I haven't studied this yet, but I'm guessing for the first one is,

Like there are 5 hotels, lets say _ _ _ _ _ , for each underline (hotels) there can 5 possibilities, and like there can be repetitions, then:

5^5 = 3125

and for the second one, im not sure but, like the couple needs to be together you can count there are only 4 elements now, and the other 3 people can only go to the other 4 hotels so,

_ _ _ _ _ , again for the first one this time will be 4 elements, but like the other 3 only can stay in the other hotels, then for the first one 4 possibilities, and the next 4 only 3 possibilities:

4(3^4) = 324

Maybe im not correct, please someone else answer.

thanks for trying to help me. Still looking for someone who knows how to do this question which i find very confusing

[quote="Misa-Campo"]Q3) 5 travelers arrive in a town where there are 5 hotels

a) how many different arrangements if there are no restrictions on where the travelers wish to stay?

b) Suppose 2 travelers are husband and wife and must go to the same hotel. how many different accommodation arrangements are there if the other 3 can go to any of the other hotels?

Please give me some tips for these questions, i dont have a clue on how to do these, thanks alot.[/quote]

a)Each can go for 5 different hotels hence total options being $ 5^5$
b)Consider the couple as a single entity since they cannot stay separate, hence we they have $ 5$ possible choices. The other three can stay in any of the 4 other hotels , so $ 4^3$ choices. Hence a total of $ 5\cdot 4^3$ choices

[quote="xcenario"]I haven't studied this yet, but I'm guessing for the first one is,

Like there are 5 hotels, lets say _ _ _ _ _ , for each underline (hotels) there can 5 possibilities, and like there can be repetitions, then:

5^5 = 625

and for the second one, im not sure but, like the couple needs to be together you can count there are only 4 elements now, and the other 3 people can only go to the other 4 hotels so,

_ _ _ _ _ , again for the first one this time will be 4 elements, but like the other 3 only can stay in the other hotels, then for the first one 4 possibilities, and the next 4 only 3 possibilities:

4(3^4) = 324

Maybe im not correct, please someone else answer.[/quote]
for 5^5, check your math there: $ 5^5\equal{}3125$.

For the second part,
I just think of it this way:
for each element it can be a hotel, so
_ _ _ _, 1st one (husband + wife) can be any of the 5 hotels, and the other 3 can go to any of the other four hotels.
So it should be $ 5\cdot(4^3) \equal{} 320$

Sorry about that one  :(
But yeah I guess I did the other way around in the second one.

Thanks tho.


Find all pairs $ (a,b)$ positive integer number such that $ a^2\minus{}1001b\equal{}1$ and $ b$ is a three digit number

[hide="hint"]
$ 1001|a^2 \minus{} 1$
$ 1001.100\leq  a^2 \minus{} 1\leq 1001.1000$
Use chinese remainder law.
[/hide]
I think it is only cacurlas.

[quote="bambang"]Find all pairs $ (a,b)$ positive integer number such that $ a^2 \minus{} 1001b \equal{} 1$ and $ b$ is a three digit number[/quote]
[hide]
$ a^{2}\equal{}1001b\plus{}1<1001001$
$ a<1001$
$ a^{2}\equal{}1001b\plus{}1>100101$, $ a>317$
Mod 7 tells us a=1/-1
Mod 11 tells us a=1/-1
Mod 13 tels us a=1/-1
Mod 1001 there are 8 cases by CRT.
1/1/1 doesn't work.
1/1/-1 gives us $ 155$, which doesn't work.
1/-1/1 gives us $ 273$ which also doesn't work.
1/-1/-1 gives us $ 428$, so $ b\equal{}183$.
-1/1/1 gives us $ 573$, so $ b\equal{}328$.
-1/1/-1 gives us $ 727$, so $ b\equal{}528$
-1/-1/1 gives us $ 846$, so $ b\equal{}715$
-1/-1/-1 gives us $ 1000$, so $ b\equal{}999$.
So our answer is $ (428,183)$, $ (573,328)$, $ (727,528)$, $ (846,715)$, and $ (1000,999)$.[/hide]

try using a programming language to find it.
The range of b is just from 100 to 999.

My solution is like this:

From the equation, we have

$ 1001b \equal{} (a \minus{} 1)(a \plus{} 1)$

$ (a \minus{} 1,a \plus{} 1) \leq 2$

Then (remember 1001 = 7.11.13)
a-1=7.p and a+1=11.13.q
a-1=7.11.p and a+1 = ...

Then the answer easily follows.


[quote="math92"]Three ducks and two ducklings weigh 32 kg. Four ducks and three ducklings weigh 44kg. All ducks weigh the same and all ducklings weigh the same. What is the weight of two ducks and one duckling?[/quote]

[hide]
d=duck, g=ducklings
    4d+3g=44
 -  3d+2g=32
    _________

    d+g=12
So:
3(12-g)+2g=32
36-3g+2g=32
-g=-4
g=4
Which means that
d=8
8+8+4=$\framebox{20}$[/hide]

[quote="math92"]Three ducks and two ducklings weigh 32 kg. Four ducks and three ducklings weigh 44kg. All ducks weigh the same and all ducklings weigh the same. What is the weight of two ducks and one duckling?[/quote]

[hide]Let x be the number of ducks and y be the number of ducklings.

$3x+2y=32$
$4x+3y=44$

$9x+6y=96$
$8x+6y=88$

$x=8$
$y=4$

2x+y=16+4=20[/hide]

[quote="math92"]Three ducks and two ducklings weigh 32 kg. Four ducks and three ducklings weigh 44kg. All ducks weigh the same and all ducklings weigh the same. What is the weight of two ducks and one duckling?[/quote]

[hide]
D=ducks d=ducklings

3D+2d=32
4D+3d=44

D+d=12
7D+5d=76
5D+5d=60
2D=16
D=8
d=4

2D+d=16+4=20
[/hide]


function
calculus
integration
logarithms
advanced fields
advanced fields unsolved

I am looking to show that the maximal operator doesn't necessarily map L^1 (R) to L^1 (loc (R)). In order to do this, i need a function that is in L^1, but not L^p for any p >1. I have one such function, f(x)= 1/(x(log(2/x))^2), but am having trouble proving that that the integral Mf(x) diverges. If anyone can give me another function of this sort, or perhaps give me an idea of how to show that Mf(x)'s integral diverges, it would be greatly appreciated.

Let's take the example you have and make it a little simpler:

$f(x)=\frac{1}{x(\ln x)^{2}}$ for $x\in\left(0,\frac12\right)$ and zero otherwise.

Now assume $x<\frac12.$ One interval that contains $x$ is the interval $[0,x].$ Hence

$M(f)(x)\ge\frac1x\int_{0}^{x}\frac{dt}{t(\ln t)^{2}}=\frac1{x|\ln x|}$

And $\frac1{x|\ln x|}\not\in L^{1}(\text{loc})$ in a neighborhood of zero.

If your maximal function was based on centered intervals, that doesn't change much; we just have to integrate from $0$ to $2x$ getting an estimate of $M(f)(x)\ge\frac1{2x|\ln 2x|}.$

It has been shown (I think this is Marcinkiewicz) that if $f\in L\log L,$ then $M(f)\in L^{1}(\text{loc}).$ Note that for $f(x)=\frac1{x(\ln x)^{2}},$ $f(x)|\ln f(x)|$ is comparable to $\frac1{x|\ln x|}$ as $x\to0,$ hence $f\not\in L\ln L.$ (And it's certainly not in $L^{p}$ for any $p>1.$)


Two points A and B are 1 unit apart. Let region S be the set of all disks with radius 1 that completely cover AB. What is the area of region S?

I tried looking at the solution, but it didn't make sense.

Well, the radius of the disk can be anywhere in the intersection of the two circles of radius 1 drawn with centers A and B. (Think about why.)

Well I'm pretty sure it doesn't mean octagon.

lol.. if you want the definition of a "disk" here ya go o.O
disk-a thin, flat, circular object

ok.. the center cannot be 1 away from both point.. A and B.. so you draw a line AB with length 1. and you use your compass to draw two circles centered at A and B with radius 1. The intersection is the set of centers it can be. Therefore.. the set of [b]DISKS[/b], is two sector radius 2 and with the angle 120 degrees.. and there are two other sectors with radius 1 on the top and bottom.. ya.. just find the area....

$2*\pi*2^{2}*\frac{120}{360}+2*\pi*1^{2}*\frac{60}{360}-1*1*\sin120^\circ=3\pi-\frac{\sqrt{3}}{2}$

Since the disks can have a radious of one, a circle w/ a center $A$ or $B$ will be 2 of the discs the maximum distance apart of the two.
And if $A$ and $B$ are 1 unit apart, then u can form an quilateral triangle that contains more area not contained in the the two other circles.

I hope that amkes sence, but I don't want to cruch out the numbers.

[hide]
Just from doing this in my head, I think it'll be a circle of radius 1.5. So the area would be $\frac{9pi}{4}$.

Your answer's incorrect.  Allan Z's answer is correct.

[quote="1=2"]thnx. :)

reason of posting so many AMC questions:

I've recently had them as homework.[/quote]

haha lol... gl to every1 in the AMC .. let's rock! =p..


A triangular array of squares has one square in the first row, two in the second, and in general, $k$ squares in the $k$th row for $1 \leq k \leq 11.$ With the exception of the bottom row, each square rests on two squares in the row immediately below (illustrated in given diagram). In each square of the eleventh row, a $0$ or a $1$ is placed. Numbers are then placed into the other squares, with the entry for each square being the sum of the entries in the two squares below it. For how many initial distributions of $0$'s and $1$'s in the bottom row is the number in the top square a multiple of $3$?
[asy]
defaultpen(linewidth(0.7));
path p=origin--(1,0)--(1,1)--(0,1)--cycle;
int i,j;
for(i=0; i<12; i=i+1) {
for(j=0; j<11-i; j=j+1) {
draw(shift(i/2+j,i)*p);
}}[/asy]

Ah, thanks worthawholebean! That helps a lot.  :)

[hide="question"]Is it not 639?  I don't think 0 is a multiple of 3...[/hide]

Uh... yeah it is. $3\cdot0=0$. Unless of course you were joking...

Huh, that sucks then.  (I wasn't joking, I answered 639 on my test.)

[quote="samath"]There's GOT to be a better solution than this ... it took too long counting to be the best way.

[hide]I just found the number of times each number on the bottom appeared in the sum at the top and took those mod 3.  It turns out the number at the top is equal to $k_{1}+k_{2}+k_{10}+k_{11}$ mod 3, where $k_{i}$ is the number in the $i^{th}$ box.  There are 5 solutions to $k_{1}+k_{2}+k_{10}+k_{11}\equiv 0\bmod{3}$ and $2^{7}=128$ ways to choose the middle 7 boxes, for a total of $5\times 128=640$.[/hide][/quote]

that's the best solution I've seen (and the way I solved it), and it took 3 just under 5 minutes.

Yeah. All it is is the tenth row of Pascal's triangle.

Can you please provide an explanation of your numbers?

As stated, it's the bottom row of Pascal's triangle, which is gonna be $1, 10$, lots of stuff divisible by $3$, $10, 1$ and the above solution follows.

It's the 10th row, not the 11th. There are 11 numbers in the 10th row of Pascal's triangle.

[quote="JSteinhardt"]Can you please provide an explanation of your numbers?

As stated, it's the bottom row of Pascal's triangle, which is gonna be $1, 11, 55$, lots of stuff divisible by $3$, $55, 11, 1$ and the above solution follows.[/quote]
It's the tenth row. Not the eleventh. So then $a_{1}+a_{2}+a_{10}+a_{11}\equiv 0\mod 3$. There are $2^{7}$ solutions for the middle ones and the outer ones can either be all $0$ or have three 1s and one 0. There are $5$ ways to do this, so the answer is $5(2^{7})=\boxed{640}$.

Yeah, that's what I did. At first, I used the 11th, and got like 1280 and was like WTF so I checked and found out, OH IT'S THE 10TH ROW HOW COOL

and since $3|\binom{10}{2}\rightarrow\binom{10}{8}$, the top box = $a_{1}+a_{2}+a_{1}0+a_{1}1$ and there's 1 way for them all to be 0 mod 3 and 4 ways for 3 to be 1 and 1 to be 0 mod 3; multiply by 2^7 and you get 640.

It really surprised me since I got 640 on #6 too and was like wtf? 2 of the same answer?

[quote="dasherm"][quote="samath"]There's GOT to be a better solution than this ... it took too long counting to be the best way.

[hide]I just found the number of times each number on the bottom appeared in the sum at the top and took those mod 3.  It turns out the number at the top is equal to $k_{1}+k_{2}+k_{10}+k_{11}$ mod 3, where $k_{i}$ is the number in the $i^{th}$ box.  There are 5 solutions to $k_{1}+k_{2}+k_{10}+k_{11}\equiv 0\bmod{3}$ and $2^{7}=128$ ways to choose the middle 7 boxes, for a total of $5\times 128=640$.[/hide][/quote]

Too long?  This is probably the most elegant solution you could hope for on an AIME #13.  Seeing all of those "choose"s disappear was both relieving and cool.[/quote]

His solution is too long because it omits the part where he actually adds up all the numbers. This can be shortened by recognizing that it is pascal's triangle.

I think this is a bit too easy for problem 13 on the AIME. I knew it was necessary to determine how many times each of $a_{1}$, $a_{2}$, ..., $a_{11}$ was counted, so I played around with the first and second row. It looked like Pascal's Triangle. Then used induction with Pascal's Identity, and the answer was pretty straightforward. Took about 5 minutes.

BAHT


(1) If $ 0^\circ < x < 180^\circ$ and $ \cos{x}\plus{}\sin{x}\equal{}{1}/{2}$, then find $ (p,q)$ such that $ \tan{x}\equal{}\minus{}\frac{p\plus{}\sqrt{q}}{4}$.
[hide]$ (4,7)$[/hide]

(2) If $ A\equal{}20^\circ$ and $ B\equal{}25^\circ$, then find the value of $ (1\plus{}\tan{A})(1\plus{}\tan{B})$.
[hide]$ 2$[/hide]

Thanks in advance!

1) Fist, square both sides: $ \cos^2 x \plus{} 2\sin x \cos x \plus{} \sin^2 x \equal{} 1/4 \iff \sin 2x \equal{} \minus{} 3/4$.  Thus we have $ \cos 2x \equal{} \pm \sqrt {1 \minus{} ( \minus{} 3/4)^2} \equal{} \pm \frac {\sqrt {7}}{4}$.  Now since $ \tan x \equal{} \frac {1 \minus{} \cos 2x}{\sin 2x}$, we have $ \tan x \equal{} \frac {1\mp \frac {\sqrt {7}}{4}}{ \minus{} \frac {3}{4}}$ so $ \tan x \equal{} \minus{} \frac {4 \pm \sqrt {7}}{3}$

We could just stop here, since the problem only gives the positive.  But to determine that the positive is correct: in the original equation, we know that $ 90 < x < 120$ because $ \cos 90 \plus{} \sin 90 \equal{} 1$ and $ \cos 120 \plus{} \sin 120 \equal{} \minus{} 1/2 \plus{} \sqrt {3}/2 \approx \minus{} .5 \plus{} 1.7/2 \equal{} \minus{} .5 \plus{} .85 \equal{} .35$ and since the function is continuous, $ 90 < x < 120$ by intermediate value theorem.  Also, no other possible solutions for $ x$ exist in $ 0 < x < 90$ or $ 120 < x < 180$ because $ f(x) \equal{} \cos x \plus{} \sin x \ge 1$ for $ 0 < x < 90$ and $ f$ decreases from $ 45$ to $ 225$, and thus is one-to-one on $ (45,180]$.  Since the tangent function is strictly increasing on the interval $ (90,120]$, $ \tan x < \tan 120 \equal{} \minus{} \sqrt {3} \approx \minus{} 1.7$.

