ratio
function
MATHCOUNTS
binomial coefficients

Is there a way to figure out the number of ways to make 65 cents without writing them all down?  Is there a formula I can use?  Thanks!!

the only way that i know of is being really organized...like put up a table so you can always keep track of which coins you are using/have already used..

Yes, Mihail911 is right (at least that I know of).

Here: [url]http://mathforum.org/library/drmath/view/57912.html[/url]

That should help somewhat. ;)

Theres a "elementary" way using extended balls and urns..but you have to realize some things and in the end, its almost just listing out (but a lot quicker)

Or...you compute the coefficient of $ x^{65}$ in the following:

$ (1+x+x^{2}\cdots+x^{65})(1+x^{5}+x^{10}+\cdots+x^{65})(1+x^{10}+x^{20}+\cdots+x^{60})(1+x^{25}+x^{50})(1+x^{50})$,

which I believe can be done with like negative binomial coefficients, I'm not exactly sure.

Uh being "organized" will actually (since so many numbers) probably lead to a higher mistake ratio compared to generating functions.

However, can someone get an answer for this?  Ill try it my way and see if I get the same answer.

[hide="probably very wrong"]We can greatly simplify this problem. We know that there must be a number of pennies that is a multiple of 5(clearly), so let's convert pennies to nickels. To account for this in counting, we will multiply each case's possibilities by $ 2^{n}$, where it includes $ n$ nickels. Now we must analyze the cases...
Case 1: all nickels. 14.
Case 2: 12 nickels. none.
Case 3: 11 nickels. (1 dime). 12.
Case 4: 10 nickels. none.
Case 5: 9 nickels. (2 dimes). 10.
Case 6: 8 nickels. (1 quarter). 9.
Case 7: 7 nickels. (3 dimes). 8.
Case 8: 6 nickels. (1 dime, 1 quarter). 7.
Case 9: 5 nickels. (4 dimes). 6.
Case 10: 4 nickels. (2 dimes, 1 quarter).5.
Case 11: 3 nickels. The first relatively complex one...(5 dimes), (2 quarters), (1 halfdollar). 3x4=12.
Case 12: 2 nickels. (3 dimes, 1 quarter). 3.
Case 13: 1 nickel. (6 dimes), (1 dime, 2 quarters), (1 dime, 1 halfdollar). 3x2=6.
Case 14: no nickels. (4 dimes, 1 quarter). 1.

We add these to get a final answer of $ \boxed{93}$[/hide]

I'm ~70% sure that's wrong.

Perfect I dont understand your solution.  If its all nickels how is it 2^13?  Isnt it only 1 case for all nickels since we are not picking the exact order for the nickels?

Ooh i just realized what a total moron i was it's not $ 2^{n}$ i must not have been thinking. i'll edit later.

[hide]Okay. I just used generating functions and got 93 as the answer.[/hide]

[quote="#H34N1"]Okay. I just used generating functions and got 93 as the answer.[/quote]

I seriously doubt you did.  I dont even think you know generating functions.

If you did, then post a solution of how you did it.

Thanks for the replies..but I am still confused about the about the correct way to calculate this answer.  #H34N1 can you show the work you did?  Perfect628 were you going to change your work?

Ordinary Boy (or anybody else who understands) , can you also post your reasoning/method for the extended   version of the balls and urns?

[quote="13375P34K43V312"]But the larger the number, and if you know what you're doing, my method is much, much quicker, and less mistake-prone. Except I don't know exactly how to carry out the computations in my method, unfortunately...[/quote]
One trick that greatly simplifies this generating function is to realize that the number of pennies has to be a multiple of 5.  We can divide the entire equation by  and look for the coefficient of $ x^{13}$.

I asked my dad (Ravi B) how to solve that generating function, and he said that there is a complicated solution in the book [i]Concrete Mathematics[/i] by Graham, Knuth, and Patashnik.  However, it is only simpler than the approaches already given for very large numbers.

[quote="ordinary boy"]I get an answer of $ 93$ using an extended form of balls and urns.
$ 2\binom{8}{2}+\binom{6}{2}+\binom{5}{2}+4\binom{3}{2}=93$[/quote]

Umm this way is a lot easier to understand...
You can always sum it up in terms of combinations...

Now how did I get there...heres a start

$ p+5n+10d+25q+50h=65$, now we see that pennis (p) must be a multiple of 5 (why?) so we can rewrite as $ 5x+5n+10d+25q+50h=65$, simplifying to $ x+n+2d+5q+10h=13$...a lot easier...
Now we can get rid of that 10h by creating 2 cases, $ h=0$ or $ h=1$
we get if $ h=0$
$ x+n+2d+5q=13$, we can easily do this with parity (kind of that has to be a multiple of 5 trick) and you will end up getting all coefficients equal to 0 and you can do easy balls and urns (thats why for the combinations)...
that was just a start but it only takes like 5 mins..
someone can finish this off...you can make subcases such as q=0..and all that..

[hide]
1 Half dollar (50) 15 cents needed
:1 dime (60)
::1 nickel (65)
::5 pennies (65)
:2 nickels (60)
::1 nickel (65)
::5 pennies (65)
:1 nickel (55)
::10 pennies (65)
:0 nickels (50)
::10 pennies (65)

6 ways

2 Quarters

see above: 6 ways

1 Quarter

40 cents left, see above: 6 ways, 25 cents left

2 dimes (20)
:1 nickel (25)
:5 pennies (25)
1 dime (10)
:3 nickels (25)
:2 nickels (20)
::5 pennies (25)
:1 nickel (15)
::10 pennies (25)
:0 nickels (10)
::15 pennies (25)
0 dimes (0)
6 ways to make 25 cents with pennies and nickels
6*6=36

6 dimes:2 combinations(1 nickel, 0 nickel)

5 dimes: 4 ways

4 dimes: 6 ways

3 dimes: 8 ways

2 dimes: 10 ways

1 dime: 12 ways

0 dimes: 14 ways

6+6+36+56=104 ways according to my calculations
[/hide]

[quote="preet patel"][quote="AnonAmoz"]I'm not so good at this stuff... can someone finish it?[/quote]

ya sure you can see from that that all of the variables must satisfy 13.  Since there are p,n,d,q,h 5 variables, its 13*5 but you have to add 7*4 to it because you have extra 2 and 5 coeffiticnets left on d and q.  And there are 4 total variables left in all after h has been set to 0 (p,n,d,q).  So its (2+5)(4)+13(5)=65+28=93[/quote]

... umm, I still don't really get it. Can you give more details? (Maybe take us through the whole process, and why do you multiply by 5 because there are 5 variables?)

Um guys, you all are making this muuuuuuuuuuuuuuuuch harder than it actually seems.  In a real Mathcounts competition, I would just use the safest and easiest method of making a table and listing out the possibilities in a systematic fashion.
[hide]I got 93 as my answer using the aforementioned listing method, and it took me about 4 minutes. :D [/hide]

ok, but as this is not competition, we can still explore some other possibilities that could be faster if you got good at them, right?

Since 65 is not such a big number, especially in coins, I'm going to have to say that listing is the quickest method...

ok... but it would still be interesting to know the balls and urns method, so could someone explain it in more depth?

Yes, I would like to know too, it sounds very useful.

you have

50h+25q+10d+5n+p=65

but i don't know how to do it with ball and urns lol.

this is sort of an old thread, but how to do it with balls and urns was never explained, so if someone could offer some thoughts, it would be extremely helpful.

I think what this turned into would more be appropriate in the High School Topic.  In Mathcounts you [b][u][i]really[/i][/u][/b] don't have time to remember HUGE formulas like that.  Listing is the easiest with small numbers.  Big numbers would be better for those formulas.


I having trouble with this problem -

There are a bunch of fields with names A through Y and a through z. There is also a person on each field that has a capital letter (i.e. A, K). The capital letter Z is reserved for the house. There is also a list of paths connecting the different fields, and paths connecting some of the fields to the barn. The first charectar is one of the fields, the second charectar is the other field, and the number is the distance of the path between them. What is the shortest path the person on the field S can take to get to the house, using the following list of paths?

[code]A C 186
A E 6
A G 135
A J 98
A O 94
A Q 161
A W 12
B C 25
B H 33
B K 199
B O 9
B Q 151
B V 96
B W 162
C F 188
C G 19
C S 175
C X 77
C Y 144
D E 86
D T 105
E F 73
E T 100
E V 3
F I 199
F K 149
F M 52
F P 199
F Q 165
F R 168
F U 128
G H 116
G J 126
G O 126
G R 13
G W 111
H I 120
H J 90
H L 2
H N 178
H X 29
H Y 183
I M 25
I Q 173
I R 86
I T 29
I W 143
I X 65
J T 182
J V 60
K R 50
L M 115
M O 183
M S 147
M X 95
N S 31
N V 149
O R 1
P S 112
Q S 87
S V 63
T X 160
T Y 131
U V 187
a d 168
a f 105
a j 109
a r 171
b h 78
b j 198
b n 150
b o 32
b r 125
b v 24
b y 79
c e 9
c n 182
c o 28
d f 198
d j 118
d l 201
e f 65
e g 67
e i 17
e v 190
e w 173
e x 105
e y 116
f k 152
f v 192
f x 26
g l 169
g n 56
g o 110
g t 30
g v 120
g x 101
h i 18
h q 123
h r 49
h y 180
i k 13
i l 185
i q 17
i v 195
j m 199
j q 175
j w 110
k m 187
k o 91
k r 62
k w 10
k y 64
l n 40
l q 183
l t 63
l y 55
m r 196
m u 38
m x 98
n t 202
n y 7
o v 122
o w 75
p r 156
p s 77
p y 44
q r 107
q s 97
q t 176
r w 32
s u 116
s y 51
t y 157
u v 36
D d 136
F f 177
L l 151
N n 145
S s 19
X x 116
b Z 54
m Z 196
q Z 145
u Z 91
[/code]

[i]Note: I took out all of the paths that are more than 203 long out of this list[/i]

I keep getting 226, but the correct answer is 203. Can anyone explain how to get 203?

Is this a solve-by-hand problem or a solve-by-computer problem? I don't really think theres any nice way of solving by hand (apart from trial + error using one of those algorithms).

But if you want the answer, I wrote a quick program and got:
S-s-y-b-Z gives 203.

Hi

I think it would be better if you could post your code.

braineater

It's actually a computer program, but I just want to know how the answer is 203, becuase when I tried it by hand, I kept getting 226.


function
inequalities unsolved
inequalities

Let $ x,y,z$ be positive integers such that $ x^4 \plus{} y^4 \plus{} z^4 \equal{} 1$
find the minimum value of $ \frac {x^3}{1 \minus{} x^8} \plus{} \frac {y^3}{1 \minus{} y^8} \plus{} \frac {z^3}{1 \minus{} z^8}$

[quote="skywalkerJ.L."]Let $ x,y,z$ be positive integers, and $ x^4 \plus{} y^4 \plus{} z^4 \equal{} 1$
[/quote]
What? :maybe:

[quote="BG Yoda"]Obviously $ x,y,z$ must be just reals. Nevertheless, the problem is still elementary even then.

$ f(x) \equal{} \frac {x^3}{1 \minus{} x^8}$.
$ f'(x) \equal{} \frac {3x^2(1 \minus{} x^8) \plus{} x^38x^7}{(1 \minus{} x^8)^2} \equal{} \frac {5x^{10} \plus{} 3x^2}{(1 \minus{} x^8)^2}$
$ f''(\sqrt {x}) \equal{} \frac {(25x^4 \plus{} 3)(1 \minus{} x^4)^2 \plus{} (5x^5 \plus{} 3x)2(1 \minus{} x^4)4x^3}{(1 \minus{} x^4)^4} > 0$ for $ x < 1$ and therefore the function is convex.

And it leaves to prove $ (x,y,z)\succ (\frac {1}{\sqrt [4]{3}},\frac {1}{\sqrt [4]{3}},\frac {1}{\sqrt [4]{3}})$, which is trivial, before using Karamata.[/quote]
Using Am-Gm for 9 numbers, we are easily to see that:
$ \frac {x^3}{1 \minus{} x^8} \geq\ \frac {9\ \sqrt [4]{3}x^4}{8}$
=> q.e.d  :)

[quote="skywalkerJ.L."][quote="nguoivn"]Using Am-Gm for 9 numbers, we are easily to see that:
$ \frac {x^3}{1 \minus{} x^8} \geq\ \frac {9\ \sqrt [4]{3}x^4}{8}$
=> q.e.d  :)[/quote]

 :huh:
Sorry ,I don't know how to do it.  :?:
Which numbers should I use?[/quote]
$ 8x^3 \plus{} 9\sqrt [4]{3}x^{12} \equal{} x^3 \plus{} x^3 \plus{} ... \plus{} x^3 \plus{} 9\sqrt [4]{3}x^{12}\geq 9\sqrt [4]{3}x^4$


What is the smallest whole number such that if it is divided by 2, 3, and 4, the remainders will be 1, 2, and 3 respectively?

If they leave a remainder of 1, 2 and 3 when divided by 2, 3 and 4, respectively, then they must be one less than a multiple of each of those.  The smallest whole number with this property is then $ \text{LCM}(2,3,4)\minus{}1\equal{}\boxed{11}$


For a collection of sets $ F$ in $ \mathbb{R}^{n}$ define $ S \equal{} \cup_{A \in F} A$ and $ T \equal{} \cap_{A \in F} A$.
If $ x$ is a limit point of $ S$, prove or disprove that it is an limit point of each set in $ F$.
If $ x$ is a limit point of $ T$, prove or disprove that it is an limit point of each set in $ F$.

Larger sets have more limit points. If $ X$ contains $ Y$, the set of limit points of $ X$ contains the set of limit points of $ Y$.

That's really all you need here.


The question: You've drawn a five card hand from a standard 52 card deck. You have a pair and three "worthless" cards. You discard those three cards and draw three new ones. What is the probability (ignoring the high card concept) that your hand will improve? (You get a three of a kind, another pair, or four of a kind.)

The answer: Between 28 and 29 percent.

The challenge: Find and describe ALL of the cases.

Have fun. :)

Odd coincidence: two of my friends came to me over the past two days asking for help on this exact same problem!  Well, if you want to find the probability, you can use a [hide]set counting argument: ie, what do you improve it to?  two pair?  three of a kind?  full house?  four of a kind? and the cases become easier[/hide] but i'll leave it to someone else...
Considering this is just listing out of cases, I don't know if it belongs in Pre-Olympiad...

Er...This may come as a surprise to you, Krishanu, but this is Katie. :)
I like posting the problems here after I'm done with them, since the kids at math team seem to have so much fun.

(Darn it! Now no one else will try...) :)

Though, of course, you have the upper hand, since we usually discuss these with you -before - we hand them in...

