD,BE,CF$ be the altitudes and $H$ be the orthocenter of an acute triangle $ABC$. Let $M,N$ be respectively the points of intersection of $DE$ and $CF$ and of $DF$ and $BE$. Prove that the line passing through $A$ perpendicular to the line $MN$ passes through the circumcenter of triangle $HBC$.

[quote="edriv"]Hmmm... the point of intersection of $DE$ and $BE$ is $E$.
Please correct it![/quote]
I edited it , thanks  :oops:

[quote="PDatK40SP"]I think the line passing through $A$ perpendicular to the line $MN$ must pass through the circumcenter of triangle $DEF$, not $ABC$[/quote]
No, I edited it, This is wrong from version English of M&Y . :D

Here is my proof!

Let K the be nine-point-circle centre. Let T be the circumcentre of $BHC$.
It is known that the circumcircle of $BHC$ is just the reflection on $BC$ of the circumcircle of $ABC$. Then, $AH \parallel OT$. And also, the midpoint of $AH$ and the midpoint of $OT$ = midpoint of $BC$ are antipodes wrt. the Feuerbach circle. Using these facts it's very easy to prove that $AOTH$ is a parallelogram and that $K$ is the midpoint of $AT$.

Then, it is sufficient to prove that $MN$ is perpendicular to $AK$.

And now I'll rewrite the problem, mapping
- ortocenter ---> incenter
- D,E,F -> A,B,C
- A,B,C -> excenters
The problem is this:
Let $ABC$ be a triangle, $I$ its incenter, $D,E,F$ the feet of the internal angle bisectors, $O$ the circumcircle, $I_{A}$ the A-excenter. Prove that $OI_{A}\perp EF$.
(please see fig.2)

For this I have a short metric solution... short but still ugly  :(

Using the fact that $\frac{AF}{FB}= \frac{CA}{CB}$ and $AF+FB = AB$ we immidiately find that:
$AF = \frac{bc}{a+b}$ and $AE = \frac{bc}{a+c}$.

Now, let the radical axis of the circumcenter of ABC and the A-excenter meet AB at P, let the A-excenter meet AB at T.
We have that $PB \cdot PA = PT^{2}$ or $(PA-c) \cdot PA = (p-PA)^{2}$, with p semiperimeter, or $PA^{2}-c \cdot PA = PA^{2}-2p\cdot PA+PA^{2}$ or $PA(2p-c) = p^{2}$ or $PA= \frac{p^{2}}{a+b}$.
If the radical exis meets AC at Q, in the same way we prove that $QA = \frac{p^{2}}{a+c}$.

Finally, just apply Thales'theorem : $\frac{AF}{AE}= \frac{a+c}{a+b}= \frac{AP}{AQ}$ => $EF \parallel PQ$.
But $PQ \perp OI_{A}$, being the radical axsis.
Then $OI_{A}\perp EF$.

Bye bye. I would like to see a non-metric proof.  :bye:

see here for other solutions to this nice problem : [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=51168[/url]

[quote="PDatK40SP"]Here's my proof
Let $O,S$ and $E$ are the circumcenter of triangle $BAC,BHC$ and $DEF$, respectively. $S$ is reflect $O$ about $BC$ so $BHSO$ a parallelogram. It means that the midpoint $E$ of $OH$ is the midpoint of $AS$
$BDHF$ is a cyclic quadrilateral so $\bar{MH}.\bar{MB}= \bar{MD}.\bar{MF}$. Note that $\bar{MH}.\bar{MB}=pow_{(BHC)}M$ and $\bar{MD}.\bar{MF}= pow_{(DEF)}M$. So $M$ lies on the radical axis $d$ of two circles $(HBC)$ and $(DEF)$. $N$ lies on $d$ can be proved similarly. So $MN$ is the radical axis of two circles $(HBC)$ and $(DEF)$, it means that $MN \perp SE$, or $MN \perp AS$[/quote]
---------------------------
why $S$ is reflect $O$ about $BC$?

[b]@gearss[/b]
Let $R,R'$ be the circumradii of $\triangle ABC,\triangle BHC$ respectively,then $R=\frac{BC}{2\sin \angle BAC}=\frac{BC}{2\sin (180^\circ-\angle BAC)}=\frac{BC}{2\sin \angle BHC}=R'$,so the two circumcircles are symmetric with respect to $BC$,and hence $S$ is the reflection of $O$ with respect to $BC$.

Dear Mathlinkers,
this problem is another vresion of
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=576790
where A, B, C are the excenters of DEF...
Sincerely
Jean-Louis


An airplane's departure is delayed half an hour, but the flight time is reduced by $ 10\%$ due to favorable winds. If the flight ends up landing at its destination at the same time it would have landed if there had been no winds and no departure delay, how many hours was the flight originally scheduled to take?

Because the flight arrives at the same time with the 10% help as it would have if it had departed on time, half and hour is 10% of the original flight time. That means that the flight would have taken 30/0.1 or 300 min, which equals [b]5[/b] hours


Hi,
 please , can someone help me with this problem.
I need to know the procedure.

[b]Let f(x,y)=x^3-xy. set a(0,1) and b(1,3).
Find a point c on the line sement[ab] for which

f[b]-f[a]= gradient(f[c]) * (b-a)[/b]

Thank you

B.


AMC
AIME
calculus
integration
number theory
prime factorization

Let $n$ be the smallest positive integer that is a multiple of 75 and has exactly 75 positive integral divisors, including 1 and itself. Find $\frac n{75}.$

[hide] 75 has 3 factor pairs: {1,75}, {3,5 :^2:}, and {5,15}. subtracting 1 from each, the exponents of the primes in n's factorization could be 0,74 (impossible), 2,24 (impossible), 2,4,4 (possible), and 4, 14 (maybe possible). so if we let n=75k, then n=

3 . 5 :^2: . (prime factorization of k) since n must have 1 3 and 2 5's in its prime factorization, 4,14 is impossible for smallest candidate, as is 2,24. so n must be 2,4,4. we want to minimize this, so we let the mandatory 5 be in the squared spot (which still accounts for the 5 :^2:) and the 3 in one of the 4's spot. then the number is 25(81)k^4, to minimize k^4, minimize k, and since its prime let k=2. so then the number is 75(27)(16), and divided by 75 is 27(16), or 432?[/hide]

lol, the key was knowing the number of facts tingy


These kind of problems have been bugging me for a while now. An example -

How many arrangements of quarters, dimes, nickels, and/or pennies can be made to total 35 cents?

Is there a trick to these? Or do you [i]have[/i] to list them out and count the number of results?  :o

Thank you for the spam ernie.

For 0-4 cents-1 way
5-9 2 ways
10-49:
$ \frac{\lfloor\frac{n}{5}\rfloor(\lfloor\frac{n}{5}\rfloor\minus{}1)}{2}\plus{}3$

$ \lfloor x \rfloor$ means the greatest integer less than or equal to x.


geometry
trapezoid
quadratics
algebra
geometry unsolved

Let $ ABCD $ (AB//CD) be a trapezium circumscribed about a circle $ (I,r) $ . The lines $ AB, BC, CD, DA $ contact (I) at $ M,N,P,Q $ . The diagonal $ AC $ intersects $ (I) $ at $ K,L $ ($ K\in AL $) . Put $ AM = m , CN = n $
 Prove that :
 $ AK.LC = mn + 2r^2-2r\sqrt{mn+r^2} $;
 $ AL.KC = mn + 2r^2+2r\sqrt{mn+r^2} $.

Denote $p = AK \cdot CL$ and $q = AL \cdot CK$.

$AK \cdot AL = m^2$ (power of the point $A$ to the incircle $(I, r)$)
$CK \cdot CL = n^2$ (power of the point $C$ to the incircle $(I, r)$)

$pq = (AK \cdot CL)(AL \cdot CK) = (AK \cdot AL)(CK \cdot CL) = (mn)^2$

On one hand,

$p + q = AK \cdot (AC - AL) + AL \cdot (AC - AK) = AC \cdot (AK + AL) - 2m^2$

On the other hand,

$p + q = CK \cdot (AC - CL) + CL \cdot (AC - CK) = AC \cdot (CK + CL) - 2n^2$

As a result,

$p + q = \frac 1 2 [AC \cdot (AK + AL) - 2m^2 + AC \cdot (CK + CL) - 2n^2] =$

${= \frac 1 2 AC \cdot (AK + CK + AL + CL} - m^2 - n^2 = AC^2 - m^2 - n^2$

WLOG, we can assume $AB > CD$, i.e., $m = 2AB > n = 2CD$. Drop a normal from the vertex $A$ to the line $CD$ and let $H$ be the foot of this normal. From the right angle triangle $\triangle ACH$,

$AC^2 = (CD + DH)^2 + AH^2 = (m + n)^2 + 4r^2$

Therefore,

$p + q = (m + n)^2 + 4r^2 - m^2 - n^2 = 2(mn + 2r^2)$

Consequently, $p, q$ are the roots of a quadratic equation

$x^2 - (p + q)x + pq = x^2 - 2(mn + 2r^2)x + (mn)^2 = 0$

$p, q = mn + 2r^2 \pm \sqrt{(mn)^2 + 4mnr^2 + 4r^2 - (mn)^2} = mn + 2r^2 \pm  2r \sqrt{mn + r^2}$

Since $p = AK \cdot CL < q = AL \cdot CK$ by definition, the minus sign corresponds to the root $p$ and the plus sign to the root $q$.

...


Hi i am reading the Euclid's elements now and i get confused from the very beginning.
How do you prove postulate 4?
$\text{All right angles equal one another.}$

A Postulate is used as an axiom.
It does not need to be proved.

[quote="shobber"]Hi i am reading the Euclid's elements now and i get confused from the very beginning.
How do you prove postulate 4?
$\text{All right angels equal one another.}$[/quote]

the best book :lol:

[quote="denizuzun"]not angel: angle  :P[/quote]
 :blush:  :P edited.......

oh wow
heres the solution
lets say x=90. So x is a right angle, right?(no pun intended) :rotfl:
lets say y also equals 90. So it is also a right angle.
Then, since by the reflexive prop, 90=90, so by the law of substitution, x=y.
So any and all right angles are congruent to each other


[quote="cancer"]Let $a,b,c$ be positive integers such that $(a,b)=(a,c)=1$. Show that the equation $x^{a}+y^{b}=z^{c}$ has infinitely many solutions in positive integers $x,y,z$ where $x,y,z$ are all distinct.[/quote]

$(a,b)=(a,c)=1$ $\implies$ $(a,bc)=1$ $\implies$ exist $\alpha,\beta$ such that $\alpha a+\beta bc=1$. In this equation :
- Iif $\beta=0$ then $a=1$ and $x+y^{b}=z^{c}$ obviously has infinitely many solutions in positive integers $x,y,z$ where $x,y,z$ are all distinct.
- If $\alpha=0$ then $bc=1$ and $b=c=1$ and $x^{a}+y=z$ obviously has infinitely many solutions in positive integers $x,y,z$ where $x,y,z$ are all distinct.
- If $\alpha<0$ then it exists $k$ such that $\alpha+kbc>0$ and so $(\alpha+kbc)a+(\beta-ka) bc=1$

So we can consider $\alpha>0$ and $\beta <0$. Then let $\gamma=-\beta$; we have $\alpha a=1+\gamma bc$ with $\alpha$ and $\gamma$ positive integers.

Then Let $u$ and $v$ such that $u^{c}>v^{b}$, ve have :
$(u^{c}-v^{b})^{\alpha a}=(u^{c}-v^{b})^{1+\gamma bc}=(u(u^{c}-v^{b})^{\gamma b})^{c}-(v(u^{c}-v^{b})^{\gamma c})^{b}$

And so  $x^{a}+y^{b}=z^{c}$  with :

$x=(u^{c}-v^{b})^{\alpha}$
$y=v(u^{c}-v^{b})^{\gamma c}$
$z=u(u^{c}-v^{b})^{\gamma b}$

Q.E.D.


conics
ellipse
geometry
cyclic quadrilateral
geometry theorems

Recently I noticed that in one of my solutions for geometry lessons I unconsciously used theorem which says that it is possible to circumscribe an ellipse on every convex quadrangle.