Now we know $ 2 < \sqrt {7} < 3$.  Testing our solutions: With the positive sign, $ \minus{} \frac {4 \plus{} 3}{3} \equal{} \minus{} 2 \frac {1}{3} < \minus{} \frac {4 \plus{} \sqrt {7}}{3} < \minus{} \frac {4 \plus{} 2}{3} \equal{} \minus{} 2$ so the positive sign is viable as far as we can tell.  But for the negative: $ \minus{} \frac {4 \minus{} 3}{3} \equal{} \minus{} \frac {1}{3} > \minus{} \frac {4 \minus{} \sqrt {7}}{3} > \minus{} \frac {4 \minus{} 2}{3} \equal{} \minus{} 2/3$, and hence if the negative holds $ \tan x > \minus{} \sqrt {3}$, which we determined to be impossible.

(2)
$ \tan(A \plus{} B) \equal{} \tan(45) \equal{} 1$, but also $ \tan(A \plus{} B) \equal{} \frac {\tan A \plus{} \tan B}{1 \minus{} \tan A \tan B}$, so we have $ 1 \minus{} \tan A \tan B \equal{} \tan A \plus{} \tan B \iff 1 \equal{} \tan A \tan B \plus{} \tan A \plus{} \tan B$.  Expanding the expression whose value we want, we have $ 1 \plus{} \tan A \plus{} \tan B \plus{} \tan A \tan B \equal{} 1 \plus{} 1 \equal{} 2$.


Fill in the black:
In a bond between oxygen and carbon,
a.) electrons are equally shared
b.) carbon develops a partial negative charge
c.) oxygen develops a partial negative charge
d.) electrons are completely transferred

How would you describe a molecule of carbon dioxide?
a.) nonpolar, with nonpolar bonds
b.) nonpolar, with polar bonds
c.) polar, with nonpolar bonds
d.) polar, with polar bonds

Given the following properties for a substance:
Soluable in water = yes
Conductive in solution = yes
Relative melting point = high
What type of bonding do you think is present in this substance?
a.) ionic
b.) covalent
c.) metallic

THANKS!!!

All three of these should be pretty obvious - what do you think the answers are?

Probably the least obvious one in the second one. One thing to note is that $\text{CO}_{2}$ has a $\text{O}=\text{C}=\text{O}$ configuration that is a straight line. The orbitals of the central carbon are $sp$ hybridized, which accounts for the linear shape.

my guesses are c for #1, b for #2, and a for #3

1. Oxygen is more electronegative than carbon.

2. Due to the linear and symmetric shape of carbon dioxide, it is nonpolar. But it does have polar bonds.

3. Ionic substances are generally soluble in water (the attraction between each ion and nearby water molecules outweighs the electrostatic forces bewteen the cations and anions). When the compounds dissociate, there will be ions in solution so it will conduct a current. I think it's (A), but I'm not 100% sure.

Yes, the answer for #3 is a. That list of properties is essentially diagnostic for an ionic compound (that is, a salt.)

Covalently bonded compounds may or may not be water soluble. But their solutions do not conduct electricity or do so only slightly. The melting points of water-soluble covalently bonded compounds vary widely. The melting point of ethanol is considerably below the melting point of water - it would be a liquid in an ordinary household freezer. The melting point of sucrose is so high that that it tends to char and decompose rather than melt.

And metals normally aren't water-soluble at all, rendering the question of whether their solutions would conduct electricity moot. To "dissolve" a metal in an aqueous solution normally means to oxidize it - the species in solution is no longer the metal, but cations formed from it.


Hi all.

I'm having some difficulty with this problem.  I'm guessing that the Mean Value Theorem and/or Rolle's Theorem is involved, but I'm not quite sure how to apply it/them.

Given:

	- $f(x)$ is continuous on the closed interval [a,b]
	- $f'(x)$ and $f''(x)$ exists on the interval $a < x < b$
	- $f(a) = f(b) = 0$
	- there is a number $x_o$ such that $a<x_o<b$ and $f(x_o)>0$

Prove:
	There is a number $\xi$ where $a<\xi<b$ such that $f''(\xi)<0$

From drawing a rough picture and fiddling around, I would guess that I want to get it into the form:

	\[ f''(\xi)=\dfrac{f'(x_2)-f'(x_1)}{x_2-x_1} \]
where the numerator is < 0

I'd greatly appreciate any help on this.

Thanks,
Bailey

Yes, your find a correct way.

Apply Lagrange's Mean-value Theorem on $[a,x_0]$ and $[x_0,b]$
we have $x_1,x_2$ such that
$f'(x_1) = \dfrac{f(x_0)-f(a)}{x_0-a} = \dfrac{f(x_0)}{x_0-a}>0$
and
$f'(x_2) = \dfrac{f(b)-f(x_0)}{b-x_0} = -\dfrac{f(x_0)}{b-x_0}<0$

Apply Lagrange's Mean-value Theorem again to $f'(x)$ on $[x_1,x_2]$, that is what you desired.

Thanks a bunch!

Looks simple now that I see it.  Thanks again!

Bailey


I got an email from someone named armlregister@gmail.com:

[quote]Dear (name omitted for privacy reasons)

This year, ARML will take place at three locations: Penn State, University of Iowa, and University of Las Vegas – Nevada on June 1st and 2nd, 2007.The competition promises to a very enriching experience for all.

ARML regularly receives requests from scholarly organizations for the names, addresses and grade levels of high scoring students of the ARML competition. This information is used for recruiting and other academic purposes. D.E. Shaw is one such organization that has recruited former ARML students in the past and intends to in the future. In addition, the TOP TEN SCORERS of the competition will be awarded substantial sums of money this year. All awards are typically mailed after the competition.

We are requesting that you provide some contact information to ARML for the purposes outlined above. Providing your contact information in advance of the competition helps ARML to help scholarly organizations connect with our ARML participants and simplifies the distribution of awards following the ARML competition.  If you decide not to share your contact information, ARML will honor that request.

We thank you in advance for your participation in ARML. Please follow the link below to the student information page where you will be asked to provide contact information such as:
student name
mailing address
phone
gender
Please note that you are given a team code in order to register This team code is to be used only once - you will be asked to choose a password once you log in.

Your team code is: (team code omitted for privacy reasons)

The link is: http://www.armlregister.org/studentregister/index.php
Thank you,
ARML[/quote]

Well, a few things stand out:

1. The email address isn't on the arml.com domain. It's a gmail.com address.
2. The website to enter the information isn't on the arml.com domain. It's a separate website.
3. arml.com doesn't have a link to that page.

Five bucks says I'm just really paranoid. (I am.) But I've read way too many Internet horror stories, and I've started questioning every single thing I see.

Could someone please verify that this is actually ARML and not a scam? (A pretty harmless scam at that, but hey, I don't want to give my personal information out to a complete stranger.)

ARML is in just 2 weeks. Isn't it a bit late to be getting emails for it?

There is a link to http://www.armlregister.org from the http://www.arml.com website. The link is through the "2007 ARML Contest Information" and is the link in Click [u]here[/u] to begin the registration process for ARML 2007.

I'm a coach and I don't remember receiving that particular email, but then I used the site to register my team some time ago.

OK. Call me paranoid then. :blush:

Thanks.

Your coach put your email address in along with your registration. Afterwards, ARML will send an email to you and request permission to send advertisements/announcements to you.


proof that the product of 4 consecutive numbers is one less than square number, in as much ways as possible. I know 2 ways, and I thought of induction, but I didn't succeed. help!
is there any generalisation here?

[quote="nutz_for2.718281828"][hide="I will try."]$n(n+1)(n+2)(n+3)=x^{2}-1$
$(n^{3}+3n^{2}+2n)(n+3)=x^{2}-1$
$n^{4}+6n^{3}+11n^{2}+6n=x^{2}-1$
$n^{4}+6n^{3}+11n^{2}+6n+1=x^{2}$

We see now that x has to be odd, or else n isn't an integer, which it is. Therefore, $n^{4}+6n^{3}+11n^{2}+6n+1$ has to be a perfect square. We try $(n^{2}+2n+1)^{2}$. That doesn't work. We try $(n^{2}+3n+1)^{2}$. That happens to work. Therefore, $n^{2}+3n+1=x$. Since n is an integer, x^2 is a perfect square. We plug it into the formula $n(n+1)(n+2)(n+3)=x^{2}-1$ and expand:

$n^{4}+6n^{3}+11n^{2}+6n=(n^{4}+6n^{3}+11n^{2}+6n+1)-1=n^{4}+6n^{3}+11n^{2}+6n$

QED[/hide][/quote]
this is your way http://mathschallenge.net/index.php?section=problems&show=true&titleid=consecutive_product_square
try more!

I'm only 11! I don't know any more ways!

[hide]assume $x(x+3)(x+1)(x+2)+1=y^{2},(x^{2}+3x)(x^{2}+3x+2)=y^{2}-1=(y+1)(y-1)$ which is true. so we only need to proof that the product of 2 numbers with difference of 2 cannot be a square: $(x+1)^{2}-x(x+2)=1$[/hide]


Let $a,b,c$ be positive real numbers satisfying $ab+bc+ca=1$.
Find the least value of the expression:
$P=\sqrt{a^2+ab+b^2} \sqrt{b^2+bc+c^2}+\sqrt{b^2+bc+c^2} \sqrt{c^2+ca+a^2}+\sqrt{c^2+ca+a^2} \sqrt{a^2+ab+b^2}$
 :)  :?  :D

[quote="probability1.01"]Hmmm... it can be done geometrically with areas I think.[/quote]
hmm...  more...
!!! :idea: !!!


geometry
trapezoid
trigonometry
function
geometry unsolved

[quote="Contpuzzles"][color=darkred]Let $ ABC$ be a triangle for which $ 2A = 3C$ and exists the point

$ E\in (AC)$ such that $ \|\begin{array}{c} AE = BC \\
\ EB = EC\end{array}$ . Prove that $ A = 30^{\circ}$ .[/color][/quote]

[color=darkblue][b][u]Proof I.[/u][/b] Consider the point $ D\in BE$ for which $ AD\parallel BC$ . Observe that

the quadrilateral $ ABCD$ is an isosceles trapezoid with $ AE = DE = BC$ .

Therefore, $ \frac {EB}{ED} = \frac {BC}{AD}$ $ \implies$ $ \frac {EB}{BC} = \frac {BC}{AD}$ $ \implies$ $ \frac {\sin C}{\sin 2C} = \frac {\sin A}{\sin (A + 2C)}$ $ \implies$

$ \sin (A + 2C) = 2\sin A\cos C$ $ \implies$ $ \sin (A + 2C) = \sin (A + C) + \sin (A - C)$ $ \implies$

$ 2\sin \frac C2\cos \frac {2A + 3C}{2} = \sin (A - C)$ $ \implies$ $ 2\sin\frac C2\cos 2A = \sin\frac C2$ $ \implies$ $ A = 30^{\circ}$ .[/color]

Actually, if you are to use trigonometric functions, there is no need to add another line or a trapezoid.
In your notation you can consider ABE and BEC and write that sin(7x) / sin(3x) = sin(4x) / sin(2x), after which you can easliy conclude that x = 10 degrees.

However, I was told that this question was actually asked at an Olympiad for students who have not yet learned trigonometry.

[color=darkblue][b]Contpuzzles[/b] , I constructed the trapezoid because I found another proof
(synthetical) for this problem and so that the trapezoid remained ... Here is ![/color]

[quote="Virgil Nicula"][color=darkred][b][u]An equivalent enunciation.[/u][/b] Let $ ABCD$ be an isoceles trapezoid ($ AD\parallel BC$) .
Denote $ E\in AC\cap BD$ . Prove that $ AE\equal{}BC$ $ \Longleftrightarrow$ $ m(\widehat {BAC})\equal{}30^{\circ}$ .[/color][/quote]

[quote="contpuzzles"]Point D has been chosen on the side AC of a triangle ABC such that AD = BC and BD = DC.  Angle BAC =3x, angle BCA = 2x.  Find x.
[/quote]

$ B^{2}C^{2} \equal{} D^{2}B{2} \plus{} D^{2}C^{2} \minus{} 2DB.DC.cos( \pi \minus{} 4x)$
 $ \Rightarrow AD\equal{} BC \equal{} 2BD.cos2x$
Alternatively, $ \frac{AD}{sin( \pi \minus{} 7x)} \equal{} \frac{BD}{sin3x}$
Therefore, $ 2sin3xcos2x \equal{} sin7x$

I afraid that $ x \neq 30$

No, x is not 30 degrees, it is 10.  Angle A is 30 degrees.

[quote="contpuzzles"]Point D has been chosen on the side AC of a triangle ABC such that AD = BC and BD = DC.  Angle BAC =3x, angle BCA = 2x.  Find x.
I have been thinking about it for 2 weeks to no avail...[/quote]

[b]Only the figure ![/b]


The same problem was posted before (by me :D ):
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45891
(See #1)

let the number be $aabb$.
$1'st$ $ Observation$:-$11$ $divides$ $ aabb$
$lemma$-- any perfect square has even powes of it's prime factors$
.$By the lemma 121 must divide $aabb$
-> as $aabb=11(100a+b)$[because digits are repeting] we have $11$ divides $100a+b$
Now a perfect square cant end in $3,7,8,2$ so these valuesof b are ruled out..
also note $100a+b$ is congruent to $a+b$ $mod 11$ so check b=$(0,1,4,5,6,9)$
clearly only $7744$satifies.
QED

[quote=Rushil]A four-digit number has the following properties:

(a) It is a perfect square;

(b) Its first two digits are equal

(c) Its last two digits are equal.

Find all such four-digit numbers.[/quote]

We know,
	102 = 100
	1002 = 10000
so, it is two digit number
(xy)2 = aa bb
aabb is multiple A 11
being a perfect square it is multiple of 112 so possible numbers are
(11)2 x 1 = 121
(11)2 x 4
(11)2 x 9
(11)2 x 16
(11)2 x 25
(11)2 x 36
(11)2 x 49
(11)2 x 64		=	 7744 in of form aabb
(11)2 x 8

Let the number be $N$
Then
$11(100x+y)=N=Q^2$
Then $100x+y=11q$, where $q$ is a perfect square.
Then $x+y=11$ and $100x+y=x0y$. Then possible values of $(x,y)$ are $(7,4),(6,5),(5,6),(7,4)$(neglected (9,2) and (8,3) because a perfect square cannot end with 2 or 3)
By hit and trial we found out that... $(7,4)$ is the only possible pair.

[hide=nice]so we have the number as $aabb$ , now $11(100a+b)=k^2$\\
so $100a+b=11j^2$\\
notice $100a+b\equiv a+b (\mod 11)$\\
so we should have $a+b \equiv 0 (\mod 11)$\\
notice a and b are digits this reduces significantly our casing\\
notice $(7,4)$ is the only possible choice for $(a,b)$ that makes it a perfect square hence $7744$ is the required number $\blacksquare$[/hide]


How many prime numbers between 30 and 65 have a prime remainder when divided by 10?

We want numbers between 30 and 65 which are both prime, and have a units digit of 2, 3, 5, or 7. 2 isn't possible, as any number of this type is divisible by two.

The only way to do this is through a list.

Prime numbers: 31, 37, 41, 43, 47, 53, 59, 61

With units digit requirement: 37, 43, 47, 53

There are $ \boxed{4}$ numbers.

EDIT: Yeah, thats probably faster. Sorry, forgot to mention 5 impossible along with 2.

They have given us a very small range

the numbers have to end with 2, 3, 5, or 7.