Anyone else is still free to try the problem! It's fun...after you get past the frustration of not getting it the first few times.  :P


Hi all i am ur new mod please pour in ur suggestions...all of them will be duly answered :)

will work with problems taken in the Iberoamerican Physics Olympiad and Olympic issues from several countries in Latin America,one  PROBLEM FOR WEEK

  I am coach of Peru Physics Olympics


modular arithmetic
number theory proposed
number theory

Show that there are infinitely many natural numbers $ k$ such that for each $ n\in N^{*}$, the number $ k\cdot 2^{n}\plus{}1$ is composite.

if $ n\equiv 1\pmod{4}$ $ \implies$
$ k\equal{}2^{4s\plus{}1}$ $ \implies$ $ 4.2^{4q}\plus{}1$  (Sophie Germain)
$ 2^{q}\equal{}m$ $ \implies$ $ 4.m^{4}\plus{}1\equal{}(2m^{2}\plus{}1)^{2}\minus{}4m^{2}\equal{}(2m^{2}\minus{}2m\plus{}1)(2m^{2}\plus{}2m\plus{}1)$
$ 2m^{2}\minus{}2m\plus{}1\equal{}1$ $ \implies$  $ m^{2}\minus{}m\equal{}0$ , $ m\equal{}1$ $ \implies$ $ q\equal{}0$ contradiction!
if $ n\equiv 2\pmod{4}$  $ \implies$  $ k\equal{}2^{4s}$
if $ n\equiv 3\pmod{4}$  $ \implies$  $ k\equal{}2^{4s\minus{}1}$
if $ n\equiv 0\pmod{4}$  $ \implies$   $ k\equal{}2^{4s\minus{}2}$


calculus
integration
trigonometry
logarithms
function
calculus computations

Compute:
$ \int_{0}^{e}\ {cosx.lnx}\ {dx}$

$ \frac {d}{dx} \sin x \ln x \equal{} \cos x \ln x \plus{} \frac {\sin x}{x}$
So $ \int \cos x \ln x \ dx \equal{} \sin x \ln x \minus{} \mathrm{Si} x$.

Thank you,Kouichi.But i'm sorry,i don't know what's Si(x).

It's a name for the non-elementary function $ \int_0^x \frac {\sin t}{t}\,dt$. Your integral cannot be evaluated exactly in closed form.
Of course, there should be a $ \plus{} C$ in that antiderivative.

hello, after all we get
$ \int_{0}^{e}\cos(x)\ln(x)\,dx\equal{}\sin(e)\minus{}Si(e)$.
Sonnhard.


integration
inequalities
limit
inequalities unsolved

Prove that $ a \cdot e^a \plus{} 1 \ge e^a$ for $ a > 0$ by elementary methods.

Do [b]not[/b] use this solution:
[hide]
$ f(x) \equal{} e^x$ then $ f'(x) \equal{} e^x$.
Then since $ f'(x)$ is strictly increasing:
$ \frac {e^x \minus{} e^{x_0}}{x \minus{} x_0} \le e^x$. (when $ x > x_0$)
Set $ a \equal{} x \minus{} x_0$ and then:
$ e^{x_0}(e^a \minus{} 1) \le ae^{x_0 \plus{} a}$ $ \iff$
$ ae^a \plus{} 1 \ge e^a$
[/hide]

[quote="Altheman"]You can prove this using the taylor series for $ e^x$...[/quote]
That isn't really elementary is it?

I want a proof using the definition of $ e$. (i.e. $ \int_e^1 \frac{1}{x} dx \equal{}  1$)

There is a really nice proof using Am-Gm inequality and $ e^a\equal{}\lim_{n\to \infty}\left(1\plus{}\frac{a}{n}\right)^n$

I don't see how your proof is any more elementary. There are alternate ways of defining $ e$ which use the fact that I cite.


geometry
3D geometry
sphere
analytic geometry

The half a sphere with a 'm' mass and 'r' radius is equilibrated on a plan absolutely polished horizontal with the side of the AB wire. So the surface of the half a sphere has with the horizontal plan an $\alpha$ angle. Find, after the break of the wire, the velocity of the O center and her maximal value too, also the maximal value of the pressure of the half-sphere over the horizontal. Find that for small angles it behaves like a mathematical pendulum.
Illustration:
[url=http://img135.imageshack.us/my.php?image=fiz1rf6.jpg][img]http://img135.imageshack.us/img135/9526/fiz1rf6.th.jpg[/img][/url]

The two arrows in the illustration to the right (at the first post) are gravity $mg$ and counteraction $N$.
The kinetic energy is :
$E_{k}= \frac{{I\omega^{2}}}{2}$
$OC = \frac{3}{8}r$
The generalized coordinate is the angle $\alpha$. Any help appreciated.


Hi, everyone!

If you were waiting to apply to HCSSiM 2007 until the Interesting Test became available, wait no longer---it's ready, and you'll receive it within a day or so of submitting your online application.

Alumns who aren't eligible to reapply to HCSSiM can get a copy of IT by emailing me and asking *grin.*

Also, we're hosting an informational Math Jam next week (Thursday the 29th), and alumns are welcome to attend---it was super-helpful last year to have y'all answering questions.

Is there any way I can see the problems before actually applying?

Just apply, you just fill out a short thing about yourself on the website, and they email you the IT.  The online part is fairly short, it wont be any trouble at all.  The IT is extremely fun and definitely worth doing even if you decide later you dont really want to apply, it is very different than most other application exams, so dont expect it to be as similar to contest math as many of them are.

Also, the IT is designed to be short (I think the guidelines said 3-4 hours), and also doesn't require prior knowledge of any concept or formula, just ingenuity. It's also quite fun.

I also encourage anyone interested in HCSSiM to attend the MathJam. Last year, I wasn't even considering applying for 2006 (1700+17*17+17) until I attended the MathJam, where sarah-marie, snorkack, and fermatprime convinced me to apply. I ended up going, and it [i]really was[/i] the best six weeks of my life.

Unfortunately, I can't attend the entire MathJam - I'll be at a masterclass/recital - but I'm going to try to stop in for a few minutes.

Sounds cool, I'll check out the MathJam tomorrow.  I've been very interested in HCSSiM since I heard about it.

Unfortunately I wont be able to attend the Math Jam either, as I will be in michigan for the rest of the week.


limit
search
logarithms
real analysis
real analysis unsolved

{[(n + 1)!]^[1/(n+1)] - (n!)^(1/n)}  . . . . . . n starts out with 1,

but you're taking the limit of the whole expression that is in the

braces as (the integer) n approaches infinity.

Limited evidence:

At n =  1, the value is sqrt(2) - 1 ~ 0.4142
At n = 68, the value is about ____0.3703

Is this limit converging to a positive number, as suggested
by these two values being relatively close to each other,
or is it converging to zero?

Assume $ a_n \equal{} \sqrt[n\plus{}1]{(n\plus{}1)!} \minus{} \sqrt[n]{n!}$ converges. Then

$ \lim_{n\to\infty} a_n \equal{} \lim_{n\to\infty} \frac{a_1 \plus{} \cdots \plus{} a_n}{n} \equal{} \lim_{n\to\infty} \frac{\sqrt[n]{n!}}{n} \equal{} \frac{1}{e}$.

This is a famous sequence called Lalescu's sequence, after the Romanian mathematician Traian Lalescu.
You cand find a discussion [url=http://www.mathlinks.ro/viewtopic.php?search_id=400769073&t=52803]here[/url], or you can also find some information on wiki.

[quote="sos440"]Assume $ a_n \equal{} \sqrt [n \plus{} 1]{(n \plus{} 1)!} \minus{} \sqrt [n]{n!}$ converges. Then

$ \lim_{n\to\infty} a_n \equal{} \lim_{n\to\infty} \frac {a_1 \plus{} \cdots \plus{} a_n}{n} \equal{} \lim_{n\to\infty} \frac {\sqrt [n]{n!}}{n} \equal{} \frac {1}{e}$.[/quote]

OK, but [b]why[/b] the limit $ \lim a_n$ exists? Could you explain, please?

$ n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n$, hence $ \sqrt[n]{n!} \sim \frac{n}{e}$.

We need a little more precision- since we're subtracting, we need $ \sqrt[n]{n!}\equal{}?\plus{}o(1)$, rather than $ \frac{n}{e}\cdot (1\plus{}o(1))$. sos440's argument essentially gives the latter.

Dealing with the $ \sqrt{2\pi n}$ does the trick. $ \sqrt[n]{\sqrt{2\pi n}}\equal{}\exp\left(\frac{\ln(2\pi n)}{2n}\right)\equal{}1\plus{}\frac{\ln n}{2n}\plus{}\frac{\ln(2\pi)}{2n}\plus{}O\left(\frac{\ln^2 n}{n^2}\right)$, so
$ (n!)^{1/n}\equal{}(1\plus{}o(1))^{1/n}\cdot\sqrt{2\pi n}^{1/n}\cdot\left(\left(\frac{n}{e}\right)^n\right)^{1/n}$
$ \equal{} (1\plus{}o(\frac1n))\cdot \left(1\plus{}\frac{\ln n}{2n}\plus{}\frac{\ln(2\pi)}{2n}\plus{}o(\frac1n)\right)\cdot\frac{n}{e}\equal{} \frac{n}{e}\plus{}\frac{\ln n}{2e}\plus{}\frac{\ln(2\pi)}{2e}\plus{}o(1)$

Differencing this gives the desired result.


Let a and b be two distinct positive real numbers, n a positive integer, and x0, x1, x2,... , xn integers. Suppose that (ax0)(ax1) ... (axn) = (bx0)(bx1) ... (bxn), and x1 + x2 + ... + xn = 1. Find x0.

(This may look hard but try it, it's really not)

[hide]well we have (ax0)a=b(bx0) since b and a are different x0 must be -1. [/hide]


A graph is [i]planar[/i] if it can be drawn in a plane with no intersecting edges. It is well known that if a graph is planar and has $ V$ vertices and $ E$ edges, then $ E\leq 3V \minus{} 6$. Prove that given any positive number of vertices $ V$, there exists a planar graph for any number of edges less than or equal to $ 3V \minus{} 6$.

[hide="maybe i'm wrong"]
what if V=2? then you can connect them with 1 edge, but $ 1 > 3(2)\minus{}6$
is there a lower bound for V?
is there a part in the definition of a planer graph that states that there are at most 1 segments connecting any 2 points?
[/hide]
but here is how you might go about proving the statement
show that you can have 3V-6 edges for V vertices
then state that any lower number can be made by removing one or more edges

That condition is characteristic of all simple, connected, planar graphs. In fact, that inequality is derived from these three properties.

[quote="stevenmeow"]
what if V=2?[/quote]

Yes, you are correct; I can no longer edit the post, but you can assume that $ V > 2$ for this problem. In fact, the derivation of the inequality $ E\leq 3V \minus{} 6$ assumes that there exists a face other than the unbounded face, as it uses the fact that such a face is bounded by at least three edges, which yields $ 2E\geq 3F$: since you need at least three vertices to have a face other than the unbounded face, the inequality is only valid for $ V\geq 3$.

[quote="stevenmeow"]but here is how you might go about proving the statement
show that you can have 3V-6 edges for V vertices
then state that any lower number can be made by removing one or more edges[/quote]

This is exactly what I had in mind.  :)

[quote="DiscreetFourierTransform"]This is a good exercise in graph-theoretical inductive proof.
Try to set up a graph that works, then show that you can always add a vertex in a particular way so as to satisfy the inequality. Then induct :)[/quote]

But of course an inductive proof is possible also.


Let $ \pi(n)$ be the number of primes not greater than $ n$. Prove that $ \pi(n)$ divides $ n$ for infnitely many $ n$.

I am confused. What does the brackets around $ n$ in $ P_{n}[n]$ mean?

[quote="aviateurpilot almost"]Let $ p_{n}$ be the $ n$th prime number, so $ p_{1}=2,\;p_{2}=3,\;p_{3}=5,\;p_{4}=7,\ldots$.
Then for any $ k$ in $ \{0,\;1,\;2,\;\ldots,\;p_{n+1}-p_{n}-1\}$ we have that $ \pi(p_{n}+k)=n$.

Choose $ r_{n}$ in the set $ \{1,2,...,n-1\}$ so that $ r_{n}\equiv p_{n}\pmod{n}$.

$ {\exists k\in\{0,1,2,...,p_{n+1}-p_{n}-1\}: \ n=\pi(p_{n}+k)|p_{n}+k\ \iff \ r_{n}\in [|n-p_{n+1}+p_{n}+1,n-1|]}$

Then if $ \left|\{n\in \mathbb{N}^{*}|\ p_{n+1}-p_{n}\ge n\}\right|=+\infty$ we'll have that $ \pi(n)|n$ for infnitely many $ n$.[/quote]

I hope that is correctly transcribed.


Let  $f: N^*\to N^*$ such that $f(m+\mu(n))=f(m) \mod n$.find all $f(n)$

it  is ve ry easy but can you post solution?


Congratulations to the romanian team. I would like to know who was who in the team:
ROU1=?...ROU6=?

[quote="Stijn"]ROU1 Gabriel Kreindler (Gold)
ROU2 Adrian Zahariuc (Gold)
ROU3 Andrei Stefanescu (Silver)
ROU4 Marius Pachitariu (Bronze)
ROU5 Dragos Michnea (Gold)
ROU6 Andrei-Bogdan Ungureanu (Gold)[/quote]
Do you know if there are mathlinkers between them?

All of the are (perhaps Ungureanu Andrei doesn't visit the site often). Their usernames are in order Kappa, Dzeta, Andreis, Ikap, Micnea Dragos, Warlord (?).


probability
Ross Mathematics Program
probability and stats

A crime has been committed by a solitary individual, who left a partial amount of his or her Dna at the scene of the crime. Forensic scientists who studied the recovered DNA noted that only five strands could be indentified and that each innocent person, independently, would have probability $ 10^{\minus{}5}$ of having their DNA match on all five strands. The district attorney supposes that the perpetrator of the crime could be any o fthe one million residents of the town . Ten thousand of these residents have been released from prison within the past ten years; consequently, a sample of their DNA is on file. Before any checking of the DNA file, the districk attorney feels that each of the ten thousand ex-criminals has probability a of being guilty of the new crime, while each of the remaining 990,000 residents has probability b, where a=cb.(That is , she supposes that each recently released convict is c times as likely to be the ccrime's perpetrator as is each town member who is not a recently released convict.) When the analyzed DNA is compared with the data base of the ten thousand ex-convicts,it turns out that AJ Jones is the only one whose DNA matches the profile.Assuming that the district attorney's estimate of the relationship between a and b is accurate, what is the probability that AJ is guilty?

This is an example problem from Sheldon Ross's <A first course in probability>.
The solution in the book is a little complex and Ross skips the most important step of the solution,which makes me very confused...