I dont even know if its true, but it look so.

Could someone show me a proof or a counterexample?

just find a linear transformation that takes your quadrilateral $ Q$ into a cyclic quadrilateral $ Q'$. inverting the transformation takes the circle circumscribing $ Q'$ to an ellipse circumscribing $ Q$

Exactly. The only problem is that question about ellipse grow from problem of finding such transformation. I had no idea about it, so i tried (and failed) to prove an existence of ellipse.

Edit: I found in web this article:

http://resources.metapress.com/pdf-preview.axd?code=av15w8n283141657&size=largest

The author mention that this problem is obvious :| but I dont think so.


Show that if $ U \subset \mathbb{R}^n$ is an open connected set then given $ a,b \in U$ there exists a differentiable mapping $ \phi: [0,1] \to U$ such taht $ \phi(0) \equal{} a$ and $ \phi(1) \equal{} b$.

The definition of connected I have is from Rudin. A set is connected if it is not separated. A set $ E$ is separated if $ A \cup B \equal{} E$ and $ A \cap \overline{B} \equal{} \emptyset$ and $ \overline{A} \cap B \equal{} \emptyset$

Let $ V$ be the set of all points in $ U$ that can be connected to $ a$ by means of such $ \phi$. Prove that $ V$ is open and nonempty. Then prove that $ U\setminus V$ is also open. Use the definition of connectedness to conclude that $ U\setminus V$ is empty.


linear algebra
matrix
algebra proposed
algebra

Let $  A $  be a $  n \times  n $   real matrix such that $ \det(I-A)\ne 0 . $ The [i]Cayley transform[/i]  of $  A $  is  the matrix \[ C(A)=(I-A)^{-1}(I+A). \]
[b]Question:[/b] [i] How we can describe all matrices $ A $ for which $ C(A) $ is orthogonal [/i] ?

i think it should be skew symmetric matrix :maybe:


calculus
derivative
function
real analysis
real analysis solved

OK, this one is much less classical but much tougher too  ;)

Suppose that $f\in C^{\infty}(\mathbb R)$ is not identically $0$ and $f$ vanishes for $x\leq 0$. Prove that there exists $x\in \mathbb R$ such that $\sup_{n\geq 0}|f^{(n)}(x)|=+\infty$.

If harazi knows this one too, I'll give up until the next week...

Don't give up! I don't know this one. I will try to solve it is very interesting. Thanks.

Well, fedja, you have killed me with this one.  :D  :(

Please, don't post solution yet, I said that as admiration to the beauty and difficulty of the problem. By the way, what is SPRA?

I just want to kill myself :( .
 Fedja, can you please give me a hint, (probably hiding it from others, (for harazi :D) )

[quote="dickchimney"]
 Fedja, can you please give me a hint, (probably hiding it from others, (for harazi :D) )[/quote]
This is one of the hardest problems in the entire book :D. Here is a hint
[hide="hidden from harazi"]
Let $S(x)=\sup_{n\geq 0}|f^{(n)}(x)|$. Show that if $S(x)<+\infty$ for every $x$ in some closed interval $I$, then $S$ is uniformly bounded on $I$. To this end show that otherwise there exists a sequence of nested open intervals $J_k\subset I$ such that $S> 2^k$ on $J_k$, $S\leq 2^k$ at the endpoints of $J_k$ and $S(x)>2^{k+1}$ for some point $x\in J_k$.
[/hide]

You're evil, fedja!  :D . I spent hours on this one playing arround with Baire's theorem aand I got nothing.  :(

If it's about me, no, I'm still trying to solve it, among other things. Myth, do you have a solution for this problem?

[quote="fedja"][quote="Myth"]:([/quote]
Hmm... I thought the hint was pretty revealing... BTW, harazi, you killed me with your $e^f+e^g+e^h=1$ problem. Still haven't the slightest idea of how to do it...  :)[/quote]
I need to time to think about it. Unfortunately, I still haven't it. Maybe when I will be sleeping  :lol:

So, Fedja, we are speaking to each other, one from the heaven and the other one from hell ( I am in hell, since grobber said I am evil :mad: ). And by the way, the score is 2-0 for you. I also have no idea to solve your problem or the problem I posted.  :(

It seems to me that I solved it, but the solution is quite hard,
so I apologize for that. May be it can be simplified.

We will prove it by contradiction, so assume that for every $ x \in \mathbb{R} $, we have
$ sup_{n \geqslant 0} |f^{(n)}(x)| < + \infty $. Denote $ M(x) := sup_{n \geqslant 0}
|f^{(n)}(x)| $.
Observe first of all that then for every $ x_{0} \in \mathbb{R} $, the series
$ \sum_{n=0}^{\infty} \frac{f^{(n)}(x_{0})}{n!} (z-x_{0})^{n} $
converges for every $ z \in \mathbb{C} $, and gives us an entire function.
( because $ M(x_{0}) < \infty $ ).

Let us prove some Lemma's:

$ \underline{Lemma1} $

There exists some $ x_{0} > 0 $ such that $ f(x) $ is not analytic at $ x = x_{0} $,
that is, if we define $ g(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_{0})}{n!} (x-x_{0})^{n} $,
then $ g(x) \ncong f(x) $ in any neighborhood of $ x_{0} $.

$ \underline{proof of Lemma1} $

Suppose, in the contrary, that $ f $ is analytic at each point $ x_{0} \in (0,+ \infty) $. Take
for example $ g(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!} (x-1)^{n} $, which is entire
function. Then because of uniqueness of analytic continuation, $ g(x) = f(x) $ for every
$ x > 0 $, so $ g^{(n)}(x) = f^{(n)}(x) $ for every $ x > 0 $, then by continuity,
$ g^{(n)}(0) = f^{(n)}(0) = 0 $ for every $ n \geqslant 0 $, so because $ g $ is entire,
we get that $ g(x) \equiv 0 $, so $ f(x) = 0 $ for every $ x > 0 $, and we come to
contradiction with the fact that $ f $ is not identically zero.

$ \underline{Lemma2} $

If $ f $ is analytic in $ (a,b) $ ( i.e in some complex neighborhood of $ (a,b) $ ),
then $ M(a) e^{a-b} \leqslant M(b) \leqslant M(a) e^{b-a} $.

$ \underline{proof of Lemma2} $

Take $ c \in (a,b) $, and define two entire functions:
$ g(x): = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^{n} $
and $ h(x): = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^{n} $
Then because of analyticity of $ f $ on $ (a,b) $ and uniqueness of
analytic continuation, we have for every $ x \in(a,b) $, $ g(x) = f(x) $.
Therefore for every $ x \in(a,b) , k \geqslant 0 $, $ g^{(k)}(x) = f^{(k)}(x) $.
By the fact that $ f, g $ are $ C^{\infty} $, we get that
$ g^{(k)}(a) = f^{(k)}(a) , k \geqslant 0 $. On the other hand, we see that
$ h^{(k)}(a) = f^{(k)}(a) $, so $ g^{(k)}(a) = h^{(k)}(a) , k \geqslant 0 $,
and because of $ g, h $ are entire, we get that they are equal. By the same
argument that $ f, g $ are $ C^{\infty} $, we get that
 $ g^{(k)}(b) = f^{(k)}(b) $, and finally we
get that $ f^{(k)}(b) = g^{(k)}(b) = h^{(k)}(b) =
\sum_{n=0}^{\infty} \frac{f^{(n+k)}(a)}{n!} (b-a)^{n} $. Therefore
$ |f^{(k)}(b)| \leqslant M(a) \sum_{n=0}^{\infty}  \frac{(b-a)^{n}}{n!} = M(a) e^{b-a} $.
This is true for every $ k \geqslant 0 $, so $ M(b) \leqslant M(a) e^{b-a} $.
By analogy, we can prove that $ M(a) e^{a-b} \leqslant M(b) $.

$ \underline{Lemma3} $
If $ f $ is not analytic in $ x_{0} $, then in any neighborhood $ U $ of $ x_{0} $,
the function $ M(x) $ is not bounded in $ x \in U $.

$ \underline{proof of Lemma3} $
Take some $ A > x_{0} $, and we can assume that $ U \subset (0,A) $.
Let us define $ h(x): = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_{0})}{n!} (x-x_{0})^{n} $
and $ g(x): = f(x) - h(x) $. Then we have $ h^{(k)}(x) \leqslant e^{A} M(x_{0}) $ for
$ x \in U $ and of course $ g^{k}(x_{0}) = 0 , k \geqslant 0 $. Then, because of $ g $ is
not analytic in $ x_{0} $, we get that there exists $ a \in U $,
such that $ C:= |g(a)| > 0 $.
Then, by remainder formula for Taylor series for $ g $ at $ x_{0} $, we see that
for every $ k \geqslant 1 $ we have that there exists $ \theta_{k} $ in $ U $, such that
$ g(a) = \frac{(a-x_{0})^{k}}{k!} g^{(k)}(\theta_{k}) $, so $ |g^{(k)}(\theta_{k})| =
\frac{k!C}{|a-x_{0}|^{k}} \rightarrow \infty $, when $ k \rightarrow \infty $. But
$ |f^{(k)}(\theta_{k})| \geqslant |g^{(k)}(\theta_{k})| - e^{A} M(x_{0}) $, so
$ |f^{(k)}(\theta_{k})| \rightarrow \infty $, so $ M(x) $ is not bounded in $ x \in U $.

Now let's solve the problem.
By lemma1, we can find $ y_{1} > 0 $, such that $ f $ is not analytic at $ y_{1} $,
and take also some neighborhood $ U_{1} $ of $ y_{1} $.
Take also some $ A > 0 $, such that the interval $ (0,A) $ contains $ U_{1} $.
Then for each $ n \geqslant 1 $ suppose we have found $ y_{n} $, at which $ f $ is not
analytic, and a neighborhood $ U_{n} $ of $ y_{n} $. By lemma3, the function $ M(x) $
is not bounded in $ x \in U_{n} $, so there exists $ z_{n+1} \in U_{n} $, such that
$ M(z_{n+1}) \geqslant n $. Take now the set of all $ y $'s in the closed interval between
$ y_{n} $ and $ z_{n+1} $, at which $ f $ is not analytic. This set is clearly non-empty,
because it contains $ y_{n} $, and denote by $ y_{n+1} $ the element of this set,
which is closest to $ z_{n+1} $. Then is we have $ y_{n+1} = z_{n+1} $, stop.
If not, then we have that $ f $ is analytic in the interval between $ y_{n+1} $ and
$ z_{n+1} $, so because of lemma2, we get that $ M(y_{n+1}) \geqslant
M(z_{n+1}) e^{- |y_{n+1} - z_{n+1}|} \geqslant M(z_{n+1}) e^{-A} \geqslant ne^{-A} $.
In any case we will have $ M(y_{n+1}) \geqslant ne^{-A} $. Therefore there exists
some integer number $ k_{n+1} \geqslant 0 $, such that
$ f^{(k_{n+1})}(y_{n+1}) \geqslant ne^{-A-1} $. Now, because $ f $ is $ C^{\infty} $, there
exists some neighborhood $ U_{n+1} $ of $ y_{n+1} $, such that $ |f^{(k_{n+1})}(y)| \geqslant
ne^{-A-2} $ for every $ y \in U_{n+1} $. Of course we can
assume also that $ \overline{U_{n+1}} \subset U_{n} $.
So by such procedure we have constructed a sequence $ y_{1},y_{2}, \ldots $ of points,
at which $ f $ is not analytic, and a sequence of neighborhoods $ U_{1},U_{2}, \ldots $,
such that $ \overline{U_{n+1}} \subset U_{n} $ , and a sequence of nonnegative
integers $ k_{2},k_{3}, \ldots $, and we know that $ |f^{(k_{n})}(y)| \geqslant
(n-1)e^{-A-2} $ for every $ y \in U_{n} $, for every $ n \geqslant 2 $. Now if we take
an element $ p $ in the non-empty intersection $ \bigcap_{n=2}^{\infty} U_{n} $, we
will get that $ |f^{(k_{n})}(p)| \geqslant (n-1)e^{-A-2} $ for every $ n \geqslant 2 $,
so $ M(p) = +\infty $ and we come to contradiction.