2 is impossible; no number ending in 2 greater than 2 is prime.
5 is impossible for similar reasons

33, 43, 53
37, 47,57

33 and 57 aren't prime

so the answer is 4


geometry
3D geometry
sphere
geometry open

hello
is there any connection between a hyperboloid and  a sphere?
if yes how do these 2 connect,i mean how do i build a certain one
from the other.
can one be used and a kind of geometric locus to form the other ?
what is the relationship between these two ?

i conclude that if noone replied either i am boring,or
i am totally wrong


Is gasoline soluble in water? $MgSO_4$? $CaCO_3$? Why?

Write an equation to show how these substances ionize or dissociate in water:

$HC_2H_3O_2$, $HgCl_2$.

What happens if you added a nonpolar molecular liquid to water and shook the mixture vigorously?

Gasoline is not soluble, $CaCO_3$ is not, and $MgSO_4$ is a little bit soluble. Why? Just drop it on water and see  ;) .

One accepted criteria for solubility prediction os the polarity. Polar substances are soluble on polar solvents, apolar are soluble on apolar.

Now the asked equations...
$HC_2H_3O_2 \thickspace \xrightarrow{H_2O} \thickspace H^{+} + C_2H_3O_2^{-}$

$HgCl_2 \thickspace \xrightarrow{H_2O} \thickspace Hg^{++} + 2Cl^{-}$

Look that! $K_{ps} = [Hg^{++}].[Cl^{-}]^2 \approx 0$


In the triangle ABC , AH is the altitude and AM is the median drawn from vertex A. Let P be a point in the plane of triangle (i.e. inside the triangle) and Q and R the foot of altitudes drawn from P to AH and BC respectively. Prove that if  $ \frac{\overline{AQ}}{\overline{QM}}\equal{}\frac{\overline{HR}}{\overline{RM}}\equal{}2$  than point p is the center of gravity of the triangle ABC.

But i think that the question is wrond and it should be $ \frac{\overline{AQ}}{\overline{QH}}\equal{}\frac{\overline{HR}}{\overline{RM}}\equal{}2$  not $ \frac{\overline{AQ}}{\overline{QM}}\equal{}\frac{\overline{HR}}{\overline{RM}}\equal{}2$.
I need help.

[hide="Easy!"]Only use [b]Thales[/b]'s theorem[/hide]

Yes using Thales Theorem you get the wanted result but the question should say that $ \frac{\overline{AQ}}{\overline{QH}}\equal{}\frac{\overline{HR}}{\overline{RM}}\equal{}2$


Prove the following inequality for $ a,b,c>0$

$ \frac{1}{a^3\plus{}b^3\plus{}abc}\plus{}\frac{1}{b^3\plus{}c^3\plus{}abc}\plus{}\frac{1}{a^3\plus{}c^3\plus{}abc} \leq \frac{1}{abc}$

The inequality is equivalent to
\[ \sum\frac{a^3+b^3}{a^3+b^3+abc}\ge 2\]
Verify that
\begin{align*}\frac{a^3+b^3}{a^3+b^3+abc}&\ge \frac{a+b}{a+b+c}\\
\iff c(a^2+b^2-ab)&\ge abc\\
\iff c(a-b)^2&\ge 0
\end{align*}
which is obviously true. Hence we conclude that
\[ \sum\frac{a^3+b^3}{a^3+b^3+abc}\&\ge \sum\frac{a+b}{a+b+c}=2\]

[quote="Nemion"]Prove the following inequality for $ a,b,c > 0$

$ \frac {1}{a^3 \plus{} b^3 \plus{} abc} \plus{} \frac {1}{b^3 \plus{} c^3 \plus{} abc} \plus{} \frac {1}{a^3 \plus{} c^3 \plus{} abc} \leq \frac {1}{abc}$[/quote]

$ a^3 \plus{} b^3 \geq ab(a \plus{} b)$


trigonometry
geometry
3D geometry
sphere

A projectile of mass m is fired from the surface of the earth of radius R at an angle a from the vertical with initial speed equal to (1/[sqrt 2]) the escape velocity.How high does the projectile rise?What is its range on earth's surface?Neglect air resistance and earth's rotation.

I solved the first part easily by using work-energy theorem.The answer is 2R/3.I hope it is correct.

But I am stuck to solve the 2nd part.Will the normal formulas for the projectile problem give the answer?

[quote="kolahalb"]at an angle a from the vertical[/quote]
Is the numerical value of this angle given :?:

No...
OK,perhaps I am heading towards the solution.
I will post it if I be able to get it.

So $\frac{2R}{3}$ is not the correct answer

You must combine conservation of energy and conservation of angular momentum.  :blush:

Yes.That was a calculation mistake.

Astronum is correct.

As mentioned above your result does not take the angle $\alpha$ into account, which is wrong.

[hide="Hint"]
We have central motion, and by definition of a central force field, we have no torque.
Thus $M=0=\frac{d L}{dt}$ and we have conservation of angular momentum $L$.

Now since we know $\alpha$ the ang. momentum $l_{1}$ at start is known, also at the most distant point, the apoapsis, the ang. momentum $l_{2}$ is easily calculated since we have right angles there.
This yield one eq.
\[mr_{1}v_{1}=mR\frac{1}{\sqrt{2}}v_{e}=l_{1}=l_{2}=mr_{2}v_{2}\]
Now two variables are unknown, namely $v_{2}$ and $r_{2}$
Use conservation of energy to solve these, for example
\[\frac{mv_{1}^{2}}{2}-\frac{GMm}{r_{1}}=\frac{mv_{2}^{2}}{2}-\frac{GMm}{r_{2}}\]

[/hide]

Dear buddy,I have done the first part already.
What I am worrying about is the second part.
I already got the value of eccentricity,angular momentum...
Now I nwwd to employ the orbit equation:

r=(L^2/mk)/(1+ e cos (theta-theta_0))

But I am stuck with it and cannot understand that what should I do after that.

So you want to employ something like this
$r=\frac{L^{2}}{mk}\frac{1}{1+e \cos{ (\theta)}}$

[hide="hint"]
What is $r$ when the projectile hits ground?
(You modelled earth by a sphere of radius $R$)
Try to solve for $\theta$, which gives where it crashed.
[/hide]

i)As pointed before, from energy and angular momentum conservation we get the distance of the apoapsis or apogee to earth center as

   R(1+cos a)   (the height could be R.cos a ?)

ii)The semimajor axis equals the earth radius R, so, in launching (and landig) point

the projectile is at the minor axis, and the straight line (tunneling) distance would be it's length, 2R.sin a.

Measuring on earth's surface, it is the arc length 2.a.R  ( remember: a is the launching angle from vertical)

Immanuel Bonfils,I did not understand on first time reading...
It's simply superb!!!

[quote]The semimajor axis equals the earth radius R[/quote]

You can see this from 2*semi-major axis=(-GMm)/E

The rest portion is just geometry.

Thank you very much.


topology
algebra
function
domain
advanced fields
advanced fields unsolved

Hi,

I'm new to algebraic topology and am grappling with some of the ideas in the field. I'm hoping someone will be able to help me with ideas and solutions to some or all of the problems in the PDF file I have attached. Complete solutions will be great, but I would also like to get a sampling of some key ideas that will be useful in general.

For #1, the pictures are attached as a separate file.

Many thanks!

1.1) The first two spaces are contractible to a point (their fundamental group is trivial) the other two aren't, so no one of the first two can be homeomorphic to one of the other two; now, call X,Y the first two spaces and suppose $f: X\to Y$ is homeo, let $x_0$ be the node of the space X, then $f(X\setminus\{x_0\})=Y\setminus\{f(x_0)\}$ and $f$ is also an homeo between $X-\{x_0\}$ and $Y-\{f(x_0)\}$, but the first space has 4 connected components and the second can have at most 3 c.c., so they cannot be homeomorphic and therefore X,Y cannot too.
Now, call $X_1,Y_1$ the other two spaces: you can choose a point $x_1\in X_1$ so that $X_1-\{x_1\}$ is no more connected, but such a point does not exist in $Y_1$, so (by an argument similar to the previous one) the two spaces cannot be homeomorphic.

1.2) (too tedious to write)

1.3) Path from $(x_0,y_0)$ to $(x_1,y_1)$ : from $(x_0,y_0)$ to $(x_1,y_0)$ (in the section $X\times \{y_0\}$ which is path connected, 'cause it's homeomorphic to X), from $(x_1,y_0)$ to $(x_1,y_1)$ (in the sectione $Y\times\{x_1\}$ which blablabla as before).

1.4) I suppose you know that the 1-point compactification is unique. Then, $S^n$ is compact (closed and bounded in $\mathbb{R}^{n+1}$) and the stereographic projection is a homeomorphism between $S^n-\{p\}$ and $\mathbb{R}^n$.
It's easy to verify that the open sets in $S^n$ are generated by the preimages under the st.proj. of the open sets of $\mathbb{R}^n$ and by the sets $f^{-1}(\mathbb{R}^n\setminus K)\cup \{p\}$ where $K$ is compact and f is the st.proj.

1.5) Fix two points $P,Q$ in $S^n$ and let $f_1: S^n\setminus\{P\}\to\mathbb{R}^n$ $f_2: S^n\setminus\{Q\}\to\mathbb{R}^n$be the st.prj. The path between $A,B$ is the preimage of the path between $f_i(A)$ and $f_i(B)$ in $\mathbb{R}^n$, where $i$ is suitably chosen so that $A,B$ are both in the domain of definition of $f_i$.

1.6) Choose $x_0$ in $X$, convex in $\mathbb{R}^n$. Define $F(x,t)=(1-t)x+tx_0$; this is the homotopy between id and the constant map in $x_0$.
$f,g: Y\to X$ are both homotopic to the constant map, so they are homotopic.


algebra polynomial limit calculus derivative function

prove that if we have 2 polinomials, and one is higher than the other. like $a=x^{5}$ > $b=x^{3}+x^{2}+123$. so prove then when
$\lim_{x\to\infty}\(a>b)$
[hide]we just derive it until b is 0?[/hide]

You need to be [b]much[/b] more precise.  What I think you were trying to say:

Let $A(x), B(x)$ be two polynomials with positive leading coefficients such that $deg(A) > deg(B)$.  Show that

$\lim_{x \to \infty}A(x)-B(x) > 0$

(Yes, you can show that some derivative of it tends to infinity and therefore that the function itself must, but that is silly when the problem is so much simpler than that.)

[hide="Discussion"] In fact, we can make the definite statement $\lim_{x \to \infty}A(x)-B(x) = \infty$.

Let $C(x) = A(x)-B(x)$ be a polynomial with a positive leading coefficient $c_{n}x^{n}$.  Then

$\lim_{x \to \infty}C(x) = \lim_{x \to \infty}x^{n}\left( c_{n}+\frac{c_{n-1}}{x}+... \right) = \lim_{x \to \infty}x^{n}\left( c_{n}\right) = \infty$ [/hide]


inequalities
function
logarithms
inequalities proposed

[quote="nhat"]here is my solution :)
this problem equavilent to prove:
        $%Error. "Longleftrighttarow" is a bad command.
\sum (\frac{a(b^3+c^3)}{a^2+bc} -\frac{ab+ac}{2})\geq \sum a^2-\sum ab$
         $%Error. "Longleftrighttarow" is a bad command.
\sum \frac{a(b+c)(3(b-c)^2+b^2+c^2-2a^2}{2(a^2+bc)} \geq 2(\sum a^2 -\sum ab)$
          $%Error. "Longleftrighttarow" is a bad command.
\sum \frac{3a(b+c)(b-c)^2}{a^2+bc}+\sum \frac{(a^2-b^2)(ab-c^2)}{(a^2+bc)(b^2+ac)} \geq 2\sum (a-b^2$
        $%Error. "Longleftrighttarow" is a bad command.
\sum (b-c)^2[\frac{3a(b+c)}{a^2+bc}+\frac{(b+c)^2(bc-a^2)}{(b^2+ac)(c^2+ab)} \geq 2\sum (a-b)^2$
Put:$S_{a}=\frac{3a(b+c)}{a^2+bc}+\frac{(b+c)^2(bc-a^2)}{(b^2+ac)(c^2+ab)}-2$
       $S_{b}=\frac{3b(a+c)}{b^2+ac}+\frac{(a+c)^2(ac-b^2)}{(a^2+bc)(c^2+ab)}-2$
       $S_{c}=\frac{3c(b+a)}{c^2+ba}+\frac{(b+a)^2(ba-c^2)}{(b^2+ac)(a^2+bc)}-2$
We need to prove $\sum S_{a} (b-c)^2 \geq 0$
WLOG $a \geq b \geq c$.
By expanding we have $S_{b},S_{c} \geq 0$
And you can check that $a^2S_{b}+b^2S_{a} \geq 0$
So from the criterion of the SOS method we have $S_{a} (b-c)^2 \geq 0$
So the orginal problem is proved :D  :D[/quote]

What an ugly and long solution. Anyone have another solution?

[quote="ductrung"]For any positive real numbers $a,\;b$ and $c$, prove that
\[ \frac{a(b^3+c^3)}{a^2+bc}+\frac{b(c^3+a^3)}{b^2+ca}+\frac{c(a^3+b^3)}{c^2+ab}\ge a^2+b^2+c^2  \][/quote]

 It isn't hard :D. Here is one-line solution for it:
L-R=
$\frac{(a^2+b^2+c^2)(a-b)^2(b-c)^2(c-a)^2+\sum a^2b^2(a^2-b^2)^2}{2(a^2+bc)(b^2+ac)(c^2+ab)}\geq 0$  :)

[quote="ductrung"]What about this one
\[ \frac{a(b^4+c^4)}{a^2+bc}+\frac{b(c^4+a^4)}{b^2+ca}+\frac{c(a^4+b^4)}{c^2+ab}\ge a^3+b^3+c^3  \][/quote]

  This one is definitely true. But I have a very ugly sols. Do you have a nicer one, Duc Trung?
  L-R=

$\frac{(a^3+b^3+c^3)(a-b)^2(b-c)^2(c-a)^2+[7,2,0]-[5,4,0]}{2(a^2+bc)(b^2+ac)(c^2+ab)}$
${+\frac{a^2b^2c^2(a^3+b^3+c^3-3abc+\sum a(a-b)(a-c))}{2(a^2+bc)(b^2+ac)(c^2+ab)}} \geq 0$
 :)

   After practising Ductrung's problem, I suffer pbD (permanent brain damage)! :D

find all n st for any positive real number a,\;b and c this inequality is true :
\frac{a(b^n+c^n)}{a^2+bc}+\frac{b(a^n+c^n)}{b^2+ac}+\frac{c(a^n+b^n)}{c^2+ab}\ge a^(n-1)+b^(n-1)+c^(n-1)

[quote="Gibbenergy"][quote="ductrung"]For any positive real numbers $a,\;b$ and $c$, prove that
$\frac{a(b^3+c^3)}{a^2+bc}+\frac{b(c^3+a^3)}{b^2+ca}+\frac{c(a^3+b^3)}{c^2+ab}\ge a^2+b^2+c^2$

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

[/quote]

  Duc trung, now it is my turn.
For any positive real numbers $a,\;b$ and $c$, prove that
$\frac{a(b^5+c^5)}{a^2+bc}+\frac{b(c^5+a^5)}{b^2+ca}+\frac{c(a^5+b^5)}{c^2+ab}\ge a^4+b^4+c^4$
and
$\frac{a(b^6+c^6)}{a^2+bc}+\frac{b(c^6+a^6)}{b^2+ca}+\frac{c(a^6+b^6)}{c^2+ab}\ge a^5+b^5+c^5$
 :D  :D  :P[/quote]

I have proved that for all $n\ge 3$
\[ \frac{a(b^n+c^n)}{a^2+bc}+\frac{b(c^n+a^n)}{b^2+ca}+\frac{c(a^n+b^n)}{c^2+ab}\ge a^{n-1}+b^{n-1}+c^{n-1} \]
See also
[url]http://www.artofproblemsolving.com/Forum/topic-83112.html[/url]

The following are nice but easy
\[ ab+bc+ca\le\frac{a^3(b+c)}{a^2+bc}+\frac{b^3(c+a)}{b^2+ca}+\frac{c^3(a+b)}{c^2+ab}\le a^2+b^2+c^2 \]

[quote="ductrung"]
I have proved that for all $n\ge 3$
\[ \frac{a(b^n+c^n)}{a^2+bc}+\frac{b(c^n+a^n)}{b^2+ca}+\frac{c(a^n+b^n)}{c^2+ab}\ge a^{n-1}+b^{n-1}+c^{n-1}  \]
See also
[url]http://www.artofproblemsolving.com/Forum/topic-83112.html[/url]

[/quote]

  Yes, that is true for all $n \geq 3$.  :D

[quote="Vasc"]Write the inequality as
$a^{n-3}A+b^{n-3}B+c^{n-3}C \geq 0$,
where $A=\frac{a^3b}{b^2+ca}+\frac{a^3c}{c^2+ab}-a^2$ etc.
Then, for $a \geq b \geq c$ we have
$a^{n-3}A+b^{n-3}B+c^{n-3}C \geq b^{n-3}(A+B)+c^{n-3}C \geq c^{n-3}(A+B+C) \geq 0$.[/quote]

Very nice! Thank you.