Anyone could give a reasonable solution..Thanks

The answer Ross gives is $ \frac{1}{0.9\plus{}\frac{10^{\minus{}5}}{a}}$


quadratics
algebra
quadratic formula
calculus
calculus computations

Find the 4 roots of the equation. The roots belong to a circle. Calculate the center and r of the circle...

$z^{4}+\lambda z^{2}+1=0;\ \lambda^{2}<4,\lambda\in\mathbb{R}$

Well using the quadratic formula we get

$z^{2}=-\frac{\lambda}{2}\pm \left(\frac{\sqrt{4-\lambda^{2}}}{2}\right) i$.

Since the RHS always has magnitude $1$, all the solutions are on the unit circle centered at the origin.


A person who has moderation powers for a particular forum can go to the You can moderate this forum link (near the bottom of the screen, at the bottom of the list of the things you can do in the forum) and follow that link. That link opens a page for mass moderation operations. Handle with care, but that will get your work done faster.

Hope this helps!

Max, click on the thread. At the bottom, u should see four lil buttons (split, move, lock, and delete). Click on the one that says delete and it'll ask you if you're sure if u wanna delete the thread and etc.

Or if u wanna delete large amounts of threads, use tolkenadult's method.

We'd prefer moderators not delete past math content.  You may find that old stuff not useful to you now, but it's a good path for others to follow in the future.

Thanks!


When three spherical balls, each of radius 10cm are placed in a hemispherical dish, it is noticed that the tops of the balls are all exactly level with the top of the dish. What is the radius, in centimetres, of the dish?

[hide]The center of the dish and the centers of the spheres are the vertices of a triangular pyramid whose base is an equilateral triangle with the side $ 2r$, its height is $ r$, and the side edge is $ R\minus{}r$. Hence

$ (R\minus{}r)^2\equal{}\left(2r{\sqrt{3}\over 3}\right)^2\plus{}r^2\implies (R\minus{}r)^2\equal{}{7r^2\over 3}\implies R\equal{}r\left(1\plus{}{\sqrt{21}\over 3}\right)$

For $ r\equal{}10\,\mathrm{cm}$, we get $ R\equal{}10\left(1\plus{}{\sqrt{21}\over 3}\right)\approx 25.275\,\mathrm{cm}$[/hide]


Is it possible to find all solutions of functional equation $ f(f(x)) \plus{} x^2 \equal{} x^2f(x)$ from R to R?

Well, there is one solution $ f(x)\equal{}x^2\minus{}1$

Well, here is what I have been able to show:

Let $ f(0) \equal{} c$.  Then setting $ x \mapsto 0$ in the original equation tells us that $ f(c) \equal{} 0$, and setting $ x \mapsto c$ in the original equation tells us that $ f(0) \plus{} c^2 \equal{} 0$ and thus $ c \plus{} c^2 \equal{} 0$.  This implies $ f(0) \equal{} 0$ or $ f(0) \equal{} \minus{} 1$.  If $ f(0) \equal{} \minus{} 1$ then $ f( \minus{} 1) \equal{} 0$.

If we assume that $ f(0) \equal{} \minus{} 1$ and $ f( \minus{} 1) \equal{} 0$ and $ f(x) \equal{} \minus{} 1 \plus{} a_1 x \plus{} a_2 x^2 \plus{} \ldots \equal{} b_1 (x \plus{} 1) \plus{} b_2(x \plus{} 1)^2 \plus{} \ldots$ can be expanded in power series around $ 0$ and $ \minus{} 1$ with overlapping intervals of convergence, then we can show (by applying the functional equation to the power series and equating coefficients) that we must have $ f(x) \equal{} x^2 \minus{} 1$.

However, this is not actually the only solution.  Here's another one: define a sequence $ \{a_n\}_{n \in \textbf{Z}_{\geq 0}}$ by $ a_0 \equal{} \minus{} 2$, $ a_1 \equal{} \minus{} 3$ and $ a_{n \plus{} 1} \equal{} a_{n \minus{} 1}^2 \cdot (a_{n} \minus{} 1)$.  Thus, $ a_2 \equal{} \minus{} 16$, $ a_3 \equal{} \minus{} 153$, etc.  Note that $ a_n < \minus{} 1$ for all $ n$.  Then define a function $ f(x)$ by
\[ f(x) \equal{} \begin{cases} a_{n \plus{} 1} & \text{if } x \equal{} a_n, \\
x^2 \minus{} 1 & \text{if } x \neq a_n \text{ for any } n.\end{cases}\]
I don't think it's difficult to verify that this function satisfies the given equation.  And, of course, it's straightforward to create all sorts of other solutions based on this same idea.  Perhaps if we required continuity, then the solution would be unique (but I'm not absolutely convinced of this).


Of the first 100 natural numbers, 'k' numbers are randomly chosen, if the sum of the 'k' numbers is even A wins, if it is odd 'B' wins, find the values of 'k' for which the game is fair.

The game is always fair when k is odd.

Pair the number $x$ with $100-x$.  If we chose the k numbers $x_{1}, \dots x_{k}$, then modulo 2, $\sum_{k}x_{i}\equiv-\sum_{k}100-x_{i}$ , so by changing each number with it's pair, the player that wins changes.

However, this is not exactly correct:  some choices of k numbers do not change when you change each number with it's pair.  This occurs only when k is divisible by 2.  When k is 0 mod 4,  then we have extra (unpaired) "even" wins, and the game is biased for A.  When k is 2 mod 4, we have extra (unpaired) "odd" wins, and the game is biased for B.

In total, the game is fair when k is odd.


geometry
rectangle
combinatorics unsolved
combinatorics

A rectangle of size $ m \times n$ has been filled completely by trominoes (a tromino is an L-shape consisting of 3 unit squares).

There are four ways to place a tromino
1st way: let the "corner" of the L be on top left
2nd way: let the "corner" of the L be on top right
3rd way: let the "corner" of the L be on bottom left
4th way: let the "corner" of the L be on bottom right

Prove that the difference between the number of trominoes placed in the 1st and the 4th way is divisible by $ 3$.


Prove that
$ \frac {1}{a(1 \plus{} b)} \plus{} \frac {1}{b(1 \plus{} c)} \plus{} \frac {1}{c(1 \plus{} d)} \plus{} \frac {1}{d(1 \plus{} a)}\geq2$ if  $ a,b,c,d > 0$ and $ abcd \equal{} 1$.

Inequality can be easily solved after taking substitution:

$ a \equal{} \frac {x}{y}$ , $ b \equal{} \frac {z}{x}$ $ c \equal{} \frac {t}{z}$ and $ d \equal{} \frac {y}{t}$

Then it becomes equivalent to:

$ \frac {y}{x \plus{} z} \plus{} \frac {x}{z \plus{} t} \plus{} \frac {z}{y \plus{} t} \plus{} \frac {t}{x \plus{} y}\geq 2$

By Cauchy-Shwarz inequality we can easily get that:
$ \frac {y}{x \plus{} z} \plus{} \frac {x}{z \plus{} t} \plus{} \frac {z}{y \plus{} t} \plus{} \frac {t}{x \plus{} y}\geq \frac {(x \plus{} y \plus{} z \plus{} t)^{2}}{yx \plus{} yz \plus{} xz \plus{} xt \plus{} zy \plus{} zt \plus{} tx \plus{} ty} \equal{} \frac {(x \plus{} y \plus{} z \plus{} t)^{2}}{xy \plus{} xz \plus{} xt \plus{} yz \plus{} yt \plus{} zt \plus{} yz \plus{} tx}$

So it is enough to prove that:

$ \frac {(x \plus{} y \plus{} z \plus{} t)^{2}}{xy \plus{} xz \plus{} xt \plus{} yz \plus{} yt \plus{} zt \plus{} yz \plus{} tx}\geq 2$ which is equivalent to:

$ x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} t^{2}\geq 2(yz \plus{} xt)$ and this is equivalent to:

$ (x \minus{} t)^{2} \plus{} (y \minus{} z)^{2}\geq 0$ which is obvious.

For equality case to occur we must have:$ x \equal{} t$ abd $ y \equal{} z$ which directly implies $ a \equal{} c$ and $ b \equal{} d$ so $ ab \equal{} 1$ and therefore [b]$ b \equal{}d\equal{} \frac {1}{a}$ and $ c\equal{}a$[/b] is  equality case.

[url=http://www.artofproblemsolving.com/community/c6h622658p3724475]Prove that[/url], for all positive reals $ a,b,c,d $ satisfying $ abcd =1 $, we have
\[ \frac{1}{b(1+a)}+\frac{1}{c(1+b)}+\frac{1}{d(1+c)}+\frac{1}{a(1+d)}\geq\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d} \]

[quote=FERZIO]By Modified C-S
$\sum\frac{1}{a(1+b)}\geq\frac{(1+1+1+1)^2}{a+b+c+d+ab+bc+cd+da}$
and by Am-Gm
$\frac{(1+1+1+1)^2}{a+b+c+d+ab+bc+cd+da}\geq\frac{16}{4(1)+4(1)}\geq2$
DONE!!![/quote]

FERZIO your solution is wrong.
$$\frac{(1+1+1+1)^2}{a+b+c+d+ab+bc+cd+da}\leq\frac{16}{4(1)+4(1)}$$


[hide]901?[/hide]

Oh, I mis-divided. argh.

[hide]Notice that when we multiply the entire equation $19!$, we obtain $\binom{19}{2}+\binom{19}{3}+\cdots+\binom{19}{9}=19N$. The identity $\binom no+\binom n1+\cdots+\binom nn=2^n$ now sticks out like a sore thumb.  :D

Simplifying, the left hand side becomes $\binom{19}{0}+\binom{19}{1}+\binom{19}{2}+\binom{19}{3}+\cdots+\binom{19}{9}-(\binom{19}{0}+\binom{19}{1})=\frac{2^{19}}{2}-20$.

Some simple algebra now leads us to the answer: $19N=\frac{2^{18}-20}{19}=\frac{(2^9+1)(2^9-1)}{19}-1=13796\implies \lfloor \frac{N}{100}\rfloor=\boxed{137}$.[/hide]


The medians of a right triangle which are drawn from the vertices of the acute angles are $ 5$ and $ \sqrt {40}$. The value of the hypotenuse is:

$ \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 2\sqrt {40} \qquad\textbf{(C)}\ \sqrt {13} \qquad\textbf{(D)}\ 2\sqrt {13} \qquad\textbf{(E)}\ \text{none of these}$

[hide]

Let the legs be $ 2a$ and $ 2b$. Applying the Pythagorean theorem twice, we have $ 4a^2\plus{}b^2\equal{}40$ and $ a^2\plus{}4b^2\equal{}25$. We want $ \sqrt{a^2\plus{}b^2}$, so we add the two equations and divide by 5 to get $ a^2\plus{}b^2\equal{}13$. Hence, the hypotenuse is $ \sqrt{13}$ or $ \mathrm{ (C) }$

[/hide]

[quote="mathwizarddude"]Since the legs are $ 2a$ and $ 2b$, you want twice that length.[/quote]mathwizarddude is correct and so for the purpose of resources section...
[hide="the answer is"]$ \bf (D)$[/hide]


induction
inequalities
vector
pigeonhole principle
trigonometry
complex numbers
triangle inequality

Let $ z_1 ,..., z_n$  be complex numbers with $ |z_k|<\equal{}1, k \in \overline { 1,n }$.Prove the existence of the numbers $ \epsilon_1,...,\epsilon_n \in$ {$ \minus{}1,1$} such as \[ | \epsilon_1z_1\plus{}...\plus{}\epsilon_pz_p|<\equal{}2\] for any $ p \in \overline{1,n}$.

[hide="Solution"]Use induction on $ n$.  For $ n\equal{}1$, the result is obvious.  For $ n\equal{}2$, the result is true by the Triangle Inequality - $ |\epsilon_1z_1\plus{}\epsilon_2z_2|\le |\epsilon_1z_1|\plus{}|\epsilon_1z_1|\le 2$.  Now assume the result is true for $ n\minus{}1$.  We will prove the result for $ n$, where $ n\ge 3$.

Draw each $ z_{i}$ and $ \minus{}z_{i}$ as vectors from the origin in the complex plane.  We have at least six vectors, and the total angle between them is $ 2\pi$, so by Pigeonhole the acute angle between some pair of vectors is at most $ \pi/3$, WLOG $ z_{1}$ and $ z_{2}$.  We want to show that $ |z_{1}\minus{}z_{2}|\le 1$.

Now draw a vector $ v$ from the head of $ z_{2}$ to the head of $ z_{1}$.  Then $ z_{2}\plus{}v\equal{}z_{1}$, so $ v\equal{}z_{1}\minus{}z_{2}$.  Now we have a triangle with side lengths $ |z_{1}|,|z_{2}|,|z_{1}\minus{}z_{2}|$, and the angle between the first two sides is at most $ \pi/3$.  By the Law of Cosines,

$ |z_{1}\minus{}z_{2}|^{2}\equal{}|z_{1}|^{2}\plus{}|z_{2}|^{2}\minus{}2|z_{1}|\cdot|z_{2}|\cos\theta$
$ |z_{1}\minus{}z_{2}|^{2}\le |z_{1}|^{2}\plus{}|z_{2}|^{2}\minus{}2|z_{1}|\cdot|z_{2}|\cdot 1/2$
$ |z_{1}\minus{}z_{2}|^{2}\le |z_{1}|^{2}\plus{}|z_{2}|^{2}\minus{}|z_{1}|\cdot |z_{2}|$
$ |z_{1}\minus{}z_{2}|^{2}\le |z_{1}|(|z_{1}|\minus{}|z_{2}|)\plus{}|z_{2}|^{2}$

WLOG $ |z_{1}|\ge|z_{2}|$, so $ |z_{1}|\minus{}|z_{2}|$ is positive.  Then

$ |z_{1}\minus{}z_{2}|^{2}\le 1\minus{}|z_{2}|\plus{}|z_{2}|^{2}$
$ |z_{1}\minus{}z_{2}|^{2}\le 1\plus{}|z_{2}|(|z_{2}|\minus{}1)$

But $ |z_{2}|$ is positive and $ |z_{2}|\minus{}1$ cannot be positive, so $ |z_{1}\minus{}z_{2}|^{2}\le 1$.  Therefore we can replace $ z_{1}$ and $ z_{2}$ in the original problem with $ z_{1}\minus{}z_{2}$, to get the same problem with only $ n\minus{}1$ variables, which is true by our inductive hypothesis.  Therefore the induction is complete, and we're done.  :) [/hide]


1984 AHSME

The largest integer n for which $n^{200} < 5^{300}$ is:

A) 8
B) 9
C) 10
D) 11
E) 12

Stete your answer with explanation. Don't just put down the letter like "A?"