I have read it.

You don't need to apologize, Leva! You wrote it very clear. And things you used are not hard. Just you gave a detailed explanation for your thoughts.

It is a great proof, like all of your proofs, Leva!

what's the solution in the book, fedja??!! I can't derive the solution from the hint :(

It is quite close to Leva's solution though some technical things are different. What is your difficulty: you cannot prove the statement in the hint about existence of a nested sequence of intervals or you cannot derive the solution from the statement?

I can prove tthe state ment in the hint, but cannot derive the uniform boundedness of the $\sup |f^{(n)}(x)|$


Fie n,k din N si A din M k (R) astfel incat A^(n+1)=Ok

Sa se arate ca: det(Ik +1/f(1)*A+1/f(2)*A^2+...+1/f(n)*A^n )=1

si f(n)=n factorial

a mai aparut problema aici http://www.mathlinks.ro/Forum/viewtopic.php?t=72576


algebra polynomial AMC AIME Stanford college

Suppose x, y, and z satisfy

x+y+z=3
x^2+y^2+z^2=5
x^3+y^3+z^3=7.

Find x^4+y^4+z^4.

hmmm i think im very close, but im stuck.

what i have so far is

[hide] -15= -3(x^4+y^4+z^4) + 6(x^2*y^2 +y^2*z^2 +x^2*z^2)  .  but i cant seem to figure out how to get rid of x^2*y^2 +y^2*z^2 +x^2*z^2.  Knowing xyz=1/3, xy+xz+xy = 2, how can I get rid of this?  thanks [/hide]

I'll post the rest after i see if i got the answer right later.

Edit: I just thought that i should point out that some of the numbers that i posted in spoiler are wrong, based on the fact that xyz is not 1/3, but it is 4/3, as fierytycoon pointed out.

cats...:

[hide]How did you get the value of xyz? Anyways, since you know the values of x+y+z=a, xy+yz+zx=b, and xyz=c, you can let x, y, z be the roots of a polynomial x^3 - ax^2 + bx - c =0 and then find the roots of this cubic.[/hide]

Unfortunately x,y, and z do not come out to be nice numbers at all.. it would be impossible to find the sum of their 4th powers numerically. I think cats is on the right track, you basically just keep [hide]multiplying equations together[/hide] until everything else cancels out.

- thanks.  i didnt think of that.  Ok so its pretty much done then.  I'll type what i have so far, then go back and finish w/ what u said.

[hide]

(x+y+z)^2 = 9 = x^2+y^2+z^2 + 2(xy + xz + yx)

                  =9 = 5 +2(xy +xz +yx)

                    4 = 2(xy+xz+yx)

                    2 = (xy+xz+yx)

(x+y+z)^3 = 27 = x^3 +y^3 +z^3 = 3(x^2*y +y^2*x + z^2*x + z^2*y + x^2*z + y^2*z) + 6xyz

27 = 7 + 3x(3-x) + 3y(3-y) + 3z(3-z) + 6xyz

27 = 7 +6(x+y+z) + 6xyz

27 = 7 + 18

2 = 6xyz

1/3 = xyz.

Stupidly, i went further and expanded (x+y+z)^4, but i'll go back and do what u said now.  thanks a lot

[/hide]

theres a good problem you should try, i'll post it in a few, but i think its a little easier than this.

TripleM, do you know how i can find 6(x^2*y^2 +y^2*z^2 +x^2*z^2), which is also 2/3*(1/z^2+1/x^2+1/y^2)   ?  Also, did you already get the answer to the problem?   If x, y, z are not good numbers i will not try the polynomial then.

The other problem i was talking about is this

Given that a,b, c are positive real numbers, and log_a b + log_b c + log_c a = 0, find the value of (log_a b)^3 + (log_b c)^3 + (log_c a) ^3 .

cats...:

[hide]Could you explain how you got from

27 = 7 + 3x(3-x) + 3y(3-y) + 3z(3-z) + 6xyz

to

27 = 7 +6(x+y+z) + 6xyz

27 = 7 + 18 + 6xyz?

I calculated and found

27 = 7 + 9(x+y+z) - 3(x^2+y^2+z^2) + 6xyz

27 = 7 + 9(3) - 3(5) + 6xyz

27 = 7 + 12 + 6xyz

My 12 differs from your 18.[/hide]

hah yeah you're right.  When i did out 3x(3-x) etc., i thought it was 9x-3x=6x, and 6y, 6z for the other terms.  So you're right, maybe that'll make the numbers nicer.  S. that means xyz = 4/3.  good catch.  Hm yeah, that changes the numbers in my expansion of (x+y+z)^4, but still doesnt change the fact that i cant get around finding 1/z^2+1/x^2+1/y^2.

Hmm, yeah, the roots of the polynomial will still be messy. But why are you trying to find 6(x^2*y^2 +y^2*z^2 +x^2*z^2)?

Because of the expansion of (x+y+z)^4.  From that expansion, i got everything to cancel out besides the x^4+y^4+z^4 and the 6(x^2*y^2 +y^2*z^2 +x^2*z^2).  If we can get rid of that, we have the answer.

i think u guys should give TRipleM's hint a good try

it works

this question is actually very similar to an AIME question....

did you actually do this darkquantum?  Cause you realize that's what we've been doing the whole time.

No, i believe ur expanding )x+y+z)^4 and looking for cancellations

TripleM is telling u to multiply the given expressions and looking for cancellations there

try (x:^2:+y:^2:+z:^2:):^2: instead.  you'll get some cross terms that you can get from (xy+yz+xz):^2:, and from that you'll get some more cross terms that should be the product of two things you know already.

This was a really long problem.

yes, but, see, its a really nice problem

so mystic, i did what you suggested and maybe i messed up, cause i didnt really get anywhere.  Can you show me exactly what you did?  thanks.

[quote="cats..."]did you even do this quantum?

mystic, when we do (x^2+y^2+z^2)^2 we still ran into the problem of finding a way to cancel x^2*y^2 + y^2*z^2 + x^2*z^2, but i did not try expanding (xy+yz+xz)^2 as you said.  So maybe that was what we needed to do, i'll see what happens.[/quote]

okay, as you said, you need to cancel these terms: x^2*y^2 + y^2*z^2 + x^2*z^2. You already have the value of xy+yz+xz, so squaring that, you have the terms we're looking for, plus the cross-terms: xy:^2:z+x:^2:yz+xyz:^2:. We can factor a xyz out of this to get (x+y+z). You have the values of both xyz and x+y+z, I believe, so put all of your values together, and there you go! (hopefully, if I didn't make a stupid mistake ;))

yeah ok, i think i expanded it wrong.  But that would lead us to our answer, thanks.

There was a question from AIME 2001 like this. I think it was #9 or 10 or somewhere around there, but I'm not quite sure. Unfortunately, AIME 2001 isn't on Kalva, but if someone is psychic and knows the problem I'm talking about, please post it!

where did you get that question from zeta?

Someone should try the log question i posted.

For the log question:

[hide]

log_a b , log_b c , log_c a = x,y,z respectively. Then a^(xyz) = a so xyz=1, and we are given x+y+z=0, trying to find x^3 + y^3 + z^3..

Then we can use the fact that x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx) = 0. So x^3 + y^3 + z^3 = 3.

[/hide]

is their another method you're looking for?  the one i went for took a while, but its basically solved.  I know i messed up in calculating xyz, and i need to go back to fix it, but my work is lost cause my dad threw away the papers, and i dont really want to go through it again.  But is there a better method that you had in mind?


Can Holder be used here, as $a,b,c$ might be negative?

[quote="silouan"]Because LHS is positive  and RHS is smaller when one or two variables are negative and when all are negative you can put    -a=x  etc his solution is true  :wink:  and very nice .[/quote]

Oh, I see  :lol:


Prove that if $c$ is a rational number, then the equation $x^3-3cx^2-3x+c=0$ has at most one rational solution.

If the cubic equation has no rational root, we are done. Assume it has a rational root $r = \frac p q$. Substituting the root to the cubic equation, we can calculate $c$ in terms of this root:

$r^3 - 3cr^2 - 3r + c = 0$

$c = r \frac{r^2 - 3}{3r^2 - 1}$

The cubic equation becomes

$x^3 - 3r \frac{r^2 - 3}{3r^2 - 1} x^2 - 3x + r \frac{r^2 - 3}{3r^2 - 1} = 0$

Factoring the root $r$ out by a long division,

$x^3 - 3r \frac{r^2 - 3}{3r^2 - 1} x^2 - 3x + r \frac{r^2 - 3}{3r^2 - 1} = (x - r) \left( x^2 + \frac{8r}{3r^2 - 1}x + \frac{3 - r^2}{3r^2 - 1}\right)$

The discriminant of the remaining quadratic factor is

$D = \frac{64r^2 - 4(3 - r^2)(3r^2 - 1)}{(3r^2 - 1)^2} = \frac{64r^2 - 4(9r^2 - 3 - 3r^4 + r^2)}{(3r^2 - 1)^2} =$

$= \frac{12(r^4 + 2r^2 + 1)}{(3r^2 - 1)^2} = 3 \cdot \left( \frac{2(r^2 + 1)}{3r^2 - 1}\right)^2$

As a result, the discriminant cannot be a complete square and consequently, the remaining quadratic factor with rational coefficients cannot have rational roots.


I had issues with my Photobucket album. That's why the links didn't work.

[url=http://photobucket.com/albums/v259/drspyder/?action=view&current=ownage.swf]Special Olympics - Revised[/url]

[url=http://www.putfile.com/media.php?n=LinkGuitar]First frame of a new animation.[/url]


A 24-foot by 72-foot rectangular dance floor is completely tiled with 1-foot square tiles. Two opposite corners of the dance floor are connected by a diagonal. This diagonal passes through the interior of exactly how many tiles? :yoda:

theres actually a cool formula for questions like this:

number of tiles crossed by a diagonal with dimensions $ a$ by $ b$ is $ a \plus{} b \minus{} \text{GCF}(a,b)$. therefore, this is $ 24 \plus{} 72 \minus{} \text{GCF}(24, 72)$, which is $ 24 \plus{} 72 \minus{} 24$, or $ \boxed{72}$

EDIT: nice call, xpmath. thanks.


inequalities
calculus
integration
inequalities proposed

[quote="soruz"]If  $ a,b \in[0,1]$ , then

$ \sqrt {[a^2 \plus{} (1 \minus{} b)^2][(1 \minus{} a)^2 \plus{} b^2]} \plus{} \sqrt {[a^2 \plus{} b^2][(1 \minus{} a)^2 \plus{} (1 \minus{} b)^2]} \ge 1$.[/quote]
Let $ A(0,0),$ $ B(0,1),$ $ C(1,1),$ $ D(1,0),$ $ E(a,b),$ $ F(a,b\minus{}1).$
Hence, $ BE\equal{}AF$ and $ EC\equal{}FD.$
For quadrilateral $ AEDF$ we obtain: $ AE\cdot FD\plus{}AF\cdot ED\geq AD\cdot EF,$  which gives
$ \sqrt {[a^2 \plus{} (1 \minus{} b)^2][(1 \minus{} a)^2 \plus{} b^2]} \plus{} \sqrt {[a^2 \plus{} b^2][(1 \minus{} a)^2 \plus{} (1 \minus{} b)^2]} \ge 1.$

Just Cauchy and AM-GM, and it becomes obvious that LHS >= RHS

Please detail,Qwerty414!Post integral solution!

After solution of qwerty, inequality becomes strong, after me computation:

If  $ a,b \in[0,1]$ , then
$ \sqrt{[a^2\plus{}(1\minus{}b)^2][(1\minus{}a)^2\plus{}b^2]}\plus{}\sqrt{[a^2\plus{}b^2][(1\minus{}a)^2\plus{}(1\minus{}b)^2]}  \ge 2$.

It's right?