[quote="Gibbenergy"][quote="ductrung"]For any positive real numbers $ a,\;b$ and $ c$, prove that
\[ \frac{a(b^{3}\plus{}c^{3})}{a^{2}\plus{}bc}\plus{}\frac{b(c^{3}\plus{}a^{3})}{b^{2}\plus{}ca}\plus{}\frac{c(a^{3}\plus{}b^{3})}{c^{2}\plus{}ab}\ge a^{2}\plus{}b^{2}\plus{}c^{2}\]
[/quote]

 It isn't hard :D. Here is one-line solution for it:
L-R=
$ \frac{(a^{2}\plus{}b^{2}\plus{}c^{2})(a\minus{}b)^{2}(b\minus{}c)^{2}(c\minus{}a)^{2}\plus{}\sum a^{2}b^{2}(a^{2}\minus{}b^{2})^{2}}{2(a^{2}\plus{}bc)(b^{2}\plus{}ac)(c^{2}\plus{}ab)}\geq 0$  :)[/quote]

Does it exist a solution without SOS and full expansion?

I cannot find it. :blush:

Thank you very much.

[quote="Vasc"]Write the inequality as
$ a^{n\minus{}3}A\plus{}b^{n\minus{}3}B\plus{}c^{n\minus{}3}C\geq 0$,
where $ A \equal{}\frac{a^{3}b}{b^{2}\plus{}ca}\plus{}\frac{a^{3}c}{c^{2}\plus{}ab}\minus{}a^{2}$ etc.
Then, for $ a\geq b\geq c$ we have
$ a^{n\minus{}3}A\plus{}b^{n\minus{}3}B\plus{}c^{n\minus{}3}C\geq b^{n\minus{}3}(A\plus{}B)\plus{}c^{n\minus{}3}C\geq c^{n\minus{}3}(A\plus{}B\plus{}C)\geq 0$.[/quote]

I tried to prove, as Vasc said, that $ A\geq 0$ and $ A\plus{}B\geq 0$.

We can assume WLOG that $ a\geq b\geq c$

We have

$ A \equal{}\frac{a^{3}b(a\minus{}b)(b\minus{}c)\plus{}a^{3}c^{2}(b\minus{}c)\plus{}a^{2}c^{2}(a^{2}\minus{}b^{2})}{(b^{2}\plus{}ca)(c^{2}\plus{}ab)}$

so $ A\geq 0$

For $ A\plus{}B \equal{}\frac{(a^{2}\plus{}bc)[a^{3}b(a\minus{}b)(b\minus{}c)\plus{}a^{3}c^{2}(b\minus{}c)\plus{}a^{2}c^{2}(a^{2}\minus{}b^{2})]\plus{}(b^{2}\plus{}ca)[b^{3}c(b\minus{}c)(c\minus{}a)\plus{}b^{3}a^{2}(c\minus{}a)\plus{}b^{2}a^{2}(b^{2}\minus{}c^{2})]}{(a^{2}\plus{}bc)(b^{2}\plus{}ca)(c^{2}\plus{}ab)}$

I tried to use $ a^{2}\plus{}bc\geq b^{2}\plus{}ca$ and it is sufficient to prove that

$ a^{3}b(a\minus{}b)(b\minus{}c)\plus{}a^{3}c^{2}(b\minus{}c)\plus{}a^{2}c^{2}(a^{2}\minus{}b^{2}))\plus{}b^{3}c(b\minus{}c)(c\minus{}a)\plus{}b^{3}a^{2}(c\minus{}a)\plus{}b^{2}a^{2}(b^{2}\minus{}c^{2})\geq 0$

but I cannot prove this one.  :maybe: Can anyone help me, please? I don't know if it is correct to use  $ a^{2}\plus{}bc\geq b^{2}\plus{}ca$

Thank you very much for your help.

[quote="Vasc"]Write the inequality as
$ a^{n\minus{}3}A\plus{}b^{n\minus{}3}B\plus{}c^{n\minus{}3}C\geq 0$,
where $ A \equal{}\frac{a^{3}b}{b^{2}\plus{}ca}\plus{}\frac{a^{3}c}{c^{2}\plus{}ab}\minus{}a^{2}$ etc.
Then, for $ a\geq b\geq c$ we have
$ a^{n\minus{}3}A\plus{}b^{n\minus{}3}B\plus{}c^{n\minus{}3}C\geq b^{n\minus{}3}(A\plus{}B)\plus{}c^{n\minus{}3}C\geq c^{n\minus{}3}(A\plus{}B\plus{}C)\geq 0$.[/quote]

Vasc or Ductrung, can you please show me why $ A\plus{}B\geq 0$ ?

Thank you very much.

P.S. See my previous post.

Manlio,

$ A\plus{}B \equal{} (A\plus{}B\plus{}C)\plus{}(\minus{}C)\ge 0$,

since $ A\plus{}B\plus{}C\ge 0$ and $ C\le 0$.

[quote="Vasc"][quote="ductrung"] I have proved that for all $ n\ge 3$
\[ \frac{a(b^{n}\plus{}c^{n})}{a^{2}\plus{}bc}\plus{}\frac{b(c^{n}\plus{}a^{n})}{b^{2}\plus{}ca}\plus{}\frac{c(a^{n}\plus{}b^{n})}{c^{2}\plus{}ab}\ge a^{n\minus{}1}\plus{}b^{n\minus{}1}\plus{}c^{n\minus{}1}\]
[/quote]

Indeed, the inequality doesn't hold for $ n < 3$ , but for $ n \equal{} 1$ the following nice inequality holds
\[ \frac{a(b\plus{}c)}{a^{2}\plus{}bc}\plus{}\frac{b(c\plus{}a)}{b^{2}\plus{}ca}\plus{}\frac{c(a\plus{}b)}{c^{2}\plus{}ab}\ge\frac{(a\plus{}b\plus{}c)^{2}}{a^{2}\plus{}b^{2}\plus{}c^{2}}.\]
Is it known?[/quote]

Vasc,
It appears in my .pdf document posted at:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=116853[/url]

Manlio,
Please see the .pdf file attached below:

[quote="ductrung"] Vasc,
It appears in my .pdf document posted at:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=116853[/url][/quote]
Yes, you are right, Ductrung. :lol:
But your proof is valid if replace $ a\ge b\ge c$ with $ a\le b\le c$. :P
On the other hand, this inequality is equivalent to

\[ \frac{a(b^{3}\plus{}c^{3})}{a^{2}\plus{}bc}\plus{}\frac{b(c^{3}\plus{}a^{3})}{b^{2}\plus{}ca}\plus{}\frac{c(a^{3}\plus{}b^{3})}{c^{2}\plus{}ab}\ge a^{2}\plus{}b^{2}\plus{}c^{2}.\]

[quote="Vasc"]But your proof is valid if replace $ a\ge b\ge c$ with $ a\le b\le c$. :P [/quote]
Do you mean that the solution I posted there is [b]invalid[/b] with the assumption $ a\ge b\ge c$?
[quote="Vasc"]On the other hand, this inequality is equivalent to
\[ \frac{a(b^{3}\plus{}c^{3})}{a^{2}\plus{}bc}\plus{}\frac{b(c^{3}\plus{}a^{3})}{b^{2}\plus{}ca}\plus{}\frac{c(a^{3}\plus{}b^{3})}{c^{2}\plus{}ab}\ge a^{2}\plus{}b^{2}\plus{}c^{2}.\]
[/quote]
Yes, I see it now. Thank you. So the solution can be shortened by referring to the other's shorter solution ([url]http://www.mathlinks.ro/Forum/viewtopic.php?t=116853[/url]).
But the identity obtained is so nice that my effort is not totally wasted  :lol:
\[ \sum_{cycl}\frac{a(b^{3}\plus{}c^{3})}{a^{2}\plus{}bc}\minus{}\sum_{cycl}a^{2}\equal{}\frac{(a\minus{}b)^{2}(b\minus{}c)^{2}(c\minus{}a)^{2}}{(a^{2}\plus{}bc)(b^{2}\plus{}ca)(c^{2}\plus{}ab)}\plus{}\sum_{cycl}\frac{a^{2}(a\minus{}b)(a\minus{}c)}{a^{2}\plus{}bc}\]

Thank you very much Vasc and Ductrung. :)

Very nice inequalities and solutions. :wink:

[quote="Gibbenergy"][quote="ductrung"]For any positive real numbers $ a,\;b$ and $ c$, prove that
\[ \frac{a(b^{3}\plus{}c^{3})}{a^{2}\plus{}bc}\plus{}\frac{b(c^{3}\plus{}a^{3})}{b^{2}\plus{}ca}\plus{}\frac{c(a^{3}\plus{}b^{3})}{c^{2}\plus{}ab}\ge a^{2}\plus{}b^{2}\plus{}c^{2}\]
[/quote]

 It isn't hard :D. Here is one-line solution for it:
L-R=
$ \frac{(a^{2}\plus{}b^{2}\plus{}c^{2})(a\minus{}b)^{2}(b\minus{}c)^{2}(c\minus{}a)^{2}\plus{}\sum a^{2}b^{2}(a^{2}\minus{}b^{2})^{2}}{2(a^{2}\plus{}bc)(b^{2}\plus{}ac)(c^{2}\plus{}ab)}\geq 0$  :)[/quote]

[quote="ductrung"]...But the identity obtained is so nice that my effort is not totally wasted  :lol:
\[ \sum_{cycl}\frac{a(b^{3}\plus{}c^{3})}{a^{2}\plus{}bc}\minus{}\sum_{cycl}a^{2}\equal{}(\sum a^{2})\frac{(a\minus{}b)^{2}(b\minus{}c)^{2}(c\minus{}a)^{2}}{(a^{2}\plus{}bc)(b^{2}\plus{}ca)(c^{2}\plus{}ab)}\plus{}\sum_{cycl}\frac{a^{2}(a\minus{}b)(a\minus{}c)}{a^{2}\plus{}bc}\]
[/quote]

so it is valid

\[ (\sum a^{2})\frac{(a\minus{}b)^{2}(b\minus{}c)^{2}(c\minus{}a)^{2}}{(a^{2}\plus{}bc)(b^{2}\plus{}ca)(c^{2}\plus{}ab)}\plus{}2\sum_{cycl}\frac{a^{2}(a\minus{}b)(a\minus{}c)}{a^{2}\plus{}bc}\equal{}\frac{\sum a^{2}b^{2}(a^{2}\minus{}b^{2})^{2}}{(a^{2}\plus{}bc)(b^{2}\plus{}ac)(c^{2}\plus{}ab)}\]

which is very hard to create

[quote="ductrung"]
\[ \sum_{cycl}\frac{a(b^{3}\plus{}c^{3})}{a^{2}\plus{}bc}\minus{}\sum_{cycl}a^{2}\equal{}\frac{(a\minus{}b)^{2}(b\minus{}c)^{2}(c\minus{}a)^{2}}{(a^{2}\plus{}bc)(b^{2}\plus{}ca)(c^{2}\plus{}ab)}\plus{}\sum_{cycl}\frac{a^{2}(a\minus{}b)(a\minus{}c)}{a^{2}\plus{}bc}\]
[/quote]
The above identity (and also in the attached .pdf file from above) should read
\[ \sum_{cycl}\frac{a(b^{3}\plus{}c^{3})}{a^{2}\plus{}bc}\minus{}\sum_{cycl}a^{2}\equal{}\frac{(a^{2}\plus{}b^{2}\plus{}c^{2})(a\minus{}b)^{2}(b\minus{}c)^{2}(c\minus{}a)^{2}}{(a^{2}\plus{}bc)(b^{2}\plus{}ca)(c^{2}\plus{}ab)}\plus{}\sum_{cycl}\frac{a^{2}(a\minus{}b)(a\minus{}c)}{a^{2}\plus{}bc}\]
Thank Manlio for correcting it.

[quote="ductrung"]I have proved that for all $ n\ge 3$
\[ \frac {a(b^n \plus{} c^n)}{a^2 \plus{} bc} \plus{} \frac {b(c^n \plus{} a^n)}{b^2 \plus{} ca} \plus{} \frac {c(a^n \plus{} b^n)}{c^2 \plus{} ab}\ge a^{n \minus{} 1} \plus{} b^{n \minus{} 1} \plus{} c^{n \minus{} 1}
\]
[/quote]
We will show that it suffices to consider the case $ n \equal{} 3$ :)
Consider the function $ f(x) \equal{} \frac {1}{a^{x \minus{} 1} \plus{} b^{x \minus{} 1} \plus{} c^{x \minus{} 1}}\sum\frac {a(b^x \plus{} c^x)}{a^2 \plus{} bc}$, we have
\[ (\sum a^{x \minus{} 1})^2 f'(x) \equal{} \sum a^{x \minus{} 1}b^{x \minus{} 1}(a \minus{} b)(\ln a \minus{} \ln b)(\frac {c}{c^2 \plus{} ab} \plus{} \frac {ab(a \plus{} b \minus{} c)}{(a^2 \plus{} bc)(b^2 \plus{} ca)}) \ge \sum ca^{x \minus{} 1}b^{x \minus{} 1}(a \minus{} b)(\ln a \minus{} \ln b)(\frac {1}{c^2 \plus{} ab} \minus{} \frac {ab}{(a^2 \plus{} bc)(b^2 \plus{} ca)}) \ge 0
\]
Hence $ f(x)$ increase for $ x > \equal{} 3$, it follows that $ f(x) > \equal{} f(3)$ and it suffices to prove the inequality in the case $ n \equal{} 3$. :)
PS: The case $ n\equal{}3$, there is a very nice solution with Vornicu Schur. :)


algebra
polynomial
calculus
integration
algebra unsolved

Find all polynomials
\[P_{n}(x)=n!x^n + a_{n-1}x^{n-1}+...+ a_{1}x +(-1)^n(n+1)n\]
with integers coefficients and with $n$ real roots $x_1, x_2, ..., x_n$, such that $k\leq x_k\leq k+1$, for $k=1,2...,n$.

[hide]Well if $n > 3$, then $n! \ge n(n+1)$
and because the polynomial is $n!(x-x_1)(x-x_2)...(x-x_n)$,  we have $\boxed{n(n+1)=n!x_1x_2...x_n}$, but because all of the roots are greater than or equal to 1, and some greater, we must have $n! < n(n+1) \implies n+1 > (n-1)!$  Thus, $n \le 3$.