[hide]Take the 200th root of both sides:
$n < \sqrt[200]{5^{300}}$
$n < 5^{3/2}$
$n < 5\cdot\sqrt{5}$
The RHS is 11.18033989... so the answer is D) 11.
[/hide]

use logs.  Makes things a lot easier.  first set it equal
n^200=5^300

I don't have the equation program (I'm planning on downloading it) so Ill use ` before and after a number to denote that is the base in the logarithm.  Like log base 5 of 25 is: log `5` 25
... k?

so log `5` n^200=log `5` 5^300

200*log `5` n=300
log `5` n=1.5
then changing back to exponential... 5^1.5=n and I don't have a calc on me but say you get like 7.1245... test 7 and 8 and the one that makes the inequality true is the value of n.

You don't need TeXnicCenter to type LaTeX on the forum. You also don't need to test a number on both sides of 5^(3/2) because 5^(3/2) makes equality, so n must be less than it.

$\text{ use logs.  Makes things a lot easier.  first set it equal }
n^{200}=5^{300}\\\\
\text{ I don't have the equation program (I'm planning on downloading it) so Ill use ` before and after a number to denote that is the base in the logarithm.  Like log base 5 of 25 is: } \log_5 25\text{
... k? }\\\\
\text{ so } \log_5 n^{200}=\log_5 5^{300}\\\\
200*\log_5 n=300
\log_5 n=1.5
\text{ then changing back to exponential... } 5^{1.5}=n \text{ and I don't have a calc on me but say you get like } 7.1245... \text{ test 7 and 8 and the one that makes the inequality true is the value of n }$

You could use this code.


a,b,c,d >0 :
prove :
$3.(\frac{a+b+c+d}{4})^3 + \frac{abc+bcd+cda+dab}{4} \ge 4. \frac{a+b+c+d}{4}.\frac{ab+bc+cd+da+ac+bd}{6}$

$3\cdot(\frac{a+b+c+d}{4})^3 + \frac{abc+bcd+cda+dab}{4} \ge 4\cdot \frac{a+b+c+d}{4}\cdot\frac{ab+bc+cd+da+ac+bd}{6}\Leftrightarrow$
$\Leftrightarrow9\cdot\sum a^3-5\cdot\sum{a^2b}+6\cdot\sum{abc}\geq0\Leftrightarrow$
$2\cdot\sum(a^3+b^3+c^3-a^2b-a^2c-b^2a-b^2c-c^2a-c^2b+3abc)+3\sum a^3-\sum a^2b\geq0.$
Schur and Murhead. :)

By the way $\frac{32}{11}\cdot(\frac{a+b+c+d}{4})^3 + \frac{12}{11}\cdot\frac{abc+bcd+cda+dab}{4} \ge 4\cdot \frac{a+b+c+d}{4}\cdot\frac{ab+bc+cd+da+ac+bd}{6}$ :)


Most (if not all) regional science fairs are over by now, so I suppose this is a good time to post this question:  who here is going to be going the the International fair in Portland, and in what category?

I'm going, in Mathematics.  :)

I'm going to view the projects cuz i'm too young to enter. PM me your project so i can see you there.

I was wondering, how do people manage to get project titles like "Isolation and Expression of an Epinephrine-synthesizing Enzyme (PNMT) from Entamoeba Parasites" (actual title of project for ISEF)? Where do they learn that kind of stuff? Is this something taught in a standard biology class? I think the only words that I actually know is "parasites" and "enzyme."


search
real analysis
real analysis unsolved

$u(x)=e^{\frac{x^2}{2}}$

$u^{(n)}(x)=B_n(x)u(x)$

for x>0 give an equivalent of $B_n(x)$ when n tends +oo

I posted it some time ago ( actually, for $e^{x^2}$ or something like this) and Fedja gave a nice solution.

but where is it on the forum ?
(le forum maintenant c'est un peu comme la caverne d'Alibaba)

No, other forums are like that, Moubinool. But here it suffices for example to tape on search button $difficult equivalent$ and author harazi and you'll have the post.  ;)


http://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf

Thanks to Vasc for such a nice theorem with many applications!

I was reading along and came across page 17, where I believe there is a typo/mistake.
At the bottom of the page it says:
$ \text{We have}$\[ f'(x)\equal{}\frac{r_0\plus{}1}{2}\left(1\plus{}\frac{x}{2}\right)^{r_0}\minus{}r_0x^{r_0\minus{}1}\minus{}1\\
\frac{1}{r_0}f''(x)\equal{}\frac{r_0\plus{}1}{4}\left(1\plus{}\frac{x}{2}\right)^{r_0}\minus{}\frac{r_0\minus{}1}{x^{2\minus{}r_0}}
\]
Shouldn't the correct second line state\[ \frac{1}{r_0}f''(x)\equal{}\frac{r_0\plus{}1}{4}\left(1\plus{}\frac{x}{2}\right)^{{\bf r_0\minus{}1}}\minus{}\frac{r_0\minus{}1}{x^{2\minus{}r_0}}
\]
Sorry, I'm just quite confused

[quote="me@home"] I was reading along and came across page 17, where I believe there is a typo/mistake.
At the bottom of the page it says:
$ \text{We have}$
\[ f'(x) \equal{} \frac {r_0 \plus{} 1}{2}\left(1 \plus{} \frac {x}{2}\right)^{r_0} \minus{} r_0x^{r_0 \minus{} 1} \minus{} 1 \\
\frac {1}{r_0}f''(x) \equal{} \frac {r_0 \plus{} 1}{4}\left(1 \plus{} \frac {x}{2}\right)^{r_0} \minus{} \frac {r_0 \minus{} 1}{x^{2 \minus{} r_0}}
\]
Shouldn't the correct second line state
\[ \frac {1}{r_0}f''(x) \equal{} \frac {r_0 \plus{} 1}{4}\left(1 \plus{} \frac {x}{2}\right)^{{\bf r_0 \minus{} 1}} \minus{} \frac {r_0 \minus{} 1}{x^{2 \minus{} r_0}}
\]
Sorry, I'm just quite confused[/quote]
Indeed, you are right. But for $ r_0\equal{}1.71$, this mistake changes nothing in the proof. Thank you. :wink:


limit
real analysis
real analysis unsolved

Let $a,b$ be coprime numbers, $b$ is even.For every natural $q$ the number $p=p(q)$ is constructed, such that $|\frac{p}{q}-\frac{a}{b}|$ is minimal, $\frac{p}{q}\neq \frac{a}{b}$. Find $\lim_{n\to\infty}\sum_{q=1}^n \frac{q|\frac{p}{q}-\frac{a}{b}|}{n}$

But isn't it clear that if $q$ is not divisible by $b$ then the corresponding term in the sum the limit of whom we must compute is $\frac{1}{bn}$ and if $q$ is a multiple of $b$ then the term is 0? Thus practically the sum is equal to the number of terms in the sequence $ 1,2,...,n$ that are not multiple of b divided by $nb$ and the limit is $\frac{1-\frac{1}{b}}{b}$. Am I missing obvious things?

[quote="harazi"]But isn't it clear that if $q$ is not divisible by $b$ then the corresponding term in the sum the limit of whom we must compute is $\frac{1}{bn}$ [/quote]
Harazi,could you please explain why it is? For example, if $a=3,b=4,q=2$ how could we find such a $p$ that $|4p-6|=1$ ?
For the another part i definitely agree with you, sorry i've forgotten to state that the approximating $\frac{p}{q}$ must be different from $\frac{a}{b}$

Sorry, I misunderstood the statement. Of course, the solution is wrong.


Prove that exist limit of sequence: $ u_n \equal{} \left(1 \plus{} \frac {1}{n}\right)^n$.

I have got the solution of this problem. But I think it can done by many other ways. So you help me, please.

[hide="One Approach"]
Using the Binomial Theorem, we have $ u_n \equal{} \sum_{k \equal{} 0}^n \binom{n}{k} \frac {1}{n^k} \equal{} \sum_{k \equal{} 0}^n \frac {n(n \minus{} 1) \ldots (n \minus{} k \plus{} 1)}{n^k} \cdot \frac {1}{k!}$

It is clear that the coefficient on each $ \frac {n(n \minus{} 1) \ldots (n \minus{} k \plus{} 1)}{n^k}$, for given $ k$, is increasing with $ n$, but is bounded above by $ 1$. This establishes that $ u_n$ is monotone increasing -- the coefficient on each $ 1/k!$ increases with $ n$, [b]and [/b]more such terms are added as $ n$ increases.

It suffices to show that the sequence is bounded above. Observe that $ u_n < \sum_{k \equal{} 0}^n \frac {1}{k!} < \sum_{k \equal{} 0}^{\infty} \frac {1}{k!} < 1 \plus{} \sum_{k\equal{}1}^{\infty} \frac{1}{2^{k\minus{}1}} \equal{} 3$[/hide]

[color=darkblue]Thank TZF for your post. Who can help me too?[/color]


Well for this problem it didnt take me too long to list everything out but if the grid was larger is there a formula for this?
Problem: How many rectangles are in a two by three grid?
My method: there are 6 1x1's
7 2x1's
2 3x1"s
2 2x2's
1 2x3
adding,6+7+2+2+1 we get [b]18.[/b]
[/b]

I don't think there is a formula, just use common sense
and add them up. That's what I did.

this topic was a triple post.  you can find the other thread in this forum with more discussion in it that might be helpful

Formula:
If $ x$ is length and $ y$ is width of the grid, then #of rectangles is:
($ x$th triangular number) x ($ y$th triangular number)

So for that problem
$ ((3(3 \plus{} 1)/2))((2(2 \plus{} 1))/2)$
$ \equal{} 6x3 \equal{} 18$


[url]http://www.dhmo.org/[/url]

Read and discuss.

[hide]Please tell me you know what DHMO really is...[/hide]

[hide] I'm DHMO intolerant.  :o I saw this in 6th grade when we learned about reliable internet sources. :D [/hide]

Behold, the power of the marvelous [url=http://www.artofproblemsolving.com/Forum/search.php]Search Function[/url]...

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28632[/url]

Good point, however, that was...a year or so ago.

Look on nutrition labels for DHMO, it's widespread use is appalling to many.

Ooo...I'd like to add, anybody want to ban the addictive stimulant trimethylxanthine?

...And... I did a similar advertisement on Hydrogen Chloride  :D  it's like salt (chlorine) but it's low sodium, so it's better for you  :lol: .

Are you kidding me? We had HCl and NaOH for one of our labs and some kid knocked over the HCl and it corroded a chunk out the table.

Haha, some kid in our school spilled HCl over another kid's glove.  Nobody was hurt, probably a low concentration hydrochloric acid solution.  NaOH is lye?

[quote="Ignite168"][quote="kyyuanmathcount"][quote="funcia"]kids shouldn't play with chemicals even if they are learning
 :lol:[/quote]

When was the last time you got excited for chemistry class?  We had the coolest experiment ever that day.  We just took random things (that the teacher set out) and either mixed em or heated them (or both :P).  The HCl+Mg was coolest though.[/quote]

Maganese is more fun to play with than Magnesium could ever be.[/quote]

espeacially w/ a bunsen burner :D

-jorian

Hmm, our labs this year involve testing wether things are acids or bases  :rotfl:

ANYWAY, when we were doing a PPM lab, the bottle disagreed with the ppm card (bottle said 50,000 ppm - card said about 300-500 ppm).

Some people try to avoid foods because their additives have long, chemical-sounding names, or because of the latest fad or obscure study. I think the point of the DHMO thing is to point out how foolish that is, because even water can be made to sound bad if we give it a long name and scary-sounding properties.

Once I saw someone on TV complaining about girl scout cookies. First he was talking about how the girl scouts don't make the cookies, and that's a valid complaint. But then he started talking about the additives in the cookies, things like niacin and riboflavin. Now, I knew that those were vitamins B3 and B2, so I was cracking up.

True.  If a commercial for bottled water says it includes "12 ounces of pure dihydrogen monoxide," the sales would most definitely not go up.

Or...if you advertised pop with extra trimethylxanthine...yeah...

It was given to vicious dogs involved in recent deadly attacks.
Yes it is dangerous

[quote="lingomaniac88"]I've seen the DHMO thing before.

I'm just wondering one thing:
What's their purpose?  To maliciously fool others, or to entertain people and give them a laugh?[/quote]

http://www.snopes.com/science/dhmo.asp

Nathan Zohner surveyed fifty ninth-grade students in 1997. Forty-three of them favored banning dihydrogen monoxide and six were undecided.

hey snopes is my favorite website! :lol:


AMC
AIME
number theory
prime factorization

[quote="machack"]2000x^6  +   100x^5  +  10x^3  +  x  -  2

how do factor this?
i have the answer but im wanna know a step by step kinda way to get this[/quote]
Very carefully.

Basically you just analyze the prime factorization of terms and signs, or you can just guess and check for roots.

Yes that would work perfectly as im sitting next to a smart person in the AIME test room.

 :mad:

It might help to see that the equation is symmetric around $x^{3}$.


For the first one, $ f(x)\equal{}\frac{x}{\sqrt{x\minus{}4}}$ so the denominator must exceed 0.  Therefore, $ \{x\in \mathbb{R}\mid x> 4\}$.  For the second one, $ p(x)\equal{}\sqrt{\frac{3}{x\minus{}2}}$ so the entire radicand must exceed 0 but it can't actually be equal to zero.  So 2 won't work, but everything greater will so $ \{x\in \mathbb{R}\mid x>2\}$.

Can you explain using words?  I don't understand your math symbols.

(1)
The only way for the value not to be a real number, is that $ \sqrt {x \minus{} 4}$ is less than or equal to 0. The value that makes $ \sqrt {x \minus{} 4}$ equal to zero is 4. Thus the answer is $ \boxed{(4,\infty)}$.

(2)
Taking the same idea from (1) the smallest value that x could be that is equal to 0 is 2, so the answer is $ \boxed{(2,\infty)}$

Note: I am using simple interval notation.

[quote="sharkman"]Find the domain of each function.

(1) f(x) = (x)/(sqrt{x - 4})

(2) p(x) = sqrt{3/(x - 2)}

NOTE: For question 2, the fraction 3/(x - 2) lies in the radicand.[/quote]
[hide]1. x>4
2. x>2[/hide]

Basically, when finding the domain of square root rational functions, you just have to ensure the denominator is greater than 0, and not equal to it or less than 0.


Hi,

The point P(x,y,z) is mapped on the point Q(X,Y,Z) by the relation

(X,Y,Z) (column)=M(x,y,z) (column),

$M = \left(\begin{array}{ccc}-1 & 2 & 0 \\ 2 & 0 &-2 \\ 0 &-2 & 1 \end{array}\right)$

Show that Q lies on the plane with equation 2X+Y+2Z=0.
Show also that M^3 =lamdaM and give the value of lambda.