[quote="soruz"]After solution of qwerty, inequality becomes strong, after me computation:

If  $ a,b \in[0,1]$ , then
$ \sqrt {[a^2 \plus{} (1 \minus{} b)^2][(1 \minus{} a)^2 \plus{} b^2]} \plus{} \sqrt {[a^2 \plus{} b^2][(1 \minus{} a)^2 \plus{} (1 \minus{} b)^2]} \ge 2$.

It's right?[/quote]
No! Try $ a\equal{}b\equal{}0.$

In fact ,I found
$ 4 \sqrt{[ab(1\minus{}b)(1\minus{}a)]} \ge 1$
How prove this?

it is $ 4\sqrt{ab(1\minus{}b)(1\minus{}a)}\geq1$ wrong . $ 4\sqrt{ab(1\minus{}b)(1\minus{}a)}\equal{}4\sqrt{a(1\minus{}a)}\sqrt{b(1\minus{}b)}\leq4\frac{a\plus{}1\minus{}a}{2}\frac{b\plus{}1\minus{}b}{2}\equal{}1$


Put multiplication signs, addition signs, subtraction signs or nothing between the digits to make this equality true.

                            123456789=2004

Ooo, I know this one. I'll wait to see if someone else gets it.

That's quite the interesting algorithm you have there. I sure wonder who the cunning person was that created that question.

Please do not post useless messages such as this one. These belong in the "Other Topics" section. Thank you.

I belive there are multiple solutions to this question 4everwise, so I dont think it would hurt to post your answer.....

[hide]1+2*3*4-5-6-7+8+9=20+04[/hide]
i love smart-alecky answers
Edit: I was wondering if you could use parentheses too. That would create lots more possibilities.


$ a$ and $ b$ are strictly positive real numbers .
prove that:
$ (a^2\plus{}b\plus{}\frac{3}{4})(b^2\plus{}a\plus{}\frac{3}{4})\geq (2a\plus{}\frac{1}{2})(2b\plus{}\frac{1}{2})$
from moroccan TST 2007

[quote="greatestmaths"]$ a$ and $ b$ are strictly positive real numbers .
prove that:
$ (a^2 \plus{} b \plus{} \frac {3}{4})(b^2 \plus{} a \plus{} \frac {3}{4})\geq (2a \plus{} \frac {1}{2})(2b \plus{} \frac {1}{2})$
from moroccan TST 2007[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103170

thanks i don't see it before but it 'isn't the same exactly :lol:

[quote="FOURRIER"]This is from bulgaria 97 it was posted a lot before[/quote]
can you give  links .

it was also at  some olympiad in Georgia republic  2005-2006.( I don't remember well)

http://www.mathlinks.ro/viewtopic.php?t=140671
kunny's solution is the best..
 :wink:


Three congruent rectangular banners are sewn so that each banner
consists of three adjacent, congruent, vertical stripes-a middle white
stripe bordered on each side by a gray stripe. The three banners are
displayed so that each overlaps the other by half their width and half
their length. Find the ratio of the visible white areas to the visible
gray areas. Express your answer as a common fraction.

[asy]path p1 = (0,0)--(10,0)--(10,40)--(0,40)--cycle;
path p2 = (10,0)--(20,0)--(20,40)--(10,40)--cycle;
path p3 = (20,0)--(30,0)--(30,40)--(20,40)--cycle;

fill(p1,gray(.4));
fill(p2,white);
fill(p3,gray(.4));
draw(p1);
draw(p2);
draw(p3);

fill(shift((15,-20))*p1,gray(.4));
fill(shift((15,-20))*p2,white);
fill(shift((15,-20))*p3,gray(.4));
draw(shift((15,-20))*p1);
draw(shift((15,-20))*p2);
draw(shift((15,-20))*p3);

fill(shift((30,-40))*p1,gray(.4));
fill(shift((30,-40))*p2,white);
fill(shift((30,-40))*p3,gray(.4));
draw(shift((30,-40))*p1);
draw(shift((30,-40))*p2);
draw(shift((30,-40))*p3);[/asy]

You count the stripes. By looking at the diagram, you can see that there are 4 complete gray stripes, and 2 halves of gray stripes. $ 4\plus{}2\left(\frac12\right) \equal{} 5$. Now the white stripes; there is one complete one and 2 $ \frac34$ ones. $ 1\plus{}2\left(\frac34\right)\equal{}2\frac12$. $ \frac{2\frac12}{5}\equal{}\boxed{\frac12}$


After I finish a problem, at the bottom, it says :

Related Readings:
Introduction to Counting & Probability - Chapter 7

Video Lessons:
Probability with Combinations Part 1

Probability with Combinations Part 2

Probability with Combinations Part 4

or some other videos. Do I get the videos, (I dont think so, I tried, and couldnt see them) or does it explain the problems in the video?

Also, I looked at a video that was Probobility w/ Combinations Part 3, and when I went to to see the Videos I could watch (I wanted to watch it again) the video wasnt there!

How can that be?????? I just watched a few days ago!! Now it isnt there??? (It wasnt there to begin, I clicked on the Suggested Video button)

Please Help!!

Thx for reading this though :)

You can only view videos for subjects which you are ready for or have passed. It is possible that a subject you were thought to be ready for was found to be too difficult for you so Alcumus no longer has you set as ready for that topic. If that's the case, then you will no longer be able to see the videos for that topic until you are once again ready.

The problem I see with basing access to the videos in this way is that it doesn't take into account why someone missed a particular problem. It could be because they made a calculation error, or misread the problem, instead of because they weren't ready for the concept. Also the amount of data collected on a particular category of problems may be small until they do a large number of problems.

So maybe it would be better to just give people access to all the videos, and let them decide on their own if they are ready for any particular one.

Also, it appears that the list of topics "passed, mastered, and ready" fluctuate in some way that is unclear to me. Even which category a particular problem fits in changes from day to day. Is it according to how other people are doing/have done in those categories? If so, then that doesn't make sense to me. How could an individuals "readiness" in a certain category be affected by how others are doing/have done in that category?

The only categories that change from day to day are the Mastery categories, and these change based on our learning more about which problems are hardest.  The others are much more stable.

As for basing the video access on what subjects are passed, there are a lot of reasons we chose this, but there are plenty of reasons to go the other way.  I'm interested to hear what more people think of this.

Other than the Mastery subjects, there shouldn't be wild fluctuations (unless you're doing lots of problems) in whether or not you've passed a subject.  Every once in a while there may be a huge shift, as we're still making occasional changes to the model now that we have a *lot* more data.


search
inequalities unsolved
inequalities

Try Kalva. Basically you do a substituion and use CauchySwcharz

Bomb

[quote="Gohan"]can you write down your solution[/quote]
check 42th imo no.2 or search on the forum. It's very well-known.

[quote="Gohan"]Prove that for $a,b,c>0$ and $abc=1$
$\frac{1}{\sqrt{1+8a}}+\frac{1}{\sqrt{1+8b}}+\frac{1}{\sqrt{1+8c}} \geq1$[/quote]

 Gohan, I think the substitutions $\frac{1}{\sqrt{1+8a}}=x$ and so on.... I'm sure this will work. ;)


geometry
ratio
trigonometry
angle bisector
geometry proposed

E is the excenter of AHB, due to <BEA=(1/2)<AHB=(1/2)(90)=45
and the point E is on the bisector of <B. Now HE is the external bisector
of <BHA,i.e. <EHC=(1/2)<AHC=(1/2)(90)=45. QED.

Actually it IS easy!

We have

  45 = < BEA = 180 - < BAE - < ABE = 180 - A - B/2
        = 90 + 90 - A - B/2 = 90 + (A + B + C)/2 - A - B/2
        = 90 + (A + B + C - 2A - B)/2
        = 90 + (C - A)/2,

and thus (C - A)/2 = 45 - 90 = - 45, and C - A = 2 $\cdot$ (- 45) = - 90, and A - C = 90, so that A = 90 + C.

Now, we have AE / EC = c / a (since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides). But after the Sine Law in triangle ABC, we have c / a = sin C / sin A = sin C / sin (90 + C) = sin C / cos C = tan C. Thus, AE / EC = tan C. On the other hand, in the right-angled triangle AHC, we have AH / HC = tan < HCA = tan C. Thus, AH / HC = AE / EC. And this yields that the line HE is the angle bisector of the angle AHC. Since < AHC = 90, we therefore get < EHC = 90 / 2 = 45.

I see Juan Carlos's solution is much simpler, but I have already written down mine...

  Darij

[quote="grobber"]In triangle $ABC$ $AH$ is an altitude and $BE$ is the internal angle bisector of $\angle B$. Show that if angle BEA has 45 degrees, then so does angle EHC.[/quote]

If K is the second point where c(AEH) cut BC, it holds that <AEK = 90. Therefore <EHC = <EAK = 45.

grobber i have the another solution about it .


integration
trigonometry
calculus
calculus computations

Let $ S_{1}\equal{}1\plus{}\frac{1}{3}\minus{}\frac{1}{5}\minus{}\frac{1}{7}\plus{}\frac{1}{9}\plus{}\frac{1}{11}\minus{}\dots$

And $ S_{2}\equal{}1\minus{}\frac{1}{3}\plus{}\frac{1}{5}\minus{}\frac{1}{7}\plus{}\frac{1}{9}\minus{}\frac{1}{11}\plus{}\dots$

Observe that $ S_{2}\equal{}\sum_{k\equal{}0}^{\infty}\frac{(\minus{}1)^{k}}{2k\plus{}1}\equal{}\int_{0}^{1}\left[\sum_{k\equal{}0}^{\infty}(\minus{}x^{2})^{k}\right]dx\equal{}\int_{0}^{1}\frac{dx}{1\plus{}x^{2}}\equal{}\frac{\pi}{4}$

Furthermore, $ S_{1}\equal{}\sum_{k\equal{}0}^{\infty}(\minus{}1)^{k}\left[\frac{1}{4k\plus{}1}\plus{}\frac{1}{4k\plus{}3}\right]\equal{}\int_{0}^{1}\left[(1\plus{}x^{2})\sum_{k\equal{}0}^{\infty}(\minus{}x^{4})^{k}\right]dx$

$ S_{1}\equal{}\int_{0}^{1}\frac{1\plus{}x^{2}}{1\plus{}x^{4}}dx\equal{}\frac{1}{2}\int_{0}^{1}\left[\frac{1}{x^{2}\minus{}\sqrt{2}x\plus{}1}\plus{}\frac{1}{x^{2}\plus{}\sqrt{2}x\plus{}1}\right]dx\equal{}\sqrt{2}\frac{\pi}{4}$

$ \therefore 1\plus{}\frac{1}{3}\minus{}\frac{1}{5}\minus{}\frac{1}{7}\plus{}\frac{1}{9}\plus{}\frac{1}{11}\minus{}\dots\equal{}\sqrt{2}\left(1\minus{}\frac{1}{3}\plus{}\frac{1}{5}\minus{}\frac{1}{7}\plus{}\frac{1}{9}\minus{}\frac{1}{11}\plus{}\dots\right)$ :)

I wonder if we can proof it implicitly or not :?:

[b]Timestopper_STG[/b], I don't know any such solution. Wrote another below:

If we develop $ f(x) \equal{} 1$ as a Fourier Series

$ f(x) \equal{} a_1\sin\;x \plus{} a_2\sin\;2x \plus{} \dots \plus{} a_m\sin\;mx\plus{}\dots$

where

$ a_m \equal{} \frac {2}{\pi}\int_{0}^{\pi}f(x)\sin\;mx\;dx$

we note that

$ \int_{0}^{\pi}\sin\;mx\;dx \equal{} \frac {1}{m}(1 \minus{} \cos\;m\pi) \equal{} 0$ or $ \frac {2}{m}$

according as $ m$ is even or odd. Hence,

$ 1 \equal{} \frac {4}{\pi}\left(\frac {\sin\;x}{1} \plus{} \frac {\sin\;3x}{3} \plus{} \frac {\sin\;5x}{5} \plus{} \dots\right)$

Letting $ x \equal{} \frac {\pi}{4}$, we have

$ \frac {\pi}{4} \equal{} \left(\frac {\sqrt {2}}{2} \plus{} \frac {\sqrt {2}}{2}\cdot\frac {1}{3} \minus{} \frac {\sqrt {2}}{2}\cdot\frac {1}{5} \minus{} \frac {\sqrt {2}}{2}\cdot\frac {1}{7} \plus{} \dots\right) \equal{}$
$ \equal{} \frac {\sqrt {2}}{2}\cdot\left(1 \plus{} \frac {1}{3} \minus{} \frac {1}{5} \minus{} \frac {1}{7} \plus{} \dots\right)$

Other is simple...