So for $n=1$, we have $x-2$, which is a valid polynomial, I suppose , because the root lies in $[1,2]$...

$n=2 \implies$ we have $2x^2+a_1x+6$ which factors to $2(x^2+\frac{a_1}{2}x+3)$.  Thus, because $x_1x_2=3, -x_1-x_2=\frac{a_1}{2}$, we have the roots $-1, -3$, which work

Finally, we consider $n=3$  But then we have $6x^3+a_2x^2+a_1x-12=6(x^3+\frac{a_2}{6}x^2+\frac{a_1}{6}x-2)$, which implies the product of three roots in the intervals $[1,2], [2,3], [3,4]$ is 2, which is impossible.

So our only two polynomials are $P_1(x)=x-2$ and $P_2(x)=2(x-1)(x-3)$[/hide]

[quote="costantin07"]for $n=2$ It's about an infinity of solutions defined by $P_n(x)=2(x-\alpha)(x-\frac{3}{\alpha})$ where $\alpha\neq 0$
and we should have $\alpha\in [1,2]$ and $\frac{3}{\alpha}\in [2,3] $ so $\alpha\in [1,\frac{3}{2}] $[/quote]
no your are wrong !
for example $ \alpha = 1.25 $ then $ \frac{3}{\alpha}=2.4 $
now $ x_{1}+x_{2}=1.25+2.4=3.65 $ it means $\frac{-a_{1}}{2}=3.65 $ and $ {a_{1}}= -7.3$ but we know $a_{1} $ is integer :)

Consider $f(x)=2x+\frac{6}{x}=a_1$.  We need a value of $x\in\left[1, \frac{3}{2}\right]$ where $f(x)$ is an integer.  But consider $f(x)-f(y)$, when $0<x<y\leq\sqrt{3}$.  This equals

$f(x)-f(y)=2x-2y+\frac{6}{x}-\frac{6}{y}=2(x-y)+6\left(\frac{y-x}{xy}\right)=(y-x)\frac{6-2xy}{xy}>0$.

Thus $f(x)>f(y)$ if $x>y$ and they are between 0 and $\sqrt{3}$.  In particular, $f(x)$ is decreasing on $\left[1, \frac{3}{2}\right]$.  The value at the endpoints are $8$ and $7$ respectively, so there are only 2 possible integral values of $a_1$, those which have already been found.


Is there a formula for $ 1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{4} \plus{} \cdots \plus{} \frac {1}{n}$?

I don't think so: Its not an arithmetic or geometric series.

But here is a formula for $ \frac{1}{1\cdot2}\plus{}\frac{1}{2\cdot3}\plus{}\frac{1}{3\cdot4}\plus{}......\plus{}\frac{1}{n(n\plus{}1)}$

Also, I think the sum of the reciprocal of perfect squares appraoches $ \frac{\pi^2}{6}$ as you add more and more

And to the 4th power, it approaches $ \frac {\pi^4}{90}$

Isn't it amazing how these things work? pi isn't even related to squares (not really, anyways)

i don't wanna graph nutin'...

lol I thought there would be a simpler formula or way to find one...

$ \sum_{n\equal{}1}^{\infty}\frac{1}{n}\rightarrow\infty$...

Hard to prove...

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

If the sum of reciprocals of the primes diverges, obviously the above sequence diverges... Wait until T0rajir0u comes over here and owns this.

Or you could take the limit as $ n$ approaches $ \infty$ to find that the limit itself is $ \infty$.

The sum of the reciprocals of squares converges!!

We're getting a little off-topic here:  showing off how much you know about series is nice, but not if it doesn't really answer the original question  :wink:

The quantity $ H_n \equal{} 1 \plus{} \frac{1}{2} \plus{} ... \plus{} \frac{1}{n}$ is known as the $ n^{th}$ [url=http://mathworld.wolfram.com/HarmonicNumber.html]harmonic number[/url], and does [b]not[/b] have a simple formula the way arithmetic or geometric summations do.  When problems like this appear on competitions, you really are just supposed to add the fractions.  (For this reason, it can be useful to memorize the value of the first few harmonic numbers as fractions in lowest terms.)

What's really interesting about the harmonic numbers is the other stuff you can figure out about them; see the link for details.  :)

It is also the approximation of the nth prime number, I'm pretty sure.
.

dynamo, that's the reason i wanted to know if there was an easy way to do this. the nth prime number approximation is n(1+1/2+...+1/n)

See (15) on [url=http://mathworld.wolfram.com/PrimeFormulas.html]the prime formula page[/url].  That approximation derives in part from the fact that $ H_n$ is approximately $ \ln n$, although there's a lot of other theory there.

I'm going to lock this topic as the topic has been fully answered.

Just in addition to what t0rajir0u said, it diverges and has name called harmonic series.  I don't think I've really seen it until I started to study for Taylor Series for AP Calculus BC last year so you can look up on the calculus book to see better proof than wikipedia (if you'd like, of course).

Done.


Here is the IMO shortlist problems 1991, I got it from internet.


Prove that phi(a*b)=phi(a)*phi(b)*d/(phi(d)) where phi(n) signifies the number of integers k<n such that (k,n)=1.

[hide]Seems to work, even if not very elegant :

phi(mn) = mn :prod:(p|mn)  ((1-1/p))

I asuume d = gcd(m,n).

There are 3 exclusive cases if p | mn : p|n, p|m, p|n and p|m i.e. p | gcd(m,n) so

:prod:(p|mn)(1-1/p) = :prod:(p|n)(1-1/p) * :prod:(p|m)(1-1/p) / :prod:(p|gcd(m,n))(1-1/p) = phi(n)/n *phi(m)/m / phi(d)/d[/hide]

While we're at it....
Prove that:

[tex]\dsplaystyle{\frac{1}{\phi{(n)}}=\frac{1}{n}\Sigma_{d|n}{\frac{\mu^2{(d)}}{\phi{(d)}}}}[/tex]

Where mu is the mobius mu function.

[quote="JBL"]Also, there is a nice proof that [tex]\displaystyle n = \sum_{d|n} \phi (d)[/tex] which you might try to find.[/quote]

Write down $n$ fractions,
\[
\frac{1}{n} , \,\frac{2}{n} , \,\frac{3}{n} , \,\dots,\frac{n-1}{n} , \,\frac{n}{n} , \,
\]
Simplify them into the irreducible form, the number of fractions with denominator $m$ is $\phi(m)$, and those $m$s are all divisors of $n$.

My atttempt at the second question :

Prove that
$\frac{1}{\phi(n)} = \frac{1}{n} \sum_{d|n} \frac{\mu^2(d)}{\phi(d)}$   $\forall n \in Z_+$

Solution :

By definition of the Mobius function, [tex]\mu(d) \neq 0[/tex] for only those divisors of $n$ that are of the form $\prod_i p_i$ where $p_i | n$, with all distinct $p_i$'s in the divisor $d$.

For these divisors, $\mu(d) = \pm 1 \Rightarrow \mu^2(d) = 1$, and $\phi(d) = \prod_i (p_i - 1)$

Thus,

$\sum_{d|n} \frac{\mu^2(d)}{\phi(d)} = \sum \frac{1}{\prod_i (p_i - 1)}$
        $= \mu(1) + \sum \frac{1}{(p_i - 1)} + \sum \frac{1}{(p_i - 1) (p_j - 1)} + \sum \frac{1}{(p_i - 1) (p_j - 1) (p_k - 1)} + .........$
        $= \prod (1 + \frac{1}{(p_i - 1)})$
        $= \prod \frac{p_i}{p_i - 1}$
        $= \frac{n}{\phi(n)}$

since for $n = \prod_i p_i^{\alpha_i}, \phi(n) = \prod_i p_i^{\alpha_i - 1} (p_i - 1)$

Thus, adding all terms for which $\mu(d) = 0$, $(d | n)$, we have

$\frac{1}{\phi(n)} = \frac{1}{n} \sum_{d|n} \frac{\mu^2(d)}{\phi(d)}$  $\forall n \in Z_+$

**********************************************************************************************
Sambit

Attempt at First Problem :

Prove that for integers $a$ and $b$,
$\phi(ab) = \phi(a) \phi(b) d / \phi(d)$
where $d = \gcd(a,b)$.

Solution :

First, let us prove first that for every integer n,
      $\frac{\phi(n^{c+1})}{\phi(n^c)} = n$

Case 1 : If n is prime, then $\frac{\phi(n^{c+1})}{\phi(n^c)} = \frac{n^c (n-1)}{n^{c-1} (n-1)} = n$

Case 2 : If n is prime power, say $n = p^{\alpha}$,
         $\frac{\phi(p^{\alpha (c+1)})}{\phi(p^{\alpha c})}
	 = \frac{p^{\alpha (c+1) - 1} (p-1)}{p^{\alpha c - 1} (p-1)}
         =  p^{\alpha} = n$

Case 3 : If $n = \prod_i p_i^{\alpha_i}$, then

         $\frac{\phi(n^{c+1})}{\phi(n^c)}
         = \frac{\phi(\prod_i p_i^{\alpha_i (c+1)})}{\phi(\prod_i p_i^{\alpha_i c})}
         = \frac{\prod_i \phi(p_i^{\alpha_i (c+1)})}{\prod_i \phi(p_i^{\alpha_i c})}
	 = \prod_i \frac{\phi(p_i^{\alpha_i (c+1)})}{\phi(p_i^{\alpha_i c})}
         = \prod_i p_i^{\alpha_i} = n$          (by Case 2)

Thus $\forall n \in \mathcal{Z}_+$,  $\frac{\phi(n^{c+1})}{\phi(n^c)} = n \Rightarrow \phi(n^{c+1}) = n \phi(n^c)$

Now, to the first problem, $\phi(ab) = \frac{\phi(a) \phi(b) d}{\phi(d)}$

Let $\gcd(a,b) = d \Rightarrow a = m*d^c, b = n*d, \gcd(m,n) = \gcd(m,d) = \gcd(n,d) = 1$ as $m$ and $n$ cannot have any common factor, and $d^c$ is the maximum power of $d$ that divides $a$. Here, $c \geq 1$.

Now, $\phi(ab) = \phi(m* d^c *n*d) = \phi(m) \phi(n) \phi(d^{c+1})
                    = \phi(m) \phi(n) \phi(d^c) d$         (by previous result)
                    $= \frac{\phi(m) \phi(d^c) \phi(n) \phi(d) d}{\phi(d)}
                    = \frac{\phi(m*d^c) \phi(n*d) d}{\phi(d)}
		    = \frac{\phi(a) \phi(b) d}{\phi(d)}$

		                                                    (Q.E.D)
*****************************************************************************************************
Sambit

A small point to note about the solution to the first problem.

One of the results used is :
for every integer $n$, $\frac{\phi(n^{c+1})}{\phi(n^c)} = n$.

Here, please note that the exponent $c >= 1$, else if we take $c = 0$, then
$\phi(n^1) / \phi(n^0) = n$
$\Rightarrow \phi(n) = n$ (Absurd unless $n=1$)

Regards,
Sambit


linear algebra
matrix
linear algebra unsolved

Let $A$ be complex matrix, and $X$ invertible complex matrix of sizes $n \times n$. Prove equivalence \[A^{2}X A X^{-1}= A \quad \leftrightarrow \quad A^{2}X^{-1}A X = A.\]

Hmmm, someone has told me about a problem in the last issue of AMM, saying that if $A,B$ have the same rank and $A^{2}B=A$ then $B^{2}A=B$. This one is a very particular case. I will not post the proof of that result, because it is not fair, but take it as a hint: that problem is very easy!


[hide="solution"]consider the function $ \Psi(t)\equal{}f(t)g*(t)\minus{}g(t)f*(t)$ clearly $ \Psi(0)\equal{}\minus{}5 < 0$ and $ \Psi(1)\equal{}32 >0$ so there exists a pont $ c \in (0,1)$ where $ \Psi(t)\equal{}$ so done[/hide]

[quote="hell_ever"]is this question true????

consider f(t)=sin t
f*(t)=cos t
g(t)=t
g*(t)=-t

i can't seem to find any t satisfying the condition[/quote]

I don't think you should really care about the functions f and g..... Using the given data, just solve.

Pardesi s got the essence of it.  :D
it is indeed an application of the Intermediate Value Theorem  :D


For real a, b, c ---->

$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} abc) \ge 2[ab(a\plus{}b) \plus{} bc(b\plus{}c) \plus{} ca(c\plus{}a)]$

I'm sorry, there is a little correction(not + abc   -   it's + 2abc)

$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc) \ge 2[ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a)]$

The easiest and fastest method to do this is that we consider this inequality as a f(x)=ax^2+bx+c(x=a).Then we just need to prove:b^2-4ac=<0(since in this case a=b^2+c^2+3-2b-2c>0)!


If $ x$ is postive,prove that$ (1\plus{}x)(1\plus{}x^2)(1\plus{}x^3)>\equal{}8x^3$

AM-GM each term: 1+x >= 2rt2, 1+x^2 >= 2x, 1+x^3 >= 2xrtx.

Then multiply to get desired result.

[quote="aznlord1337"]AM-GM each term: [b]1+x >= 2rt2[/b], 1+x^2 >= 2x, 1+x^3 >= 2xrtx.

Then multiply to get desired result.[/quote]

That is actually $ 1\plus{}x \ge 2 \sqrt x$. :lol:


The sequence $ a_n$ is defined as follows: $ a_1\neq 2,a_2 \equal{} \frac {a_1}{a_1 \minus{} 2},a_{n \plus{} 1} \equal{} \frac {na_n^2}{na_n^2 \minus{} (n \plus{} 1)a_n \plus{} (n \plus{} 1)}$, if $ n\geq2$. Prove that the equality $ a_1 \plus{} 2a_2 \plus{} 3a_3 \plus{} ... \plus{} na_n \equal{} a_1.a_2.a_3.....a_n$ is true for all $ n$.

Define the sequence $ b_1$, $ b_2$, $ b_3$, ... as follows: $ b_1 \neq 2$ is an arbitrary real number,
\[ b_{k + 1} = \frac{b_1 b_2 \dotsm b_k}{b_1 b_2 \dotsm b_k - (k + 1)}\]
for all $ k \ge 1$.

Then $ b_1 b_2 \dotsm b_k b_{k + 1} - b_1 b_2 \dotsm b_k = (k + 1) b_{k + 1}$ for all $ k \ge 1$.  Summing from $ k = 1$ to $ n - 1$, we get
\[ b_1 b_2 \dotsm b_n - b_1 = 2b_2 + 3b_3 + \dots + nb_n,\]
so $ b_1 + 2b_2 + \dots + nb_n = b_1 b_2 \dotsm b_n$.

Shifting $ k$, we get
\[ b_k = \frac{b_1 b_2 \dotsm b_{k - 1}}{b_1 b_2 \dotsm b_{k - 1} - k},\]
so
\[ b_1 b_2 \dotsm b_{k - 1} = \frac{kb_k}{b_k - 1}.\]
Therefore,
\begin{align*}
b_{k + 1} &= \frac{b_1 b_2 \dotsm b_k}{b_1 b_2 \dotsm b_k - (k + 1)} \\
&= \frac{\frac{kb_k}{b_k - 1} \cdot b_k}{\frac{kb_k}{b_k - 1} \cdot b_k - (k + 1)} \\
&= \frac{kb_k^2}{kb_k^2 - (k + 1)b_k + (k + 1)}.
\end{align*}
Thus, the sequences $ (a_n)$ and $ (b_n)$ coincide.