Could you please help me with this qustion.

Thank you.

this probably doesn't belong in HSB  :maybe:
or are matricies allowed?


[quote="super_math"]Let a,b,c, are positive real numbers, prove:
 a\sqrt[3]{1+b-c}+b\sqrt[3]{1+c-a}+c\sqrt[3]{1+a-b}\le1[/quote]
I think, it's wrong. Try $ a\equal{}b\equal{}0.4$ and $ c\equal{}0.2.$ :wink:

[quote="arqady"][quote="super_math"]Let a,b,c, are positive real numbers, prove:
 a\sqrt[3]{1+b-c}+b\sqrt[3]{1+c-a}+c\sqrt[3]{1+a-b}\le1[/quote]
I think, it's wrong. Try $ a \equal{} b \equal{} 0.4$ and $ c \equal{} 0.2.$ :wink:[/quote]
 $ a\sqrt[3]{1\plus{}a\minus{}c}\plus{}a\sqrt[3]{1\plus{}c\minus{}a}\plus{}c\le1$
 $ 0,4\sqrt[3]{1,2}\plus{}0,4\sqrt[3]{0,8}\le0,8$
 $ \sqrt[3]{1,2}\plus{}\sqrt[3]{0,8}\le2$
It's correct.

[quote="super_math"]Let a,b,c, are positive real numbers, prove:
 $ a\sqrt [3]{1 \plus{} b \minus{} c} \plus{} b\sqrt [3]{1 \plus{} c \minus{} a} \plus{} c\sqrt [3]{1 \plus{} a \minus{} b}\le1$[/quote]
We have
$ 3a\sqrt [3]{1 \plus{} b \minus{} c}\equal{}3a\sqrt[3]{a\plus{}2b}\le a(a\plus{}2b\plus{}2)$
So
$ LHS\le \frac{(a\plus{}b\plus{}c)^2\plus{}2(a\plus{}b\plus{}c)}{3}\equal{}1$
 :blush:

By holder,
$ (\sum a)(\sum a)(\sum a(1 \plus{} b \minus{} c)) \geq (\sum a\sqrt [3]{1 \plus{} b \minus{} c})^{3}$. :D

[quote="super_math"][quote="arqady"][quote="super_math"]Let a,b,c, are positive real numbers, prove:
 a\sqrt[3]{1+b-c}+b\sqrt[3]{1+c-a}+c\sqrt[3]{1+a-b}\le1[/quote]
I think, it's wrong. Try $ a \equal{} b \equal{} 0.4$ and $ c \equal{} 0.2.$ :wink:[/quote]
 $ a\sqrt [3]{1 \plus{} a \minus{} c} \plus{} a\sqrt [3]{1 \plus{} c \minus{} a} \plus{} c\le1$
 $ 0,4\sqrt [3]{1,2} \plus{} 0,4\sqrt [3]{0,8}\le0,8$
 $ \sqrt [3]{1,2} \plus{} \sqrt [3]{0,8}\le2$
It's correct.[/quote]
OK. I have thought about the following: $ \sum_{cyc}\frac{a}{\sqrt[3]{1\plus{}b\minus{}c}}\leq1.$

I think the easiest way is Jensen:

$ \frac{\sum a\sqrt[3]{a\plus{}2b}}{a\plus{}b\plus{}c} \leq \sqrt[3]{\frac{\sum a^2 \plus{} 2ab}{a\plus{}b\plus{}c}} \equal{} 1$


[quote="jackie-chan01"]For $ n\geq 1,p\geq 1$,
Prove $ \frac {1}{n \plus{} 1} \minus{} \frac {1}{n \plus{} p \minus{} 1} < \frac {1}{(n \plus{} 1)^2} \plus{} \frac {1}{(n \plus{} 2)^2} \plus{} ..... \plus{} \frac {1}{(n \plus{} p)^2} < \frac {1}{n} \minus{} \frac {1}{n \plus{} p}$[/quote]

just ues
$ \frac {1}{n\plus{}k}\minus{}\frac {1}{n \plus{} k\plus{}1}<\frac {1}{(n \plus{} k)^2}<\frac {1}{n\plus{}k\minus{}1}\minus{}\frac {1}{n \plus{} k},k\equal{}1,2,...,p$
?


See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164077 .  Moderator, please lock or delete this thread.


modular arithmetic
number theory unsolved
number theory

Let be given a natural number $n(n\ne 0)$.Denote $A=(n+1)^{2005}-n^{2005}$.Prove that all prime divisors of $A$ have one of this two forms:$10k+1$,$802k+1$

Note that $n, n+1$ are of different parity, so $A$ is odd and all its prime divisors are odd. Then we have to prove they're of one of the forms $5k+1, 401k+1$.

Suppose that some prime $p$ is of neither of those forms, so that $(p-1, 2005)=1$. Let $g$ be a primitive root $\pmod{p}$. Let $a$ be a non-zero residue and $g^b \equiv a$. Then the equation $x^{2005} \equiv a$ has a solution iff $g^{2005c} \equiv g^b$ has a solution, that is, if there are $b, c$ such that $p-1 \mid 2005c-b$. Since $(p-1, 2005)=1$, there are, so for any $a=1, \ldots, p-1$ there's some $x$ such that $x^{2005} \equiv a$. Since there are $p-1$ possible values of $x$, this implies that for any $a$ there's a unique solution.

In particular, for $a=1$ there's a unique $x$ such that $x^{2005} \equiv 1$. Since $x=1$ works, this is the only solution. If $(n+1)^{2005} \equiv n^{2005}$, then $p$ is obviously coprime with both $n, n+1$ so $n$ is inversible and we have $(\frac{n+1}{n})^{2005} \equiv 1$. Then $\frac{n+1}{n} \equiv 1$, so $n+1 \equiv n$, absurd.


cosα=-0.1392, Quadrant 3. Find angle α. My teacher let us use the calculator to find the reference angle first, but he told us not to put (-) negative sign in it, because the negative sign just tells you the angle in which quadrant. But the problem is the one that i put the (-) into it equals to 81, the one that i didn't put the negative sign into it equals to 98. How come?!
Hope you can tell me, thanks.

[quote="sinajackson"]If you don't put the (-) then it gives $81.99844365$ and if you put it , it gives $98.00155635$. And  $81.99844365+98.00155635 = 180$. that's obvious , because : $Arccos(\alpha)+Arccos(-\alpha) = 180$ :)
Is this what you wanted?[/quote]

Hi, sinajackson, thanks for your answer. But as i said cos[i]α[/i]=-0.1392, Quadrant 3. Find angle [i]α[/i]. My teacher let us use the calculator to find the reference angle first. But if there are two different answers for the reference angles, then there must be two different answers for the angle [i]a[/i] also. As in, if reference angle is 82, then angle [i]a[/i]= 82+180=262. If reference angle is 98, then angle [i]a[/i]= 98+180=278. That's my point, thanks.


A strip of paper consists of eight squares as shown. The strip is
folded in half so that the right-most square (8) lands face-down
and on top of the left-most square (1). Then, the new rightmost
square is again folded over on top of the new left-most
square, and then again one more time, so the strip has been
folded into a stack of single squares with the square labeled 1
still on the bottom. What number is in the square on the top of
the stack?

[asy]for(int i = 0; i<8; ++i)
{
draw(shift(i*right)*unitsquare,linewidth(1));
label(string(i+1),(i+.5,.5),fontsize(10pt));
}[/asy]

AHHH THIS PROB HAD A PICTURE IN IT?!

hmm hmm.

after 8 is folded to 1, 4 becomes the rightmost.

after 4 is folded to 8 and 1, 2 becomes the rightmost.

fold 2 to 1,4,8. so 2 becomes the number on top of the stack.

answer : 2


The sum of each pair of positive integers $ x,y$ that satisfy $ 3x\plus{}4y\equal{}50$ is computed. What is the largest such sum?

Since $ x\plus{}y\equal{}\frac{50\minus{}y}3$, it is clear that we want to minimize $ y$. Thus the maximum sum occurs at $ y\equal{}2\implies x\plus{}y\equal{}\boxed{16}$ (this corresponds with the ordered pair $ (x,y)\equal{}(14,2)$).

Merely a clarification.

a minimized y=2 because both x and y are integers, y=1 would yield x= 46/3


calculus
derivative
calculus computations

That is pretty much what I did, jmerry.  I guess that IS the best way to do it... are there any other ways that could be done to calculate the summation I gave?


algebra polynomial calculus integration quadratics

Let $ P(x)$ be a polynomial of degree $ 16$ with integer coefficients.

(a) Knowing that $ P(x)$ assumes odd values for $ x \equal{} 0$ and $ x \equal{} 1$, prove that $ P(x)$ has no integer roots.

(b) Knowing that $ P(x) \equal{} 7$ for four distinct integer values of $ x$, for how many integer values of $ x$ does $ P(x)$ assume the value $ 14$?

[hide="Solution to part a"]Assume that $ P(x)$ has an integral root $ n$. If $ P(0)$ is odd, that implies that $ n$ is odd; we now prove this.

Factor $ P(x)$ into a product of irreducible quadratics and binomials. When we evaluate $ P(x)$ at $ x \equal{} 0$, we get an integer because when we evaluate it into the original polynomial, we get an integer. Note that the irreducible quadratics and binomials must all evaluate to integers. Consider the binomial $ x\minus{}n$. If $ n$ were even, than $ x\minus{}n \equal{} \minus{}n$, which is even, and the product would be even. This is a contradiction. Therefore, $ n$ must be odd.

By similar logic, if $ P(1)$ is odd, that implies $ n$ is even. Contradiction, $ n$ cannot be both odd and even. Therefore, $ P(x)$ has no integral roots.
[/hide]

Simpler: use the classical fact that if $ P$ has integer coefficients, and $ a$, $ b$ are integer, then $ a\minus{}b \mid P(a)\minus{}P(b)$.

If $ n$ is an integer root, it follows $ n\minus{}0 \mid P(n)\minus{}P(0)$, so $ n \mid P(0)$, and $ n\minus{}1 \mid P(n)\minus{}P(1)$, so $ n\minus{}1 \mid P(1)$, therefore both $ n$ and $ n\minus{}1$ must be odd, absurd.

Apply same idea for the other question.

Question (c): why should $ P$ be of degree $ 16$ ?

Sorry, mavropnevma, the conditions in (a) do not apply to (b).

Let's have $ P(x) \equal{} 7$ for four distinct values $ x\equal{}a,b,c,d$, and let $ P(t) \equal{} 14$.

Then $ t\minus{}a \mid P(t)\minus{}P(a) \equal{} 14\minus{}7 \equal{} 7$, and the same for $ t\minus{}b, t\minus{}c$ and $ t\minus{}d$.
Since $ 7$ has only four divisors: $ \pm 1$ and $ \pm 7$, it means we may have at most one such value $ t$.

Does it work?

Let the four distinct values be $ a,b,c,d$.

Therefore, $ P(x) \equal{} (x \minus{} a)(x \minus{} b)(x \minus{} c)(x \minus{} d)Q(x) \plus{} 7$. Suppose there's a $ t \in Z$ so that $ P(t) \equal{} 14$

Then $ (t \minus{} a)(t \minus{} b)(t \minus{} c)(t \minus{} d)Q(t) \equal{} 7$. This way, we need to represent the number 7 as a product of five integers, with at least four of them distinct. However, this is impossible, so no such $ t$ exists.


algebra
polynomial
function
algebra unsolved

find all polynomials with real coefficients f(x) such that cos(f(x)),x in R,is a periodic fonction

Let T be the periodic of the function $cos(f(x))$, we have $cos(f(x)) = cos(f(x+T))$ for all the real number $x$. Then we have $f(x+T) = f(x)+k2\pi$ or $f(x+T)=-f(x)+k2\pi$, where $k$ is an integer. Note that $f$ is a polynimial, hence it is continuous. Hence we have $f(x+T) = f(x)+k2\pi$ or  $f(x+T)=-f(x)+k2\pi$ for all $x$. In the first case, it imply that $f$ is a polynomial of degree at most $1$. In the second case, it implies that $f$ is a constant. Thus the polynomials are polynomials of degree at most one. QED


calculus
integration
derivative
logarithms
trigonometry
function
calculus computations

I don't know about the other ones but

$ \int \frac {e^x}{x} dx$

requires infinite series.

[quote="Fitim"]$ \int \frac {e^x \plus{} 1}{e^{2x} \plus{} e^x \plus{} 1} dx$

$ \int \frac {1}{2 \minus{} sinx} dx$

$ \int \frac {x}{\sqrt {6x^2 \minus{} x \minus{} 1}} dx$

$ \int \frac {e^x}{x} dx$

thanks,

Fitim[/quote]

third one:
observe derivative of $ 6x^2 \minus{} x \minus{} 1$ is of the form $ cx \plus{} d$. keep cx and seperate d. the second one rearranges to a form $ (x \plus{} a)^2 \pm b^2$  .

second:
write 2 as $ 2(sin^2\frac {x}{2} \plus{} cos^2\frac {x}{2})$
$ sinx$ as $ 2 sin\frac {x}{2} cos\frac {x}{2}$
multiply above and below by $ sec^2\frac {x}{2}$

[quote="Fitim"]$ \int \frac {x}{\sqrt {6x^2 \minus{} x \minus{} 1}} dx$[/quote]

We have:

$ \int \frac {x}{\sqrt {6x^2 \minus{} x \minus{} 1}} dx \equal{} p(x) \sqrt {6x^2 \minus{} x \minus{} 1} \plus{} \int \frac {B}{\sqrt {6x^2 \minus{} x \minus{} 1}} dx$,

where $ \deg p(x) < \deg x \Rightarrow p(x) \equal{} A; A$ and $ B$ are constants.

Differentiating both sides gives $ A \equal{} \frac {1}{6}$ and $ B \equal{} \frac {1}{12}$.

It remains to solve $ \int \frac {1}{\sqrt {6x^2 \minus{} x \minus{} 1}} dx$:

$ \int \frac {1}{\sqrt {6x^2 \minus{} x \minus{} 1}} dx \equal{} \int \frac {1}{\sqrt {6} \sqrt {x^2 \minus{} \frac {x}{6} \minus{} \frac {1}{6} } } dx \equal{} \\
\int \frac {1}{\sqrt {6}\sqrt {(x \minus{} \frac {1}{12})^2 \minus{} \frac {25}{144}}} dx \equal{} \\
\int \frac {1} {\sqrt {6} \frac {5}{12} \sqrt { (\frac {12}{5}( x \minus{} \frac {1}{12} ))^2 \minus{} 1 } } dx \equal{} ^{\circ} \\
\int \frac {1} {\sqrt {6} \sqrt {t^2 \minus{} 1}} dt \equal{} \frac {1}{\sqrt {6}} \mathrm{arcosh}{t} \equal{} \\
\equal{} \frac {1}{\sqrt {6}} \mathrm{arcosh}{\frac {12}{5}(x \minus{} \frac {1}{12})}$

$ \circ:$ We have used $ t \equal{} \frac {12}{5}(x \minus{} \frac {1}{12})$

Result: $ \int \frac {x}{\sqrt {6x^2 \minus{} x \minus{} 1}} dx \equal{} \frac {1}{6}\sqrt {6x^2 \minus{} x \minus{} 1} \plus{} \frac {1}{12\sqrt { 6}} \mathrm{arcosh}{\frac {12}{5}(x \minus{} \frac {1}{12})}$

[quote="Fitim"]$ \int \frac {e^x \plus{} 1}{e^{2x} \plus{} e^x \plus{} 1} dx$[/quote]

We use substitution $ u \equal{} e^x \Rightarrow \frac {du}{u} \equal{} dx$.