Hey, this is a classic problem, but i don't seem to remember the solution; would be awesome if somebody could post it:

Does there exist a natural $k$ such that the four leftmost digits of $k!$ are 2003...?

Thank you for your time and effort

See http://www.mathlinks.ro/Forum/viewtopic.php?t=23417 (and http://www.mathlinks.ro/Forum/viewtopic.php?t=39366).


geometry
circumcircle
incenter
inradius
perimeter
geometry unsolved

Let $ABC$ be a triangle . $O, I, H$ are respectively the circumcenter , the incenter , the orthocenter of $\triangle ABC$ . Suppose $IO=IH=r$ . Prove that : among three angles of $\triangle ABC$ , there is exacly one angle equal $60$ degree .
  Erect $\triangle ABC$ when having $I , O$ ?

What do you mean by $r$? Is this any special value??? Inradius, for example??

For the triangle $ABC$ denote: the perimeter $P=2p=a+b+c$; the circumcircle $C(O,R)$; the incircle $C(I,r)$; the orthocentre $H$. The properties $(1),(2),(3)$ are wellkonown or you can prove them:

$1^{\circ}.\blacktriangleright\ OI^2=R^2-2Rr,\ IH^2=4R^2+4Rr+3r^2-p^2\ .$
$2^{\circ}.\blacktriangleright\ \cot \frac A2+\cot \frac B2+\cot \frac C2=\cot \frac A2\cot \frac B2\cot \frac C2=\frac pr\ ,$
$\cot \frac A2\cot \frac B2+\cot \frac B2\cot \frac C2+\cot \frac C2\cot \frac A2=\frac{4R+r}{r}$, i.e. the equation
$f(t)\equiv rt^3-pt^2+(4R+r)t-p=0$ has the roots $\left\{\ \cot \frac A2,\ \cot \frac B2,\ \cot \frac C2\ \right\}.$
$3^{\circ}.\blacktriangleright\ 60^{\circ}\in \{A,B,C\}\Longleftrightarrow f(\sqrt 3)=0\Longleftrightarrow p=(R+r)\sqrt 3\ .$

[hide="Here is the my solution."]$\underline {\underline {\mathrm {Proof}}}\ .\ IO=IH=r\Longleftrightarrow R^2=2Rr+r^2\ \wedge \ 4R^2+4Rr+3r^2-p^2=r^2\Longleftrightarrow$
$R^2=2Rr+r^2\ \wedge\ R^2-2Rr=4R^2+4Rr+3r^2-p^2\Longleftrightarrow$
$R^2=2Rr+r^2\ \wedge\ p^2=3(R+r)^2=3R^2+3(R^2-r^2)+3r^2\Longleftrightarrow$
$p=R\sqrt 6=(R+r)\sqrt 3$. From the property $3^{\circ}$ results:
$\boxed {\ IO=IH=r\Longleftrightarrow p=R\sqrt 6\ \wedge\ 60^{\circ}\in \{A,B,C\}\ }\ .$
The oneness prove easily because otherwise $A=B=C=60^{\circ}\Longleftrightarrow$
$I\equiv H\equiv O\Longrightarrow r=0$, what is false.
In a different way, $f'(\sqrt 3)\ne 0$. Indeed, $f'(\sqrt 3)=0$ $\Longrightarrow$
$9r-2p\sqrt 3+4R+r=0$ $\Longrightarrow$ $p\sqrt 3=5r+2R$ $\Longrightarrow$
$2R+5r=3(R+r)$ $\Longrightarrow R=2r\Longrightarrow I\equiv O\equiv H\Longrightarrow r=0$, what is false.

[b]Remark.[/b] Generally, $IO=IH\ne 0\Longleftrightarrow O\not \equiv I,\ R^2-2Rr=4R^2+4Rr+3r^2-p^2\Longleftrightarrow$
$O\not\equiv I,\ p^2=3(R+r)^2\Longleftrightarrow O\not \equiv I,\ p=(R+r)\sqrt 3\Longleftrightarrow$
$60^{\circ}\in \{A,B,C\}$ and is unique.[/hide]

Can you erect this triangle when knowing $O , I$ ?

If $\{A,A'\}\subset AI\cap C(O,R)$ then $A'B=A'C=A'I$ a.s.o.

But you don't know the position of $A'$ and the circumradius of $\triangle ABC$ , do you ?

Know we $A=60^{\circ}$, $r=IO$ and the positions of the centres $O$, $I$ ? Yes ?


Let $f(x,y,z)=g(r)$ where $r=\sqrt{x^{2}+y^{2}+z^{2}}$.

If
$\frac{\partial^{2}f}{\partial x^{2}}+\frac{\partial^{2}f}{\partial y^{2}}+\frac{\partial^{2}f}{\partial z^{2}}=0$

then find $f$ in terms of $r$. :)

This is the sphericallly symmetric solution to Laplace eq.
we have that
\[\frac{\partial f }{\partial x}=f'(r)\cdot\frac{\partial r }{\partial x}=f'(r)\cdot\frac{x }{r}\]
and
\[\frac{\partial^{2}f }{\partial x^{2}}=\frac{\partial}{\partial x}\left(f'(r)\cdot\frac{\partial r }{\partial x}\right)=f''(r)\frac{x}{r}\cdot\frac{x}{r}+f'(r)\cdot(\frac{\partial}{\partial x}\frac{x}{r}) \]

\[= f''(r)\frac{x^{2}}{r^{2}}+f'(r)\cdot(\frac{1}{r}-\frac{x}{r^{2}}\frac{x}{r})=f''(r)\frac{x^{2}}{r^{2}}+f'(r)\cdot(\frac{1}{r}-\frac{x^{2}}{r^{3}}) \]
Now if we sum up corresponding terms for $y$ and $z$ we end up with
an ODE
\[f''+f'\frac{2}{r}=0 \]
The integrating factor becomes $r^{2}$
\[(f'r^{2})'=0 \]
Thus we have the solution
$f(r)=\frac{C}{\sqrt{x^{2}+y^{2}+z^{2}}}+D=\frac{C}{r}+D$

It is beautiful  :)


geometry
inradius
inequalities proposed
inequalities

Let $ m_a,m_b,m_c$ the medians, $ r$ the inradius and $ r_a,r_b,r_c$ the exradii of an scalene triangle. Find the maximum value of $ k$

$ \sum\frac{m^2_a}{r_a}\geq kr$

[quote="Ligouras"]Let $ m_a,m_b,m_c$ the medians, $ r$ the inradius and $ r_a,r_b,r_c$ the exradii of an scalene triangle. Find the maximum value of $ k$

$ \sum\frac {m^2_a}{r_a}\geq kr$[/quote]
A computer program shows that $ k_{max}\equal{}9.$


Να εξετασετε αν η σειρα:

$ \sum_{n = 1}^{ + \infty}{\frac {1}{a^{n^2}}}$,
$ a \in N^{*} - \{1\}$

συγκλινει σε ρητο αριθμο.

Hehe, quick and cute :)

[hide]Ο αριθμός $ 0.1001000010000001\ldots$ (με $ 1$ στις θέσεις $ 1, 4, 9\ldots$ και $ 0$ παντού αλλού) είναι σίγουρα άρρητος. Hence... :)[/hide]

Cheerio,

Durandal 1707


Let P and Q be points on sides BC and CD of a square ABCD such that PQ is a tangent of the circle with the center A and radius AB. Segments AP and AQ intersect the diagonal BD at R and S. prove that points C, P, Q, R and S lay on a circle.

Let PQ be tangent to the circle at X. Drop the perpendicular from Q to PA; let it meet AP at R' and BC at T. Then, as angles APQ and APT are the same (equal tangents PX and PB), we see that right triangles PR'Q and PR'T are congruent. Now, angles QAX and XAP are half of DAX and XAB respectively, so we see that angle QAR' = QAP = QAX + XAP = (1/2)(DAX + XAB) = (1/2)(DAB), i.e. QAR' is 45 degrees. Thus, QAT is an isosceles right triangle, so a spiral similarity about A takes it to DAB. Thus, there also exists a spiral similarity taking ABT to ADQ, so as AB = AD, BT = DQ. Thus, as (QR'/R'T) * (TB/BC) * (CD/DQ) = 1 * (TB/DQ) * (CD/BC) = 1*-1*1 = -1, we have that B, D, and R' are collinear, or R' = R, so angle QRP is 90 and thus R, C, and analagously S lie on the circle with PQ as diameter.

Since $QT$, $QD$, $PT$, $PB$ are all tangants, so $\angle{DAQ}=\angle{TAQ}$, $\angle{TAP}=\angle{BAP}$.
So $\angle{PAQ}=45^{o}$

$\angle{PAQ}=45^{o}=\angle{PBS}$. So $P$, $B$, $A$ and $S$ are cyclic. Hence: $\angle{QSP}=90^{o}$.

Similarly we get: $\angle{QRP}=90^{o}$.

Therefore $C$, $Q$, $S$, $R$, $P$ lie on the same circle with diameter $PQ$.

I knew there must be some easy way  ;) couldn't find it though  :mad: oh well :P smileys are good :lol:

Dear Mathlinkers,
have a look at

http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20II.pdf  p. 30-31.

Sincerely
Jean-Louis


A large fraction of all posts to this forum in the last few weeks have been 100% content-free.

This forum already has a rule against "spam".  I have been tolerant because I know many of you have pre-contest jitters.  But the phenomenon is completely out of control.  Don't be surprised if your content-free posts disappear.

I understand, but "100%-content free," as mathematical as it sounds, is relative. For example, would you consider my post about "one week until the big finale" asking people about their feelings now that we've come down to the final stretch to be "100%-content free?"

A question about feelings prior to the big MATHCOUNTS competition is a legitimate topic for this Forum.  I don't intend to remove it.  A response that said "toasted" would be fine.  A series of 3 consecutive responses that said "toasted", "toasted", "toasted" or "lightly", "browned", "toast" would probably be removed.

I removed about ten posts, yesterday.  Unless you like off-topic, content-free spam, I don't think you'll miss them.


Hello. For the new year, I have decided to have an AoPS Countdown that I will display the results to on January 1st.

Directions;

I am going to do a countdown for AoPS polls, AoPS books, AoPS users, AoPS games, AoPS mafia games, AoPS mafia players, AoPS sections of the forum, AoPS threads, and AoPS fun factory games.

So for the AoPS polls, you should pm me as many AoPS polls as you can in the order that you like them, so your number one favorite would be at the top of the list,... etc.

For AoPS books, look in the bookstore, and pick your top five favorites, pm me the order of these five in the order that you like them.

For AoPS users, pick as many AoPS users as you can and list them out in the order that you like them, and then pm me the list. (you may put yourself on the list, but it must be last on the list, to make it fair)

For AoPS games, this means games like the Mole, or Big Brother, or Survivor, or Elect the Executioner, or AoPS City... etc. This is an example of a list you might pm me;
1. Mole Season 1
2. Big Brother Season 2
3. Mole Season 3
...etc. You can put how many games on the list as you want.

For AoPS mafia games, look through the mafia section and glance at some of the games. You might rate them based on how many players joined, how active the players were, how long it was... etc. You may put all of them in your list, but put them in the order you like them.

For AoPS mafia players, look through the mafia section again, and see what players did well, or played a lot, or even people that stood out to you.

For AoPS sections of the forum, you must put them in the order that you like them, and you can include as many as you want. This would be a decent example;
1. MathCounts
2. Intermediate Topics
3. Highschool Basics
...etc.