A bit different thinking:
By induction, call $ S_n\equal{}\sum_{k\equal{}1}^nka_k\equal{}\prod_{k\equal{}1}^na_k$, we need to prove $ S_n\plus{}(n\plus{}1)a_{n\plus{}1}\equal{}S_n.a_{k\plus{}1}\Leftrightarrow S_n\equal{}\frac{(n\plus{}1)a_{n\plus{}1}}{a_{n\plus{}1}\minus{}1}$
Continue by induction we need to prove $ \frac{(n\plus{}1)a_{n\plus{}1}}{a_{n\plus{}1}\minus{}1}\cdot a_{n\plus{}1}\equal{}\frac{(n\plus{}2)a_{n\plus{}2}}{a_{n\plus{}2}\minus{}1}$. It's OK by definition of sequence $ (a_n)$!


trigonometry
inequalities
geometry
inequalities unsolved

Prove for all traingle ABC :
$SinA+SinB+SinC \geq Sin^{3}A+Sin^{3}B+Sin^{3}A+SinA.SinB.SinC$
To everyone : My proof is very long And I don't satisfy about it . I need some help

use $\sin{A}=\frac{a}{2R}$, etc... and the inequality becomes
\[4R^{2}(a+b+c)\geq a^{3}+b^{3}+c^{3}+abc \]
i.e.,
\[4(abc)^{2}\geq (a^{3}+b^{3}+c^{3}+abc)(a+b-c)(a+c-b)(b+c-a) \]
which is, under Ravi substitution equivalent to prove that
\[[(x+y)(x+z)(z+y)]^{2}\geq 2xyz[(x+y)(x+z)(z+y)+(x+y)^{3}+(x+z)^{3}+(z+y)^{3}] \]
if $x,y,z \geq 0$.
Expanding gives the equivalent inequality \[[ \sum_{sym}x^{4}y^{2}-\sum_{sym}x^{4}yz ]+[\sum_{sym}x^{2}y^{2}z^{2}+\sum_{sym}x^{3}y^{3}-2\sum_{sym}x^{3}y^{2}z]\geq 0.\] First term is killed by Muirhead $(4,2,0)\succ(4,1,1)$ and second one is just Schur applied with $a=xy$,...

ok first we have these identities:

$\sin A+\sin B+\sin C=\frac{p}{R}$

$\sin A.\sin B.\sin C=\frac{pr}{2R^{2}}$

$\sin^{3}A+\sin^{3}B+\sin^{3}C=\frac{a^{3}}{8R^{3}}+\frac{b^{3}}{8R^{3}}+\frac{c^{3}}{8R^{3}}=\frac{2p(p^{2}-3r^{2}-6Rr)}{8R^{3}}$

(take a look at my posts in the second page of [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=151066&start=20[/url])

now if we put these in the inequality we have to prove that:

$\frac{p}{R}\geq \frac{p(p^{2}-3r^{2}-6Rr)}{4R^{3}}+\frac{pr}{2R^{2}}$

so we have to rpove that:

$4R^{2}\geq p^{2}-3r^{2}-6Rr+2Rr$

so we have to rpove that:

$4R^{2}+4Rr+3r^{2}-p^{2}\geq 0$

but we know that $HI^{2}=4R^{2}+4Rr+3r^{2}-p^{2}$ where $H$ is the orthocenter and $I$ is the incenter...

which is trivialy positive  :!:


In the math exam,we often meet the kinnd of problems with the following asumption:
Let $\{a_n\}_{n=1}^\infty$ be the sequence satisfying the condition $a_{i+j}\leq a_i+a_j,\forall i,j\geq1$
When you meet them,how will solve them?
As for me, I will do as follow:
For fixed point n,put $t=\min\{\frac{a_i}{i}/i=\overline{1,n}\}$,consider $b_i=a_i-it,\forall i=\overline{1,n}$.We have the new sequence$\{b_n\}_{n=1}^\infty$ satisfying:
[b]a)[/b]$b_i\geq 0, \forall \overline{1,n}$
[b]b)[/b]$\exists 1\leq i_0\leq n $ such that:$b_{i_0}=0$
[b]c)[/b]$b_{i+j}\leq b_i+b_j, \forall i,j,i+j \leq n$
I think the new problem is easier than the old one.Here are the example:
[b]1)[/b]Let $\{a_n\}_{n=1}^\infty$ be the sequence satisfying the condition $a_{i+j}\leq a_i+a_j,\forall i,j\geq 1$
Prove that:
\[\sum_{i=1}^n\frac{a_i}{i}\geq a_n\]
Source: APMO 1999/Problem 2
[b]2)[/b]Let $x\in \mathbb {R^*},n\in \mathbb {N^*}$.Prove that:
\[[nx]\geq \sum_{i=1}^n \frac{[ix]}{i}\]
Source:USAMO1981/Problem5
[b]3)[/b]Let $f:\mathbb {R^*}\to \mathbb {R^*}$ satisfying:$f(xy)\leq f(x)\cdot f(y),\forall x>0,y>0$
Prove that:
\[f(x^n)\leq f(x)\cdot f(x^2)^{\frac{1}{2}}\cdots f(x^n)^{\frac{1}{n}},\forall x>0,n\in \mathbb {N^*}\]
Source:China MO 1995/Problem 6
[b]4)[/b]Let $\{a_n\}_{n=1}^\infty$ be the sequence satisfying the condition:$a_{i+j}\leq a_i+a_j,\forall i,j\geq 1$
Prove that:
\[ a_n\leq ma_1+(\frac{n}{m}-1)a_m,\forall n\geq m\]
Source:China MO 1997 /Problem 6
[b]5)[/b]Let $a_1,a_2,...,a_n\in \mathbb R$satisfying:$a_{i+j}\leq a_i+a_j,\forall i,j\geq1,i+j\leq n$.Prove that:
\[\sum_{i=1}^n \frac {a_i}{i^2}\geq \frac {a_n}{n}\cdot \sum_{i=1}^{n}\frac{1}{i}\]
Source:AMM
Valentin, I don't intent posting many problem in the topic.When making this thread my purpose is to discuss the method of solving these problems.So please don't discuss the solution here.If you wish to do so discuss them on the other topic .Remember that:Discuss the method!Thanks :)

:D [b]PIG-FLY[/b]: i agree with your note, but how can you tell me about [b]"problems with same topic"[/b] and [b]"problems could be solved by the same method [/b] :D

[quote="pigfly"]
[b]4)[/b]Let $ \{a_n\}_{n \equal{} 1}^\infty$ be the sequence satisfying the condition:$ a_{i \plus{} j}\leq a_i \plus{} a_j,\forall i,j\geq 1$
Prove that:
$ a_n\leq ma_1 \plus{} (\frac {n}{m} \minus{} 1)a_m,\forall n\geq m$
[/quote]
By induction on $ k$,$ a_n \leq ka_m \plus{} a_{n \minus{} mk}$ for $ k < \frac {m}{n}$
Put $ n \equal{} mk \plus{} r$ with $ r \in \{1;...;m\}$ then
$ a_n \leq ka_m \plus{} a_r \equal{} \frac {n \minus{} r}{m}a_m \plus{} a_r \leq \frac {n \minus{} m}{m}a_m \plus{} ma_1$
(Note $ a_m \leq ma_1$ and $ a_r \leq ra_1$)


superior algebra
superior algebra unsolved

Let $ G$ be an nonabelian finite group, $ Z(G)$ is cyclic and $ G'\subset Z(G)$
If $ \xi: G\to C$ is an irreducible faithful character and $ \xi^2(1) \equal{} \sum_{i \equal{} 1}^{k}\lambda_i(1) \equal{} k$ , $ \lambda_i \in LIN(G)$
show that $ \xi(G \minus{} Z(G) \equal{} 0$

This is Corollary 2.30 of Isaacs's Character Theory of Finite groups.

k is the index of G' in G.  For any irreducible character z (faithful or not, cyclic center or not), z(1)^2 is at most the index of Z(G) in G.  Since G' is contained in Z(G), the index of Z(G) in G is at most k.  Putting this together with z(1)^2 = [ G:G' ] >= [G:Z(G)], you get that G'=Z(G).  Again since z(1)^2 = [G:Z(G)], z vanishes outside of Z(G).


inequalities
algebra
polynomial
inequalities proposed

If $ a,b,c>0$ and $ abc\equal{}1$. Prove that \[ \frac {a^3}{(b\plus{}1)(c\plus{}1)} \plus{}\frac {b^3}{(c\plus{}1)(a\plus{}1)}\plus{}\frac {c^3}{(a\plus{}1)(b\plus{}1)} \ge \frac 34 \]

[quote="mathson"]If $ a,b,c > 0$ and $ abc \equal{} 1$. Prove that
\[ \frac {a^3}{(b \plus{} 1)(c \plus{} 1)} \plus{} \frac {b^3}{(c \plus{} 1)(a \plus{} 1)} \plus{} \frac {c^3}{(a \plus{} 1)(b \plus{} 1)} \ge \frac 34
\]
[/quote]
Nice.
My profe is:
$ \sum \frac {a^3}{(b \plus{} 1)(c \plus{} 1)} \equal{} \sum \frac {a^3}{(abc \plus{} b)(abc \plus{} c)} \equal{} \sum \frac {a^3}{bc(ab \plus{} 1)(ac \plus{} 1)} \equal{} \sum \frac {a^3}{\frac {1}{a}(ab \plus{} 1)(ac \plus{} 1)} \equal{} \sum\frac {a^4}{(ab \plus{} 1)(ac \plus{} 1)}$
So by andreescu we have
$ \sum\frac {a^4}{(ab \plus{} 1)(ac \plus{} 1)} \geq \frac {(a^2 \plus{} b^2 \plus{} c^2)}{a \plus{} b \plus{} c \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} 3}$
So we have to prove that:
$ 4[a^4 \plus{} b^4 \plus{} c^4 \plus{} 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)] \geq 3(a \plus{} b \plus{} c) \plus{} 6(ab \plus{} bc \plus{} ca) \plus{} 9$
$ 3(a^4 \plus{} b^4 \plus{} c^4) \geq 3(a \plus{} b \plus{} c)$(it can be easily killed by power ineq..
And $ 6(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \geq 6(ab \plus{} bc \plus{} ca)$
This is easy to prove by power inequality again...
So we have to prove that:
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \geq 9$
But $ a^4 \plus{} b^4 \plus{} c^4 \geq 3$
and $ 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \geq 2*3 \equal{} 6$ So the ineq ended :)

Another proof.
[hide]
By Holder's inequality we have that
$ \left(\sum{a \plus{} 1}\right)\left(\sum{b \plus{} 1}\right)\left(\sum{\dfrac{a^3}{(a \plus{} 1)(b \plus{} 1)}}\right)\ge (a \plus{} b \plus{} c)^3$

This is equivalent to $ \sum{\dfrac{a^3}{(a \plus{} 1)(b \plus{} 1)}}\ge \dfrac{x^3}{(x \plus{} 3)^2}$ where $ x \equal{} a \plus{} b \plus{} c$.

So we would like to prove $ \dfrac{x^3}{(x \plus{} 3)^2}\ge\dfrac{3}{4}$.

The former is equivalent to $ 4x^3 \minus{} 3x^2 \minus{} 18x \minus{} 27\ge 0$. The left side is a polynomial in $ x$ wich has a zero in $ x \equal{} 3$, from here we factor it as $ (x \minus{} 3)(4x^2 \plus{} 9x \plus{} 9)$.

Since, by AmGm, $ x\ge 3$ and $ 4x^2 \plus{} 9x \plus{} 9 > 0$ we have that $ (x \minus{} 3)(4x^2 \plus{} 9x \plus{} 9)\ge 0$ and this concludes the proof.

*(all sums above are cyclic)
[/hide]

:)


Simplify
   $sin\frac{\pi}{2n}sin\frac{2\pi}{2n}.........sin\frac{(2n-1)\pi}{2n}$

Define f(n)= :prod: sin(i:pi:/n), where n is a positive integer while i ranging from 1 to n-1. We will show that f(n)=n/2n-1.

From De Moivre's Theorem, we have cos(n:theta:)+isin(n:theta:)=(cos(:theta:)+isin(:theta:))n. For n=2m+1 (n is odd) and :theta: is not a multiple of :pi:, equating the imaginary parts gives: sin(n:theta:)/sin(:theta:)=:Sigma:(-1)k(2m+1 C 2k+1)*cos2m-2k(:theta:)*sin2k(:theta:)=:Sigma:(-1)k(2m+1 C 2k+1)*(1-sin2(:theta:)) m-k*sin2k(:theta:) (#), where both sums are taken over k=0 to k=m and (2m+1 C 2k+1) denotes the binomial coefficients. Realizing that RHS of (#), when expanded, is a polynomial of variable sin(:theta:) with (n-1) being the highest degree, together with sin(n:theta:)=0 when :theta:=j:pi:/n for j=1,2,...,n-1. It is plain to see that sin(j:pi:/n) (with j=1,2,...,n-1) are roots of the polynomial on the RHS. Their product, therefore, equals n/2n-1.

Now for n=2m (n is even), equating the real parts will arrive at the same conclusion. So in treegoner's problem, n is replaced by 2n, hence the answer is n/22n-2.

Of course, to be completely rigorous, it requires a bit of work to justify that the constant and the leading coefficient equal, respectively, n and 2n-1 on the RHS of (#). But by my speed of typing, it would probably take days. (I am just not used to type huge formulas and keep track of the terms on computer...:cool: ) Besides, they are "not difficult".  :D

Look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=4105&highlight=sin

Pierre.

Yes, complex numbers work too. Well, I just gave the "non-complex-number" solution that A1lqdSchool wanted,  :lol:

That 's right. You can also use the Chebychev 's polinomial to prove it by using the Viette theorem to the equation $con(n+1)x = cos(n-1)x$


[quote="hoangclub"]Find all possible value of product of prime numbers if that product divisible by each that prime number subtract 1.[/quote]

I think there is something missing in your problem since a lot of such products exist :

$ 2^n$

$ 2^m(2^n\plus{}1)^q$ with $ m\geq$ if $ 2^n\plus{}1$ is prime :
$ 2^m5^q$ (with $ m>1$),
$ 2^m17^q$ (with $ m>3$),
$ 2^m257^q$ (with $ m>15$) and so on

$ 2^m5^q11^r$ with $ m>1$ and $ q>0$,
$ 2^m5^q41^r$ (with $ m>2$ and $ q>0$)
and so on


[color=blue] [size=150]$15{th}$  Iberoamerican Olympiad[/size]
Merida, VENEZUELA. [2000]

Edited by djimenez

Carlos Bravo ;)

[/color]

[quote][b]Problema 1[/b]. Un polígono regular de $n$ lados $(n\geq3)$ tiene sus vértices numerados del $1$ a $n.$ Se trazan todas las diagonales del polígono. Mostrar que si $n$ es impar, entonces es posible asignar a cada lado y a cada diagonal un número entero entre $1$ y $n,$ tal que el número asignado a cada lado o diagonal es diferente al número asignado a cualquiera de los vértices de sus puntos extremos y que además, para cada vértice, todos los lados y diagonales que concurren en él tienen diferente número asignado. [/quote]
Desígnese cada lado con puntos extremos  $i,j\in\overline{1,n-1}$ con el número en $\overline{1,n}$ congruente a $i+j\pmod n.$ Ahora, para cada $i\in\overline{1,n-1}$ todos los números en $\overline{1,n}$ aparecen en los lados adyacentes a $i$ excepto por uno, el cual es diferente de $n$ (de otro modo, se tendría que $2i\equiv 0\pmod n\Longrightarrow i\equiv 0\pmod n,$ lo cual es contradicción. La implicación es pues una consecuencia del hecho que $n$ es impar). Asígnese el lado $in$ con este número (asignación buscada).