Hence:

$ \int \frac {e^x \plus{} 1}{e^{2x} \plus{} e^x \plus{} 1} dx \equal{} \int \frac {u \plus{} 1}{(u^2 \plus{} u \plus{} 1)u}du \equal{} \\
A \ln{u} \plus{} B \ln{(u^2 \plus{} u \plus{} 1)} \plus{} C \arctan{\frac {2u \plus{} 1}{\sqrt {3}}}.$

It remains to differentiate both sides and calculate constants $ A, B$ and $ C$.

[quote="Fitim"]
$ \int \frac {1}{2 \minus{} sinx} dx$

[/quote]
Try $ \tan\frac{x}{2}\equal{}t.$ :wink:

[quote="tmrfea"]

It remains to differentiate both sides and calculate constants $ A, B$ and $ C$.[/quote]

how do you know that A, B, C are constants?

Because:
$ \frac {u \plus{} 1}{(u^2 \plus{} u \plus{} 1)u} \equal{} \frac {A_1u \plus{} A_2}{(u^2 \plus{} u \plus{} 1)} \plus{} \frac {A_3}{u}$

We know that for rational functions of the form $ \frac{A_1u \plus{} A_2}{(u^2 \plus{} u \plus{} 1)}$,  with discriminant $ D < 0$ in the denominator:$ \int {\frac {A_1u \plus{} A_2}{(u^2 \plus{} u \plus{} 1)}du} \equal{} B_1 \ln(u^2 \plus{} u \plus{} 1) \plus{} B_2 \arctan{\frac {2u \plus{} 1}{\sqrt { \minus{} D}}}$

from tmrfea

Fitim

how do you get the $ \arctan \frac {2u \plus{} 1}{\sqrt { \minus{} D}}$ stuff?
I'm new to this method  :(

It is generally true that if discriminant $ D$ of $ au^2 + bu + c$ is negative, then:

$ \int {\frac {A_1u + A_2}{(au^2 + bu + c)}du}=B_1 \ln(au^2 + bu + c) + B_2 \arctan{\frac {2au + b}{\sqrt { - D}}}$

You derive this by first using the substitution $ au^2 + bu + c=t \Rightarrow (2au+b) \; du=dt$

So

\begin{eqnarray*}
\int {\frac {A_1u + A_2}{(au^2 + bu + c)}du}&=&\int {\frac {\frac{A_1}{2a}(2au + b)+A_2-\frac{A_1 b}{2a}}{(au^2 + bu + c)}du}=\\
&=&\int {\frac {\frac{A_1}{2a}(2au + b)}{(au^2 + bu + c)}du}+  \int {\frac {A_2-\frac{A_1 b}{2a}}{(au^2 + bu + c)}du}=\\
&=&\frac{A_1}{2a}\int {\frac {dt}{t}}+ \left(A_2-\frac{A_1 b}{2a}\right) \int {\frac {1}{(\sqrt{a}u + \frac{b}{2\sqrt{a}})^{2} + (c-\frac{b^{2}}{4a})}du}=\\
&=&\frac{A_1}{2a} \ln{t}+ \left(A_2-\frac{A_1 b}{2a}\right)\int {\frac {4a}{(2au + b)^{2} -D} du}
\end{eqnarray*}

To solve $ \int {\frac {4a}{(2au + b)^{2} -D} du}$, we use substitution $ v=2au+b \Rightarrow dv=2a \;du$:

$ \int {\frac {4a}{(2au + b)^{2} -D} du}=2 \int {\frac {dv}{v^{2} -D} du}=2\frac{1}{\sqrt{-D}} \arctan{\frac{v}{\sqrt{-D}}}$

Hence the final result:

\begin{eqnarray*}
\int {\frac {A_1u + A_2}{(au^2 + bu + c)}du}= \frac{A_1}{2a} \ln{(au^2 + bu + c)}+\left(A_2-\frac{A_1 b}{2a}\right)2\frac{1}{\sqrt{-D}} \arctan{\frac{2au+b}{\sqrt{-D}}}.
\end{eqnarray*}

[quote="Fitim"]....

$ \int \frac {e^x}{x} dx$

Fitim[/quote]

hello :) !!

the function $ \int_{ \minus{} \infty}^x \frac {e^t}{t}dt \equal{} Ei(x)$ with $ x \to Ei(x)$ is exponentiel integral and also $ Ei(x) \equal{} \gamma \plus{} ln(x) \plus{} \sum_{k \equal{} 1}^\infty \frac {x^k}{kk!}$

Very nice mathema:

We can solve this also from Leibniz: $ e^x \equal{} 1 \plus{} \frac {x}{1!} \plus{} \frac {x^2}{2!} \plus{} .... \plus{} \frac {x^n}{n!} ..$; now we just find the function $ \frac {e^x}{x}$  :)

Fitim

[quote="Fitim"]$ \int \frac {e^x \plus{} 1}{e^{2x} \plus{} e^x \plus{} 1} dx$

$ \int \frac {1}{2 \minus{} sinx} dx$

$ \int \frac {x}{\sqrt {6x^2 \minus{} x \minus{} 1}} dx$

$ \int \frac {e^x}{x} dx$

thanks,

Fitim[/quote]
$ \int \frac {e^x \plus{} 1}{e^{2x} \plus{} e^x \plus{} 1}dx \equal{} \int \frac {e^x}{e^{2x} \plus{} e^x \plus{} 1}dx \plus{} \int \frac {1}{e^{2x} \plus{} e^x \plus{} 1}dx \equal{} I_{1} \plus{} I_{2}$
$ I_{1} \equal{} \int \frac {e^x}{e^{2x} \plus{} e^x \plus{} 1}dx \equal{} \int\frac {1}{(e^x \plus{} \frac {1}{2})^2 \plus{} \frac {3}{4}}d(e^x \plus{} \frac {1}{2}) \equal{} \frac {2}{\sqrt {3}}\arctan \frac {2}{\sqrt {3}}(e^x \plus{} \frac {1}{2})$
Who can solve $ I_{2}$,thank you!

hello, setting $ z=\tan(\frac{x}{2})$ we get $ dx=2\frac{dz}{1+z^2}$ and $ \sin(x)=\frac{2z}{1+z^2}$ so our integral will be ${ \int\frac{1}{2-\frac{2z}{1+z^2}}}\frac{2dz}{1+z^2}=\int\frac{1}{z^2-z+1}\,dz=\frac{2\sqrt{3}}{3}\arctan\left(\frac{\sqrt{3}}{3}(2z-1)\right)$
Sonnhard.

$ \int \frac1{2 - \sin x}dx$
$ = \int \frac {2 + \sin x}{4 - \sin^2 x}dx$
$ = \int \frac {2}{4 - \sin^2 x}dx + \int \frac {\sin x}{4 - \sin^2 x}dx$
$ = I_1 + I_2$
$ I_1 = \int \frac {2}{4 - \sin^2 x}dx$
$ = \int \frac {2}{3\sin^2 x + 4\cos^2 x}dx$
$ = 2\int \frac {\sec^2 x}{4 + 3\tan^2 x}dx$
$ = 2\int\frac {d\tan x}{4 + 3\tan^2 x}$
$ = \frac {1}{\sqrt 3}\arctan(\frac {\sqrt 3}{2}\tan x) + C$
$ I_2 = \int \frac {\sin x}{4 - \sin^2 x}dx$
$ = \int \frac {\sin x}{3 + \cos^2x}$
$ = - \int \frac {d\cos x}{3 + \cos^2 x}$
$ = \frac {\sqrt3}{3}\arctan\frac {\cos x}{\sqrt3} + C$


What are the last three digits of     $2003^{2002^{2001}}$? Take a try for fun!!!!! :)

After a lot of tedious "modding," I get [hide]49[/hide].  But it is 3 AM so I could very well be wrong.  I should go study for my exams.

maybe[hide]41[/hide]

EDIT: d'oh! [b]three[/b] digits... I misread... Ok, I'll try later.

The remainder of $2003^{{2002}^{2001}}$ when divided by $1000$ is the same as the remainder of $3^{{2002}^{2001}}$ when divided by $1000$, since $2003\equiv 3(\bmod 1000)$. We will try to find positive integer $n$ such that $3^n\equiv 1(\bmod 1000)$ and then express $2002^{2001}$ in the form of $nk+r$ so that\[2003^{{2002}^{2001}}\equiv 3^{nk+r}\equiv (3^n)^k.3^r\equiv 3^r(\bmod 1000)\]Now,\[3^{2m}=(10-1)^m=(-1)^m+10m(-1)^{m-1}+100\frac{m(m-1)}{2}(-1)^{m-2}+...+10^m\]After the first 3 terms of the expansion, all the remaining terms are divisible by $1000$, so letting $m=2q$, we have\[3^{4q}\equiv 1-20q+100q(2q-1)(\bmod 1000)\]Using this, we can check that $3^{100}\equiv 1(\bmod 1000)$, now we want to find the remainder when $2002^{2001}$ divided by $100$.
Now, $2002^{2001}\equiv 2^{2001}(\bmod 100)\equiv 4.2^{1999}(\bmod 4.25)$, so we will investigate powers of $2$ modulo $25$. Note that $2^{10}\equiv -1(\bmod 25)$, so we have\[2^{1999}\equiv (2^{10})^{199}.2^9\equiv 13(\bmod 25)\]Thus,$2^{2001}=4.13\equiv 52(\bmod 100)$. Therefore, $2002^{2001}$ can be written as $100k+52$ for some integer $k$, so\[2003^{{2002}^{2001}}\equiv 3^{52}(\bmod 1000)\equiv 1-20.13+1300.25\equiv 241(\bmod 1000)\]using the equation above.
Hence, the last 3 digits is $241$

Is that the exact same solution as they gave at the CMO?  :?

It's almost the exact solution on Kalva, though worded differently :D


geometry
3D geometry
tetrahedron
sphere
geometry proposed

A sphere is tangent to the sides $AB, BC, CD, DA$ of a tetrahedron in the points $E, F, G, H$ respectively, and these points are the vertices of a square. Prove that if, additionally, the sphere is tangent to $AC,$ then it is also tangent to $BD.$

Could someone post a hint?

Hint: A sphere is tangent to all edges of a tetrahedron iff all three sums of the lengths of the opposite edges are equal.

I think it can be used.. at least we learned this at school.


probability
function
conditional probability

In a tournament club C plays 6 matches, and for each match the probabilities of a win, draw and loss are equal. Find the probability that C finishes with more wins than losses is .

ahh tricky conditional probability
there is a $ \frac {\binom{6\plus{}n}{n}}{2^n} \equal{} \frac {\binom{6\plus{}n}{6}}{2^n}$ probability that from 6+n games, team c wins 6 of them
the sum
$ \sum^{\infty}_{k \equal{} 0}\frac {\binom{6\plus{}n}{6}}{2^n}$ is the sum of all these probabilities (total)
and the sum
$ \sum^{5}_{k \equal{} 0}\frac {\binom{6\plus{}n}{6}}{2^n}$ is the sum of the probabilities where n ranges from 0 to 5 (such that c has more wins than losses) and this is the 'good' part
you want to find $ \frac {\text{good}}{\text{total}}$
but i cant really figure out how to do that
hope i helped

I know I'm wrong, but can someone help point out how?

Odd of wins>losses = Odd of losses>wins.
So just calculate the number of ties?

3-3
20

2-2
15*6/6

1-1
30

0-0
1

Total possibilities = 3^6

So subtracting and dividing to get 221 / 486?

@alex: Your first three statements I agree with. The next one should actually be $ \binom{6!}{2,2,2} \equal{} \frac {6!}{2!^3} \equal{} 90$. The next one is $ \binom{6!}{1!^2*4!} \equal{} 30$, so you are right there, and the next one is also right.

$ 20 \plus{} 90 \plus{} 30 \plus{} 1$

BUT the next statement is false, because that takes order into account. We need to find the number of possibilities WITHOUT order since you didn't use order on the others.

P.S. Hi alex!

So the method works, but the calculatinos are wrong?  Dang, I wish I hadn't lost my intro C&P book a couple months ago...

'Tis O.K. It's a mistake that lots of people make. Don't mix ordered and not ordered.

[b]Related Problem:[/b] Find the constant term in the expansion of:

$ \blacktriangleright \left( x \plus{} 1 \plus{} \frac {1}{x} \right)^6$

$ \blacktriangleright \left( x \plus{} 1 \plus{} \frac {1}{x} \right)^n$

oh god generating functions :ewpu:

But how would you do that without expanding the whole thing?

$ \left( x \plus{} 1 \plus{} \frac {1}{x} \right)^6$
We know that every term of this expansion can be expressed as $ \dbinom6{a,b,c}(x)^a(1)^b\left(\frac1x\right)^c \equal{} \frac {6!}{a!b!c!}x^{a \minus{} c}$.
Thus, $ a \minus{} c \equal{} 0$ iff $ a \equal{} c$. $ 0\leq a \equal{} c\leq3$, and we get a sum of $ 1 \plus{} 30 \plus{} 90 \plus{} 20 \equal{} \boxed{141}$.

Refer to [url=http://en.wikipedia.org/wiki/Multinomial_theorem]this[/url].


Determine the number of arrangements $ a_1,a_2,...,a_{10}$ of the numbers $ 1,2,...,10$ such that $ a_i>a_{2i}$ for $ 1 \le i \le 5$ and $ a_i>a_{2i\plus{}1}$ for $ 1 \le i \le 4$.

$ a_1>a_2,a_3$
$ a_2>a_4,a_5$
$ a_3>a_6,a_7$
$ a_4>a_8,a_9$
$ a_5>a_{10}$

Therefore $ a_1\equal{}\max\{a_1,a_2,\ldots,a_{10}\}\equal{}10$. Note that the two subsets $ \{a_2,a_4,a_5,a_8,a_9,a_{10}\}$ and $ \{a_3,a_6,a_7\}$ do not affect each other. The number of possible distributions of 1,2,3,...,9 into these subsets is $ \binom{9}3\equal{}63$.