For AoPS threads, you can include as many as you want, and they must be in the right order.

For AoPS fun factory games, you should include games like Corrupt a Wish, Public Shaming, Vending Machine, Ctrl+V, Point game...etc.

Please pm me as many as you can before January. Thank you.  :)

Alright, I will include AoPS Blogs. Same rules as the other ones.

Please don't make random posts like that. It's just spam.


conics
hyperbola
function
linear algebra
matrix
induction
superior algebra

(1)
Consider the hyperbola C={M(x,y) in R 2 |
x 2 /4 - y 2 = 1} and a conic C' which has no common points with C. Let n(C, C') the maximum number of pairs of points (A, A') in C x C' for which AA'<=BB' for any (B, B') in C x C'. For every p in {0, 1, 2, 4} write the equation of a conic C' for which n(C, C')=p. Justify your answer.

(2)
Let f:R->R be a function which has a limit in any real point and has no local extreme points. Show that:

a)f is continuous.
b)f is strictly monotonous.

(3)
A is in M_4(C), A=/=0_4.

a) if rank A=r<4 show that there exist U, V in M_4(C), invertible, s.t.
UAV=
(I_r 0), where I_r is the unit matrix of order r.
(0    0)
b) Prove that if A and A  2 have the same rank k, then the matrix A n has rank k for all n>=3.

(4)
f:[0,1]->[0,1] is continuous and bijective. Determine the set A={f(x)-f(y) | x, y are in [0,1]\Q}.

(Students may use the result: there are no injective functions from [a,]\Q to Q)

[quote="grobber"]
(3)
A is in M_4(C), A=/=0_4.

a) if rank A=r<4 show that there exist U, V in M_4(C), invertible, s.t.
UAV=
(I_r 0), where I_r is the unit matrix of order r.
(0    0)
b) Prove that if A and A  2 have the same rank k, then the matrix A n has rank k for all n>=3.
[/quote]

a) Can be find in text book in the  rank
b)
If rgA=rgA^2 with
dim(_C)C^4 = rgA + dimkerA=rgA^2 + dimkerA^2 => dimkerA=dimKerA^2 but kerA is included in kerA^2 so
kerA=kerA^2

(*) let x \in kerA^3=> A^3(x)=0 => A(x)\in kerA^2=kerA=>A^2(x)=0
this prove kerA^3 is included in kerA^2 but kerA^2 is included in kerA^3
so kerA^2=kerA^3  => rgA^2=rgA^3

We apply the same idea with  rgA^2=rgA^3 => kerA^2=kerA^3=>
working like in (*) with x \in kerA^4 => A(x) \in  kerA^3=kerA^2=>
A^3(x)=0 so kerA^4=kerA^3 we deduce ImA^4=ImA^3 and
rgA^3=rgA^4

By induction  if rgA=rgA^2 then for any n \in N n \geq 2 rgA^n=rgA^2


A team won 20 of its first 50 games. How many of the remaining 40 games must this team win in order to win exactly $ 50\%$ of its games for the season?

A total of 50+40=90 games will be played, so to get exactly 50% of the wins, you must win 45. Since this team already won 20, the team must win 45-20=25 of the remaining games, or 25/40 or 5/8 of the remaining 40 games.

Well, we want the number of games to be won, not the fraction, so $ \boxed{25}$ is the final answer.


calculus
integration
quadratics
Euler
geometry
geometric transformation
reflection

If you want to keep it in real terms, here's the appropriate factorization:

$x^4+1=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}+1).$

Using that - partial fractions, completing the square in the irreducible quadratics - it's all do-able but it's going to get ugly and tedious. You'd have a little more notational freedom if you were to use complex numbers.

And I'd much rather compute the definite integral

$\int_{-\infty}^{\infty}\frac{dx}{x^4+1}$

than try to find the antiderivative.

I had to do that second one before.  It took like 3 pages of tedious work (it's not really difficult, just very long).  So I guess if you don't [i]have[/i] to do it, it might not be worth it.

And I thinks Euler Reflection formula can be used here
\[ \int_0^{\infty} \frac{s^{p-1}}{s+1} \, ds = \frac{\pi}{\sin p \pi} \]

the second integral(definite integral) can be reduced to that form.

ok for the $\int{\frac{1}{x^4+1}}$ sub in $x^2=tan\theta$ and u will get that ${dx=\frac{sec^2\theta}{2\sqrt{tan\theta}}}d\theta$

this produces $\frac{1}{2}\int{\sqrt{cot\theta}}=\frac{1}{2}\int{\frac{\sqrt{cos\theta}}{\sqrt{sin\theta}}d\theta}$

then you sub $u=\sqrt{sin\theta}$ and $du=\frac{cos\theta}{2\sqrt{sin\theta}}d\theta=\frac{1-u^2}{2u}$

so you get $\frac{1}{2}\int{\frac{(\sqrt{1-u^2})(1-u^2)}{2u}}du$

i assume that you can solve this last integral....once you have i leave it to you to work back through all of the substitution...have fun.  :rotfl:
i know it will be ugly


geometry
3D geometry
tetrahedron
IMO Shortlist
combinatorics

Let $ P$ be a point inside a regular tetrahedron $ T$ of unit volume. The four planes passing through $ P$ and parallel to the faces of $ T$ partition $ T$ into 14 pieces. Let $ f(P)$ be the joint volume of those pieces that are neither a tetrahedron nor a parallelepiped (i.e., pieces adjacent to an edge but not to a vertex). Find the exact bounds for $ f(P)$ as $ P$ varies over $ T.$

Positive numbers $a,b,c,d$ sum to $1$.
We need to find the range of $1-(a^3+b^3+c^3+d^3+6(abc+bcd+cda+dab))$


This is a cute one...

Let A be the sum of digits for 4444^4444, B the sum of digits for A and C the sum of digits for B. We are looking for C. We are just going to look for the upper bounds for A,B,C.
we have that 4444^4444<10000^4444<10000^5000<10^20000. So the number has no more than 20000 digits and A cannot be bigger than 9*20000<200000, so A<=200000 or A has less than 6 digits. SO B<=9*6=54 so we get that C<=13.

on the other hand we know that the rest of a division with 9 is the same for a division on the sum of the digits.
SO all numbers 4444^4444, A,B, and C have the same rest on the division with 9, which is 7. 4444^4444=(4444^4444-7^4444)+7(7^(3*1481-1)+7 so we have that for every positive number n (a-b) | (a^n-b^n).
IN conclusion 4444-7=9*493 and 7^3-1=9*38 which are divisible with 9. So C<=13, if is divided by 9 has the rest 7, which implies C=7.

cheers!

Yeah, I like it too. That's more or less my soln too. I found these problems in Cristinel Mortici's book "600 problems" (he's a Romanian teacher, for those who don't know that).

This is IMO 1975/4 I think. (I'm not sure about the year.)


farz konin x,y,z>0. va darim $ x \plus{} y \plus{} z \equal{} 3$
sabet konid

$ \sum \frac {x^3}{y^3 \plus{} 8} \geq \frac {1}{9} \plus{} \frac {2}{27} \times(xy \plus{} yz \plus{} zx)$

Ebteda leme zir ke esbatesh ba shoma
\[ (a^2 \plus{} b^2 \plus{} c^2)^2\ge 3(a^3b \plus{} b^3c \plus{} c^3a)
\]
baraye hale soal ebteda sabet mikonim
\[ \sum_{cyc}\frac {x^3}{y^3 \plus{} 8}\ge \frac {1}{3}
\]
Ba cauchy-schwarz darim
\[ \sum_{cyc}\frac {x^3}{y^3 \plus{} 8} \equal{} \sum_{cyc}\frac {x^4}{y^3x \plus{} 8x}\ge \frac {(x^2 \plus{} y^2 \plus{} z^2)^2}{\sum_{cyc}x^3y \plus{} 24}
\]
pas ba estefade az lem kafie sabet konim
\[ (x^2 \plus{} y^2 \plus{} z^2)^2\ge 9
\]
ke badihiye bagiyash ham assone :wink:

In ba AM-GM e tanha ham sabet mishe kafie tavajoh konin $ y^3\plus{}8\equal{}(y\plus{}2)(y^2\minus{}2y\plus{}4)$
hala baraye $ \frac{x^3}{y^3\plus{}8}$ , $ \frac{y\plus{}2}{27}$ va $ \frac{y^2\minus{}2y\plus{}4}{27}$ AM-GM bezanin .
in karo baraye dota kasre dige ham bokonin bad age jam konin daghighan namosavie masale bedast miad .

also we have:
$ \frac {x^3}{y^3 \plus{} 8} \plus{} \frac {y^3}{z^3 \plus{} 8} \plus{} \frac {z^3}{x^3 \plus{} 8}\geq\frac {1}{3}\geq\frac {1}{9} \plus{} \frac2{27}(xy \plus{} xz \plus{} yz)$
$ \sum \frac {x^3}{y^3 \plus{} 8} \geq \frac {{(x^3 \plus{} y^3 \plus{} z^3)}^2}{(x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3) \plus{} 8(x^3 \plus{} y^3 \plus{} z^3)}\geq\frac {{(x^3 \plus{} y^3 \plus{} z^3)}^2}{\frac {1}{3}{(x^3 \plus{} y^3 \plus{} z^3)}^2 \plus{} 8(x^3 \plus{} y^3 \plus{} z^3)}\geq \frac {1}{3}\longleftrightarrow(x^3 \plus{} y^3 \plus{} z^3)\geq3$.
 you can get the first inequality by couchy or generalized jensen(convexity of $ \frac {1}{x}$)( albate hamarzan)

vasc inequality:$ (a^2 \plus{} b^2 \plus{} c^2)^2\geq3(a^3b \plus{} b^3c \plus{} c^3a)$

In namosavi ye halat tasavi dare ke kheyli nabadihio bad daste . bekhatere hamin fek konam ye khorde ba Cauchy o makhsoosan AM-GM roosh fek kardan kare sakhtie .


Got back from the NEAML contest last night late.  Pretty tough contest this year, but there were some pretty nice problems to check out.  Only one perfect score this year-way less than the flood in some years.  And let's just say the beginning of the meet started out-interestingly.
Anyone here there? :D

We ended up with 9th place out of the large schools (we were the only team from Vermont btw).  Same as last year actually.  We could have been higher but in the team round one kid on one side of the table figured out the first two and never told anyone, so we only ended up getting 2/6 on that test. :blush: I ended up with a 16, which I was happy about but I didn't get stuck doing one of the really hard tests. :D
Maybe next year we can do a little better on team-that has always been our weak spot.  I don't know, we are just disorganised I guess.  And our league in Vermont has completely different rules, about timing for the tests and number of questions, etc, so we have to re-learn some stuff when we go down(not that that is much of a problem, but still, it certainly doesn't help things).  Hopefully we'll be able to come back next year.

I was there.

I was the kid who just stood at the side and who looked like he was lost when all the other people who got 13 were looking through the books. Yeah, that was me.

CRH, first medium.

What book did you pick?  I was going to get one of the programming ones but then I realised I din't really need it so I got the discrete mathematics one instead... it'll probably sit on a shelf for a few years and then I'll find it again someday.

i took a look at the discrete math book, and it looked kind of easy, just like the other books.  but hey its a prize (they didnt have to give anything), so cant complain.  i got the calculus book.


The following is an incomplete list of local math competitions. If any of you are aware of other competitions in the state, then feel free to add to the list.

GPML (Great Plains Math League)
Forest Park Excellence in Mathematics
Missouri Math League
Florissant Valley Excellence in Mathematics (grades 7-9)
MCTM (Missouri Council of Teachers of Mathematics), although I believe this has now merged with GPML or some other organization.

Of course, various national contests are also available in Missouri-

AMC/AIME/USAMO
ARML (Missouri has an ARML team with one branch in St. Louis under Rick Armstrong in St. Louis, and another branch operating out of Springfield)
Mandelbrot
Mu Alphta Theta/National Assessment and Testing contests

I hope you don't mind my posting in a two-year-old topic :blush:

I'm looking at a math team sheet and apparently there's also one that takes place in Warsaw.