[quote][b]Problema 2.[/b] Sean $ \mathcal{S}_1$ y $ \mathcal{S}_2$ dos circunferencias con centros $ O_1$ y $ O_2,$ respectivamente secantes en $ M,N.$ La recta $ t$ es la tangente común de  $ \mathcal{S}_1$ y $ \mathcal{S}_2$ mas cerca de $ M.$ $ t$ toca a $ \mathcal{S}_1$ y $ \mathcal{S}_2$ en $ A,B.$ $ C$ es un punto tal que $ BC$ es un diametro de $ \mathcal{S}_2$ y $ D$ es la intersección de la perpendicular a $ AM$ desde $ B$ con $ O_1O_2.$ Probar que $ C,D,M$ están alineados.[/quote]
Primeramente demostremos el siguiente resultado

[b]Teorema: [/b] En el triángulo $ \triangle ABC,$ sea $ A'$ el antipodal de $ A$ en su circuncírculo $ (O).$ La recta tangente $ \lambda$ a $ (O)$ por $ A'$ corta a $ BC$ en $ P$ y la recta $ OP$ corta a $ AC,AB$ en $ D,E.$ Entonces $ O$ es punto medio de $ DE.$

Basta considerar el antipodal $ C'$ de $ C$ en $ (O)$ y $ E' \equiv A'C' \cap AB.$ Luego por el teorema de Pascal en el hexagono no convexo y degenerado $ A'A'ABCC',$ se tiene que las intersecciones $ P\equiv BC \cap \lambda ,$ $ O \equiv AA' \cap CC'$ y $ E' \equiv A'C' \cap AB$ están alineadas $ \Longrightarrow$ $ E \equiv E'.$ Entonces como $ \angle C'A'C$ es recto, resulta $ A'E \parallel AC$ $ \Longrightarrow$ $ AEA'D$ es un paralelogramo $ \Longrightarrow$ $ O$ es punto medio de $ ED.$
__________________________________________

Ahora, en la figura problema considérese $ T \equiv AB \cap O_1O_2.$ Por ser $ T$ centro de la homotecia positiva que transforma $ \mathcal{S}_1$ en $ \mathcal{S}_2,$ entonces $ TM$ corta por segunda vez a $ \mathcal{S}_2$ en el homologo $ M'$ de $ M,$ y siendo $ P$ el antipodal de $ A$ en el círculo $ \mathcal{S}_1,$ resulta $ BD \parallel MP \parallel M'C.$ Consecuentemente, si $ E \equiv CM' \cap O_1O_2,$ entonces $ D$ ha de ser el simétrico de $ E$ con respecto a $O_2.$  Por otro lado, si $ D' \equiv CM \cap O_1O__2,$ aplicando el resultado previo a la configuración de $ \triangle CMM'$ con la tangente $ t$ por el antipodal $ B$ de $ C,$ resulta que $ D'$ es el simétrico de $ E$ con respecto a $ O_2$ $ \Longrightarrow$ $ D' \equiv D',$  por lo tanto $ C,D,M$ están alineados.

[quote]Problema 3. Hallar todas las soluciones de $(m+1)^a=m^b+1,$ para enteros mayores a 1.[/quote]
Tomando la ecuación $\mod (m+1),$ tenemos $(-1)^b = -1$ $\Longrightarrow$ $b$ es impar. Entonces, podemos dividir el lado derecho por $m+1$ para obtener $m^{b-1} - m^{b-2} + ... - m + 1.$ Esta tiene un número impar de terminos. Si $m$ es impar, entonces cada termino es impar y por ende el total es impar, pero $(m+1)^{a-1}$ es par (ver que a>1), lo cual es contradicción, así $m$ es par. Tenemos $m^b = (m+1)^a - 1.$ Luego, expandiendo el lado derecho de la ecuación usando el teorema del binomio y usando $b>1,$ vemos que $m$ debe dividir $a.$ Así, $a$ es también par. Denotemos $a=2A,$ $m=2M.$ Factorizamos $(m+1)^a - 1$ como $[(m+1)^A + 1] \cdot [(m+1)^A - 1].$ Los dos factores tiene diferencia $2,$ entonces su MCD divide $2,$ pero ambos factores son pares, así su MCD  es exactamente $2.$

Si $M=1$ o es una potencia de $2,$ entonces el menor factor $3^A-1$ debe ser $2,$ así $A=1$ y resulta $3^A+1=4$ $\Longrightarrow$ $(2M)^b=8$ $\Longrightarrow$ $M=1$ y $b=3.$ Así se tiene la solución $(m,a,b)=(2,2,3).$ Si $M$ no es una potencia de $2,$ entonces $M^b>2^b,$ así debemos tener el factor mayor $2M^b$ y factor menor $2^b-1.$ Pero el factor mayor es ahora mayor a $2^{b+1},$ por tanto la diferencia entre los factores es al menos $3 \cdot 2^{b-1}>2.$ Contradicción.

[quote][b]Problema 4.[/b] Algunos términos son borrados de una progresión aritmética infnita $1,x,y,...$ de números reales para producir una progresión geométrica infinita $1,a,b,...$ Hallar todos los valores posibles de $a.$[/quote]
Si $a$ es negativo, entonces los términos de la PG son alternamente positivos y negativos, mientras todos los términos en la PA desde un cierto punto son positivos o todos los términos en la PA desde un cierto punto son negativos. Por ende, $a$ no puede ser negativo. Si $a$ es cero, entonces todos los términos de la PG son cero, excepto el primero, pero a lo mas un término de la PA es cero, así $a$ tampoco puede ser cero. Es decir, que la PA debe tener infinitos términos positivos y por tanto $x \ge 1.$

Sea $d=x-1,$ así todos los términos de la PA tienen la forma $1+nd$ para $n$ natural. Supongamos que $a=1+md,$ $a^2=1+nd$ $\Longrightarrow$ $(1+md)^2=1+nd$ $\Longrightarrow$ $d=(n-2m)m^{-2},$ el cual es irracional. Así, $a$ es irracional. Supongamos que $a=b/c,$ donde $b$ y $c$ son naturales primos relativos y $c>1.$ Luego el denominador del enésimo término de la PG es $c^n,$ el cual se vuelve arbitrariamente grande a medida que $n$ incrementa. Pero si $d=h/k,$ entonces todos los términos de la PA tienen denominador a lo mas $k.$ Entonces, no se puede tener $c>1.$ Por tanto $a$ debe ser entero positivo. Por otro lado, es facil ver que cualquier entero positivo funciona. Tómerse $x=2,$ luego la PA incluye todos los enteros positivos y por tanto incluye cualquier PG con términos enteros positivos.


inequalities
geometry
inequalities proposed

For any accute triangle prove that
$cotA.a^{2}+cotB.b^{2}+cotC.c^{2}\ge 4S_{ABC}$

The following is true
Given two triangles $ABC,A'B'C',S$ is area of $ABC$ then
$\cot A'a^{2}+\cot B'b^{2}+\cot C'c^{2}\ge 4S$

[quote="The soul of rock"]Yes, it's just the same way to prove the first inequality :wink:[/quote]
Really ? What is the way you said ? :D  :D

Because $cotAcotB+cotBcotC+cotCcotA=1$ for all triangle $ABC$ :wink:  :lol:

We have $xa^{2}+yb^{2}+zc^{2}\geq 4S\sqrt{xy+yz+zx}$.

Yes, it's exactly the way I haved this problem.


Is it true that the squareroot of an integer can only have an integer solution, or an irrational solution? Could it be a rational solution, but not an integer?

Yes it can be.
For example, sqrt(4/9)=2/3
sqrt2=1.4142...
sqrt4=2

@[b]mathq[/b]: $ \frac{4}{9}$ is not an integer.

Suppose $ \sqrt{n}\equal{}\frac{p}{q}$ where $ (p,q)\equal{}1$ and $ q>1$. Then $ n\equal{}\frac{p^2}{q^2}$, but $ (p,q)\equal{}1$, so this is only possible if $ q\equal{}1$: contradiction.


The convex quadrilateral $ABCD$ is inscribed in a circle. The tangents to the circumscribed circle in $A$ and $C$ intersect at $P$, such that $P$ does not lie on $BD$ and $PA^2=PB\cdot PD$. Prove that $BD$ intersects $AC$ at the middle of $AC$.

1. My native country is [b]MOLDOVA[/b], thus the Olympiad is called[b] MOLDOVAN [/b]Mathematical Olympiad.
2. This problem was already posted(and solved) here: http://www.mathlinks.ro/Forum/viewtopic.php?t=77685


Call a positive real number special if it has a decimal representation that consists entirely of digits $ 0$ and $ 7$. For example, $ \frac{700}{99} \equal{} 7.\overline{07} \equal{} 7.070707\cdots$ and $ 77.007$ are special numbers. What is the smallest $ n$ such that $ 1$ can be written as a sum of $ n$ special numbers?

$ \textbf{(A)}\ 7\qquad
\textbf{(B)}\ 8\qquad
\textbf{(C)}\ 9\qquad
\textbf{(D)}\ 10\qquad\\
\textbf{(E)}\ \text{The number 1 cannot be represented as a sum of finitely many special numbers.}$

[hide]Let's try to find some list that works.
Suppose we have some finite list of special numbers. Let the total number of 7s in the tenths digits be $ x_1$, in the hundredths digit be $ x_2$ etc. If these number add to 1, then
\[ \sum_{k \equal{} 1}^\infty\frac {7x_k}{10^k} \equal{} 1
\]

\[ \sum_{k \equal{} 1}^\infty\frac {x_k}{10^k} \equal{} 1/7
\]
But this is equivalent to looking for a decimal representation of $ 1/7 \equal{} .142857142857\ldots$ Since each number has at most one 7 per digit we can find a sequence of $ x_i$ that works and we need $ \boxed{8}$ numbers to make it work. So (B) is the answer.[/hide]


geometry
rectangle
limit
integration
calculus
logarithms
real analysis

Evaluate: lim_{k→∞} e^{-k} ∫_R (e^x - e^y) / (x - y) dx dy, where R is the rectangle 0 ≤ x, y, ≤ k.

\[ \displaystyle\lim\limits_{k\rightarrow \infty} e^{-k} \int\limits_{0 \le x, y \le k} \frac{e^x - e^y}{x - y} dx dy  \]

Think I solved it:
\[ \lim\limits_{k\rightarrow \infty} e^{-k} \int\limits_{0 \le x, y \le k} \frac{e^x - e^y}{x - y} dx dy \]
We can reduce area of integration to triangle above line $x = y$, and use substitution $u = x + y, v = x - y$.
Let \[ f (k) = \int\limits_{0 \le x, y \le k} \frac{e^x - e^y}{x - y} dx dy \]
\[{ f (k) = 2 \int\limits_{0 \le x \leq y \le k} \frac{e^x - e^y}{x - y} dx dy = 2 \int_{0}^{k} dv \int_{v}^{2k-v} du \cdot \frac{1}{2} \frac{e^{\frac{u+v}{2}} - e^{\frac{u-v}{2}}}{v}} \]
\[ f (k) = \int_{0}^{k} \frac{dv}{v} (e^{\frac{v}{2}} - e^{-\frac{v}{2}}) \int_{v}^{2k-v} e^{\frac{u}{2}}du = \int_{0}^{k} \frac{2dv}{v} (e^k - e^v - e^{k -v} + 1) = 2(e^k+1)\ln x - 2 \int_0^k {\frac{e^v + e^{k - v}}{v} dv} \]

So, $f'(k) = 2e^k \ln k$. By L' Hospital Rule we get:
\[ \lim \limits_{k\rightarrow \infty} \frac{f(k)}{e^k} = \lim \limits_{k\rightarrow \infty} \frac{2e^k \ln k}{e^k} = \lim \limits_{k\rightarrow \infty} 2 \ln k = \infty \]


modular arithmetic
quadratics
number theory proposed
number theory

find all solution in integers $m,n$ such that:
               $1997^{1001}*(m^2-1)-(2*m-5)=3*\binom {2003^{2004}}{n}$
So nice and very attractive :)

My recent contribution in http://www.mathlinks.ro/Forum/viewtopic.php?t=34664 helps greatly :)

Note, that if $n\neq 0$ and $n\neq 2003^{2004}$, then $2003\mid \binom{2003^{2004}}{n}$ (do you know why? ;) ). Using mresults from the mentioned topic, we see that $1997^{1001}=\left(\frac{1997}{2003}\right)=1$ (we use Legendre symbol here). Thus we get $(m^2-1)-(2m-5)=0\pmod{2003}$, so $(m-1)^2=-3\pmod{2003}$, so $3$ is a quadratic residue modulo 2003. Using my great topic again we see that $\left(\frac{-3}{2003}\right)=-1$. Contradiction!!!

Therefore $n=0$ or $n= 2003^{2004}$. In both cases we obtain \[1997^{1001}(m^2-1)-(2m-5)=3.\] The only integer solution is $m=1$.

I think we can prove $\binom {2003^{2004}}{n}$ dvisible to $2003$ by use $Lucas's  theorem$.It's very useful and nice.Your solution is similar to my solution :)

It is an immediate consequnce from formula of maximal power of $p$ in $n!$.


geometry
3D geometry
sphere
analytic geometry
symmetry
vector

We have an insolated conductive sphere of radius $R$ and of charge $q$. We put a very tiny ball of charge $q$ in distance $d$ from the center of the sphere ($d \gneqq R$). In what distance from the surface of the sphere will there be attraction, and where will be repulsion between the sphere and the charge?

here is a tip:
Calculate were the mirror charge of the small ball will be, and from here calculate the induced charge on the sphere.

I was afraid that would be the tip :o Let's see (I'm not very good at this): when you bring the tiny charge, the image will appear so as to retain the previous potential of the surface of the sphere, right? So the image should be placed symetrically with respect to the center of the sphere... Now you have two charges, the image $-q$ and the sphere $q$. I think that it's not important how the induced charges on the sphere position themselves as long as you have the image placed where it should be. Where am I wrong so far  :maybe:

The image will not be symmetrical with respect to the centre of the sphere.
to make it easier, calculate where the mirror charge will be and how big it will be if the sphere is grounded conductive and its initial charge is 0. After this I will help more.

Since nobody gave a solution, here is mine:
first of all lets calculate the mirror charge if the sphere was grounded.
Lets fix a coordinate system to the center of the sphere.
because of symmetry the mirror charge will be on the line that connects the charge and the center of the sphere.
lets examine a point outside the sphere and calculate the potential there:
the position vector of the charge is $\vec d$ the position vector of the mirror charge $\vec a$ the position vector of the point $\vec x$.
so the potential:
$\Phi (x)=\frac{1}{4\pi \varepsilon_{0}}\frac{q}{|\vec x-\vec d |}+\frac{1}{4\pi \varepsilon_{0}}\frac{q_{m}}{|\vec x-\vec a|}$ (1)
We need to find $q_{m}$ and $a$ so that if $|\vec x|=R$ then $\Phi (x)=0$
if $\vec n$ is the unit vector in the $\vec x$ direction and $\vec{n_{1}}$ is the unit vector in the $\vec d$ direction then:
$\Phi (x=R)=\frac{1}{4\pi \varepsilon_{0}}\frac{q}{R| \vec n-\frac{d}{a}\vec{n_{1}}|}+\frac{1}{4\pi \varepsilon_{0}}\frac{q_{m}}{a|\vec{n_{1}}-\frac{R}{a}\vec n|}=0$
from here:
$q_{m}=-\frac{R}{d}q$ and $a=\frac{R^{2}}{d}$
So the total induced charge on the sphere if it is grounded is $q_{m}$.
now we have a insolated conductive sphere of charge $Q$.
Lets assume that the sphere was at first grounded, and we saw that the induced charge on it was $q_{m}$. Now lets insolate it and put $Q-q_{m}$ charge on it. the total charge of the sphere will be now $Q$.So the later put on $Q-q_{m}$ charge spreads uniformly around the surface of the sphere, because the field caused be the $q_{m}$ charge is wholly compensated be the field of $q$. So the potential outside the sphere by the $Q-q_{m}$ sharge will be the same as if we put this charge in the center of the sphere.
So according to coulombs law the force on the small $q$ charge by the sphere:
$F=\frac{1}{4\pi \varepsilon_{0}}\frac{q}{d^{2}}\bigg [Q-\frac{qd^{3}(2d^{2}-R^{2})}{d(d^{2}-R^{2})^{2}}\bigg ]$
So to solve the problem we have to calculate were this force will be $0$.
In the posted problem $Q=q$ so :
$1-\frac{d^{3}(2d^{2}-R^{2})}{d(d^{2}-R^{2})^{2}}=0$
from here we get the following equation:
$(\frac{d}{R})^{5}-2(\frac{d}{R})^{3}-2(\frac{d}{R})^{2}+(\frac{d}{R})+1=0$
because $d \gneqq R$ the only solution is :
$\frac{d}{R}=\frac{\sqrt{5}+1}{2}$
So the distance from the surface of the sphere where the force will be $0$:
$d_{s}=\frac{\sqrt{5}-1}{2}R$
below this distance there will be attraction above this distance repulsion.


show that given any prime p, there exist integers x, y, z, w
satisfying x^2 + y^2 + z^2 - wp = 0 and 0 < w < p.