Now consider $ a_2,a_4,a_5,a_8,a_9,a_{10}$. $ a_2$ is clearly the greatest. As before, $ (a_4,a_8,a_9)$ and $ (a_5,a_{10})$ do not affect each other, so there are $ \binom53\equal{}10$ arrangements. But $ a_8$ and $ a_9$ are interchangeable, so this makes two arrangements. Now we have $ 63\times10\times2\equal{}1260$ arrangements.

Lastly, consider $ a_3,a_6,a_7$. $ a_3$ is the greatest, but $ a_6$ and $ a_7$ are interchangeable. So the total number of arrangements is $ 1260\times2\equal{}2520$.

Since these Irish problems are all pretty basic as far as Olympiad problems go, I figure it might be worthwhile to mention on a few of them their connection to actual branches of mathematics.  So: this question may be considered either to be asking for the number of linear extensions of a (fairly simple) poset or for the number of decreasing labellings of a particular (binary) tree.


Prove for reals $a,b,c,d,e,f$ that $ad+be+cf=\sqrt{(a^{2}+b^{2}+c^{2})(d^{2}+e^{2}+f^{2})}\ \Longrightarrow \ \frac{a+b+c}{\sqrt{a^{2}+b^{2}+c^{2}}}=\frac{d+e+f}{\sqrt{d^{2}+e^{2}+f^{2}}}$.

[hide="Read only if you must, big hint (I think, at least)"] Vectors[/hide]

I can't think of other ways to do this (maybe since I'm really tired right now and can't think straight), but im sure there's a quick way to do this. Thanks in advance  :) .

If you do choose to use vectors, the hypothesis is equivalent to $\mathbf{a}\cdot\mathbf{b}=\left|\mathbf{a}\right|\left|\mathbf{b}\right|\implies\cos\theta =1$.  In other words, $\mathbf{a}=k\mathbf{b}$.  Therefore \[\frac{a+b+c}{\sqrt{a^{2}+b^{2}+c^{2}}}=\frac{kd+ke+kf}{\sqrt{k^{2}d^{2}+k^{2}e^{2}+k^{2}f^{2}}}=\frac{d+e+f}{\sqrt{d^{2}+e^{2}+f^{2}}}.\]

We could also use the equality case in Cauchy, from which we immediately get $\frac{a}{d}= \frac{b}{e}= \frac{c}{f}: = k$. Therefore, $a=kd, b=ke, c=kf$ and we proceed as [b]amcavoy[/b]'s solution.

[quote="Kurt Gödel"]We could [b]also[/b] use the equality case in Cauchy[/quote]In fact, Cauchy is precisely the the inproduct inequality, and its equality case comes precisely from that vector relation. :)

yeah.. In cauchy, the equality holds if the numbers are in ratios.


group theory
abstract algebra
superior algebra
superior algebra unsolved

assume that $S$ is arbitrary subset of group $G,n=|G|$ and $S_{n}=\{ a_{1}\dots a_{n}: a_{i}\in S\}$ is $S_{n}$ subgroup?.

$S_{n}$ is not a group, it is just a set. In a group you have to define the law of composition.

It inherites a structure since it is a subset of a group.

uhm.. have i got it right? :huh:

is $S_{n}$ the set of all $a_{1}\dots a_{n}$ as $(a_{1}, \dots , a_{n})$ varies in the whole $G^{n}$?
if so, clearly $e = e^{n}= e\dots e \in S_{n}$; $a = a_{1}\dots a_{n}\in S_{n}$ gives $a^{-1}= a_{n}^{-1}\dots a_{1}^{-1}\in S_{n}$, and finally, given $a=a_{1}\dots a_{n}, b=b_{1}\dots b_{n}$, you can see $ab = (a_{1}a_{2})\dots(b_{n-1}b_{n})\in S_{n}$, where there's a central term $a_{n}b_{1}$ if $n$ is odd, and two central terms $a_{n-1}a_{n}, b_{1}b_{2}$ otherwise...

EDIT: uhm.. after writing this silly and obvious thing, something just tells me that you meant that $\{a_{i}\}= G$...

never does typo error leave me alone :D I have edited the post

Let $G$ be a group of order $n$, $S \subseteq G$ nonempty, and define $S_{0}: = S$, $S_{i+1}: = S_{i}S, i \in \mathbb N_{0}$. Then $S_{n}$ is a subgroup of $G$.

EDIT: The following solution implicitly assumes $e \in S$. :-(

Let $H$ be the subgroup generated by $S$. The union of all the $S_{i}, i \in \mathbb N_{0}$ equals $H$, because all elements of $S$ have finite order and the neutral element and every inverse of $S$ is eventually generated and so  the semigroup generated by $S$ (which is exactly the union of the $S_{i}$) equals the subgroup generated by $S$. The sequence $S_{0}\subset S_{1}\subset ... \subset S_{n}\subset ...$ is strictly increasing up to a point where it becomes constant, i.e. we have $S_{0}\subsetneq ... \subsetneq S_{k}= ... = S_{n}= ...$, because once we have $S_{i}= S_{i-1}$, we also have $S_{i+1}= S_{i}S = S_{i-1}S = S_{i}$ (and the sequence can increase at most $n-1$ steps). So $\bigcup_{i \in \mathbb N_{0}}S_{i}= S_{n}= H$ is a subgroup of $G$.

LiMa

PS: In case $S$ is empty, every $S_{i}$ is of couse empty.

it is not obvious that your sequence is increasing.

That is exactly the point where I assume $e \in S$. In that case $S_{i}= S_{i}{e}\subseteq S_{i}S = S_{i+1}$.
If we have $e \notin S$, then the sequence is not necessarily increasing and my argument fails.

In the case of $|S|=1$, we get $S_{n}= \{e\}$.
We always have $e \in S_{n}$, but I don't see anything more yet.

LiMa

I read the hint to look at the cardinalities of the $S_{i}$.

Let $a \in S$, then $S_{i}a \subset S_{i+1}$ and $|S_{i}| = |S_{i}a|$, so the sequence of cardinalities is increasing: $|S_{0}| \leq |S_{1}| \leq ...$. And of course it is eventually constant. If we can show that $|S_{i}| = |S_{i+1}|$ implies $|S_{i+1}| = |S_{i+2}|$, we get $|S_{n}| = |S_{n+1}| = ... = |S_{2n}|$. We have $S_{n}\subseteq S_{2n}$, because $e \in S_{n}$. So $S_{n}= S_{2n}= S_{n}S_{n}$, and $S_{n}$ is a group.

LiMa

[quote="LiMa"] If we can show that $|S_{i}| = |S_{i+1}|$ implies $|S_{i+1}| = |S_{i+2}|$, we get ...[/quote]

this reduces to $S_{i+1}= a S_{i}\Rightarrow S_{i+2}= a S_{i+1}$, which is clear from $a S_{i+1}= a S_{i}S = S_{i+1}S = S_{i+2}$.

this completes the proof, $S_{n}$ is a subgroup for every non-empty $S$.

A perfect problem and a perfect solution, thank you all guys! I was trying this: Suppose $G$ is abelian and $|G|=n$, then according to a famous theorem (although sharper bounds seem to exist) In every sequence of $2n-1$ elements of $G$ there exist $n$ elements such that their product
is the identity.  Does this special case lead us to the general one? Does any one have an idea?

[quote="The Captain"]I was trying this: Suppose $G$ is abelian and $|G|=n$, then according to a famous theorem (although sharper bounds seem to exist) In every sequence of $2n-1$ elements of $G$ there exist $n$ elements such that their product
is the identity.  Does this special case lead us to the general one? Does any one have an idea?[/quote]
Does anyone have noticed this nice problem,I think I have show that we should only consider the special case when $G=Z_{p}$.


Euler
geometry
circumcircle
LaTeX
parallelogram
projective geometry
perpendicular bisector

Let $AA_1$ and $BB_1$ be the altitudes of an acute-angled, non-isosceles triangle $ABC$. Also, let $A_0$ and $B_0$ be the midpoints of its sides $BC$ and $CA$, respectively. The line $A_1B_1$ intersects the line $A_0B_0$ at a point $C'$. Prove that the line $CC'$ is perpendicular to the Euler line of the triangle $ABC$ (this is the line that joins the orthocenter and the circumcenter of the triangle $ABC$).

[hide]As usual, let $H$ denote the orthocentre of triangle $ABC$.  Let $X = A_0 B_1 \cap A_1 B_0$.  We claim that $X$ lies on the Euler line of triangle $ABC$.

Let $P$ and $Q$ be the midpoints of $AH$ and $BH$, respectively.  Then $P$, $Q$, $A_0$, $B_0$, $A_1$, and $B_1$ all lie on the nine-point circle of triangle $ABC$, centred at $N$, the nine-point centre of triangle $ABC$.  Furthermore, $H = PA_1 \cap AB_1$, $X = A_0 B_1 \cap A_1 B_0$, and $N = PA_0 \cap QB_0$.  Therefore, by Pascal's theorem, $H$, $X$, and $N$ are collinear.  Since $H$ and $N$ lie on the Euler line, so does $X$.

Then, $C' = A_0 B_0 \cap A_1 B_1$, $C = A_0 A_1 \cap B_0 B_1$, and $X = A_0 B_1 \cap A_1 B_0$.  Therefore, by results on poles and polars, $NX \perp CC'$.  But $X$ and $N$ lie on the Euler line, so the result follows.[/hide]

By Pappus's Theorem applied to the lines $(A,B_1,B_0)$ and $(B,A_1,A_0)$ we conclude that $Q=A_1B_0\cap A_0B_1$ lies on the Euler line of $ABC$. Since $CC'$ is the polar of $Q$ wrt the nine-point circle, it follows that it must be perpendicular to the diameter of this circle which passes through $Q$, which, by the observation before, is precisely the Euler line.

I'll state a more general statement (which appears in a ISL problem, from 1995 or smth... ):

Given a quadrilateral $ABCD$ inscribed in a circle of center $O$, denote by $E,F$ intersections of extensions of opposite sides. We know that the circles circumscribed about $EAB,ECD,FBC,FAD$ have a common point $M$ (Miquel point). Then $OM\perp EF$.
For a proof see we can show that the quadrilateral formed by circumcenters of $ABCD$,$ECD$,$EAB$ and the point $E$ is a parallelogram (*) because each pair of opposite edges is parallel with $DC$ and $AB$. Now, the line joining cicumcenters of $EAB$ and $ECD$ is the perpendicular bisector of the common chord $EM$ of those circles. Because that parallelogram (*), we know that this line pass trough midpoint of $EO$. Hence triangle $EMO$ is rectangular. And we are done.

It is easy to prove that $CA_0HB_0$ is inscriptible and the circumcenter is the midpoint $T$ of $CH$. It is also easy to prove that $CA_1OB_1$ is inscriptible and the circumcenter is the midpoint $S$ of $CO$. It is obvious that $ST\parallel OH$
Aplying this statement for this problem we get that $OH\parallel ST\perp CC'$

Sorry for makin lots of mistakes in my previous post. I will correct it.

We need only that miquel point $M$ of the inscriptible quadrilateral $A_0A_1B_1A_0$ (the circumcircle in the nine point circle) lies on the line joining the points of intersetions of extensions of edges: $CC'$. As I proved in the previous post $OH$ is parallel with the line joining centers of the triangles $CA_0B_0$ and $CA_1B_1$. This line is perpendicular to the radical axis: $CM=CC'$. And we are done.

Why if $X$ lies on euler line,we get $NX\perp CC'$ ?
Thx...

Let $\omega_1$ denote the nine-point circle of triangle $ABC$.  Let $\omega_2$ denote the circle with diameter $AO$; note that $A_0$ and $B_0$ lie on $\omega_2$.  Let $\omega_3$ denote the circle with diameter $AH$; note that $A_1$ and $B_1$ lie on $\omega_3$.

The radical axis of $\omega_1$ and $\omega_2$ is $A_0 B_0$.  The radical axis of $\omega_1$ and $\omega_3$ is $A_1 B_1$.  Hence, $C' = A_0 B_0 \cap A_1 B_1$ is the radical center of $\omega_1$, $\omega_2$, and $\omega_3$.

Let $D$ be the projection of $C$ onto $HO$.  Since $\angle CDO = \angle CDH = 90^\circ$, $D$ lies on both $\omega_2$ and $\omega_3$.  Hence, $CD$ is the radical axis of $\omega_2$ and $\omega_3$, which means $C'$ lies on $CD$.  It follows that $CC'$ and $HO$ are perpendicular.

[asy]
unitsize(0.6 cm);

pair A0, B0, C0, D, H, O;
pair[] A, B, C;

C[0] = (5,12);
A[0] = (0,0);
B[0] = (14,0);
A0 = (B[0] + C[0])/2;
B0 = (A[0] + C[0])/2;
A[1] = (A[0] + reflect(B[0],C[0])*(A[0]))/2;
B[1] = (B[0] + reflect(C[0],A[0])*(B[0]))/2;
C[1] = extension(A[1],B[1],A0,B0);
H = orthocenter(A[0],B[0],C[0]);
O = circumcenter(A[0],B[0],C[0]);
D = extension(C[0],C[1],O,H);

draw(circumcircle(C[0],A0,B0),red);
draw(circumcircle(A0,B0,(A[0] + B[0])/2),red);
draw(circumcircle(C[0],A[1],B[1]),red);
draw(A[0]--B[0]--C[0]--cycle);
draw(A[0]--A[1]);
draw(B[0]--B[1]);
draw(C[0]--D);
draw(A0--B0);
draw(A[1]--B[1]);
draw(O--H);

label("$A$", A[0], SW);
label("$A_0$", A0, E);
label("$A_1$", A[1], NE);
label("$B$", B[0], SE);
label("$B_0$", B0, W);
label("$B_1$", B[1], W);
label("$C$", C[0], N);
label("$C'$", C[1], NW);
label("$D$", D, S);
dot("$H$", H, S);
dot("$O$", O, SE);

label("$\omega_1$", circumcenter(A0,B0,(A[0] + B[0])/2) + abs(A0 - circumcenter(A0,B0,(A[0] + B[0])/2))*dir(-50), dir(-50));
label("$\omega_2$", (C[0] + O)/2 + abs(C[0] - O)/2*dir(30), dir(30));
label("$\omega_3$", (C[0] + H)/2 + abs(C[0] - H)/2*dir(150), dir(150));
[/asy]

Dear Mathlinkers,
you can see at

http://perso.orange.fr/jl.ayme  vol. 2  Schroeter p. 5

Sincertely
Jean-Louis


geometry
trapezoid
trigonometry
power of a point
radical axis
geometry proposed

Let $ABCD$ be a trapezium $(AB \parallel CD)$ and $M,N$ be the intersection points of the circles of diameters $AD$ and $BC$. Prove that $O \in MN$, where $O \in AC \cap BD$.