Wow. That's scary. Another 2 years now. And also, the Flo Valley one now has 6th graders participate too. But this is an ancient thread...

Any of these events for individual, ie. not school registration? I live in KS right across state-line, but my school isn't exactly mathy, so school team events are hard to do. I already do MO ARML and kinda do GPML.

BTW, KU has a competition usually in April, that might not be too far from some MO people. There's some prize money in for it, $100 for first, 75 for second, 50 for third:
http://www.math.ku.edu/info/info-mathawareness.html

And KSU also has one, the manhattan math olympiad, but that's probably too far away for most of you:
http://www.k-state.edu/media/newsreleases/march08/olymath31408.html

BTW does anyone do this?

http://www.dehn.wustl.edu/~blake/circles/

Washington Uni Math Circle.

[quote="math154"]No, in Missouri it's usually advisable to stay home. :wink:  :P[/quote]

Haha I see  :P

Wait why is it advisable to stay home? (Mod edit: If mewto is out and about, then why do you think you [i]should stay home?[/i])

Oh, I forgot, the STD rate is quite high.

That's why all you guys should get your HPV vaccines.


function
analytic geometry
graphing lines
slope
real analysis
real analysis unsolved

Is this a sufficient sort of argument for the continuity of convex functions on open intervals?

If the function isn't continuous somewhere, say at a point a in the interval, we have either, f(x) > f(a)+e  or f(x)<f(a)-e in every neighborhood of a,  where e is some positive number. Lets say the former is the case;

given that a<b<c,  we have (due to convexity)

[f(b)-f(a)]/(b-a) < [f(c)-f(a)]/(c-a)

But we can choose b arbitrarly close to a with f(b)>f(a)+e, which means

e/(b-a) < [f(c)-f(a)]/(c-a)

but this is absurd, considering that the expression e/(b-a) is not bounded by any number as we let b approach a.

Your proof is correct, but not complete. You still have to eliminate the possibility $\liminf_{x\to a}f(x)<f(a)$, which requires a different argument.

Here's a slightly different proof of the part that you already have: given a point $a$, choose $b<a$ so that $b$ lies in the interval. For any $x\in (b,a)$ we have $f(x)\le \frac{f(a)(x-b)+f(b)(a-x)}{a-b}$ by convexity. Letting $x\to a$ we obtain $\limsup_{x\to a-}f(x)\le f(a)$. For the same reason $\limsup_{x\to a+}f(x)\le f(a)$.

I'm not familiar with the "lim sup"  and "lim inf" notation; how do they differ from usual limits? (though I DO know what least upper bounds and greatest lower bounds are)

The forget about $\limsup$ and $\liminf$. My point was that you have identified two ways for $f$ to be discontinuous:
[quote="alyosha"]we have either, f(x) > f(a)+e  or f(x)<f(a)-e in every neighborhood of a,  where e is some positive number.[/quote]
and your proof rules out only one of them.

Yes; the purpose of the post was check the general idea rather than present a complete proof.

I see. The second case requires a different idea, though.

If we have that f(x)<f(b)-e for x arbitrarily close to b,   (I've switched letters to preserve alphabetical order)

and a<b<c, we have

[f(b)-f(a)]/(b-a)< [f(c)-f(a)]/(c-a)

but again we can find a close to b  with f(a)<f(b)-e , and

e/(b-a) < [f(c)-f(a)]/(c-a)

with an analogous argument as before; how different of an argument did you mean?

Well, the second case is not very different, except that it is the only place where you use the assumption that $f$ is defined on an open interval. The first part worked for closed intervals as well.

If you're working in an open interval $(a,b)$, pick points $c,d$ and $e,f$ such that

$(e,f)\subset [c,d]\subset (a,b)$ (normal spaces are fun).

Using the fact that slopes of convex functions are increasing and bounding the slopes between $c,e$ and $f,d$, you can show that your convex function $\varphi$ actually satisfies a Lipschitz condition on $(e,f)$, which is certainly enough for continuity.

Of course, there is no hope of extending the Lipschitz condition to hold on all of $(a,b)$ (consider the bottom half of a circle).


IMO Shortlist
binomial coefficients
Divisibility
number theory
Poland

For every integer $ k \geq 2,$ prove that $ 2^{3k}$ divides the number
\[ \binom{2^{k \plus{} 1}}{2^{k}} \minus{} \binom{2^{k}}{2^{k \minus{} 1}}
\]
but $ 2^{3k \plus{} 1}$ does not.

[i]Author: Waldemar Pompe, Poland[/i]

[quote="orl"]For every integer $ k \geq 2,$ prove that $ 2^{3k}$ divides the number
\[ \binom{2^{k + 1}}{2^{k}} - \binom{2^{k}}{2^{k - 1}}
\]
but $ 2^{3k + 1}$ does not.

[i]Author: unknown author, Poland[/i][/quote]

It is from Russian journal [i]Qvant - 1970-1975[/i].

$ p^{3n + t_p}|\binom{p^{n + 1}}{p^n} - \binom{p^n}{p^{n - 1}}$.
Where $ t_2 = 0$, $t_3 = 1$, $t_p = 2$ if $p$ is not a Wolstenholme's prime. If $p$ is a Wolstenholm's prime, then $ t_p\ge 3$. For 2 known Wolstenholme's primes, $ t_p = 3$.

In email exchange Rustem says the problem is based on the article by Fuchs, D. B. and Fuchs., M.B: Arithmetic of binomial coefficients, Kvant, #6,1970 whose Russian Фукс Д., Фукс М. Фрифметика биномиальных коэффициентов. Квант №6, 1970. which can be found at http://kvant.mirror1.mccme.ru/1970/06/arifmetika_binomialnyh_koeffic.htm.

[quote="Rust"]It is not hard to prove:
$ v_p(\binom{pc}{pa} \minus{} \binom{c}{a}) \equal{} 2 \plus{} t_p \plus{} v_p(c) \plus{} v_p(a) \plus{} v_p(b),b \equal{} c \minus{} a$.
In these case $ c \equal{} p^n,a \equal{} p^{n \minus{} 1}$.[/quote]
Hey, can anyone furnish a proof of this proposition? (BTW I can't even understand the definition of $t_p$ given in post #2...)

[hide="P.S."]
Very nice proofs, Zhero and math154, I got the same :) [/hide]

[quote="Rust"]It is not hard to prove:
$ v_p(\binom{pc}{pa} \minus{} \binom{c}{a}) \equal{} 2 \plus{} t_p \plus{} v_p(c) \plus{} v_p(a) \plus{} v_p(b),b \equal{} c \minus{} a$.
In these case $ c \equal{} p^n,a \equal{} p^{n \minus{} 1}$.[/quote]
I don't think the solving this lemma is simple! Can anyone solve it? :blush:  :blush:

[quote="dinoboy"]I think Rust's statement is wrong :?:

$\dbinom{14}{8} - \dbinom{7}{4} = 2968$, and $v_2(2968) = 3$. However, his formula says it's should be $4$.[/quote]
But i was taking $c=p^n,a=p^{n-1}$. I think this lemma is true in this case!

Just an useful step towards the proof of what Rust claimed is that $v_p\binom{pc}{pa}=v_p\binom{c}{a}$.

I am giving the proof(though its easy to prove):
Let $a=(a_1a_2 \cdots a_k)_p,c-a=(c_1c_2 \cdots c_k)_p$,Then clearly we have $pa=(a_1a_2 \cdots a_k0)_p$ and $pc-pa=(c_1c_2 \cdots c_k0)_p$ so the result follows directly from Kummer's theorem.

BTW Rust can you tell me what is $t_p$ here?I think it would be of some help.

I have a long proof. It is equivalent to proving $v_2\left( \frac{2^{k+1}!}{2^k!} - \left( \frac{2^k!}{2^{k-1}!} \right)^2 \right) = 3k+2^k-1$, which is equivalent to proving $v_2((2^{k+1})!!/(2^k)!! - 2^k!!) = 3k-1$, next reduce it to proving $v_2\left( \sum 1/i + 2^k\left( \sum 1/ij \right) \right)=2k-1$ where the sums run over $i \neq j$ odd numbers $\le 2^k-1$. Now use squaring identity to reduce it to $v_2\left( \sum_{i \text{ odd}}^{2^k-1} \frac{1}{i} - \sum_{i \text{ odd}}^{2^k-1} \frac{2^{k-1}}{i^2} \right) = 2k-1$ which is equivalent to

$\left( \sum_{i \text{ odd}}^{2^{k-1}-1} \left( \frac{1}{i} +\frac{1}{(2^k-i)} - \frac{2^{k-1}}{i^2}- \frac{2^{k-1}}{(2^k-i)^2} \right) \right) = 2^{2k-1} (\text{mod }2^{2k})$

which is equivalent to $\sum_{i=1}^{2^{k-1}-1} \frac{i^2}{i^4} = 2^{k-2} (\text{mod } 2^{k-1})$, which can be done immediately by induction on $k$.

First, note that for all positive integers $n$, we have\[v_2\left[\binom{2^{n+1}}{2^n}\right] = v_2[2^{n+1}!] - 2v_2[2^n!] = (2^{n + 1} - 1) - 2(2^n - 1) = 1\]for all positive integers $n$ by Legendre. So it remains to show that\[v_2\left[\frac{\binom{2^{n+1}}{2^n}}{\binom{2^{n}}{2^{n-1}}} - 1\right] = v_2[N] - v_2\left[\binom{2^{n}}{2^{n-1}}\right] = 3n-1.\]Using the property that $2k! = k! \times 2^k \times (2k-1)!!$ for all positive integers $k$ where we let $m = 2^{n-1}$, we may rewrite\begin{align*}
    \frac{\binom{2^{n+1}}{2^n}}{\binom{2^{n}}{2^{n-1}}} - 1&= \frac{(4m)!}{(2m)!^2} \times \frac{m!^2}{(2m)!} - 1\\&= \frac{(2m)! \times 2^{2m} \times (4m - 1)!!}{(2m)!^2} \times \frac{m!^2}{m! \times 2^m \times (2m - 1)!!} - 1\\&= \frac{2^m \times m!}{(2m)!} \times \frac{(4m-1)!!}{(2m-1)!!} - 1\\&= \frac{(4m - 1)!!}{(2m - 1)!!^2} - 1\\&= \frac{(2^{n + 1} - 1)\ldots (2^n + 1)}{(2^n - 1)!!} - 1\\&= \frac{(2^{n + 1} - 1)\ldots (2^n + 1) - (2^n - 1)!!}{(2^n - 1)!!}
\end{align*}
and since the denominator is odd, it suffices to show that $v_2[(2^{n + 1} - 1)\ldots (2^n + 1) - (2^n - 1)!!] = 3n - 1$. Define the polynomial $P(x) = (x + 1)(x + 3)\ldots (x + (2^n - 1)) - (x - 1)(x - 3)\ldots (x - (2^n - 1))$. Note that $P(2^n)$ is the expression in question so we wish to prove that $v_p[P(2^n)] = 3n-1$. Note that $P$ is odd since even degree terms cancel out. Clearly, we need only prove that $v_2[C \times 2^n] = 3n - 1 \implies v_2[C] = 2n - 1$ where $C$ is the coefficient of the $x$ term since all other terms have degree $\geq 3$ and hence $v_2$ of them is $\geq 3n$. Upon a little bit of expansion we see that\[C = 2\times (2n-1)!! \times \left(\frac11 + \frac13 + \ldots + \frac{1}{2^n - 1}\right)\]so it remains to show that $v_2\left[\frac11 + \frac13 + \ldots + \frac{1}{2^n - 1}\right] = 2n - 2$. We can pair terms whose denominators add to $2^n$ as follows:\[\frac11 + \frac13 + \ldots + \frac{1}{2^n - 1} = 2^n \left(\frac{1}{1 \times (2^n - 1)} + \ldots + \frac{1}{(2^{n-1} - 1) \times (2^{n - 1} + 1)}\right) = 2^{n-1}\left(\frac{1}{1 \times (2^n - 1)} + \ldots + \frac{1}{(2^n - 1) \times 1}\right)\]so it suffices to show $v_2\left[\frac{1}{1 \times (2^n - 1)} + \ldots + \frac{1}{(2^n - 1) \times 1}\right] = n - 1 \iff \frac{1}{1 \times (2^n - 1)} + \ldots + \frac{1}{(2^n - 1) \times 1} \equiv 2^{n-1} \pmod{2^n}$. Rewrite this as\[\sum_{\text{odd } k = 1}^{2^n - 1} \frac{1}{k \times(2^n - k)} \equiv \sum_{\text{odd } k = 1}^{2^n - 1} -(k^{-1})^2 \equiv -(1^2 + 3^2 + \ldots + (2^n - 1)^2) \pmod{2^n}\]Upon evaluation using sum of consecutive squares we get $\tfrac{(2^n - 1)(2^n)(2^{n+1} - 1)}{6} - \tfrac{4(2^{n-1} - 1)(2^{n-1})(2^n - 1)}{6} = 2^{n-1} \times \text{odd}$ hence indeed the expression is $2^{n-1}$ modulo $2^n$ so we are done. $\blacksquare$