[hide]

if p  :equiv: 1 (mod 4)

p is expressible as the sum of two squares

so a:^2: + b:^2: +0:^2: = 1p for some a,b

if p  :equiv: 3 (mod 4)

The only numbers that are not expressible as the sum of three squares are those of the form (4^k)(8j+7)

Clearly, 2p is not a multiple of 4, and 2p  :equiv: 6 (mod 8), so it is not of this form.

Hence, 2p can be expressed as the sum of three squares, so the result holds when w = 2.

If p = 2,  1 + 1 + 0 = 1p

Done

[/hide]


geometry
angle bisector
inequalities proposed
inequalities

Let triangle ABC with side length a,b,c and the angle bisector$l_{a},l_{b},l_{c}$ .Prove that
      $ab+bc+ca\leq 4S\frac{R}{2r}\sqrt{\frac{h_{a}}{l_{a}}+\frac{h_{b}}{l_{b}}+\frac{h_{c}}{l_{c}}}$

And prove
     ${ab+bc+ca\ge 8S(\frac{m_{a}}a}+\frac{m_{b}}{b}+\frac{m_{c}}{c})\ge 4\sqrt{3}S$

[color=purple]Auguste prove or disprove that : [/color]
 $\frac{m_{a}}{a}+\frac{m_{b}}{b}+\frac{m_{c}}{c}\ge \frac{9\sqrt{3}}{4(1+\frac{r}{R})}$


The problem is-
The mark-up on an item is the difference between what a store charges for an item and what the store originally payed for the item. A certain store buys table soccer games for $ 95 each and sells each of them for$399.99. How many times the original cost to the store is the store's mark-up on each game? Express your answer as a decimal to the nearest hundredth.

The mark up is the price-price payed for by the store. In this case, 399.99-95, which is 304.99. Since we are trying to find out how many times higher the price is than the original cost to the store, we do the problem, 304.99/95, which is 3.210421, which rounded to the nearest hundredth is 3.21.

A: 3.21 times :lol:


[b]If a number X consists of 66 digits of threes, and a number Y consists of 66 digits of sixes, name all digits which will appear in the product of X and Y[/b]

umm....
[hide="a guess"]ok... 33*66 has only 2, 1, 7, and 8. 333*666 has only 2, 1, 7, 8. 3333*6666 has the same thing... I'm too lazy to go further.[/hide]

EDIT: btw, Peijin and I posted at the EXACT same time, and his post blocked mine for like 30 sec... it. was. annoying.

uh yah only digits 1,2,7,8  are in the product!!!!

that is indeed the answer. any further justification would be appreciated.

[quote="ZhangPeijin"]Hmm, how do you guys get that?
And your pattern doesn't work, when there are 10 3's and 10 6's there's a 9.[/quote]

lol no there isn't.

Eh, maybe I pressed 9 6's and 10 3's or something...

well, if you say its 333...333*333...333*2
then i'm pretty sure the 3 squared part only has 0,1,8and 9, which, when doubled yields 1,2,7, and 8.

not sure, but i've found the pattern with the threes before i think


Let $(a_n)_{n\geq 1}$ be a sequence for real numbers given by $a_1=1/2$ and for each positive integer $n$
\[ a_{n+1}=\frac{a_n^2}{a_n^2-a_n+1}.  \]
Prove that for every positive integer $n$ we have $a_1+a_2+\cdots + a_n<1$.

Define $b_n = \frac{1}{a_n}$. Then $b_{n + 1} = b_n^2 - b_n + 1$ or \[\frac{b_{n + 1} - 1}{b_n - 1} = b_n.\] Multiplying these relations (the whole thing telescopes) we get \[b_{n + 1} = b_n\cdots b_2b_1 + 1 \Leftrightarrow \frac{1}{b_n} = \frac{1}{b_1b_2\cdots b_{n - 1}} - \frac{1}{b_1b_2\cdots b_{n}}.\] Telescoping again we get \[\frac{1}{b_1} + \frac{1}{b_2} + \cdots + \frac{1}{b_n} = 1 - \frac{1}{b_1b_2\cdots b_n}\] and we are done.


logarithms
linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

Let $A \in GL_n(C)$ be an invertible matrix.
Prove that there exists a complex polynomial $P$ such that:
$e^{P(A)}=A$.
Is it true in $GL_n(R)$ ?

6.2.12
6.2.13
6.2.15  page  418

Topics in Matrix Analysis
 Horn & Johnson


calculus
function
real analysis
real analysis unsolved

In Rudin's Principles of Analysis, the Ex. 6 of Ch. 1 (on p. 22) first defines $ b^r$ for $ r\in\mathbb Q$ with $ b > 1$, and tell us to prove $ b^{r \plus{} s} \equal{} b^r b^s.$ Then it defines $ b^x: \equal{} \sup B(x)$ for all real $ x$, where $ B(x) \equal{} \{e^t: t\in\mathbb Q,t\leq x\}$. Finally, it tell us to show $ b^{x \plus{} y} \equal{} b^x b^y$ for all $ x,y\in\mathbb R$.

We see that the identity is equivalent to
$ \sup B(x \plus{} y) \equal{} \sup B(x)\sup B(y).$
So, does it suffice to show that $ \sup AB \equal{} \sup A\sup B$, where $ A$ and $ B$ are sets whose elements are positive?

If the above idea is correct, is there any other method to prove it?

If you already assume the existence of $ e$, you can prove this equality using the series expansion of $ e$.

It's quite clear from the logical structure of the question (as well as it position in the book) that you can't use calculus information about exponential functions as already established.

This exercise is part of one possible way of developing and constructing exponential functions in the first place.


Prove that for any triangle $ABC$ the following inequality holds:

    $6\sqrt{3} r(l_{a}+l_{b}+l_{c})\leq (a+b+c)(h_{a}+h_{b}+h_{c}).$ ;)

We denote $S=area[ABC]$ and $p=\dfrac{a+b+c}{2}.$
We have $r=\dfrac{S}{p},$ $l_{a}\leq \sqrt{p(p-a)}$ and analogues.
Firstly, from $CS:$ $(a+b+c)(h_{a}+h_{b}+h_{c})\geq 18S,$
then $6\sqrt{3} r(l_{a}+l_{b}+l_{c})\leq \dfrac{6S\sqrt{3}}{\sqrt{p}}(\sqrt{p-a}+\sqrt{p-b}+\sqrt{p-c}).$
It remains to prove that $\sqrt{p-a}+\sqrt{p-b}+\sqrt{p-c}\leq \sqrt{3p},$ obvious from $CS$ inequality.


number theory
least common multiple
greatest common divisor

The least common multiple of two integers is 36 and 6 is their greatest common divisor. What is the product of the two numbers?

Start to list the numbers whose least common multiple is 36:

36, 2 - nope
36, 3 - nope
36, 6 - yup!

$ 36 \times 6 \equal{} \boxed{216}$.

or, there is an interesting relation, which states that for positive integers $ a$ and $ b$, \[ \text{LCM}(a,b)\cdot \text{GCF}(a,b)\equal{}ab\]
So $ ab\equal{}36\cdot 6\equal{}\boxed{216}$
This is proven using the definition of LCM and GCF, and I don't believe there exists a similar relation for 3 or more variables

[quote="5849206328x"]...
This is proven using the definition of LCM and GCF, and I don't believe there exists a similar relation for 3 or more variables[/quote]

Yeah--it doesn't, because for every $ p_i^j$ for the first number, $ p_i^k$ for the second, and $ p_i^l$ for the third, the least common multiple and greatest common factor can determine the maximum or minimum of these three values $ j,k,l$, but not the middle one. Thus, the product of the three numbers can't be represented in terms of their least common multiple and greatest common factor. A similar argument holds for more numbers.

Actually, in ANY case where one number is a factor of the other, the lower number is the GCF and the higher number is the LCM.

Did you even read my post? It has almost nothing to do with what you just said.

...uh math154...i wasn't even talking about ur post...


The number of ordered pairs of integers (m,n) for which mn >= 0 and

m^2 + n^3 + 99mn = 33^3

is equal to:

oh god not this again
its 35
[hide]
(-33,-33) works plus all the pairs for which m+n=33

[/hide]

Okay for one thing, this is not even right year. It's 1999 #30. Second, as CircleSquare exclaimed, this has been posted many times so please, please, use search button!

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1278276222&t=68389

Locked.


Let p and q be two circles, of centers P and Q respectively, which are tangent at R. Let t be the tangent at q through T. Let U  be the intersection between t and p which is inner  to the circle v which have diameter PQ. Finally let S be the intersection between t and v inner p. Prove that SQ cut a chord XY on c(TUR)  so that TU=XY.

Some equivalent thesys of that claimed in this problem can be used to give an one line solution of the Sangaku problem  treated at

http://www.cut-the-knot.org/Curriculum/Geometry/CirclesAndRegularTriangle.shtml

Starting from this result one can also solve the problem of the 17-th Iranian Mathematical Olympiad 1999/2000 posed at

http://www.mathlinks.ro/Forum/viewtopic.php?t=65854


trigonometry
function
real analysis
real analysis solved

let $a_n$ be a decreasing sequence that converges towards $0$. Prove that the 2 following assertions are equivalent:
1/$a_n=o(1/n)$
2/$\sum_0^{+\infty}a_nsin(nx)$ converges uniformly on $R$

Let's try it.

Suppose the series $\sum_{n=1}^{\infty}a_n\sin nx$ uniformly converges on $\mathbb{R}$ and $na_n\neq 0$. Then there is a sequence $\{a_{k_l}\}$ s.t. $k_la_{k_l}\geq \alpha>0$.
Since the series uniformly converges we can find such $N$ that for all $n,m>N$ we have $\left|\sum_{k=n}^{m}a_k\sin kx\right|<\frac{\alpha}{5}$.
Let $k_{L}>N$, $x=\frac{1}{k_l}$, $l>L$. We can write $\sin kx>\frac{kx}{2}$ for all $k\in\{k_L,...,k_l\}$. Therefore \[S_l=\sum_{k=k_L}^{k_l}a_k\sin kx>\frac{1}{2}\sum_{k=k_L}^{k_l}a_kkx=\frac{1}{2k_l}\sum_{k=k_L}^{k_l}a_kk.\] Note that we must have $S_l<\frac{\alpha}{5}$ for all $l>L$.
Using $a_k\geq a_{k_l}\geq \frac{\alpha}{k_l}$ for all $k\leq k_l$ we obtain \[\sum_{k=k_L}^{k_l}a_kk> \frac{\alpha}{k_l}\sum_{k=k_L}^{k_l}k=\frac{\alpha}{k_l}\cdot \frac{k_l+k_{L}}{2}\cdot (k_l-k_L+1)>\frac{\alpha k_l}{4}\] for large $l$. It follows that $S_l>\frac{\alpha}{4}$. Contradiction.

I have proved one part of the problem. Another part has a similar proof, I think. I am too lazy to write it.

[quote="alekk"]1 => 2 is more interesting ![/quote]
Ok! Let's remove my post. :(

Solution by Prof. Soloviev A.A.

Let $x\in(0,\pi)$. Choose $m$ in such way that $\frac{\pi}{m+1}<x\leq\frac{\pi}{m}$.
Then $R_r(x)=\sum_{n=r}^{\infty}a_n\sin nx=\sum_{n=r}^{r+m-1}a_n\sin nx+\sum_{n=r+m}^{\infty}a_n\sin nx=S_1+S_2$.

We have for $S_1$ the following estimate:
\[\left|\sum_{n=r}^{r+m-1}a_n\sin nx\right|\leq \sum_{n=r}^{r+m-1}a_nnx\leq \frac{\pi}{m}\sum_{n=r}^{r+m-1}na_n\leq \frac{\pi}{m}\cdot m\cdot\sup_{n\geq r}\{na_n\}=\pi\sup_{n\geq r}\{na_n\}.\]

Write for $S_2$ the following identity:
\[S_2=\sum_{n=r+m}^{\infty}a_n\sin nx=\sum_{n=r+m}^{\infty}(a_n-a_{n+1})\sum_{k=1}^{n}\sin kx-a_{r+m}\sum_{k=1}^{r+m-1}\sin kx=\] \[=\sum_{n=r+m}^{\infty}(a_n-a_{n+1})\frac{\cos(n+1/2)x-\cos x/2}{\sin x/2}-a_{r+m}\frac{\cos(r+m-1/2)x-\cos x/2}{\sin x/2}.\]

We have \[\left|a_{r+m}\frac{\cos(r+m-1/2)x-\cos x/2}{\sin x/2}\right|\leq a_{r+m}\frac{2}{\sin x/2}\leq\] \[\leq 2(m+1)\frac{1}{r+m}a_{r+m}(r+m)\leq 2\sup_{n\geq r}\{na_n\}\] and \[\left|\sum_{n=r+m}^{\infty}(a_n-a_{n+1})\frac{\cos(n+1/2)x-\cos x/2}{\sin x/2}\right|\leq \sum_{n=r+m}^{\infty}(a_n-a_{n+1})\frac{2}{\sin x/2}\leq\] \[\leq 2(m+1)a_{r+m}\leq 2\sup\{na_n\}.\]
Therefore $|S_2|\leq 4\sup\{na_n\}$. It follows that $|R_r(x)|\leq |S_1|+|S_2|\leq (\pi+4)\sup_{n\geq r}\{na_n\}.$
That's all.

I can't say that it is a well known fact, but I have just found it in the brilliant book "The Theory of Functions" by E.C. Titchmarsh (1939). The theorem belongs to Chaundy and Jolliffe (I guess it has french roots).

All I can say is that in France it is extremely well-known. They use it quite often for oral examinations at X and ENS. You must be really unlucky to have something like this in an oral examination, unless you already know the main ideas.


2n points are on plane . prove that there exists a line with these properties:
1.it bisects the 2n points (half of the points on each side of the line)
2.the sum of the distances of points on one side from the line equals the sum of the distances of points of the other side from the line
(e.g for n=1 the perpendicular bisector of the points is the line with the two properties )

Своиство 2 верно для вся прямая $ d$, проходящая через точка $ G\equal{}$точка тяжелости точки:  $ A_1,A_2,\dots ,A_{2n}$.

is it possible for you to explain more,please.
i don't understand why any of the straight lines passing through .... is the answer
also in english if possible .