I will also sketch a simple solution: $ O \in MN \Longleftrightarrow OP^{2} \minus{} \frac {AD^{2}}{4} \equal{} OQ^{2} \minus{} \frac {BC^{2}}{4}$, because $ MN$ is the radical axe of the circles, and where $ P,Q$ are respetively the midpoints of $ AD$, $ BC$.  But $ OP^{2} \equal{} \frac {2(OA^{2} \plus{} OD^{2}) \minus{} AD^{2}}{4}$, and the same for $ OQ$, and then we optain : $ O \in MN \Longleftrightarrow OA^{2} \plus{} OD^{2} \minus{} AD^{2} \equal{} OB^{2} \plus{} OC^{2} \minus{} BC^{2}$, but $ AD^{2} \equal{} OA^{2} \plus{} OB^{2} \minus{} 2 OA \cdot OB \cos{\angle{AOB}}$, and then it is obvious.

my solution is the same of yours, pohoatza!

i was posting when i see that you posted first!  :)

This problem is more accurately so :

[quote][color=darkred]Let $ABCD$ be a convex quadrilateral.  Denote $O\in AC\cap BD$.
Prove that the point $O$ belongs to the radical axis of the circles of diameters $AD$ and $BC$ $\Longleftrightarrow$ $AB\parallel CD\ \ \vee\ \ AC\perp BD$.[/color][/quote]

Let the cirles with diameter $AD,BC$ be $(O1)$ and $(O2)$
$AC,BD$ meets $(O1), (O2)$ at $F,E$
$DF,CE$ meet at $M$
We see $CF\bot MD, BD\bot MC$ so $O$ is the orthecenter of triangle $MCD$.This means $MO\bot CD$
Because of trapezium $ABCD$ and $O1,O2$ are the midpoint of $AD,BC$ we have $O1O2//CD$ so $MO\bot O1O2$ $(1)$
Otherwise $\angle CED=\angle CFD=90$ so $C,E,F,D$ are cyclic
So $MF*MD=ME*MC$.Thus $M$ is on the radical axis of $(O1)$ and $(O2)$ $(2)$
from $(1)$ and $(2)$ we have $O$ is on the radical axis of $(O1)$ and $(O2)$ $q.e.d$

It seems my solution does not belong among those found yet around here:
Let $\{O\}\in AC\cap BD$, $P$ the projection of $A$ onto $BD$, $Q, R$ the projections of $B, D$ respectively onto $AC$.
Obviously $BQPA$ is cyclic, so $\angle ABP=\angle AQP$, but $\angle ABO=\angle ODC$, so $CDPQ$ is cyclic, with $OC\cdot OQ=OD\cdot OP$. From $APDR$ cyclic, $OA\cdot OR=OP\cdot OD$. From the last 2 equalities, $O$ has equal power w.r.t. the two circles.

Best regards,
sunken rock


abstract algebra
group theory
superior algebra
superior algebra unsolved

How we could define a ring s.t. all of its elements are divisors of zero.

Answer: if we take any abelian group $ G$ with defining $ x*y\equal{}0$ I'll get a lot of ex's for finite and infinite rings.

But , I need a specific ex's  that for finite and infinite rings s.t. each of them has an identity and all of their elements ( except zero and identity ) are divisors of zero.

 :)

For any set $ S$ take $ R\equal{}\mathbb{F}_2^S$ the ring of all maps $ S\to \mathbb{F}_2$.
Then $ f^2\equal{}f$ and $ f(1\minus{}f)\equal{}0$ for all $ f\in R$. This gives all $ 1\not\equal{}f$ divisor of zero.

What is the definition of $ \mathbb F_2$ please.i.e what is it?

Thank You

 :)

It's standard notation for the field with two elements.

[quote="olorin"][quote="Hidden Scofield"] I need a specific ex's of rings s.t. each of them has an identity and all of their elements ( except zero and identity ) are divisors of zero.[/quote]
For any set $ S$ take $ R \equal{} \mathbb{F}_2^S$ the ring of all maps $ S\to \mathbb{F}_2$.
Then $ f^2 \equal{} f$ and $ f(1 \minus{} f) \equal{} 0$ for all $ f\in R$. This gives all $ 1\not \equal{} f$ divisor of zero.[/quote]

Indeed $ \mathbb{F}_2$ is such a ring, and such rings are closed under direct products.


What is the sum of all the integers between -12.1 and 3.3?


what do you mean by Phase transfer catalysis????

[url=http://www.mathlinks.ro/viewtopic.php?search_id=407591328&t=147371]here[/url] is an earlier discussion of mine with Carcul.
There is a beautiful account of this in Morrison&Boyd


let $a, b,c$ be positive real munbers . prove that :
$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+a}\geq 3 \frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}+\frac{a+b+c}{\sqrt[3]{abc}}$

Ithink it can be solve by SOS
isn't it????


geometry
calculus
integration
articles
number theory

In the recent General Paper exam for Singapore A-Level students, one of the options for the essay component was as follows:

10. Consider the view that the study of mathematics is intellectually satisfying, but of little practical use.

What are your views?  :D

Like almost everything, as time goes on fewer and fewer people perform any given task, and the inherient value in having the ability to be a generalist in a field that has such a reduced scope does decrease.

Consider raising food.  At one point almost everyone was engaged in this activity, as those that didn't, unless they were the rulers, tended to go hungry.  The ability to do mathmatical calculations used to be critical everywhere.  Now we have computers and calculators -- people that have the ability to do the calculations themselves are not allowed to as this is a variable that may lead to different results (at least in the records kept and the ability to track what occured later) and generally such self calculations consume more time and attention of the employees.

Many of the cash register operators, now just find a button to push for each item -- they generally lack the ability to enter the sales price.  Often in the intermediate and larger organizations, the transactions are being conducted via the barcodes to allow inventory updating as well as completion of the sales transaction.  Several stores are looking at lowering their costs by having the customer conduct the scanning transaction, greatly eliminating the roles for their employees.  We now have a customer overseer and conflict resolver, rather than a cash register operator.

Japan often have their students take classes in abacus operation, they become so skilled that they can operate a mental abacus, doing those tasks much faster and with less error than they can on a calculator, in many instances.  This admirable skill is useful, but in fewer and fewer instances.  As schools get more and more things to teach, the adminstrators can't help but view the hours consumed teaching this skill as potentilly something that may need to be sacrificed to allow something else to be taught.

One doesn't have to be a mechanic to drive an automobile, but sometimes it helps.  One doesn't have to be a computer programmer to operate a computer or computer program, such as excel.  However, it sometimes helps.  When something goes wrong or an unexpected result occurs, knowing how the problem might have occurred, how extensive is the problem and what might be a way to work around the problem, the greater understanding of the operations that are combined to produce the operated machine can be very useful.

I think that this understanding of the mathmatics behind integration, differential equations and number theory has a similar value, even if one isn't in a job that uses the math theory every day.  For example, what is the math behind data encryption?  If a new breakthrough occurs (when actually) that makes keys less secure, knowing how much more secure a 526 bit key is than a 128 bit key may be very uesful.

I have never found knowledge in any area useless.  While I think that people are levering intellectual abilities now, just as we once leveraged physical skills (think spreadsheet rather than a pencil and a long piece of paper to keep books verses a wagon rather than carrying the load on foot), there is at least some residual value in knowing how to do it on your own.

One used to have their own local expert for everything, even math.  One couldn't get a response, let alone a prompt response from someone in a distant location.  Now, with the internet . . . .  If one person can provide the answer and either put it in a program, article or answer questions on the internet, everyone else's ability to do the same thing becomes redundant.

Tis a shame, but we are being freed to do new things.  I don't know what they are.  Consider the same arguement from folks freed from having to raise food years ago in most of the world -- But what will I do with my time?  How can I get food to eat?  The answers today are the same as in our forfathers day - I don't know, some will not easily make the transistion, but overall, everyone will be better off.  (Also, we don't really have a choice.)

Hope this helps.

[quote="maersklandro"]Well, as long as i can ask "What is the formula for the area of overlap between whatever 2 triangles?" of any mathematician in the world and he/she will have nothing to answer me while at the same time 20 or so of the brightest mathematicians alive today haggle over the number of "universes" there are, what can I say? The plain majority of current mathematical research and modern developments in mathematics is of no practical use whatsoever. At least to one of simple mind, such as me.[/quote]

The idea that there must be a formula for every easily describable mathematical quantity is undoubtedly one of the first notions which every prospective student of mathematics should be disabused of.

Besides, what do you know of current mathematical research and modern developments in mathematics?

Most subjects would fit that question equally well.  For example: Literature.  Taking high school/college level english classes is not usually worthwhile in and of itself unless you plan to be an english teacher or writer, or some other job that is directly involved with the field.  There are benefits to learning things, though... in most societies today, there is a basic level of knowledge expected from people, including knowledge of basic government and history, some major literary works, the ability to speak 1+ languages, and the ability to do basic math problems.  The process of learning anything is also a good mental excercise, and contributes to one's overall intelligence.  With Math, there is the added benefit of helping with problem solving skills and creative approaches to situations.

So........ math might not be important in most people's practical lives, but neither is a lot of what you learn at school.


Let $ f: N^{*}\to N$ satisfy $ f(1)\equal{}2,f(2)\equal{}0,f(3k)\equal{}3f(x)\plus{}1,f(3k\plus{}1)\equal{}3f(k)\plus{}2,f(3k\plus{}2)\equal{}3f(x)$. Prove that: no exist n satisfy: $ f(n)\equal{}n\plus{}2008$

[quote="Narutovn"]Let $ f: N^{*}\to N$ satisfy $ f(1) \equal{} 2,f(2) \equal{} 0,f(3k) \equal{} 3f(x) \plus{} 1,f(3k \plus{} 1) \equal{} 3f(k) \plus{} 2,f(3k \plus{} 2) \equal{} 3f(x)$. Prove that: no exist n satisfy: $ f(n) \equal{} n \plus{} 2008$[/quote]

It's easy to see that $ f(n)$ is obtained by translating $ 0\to 1$, $ 1\to 2$ and $ 2\to 0$ in the ternary representation of $ n$.

Then $ f(n)\minus{}n$ is $ \sum_{k\equal{}0}^m a_k3^k$ with either $ a_k\equal{}1$, either $ a_k\equal{}\minus{}2$ and so the ternary representation of $ f(n)\minus{}n$ is made only with $ 0$ or $ 1$.

And since $ 2008\equal{}2202101_3$ contains some digits $ 2$, there is no $ n$ such that $ f(n)\minus{}n\equal{}2008$
Q.E.D.


In triangle $ABC$ prove that:
$\sum{asin{\frac A2}} \geq p$
$a,b,c$ are the sides of triangle and $p=\frac{a+b+b}2$

We have $sin {\frac{A}{2}} \geq \frac{cosB + cosC}{2}$.

 Hence $\sum sinAsin{\frac{A}{2}} \geq \sum \frac{sinAcosB + sinAcosC}{2} = \frac{1}{2} (\sum sinA)$

 And the inequality follows.


The smallest positive number divisible by $2004$ and whose its digits are only $2$ or $4$ is $222444$. Who is the following?

Is your problem a statement or a question???

Bomb

It's a question...

Given that 222444 is the smallest multiple of 2004 containing only digits 2 and 4, find the next smallest positive integer which is a multiple of 2004 and consists of 2's and 4's only.

its equivalent to find the smallest multiple of 1002 using only 1 and 2 . factor 1002 into primes and solve the problem for each prime then the mcm of least number of digits for each prime is the smallest number for 1002 something like that i came up with this in 5 mins so its probably wrong. but not totally i hope... :)

I just got some vague ideas...

Suppose we've got a n-digit number which consists of digits 1 and 2 only - we want this number to be divisible by 1002. This number is the sum of $\frac{10^n - 1}{9}$ and some distinct powers of 10... (and we need to include $10^0 = 1$).

Now, consider the remainder upon division of the number $a_n = 111...111$ by 167.

We get, using the fact that $a_{n + 1} = 10a_n + 1$ (mod 167)

[WARNING: this is BORING]
$a_1 = 1$
$a_2 = 11$
$a_3 = 111$
$a_4 = 109$
$a_5 = 89$
$a_6 = 56$
$a_7 = 60$
$a_8 = 100$
$a_9 = 166 = -1$ (this one might be interesting)
$a_{10} = 158 = -9$
$a_{11} = 78$
$a_{12} = 113$

I guess that will do sigh. Now, the powers of 10 (mod 167):

$10^0 = 1$
$10^1 = 10$
$10^2 = 100$
$10^3 = 165 = -2$
$10^4 = 147 = -20$
$10^5 = 134 = -33$
$10^6 = 4$
$10^7 = 40$
$10^8 = 66$
$10^9 = 159 = -8$
$10^{10} = 87 = -80$
$10^{11} = 35$

and that will do.

So now, we know that we need to look for some $a_n$ with $n > 6$ since it is given that 111222 is the smallest such integer. We add $10^0 = 1$ since the number is even, and we try to add some other powers of 10 listed above to get 0 or a multiple of 167.

So that gives you a few possibilities to check  :P

The thing is, maybe these ideas ARE useful, but it might be that I don't see how to use them, or maybe I am missing something obvious. Or maybe it's very tricky :)

And we get the answer...

Any better ideas?

If that's the best shot it gives, you may be better of with this case study:
- ending in 11, then it is my thing, but that give a suboptimal solution.
- ending in 61, which I forgot, you can easily show it ends in 0061, moreover that it's 40000k+0061, with k=8 mod 25.

This is pretty close yet, but it would also require some additional casework. [the missing step in my proof is: proving that it is actually 08 and not 33/58/83]

But after that we're done. If ther's no solution with a 8-digit factor, my 111000111 wins. And we already have the last 6 digits.
Consider the first 2 digits of the factor. It's clear that they must both be in {1,2}. So running up 11,12,21 we have a winner on the third pick. :)

So unless someone finds a way to eliminate 33/58/83, I feel forced to run down all possibilities from small to large...
1108
1133
1158
1183
1208
1233
1258
1283
2108 - bingo


The point $ P$ lies inside an equilateral triangle and its distances to the three vertices are $ 3, 4, 5$. Find the area of the triangle.

Let PA=3,PB=4,PC=5. Rotate ABC around A by 60 degrees counterclockwise, P gets into Q. Now APQ is equilateral with side 3, PCQ is the Pythagorean triangle 3,4,5. So AQC=150°,
$ AC^{2}\equal{}3^{2}\plus{}4^{2}\plus{}3\cdot4\sqrt3$
$ S\equal{}\sqrt{3}/4\cdot AC^{2}\equal{}25\sqrt{3}/4\plus{}9$