Let $O$ be the set of odd integers. Note that
\[\binom{2^{k}}{2^{k - 1}} = 2\prod_{i=1}^{k-1}\prod_{2^i < m < 2^{i+1},m\in O}m^{i-k+2}\]
and
\[\binom{2^{k+1}}{2^{k}} = 2\prod_{i=1}^{k}\prod_{2^i < m < 2^{i+1},m\in O}m^{i-k+1} = \binom{2^{k}}{2^{k - 1}} \cdot \frac{\displaystyle\prod_{2^k<m<2^{k+1},m\in O}m}{\displaystyle\prod_{m<2^k,m \in O}m},\]
so it suffices to show that the $2$-adic value of
\[\displaystyle\prod_{2^k<m<2^{k+1},m\in O}m-\displaystyle\prod_{m<2^k,m \in O}m\]
is $3k-1$. Some computation gives that
\[\displaystyle\prod_{2^k<m<2^{k+1},m\in O}m-\displaystyle\prod_{m<2^k,m \in O}m=\displaystyle\prod_{0 \le m <2^{k-1}}(2m+1+2^k)-\displaystyle\prod_{0 \le m <2^{k-1}}(2m+1) =\]
\[2^{3k}A + 2^{2k}\sum_{0\le m<n<2^{k-1}}\frac{1}{(2m+1)(2n+1)}\displaystyle\prod_{0 \le m <2^{k-1}}(2m+1)+2^{k}\sum_{0\le m<2^{k-1}}\frac{1}{2m+1}\displaystyle\prod_{0 \le m <2^{k-1}}(2m+1)\]
for some integer $A$.
It is enough to show that the $2$-adic value of
\[2^{k}\sum_{0\le m< n<2^{k-1}}\frac{1}{(2m+1)(2n+1)}+\sum_{0\le m<2^{k-1}}\frac{1}{2m+1}\]
is $2k-1$. Now, let
\[\sum_{0\le m<2^{k-1}}\frac{1}{2m+1} = S.\]
Remark that by PFTB 3.11, $\nu_2(S)=2k-2$.
Then, it is sufficient to show that
\[2^k\sum_{0\le m< n<2^{k-1}}\frac{1}{(2m+1)(2n+1)} \equiv S\pmod{2^{2k}}.\]
Observe that
\[2^{k+1}\sum_{0\le m\le n<2^{k-1}}\frac{1}{(2m+1)(2n+1)}\equiv 2^kS^2 - 2^k\sum_{0\le m<2^{k-1}}\frac{1}{(2m+1)^2}\pmod{2^{2k+1}}.\]
As $\nu_2(2^kS^2) = k+2(2k-2)=5k-4>2k+1$, it suffices to demonstrate
\[\sum_{0\le m<2^{k-1}}\frac{2^k}{(2m+1)(2^k-2m-1)} \equiv 2S\equiv -2^k\sum_{0\le m<2^{k-1}}\frac{1}{(2m+1)^2}\pmod{2^{2k+1}}.\]
Equivalently, it suffices to demonstrate
\[\sum_{0\le m<2^{k-1}} \frac{1}{2m+1}\left(\frac{1}{2^k-2m-1}+\frac{1}{2m+1}\right)\equiv 0\pmod{2^{k+1}}.\]
Alternatively, it suffices to demonstrate
\[\sum_{0\le m<2^{k-1}}\frac{1}{(2m+1)(2^k-2m-1)(2m+1)}\equiv 0\pmod{2}.\]
This final result is trivial because the sum is simply $2^{k-1}\equiv 0\pmod{2}$, so we are done.

Let $2^{k-1}=n$ and let the given expression be $N.$ Note that for any $m,$ we have $(2m)!=2^m\cdot m!\cdot (2m-1)!!.$ It follows that \[N=\binom{4n}{2n}-\binom{2n}{n}=\binom{2n}{n}\Bigg(\binom{4n}{2n}\binom{2n}{n}^{-1}-1\Bigg)=\]\[=\binom{2n}{n}\Bigg(\frac{2^{2n}\cdot (2n)!\cdot (4n-1)!!}{2^n\cdot n!\cdot (2n-1)!!}\cdot\bigg(\frac{2^n\cdot n!\cdot (2n-1)!!}{n!}\bigg)^{-1} -1\Bigg)=\]\[=\binom{2n}{n}\Bigg(\frac{(4n-1)!!}{(2n-1)!!^2}\cdot\frac{(2n)!}{2^n\cdot n!\cdot (2n-1)!!}-1\Bigg)=\binom{2n}{n}\Bigg(\frac{(4n-1)!!}{(2n-1)!!^2}-1\Bigg).\]We want to compute $\nu_2\big(N\big).$ Begin by observing that since $\nu_2(m!)=m-s_2(m)$ then because $n=2^{k-1}$ we have \[\nu_2\Bigg(\binom{2n}{n}\Bigg)=\big(2^k-s_2(s^k)\big)-2\big(2^{k-1}-s_2(2^{k-1})\big)=1.\]Therefore, to show that $\nu_2\big( N\big)=3k$ it only remains to prove the folliowing assertion \[3k-1=\nu_2\Bigg(\frac{(4n-1)!!}{(2n-1)!!^2}-1\Bigg)=\nu_2\big((4n-1)!!-(2n-1)!!^2\big)\iff\]\[\iff 3k-1=\nu_2\big((2n+1)(2n+3)\cdots (4n-1)-(2n-1)!!\big).\]Recall that $n=2^{k-1}$ and notice that $(2n+1)(2n+3)\cdots (4n-1)-(2n-1)!!$ can be rewritten as  such: \[\prod_{i=1}^{2^{k-1}}\big(2^k+(2i-1)\big)-\prod_{i=1}^{2^{k-1}}\big(2^k-(2i-1)\big)=2^{3k}\cdot C_0+2^{k}\cdot C_1\]for some constants $C_0$ and $C_1.$ But $C_1$ is the coefficient of $X$ in the above expression viewed as a polynomial. Hence $C_1$ can be computed to \[C_1=2\cdot (2n-1)!!\cdot\bigg(\frac{1}{1}+\frac{1}{3}+\cdots+\frac{1}{2^k-1}\bigg).\]Now, to show that $\nu_2\big(2^{3k}\cdot C_0+2^k\cdot C_1\big)=3k-1$ we only have to show that $\nu_2(C_1)=2k-1.$ Or, in other words, we want to show that \[\nu_2\bigg(\frac{1}{1}+\frac{1}{3}+\cdots+\frac{1}{2^k-1}\bigg)=2k-2.\]This can be easily be seen by pairing the terms $1/i$ and $1/(2^k-i),$ summing, and analysing modulo $2^k.$

Replace $k$ with $n$, just because. By Kummer's theorem we have $\nu_2\left(\binom{2^{i+1}}{2^i}\right)=1$ for all $i$. Hence by dividing out all powers of $2$ it suffices to show that the $\nu_2$ of
$$\frac{(2^1-1)!!\ldots(2^{n+1}-1)!!}{((2^1-1)!!\ldots(2^n-1)!!)^2}-\frac{(2^1-1)!!\ldots(2^n-1)!!}{((2^1-1)!!\ldots(2^{n-1}-1)!!)^2}=\frac{(2^{n+1}-1)!!}{(2^1-1)!!\ldots(2^n-1)!!}-\frac{(2^n-1)!!}{(2^1-1)!!\ldots(2^{n-1}-1)!!},$$
where we count individual terms in the product by their $\nu_2$. Since all of these terms are odd, by multiplying through it suffices to show that the $\nu_2$ of $(2^{n+1}-1)!!-((2^n-1)!!)^2$ is $3n-1$, which is equivalent to
$$\nu_2((2^n+(2^n-1))(2^n+(2^n-3))\ldots(2^n+1)-(2^n-1)(2^n-3)\ldots(1))=3n-1.$$

Considering the expansion of this product as a "polynomial in $2^n$" and then dividing by $(2^n-1)!!$, this is equivalent to saying that
$$2^n\left(\sum_{\substack{1 \leq i \leq 2^n\\i\text{ odd}}} \frac{1}{i}\right)+2^{2n}\left(\sum_{\substack{1 \leq i<j<2^n\\i,j\text{ odd}}} \frac{1}{ij}\right) \equiv 2^{3n-1} \pmod{2^{3n}} \iff \sum_{\substack{1 \leq i \leq 2^n\\i\text{ odd}}} \frac{1}{i}+2^{n-1}\left(\sum_{\substack{1 \leq i \leq 2^n\\i\text{ odd}}} \frac{1}{i}\right)^2-2^{n-1}\sum_{\substack{1 \leq i \leq 2^n\\i\text{ odd}}} \frac{1}{i^2} \equiv 2^{2n-1} \pmod{2^{2n}}.$$
We now turn to evaluating
$$\sum_{\substack{1 \leq i \leq 2^{n-1}\\i\text{ odd}}} \frac{1}{i^2} \pmod{2^n}.$$
Since $x^2=(-x)^2$, this quantity is
$$\sum_{\substack{1 \leq i \leq 2^{n-1}\\i\text{ odd}}} i^2=\sum_{j=1}^{2^{n-2}} (2j-1)^2=\sum_{j=1}^{2^{n-2}} (4j^2-4j+1)=\frac{2(2^{n-2})(2^{n-2}+1)(2^{n-1}+1)}{3}-2(2^{n-2})(2^{n-2}+1)+2^{n-2}=2^{n-1}(2^{n-2}+1)\left(\frac{2^{n-1}-2}{3}\right)+2^{n-2} \equiv 2^{n-2} \pmod{2^n}.$$
Thus,
\begin{align*}
\sum_{\substack{1 \leq i \leq 2^n\\i\text{ odd}}} \frac{1}{i}=\sum_{\substack{1 \leq i \leq 2^{n-1}\\i\text{ odd}}} \left(\frac{1}{i}+\frac{1}{2^n-i}\right)=\sum_{\substack{1 \leq i \leq 2^{n-1}\\i\text{ odd}}} -\frac{2^n}{i^2-2^ni}&=-2^{2n-2} \pmod{2^{2n}}\\
2^n \mid \sum_{\substack{1 \leq i \leq 2^n\\i\text{ odd}}} \implies 2^{n-1}\left(\sum_{\substack{1 \leq i \leq 2^n\\i\text{ odd}}} \frac{1}{i}\right)^2&\equiv 0 \pmod{2^{2n}}\\
-2^{n-1}\sum_{\substack{1 \leq i \leq 2^n\\i\text{ odd}}} \frac{1}{i^2} \equiv -2^{n-2} \pmod{2^{2n}}.
\end{align*}
Hence the desired quantity is $-2^{2n-1} \equiv 2^{2n-1} \pmod{2^{2n}}$; done. $\blacksquare$


Suppose one has to prove that there exist infinite pairs of solutions (x,y) for an equation in x,y; and I transform it into a Pell equation and find out the smallest pair of solution. Is the problem solved?

Yes. the problem is solved since every solution of a Pell's equation can be generated from the minimal solution of the equation.